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(1)Elementary Algebra Exercise Book II Wenlong Wang; Hao Wang. Download free books at.

(2) . WENLONG WANG AND HAO WANG. ELEMENTARY ALGEBRA EXERCISE BOOK II. Download free eBooks at bookboon.com 2.

(3) ELEMENTARY ALGEBRA EXERCISE BOOK II. . Elementary Algebra Exercise Book II 1st edition © 2016 Wenlong Wang and Hao Wang & bookboon.com ISBN 978-87-403-1297-3. Download free eBooks at bookboon.com 3.

(4) ELEMENTARY ALGEBRA EXERCISE BOOK II. Contents. CONTENTS. Author Biographies. 5. Preface. 6. 4. 7. Trigonometric Functions. 5 Sequences. 50. 6 Functions. 95. www.sylvania.com. We do not reinvent the wheel we reinvent light. Fascinating lighting offers an infinite spectrum of possibilities: Innovative technologies and new markets provide both opportunities and challenges. An environment in which your expertise is in high demand. Enjoy the supportive working atmosphere within our global group and benefit from international career paths. Implement sustainable ideas in close cooperation with other specialists and contribute to influencing our future. Come and join us in reinventing light every day.. Light is OSRAM. Download free eBooks at bookboon.com 4. Click on the ad to read more.

(5) ELEMENTARY ALGEBRA EXERCISE BOOK II. Author biographies. AUTHOR BIOGRAPHIES Author Biographies. Mr. Wenlong Wang is a retired mathematics educator in China. Professor Hao Wang is a faculty member in the Department of Mathematical & Statistical Sciences at the University of Alberta, an advisory board member of Centre for Mathematical Biology, an associate editor for International Journal of Numerical Analysis & Modeling - Series B, an editor for Nonlinear Dynamics and Systems Theory (an international journal of research and surveys), an editor for a special issue of The Canadian Applied Mathematics Quarterly, and an associate faculty member for Faculty of 1000 Biology. Dr. Wang has strong interests in interdisciplinary research of mathematical biology. His research group is working on areas as diverse as modeling stoichiometry-based ecological interactions, microbiology, infectious diseases, predator-prey interactions, habitat destruction and biodiversity, risk assessment of oil sands pollution. Mathematical models include ordinary differential equations, delay differential equations, partial differential equations, stochastic differential equations, integral differential/difference equations.. Download free eBooks at bookboon.com 5.

(6) ELEMENTARY ALGEBRA EXERCISE BOOK II. Preface. PREFACE Preface. The series of elementary algebra exercise books is designed for undergraduate students with any background and senior high school students who like challenging problems. This series should be useful for non-math college students to prepare for GRE general test - quantitative reasoning and GRE subject test - mathematics. All the books in this series are independent and helpful for learning elementary algebra knowledge. The number of stars represents the difficulty of the problem: the least difficult problem has zero star and the most difficult problem has five stars. With this difficulty indicator, each reader can easily pick suitable problems according to his/her own level and goal. Many thanks to Lina Zhang for translating and typing the our handwriting notes into Latex.. Download free eBooks at bookboon.com 6.

(7) ELEMENTARY ALGEBRA EXERCISE BOOK II. Trigonometric Functions. 4 TRIGONOMETRIC FUNCTIONS 4. Trigonometric Functions. 1 π < x < 0 and sin x + cos x = , (1) find the value of sin x − cos x; (2) 2 5 3 sin2 x2 − 2 sin x2 cos x2 + cos2 x2 . find the value of tan x + cot x 1 1 Solution: (1) The equation sin x + cos x = implies that sin x = − cos x. Substitute 5 5 3 2 2 2 it into sin x + cos x = 1 to obtain 25 cos x − 5 cos x − 12 = 0. Then, cos x = − 5 4 π 4 3 or cos x = . Since − < x < 0, we have cos x = and sin x = − . Hence, 5 2 5 5 7 sin x − cos x = − . 5 2 sin2 x2 − sin x + 1 = (2 − cos x − sin x) · sin x · cos x = (2) The expression is equal to sin x x + cos cos x sin x 108 4 3 −3 4 · =− . (2 − + ) · 5 5 5 5 125 4.1. Given −. 4.2. Find the range of y =. sin x + 1 . cos x + 2. Solution: The equation implies that y cos x + 2y = sin x + 1, then sin x − y cos x = y . Since sin(x − φ) 1, 2y − 1 ⇒ 1 + y 2 sin(x − φ) = 2y − 1, where sin φ = 2 1 + y that is 1 + y 2 |2y − 1|. Squaring both sides of the equation, we can obtain 4 4 3y 2 − 4y 0. Therefore, 0 y , that is, the range of y is [0, ]. 3 3 sin 52 θ 1 4.3 Given f (θ) = − + (0 < θ < π), (1) express f (θ) as a polynomial 2 2 sin 2θ of cos θ. (2) If a ∈ R, find the range of a where there is at least one intersection of the curve y = a cos θ + a with the curve y = f (θ). sin 52 θ − sin 12 θ cos 32 θ sin θ θ 3 Solution: (1) f (θ) = = = 2 cos θ cos = cos 2θ + cos θ = 1 1 2 2 2 sin 2 θ sin 2 θ 2 2 cos θ + cos θ − 1. (2) According to the given condition and (1), we have y = 2 cos2 θ + cos θ − 1 y = a cos θ + a. It is easy to figure out that (cos θ + 1)(2 cos θ − 1) = a(cos θ + 1). Since 0 < θ < π and cos θ + 1 = 0, we have 2 cos θ − 1 = a. On the other hand, −1 < cos θ < 1, thus. −1 <. a+1 < 1, which is equivalent to −3 < a < 1. 2. √ sin β − cos β , show sin β − cos β = ± 2 sin α. sin β + cos β Download free eBooks at bookboon.com sin β + cos β 7 Solution: The equation is equivalent to cot α = , then cot2 α + 1 = sin β − cos β 4.4. Given tan α =.

(8) a+1 2. ELEMENTARY ALGEBRA EXERCISEisBOOK II −1 < < 1, which equivalent. Trigonometric Functions. to −3 < a < 1.. √ sin β − cos β , show sin β − cos β = ± 2 sin α. sin β + cos β sin β + cos β , then cot2 α + 1 = Solution: The equation is equivalent to cot α = sin β − cos β 1 2 sin β + cos β 2 2 ) +1 = , hence , that is (sin β − ( = 2 sin β − cos β 1 − 2 sin β cos β (sin β − cos β)2 sin α √ cos β)2 = 2 sin2 α. Thus, sin β − cos β = ± 2 sin α. 4.4. Given tan α =. 4.5. Given ex − e−x = 2 tan θ, ex + e−x = 2 sec θ, 0 < θ <. π , solve for x. 2. 1 + sin θ = cos θ π π θ π < + < > 0, 0 < θ < , then 2 4 4 2. Proof: Adding the given equations, we obtain ex = tan θ + sec θ = 1 − cos( π2 + θ) π θ = tan( + ). Since ex π sin( 2 + θ) 4 2 π θ π θ π , tan( + ) > 0, then x = ln tan( + ). 2 2 2 4 2. π 4.6 If acute angles x and y satisfy sin y csc x = cos(x + y), x + y = , evaluate 2 the maximum value of tan y. Solution: Since sin y csc x = cos x cos y−sin x sin y ⇒ sin y(sin x+csc x) = cos x cos y ⇒ sin x cos x tan x cos x sin x cos x tan x = √ tan y = = = = 2 2 2 2 sin x + csc x 1 + 2 tan x 1 + sin x 2 sin x +√cos x 2 2 tan x √ 2 2 , the equality holds if and only if tan x = . Therefore, the maximum value is 2 √4 2 . 4 √. 2 , evaluate the range of cos α + cos β. 2 √ 2 1 and 2 Solution: Let t = cos α + cos β · · · = sin α + sin β · · · . 2 1 3 12 + 2 2 ⇒ t2 + = 2 + 2 cos(α − β), that is 2 cos(α − β) = t2 − . Since 2 √ √2 14 14 3 7 2 2 −1 cos(α − β) 1 ⇒ −2 ≤ t − ≤ 2, then t , t ∈ [− , ], hence 2 2 2 2 √ √ 14 14 , ]. (cos α + cos β) ∈ [− 2 2 4.7. Given sin α + sin β =. x−3 π > . 2x − 1 6 1 π x−3 1 x−3 1 Solution: arcsin = ⇒ arcsin > arcsin ⇒ > . 2 6 2x − 1 2 2x − 1 2 x−3 1. On the other hand, by the domain of arcsine function, we can obtain −1 2x − 1 Subsequently, we have x −2 as the solution.. 4.8. Solve the inequality arcsin. 4.9. Show arctan. Download free eBooks at bookboon.com 4 5 56 + arccot + arctan 8 = π. 3 12 33.

(9) √. √. 14 BOOK 14II ELEMENTARY ALGEBRA EXERCISE (cos α + cos β) ∈ [−. 2. ,. 2. Trigonometric Functions. ].. π x−3 > . 2x − 1 6 π x−3 1 x−3 1 1 > arcsin ⇒ > . Solution: arcsin = ⇒ arcsin 2 6 2x − 1 2 2x − 1 2 x−3 1. On the other hand, by the domain of arcsine function, we can obtain −1 2x − 1 Subsequently, we have x −2 as the solution.. 4.8. Solve the inequality arcsin. 4 5 56 + arccot + arctan = π. 3 12 33 4 + 56 56 4 π π 56 12 4 3 33 , and arctan ∈ ( , ), arctan ∈ Proof: Since tan(arctan +arctan ) = 4 56 = − 3 33 5 3 4 2 33 1 − 3 33 π π 4 56 π 5 π ( , ), then (arctan + arctan ) ∈ ( , π). In addition arccot ∈ (0, ), it im4 2 3 33 2 12 2 5 π 4 56 5 plies (π − arccot ) ∈ ( , π) ⇒ arctan + arctan = π − arccot . Therefore, 12 2 3 33 12 4 5 56 arctan + arccot + arctan = π. 3 12 33 4.9. Show arctan. nπ , (n ∈ N ∗ ), evaluate the value of f (1) + f (2) + · · · + f (2000). 5 2π Solution: Assume the period of the function is T, then T = π = 10. Thus f (1) + 4.10 . If f (n) = cos. 5. π 2π 3π f (2) + f (3) + f (4) + f (5) + f (6) + f (7) + f (8) + f (9) + f (10) = cos + cos + cos + 5 5 5 4π 5π 6π 7π 8π 9π 10π π 9π cos + cos + cos + cos + cos + cos + cos . Since cos = cos = 5 5 5 5 5 5 5 5 5 4π 6π 2π 8π 3π 7π 5π 10π − cos = − cos , cos = cos = − cos = − cos , cos = − cos , 5 5 5 5 5 5 5 5 then f (1) + f (2) + · · · + f (10) = 0. As a conclusion, f (1) + f (2) + · · · + f (2000)=0. 4.11 . Find the monotony interval of the function y = cos2 x + sin x.. Solution: Since y = cos2 x + sin x = − sin2 x + sin x + 1, let t = sin x, then y = 1 1 5 −t2 + t + 1 = −(t − )2 + . It is monotonically increasing when x ∈ (−∞, ], and it 2 4 2 1 1 5π is monotonically decreasing when x ∈ [ , ∞). Since t = sin x ⇒ 2kπ + x 2 2 6. Download free eBooks at bookboon.com 9.

(10) ELEMENTARY ALGEBRA EXERCISE BOOK II. Trigonometric Functions. 1 π 5π 13π , (k ∈ Z), t = sin x ⇒ 2kπ + x 2kπ + , (k ∈ Z). Additionally 6 2 6 6 π π since the function t = sin x is increasing in the interval [2kπ − , 2kπ + ], (k ∈ Z), and 2 2 π 3π it is decreasing in the interval [2kπ + , 2kπ + ], (k ∈ Z). As a conclusion, the in2 2 π π 5π π creasing interval y = cos2 x+sin x is [2kπ− , 2kπ+ ] [2kπ+ , 2kπ+ ] (k ∈ Z), 2 6 2 6 π π 5π 3π the decreasing interval is [2kπ + , 2kπ + ] [2kπ + , 2kπ + ] (k ∈ Z). 6 2 6 2 2kπ +. 4.12 . Find the domain and range of the function y =. . arccos(x2 + x + 1).. Solution: According to the domain of square root, we can obtain 0 x2 + x + 1 1. Since x2 + x + 1 1, then −1 x 0. Hence, the domain of the function is [−1, 0]. 3 3 1 3 Since x2 + x + 1 = (x + )2 + ⇒ x2 + x + 1 1 ⇒ 0 arccos(x2 + x + 1) 2 4 4 4 3 3 arccos ⇒ 0 arccos(x2 + x + 1) arccos . Therefore, the range of the func4 4 3 tion is [0, arccos ]. 4 4.13 For arbitrary real number x and integer n, the equation f (sin x) = sin(4n + 1)x always holds. Evaluate f (cos x). π Solution: Since f (sin x) = sin(4n + 1)x and cos x = sin( − x), then f (cos x) = 2 π π π π f [sin( − x)] = sin[(4n + 1)( − x)] = sin[2nπ + − (4n + 1)x] = sin[ − (4n + 1)x] = 2 2 2 2 cos(4n + 1)x. 4.14 . π π Given x, y ∈ [− , ], a ∈ R, and x, y are the roots of the equation system, 4 4 x3 + sin x − 2a = 0 4y 3 + sin y cos y + a = 0. find the value of cos(x + 2y). Solution: The second equation implies 4y 3 + sin y cos y = −a. Multiply the equation by −2 to obtain (−2y)3 + sin(−2y) = 2a. The first equation implies x3 + sin x = 2a, then f (x) = f (−2y). Let f (t) = t3 + sin t. Since the function f (t) is increasing where π π t ∈ [− , ] ⇒ x = −2y, therefore x + 2y = 0. As a conclusion, cos(x + 2y) = 1. 2 2. Download free eBooks at bookboon.com 10.

(11) Deloitte & Touche LLP and affiliated entities.. ELEMENTARY ALGEBRA EXERCISE BOOK II. Trigonometric Functions. 4.15 Let α, β, γ form a geometric sequence with the common ratio 2, α ∈ [0, 2π], and sin α, sin β, sin γ also form a geometrical sequence, find the values of α, β, γ. sin γ sin 2α sin 4α sinβ = ⇒ = ⇒ cos α = sin α sin β sin α sin 2α 2 cos2 α − 1 ⇒ 2 cos2 α − cos α − 1 = 0. The roots of 2 cos2 α − cosα − 1 = 0 are 1 cos α = 1 and cos α = − . When cos α = 1, then sin α = 0, it does not satisfy the 2 1 condition that the first term is nonzero. Therefore, cos α = 1. When cos α = − , since 2 2π 4π 2π 4π 8π α ∈ [0, 2π], then α = or α = . When α = , then β = ,γ = . When 3 3 3 3 3 4π 8π 16π α= , then β = ,γ= . 3 3 3 Solution: Let β = 2α, γ = 4α, and. √ 3 1 Compare arcsin , arctan 2, arccos . 3 4 1 1 π 1 Solution: Let arcsin = α, then sin α = < , thus 0 < α < . 3 3 2 √ 6 √ √ √ π π Let arctan 2 = β, then tan β = 2. Since 1 < 2 < 3, then < β < . 4 3 √ √ 3 3 3 π π 2 3 Let arccos = γ, then cos γ = . Since < < , then < γ < . 4 4 2 4 2 6 4 √ 1 3 As a conclusion, α < γ < β, therefore,arcsin < arccos < arctan 2. 3 4 4.16. 360° thinking. .. 3 3 x x π Given a = (cos x, sin x), b = (cos , − sin ), x ∈ [0, ]. (1) Solve a · b and 2 2 2 2 2 3 |a + b|. (2) If the minimum value of f (x) = a · b − 2λ|a + b| is − , compute λ. 2 3 x 3 x π Solution: (1) a · b = cos x cos − sin x sin = cos 2x, x ∈ [0, ]. 2 2 2 2 2 √ x x 3 3 π |a + b| = (cos x + cos )2 + (sin x − sin )2 = 2 cos2 x = 2 cos x, x ∈ [0, ]. 2 2 2 2 2 (2) f (x) = cos 2x − 4λ cos x = 2 cos2 x − 4λ cos x − 1 = 2(cos x − λ)2 − 1 − 2λ2 . Since π x ∈ [0, ], then cos x ∈ [0, 1]. 2 When λ < 0, cos x = 0, then f (x)min = −1. It does not satisfy the given condition. 3 1 When 0 λ 1, cos x = λ, then f (x)min = −1 − 2λ2 = − , then λ = . 2 2 5 3 When λ 1, cos x = 1, then f (x)min = 1 − 4λ = − , then λ = < 1. It does not 8 2 1 satisfy the condition λ 1. After all, λ = . 2 4.17 . 360° thinking. .. 360° thinking. .. Discover the truth at www.deloitte.ca/careers. © Deloitte & Touche LLP and affiliated entities.. Discover the truth at www.deloitte.ca/careers. © Deloitte & Touche LLP and affiliated entities.. Discoverfree theeBooks truth atatbookboon.com www.deloitte.ca/careers Download. Click on the ad to read more. 11. Dis.

(12) 3 4 √ √ 3 3 2 3 3 π π Let arccos = γ, then cos γ = . Since < < , then < γ < . 4 4 2 4 2 6 4 √ 1 3 ELEMENTARY ALGEBRA EXERCISE BOOK II Trigonometric Functions As a conclusion, α < γ < β, therefore,arcsin < arccos < arctan 2. 3 4 3 x x π 3 Given a = (cos x, sin x), b = (cos , − sin ), x ∈ [0, ]. (1) Solve a · b and 2 2 2 2 2 3 |a + b|. (2) If the minimum value of f (x) = a · b − 2λ|a + b| is − , compute λ. 2 x 3 x π 3 Solution: (1) a · b = cos x cos − sin x sin = cos 2x, x ∈ [0, ]. 2 2 2 2 2 √ x x 3 3 π |a + b| = (cos x + cos )2 + (sin x − sin )2 = 2 cos2 x = 2 cos x, x ∈ [0, ]. 2 2 2 2 2 (2) f (x) = cos 2x − 4λ cos x = 2 cos2 x − 4λ cos x − 1 = 2(cos x − λ)2 − 1 − 2λ2 . Since π x ∈ [0, ], then cos x ∈ [0, 1]. 2 When λ < 0, cos x = 0, then f (x)min = −1. It does not satisfy the given condition. 1 3 When 0 λ 1, cos x = λ, then f (x)min = −1 − 2λ2 = − , then λ = . 2 2 5 3 When λ 1, cos x = 1, then f (x)min = 1 − 4λ = − , then λ = < 1. It does not 2 8 1 satisfy the condition λ 1. After all, λ = . 2 4.17 . √. 2 , AC = 2, AB = 3, find the value of tan A 4.18 Let ABC, sin A + cos A = 2 and the area of ABC. √ √ 1 2 Solution: Since sin A + cos A = 2 cos(A − 450 ) = ⇒ cos(A − 450 ) = . Ad2 2 0 0 0 0 0 , then A − 45 = 60 ⇒ A = 105 ⇒ tan A = tan(45 + 600 ) = ditionally, 0 < A < 180 √ √ 1+ 3 √ = −2 − 3. Since sin A = sin(450 + 600 ) = sin 450 cos 600 + cos 450 sin 600 = 1− √ 3 √ √ √ √ 2+ 6 2+ 6 1 3 √ 1 , we have SABC = AC · AB · sin A = × 2 × 3 × = ( 2 + 6). 4 2 2 4 4 4.19 Find the symmetric center, the symmetric axis equation of the function y = π 3 − 2 cos(2x − ) and the value of x when y has the maximum and the minimum. 3 π Solution : Since the symmetric center of y = cos x is (kπ+ , 0) (k ∈ Z), and the sym2 π π kπ 5π metric axis equation is kπ (k ∈ Z). Thus, 2x− = kπ+ ⇒ x = + (k ∈ Z). 3 2 2 12 π kπ π Since 2x− = kπ, then x = + (k ∈ Z), therefore the the symmetric center of the 3 2 6 π kπ 5π function y = 3−2 cos(2x− ) is ( + , 3) (k ∈ Z), and the symmetric axis equation 3 2 12 kπ π π π is x = + (k ∈ Z). When 2x − = 2kπ ⇒ x = kπ + (k ∈ Z), the minimum 2 6 3 6 π π 2π of y = 3 − 2 cos(2x − ) is 1. When 2x − = (2k + 1)π ⇒ x = kπ + (k ∈ Z), 3 3 3 π the maximum of y = 3 − 2 cos(2x − ) is 5. 3 4.20 If the equation (2 cos θ − 1)x2 − 4x + 4 cos θ + 2 = 0 has two distinct positive roots, and θ is an acute angle. Find the range of θ. 0. Since ∆ = (−4)2 − 4(2 cos θ − Solution: Assume the two √ roots are 1, x Download freexeBooks at bookboon.com √2 > 3 3 4 1 Since x1 + x2 = . < cos θ < 12 1)(4 cos θ + 2) > 0, then − > 0, 2 2 2 cos θ − 1.

(13) (k ∈ Z), the minimum (k ∈ Z). When 2x − = 2kπ ⇒ x = kπ + + 6 3 6 2 2π π π (k ∈ Z), of y = 3 − 2 cos(2x − ) is 1. When 2x − = (2k + 1)π ⇒ x = kπ + 3 3 3 π ELEMENTARY ALGEBRA II Trigonometric Functions the maximum of yEXERCISE = 3 −BOOK 2 cos(2x − ) is 5. 3 is x =. 4.20 If the equation (2 cos θ − 1)x2 − 4x + 4 cos θ + 2 = 0 has two distinct positive roots, and θ is an acute angle. Find the range of θ. Solution: Assume the two √roots are x1 , x√2 > 0. Since ∆ = (−4)2 − 4(2 cos θ − 4 3 3 1 Since x1 + x2 = 1)(4 cos θ + 2) > 0, then − < cos θ < . > 0, 2 2 2 cos θ − 1 1 1 4 cos θ + 2 1 2 Since x1 x2 = 3 . > 0, then cos θ < − or cos θ > · · · . then cos θ > 2 2 cos θ − 1 2 2 √ 1 3 1 , 2 , 3 we can obtain < cos θ < According to , . Since θ is an acute angle, then 2 2 0 0 30 < θ < 60 . √ 4.21 Let the function f (x) = −a cos 2x − 2 3a sin x cos x + 2a + b, its domain is π [0, ], the range is [−5, 1]. Evaluate a and b. 2 √ π Solution: f (x) = −a cos 2x − 3a sin 2x + 2a + b = −2a cos(2x − ) + 2a + b. S3 π π π 2π 1 π ince x ∈ [0, ] ⇒ − 2x − , then − cos(2x − ) 1. 2 3 3 3 2 3 When a > 0, then b f (x) 3a + b ⇒ 3a + b = 1 b = −5 we can obtain a = 2, b = −5. When a < 0, then 3a + b f (x) b ⇒ 3a + b = −5 b=1 we can obtain a = −2, b = 1. √. 1 x) = sin(cot−1 ), find the value of x. 2 √ √ 2θ √ √ sin θ 1 − cos 1−x = = √ Solution: Let cos−1 x = θ, then cosθ = x, tan θ = . cosθ cos θ x 1 1 1 2 = Let cot−1 12 = φ, then cot φ = , sin φ = = √ . The equation is 2 2 csc φ 5 1 + cot φ √ 4 5 1−x 2 1−x equal to √ = , therefore x = . =√ ⇒ x 5 9 x 5 4.22 . Given tan(cos−1. θ Let 0 < θ < π, find the maximum value of sin (1 + cos θ). 2 √ θ θ θ θ 2 θ = 2 2 sin2 cos4 Solution: Since 0 < θ < π, then sin (1+cos θ) = 2sin cos 2 2 2 2 2 √ √ √ 2 2 sin2 2θ + cos2 2θ + cos2 2θ 3 √ 4 3 2 2 . Hence the max= 2 ( ) = 2 ( )3 = 2× × 9 3 3 3 3 √ θ 4 3 . imum value of sin (1 + cos θ) is 9 2 4.23 . Download free eBooks at bookboon.com. 4.24 . Find the value of sin4. π. 13. + sin4. 3π. + sin4. 5π. + sin4. 7π. ..

(14) x 1 1 1 2 Let = φ, then cot φ = , sin φ = = = √ . The equation is 2φ 2 csc φ 5 1 + cot √ 1−x 4 5 2 1−x ELEMENTARY EXERCISE BOOK II Trigonometric Functions √ ⇒ √ = , therefore x = . equal to ALGEBRA = x 5 9 x 5 cot−1 12. θ Let 0 < θ < π, find the maximum value of sin (1 + cos θ). 2 √ θ θ θ θ θ Solution: Since 0 < θ < π, then sin (1+cos θ) = 2sin cos2 = 2 2 sin2 cos4 2 2 2 2 2 √ 2 θ θ θ √ 2 √ 2 sin 2 + cos2 2 + cos2 2 3 √ 4 3 2 2 ) = 2 ( )3 = 2× × = . Hence the max2 ( 3 3 3 3 9 √ θ 4 3 imum value of sin (1 + cos θ) is . 2 9 4.23 . 4.24 . Find the value of sin4. π 3π 5π 7π + sin4 + sin4 + sin4 . 16 16 16 16. π π 3π π π 3π + sin4 + sin4 ( − ) + sin4 ( − ) = Solution: The quantity is equal to sin4 16 16 2 16 2 16 3π 4 π 4 3π 4 3π 4 π 2 π 2 π 2 2 π 2 π sin +sin +cos +cos = (sin +cos ) −2 sin cos +(sin2 + 16 16 16 16 16 16 16 16 16 1 1 π π 3π 3π 3π π 3π π cos2 )2 − 2 sin2 cos2 = 2 − (sin2 + sin2 ) = 2 − [sin2 + sin2 ( − )] 16 16 16 2 8 8 2 8 2 8 1 3 2 π 2 π = 2 − (sin + cos ) = . 2 8 8 2 √ Given vector m = (cos θ, sin θ), n = ( 2 − sin θ, cos θ), θ ∈ (π, 2π), and √ 8 2 θ π , find the value of cos( + ). |m + n| = 5 2 8 √ Solution: From the given condition, we have m + n = (cos θ − sin θ + 2, sin θ + cos θ), √ √ then |m + n| = (cos θ − sin θ + 2)2 + (cos θ + sin θ)2 = 4 + 2 2(cos θ − sin θ) = √ π 8 2 π π ⇒ 2 1 + cos(θ + ) = n| = 4 + 4 cos(θ + ) = 2 1 + cos(θ + ). Since |m+ 4 5 4 4 √ 8 2 π 7 π θ π θ π 16 ⇒ cos(θ + ) = . Sine cos(θ + ) = 2 cos2 ( + ) − 1 ⇒ cos2 ( + ) = . 5 4 25 4 2 8 2 8 25 We will turn 5π yourθ CVπ into 9π θ π θ π 4 Since π < θ < 2π ⇒ < + < ⇒ cos( + ) < 0. Thus, cos( + ) = − . 8 of 2 a lifetime 8 8 2 8 2 8 5 an opportunity 4.25 . α π α 4.26 Given α, β ∈ (0, ), 3 sin β = sin(2α + β), 4 tan = 1 − tan2 , 2 4 2 evaluate α + β. 4 tan α2 α 1 2 α . ⇒ Solution: Since 4 tan = 1 − tan α = 1 ⇒ 2 tan α = 1 ⇒ tan α = 2 2 2 2 1 − tan 2 1 Since 3 sin β = sin(2α + β) = sin(α + β) cos α + cos(α + β) sin α , 2 3 sin β = 3 sin(α + β − α) = 3 sin(α + β) cos α − 3 cos(α + β) sin α . 2 − 1 ⇒ sin(α + β) cos α = 2 cos(α + β) sin α ⇒ tan(α + β) = 2 tan α = 1. For π π α, β ∈ (0, ), thus α + β = . Do you like4cars? Would you like to4be a part of a successful brand? We will appreciate and reward both your enthusiasm and talent. Send us your CV. You will be surprised where it can take you.. Send us your CV on www.employerforlife.com. √ π 1 − arctan 2 cos x − 1. 4 2 1 Solution: the function is defined if and only if 2 cos x − 1 0, that is cos x . 2 π free eBooks at πbookboon.com Download Click on the ad to read more (n ∈ Z). Thus the domain of y is 2nπ − x 2nπ + 3 3 14 √ √ π π π Since 0 2 cos x − 1 1, then 0 arctan 2 cos x − 1 , thus y .. 4.27 . Find the domain and range of the function y =.

(15) sin +sin4 16 +cos +cos4 16 +cos +cos4 16 = = (sin (sin2 16 +cos +cos2 16 ))2 −2 cos2 16 +(sin +(sin2 16 + + sin4 16 +sin −2 sin sin2 16 cos 16 16 16 16 16 16 16 16 16π 1 1 π 3π 3π 3π π 3π π 2 3π 2 2 3π 2 3π 2 π 2 3π 2 π 2 π 1 1 π cos cos2 16 ))2 − cos2 16 = = 22 − − 2 (sin (sin2 8 + + sin sin2 8 )) = = 22 − − 2 [sin [sin2 8 + + sin sin2 (( 2 − − 22 sin sin2 16 cos − 8 )] )] 161 16 π 163 2 8 8 2 8 2 8 π 2 π 2 π 3 1 ALGEBRA ELEMENTARY Trigonometric Functions cos = = (sin2 8 + +EXERCISE cos2 8 ))BOOK = 2 ..II = 22 − − 2 (sin 2 8 8 2 √ √ Given vector m = (cos θ, sin θ), n = (( 22 − − sin sin θ, θ, cos cos θ), θ), θθ ∈ ∈ (π, (π, 2π), 2π), and and Given vector m = (cos θ, sin θ), n = √ √ θ π π 8 2 find the the value value of of cos( cos( θ + + ). ). m + n n|| = = 8 2 ,, find ||m + 55 22 88 √ √ 2, sin sin θθ + + cos cos θ), θ), Solution: From From the the given condition, condition, we we have have m m + + n n= = (cos (cos θθ − sin sin θθ + + 2, Solution: given − √ √ √ 2 √ 2 then ||m m + n n|| = = (cos 2(cos θθ − − sin sin θ) θ) = = then + (cos θθ− − sin sin θθ + + 2) 2)2 + + (cos (cos θθ + + sin sin θ) θ)2 = =√ 44 + + 222(cos √ 88 22 π π π π π Since |m+ π n 44 + = ⇒ 22 11 + + cos(θ cos(θ + + 4 )) = ). Since |m+ n|| = = 5 ⇒ = 22 11 + + cos(θ cos(θ + + 4 ). + 44 cos(θ cos(θ + + 4 )) = 4 5 4 4 √ √ 88 22 π π ) = 77 . Sine cos(θ + π π ) = 2 cos22 ( θθ + π π ) − 1 ⇒ cos22 ( θθ + π π ) = 16 16 ⇒ cos(θ + ⇒ cos(θ + ) = . Sine cos(θ + ) = 2 cos ( + ) − 1 ⇒ cos ( + ) = 25 .. 55 44 25 4 2 8 2 8 25 4 2 8 2 8 25 5π θθ π 9π θθ π θθ π 44 π π 9π π 5π Since =− − 5 .. + )) = < 0. 0. Thus, Thus, cos( cos( 2 + + )) < ⇒ cos( cos( 2 + < 8 ⇒ + < <2+ Since π π< < θθ < < 2π 2π ⇒ ⇒ 8 < 5 2 88 2 88 8 2 88 8 4.25 4.25 . π α α 1 − tan22 α π 3 sin β = sin(2α + β), 4 tan α 4.26 = 1 − tan 2 ,, ), 3 sin β = sin(2α + β), 4 tan 2 = 4.26 Given Given α, α, β β∈ ∈ (0, (0, 4 ), 2 4 2 evaluate evaluate α α+ + β. β. tan αα2 α 1 α = 1 − tan22 α α ⇒ 44 tan Solution: Since 4 tan = 1 .. Solution: Since 4 tan 2 = 1 − tan 2 ⇒ 1 − tan222 αα = = 11 ⇒ ⇒ 22 tan tan α α= = 11 ⇒ ⇒ tan tan α α= 2 2 22 1 − tan 22 Since 33 sin sin β β= = sin(2α sin(2α + + β) β) = = sin(α sin(α + + β) β) cos cos α α+ + cos(α cos(α + + β) β) sin sin α α , , 11 Since 33 sin sin β β= = 33 sin(α sin(α + +β β− − α) α) = = 33 sin(α sin(α + + β) β) cos cos α α− − 33 cos(α cos(α + + β) β) sin sin α α . . 22 − ⇒ sin(α sin(α + + β) β) cos cos α α = = 2 cos(α + β) sin α ⇒ tan(α + β) = 2 tan α = 1. For 22 − 11 ⇒ π π 2 cos(α + β) sin α ⇒ tan(α + β) = 2 tan α = 1. For π π α, β β∈ ∈ (0, (0, ), ), thus thus α α+ +β β= = .. α, 44 44 π √ π 11 arctan √ Find − 2 arctan 22 cos cos x x− − 1. 1. Find the the domain domain and and range range of of the the function function yy = = 4− 4 2 11 Solution: Solution: the the function function is is defined defined if if and and only only if if 22 cos cos x x− − 11 0, 0, that that is is cos cos x x 2 .. 2 π π x 2nπ + π π (n ∈ Z). Thus the domain of y is 2nπ − Thus the domain of y is 2nπ − 3 x 2nπ + 3 (n ∈ Z). 3 √ 3 √ π π π √ √ Since cos x x− − 11 1, 1, then then 00 arctan arctan 22 cos cos x x− − 11 π ,, thus thus π yy π .. Since 00 22 cos 44 88 44. 4.27 4.27 . π π Therefore, the domain of the function is x ∈ [2nπ − , 2nπ + ] (n ∈ Z), and the 3 3 π π range is y ∈ [ , ]. 8 4 a3 + b3 + c3 3 4.28 Given ABC, = c2 , and sin A sin B = . Judge the shape a+b+c 4 of ABC. a3 + b3 + c3 = c2 ⇒ a3 +b3 = ac2 +bc2 ⇒ (a+b)(a2 −ab+b2 ) = c2 (a+b). Since a+b+c 1 The cosine theorem is c2 = a2 +b2 −2ab cos C . 2 a+b = 0, then a2 −ab+b2 = c2 . 1 1 , 2 we get cos C = ⇒ C = 600 ⇒ A + B = 1200 . Since According to , 2 3 1 3 3 1 sin A sin B = ⇒ [cos(A − B) − cos(A + B)] = , then cos(A − B) = − = 1, 4 2 4 2 2 thus A − B = 0 ⇒ A = B. Therefore ABC is a right triangle. Solution:. π πDownload free eBooks at bookboon.com x , f (x) satisfies152f (− sin x) + 3f (sin x) = 4 sin x cos x. 2 2 (1) Show f (x) is an odd function. (2) Find the analytic expression of f (x). 4.29 . Let −.

(16) 1 1 , 2 we get cos C = ⇒ C = 600 ⇒ A + B = 1200 . Since According to , 2 3 1 3 1 3 sin A sin B = ⇒ [cos(A − B) − cos(A + B)] = , then cos(A − B) = − = 1, 2 2 4 2 4 ELEMENTARY ALGEBRA EXERCISE BOOK II thus A − B = 0 ⇒ A = B. Therefore ABC is a right triangle. Trigonometric Functions π π x , f (x) satisfies 2f (− sin x) + 3f (sin x) = 4 sin x cos x. 2 2 (1) Show f (x) is an odd function. (2) Find the analytic expression of f (x). 4.29 . Let −. 1 substitute − sin x into 1 to (1) Proof: Since 2f (− sin x)+3f (sin x) = 4 sin x cos x , 2 1 2 obtain 2f (sin x) + 3f (− sin x) = −4 sin x cos x . + ⇒ 5f (sin x) + 5f (− sin x) = 0 ⇒ f (sin x) = −f (− sin x), therefore f (x) is an odd function. 1 − 2 ⇒ f (sin x) − f (− sin x) = 8 sin x cos x. Since f (sin x) = (2) Solution: −f (− sin x) ⇒ 2f (sin x) = √ 8 sin x cos x, then f (sin x) = 4 sin x 1 − sin2 x, (−1 x 1). Therefore, f (x) = 4x 1 − x2 , (−1 x 1). 4.30 . Solve the equation sin x + cos x + sin x cos x = 1.. Solution 1: Multiply both sides of the equation by 2 and adding 1, we obtain 2(sin x + cos x) + 2 sin x cos x + 1 = 3 ⇒ (sin x + cos x)2 + 2(sin x + cos x) − 3 = 0 ⇒ [(sin x + cos x) − 1][(sin x + cos x) + 3] = 0 ⇒ sin x + cos x = 1 or sin x + cos x = −3. Since −1 sin x 1, −1 cosx 1, we have sin x + cos x = −3. √ √ π π π π 2 Since sin x + cos x = 1 ⇒ 2 sin(x + ) = 1 ⇒ sin(x + ) = ⇒ x + = 2kπ + , 4 4 2 4 4 π π π or x + = (2k + 1)π − ⇒ x = 2kπ, or x = 2kπ + , (k ∈ Z). Therefore, the solution 4 4 2 π of the equation is {x|x = 2kπ, k ∈ Z} {x|x = 2kπ + , k ∈ Z}. 2 u2 − 1 Solution 2: Assume sin x + cos x = u, then sin x cos x = . Substitute it into the 2 u2 − 1 = 1 ⇒ u2 + 2u − 3 = 0 ⇒ u = 1 or equation sin x + cos x + sin x cos x = 1: u + 2 u = −3. Since −1 sin x 1, −1 cos x 1, then u =√sin x + cos x = −3. Hence, √ 2 π π π π sin x + cos x = 1 ⇒ 2 sin(x + ) = 1 ⇒ sin(x + ) = ⇒ x + = 2kπ + , or 4 4 2 4 4 π π π x + = (2k + 1)π − ⇒ x = 2kπ, or x = 2kπ + , (k ∈ Z). Therefore, the solution 4 4 2 π of the equation is {x|x = 2kπ, k ∈ Z} {x|x = 2kπ + , k ∈ Z}. 2 4.31 If 0 < x < 450 , and lg tan x − lg sin x = lg cos x − lg cot x + lg 9 − lg find the value of cos x − sin x.. √. 8,. √ 8√ − lg 9, Solution: The given equation is equal to lg(sin x cos x) − lg(tan x cot x) = lg √ 2 2 4 2 then sin x cos x = . Since (cos x − sin x)2 = 1 − 2 sin x cos x = 1 − = 9 9 √ √ 1 1 √ 9−4 2 , 0 < x < 450 , cos x > sin x. Thus cos x − sin x = 9 − 4 2 = (2 2 − 1). 9 3 3. 4.32 Find all positive integer solutions which satisfy the equation tan−1 x + cot−1 y = tan−1 3. 1 = tan−1 3 ⇒ tan(tan−1 x + y 1 x + 3y − 1 10 1 y . Since x, y are = 3 ⇒ x = = 3− tan−1 ) = tan(tan−1 3) ⇒ x Download y+3 y 1 − y free eBooks at bookboon.com y+3 positive integers, then y + 3 is the divisor of16 10. Thus y = 2 or y = 7, x = 1 or x = 2. As a conclusion, the positive integer solutions are Solution: tan−1 x + cot−1 y = tan−1 3 ⇒ tan−1 x + tan−1.

(17) Solution: The given equation is equal to lg(sin x cos x) − lg(tan x cot x) = lg 8 √ − lg 9, √ √ √ 22 22 4 4 22 2 then Since (cos (cos x x− − sin sin x) x)2 = = then sin sin x x cos cos x x = = 9 .. Since = 11 − − 22 sin sin x x cos cos x x = = 11 − − 9 = √ 9 9 √ √ √ 99 − − 44 22 < x < 4500 , cos x > sin x. Thus cos x − sin x = 11 9 − 4√2 = 11 (2√2 − 1). ELEMENTARY,, 0 0ALGEBRA < x <EXERCISE 45 , cosBOOK x > IIsin x. Thus cos x − sin x = 3 9 − 4Trigonometric 2 = 3 (2 2Functions − 1). 99 3 3 −1 x + 4.32 Find Find all all positive positive integer integer solutions solutions which which satisfy satisfy the the equation equation tan tan−1 4.32 x+ −1 −1 −1 y = tan−1 3. cot cot y = tan 3.. 1 = tan−1 −1 −1 −1 −1 −1 1 −1 Solution: tan−1 33 ⇒ ⇒ tan(tan tan(tan−1 x x+ + Solution: tan tan−1 x x+ + cot cot−1 yy = = tan tan−1 33 ⇒ ⇒ tan tan−1 x x+ + tan tan−1 y = y 1 x 10 3y 1 x+ + yy1 3y − − 11 10 −1 −1 1 ) = tan(tan−1 −1 3) ⇒ Since x, x, yy are are tan = 33 ⇒ ⇒ x x = = y+3 = = 33 − − y + 3 .. Since tan y ) = tan(tan 3) ⇒ 1 − xx = y+3 y 1 − yy y+3 positive integers, integers, then then yy + + 33 is is the the divisor divisor of of 10. 10. Thus Thus yy = = 22 or or yy = = 7, 7, x x= = 11 or or x x= = 2. 2. positive As aa conclusion, conclusion, the As the positive positive integer integer solutions solutions are are x= = 11 x yy = = 22 or or. . x= = 22 x yy = = 77. sin(cos θ) Check the the sign sign of of the the formula formula sin(cos θ) when when θθ is is in in the the second second quadrant. quadrant. Check cos(sin 2θ) 2θ) cos(sin 4π π If π π< <α α+ +β β< < 4π ,, −π −π < <α α− −β β< <− − π ,, find find the the range range of of 2α 2α − − β. β. If 33 33 π Solution: (1) (1) 2kπ 2kπ + +π < < θθ < < 2kπ 2kπ + + π.(k π.(k ∈ ∈ Z) Z) ⇒ ⇒ −1 −1 < < cos cos θθ < < 0. 0. The The condition condition Solution: 22 4.33 4.33. I joined MITAS because I wanted real responsibili� I joined MITAS because I wanted real responsibili�. Real work International Internationa al opportunities �ree wo work or placements. Maersk.com/Mitas www.discovermitas.com. �e G for Engine. Ma. Month 16 I was a construction Mo supervisor ina const I was the North Sea super advising and the No he helping foremen advis ssolve problems Real work he helping fo International Internationa al opportunities �ree wo work or placements ssolve pr. Download free eBooks at bookboon.com 17. �e Graduate Programme for Engineers and Geoscientists. Click on the ad to read more.

(18) 4.33 . Check the sign of the formula. ELEMENTARY ALGEBRA EXERCISE BOOK II. sin(cos θ) when θ is in the second quadrant. cos(sin 2θ). Trigonometric Functions π 4π , −π < α − β < − , find the range of 2α − β. If π < α + β < 3 3 π Solution: (1) 2kπ + < θ < 2kπ + π.(k ∈ Z) ⇒ −1 < cos θ < 0. The condition 2 4kπ + π < 2θ < 4kπ + 2π gives −1 < sin 2θ < 0. Thus sin(cos θ) < 0, cos(sin 2θ) > 0. sin(cos θ) < 0. Therefore cos(sin 2θ) (2) Assume x = α+β, y = α−β, 2α−β = mx+ny. Then 2α−β = mα+mβ+nα−nβ = (m + n)α + (m − n)β. Comparing the coefficients, we have m+n=2 m − n = −1. Hence m = 1 −π < x + 2. 1 3 1 3 4π π , n = ⇒ 2α − β = x + y. Since π < x < , −π < y < − , then 2 2 2 2 3 3 3 π π y < . As a conclusion, the range of 2α − β is (−π, ). 2 6 6. 4.34 . √ Let tan α + sin α = m, tan α − sin α = n. Show m2 − n2 = 4 mn.. Solution: Multiplying the two equations together to obtaion mn = tan2 α − sin2 α = 1 tan2 α(1 − cos2 α) = tan2 α sin2 α . m+n 2 Adding the two equations: 2 tan α = m + n ⇒ tan α = . 2 m−n 3 Using subtraction for the two equations: 2 sin α = m − n ⇒ sin α = . 2 √ m+n 2 m−n 2 2 and 3 into 1 ⇒ mn = ( Substituting )( ) ⇒ m2 − n2 = 4 mn. 2 2 b c 2b − a − c a α = = = 0, show sin2 = . cos α cos 2α cos 3α 2 4b a+c b = = 0. Since Proof: Applying the equal radio theorem, we have cos 2α cos α + 3 cos α a+c b = = 0. In particular, b = 0, cos α = cos α = 0, cos 2α = 0 ⇒ cos 2α 2 cos α cos 2α 1 − cos α 1 a+c 2b − a − c α a+c . Therefore sin2 = = (1 − )= . 2b 2 2 2 2b 4b 4.35 . Given. 4.36 Let sin2 (n + 1)θ = sin2 nθ + sin2 (n − 1)θ, and (n + 1)θ, nθ, (n − 1)θ are the three interior angles of a triangle, find the integer value of n. Solution: sin2 (n+1)θ = sin2 nθ +sin2 (n−1)θ ⇒ sin2 (n+1)θ −sin2 (n−1)θ = sin2 nθ ⇒ [sin(n+1)θ−sin(n−1)θ][sin(n+1)θ+sin(n−1)θ] = sin2 nθ ⇒ 2 sin θ cos nθ2 sin nθ cos θ = sin2 nθ ⇒ sin 2nθ sin 2θ = sin2 nθ (∗). Since (n + 1)θ + √ nθ + (n − 1)θ = π, then 1 π π π 3 π ⇒ 2θ = ⇒ θ = . nθ = . Substituting it into (∗), then sin 2θ = tan = 3 2 3 2 3 6 π Since nθ = , we have n = 2. 3. Download free eBooks at bookboon.com 18.

(19) ELEMENTARY ALGEBRA EXERCISE BOOK II. 4.37 . Trigonometric Functions. Given sin α + 3 cos α = 2, compute. sin α − cos α . sin α + cos α. sin α − cos α 1 Denote sin α = = k ⇒ (1 − k) sin α = (1 + k) cos α . sin α + cos α 1−k 1 we 2 Applying 2 ÷ , 1 we have cos α = . Substituting it into , 2 − 3 cos α . 2−k 1+k 2 1−k 2 1+k (k = 2). Since sin2 α + cos2 α = 1, then ( ) +( ) = 1. have sin α = 2−k 2−k 2 − k√ It can be written as k 2 + 4k − 2 = 0. Solving the equation, we have k = −2 ± 6. As √ sin α − cos α = −2 ± 6. a conclusion, sin α + cos α. Solution: Let. 4.38 . Let a, b, c are the three side lengths of ABC, and they form a geometric −→−−→ 3 3 sequence, and cos B = . (1) Find the value of cot A + cot C. (2)Let BABC = , 4 2 compute a + c. √ 3 π 3 2 7 Solution: (1) cos B = ⇒ 0 < B < ⇒ sin B = 1 − ( ) = . Since a, b, c 4 2 4 4 form a geometric sequence, applying the sine theorem, we have sin2 B = sin √ A sin C. cos A cos C sin(A + C) sin B 4 7 1 Therefore cot A + cot C = + = = = . = 2 sin A sin C sin A sin C sin B 7 sin B −→−−→ 3 3 3 ⇒ ca cos B = . Additionally cos B = , thus ca = 2. S(2) BABC = 2 2 4 ince b2 = ac = 2, applying the cosine theorem b2 = a2 + c2 − 2ac cos B , we have 2 2 2 2 a2 + c2 = √ b + 2ac cos B = 7 ⇒ (a + c) = a + c + 2ac = 7 + 4 = 11. Therefore, a + c = 11. π 2 sin 2θ ). If logtan θ cos θ = , (θ ∈ (0, )), find the value of logcsc2 θ ( 3 2 2 lg cos θ = Solution: Changing the base number of the given equation, we have lg sin θ − lg cos θ 2 lg cos θ 2 2 sin 2θ ⇒ = ⇒ logsin θ cos θ = . Hence, logcsc2 θ ( ) = − logsin2 θ (sin θ cos θ) = 3 lg sin θ 5 5 2 1 + 25 1 7 1 1 + logsin θ cos θ =− =− . − logsin θ (sin θ cos θ) 2 = − logsin θ (sin θ cos θ) = − 2 2 2 10. 4.39 . √ 1 π 3 , find the 4.40 Given f (x) = 2acos x + b sin x cos x, f (0) = 2, f ( ) = + 3 2 2 set of x values that satisfy the formula f (x) > 2. 2. Download free eBooks at bookboon.com 19.

(20) ELEMENTARY ALGEBRA EXERCISE BOOK II. Trigonometric Functions. √ √ 3 3 1 1 π b = + , substiSolution: f (0) = 2a = 2 ⇒ a = 1. Since f ( ) = a + 3 2 4 2 2 tuting a = 1 into this formula, we have b = 2. Thus f (x) = 2 cos2 x + 2 sin x cos √x = π 2 ⇒ sin 2x + cos 2x + 1. Sincef (x) > 2, then sin 2x + cos 2x + 1 > 2 ⇒ sin(2x + ) > 4 2 π π 3π 2kπ + < (2x + ) < 2kπ + .(k ∈ Z). Therefore the set of x values that satisfy 4 4 4 π the formula f (x) > 2 is {x|kπ < x < kπ + , k ∈ Z}. 4 4.41 Let a, b, c are the three side lengths of ABC, and they form a geometric sequence. sin B + cos B = m2 . Find the range of m. Solution: Since a, b, c form a geometric sequence, then b2 = ac. Applying the sine the1 orem, we have sin2 B = sin A sin C. Then 1 − cos2 B = − [cos(A + C) − cos(A − C)] ⇒ 2 1 2 2 2 cos B + cos B − 1 = 1 − cos(A − C) ⇒ 2cos B + cos B − 1 0 ⇒ cos B , 2 π or cos B −1 (truncated). Hence 0 < B . Additionally since sin B + cos B = 3 √ √ √ √ π 4 2 sin(B + ) ⇒ 1 m2 2 ⇒ − 2 m −1, or 1 m 4 2. 4 4.42 Let α,√ β are the two real roots of the equation x2 + 2(sin θ + 1)x + sin2 θ = 0, and |α − β| 2 2. Find the range of θ. Solution: Since the equation has real roots, then ∆ = 4(sin θ + 1)2 − 4 sin2 θ = 1 8 sin θ + 4 0 ⇒ sin θ − . Applying the Vieta theorem, we have 2 α + β = −2(sin θ + 1) 2αβ = sin2 θ √ 1 Since |α − β| 2 2 ⇒ (α − β)2 Hence (α − β)2 = (α + β)2 − 4αβ = 8| sin θ| + 4 . 1 1 2 According to 1 and , 2 we have 8| sin θ| + 4 8 ⇒ | sin θ| 8 . ⇒ − 2 2 π π 1 (k ∈ Z). sin θ . Therefore kπ − θ kπ + 6 6 2 4.43 Let A, B, C are the three interior angles of triangle ABC, show that A B C 1 sin sin sin . 2 2 2 8 A B C 1 A+B A−B C 1 C C A−B Proof 1: sin sin sin = − (cos −cos ) sin = − (sin2 −sin cos )= 2 2 2 2 2 2 2 2 2 2 2. Download free eBooks at bookboon.com 20. Click on the ad to read more.

(21) 2 cos2 B + cos B − 1 = 1 − cos(A − C) ⇒ 2cos2 B + cos B − 1 0 ⇒ cos B , 2 π or cos B −1 (truncated). Hence 0 < B . Additionally since sin B + cos B = 3 √ √ √ √ π 4 ELEMENTARY ALGEBRA Functions ) ⇒EXERCISE 1 m2BOOK 2 sin(B + 2. II2 ⇒ − 2 m −1, or 1 m 4Trigonometric 4 4.42 Let α,√ β are the two real roots of the equation x2 + 2(sin θ + 1)x + sin2 θ = 0, and |α − β| 2 2. Find the range of θ. Solution: Since the equation has real roots, then ∆ = 4(sin θ + 1)2 − 4 sin2 θ = 1 8 sin θ + 4 0 ⇒ sin θ − . Applying the Vieta theorem, we have 2 α + β = −2(sin θ + 1) 2αβ = sin2 θ √ 1 Since |α − β| 2 2 ⇒ (α − β)2 Hence (α − β)2 = (α + β)2 − 4αβ = 8| sin θ| + 4 . 1 1 2 According to 1 and , 2 we have 8| sin θ| + 4 8 ⇒ | sin θ| 8 . ⇒ − 2 2 1 π π sin θ . Therefore kπ − θ kπ + (k ∈ Z). 2 6 6 4.43 Let A, B, C are the three interior angles of triangle ABC, show that B C 1 A sin sin sin . 2 2 2 8 A C B C 1 A+B A−B C 1 C A−B Proof 1: sin sin sin = − (cos −cos ) sin = − (sin2 −sin cos )= 2 2 2 2 2 2 2 2 2 2 2 C A−B 1 A−B C A−B 1 1 C 1 A−B 2 1 + cos2 − cos2 ) = − [(sin − cos )− − (sin2 −sin cos 2 2 2 2 4 4 4 4 2 2 2 2 A−B A−B 1 1 1 C 1 A−B 2 1 cos2 ] = cos2 − (sin − cos ) ≤ . 4 2 8 2 2 2 2 2 8 2 2 2 1 − cos A 1 b +c −a a2 − (b − c)2 a2 A = = (1 − ) = , Proof 2: Since sin2 2 2 2 2bc 4bc 4bc A a b c B C A and sin2 0 ⇒ sin √ . Similarly ,we have sin √ , sin √ . 2 2 2 2 2 ac 2 bc 2 ab b c B C a A 1 √ = . √ Hence sin sin sin √ 2 2 2 8 2 bc 2 ac 2 ab √ x π x π x x π Given vectors a = (2 cos , tan( + )), b = ( 2 sin( + ), tan( − )). 2 2 4 √ 2 4 2 4 π Let f (x) = a · b. Find the value of x when f (x) = 2, (0 < x < ). 2 √ √ √ x π x π x π x x x 2 Solution: f (x) = a·b = 2 2 cos sin( + )+tan( + ) tan( − ) = 2 2 cos ( sin + 2 2 4 2 4 2 4 2 2 2 √ 1 + tan x2 tan x2 − 1 x x x 2 x cos ) + = 2 sin cos + 2 cos2 − 1 = sin x + cos x = 2 2 1 − tan x2 1 + tan x2 2 2 2 √ √ π π π π 2 sin(x + ) = 2. Thus sin(x + ) = 1. On the other hand, 0 < x < , thus x = . 4 4 2 4 4.44 . 4.45 . Let f (x) = sin(ωx + φ)(ω > 0, 0 φ π) is an even function defined 3π in R. Its graph is symmetric about the point M ( , 0). It is monotone on the interval 4 π [0, ]. Find the values of ω and φ. 2 Solution: Since f (−x) = f (x) ⇒ sin(−ωx + φ) = sin(ωx + φ) , then 2 cos φ sin ωx = 0. π Since x ∈ R, ω > 0, then cos φ = 0. In other words, since 0 φ π, we have φ = . 2 3π 3π 3π Since its graph is symmetric about the point M ( , 0), then f ( − x) = −f ( + x), Download free eBooks at bookboon.com 4 4 4 3π 3π 3π 21 then f ( ) = −f ( ) when x = 0. Since f ( , 0) is a point of the graph, then 4 4 4.

(22) 4. 4. 2. 4. 4.45 . Let f (x) = sin(ωx + φ)(ω > 0, 0 φ π) is an even function defined 3π in R. Its graph is symmetric about the point M ( , 0). It is monotone on the interval ELEMENTARY ALGEBRA EXERCISE BOOK II Trigonometric Functions 4 π [0, ]. Find the values of ω and φ. 2 Solution: Since f (−x) = f (x) ⇒ sin(−ωx + φ) = sin(ωx + φ) , then 2 cos φ sin ωx = 0. π Since x ∈ R, ω > 0, then cos φ = 0. In other words, since 0 φ π, we have φ = . 2 3π 3π 3π Since its graph is symmetric about the point M ( , 0), then f ( − x) = −f ( + x), 4 4 4 3π 3π 3π then f ( ) = −f ( ) when x = 0. Since f ( , 0) is a point of the graph, then 4 4 4 3π 3ωπ π 3ωπ 3ωπ 3ωπ f ( ) = −f ( + ) = cos , that is cos = 0. Since ω > 0, then = 4 4 2 4 4 4 2 2 2 π π + kπ, k = 0, 1, 2, · · · ⇒ ω = (2k + 1). When k = 0, ω = , f (x) = sin( x + ) is 2 3 3 3 2 π π decreasing on the interval [0, ). When k = 1, ω = 2, ]f (x) = sin(2x + ) is decreasing 2 2 π 10 π on the interval [0, ]. When k 2, ω , f (x) = sin(ωx + ) is not a monotone 2 3 2 π 2 function on the interval [0, ). After all, ω = or ω = 2. 2 3. 4.46 c . sin C. Let sides a, b, c correspond to angle A, B, C in ABC, show. a cos B − b cos A = sin(A − B). a2 + c2 − b2 b2 + c2 − a2 1 − = Proof: Since a cos B − b cos A = (ac cos B − bc cos A) = c 2c 2c ( c sin A)2 − ( sinc C sin B)2 a2 − b 2 a cos B − b cos A a 2 − b2 , we have = = sin C = c sin(A − B) c sin(A − B) c sin(A − B) cos A+B 2 sin A+B cos A−B c2 sin A−B c(sin A − sin B)(sin A + sin B) c2 sin2 A − c2 sin2 B 2 2 2 2 = = = c sin2 C sin(A − B) sin2 C sin(A − B) sin2 C sin(A − B) c sin(A + B) c sin C c c sin(A − B) sin(A + B) . = = = 2 2 2 sin C sin C sin(A − B) sin C sin C π If θ ∈ (0, ), compare tan(sin θ), tan(tan θ), tan(cos θ). 6 sin θ π , then Solution: Since θ ∈ (0, ), then 0 < sin θ < cos θ < 1. Since tan θ = 6 cos θ sin θ = tan θ cos θ. On √ the other hand, 0 < cos √ θ < 1, we have sin θ < tan θ. Since π 3 3 π , 1 > cos θ > cos = , then 1 > cos θ > tan θ > 0. Hence 0 < tan θ < tan = 6 3 6 2 0 < sin θ < tan θ < cos θ < 1. Since y = tan x is increasing in the interval (0, 1), then tan(sin θ) < tan(tan θ) < tan(cos θ). 4.47 . 4.48 Find the value of m which satisfies the inequality cos2 α + 2m sin α − 2m − 2 < 0. Solution: cos2 α+2m sin α−2m−2 < 0 ⇒ sin2 α−2m sin α+2m+1 > 0. Let sin α = t, then −1 t 1. Assume f (t) = t2 − 2mt + 2m + 1 = (t − m)2 − m2 + 2m + 1 > 0, t ∈ [−1, 1]. (1) If m < −1, then f (t)min = 2 + 4m at t = −1. Let 2 + 4m > 0, then 1 1 m > − . It is in contradiction with m < −1. Therefore m > − should be rejected. 2 2 (2) If −1 m 1, then f (t)min =√−m2 + 2m + 1 at t = m. Let −m2 + 2m + 1 > 0, that is m2 − 2m − 1 < 0, √ then 1 − 2 < m 1. (3) If m > 1, then f (t)min = 2 > 0 at Download free eBooks at bookboon.com t = 1. After all, m > 1 − 2. 22.

(23) √ √ 3 π 3 π , then 1 > cos θ > tan θ > 0. Hence , 1 > cos θ > cos = 0 < tan θ < tan = 2 6 3 6 0 < sin θ < tan θ < cos θ < 1. Since y = tan x is increasing in the interval (0, 1), then ELEMENTARY EXERCISE II θ). Trigonometric Functions tan(sin θ) ALGEBRA < tan(tan θ) < BOOK tan(cos 4.48 Find the value of m which satisfies the inequality cos2 α + 2m sin α − 2m − 2 < 0. Solution: cos2 α+2m sin α−2m−2 < 0 ⇒ sin2 α−2m sin α+2m+1 > 0. Let sin α = t, then −1 t 1. Assume f (t) = t2 − 2mt + 2m + 1 = (t − m)2 − m2 + 2m + 1 > 0, t ∈ [−1, 1]. (1) If m < −1, then f (t)min = 2 + 4m at t = −1. Let 2 + 4m > 0, then 1 1 m > − . It is in contradiction with m < −1. Therefore m > − should be rejected. 2 2 (2) If −1 m 1, then f (t)min =√−m2 + 2m + 1 at t = m. Let −m2 + 2m + 1 > 0, that is m2 − 2m − 1 < 0, √ then 1 − 2 < m 1. (3) If m > 1, then f (t)min = 2 > 0 at t = 1. After all, m > 1 − 2. 4.49 Let the angles A, B, C of ABC form an arithmetic progression, a, b, c are the side lengths corresponding to angles A, B, C, and c − a is equal to the altitude C −A . h on the side AC. Find the value of sin 2 h h − . The eSolution: From the given condition, we have h = c − a = sin A sin C C +A C −A cos = quation is equivalent to sin C − sin A = sin A sin C. Thus 2 sin 2 2 1 [cos(C − A) − cos(C + A)] · · · (∗). Since A + C = 2B and A + B + C = 1800 , then 2 1 1 C −A A + C = 1200 . Substituting it into (∗), we have sin = [cos(C − A) + ] ⇒ 2 2 2 1 3 C −A 1 − cos(C − A) 3 C −A = − [− cos(C − A) + 1 − ] ⇒ sin = − + ] ⇒ sin 2 2 2 2 2 4 C −A 2 C −A 1 C −A 3 C −A 3 (sin ) + sin − = 0. Hence, sin = or sin = − (reject2 2 4 2 2 2 2 C −A 1 ed). Therefore, sin = . 2 2 √ π 4.50 Given the function f (x) = a sin +b cos x. (1) If f ( ) = 2 and the maximum 4 √ π value of f (x) is 10, find the value of a, b. (2) If f ( ) = 1 and the minimum value of 3 f (x) is k, find the range of k. √ √ π 2 π Solution: (1) It is easy to figure out that a sin + b cos = 2. Thus (a + b) = 4 4 √ 2 √ 2 ⇒ a + b = 2. On the other √ hand, f (x) = a sin +b cos x = √ a2 + b2 sin(x√+ θ). Since the maximum value of f (x) is 10 when sin(x + θ) = 1, then a2 + b2 = 10, that is a2 + b2 = 10. Since a+b=2 a2 + b2 = 10 we have. or. . a = −1 b=3. . √. a=3 b = −1. √ 1 3 b= 1, that is b = 2− 3a. On the other a+ at (2) From the given condition,Download we havefree eBooks bookboon.com 2 2 √ hand, f (x) = a sin +b cos x = a2 + b2 sin(x23+√θ). The condition sin(x + θ) = −1 can lead to the minimum value of f (x). Hence − a2 + b2 = k, (k < 0). For the equation.

(24) 2 4 4 √ 2 2 2 ⇒ a + b = 2. On the other √ hand, f (x) = a sin +b cos x = √ a + b sin(x√+ θ). Since the maximum value of f (x) is 10 when sin(x + θ) = 1, then a2 + b2 = 10, that is a2 + b2 = 10. Since ELEMENTARY ALGEBRA EXERCISE BOOK II Trigonometric Functions a+b=2 a2 + b2 = 10 √. we have. . or. a = −1 b=3. . √. a=3 b = −1. √ 1 3 (2) From the given condition, we have a+ b = 1, that is b = 2− 3a. On the other 2 2 √ hand, f (x) = a sin +b cos x = a2 + b2 sin(x +√θ). The condition sin(x + θ) = −1 can lead to the minimum value of f (x). Hence − a2 + b2 = k, (k < 0). For the equation system √ b = 2 − 3a a2 + b2 = k 2 √ Eliminating b, we obtain 4a2 −4 3a+4−k 2 = 0. Since a ∈ R, then ∆ = 48−64+16k 2 0 ⇒ k 2 1. Since k < 0, we have k −1.. 4.51 Evaluate the equation metric functions.. √. 1+. x2. +. √. √ 1 + x2 = 2 2 by applying the trigonox. 1 π π + Solution: Let x = tan θ, θ ∈ (− , ), θ = 0. The equation is equivalent to cos θ 2 2 √ √ √ √ π 1 = 2 2 ⇒ sin θ + cos θ = 2 2 sin θ cos θ ⇒ 2 sin(θ + ) = 2 sin 2θ ⇒ 4 sin θ π π π sin(θ + ) = sin 2θ. Hence 2θ = 2kπ + θ + or 2θ = (2k + 1)π − θ − , (k ∈ Z). Thus 4 4 4 5π π π π 2kπ π π ), then θ = or θ = − . , (k ∈ Z).AT Since θ TOP ∈ (− ,RANKED +STUDY θ = 2kπ + or θ = A 12 4 2 2 4√ 3 4 Therefore x = 1 or x = −2 − 3. INTERNATIONAL BUSINESS SCHOOL. no.1. Sw. ed. en. nine years A+B in ABC, and the three side lengths a, b, c 4.52 inIfa row sin(A + B) = tan Reach your 2 full potential at the Stockholm School of Economics, in evaluate one of the innovative cities in the world. The School form an arithmetic sequence, themost radius of its circumcircle and the radius of is ranked by the Financial Times as the number one business its incircle.. school in the Nordic and Baltic countries.. sin A+B A+B A+B A+B 2 = cos ⇒ 2 sin ⇒ A+B www.hhs.se 2Visit us atcos 2 2 2 A+B π = 1 ⇒ cos(A + B) = 0 ⇒ A + B = , we have b = c cos A, a = c sin A. 2 cos2 2 2 Since a, b, c form an arithmetic sequence, that is 2b = a + c, thus 2c cos A = c sin A + c, hence 2 cos A = sin A + 1 · · · (i). On the other hand sin2 A + cos2 A = 1 · · · (ii). Ac4 3 . Assume the radius of incircording to (i) and (ii), we have sin A = , cos A = 5 5 cle is r and the radius of circumcircle is R. Since ABC is a right triangle, then a+b−c c sin A + c cos A − c c r = = . Applying the since theorem = 2R, we 2 2 sin 900 2 c sin A + c cos A − c r c = sin A + cos A − 1 = . = have R = . Therefore, 5 c R 2 Stockholm. Solution: Since sin(A + B) = tan. 4.53 Let a, b, c are real numbers, find the sufficient and necessary condition for that a sin x + b cos x + c > 0 always any real number x. Download free holds eBooksfor at bookboon.com Click on the ad to read more 24. Solution: (1) When a, b are not zero at the same time, we have a sin x + b cos x + c >.

(25) 4.51 Evaluate the 4.51 ALGEBRA Evaluate the equation equation ELEMENTARY EXERCISE BOOK II metric functions. metric functions.. √ √. 11 + +. 2 x x2. + +. √ √. 2 11 + √ +x x2 = 2√ 22 by the = 2 by applying applying the trigonotrigonoTrigonometric Functions x x. π π θ = 0. The equation is equivalent to 11 + π π Solution: ), θ = 0. The equation is equivalent to cos θ + Solution: Let Let x x = = tan tan θ, θ, θθ ∈ ∈ (− (− 2 ,, 2 ), 2 2√ cos θ √ √ π 11 √ √ √ √ π) = √ 2 sin = 2 2 ⇒ sin θ + cos θ = 2 2 sin θ cos θ ⇒ 2 sin(θ + = 2 2 ⇒ sin θ + cos θ = 2 2 sin θ cos θ ⇒ 2 sin(θ + 4 ) = 2 sin 2θ 2θ ⇒ ⇒ sin sin θθ π 4 π π π π π sin(θ sin(θ + + 4 )) = = sin sin 2θ. 2θ. Hence Hence 2θ 2θ = = 2kπ 2kπ + + θθ + + 4 or or 2θ 2θ = = (2k (2k + + 1)π 1)π − − θθ − − 4 ,, (k (k ∈ ∈ Z). Z). Thus Thus 4 π 4 4 π π π π 2kπ π 2kπ π , (k ∈ Z). Since θ ∈ (− π , π ), then θ = π or θ = − 5π 5π θθ = = 2kπ 2kπ + + 4 or or θθ = = 3 + + 4 , (k ∈ Z). Since θ ∈ (− 2 , 2 ), then θ = 4 or θ = − 12 .. 4 3 4√ 2 2 4 12 √ Therefore Therefore x x= = 11 or or x x= = −2 −2 − − 3. 3.. A A+ +B B in ABC, and the three side lengths a, b, c 4.52 in ABC, and the three side lengths a, b, c 4.52 If If sin(A sin(A + + B) B) = = tan tan 2 2 form form an an arithmetic arithmetic sequence, sequence, evaluate evaluate the the radius radius of of its its circumcircle circumcircle and and the the radius radius of of its incircle. its incircle. A+B sin A+ +B B A+ +B B A sin A+B A+ +B B ⇒ 2 sin A A 2 2 cos = Solution: Since sin(A + B) = tan ⇒ ⇒ Solution: Since sin(A + B) = tan 2 ⇒ 2 sin 2 cos 2 = cos A+B A+B 2 2 2 cos 22 A+B π cos22 A + B = = 11 ⇒ ⇒ cos(A cos(A + + B) B) = = 00 ⇒ ⇒A A+ +B B= = π ,, we have b = c cos A, a = c sin A. 22 cos 2 22 we have b = c cos A, a = c sin A. 2 Since Since a, a, b, b, cc form form an an arithmetic arithmetic sequence, sequence, that that is is 2b 2b = =a a+ + c, c, thus thus 2c 2c cos cos A A= = cc sin sin A A+ + c, c, 2 2 2 2 hence hence 22 cos cos A A= = sin sin A A+ + 11 ·· ·· ·· (i). (i). On On the the other other hand hand sin sin A A+ + cos cos A A= = 11 ·· ·· ·· (ii). (ii). AcAc4 33 cording cos A A = = 4 .. Assume the radius of incircording to to (i) (i) and and (ii), (ii), we we have have sin sin A A = = 5 ,, cos 55 Assume the radius of incir5 cle of circumcircle circumcircle is is R. cle is is rr and and the the radius radius of R. Since Since ABC ABC is is aa right right triangle, triangle, then then c sin A + c cos A − c c a + b − c a + b − c c sin A + c cos A − c c = .. Applying rr = = = Applying the the since since theorem theorem sin 9000 = = 2R, 2R, we we 22 22 sin 90 cc rr cc sin sin A A+ + cc cos cos A A− − cc 22 have = = sin sin A A+ + cos cos A A− − 11 = = .. = Therefore, R = have R R= = 2 .. Therefore, 55 cc R 2. 4.53 Let Let a, a, b, b, cc are are real real numbers, numbers, find find the the sufficient sufficient and and necessary necessary condition condition 4.53 for that that a a sin sin x x+ + bb cos cos x x+ + cc > > 00 always always holds holds for for any any real real number number x. x. for Solution: (1) (1) When When a, a, bb are are not not zero zero at at the the same same time, time, we we have have a a sin sin x x+ + bb cos cos x x+ + cc > > Solution: √ cc √ 2 2 ⇔ a . The sufficient and √ 2 00 ⇔ a2 + + bb2 sin(x sin(x + + φ) φ) + + cc > > 00 ⇔ ⇔ sin(x sin(x + + φ) φ) > > − −√ 2 . The sufficient and a2 + a + bb2c √ c necessary < necessary condition condition for for that that the the formula formula always always holds holds is is − −√ < −1. −1. That That is is 2 2 2 2 a + b a + b √ √ 2 2 a a2 + + bb2 < < c. c. (2) When a, (2) When a, bb are are both both zero zero at at the the same same time, time, then then cc > > 0. 0. As a conclusion, for any real number x, the sufficient and necessary As a conclusion, for any real number x,√ the sufficient and necessary condition condition for for that that √ 2 2 2 2 a sin x + b cos x + c > 0 always holds is a + b < c. a sin x + b cos x + c > 0 always holds is a + b < c. √ 3 3 3 cos ωx + 1, (ω > 0), and its period 4.54 Let the function f (x) = sin ωx + 2 2 is π. If α, β are the two roots of the equation f (x) = 0, and α = kπ + β, (k ∈ Z), compute tan(α + β). π Solution: From the given condition, we have f (x) = 3 sin(ωx + ) + 1, (ω > 0). 3 2π = π, then ω = 2. Since α, β are the two roots of the equation Since the period T = ω f (x) = 0, we have 3 sin(2α + π3 ) + 1 = 0, 3 sin(2β + π3 )at+bookboon.com 1 = 0. Download free eBooks π π 25 Simplifying the equation system, we have sin(2α + ) − sin(2β + ) = 0. That is 3 3.

(26) 3 3 3 cos ωx + 1, (ω > 0), and its period sin ωx + 2 2 is π. If α, β are the two roots of the equation f (x) = 0, and α = kπ + β, (k ∈ Z), compute tan(α + β).. 4.54 . Let the function f (x) =. ELEMENTARY ALGEBRA EXERCISE BOOK II. Trigonometric Functions. π Solution: From the given condition, we have f (x) = 3 sin(ωx + ) + 1, (ω > 0). 3 2π = π, then ω = 2. Since α, β are the two roots of the equation Since the period T = ω f (x) = 0, we have 3 sin(2α + π3 ) + 1 = 0, 3 sin(2β + π3 ) + 1 = 0. π π Simplifying the equation system, we have sin(2α + ) − sin(2β + ) = 0. That is 3 3 π 2 cos(α + β + ) sin(α − β) = 0. Since α − β = kπ, (k ∈ Z), then sin(α − β) = 0. Thus 3 π π π π cos(α + β + ) = 0. Hence α + β + = kπ + (k ∈ Z) ⇒ α + β = kπ + (k ∈ Z). 3 3 2 6 √ 3 Therefore, tan(α + β) = . 3 π | sin t|, cos θ = | cos t|, and 0 θ . Find the 2 π value of t such that θ is in the interval [0, ]. 4 π sin θ π = Solution: Since 0 θ , then 0 θ ⇔ 0 tan θ 1. Since tan θ = 2 4 cos θ | sin t| = | tan t|, then 0 tan θ 1 ⇔ 0 | tan t| 1 ⇔ 0 | tan t| 1 ⇔ | cos t| π π −1 tan t 1. Since y = tan t is increasing on the interval (− , ) and the period 2 2 π π T = π, then the solution of inequality −1 tan t 1 is kπ− t kπ+ (k ∈ Z). 4 4. 4.55 . Given sin θ =. . 4.56 The side lengths a, b, c correspond to the angles A, B, C in ABC. √ c cot B , value A, B, C. = If a = ( 3 − 1)c, and 2a − c cot C cos B sin C = Solution: Applying the given equation and the since theorem, we have sin B cos C sin C ⇒ (2 sin A − sin C) cos B = sin B cos C ⇒ 2 sin A cos B = sin(B + C). 2 sin A − sin C 1 π 2π 1 On the other hand, sin(B + C) = sin A, then cos B = , B = , A + C = . 2 3 3. Download free eBooks at bookboon.com 26. Click on the ad to read more.

(27) = | tan t|, then 0 tan θ 1 ⇔ 0 | tan t| 1 ⇔ 0 | tan t| 1 ⇔ | cos t| π π −1 tan t 1. Since y = tan t is increasing on the interval (− , ) and the period 2 2 π π (kFunctions ∈ Z). kπ+ T = π, then the solution inequality −1 tan t 1 is kπ− t Trigonometric ELEMENTARY ALGEBRA EXERCISE of BOOK II 4 4 4.56 The side lengths a, b, c correspond to the angles A, B, C in ABC. √ c cot B = , value A, B, C. If a = ( 3 − 1)c, and cot C 2a − c cos B sin C Solution: Applying the given equation and the since theorem, we have = sin B cos C sin C ⇒ (2 sin A − sin C) cos B = sin B cos C ⇒ 2 sin A cos B = sin(B + C). 2 sin A − sin C π 2π 1 1 . On the other hand, sin(B + C) = sin A, then cos B = , B = , A + C = 2 3 3 √ √ 2 sin A+C cos A−C a sin A 2 2 = 3−1 ⇒ +1 = 3 ⇒ = c sin C sin C √ √ √ 2 3 cos A−C π A−C 2 = 3 ⇒ cos = sin C = cos( − C). Since A, B, C are three 3⇒ 2 sin C 2 2 C −A π 2 According interior angles of a triangle, thus = − C. That is 3C − A = π . 2 2 5 π π 1 and , 2 we have C = π, A = , B = . to 12 4 3 From the given condition. 4.57 If the positive numbers a, b, c form an arithmetic sequence, and a + b = c, 1 1 π 1 arctan + arctan + arctan = . Find the values of a, b, c. a b c 2 Solution: Since a, b, c form an arithmetic sequence, then a + c = 2b. On the oth3 b er hand a + b = c, solving the above two equations, we have a = , c = b. Let 2 2 1 1 1 arctan = α, arctan = β, arctan = γ. Since a, b, c are positive numbers, then a b c 1 1 1 1 α, β, γ are acute angles. Hence tan α = , tan β = , tan γ = . Since arctan + a b c a 1 π 1 π tan α + tan β arctan = − arctan ⇒ tan(α + β) = tan( − γ) ⇒ = cot γ ⇒ b 2 c 2 1 − tan α tan β 1 + 1b b 3 3 b a + b = · b · b − b ⇒ b3 − 4b = 0 ⇒ b = 2, b = 0 1 1 = c ⇒ a + b = abc − c ⇒ 2 2 2 2 1− ab (rejected), b = −2 (rejected). Therefore, a = 1, b = 2, c = 3. 4.58 . If x ∈ [−1, 1], show arcsin x + arccos x =. π . 2. Proof: The function arcsin x and arccos x are defined for x ∈ [−1, 1]. Applying the π induction formula and the definition of inverse cosine function, we obtain sin( − 2 arccos x) = cos(arccos x) = x. Since 0 arccos x π, thus −π − arccos x 0, then π π π π π π − − arccos x , that is − arccos x ∈ [− , ]. Applying the definition of in2 2 2 2 2 2 π π verse sine function, we have arcsin x = − arccos x. After all, arcsin x + arccos x = . 2 2 √. π π π ), x ∈ [− , ]. Find the maxi6 2 4 mum and minimum values of the function y = (sin x + 1)(cos x + 1). 4.59 . Given sin x + cos x =. 2 sin(x +. √ u2 − 1 π Solution: Let sin x + cos x = u, then sin x cos x = , and u = 2 sin(x + ). 4 2 √ Download free eBooks π π at bookboon.com When x = , umax = 2. When x = −27 , u reaches the minimum value. Since 4 6.

(28) 2 arccos x) = cos(arccos x) = x. Since 0 arccos x π, thus −π − arccos x 0, then π π π π π π − − arccos x , that is − arccos x ∈ [− , ]. Applying the definition of in2 2 2 2 2 2 π π ELEMENTARY ALGEBRA EXERCISE BOOK II Trigonometric x + arccos xFunctions = . verse sine function, we have arcsin x = − arccos x. After all, arcsin 2 2 √. π π π ), x ∈ [− , ]. Find the maxi4 6 2 mum and minimum values of the function y = (sin x + 1)(cos x + 1). 4.59 . Given sin x + cos x =. 2 sin(x +. √ u2 − 1 π , and u = 2 sin(x + ). Solution: Let sin x + cos x = u, then sin x cos x = 2 4 √ π π When x = , umax = 2. When x = − , u reaches the minimum value. Since 4 6 √ √ 2 2 π u −1 u −1 π 2− 3 3 2 ⇒ − = ⇒ u = . Since u > 0, then sin(− ) cos = 6 6 2 4 2 2 √ √ √ 2− 3 3−1 3−1 √ = , then u ∈ [ , 2]. Since y = (sin x + 1)(cos x + 1) = umin = 2 2 2 √ 1 3−1 √ 2 sin x+cos x+sin x cos x+1 = (u+1) , and y is increasing on the interval [ , 2], 2 2 √ √ 2+ 3 3+2 2 , ymax = . therefore, ymin = 4 2 4.60 θ.. Given. sin2 θ cos θ sin θ cos2 θ 6 , and = + = . Find the value of y x y2 x2 + y 2 x2. 1 tan2 θ 1 x = = = , we have sin2 θ = tan2 θ + 1 1 + cot2 θ csc2 θ y x2 1 y2 1 2 = = , and cos θ = . Substituting sin2 θ and cos2 θ into x2 + y 2 1 + tan2 θ x2 + y 2 sec2 θ x2 y 2 y2 6 x2 + = ⇒ + = the given second equation, we have 2 (x + y 2 )y 2 (x2 + y 2 )x2 y 2 x2 x2 + y 2 √ √ √ x2 x2 x2 6 ⇒ ( 2 )2 − 6( 2 ) + 1 = 0. Thus 2 = 3 ± 2 2 = ( 2 ± 1)2 . Hence tan θ = ±( 2 + 1) y y y √ √ 3π or tan θ = ±( 2 − 1). Applying tan θ = ±( 2 + 1), we have θ = nπ ± , (n ∈ Z). 8 √ π Applying tan θ = ±( 2 − 1), we have θ = nπ ± , (n ∈ Z). 8 mp − nq p m cos A sin A . = − . Show cos C = = , 4.61 Given in ABC, np − mq q n cos B sin B sin B cos C + cos B sin C m sin(B + C) = ⇒ = Solution : From the given condition, we have sin B n sin B cos(B + C) m m m 1 Since = − cos C . ⇒ cot B sin C = ⇒ cos C + cot B sin C = cos B n n n p cos B cos C − sin B sin C p p 2 Applying − ⇒ = − ⇒ tan B sin C = + cos C . q cos B q q p m mp p m 1 × , 2 we obtain sin2 C = ( − cos C)( + cos C) ⇒ 1 = + cos C( − ). q n nq q n mp − nq . Therefore cos C = np − mq Solution: Since tan θ =. 4.62 Given cos θ + cos φ = a, sin θ + sin φ = b. Compute cos(θ + φ) and sin 2θ + sin 2φ.. Download free eBooks at bookboon.com 28. Click on the ad to read more.

(29) 2. 2. 2. √ 1 3−1 √ 2 sin x+cos x+sin x cos x+1 = 2 (u+1)2 , and y is increasing on the interval [ 2 , 2], √ √ 2 + 3 3 + 2 2. ELEMENTARY II = Trigonometric Functions ymax therefore, ALGEBRA ymin = EXERCISE ,BOOK min max 4 2 4.60 θ.. 2 cos22 θ sin2 θ cos θ 6 sin θ Given x = y , and y 22 + x22 = x22 + y 22 . Find the value of. 1 tan22 θ 1 x = = = Solution: Since tan θ = , we have sin22 θ = tan22 θ + 1 1 + cot22 θ csc22 θ y x22 1 y 22 1 2 2θ = = = , and cos . Substituting sin22 θ and cos22 θ into x22 + y 22 1 + tan22 θ x22 + y 22 sec22 θ 2 2 2 2 x2 y 2 y2 6 x2 + = ⇒ + = the given second equation, we have 22 x22 + y 22 (x + y 22 )y 22 (x22 + y 22 )x22 y 22 x22 √ √ x22 x22 x22 √ 6 ⇒ ( 2 )22 − 6( 2 ) + 1 = 0. Thus 2 = 3 ± 2 2 = ( 2 ± 1)22 . Hence tan θ = ±( 2 + 1) y2 y2 y2 √ √ 3π or tan θ = ±( 2 − 1). Applying tan θ = ±( 2 + 1), we have θ = nπ ± , (n ∈ Z). 8 √ π Applying tan θ = ±( 2 − 1), we have θ = nπ ± , (n ∈ Z). 8 mp pp m cos A sin A mp − − nq nq . .. Show cos C = = − = , 4.61 Given in ABC, = − Show cos C = . np − mq q B n cos sin B cos B q np − mq sin(B sin(B + + C) C) = m m ⇒ sin sin B B cos cos C C+ + cos cos B B sin sin C C = Solution Solution :: From From the the given given condition, condition, we we have have = ⇒ = sin n sin sin B B n sin B B cos(B m m ⇒ cot B sin C = m m − cos C . cos(B + + C) C) = m ⇒ cos C + cot B sin C = m 11 Since ⇒ cos C + cot B sin C = ⇒ cot B sin C = − cos C . Since = cos B n n n n n n cos B p p cos B cos C − sin B sin C pp cos B cos C − sin B sin C = − p ⇒ tan B sin C = p + cos C . Applying − 22 Applying −q ⇒ ⇒ = − q ⇒ tan B sin C = q + cos C . cos B B q cos q q p p × , , we obtain obtain sin sin22 C = ( m − cos C)( + cos C) ⇒ 1 = mp + cos C( m − ). ). 11 × 22 we n n q nq qq mp − nq Therefore cos cos C C= = mp − nq .. Therefore np np − − mq mq 4.62 4.62 Given Given cos cos θθ + + cos cos φ φ = = a, a, sin sin θθ + + sin sin φ φ = = b. b. Compute Compute cos(θ cos(θ + + φ) φ) and and sin sin 2θ 2θ + + sin sin 2φ. 2φ. 2 sin θ+φ cos θ−φ b sin θ + sin φ sin θ + sin φ 2 2 = , on the other hand, = = θ−φ cos θ + cos φ a cos θ + cos φ 2 cos θ+φ cos 2 2 θ+φ b θ+φ 1 − t2 θ+φ , then tan = . Assume tan = t, then cos(θ + φ) = = tan 2 2 a 2 1 + t2 a2 − b 2 2t 2ab , sin(θ + φ) = = 2 (Applying trigonometric function formulas). 2 2 2 a +b 1+t a + b2 Since 2(cos θ + cos φ)(sin θ + sin φ) = 2ab ⇒ sin 2θ + sin 2φ + 2 sin(θ + φ) = 2ab, we 2 4ab = 2ab(1 − 2 ). have sin 2θ + sin 2φ = 2ab − 2 2 a +b a + b2. Solution: Since. 1 1 1 √ − tan−1 = tan−1 , evaluate the value of x. x 3 2+ 3 1 1 3 tan α − tan3 α −1 √ = α, then tan α = √ , tan 3α = Solution: Let tan = 1 − 3 tan2 α 2+ 3 2+ 3 √ √ 3√ 1 − (2+√ 3(2 + 3)2 − 1 20 + 12 3 2+ 3 3)3 √ = √ √ = 1. Hence, 3α = tan−1 1. = 3 3 − 3(2 + 1 − (2+√ 3) (2 + 3) 20 + 12 3 3)2 1 1 −1 −1 1 Download free bookboon.com √ eBooks = tan−1 is equivalent to the eTherefore, the equation 3 tan − tanat x 3 2 + 3 29 1 1 1 1 quation tan−1 1 − tan−1 = tan−1 . Since tan(tan−1 1 − tan−1 ) = tan(tan−1 ) ⇒ 4.63 . If 3 tan−1.

(30) a2 − b 2 2t 2ab , sin(θ + φ) = = (Applying trigonometric function formulas). a2 + b 2 1 + t2 a2 + b2 Since 2(cos θ + cos φ)(sin θ + sin φ) = 2ab ⇒ sin 2θ + sin 2φ + 2 sin(θ + φ) = 2ab, we 2 4ab ELEMENTARY EXERCISE Trigonometric Functions = 2ab(1 − 2 ). have sin 2θALGEBRA + sin 2φ = 2abBOOK − II2 2 a + b2 a +b 1 1 1 √ − tan−1 = tan−1 , evaluate the value of x. x 3 2+ 3 1 3 tan α − tan3 α 1 √ = α, then tan α = √ , tan 3α = = Solution: Let tan−1 1 − 3 tan2 α 2+ 3 2+ 3 √ √ 3√ 1 − (2+√ 3(2 + 3)2 − 1 20 + 12 3 2+ 3 3)3 √ √ = √ = 1. Hence, 3α = tan−1 1. = 3 3 1 − (2+√ (2 + 3) − 3(2 + 3) 20 + 12 3 3)2 1 1 1 √ − tan−1 = tan−1 is equivalent to the eTherefore, the equation 3 tan−1 x 3 2+ 3 1 1 1 1 quation tan−1 1 − tan−1 = tan−1 . Since tan(tan−1 1 − tan−1 ) = tan(tan−1 ) ⇒ 3 x 3 x 1 − 31 1 = . After all, x = 2. x 1 + 13 4.63 . If 3 tan−1. 4.64 Let sin α = p sin β, cos α = q cos β, sin α + cos α = r(sin β + cos β), show (p − r)2 (1 − q 2 ) + (q − r)2 (1 − p2 ) = 0. Solution: From the given conditions, we have p2 sin2 β + q 2 cos2 β = sin2 α + cos2 α = sin2 β + cos2 β. Dividing both sides of the equation by cos2 β, we have p2 tan2 β + q 2 = q2 − 1 1 Since p sin β + q cos β = r(sin β + cos β) ⇒ . tan2 β + 1, that is, tan2 β = 1 − p2 r−q 2 Applying 1 and , 2 we have . (p − r) sin β = (r − q) cos β, then tan β = p−r (r − q)2 q2 − 1 = . Simplifying the formula, we obtain (p−r)2 (1−q)2 +(q−r)2 (1−p)2 = 1 − p2 (p − r)2 0. 4.65 Let a, b, c are the side lengths of triangle ABC corresponding to angles A, B, C, (sin B + sin C + sin A)(sin B + sin C − sin A) = 3 sin B sin C. b, c are the two roots of equation x2 − 3x + 4 cos A = 0, and b > c. The radius of circumcircle of ABC is 1. Find the value of ∠A, a, b, c. Solution: From the given conditions, we have b + c = 3, bc = 4 cos A. Applying the sine law, b = 2R sin B = 2 sin B, c = 2R sin C = 2 sin C. Adding the two equations togethb+c 3 1 Multiplying the two equations, we obtain er, we obtain sin B + sin C = = . 2 2 bc 2 Simplifying the equation (sin B + sin C + sin A)(sin B + sin B sin C = = cos A . 4 3 sin C − sin A) = 3 sin B sin C, we get (sin B + sin C)2 − sin A2 = 3 sin B sin C . 9 1 and 2 into , 3 then − sin2 A = 3 cos A ⇒ 4 cos2 A − 12 cos A + 5 = Submit 4 1 5 0 ⇒ cos A = or cos A = (rejected). Hence ∠A = 600 . According to the equations 2 2 system b+c=3 bc = 2 √ and b > c, we have b = 2, c = 1, a = 2R sin A = 3. 1 x 1 x 1 x 1 x tan + 2 tan 2 + · · · + n tan n = n cot n − 2 cot 2x. 2 2free eBooks 2 Download 2 at bookboon.com 2 2 2 2 2 2 1 1 − tan x 30 1 − tan x = 2 cot 2x, then = 2 Proof: Since cot x − tan x = = 2. 4.66 Show tan x +.

(31) system. . b+c=3 bc = 2 √ ELEMENTARY ALGEBRA EXERCISE BOOK II and b > c, we have b = 2, c = 1, a = 2R sin A = 3.. Trigonometric Functions. x 1 1 x 1 x 1 x tan + 2 tan 2 + · · · + n tan n = n cot n − 2 cot 2x. 2 2 2 2 2 2 2 2 2 2 1 − tan x 1 1 − tan x = 2 = 2 = 2 cot 2x, then Proof: Since cot x − tan x = tan x 2 tan x tan 2x 1 x 1 x 1 x 1 x tan x = cot x − 2 cot 2x. Similarly tan = cot − cot x, 2 tan 2 = 2 cot 2 − 2 2 2 2 2 2 2 2 1 x 1 x 1 x 1 x cot , · · · , n tan n = n cot n − n−1 cot n−1 . Adding the above equations, we 2 2 2 2 2 2 2 2 1 x 1 x 1 x 1 x have tan x + tan + 2 tan 2 + · · · + n tan n = n cot n − 2 cot 2x. 2 2 2 2 2 2 2 2. 4.66 Show tan x +. x−. 4.67 . √ π 1 − x2 √ = . 4 2. If 0 x 1, show arcsin x − arcsin √ x − 1 − x2 √ Proof: Let arcsin x = α, arcsin = β. Since sin α = x, 0 x 1, then 0 2 √ π α . Hence cos α = 1 − sin2 α = 1 − x2 . Since 0 x2 1 ⇒ −1 −x2 0 ⇒ 2 √ √ √ 0 1 − x2 1 ⇒ 0 √ 1 − x2 1 ⇒ −1 − 1 − x2 √0 ⇒ −1 x − 1 − x2 1. π 1 x − 1 − x2 1 x − 1 − x2 π √ √ That is − √ √ . Since sin β = , then − β . 4√ 4 2 2 2 2 2 √ π π π 1 x− 1−x 1 √ Since sin(α − ) = sin α cos − cos α sin = x √ − 1 − x2 √ = = 4 4 4 2 2 2 π π π π π π sin β. Since 0 α , then − α − . Hence α − = β, that is α − β = . 2 4 4 4 4 4 √ x − 1 − x2 π √ Therefore, arcsin x − arcsin = . 4 2 4.68 . Solve the equation system  2   arcsin x arcsin y = π · · · 1 122   arccos x arccos y = π · · · 2 24. π π − arcsin x, arccos y = − arcsin y to rewrite the given 2 2 π π π2 2 as the formula ( − arcsin x)( − arcsin y) = equation . Let α = arcsin x, β = 2 2 24 arcsin y, then the given equation system is equivalent to  2   αβ = π 12 2   αβ − (α + β) π = − 5π 2 24 Solution: Applying arccos x =. That is,.     . π2 12 7π α+β = 12 αβ =. π π Assume α, β are the roots of equation 12z 2 − 7πz + π 2 = 0, then α = , β = , or 3 4 π π π π π Download freeπeBooks at bookboon.com α = , β = . Those are arcsin x = , arcsin y = , or arcsin x = , arcsin y = . 4 3 4 3 √ 3 31 √4 √ √ 2 2 3 3.

(32) arcsin y, then the given equation system is equivalent to  2   αβ = π 12 2 ELEMENTARY ALGEBRA EXERCISE BOOK II   αβ − (α + β) π = − 5π 2 24 That is,.     . Trigonometric Functions. π2 12 7π α+β = 12 αβ =. π π Assume α, β are the roots of equation 12z 2 − 7πz + π 2 = 0, then α = , β = , or 3 4 π π π π π π α = , β = . Those are arcsin x = , arcsin y = , or arcsin x = , arcsin y = . 4 3 4 3 √3 √4 √ √ 3 2 2 3 , y1 = or x2 = , y2 = . We can verify that The solutions are x1 = 2 √ 2√ 2 2 √ √ 3 2 2 3 x1 = , y1 = and x2 = , y2 = are both the roots of the system. 2 2 2 2 4.69 Let A, B, C are the three angles of ABC corresponding to the side lengths a, b, c, and they form a geometric sequence, and b2 − a2 = ac. Find the value of ∠B. 1 B, C = qB. q 1 qπ . SSince A + B + C = π, then B + B + qB = π, that is B = 2 q q +q+1 ince b2 − a2 = ac, according to the cosine law b2 = a2 + c2 − 2ac cos B, we have ac = c2 − 2ac cos B. Since c = 0, thus a = c − 2a cos B. Applying the sine law, we have Solution: Since A, B, C form a geometric sequence, we assume A =. Excellent Economics and Business programmes at:. “The perfect start of a successful, international career.” CLICK HERE. to discover why both socially and academically the University of Groningen is one of the best places for a student to be. www.rug.nl/feb/education. Download free eBooks at bookboon.com 32. Click on the ad to read more.

(33) 4.69 Let A, B, C are the three angles of ABC corresponding to the side lengths a, b, c, and they form a geometric sequence, and b2 − a2 = ac. Find the value of ∠B.. ELEMENTARY ALGEBRA EXERCISE BOOK II. Trigonometric Functions. 1 B, C = qB. q 1 qπ . SSince A + B + C = π, then B + B + qB = π, that is B = 2 q q +q+1 ince b2 − a2 = ac, according to the cosine law b2 = a2 + c2 − 2ac cos B, we have ac = c2 − 2ac cos B. Since c = 0, thus a = c − 2a cos B. Applying the sine law, we have Solution: Since A, B, C form a geometric sequence, we assume A =. sin A = sin C − 2 sin A cos B ⇒ sin A = sin C − [sin(A + B) + sin(A − B)] ⇒ sin A = 1 sin C − sin C − sin(A − B) ⇒ sin A = sin(B − A) ⇒ A = B − A ⇒ A = B. Hence, 2 1 2π 2π 1 B = B. After all, q = 2, B = 2 = . 2 q 2 +2+1 7 π x , show cot n − cot x n, (n ∈ N ). 2 2 1 + cos x cos x 1 x − = . Since 0 < x Proof: (1) When n = 1, cot − cot x = 2 sin x sin x sin x 1 x π , 0 < sin x 1, then 1, that is cot − cot x 1. The equation holds when 2 sin x 2 π x= . 2 x (2) Assume the inequation holds when n = k, (k ∈ N ), that is cot k − cot x k, then 2 1 + cos 2xk x 1 x cot k+1 − cot x = − cot x = + cot k − cot x when n = k + 1. Since 2 sin 2xk sin 2xk 2 x π 1 > 1, (0 < x ), k ∈ N , then cot k+1 − cot x > k + 1. Therefore, for all n ∈ N , sin 2xk 2 2 x cot n − cot x n holds. 2 4.70 . If 0 x . 4.71 ···.. Find the sum of the formula tan−1. 1 1 1 +tan−1 +tan−1 + 1+1·2 1+2·3 1+3·4. 1 (n + 1) − n = tan−1 = tan−1 (n + 1 + n · (n + 1) 1 + n · (n + 1) 1) − tan−1 n. Substituting n = 1, 2, 3, · · · , into the above equation and adding these 1 1 1 equations, we have tan−1 + tan−1 + tan−1 + · · · = (tan−1 2 − 1+1·2 1+2·3 1+3·4 tan−1 1) + (tan−1 3 − tan−1 2) + (tan−1 4 − tan−1 3) + · · · = − tan−1 1 + tan−1 ∞ = π π π − + = . 4 2 4 Solution: The nth term is tan−1. 4.72 n−1 . n+1. Let A + B + C = π, and sin(A +. C A B C ) = n sin . Show tan tan = 2 2 2 2. 1800 − A − B B−A B−A C 1 S) = sin(900 − ) = cos · · · . Proof: sin(A+ ) = sin(A+ 2 2 2 2 C C C A+B C A+B 2 ince sin(A+ ) = n sin , then sin = cos . Hence sin(A+ ) = n cos · · · . 2 2 2 2 2 2. Download free eBooks at bookboon.com 33.

(34) ELEMENTARY ALGEBRA EXERCISE BOOK II. Trigonometric Functions. A+B A B A B B−A 1 and , 2 we have cos = n cos , that is cos cos +sin sin = Applying 2 2 2 2 2 2 A B A B A B A B n(cos cos − sin sin ) ⇒ (n + 1) sin sin = (n − 1) cos cos . Therefore 2 2 2 2 2 2 2 2 A B n−1 tan tan = . 2 2 n+1 4.73 If tan α, tan β are the two roots of the equation x2 + px + q = 0, express sin2 (α + β) + p sin(α + β) cos(α + β) + q cos2 (α + β) by p and q. Solution: According to the relation between roots and coefficients, we have tan α + tan β = −p, that is p = −(tan α + tan β), q = tan α tan β. Hence the quantity is equal to sin2 (α + β) − (tan α + tan β) sin(α + β) cos(α + β) + tan α tan β cos2 (α + β) = sin2 (α + β) − (tan α + tan β) sin(α + β) cos(α + β) + tan α tan β[1 − sin2 (α + β)] = sin2 (α + β)(1 − tan α tan β) − (tan α + tan β) sin(α + β) cos(α + β) + tan α tan β = sin(α + β) (1 − tan α tan β) − (tan α + tan β)} + tan α tan β = sin(α + β) cos(α + β){ cos(α + β) tan α + tan β sin(α + β) cos(α + β){ (1 − tan α tan β) − (tan α + tan β)} + tan α tan β = 1 − tan α tan β tan α tan β = q. 4.74 . If (1 − tan θ)(1 + sin 2θ) = 1 + tan θ, evaluate the value of θ.. sin θ sin θ )(sin θ + cos θ)2 = 1 + ⇒ cos θ cos θ 2 (cos θ − sin θ)(cos θ + sin θ) = cos θ + sin θ ⇒ (cos θ + sin θ)[(cos θ − sin θ)(cos θ + 3π sin θ) − 1] = 0. If cos θ + sin θ = 0, then tan θ = −1, thus θ = nπ + (n ∈ Z). If 4 (cos θ − sin θ)(cos θ + sin θ) − 1 = 0, then cos2 θ − sin2 θ = 1, thus cos 2θ = 1, hence 3π 2θ = 2nπ, that is θ = nπ (n ∈ Z). Therefore, θ = nπ + or θ = nπ (n ∈ Z). 4. Solution: The given equation is equivalent to (1 −. B C A Let A, B, C are the interior angles of triangle ABC, and cot , cot , cot 2 2 2 A C form an arithmetic sequence. Show cot cot = 3. 2 2 cos A2 C B A = 2 cot ⇒ + Proof: From the given conditions, we have cot + cot 2 2 2 sin A2 cos B2 sin A+C sin A+C 2 sin A+C cos C2 2 2 2 =2 =2 ⇒ = . Since A, B, C are the inteA C A+C sin C2 sin B2 cos A+C sin sin cos 2 2 2 2. 4.75 . A+C A C A+C , cos , sin , sin are all nonzero. rior angles of triangle ABC, then sin 2 2 2 2 A+C A C A C A C A C Hence cos = 2 sin sin ⇒ cos cos − sin sin = 2 sin sin ⇒ 2 2 2 2 2 2 2 2 2 A C A C A C cos cos = 3 sin sin . Therefore, cot cot = 3. 2 2 2 2 2 2 π π 4.76 If α ∈ (0, ), β ∈ (0, ), and α + β = θ is a constant. Find the 2 2 minimum value of csc α + csc β. Download free eBooks at bookboon.com Solution: csc α+csc β =. 1. +. 1. =. sin α34+ sin β. =. 2 sin α+β cos α−β 2 2. =. 4 sin α+β cos α−β 2 2.

(35) rior angles of triangle ABC, then sin , cos , sin , sin are all nonzero. 2 2 2 2 A+C A C A C A C A C Hence cos = 2 sin sin ⇒ cos cos − sin sin = 2 sin sin ⇒ 2 2 2 2 2 2 2 2 2 A C A BOOK C II A C ELEMENTARY EXERCISE Trigonometric Functions cos cos ALGEBRA = 3 sin sin . Therefore, cot cot = 3. 2 2 2 2 2 2 π π 4.76 If α ∈ (0, ), β ∈ (0, ), and α + β = θ is a constant. Find the 2 2 minimum value of csc α + csc β. 2 sin α+β 4 sin α+β cos α−β cos α−β 1 sin α + sin β 1 2 2 2 2 + = = = Solution: csc α+csc β = sin α sin β sin α sin β sin α sin β 2 sin α sin β 4 sin α+β 2 sin α+β 2 sin α+β cos α−β cos α−β cos α−β α+β 2 2 2 2 2 2 = = 1+cos(α−β) 1+cos(α+β) = = sin α−β α+β 2 2 cos(α − β) − cos(α + β) 2 cos 2 − cos 2 − 2 2 1 1 + ]. Since α + β = θ is a constant, then the above [ α+β α−β α+β cos α−β − cos cos + cos 2 2 2 2 1 1 θ + ]. This function reachquantity is equal to sin [ α−β α−β θ 2 cos 2 − cos 2 cos 2 + cos 2θ α−β = 1, i.e. α = β. The minimum value of es the minimum value when cos 2 θ 1 1 θ 2 2 csc α + csc β is sin [ + ] = sin = . As a conclusion, 2 θ θ θ 2 1 − cos 2 2 sin 2 1 + cos 2 sin 2θ 2 (0 < θ < π). (csc α + csc β)min = sin 2θ 4.77 Let c be the hypotenuse length of ABC, ∠C = 900 , the area is S. Find the values of a, b, ∠A, ∠B. 1 Solution: From the given conditions, we have c2 = a2 + b2 , S = ab. Hence a2 + 2ab + 2 √ √ 1 Similarly we have a − b = c2 − 4S · · · . 2 b2 = c2 + 4S, that is a + b = c2 + 4S · · · . √ √ √ 1 1 √ 2 1 and , 2 a = ( c + 4S + c2 − 4S), b = ( c2 + 4S − c2 − 4S). According to 2√ 2 √ √ √ 1 2 + 4S + 2 − 4S) 2 − 4S)2 ( c c + 4S +we chave a (pastc2four In the years drilled 2 √ We also can obtain tan A = = 1 √ = = 2 2 b c + 4S − c + 4S ( c2 + 4S − c2 − 4S) 2 √ √ √ c2 + c4 − 16S 2 c2 + c4 − 16S 2 c2 + 4S + 2 c4 − 16S 2 + c2 − 4S = ⇒ A = arctan . 8S 4S 4S √ √ √ 1 2 ( c2 + 4S − c2 − 4S) b c2 + 4S − more 2 c4than − 16S c2 −the 4Sworld. That’s twice + around √ Similarly tan B = = 21 √ = = 2 2 2 2 c + 4S − c + 4S a ( c + 4S + c − 4S) 2 √ √ √ c2 − c4 − 16S 2 c2 Who − are c4 −we? 16S 2 2c2 − 2 c4 − 16S 2 = ⇒ B = arctan We are the world’s . 8S 4S 4S largest oilfield services company . Working globally—often in remote and challenging locations—. 89,000 km 1. we invent, design, engineer, and apply technology to help our customers find and produce oil and gas safely.. Who are we looking for?. Every year, we need thousands of graduates to begin dynamic careers in the following domains: n Engineering, Research and Operations n Geoscience and Petrotechnical n Commercial and Business. What will you be?. careers.slb.com Based on Fortune 500 ranking 2011. Copyright © 2015 Schlumberger. All rights reserved.. 1. Download free eBooks at bookboon.com 35. Click on the ad to read more.

(36) 2 1 θ 1 θ 2 csc α + csc β is sin [ + ] = sin θ θ 2 1 − cos 2 2 sin2 1 + cos 2 2 ELEMENTARY ALGEBRA BOOK = (0 II< θ < π). (csc α + csc β)min EXERCISE sin 2θ. θ 2. =. 2 . As a conclusion, sin 2θ Trigonometric Functions. 4.77 Let c be the hypotenuse length of ABC, ∠C = 900 , the area is S. Find the values of a, b, ∠A, ∠B. 1 Solution: From the given conditions, we have c2 = a2 + b2 , S = ab. Hence a2 + 2ab + 2 √ √ 1 Similarly we have a − b = c2 − 4S · · · . 2 b2 = c2 + 4S, that is a + b = c2 + 4S · · · . √ √ √ 1 1 √ 2 1 and , 2 a = ( c + 4S + c2 − 4S), b = ( c2 + 4S − c2 − 4S). According to 2 2√ √ √ √ 1 2 + 4S + 2 − 4S) ( c c a ( c2 + 4S + c2 − 4S)2 2 √ We also can obtain tan A = = 1 √ = = b c2 + 4S − c2 + 4S ( c2 + 4S − c2 − 4S) 2 √ √ √ c2 + c4 − 16S 2 c2 + c4 − 16S 2 c2 + 4S + 2 c4 − 16S 2 + c2 − 4S = ⇒ A = arctan . 8S 4S 4S √ √ √ 1 ( c2 + 4S − c2 − 4S) b c2 + 4S − 2 c4 − 16S 2 + c2 − 4S √ Similarly tan B = = 21 √ = = 2 + 4S − c2 + 4S 2 + 4S + 2 − 4S) a c ( c c 2 √ √ √ c2 − c4 − 16S 2 c2 − c4 − 16S 2 2c2 − 2 c4 − 16S 2 = ⇒ B = arctan . 8S 4S 4S. 4.78 If tan α, tan β are the two roots of the equation x2 − 2(log3 12 + log4 12)x − log3 12 log4 12 = 0, show sin(α + β) + 2 sin α sin β = 0. Proof: The Vieta’s formulas lead to tan α + tan β = 2(log3 12 + log4 12), tan α tan β = sin β sin(α + β) sin α 1 + = = 2(log3 4 + log4 3 + 2) · · · , − log3 12 log4 12. Then cos α cos β cos α cos β sin α sin β 2 divided 1 by , 2 = −(log3 4 + 1)(log4 3 + 1) = −(log3 4 + log4 3 + 2) · · · . cos α cos β sin(α + β) then = −2 ⇒ sin(α + β) + 2 sin α sin β = 0. sin α sin β cos θ cos φ2. cos φ cos 2θ 4.79 Let + = 1, show cos θ + cos φ = 1. cos(θ − φ2 ) cos(φ − 2θ ) cos θ cos φ2 cos φ cos 2θ cos(θ + φ2 ) + cos(θ − φ2 ) cos(φ + 2θ ) + cos(φ − 2θ ) Proof: + + =1⇒ = 2 cos(φ − 2θ ) cos(θ − φ2 ) cos(φ − 2θ ) 2 cos(θ − φ2 ) cos(θ + φ2 ) + cos(θ − φ2 ) 1 cos(φ + 2θ ) + cos(φ − 2θ ) 1 cos(θ + φ2 ) 1⇒ − = − + = 0 ⇒ 2 2 2 cos(φ − 2θ ) 2 cos(θ − φ2 ) cos(θ − φ2 ) cos(φ + 2θ ) cos θ cos φ2 − sin θ sin φ2 cos φ cos 2θ − sin φ sin 2θ − ⇒ = − . Cancellate the cos(φ − 2θ ) cos φ cos 2θ + sin φ sin 2θ cos θ cos φ2 + sin θ sin φ2 cos θ cos φ2 cos θ cos φ2 sin φ sin 2θ denominator to rewrite the equation as ⇒ = = cos φ cos 2θ sin θ sin φ2 2 sin 2θ cos 2θ sin φ2 2 sin φ2 cos φ2 sin 2θ 2 sin φ2 sin 2θ cos θ θ φ ⇒ cos θ cos φ = 4 sin2 sin2 = ⇒ = θ φ θ cos φ 2 2 cos φ cos 2 2 sin 2 sin 2 (1−cos θ)(1−cos φ) ⇒ cos θ cos φ = 1−cos θ−cos φ+cos θ cos φ. Hence cos θ+cos φ = 1. 4.80 . Given sin{2 cos−1 (cot 2 tan−1 x)} = 0, evaluate the value of x.. 1 − x2 1 1 − tan2 θ = . = 2 tan θ 2x tan 2θ Download free eBooks at bookboon.com 2 2 1−x 1−x 36 −1 Thus the given equation yields that sin{2 cos } = 0. Hence 2 cos−1 = 2x 2x 2. Solution: Let tan−1 x = θ, then tan θ = x, cot 2θ =.

(37) cos φ cos 2 sin θ sin 2 2 sin 2 cos 2 sin 2 φ θ 2 sin cos sin 2 sin 2 sin 2 cos θ θ φ = ⇒ cos θ cos φ = 4 sin2 sin2 ⇒ = θ φ θ 2 2 cos φ cos φ cos 2 2 sin 2 sin 2 ELEMENTARY ALGEBRAφ) EXERCISE Trigonometric (1−cos θ)(1−cos ⇒ cosBOOK θ cosII φ = 1−cos θ−cos φ+cos θ cos φ. Hence cos θ+cosFunctions φ = 1. φ 2. φ 2. 4.80 . θ 2. Given sin{2 cos−1 (cot 2 tan−1 x)} = 0, evaluate the value of x.. 1 − tan2 θ 1 − x2 1 = = . tan 2θ 2 tan θ 2x 2 2 1−x 1−x Thus the given equation yields that sin{2 cos−1 } = 0. Hence 2 cos−1 = 2x 2x 2 nπ 1−x nπ, (n ∈ N ). That is cos−1 2x = 2 . Dividing the equation by the cosine function, we 1 − x2 nπ nπ have = cos , (n ∈ N ). Since n is an arbitrary real number, then cos = 0, 2x 2 2 √ 1 − x2 1 − x2 or 1, or −1. If = 0, we have x = ±1. If = 1, then x = −1 ± 2. 2x 2x Solution: Let tan−1 x = θ, then tan θ = x, cot 2θ =. √ 1 − x2 = −1, then x = 1 ± 2. As a conclusion, the solutions are x = ±1, or 2x √ √ x = −1 ± 2, or x = 1 ± 2. If. 4.81 . Given tan−1 x +. 1 π sec−1 5x = , find the value of x. 2 4. 1 Solution: Let tan−1 x = θ, sec−1 5x = φ. Then x = tan θ, 5x = sec 2φ. Since 2 π π π 1 θ + φ = , we have 2φ = − 2θ. Hence 5x = sec( − 2θ) = csc 2θ = = 4 2 2 2 sin θ cos θ 1 1 1 1 sin2 θ + cos2 θ = (tan θ + ). That is 10x = x + ⇒ 9x2 = 1 ⇒ x = ± . 2 sin θ cos θ 2 tan θ x 3 4.82 . If a, b, c are the side lengths of triangle ABC, show. a sin(B − C) = b2 − c2. c sin(A − B) b sin(C − A) = . 2 2 c −a a2 − b 2 a b c a sin(B − C) k sin A sin(B − C) Proof: Let = = = k. We obtain that = = 2 2 sin A sin B sin C b −c b2 − c 2 k[(sin B cos C)2 − (cos B sin C)2 ] k sin(B + C) sin(B − C) = b2 − c2 b2 − c2 2 2 2 k(sin2 B − sin2 C) sin B k[sin B(1 − sin C) − (1 − sin B) sin2 C] = = . Since = 2 2 2 2 b −c b −c b 1 sin2 B − sin2 C 1 sin C = , applying the geometric theorem, we have = 2 , then 2 2 c k b −c k a sin(B − C) b sin(C − A) 1 1 c sin(A − B) 1 = . Similarly, = , = . Therefore b2 − c2 k c2 − a2 k a2 − b 2 k b sin(C − A) c sin(A − B) a sin(B − C) = = . b2 − c 2 c 2 − a2 a2 − b 2 4.83 For a triangle ABC, if tan A tan B > 1, show the triangle is an acute triangle. Proof 1: Since tan A tan B > 1 ⇒ tan A tan B − 1 > 0 ⇒. sin A sin B − cos A cos B > cos A cos B. cos(A + B) cos C 0 ⇒ − > 0 ⇒ > 0. Hence cos A cos B cos C > 0. Therecos A cos B cos A cos B fore cos A > 0. Otherwise, if cos A < 0 which means A is an obtuse angle, applying Download freeC eBooks at bookboon.com cos A cos B cos C > 0, we have cos B cos < 0 which means one of B and C is an obtuse 0 37 angle. Hence A + B + C > 180 . The conclusion is contradicting to A + B + C = 1800 . Therefore cos A > 0. Similarly, we obtain cos B > 0, cos C > 0. As a conclusion,.

(38) k c b 2 − c2 k2 a sin(B − C) 1 b sin(C − A) 1 c sin(A − B) 1 = . Similarly, = , = . Therefore b2 − c2 k c2 − a2 k a 2 − b2 k b sin(C − A) c sin(A − B) a sin(B − C) = EXERCISE . ELEMENTARY ALGEBRA BOOK = II Trigonometric Functions c 2 − a2 a 2 − b2 b2 − c2 4.83 For a triangle ABC, if tan A tan B > 1, show the triangle is an acute triangle. Proof 1: Since tan A tan B > 1 ⇒ tan A tan B − 1 > 0 ⇒. sin A sin B − cos A cos B > cos A cos B. cos C cos(A + B) > 0 ⇒ > 0. Hence cos A cos B cos C > 0. There0 ⇒ − cos A cos B cos A cos B fore cos A > 0. Otherwise, if cos A < 0 which means A is an obtuse angle, applying cos A cos B cos C > 0, we have cos B cos C < 0 which means one of B and C is an obtuse angle. Hence A + B + C > 1800 . The conclusion is contradicting to A + B + C = 1800 . Therefore cos A > 0. Similarly, we obtain cos B > 0, cos C > 0. As a conclusion, ABC is an acute triangle.. Proof 2: Since tan A tan B > 1, then tan A and tan B are the same sign. If tan A and tan B are both negative, we obtain A and B are both obtuse angles which is contradictory with the given condition. If tan A and tan B are both positive, we obtain A and B tan A + tan B tan A + tan B are both acute angles. Since 0 > 1 − tan A tan B = =− . tan(A + B) tan C On the other hand, since tan A + tan B > 0, then tan C > 0, hence C is also an acute angle. Consequently, ABC is an acute triangle. 4.84 . Given |A| < 1, sin α = A sin(α + β). Show tan(α + β) =. sin β . cos β − A. Solution: sin α = A sin(α + β) ⇒ sin[(α + β) − β] = A sin(α + β) ⇒ sin(α + β) cos β − cos(α+β) sin β = A sin(α+β) ⇒ sin(α+β)(cos β−A) = cos(α+β) sin β ⇒ tan(α+β) = sin β . cos β − A. American online 4 LIGS 4.85 GivenUniversity 0 < x < π, find the minimum value of function f (x) = sin x + . sin x is currently enrolling in the. Solution: Since 0 < sin x 1 for 0 < x < π, then the minimum value of f (x) is Interactive Online BBA, MBA, MSc, π π equal to the minimum value of f (x) for 0 < x . Assume 0 < x2 < x1 , then DBA and PhD programs: 2 2 4 4 −(sin x1 − sin x2 )(4 − sin x1 sin x2 ) )−(sin x2 + )= . f (x1 )−f (x2 ) = (sin x1 + sin x1 sin x2 sin x1 sin x2 ▶▶ enroll 30th, Since 0 < by sin September x2 < sin x1 1, 4 2014 − sin and x1 sin x2 > 0, we have f (x1 ) − f (x2 ) < 0, that is π on thef (x) tuition! f▶ (x▶ 1save ) < fup (x2to ). 16% Therefore is decreasing on the interval (0, ]. The minimum value 2 π payisin510atinstallments / 2 years the minimum value of f (x) is 5 for 0 < x < π. of▶f▶ (x) x = . Consequently, 2. ▶▶ Interactive Online education ▶▶ visit www.ligsuniversity.com to. 4.86 If α and β are two acute angles that satisfy the equation sin2 α+sin2 β = π find out more! sin(α + β). Show α + β = . 2 2 Note: LIGS University accredited + sin2 β is =not sin(α + β) ⇒by sinany α + sin2 β = sin α cos β + cos α sin β ⇒ Proof: sin2 α nationally recognized accrediting agency listed π sin α− cos β) =of sin β(cos α − sin β). Since 0 < α, β < , we have sin α > byα(sin the US Secretary Education. 2 More info here. 0, sin β > 0. Hence sin α − cos β and cos α − sin β are the same signs, or they are both zero at the same time. (1) If − cosatβbookboon.com >0 Download sin free α eBooks Click on the ad to read more cos α −38sin β > 0 then .

(39) tory with the given condition. If tan A and tan B are both positive, we obtain A and B tan A + tan B tan A + tan B . are both acute angles. Since 0 > 1 − tan A tan B = =− tan C tan(A + B) On the other hand, since tan A + tan B > 0, then tan C > 0, hence C is also an acute ELEMENTARY ALGEBRA EXERCISE BOOK is II an acute triangle. Trigonometric Functions angle. Consequently, ABC 4.84 . Given |A| < 1, sin α = A sin(α + β). Show tan(α + β) =. sin β . cos β − A. Solution: sin α = A sin(α + β) ⇒ sin[(α + β) − β] = A sin(α + β) ⇒ sin(α + β) cos β − cos(α+β) sin β = A sin(α+β) ⇒ sin(α+β)(cos β−A) = cos(α+β) sin β ⇒ tan(α+β) = sin β . cos β − A 4.85 . Given 0 < x < π, find the minimum value of function f (x) = sin x +. 4 . sin x. Solution: Since 0 < sin x 1 for 0 < x < π, then the minimum value of f (x) is π π equal to the minimum value of f (x) for 0 < x . Assume 0 < x2 < x1 , then 2 2 4 4 −(sin x1 − sin x2 )(4 − sin x1 sin x2 ) )−(sin x2 + )= . f (x1 )−f (x2 ) = (sin x1 + sin x1 sin x2 sin x1 sin x2 Since 0 < sin x2 < sin x1 1, 4 − sin x1 sin x2 > 0, we have f (x1 ) − f (x2 ) < 0, that is π f (x1 ) < f (x2 ). Therefore f (x) is decreasing on the interval (0, ]. The minimum value 2 π of f (x) is 5 at x = . Consequently, the minimum value of f (x) is 5 for 0 < x < π. 2 4.86 If α and β are two acute angles that satisfy the equation sin2 α+sin2 β = π sin(α + β). Show α + β = . 2 Proof: sin2 α + sin2 β = sin(α + β) ⇒ sin2 α + sin2 β = sin α cos β + cos α sin β ⇒ π sin α(sin α − cos β) = sin β(cos α − sin β). Since 0 < α, β < , we have sin α > 2 0, sin β > 0. Hence sin α − cos β and cos α − sin β are the same signs, or they are both zero at the same time. (1) If sin α − cos β > 0 cos α − sin β > 0 then sin α > cos β > 0 cos α > sin β > 0 ⇒ sin2 α + cos2 α > sin2 β + cos2 β, which means 1 > 1. It does not hold. (2) If sin α − cos β < 0 cos α − sin β < 0. then. . cos β > sin α > 0 sin β > cos α > 0. ⇒ sin2 β + cos2 β > sin2 α + cos2 α, which means 1 > 1. It does not hold. The above two cases are both false. Therefore we have 1 cos β − sin α = 0 · · · 2 sin β − cos α = 0 · · · 1 2 + 2 2 , we obtain sin α cos β+cos α sin β = 1 which implies that sin(α+β) = Checking π 1. Since α and β are acute angles, then α + β = . Download free eBooks at bookboon.com 2 39. π.

(40) then. ELEMENTARY ALGEBRA EXERCISE BOOK II. sin α − cos β < 0 cos α − sin β < 0 . cos β > sin α > 0 sin β > cos α > 0. Trigonometric Functions. ⇒ sin2 β + cos2 β > sin2 α + cos2 α, which means 1 > 1. It does not hold. The above two cases are both false. Therefore we have 1 cos β − sin α = 0 · · · 2 sin β − cos α = 0 · · · 1 2 + 2 2 , we obtain sin α cos β+cos α sin β = 1 which implies that sin(α+β) = Checking π 1. Since α and β are acute angles, then α + β = . 2 π Given a, b, c in the interval (0, ), and a = cos a, b = sin(cos b), c = 2 cos(sin c), compare their values. 4.87 . Solution: Their order is b < a < c. π Otherwise, assume b a. Since cosine function is decreasing on the interval (0, ), then 2 π π 0 < cos b cos a = a < . Applying the relation that is sin x < x when x ∈ (0, ), 2 2 we have 0 < sin(cos b) < cos b cos a = a which means b < a. It contradicts to the assumption. Therefore b < a. Next, assume c a. Since cosine function is decreasing π π in the interval (0, ), thus 0 < sin c < c a < , hence cos(sin c) > cos c cos a = a 2 2 which means c > a. It contradicts to the assumption. Therefore c > a. As a conclusion, b < a < c. 4.88 . Given the three side lengths a, b, c corresponding to angles A, B, C of an k obtuse triangle ABC, sin C = √ , k ∈ Z, and equation x2 − 2kx + 3k 2 − 7k + 3 = 0 2 has real roots. The formula (c − b) sin2 A + b sin2 B = c sin2 C holds. Find the values of A, B, C. Solution: Since the equation has real roots, then ∆ = 4k 2 − 4(3k 2 − 7k + 3) 0, 1 that is 2k 2 − 7k + 3 0 ⇒ k 3. Since k is an integer, then k = 1 or 2 2 √ or 3. Since k = 2 sin C, and 0 < sin C < 1 in the obtuse triangle ABC, we have √ 1 2 , ∠C = 450 or ∠C = 1350 . Since (c − b) sin2 A + b sin2 B = k = 1, sin C = √ = 2 2 c sin2 C, we apply the sine law a = 2R sin A, b = 2R sin B, c = 2R sin C to obtain that (c − b)a2 + b3 − c3 = 0. By solving the equation (b − c)(b2 + c2 − a2 + bc) = 0, we have b = c or b2 + c2 − a2 + bc = 0. When b = c, B = 450 or B = 1350 . ∠B = ∠C = 450 and ∠B = ∠C = 1350 do not hold, since they conflict with the given condition that ABC is an obtuse triangle and ∠A+∠B +∠C = 1800 . When b2 +c2 −a2 +bc = 0, we −bc 1 b2 + c2 − a2 = = − . Therefore, can apply the cosine law to obtain that cos A = 2bc 2bc 2 ∠A = 1200 , ∠B = 150 , ∠C = 450 . 4.89 If a1 , a2 , a3 , · · · , an are positive numbers which are all less than 1, n n nπ 1 − ak − arctan ak = arctan . show 4 1 + a k k=1 k=1 1 − ak Proof: Let α = arctan ak , βDownload = arctan (kbookboon.com = 1, 2, 3, · · · , n). Since 0 < ak < 1, free eBooks, at 1 + ak 40 k ak + 1−a π π 1+ak = then 0 < α < , 0 < β < , then 0 < α + β < π. Hence tan(α + β) =.

(41) ABC is an obtuse triangle and ∠A+∠B +∠C = 1800 . When b2 +c2 −a2 +bc = 0, we 1 −bc b2 + c2 − a2 = − . Therefore, = can apply the cosine law to obtain that cos A = 2 2bc 2bc 0 0 ∠A = 120ALGEBRA , ∠B =EXERCISE 150 , ∠CBOOK = 45 ELEMENTARY II . Trigonometric Functions 4.89 If a1 , a2 , a3 , · · · , an are positive numbers which are all less than 1, n n nπ 1 − ak − arctan ak = arctan . show 4 1 + a k k=1 k=1 1 − ak Proof: Let α = arctan ak , β = arctan , (k = 1, 2, 3, · · · , n). Since 0 < ak < 1, 1 + ak k ak + 1−a π π 1+ak = then 0 < α < , 0 < β < , then 0 < α + β < π. Hence tan(α + β) = 4 2 1 − ak (1−ak ) 1+ak. π 1 − ak π a2k + 1 = 1 ⇒ α + β = . Therefore arctan ak + arctan = . Substitut2 1 + ak 4 1 + ak 4 ing separately k = 1, 2, 3, · · · , n into the equation and adding these equations, we n n 1 − ak nπ . It can be shown that have arctan ak = (arctan ak + arctan ) = 1 + ak 4 k=1 k=1 n 1 − ak nπ − arctan . 4 1 + ak k=1 √. A+B A−B 5 sin +i cos , where A, B, C 2 √ 2 2 3 2 are the interior angles of ABC, and |z| = . (1) Compute tan A tan B. (2) If 4 |AB| = 6, calculate the area of ABC when ∠C reaches its maximum value. √ 5 A+B 2 A−B 2 sin ] + [cos Solution: (1) By the given condition, we have |z|2 = [ ] = 2 2 2 √ 9 3 22 5 1 − cos(A + B) 1 + cos(A − B) ] ⇒ = ⇒ 4 cos(A − B) = 5 cos(A + B) ⇒ [ + 8 4 2 4 2 1 9 sin A sin B = cos A cos B. Hence tan A tan B = . 9 tan A + tan B tan A + tan B 9 = − (2) tan C = − tan(A + B) = − = − (tan A + 1 1 − tan A tan B 8 1− 9 4.90 . Given complex number z =. .. Download free eBooks at bookboon.com 41. Click on the ad to read more.

(42) √. A+B 5 A−B +i cos sin , where A, B, C 2 2 √ 2 3 2 . (1) ComputeTrigonometric tan A tan B. Functions (2) If are the interior ofBOOK ABC, and |z| = ELEMENTARY ALGEBRAangles EXERCISE II 4 |AB| = 6, calculate the area of ABC when ∠C reaches its maximum value. √ A−B 2 5 A+B 2 2 sin ] + [cos ] = Solution: (1) By the given condition, we have |z| = [ 2 2 2 √ 3 22 9 5 1 − cos(A + B) 1 + cos(A − B) [ ] ⇒ + = ⇒ 4 cos(A − B) = 5 cos(A + B) ⇒ 4 4 2 2 8 1 9 sin A sin B = cos A cos B. Hence tan A tan B = . 9 tan A + tan B tan A + tan B 9 (2) tan C = − tan(A + B) = − = − = − (tan A + 1 1 − tan A tan B 8 1− 9 4.90 . Given complex number z =. 9√ 3 tan A tan B = − , and tan C gets the maximum value if and only if 4 4 1 tan A = tan B = . It means ABC is an isosceles triangle when ∠C reaches its 3 |AB| tan A = 1. maximum value. The value of altitude h on the side AB is h = 2 1 Therefore SABC = |AB|h = 3. 2 tan B) −. 4.91 Given a, b, c are three side lengths of ABC, a + b = 10, (a + b + c)(a + b − c) = 3ab, compute the maximal area and the minimal perimeter of ABC. Solution: From the given conditions and the cosine theorem, we have (a + b + c)(a + b − c) = 3ab c2 = a2 + b2 − 2ab cos C ⇒. . ⇒. . (a + b)2 − c2 = 3ab a2 + b2 − c2 = 2ab cos C a2 + b2 − c2 = ab a2 + b2 − c2 = 2ab cos C. π 1 ⇒ C = . Let the area is S. Since b = 10 − a, we have 2 √3 √ √ √ 1 1 25 3 25 3 3 3 2 S = ab sin C = a(10−a) =− (a−5) + . Smax = when a = b = 5. 2 2 2 4 4 √ 4 2 2 Let the perimeter of ABC is p, then p = a + b + c = 10 + a + b − 2ab cos C = 1 10 + a2 + (10 − a)2 − 2a(10 − a) = 10 + 3(a − 5)2 + 25. When a = b = 5, 2 pmin = 15. Thus cos C =. 1 1 1 + ··· + + = cot α − cot 2n α. sin 2n α sin 2α sin 4α 1 sin α sin(2α − α) sin 2α cos α − cos 2α sin α Proof: = = = = cot α − sin 2α sin α sin 2α sin α sin 2α sin α sin 2α 1 sin 2α sin(4α − 2α) sin 4α cos 2α − cos 4α sin 2α cot 2α. = = = = sin 4α sin 2α sin 4α sin 2α sin 4α sin 2α sin 4α 1 cot 2α − cot 22 α. Applying the recurrence relation, we have = cot 22 α − sin 8α 1 cot 23 α,· · · , = cot 2n−1 α − cot 2n α. Adding all above equations , we obtain n sin 2 α 1 1 1 + + ··· + = cotfree α− cot 2n α. nα eBooks at bookboon.com sin 2α sin 4α sin 2Download 4.92 . Show. 42.

(43) 4 2 2 2 4 √ 4 Let the perimeter of ABC is p, then p = a + b + c = 10 + a2 + b2 − 2ab cos C = 1 10 + a2 + (10 − a)2 − 2a(10 − a) = 10 + 3(a − 5)2 + 25. When a = b = 5, 2 ELEMENTARY ALGEBRA EXERCISE BOOK II Trigonometric Functions pmin = 15. 1 1 1 + + ··· + = cot α − cot 2n α. sin 2α sin 4α sin 2n α 1 sin α sin(2α − α) sin 2α cos α − cos 2α sin α Proof: = = = = cot α − sin 2α sin α sin 2α sin α sin 2α sin α sin 2α sin 2α sin(4α − 2α) sin 4α cos 2α − cos 4α sin 2α 1 = = = = cot 2α. sin 4α sin 2α sin 4α sin 2α sin 4α sin 2α sin 4α 1 cot 2α − cot 22 α. Applying the recurrence relation, we have = cot 22 α − sin 8α 1 cot 23 α,· · · , = cot 2n−1 α − cot 2n α. Adding all above equations , we obtain sin 2n α 1 1 1 + + ··· + = cot α − cot 2n α. sin 2α sin 4α sin 2n α 4.92 . Show. 4.93 Let y cos α − x sin α = a cos 2α, y sin α + x cos α = 2a sin 2α, show 2 2 2 (x + y) 3 + (x − y) 3 = 2a 3 . cos α a cos 2α sin α 2a sin 2α = 2a sin 2α cos α− Proof: From the two given conditions, we obtain x = cos α − sin α sin α cos α a sin α cos 2α = 2a2 sin α cos2 α − a sin α(cos2 α − sin2 α) = a(3 sin α cos2 α + sin3 α). a cos 2α − sin α 2a sin 2α cos α = a cos α cos 2α + 2a sin 2α sin α = a[cos α(cos2 α − sin2 α) + y = cos α − sin α sin α cos α 4 sin2 α cos α] = a(cos3 α − sin2 α cos α + 4 sin2 α cos α) = a(cos3 α + 3 sin2 α cos α). Thus x + y = a(sin3 α + 3 sin2 α cos α + 3 sin α cos2 α + cos3 α) = a(sin α + cos α)3 . x − y = a(sin3 α − 3 sin2 α cos α + 3 sin α cos2 α − cos3 α) = a(sin α − cos α)3 . Hence 2 2 2 2 2 (x+y) 3 +(x−y) 3 = a 3 [(sin α+cos α)2 +(sin α−cos α)2 ] = a 3 [2(sin2 α+cos2 α)] = 2a 3 . 4.94 . Let the incircle radius of triangle ABC is r, the circumcircle ra1 dius of triangle ABC is R, show r R. 2 1 Proof: Let a, b, c are the side lengths of ABC, p = (a+b+c), the area of ABC is S. 2 r S abc 4S 2 4p(p − a)(p − b)(p − c) (p − a)(p − b) (p − b)(p − c) Then = ÷ = = =4 R p 4S pabc pabc bc ab 2 2 2 2 2 2 2 2 (p − c)(p − a) c − a + 2ab − b a − b + 2bc − c b − a + 2ac − c2 = 4 = ca 4ab 4bc 4ca 2ab(1 − cos C) 2bc(1 − cos A) 2ac(1 − cos B) (1 − cos C) (1 − cos A) 4 =4 4ab 4bc 4ca 2 2 A B C (1 − cos B) = 4 sin sin sin . 2 2 2 2 A B A B A B A B A B A And sin sin = sin[( + ) + ( − )] sin[( + ) − ( − )] = [sin( + 2 2 4 4 4 4 4 4 4 4 4 A A B A B A B A B A B A B B ) cos( − )+cos( + ) sin( − )][sin( + ) cos( − )−cos( + ) sin( − 4 4 4 4 4 4 4 4 4 4 4 4 4 4 B B B B B B 2 A 2 A 2 A 2 A 2 A 2 A )] = sin ( + ) cos ( − ) − cos ( + ) sin ( − ) = sin ( + ) cos ( − 4 4 4 4 Download 4 4 4at bookboon.com 4 4 4 4 4 free eBooks A B A B A B A B A B B 2 ) − [1 − sin2 ( + )] sin2 ( − ) = sin43 ( + )[cos2 ( − ) + sin2 ( − )] − 4 4 4 4 4 4 4 4 4 4 4.

(44) a − b + 2bc − c b − a + 2ac − c (p − c)(p − a) c − a + 2ab − b = = 4 4ab 4bc ca 4ca 2ab(1 − cos C) 2bc(1 − cos A) 2ac(1 − cos B) (1 − cos C) (1 − cos A) 4 =4 4ab 4bc II 4ca 2 Trigonometric2Functions ELEMENTARY ALGEBRA EXERCISE BOOK C B A (1 − cos B) = 4 sin sin sin . 2 2 2 2 A B A B A B A B A B A = sin[( + ) + ( − )] sin[( + ) − ( − )] = [sin( + And sin sin 2 2 4 4 4 4 4 4 4 4 4 A B A B A B A B A B A B A B ) cos( − )+cos( + ) sin( − )][sin( + ) cos( − )−cos( + ) sin( − 4 4 4 4 4 4 4 4 4 4 4 4 4 4 B B B B B B 2 A 2 A 2 A 2 A 2 A 2 A )] = sin ( + ) cos ( − ) − cos ( + ) sin ( − ) = sin ( + ) cos ( − 4 4 4 4 4 4 4 4 4 4 4 4 B B B B B B 2 A 2 A 2 A 2 A 2 A ) − [1 − B sin ( + )] sin (B − ) =Asin ( B + )[cos ( B− ) + sin ( − )] − 4 2( A 4 sin24( A 4) − 4sin2 ( A − B 4 ). 4sin A 4B reaches 4 4 A B A B A sin − ) = + sin its 4maximum 2 2 B A B A B A B sin22 ( A − ) = sin ( + ) − sin ( − ). sin sin reaches its maximum 4 4 4 4 4 4 2 2 2 2 A B A B A B A B 2 B B B A B sin sin sin sin its maximum 4 − 4 )) = 4 + 4 )) − 4 − 4 ). 2 sin 2 reaches 2( C sin22 ((( A − = sin (A + − sin sin22 ((( A − ). sin sin its maximum − ) = sin + ) − − ). sin sin reaches its A sin ( 4 4 4 4 4 4 2 22 reaches Amaximum value44when44 A = B. 44Following the same we can that sin sin C = 4 44 logic, 44 22 prove 4 2 C value when A = B. Following the same logic, we can prove that sin A sin 2 = 2 A C A C value when A = B. Following the same logic, we can prove that sin sin 2 2 = A C A C A C value when A = B. Following the same logic, we can prove that sin sin = 2 Awhen 2 A Following value same logic, we prove value that sin = 22 C. 22 sin A sin C reaches sin + C )A− = sinB. − C ). sin the its can maximum when A = 2( 2( 2 2 A C A C A C sin2 ( A +C ) − sin2 ( A −C ). sin A sin C reaches its maximum value when A = C. 4 4 4 4 2 2 2 2 C C sin (( A sin − sin its maximum value A = C. 4 + 4 )) − 4 C 4 ). 2 sin 2 reaches 2B 2( C B C C sin + sin (( A − ). 2sin sinBA sin reaches its maximum maximum value value when when A= = C. C. 2 B sin +C )− −sin sin − )−sin ). sin reaches its when A ( 44sin 44 = 4 4 2 2 C B C C C B B reaches its maximum value when sin ). sin + − sin B ( ( 4 4 2 2 2 2 4 4 4 4 2 2 C sin B 2 reaches its maximum value when 2 sin C 2 = sin22 ( B 4 )−sin22 ( B 4 ). sin B 2 sin C 4 +C 4 −C C B C C C B B B C C B C B B its maximum value when sin )−sin sin sin 2 reaches 2 sin 2 = 4B 4 ). 2 sin 4+ 4− 2( 2( π C A its maximum value when sin = sin )−sin ). sin sin + − sin B ( ( sin sin = sin + )−sin − ). sin sin reaches its maximum value when ( ( 22 reaches 22 22 44 its maximum 44 sin44B sin C 44reaches A value when A = B = C = π. B =222 C. Hence sin 2 4 4 4 4 2 2 π C B reaches its maximum value when A = B = C = sin sin B = C. Hence sin A 3. 2 2 2 π C B A A B C π its maximum value when A = B = C = B = C. Hence sin 3. 2Creaches 2Bsin 2Asin r 1 1 π reaches its maximum value when A = B = C = sin sin B = C. Hence sin B = C. Hence whenrA=1 R. B = C = 33 .. r = 4sin 4 sin33itsπ maximum sin222Creaches sin222Bsin sin222Asin Therefore summary, = 1 . As avalue 3 rr = 4 sin A Therefore R 2 4 sin33 π 2 sin C 2 sin B 6 = 1211 . As a summary, r 2111 R. C π r = A B C 44 sin sin B = 44 sin sin A Therefore R R. As aa summary, summary, rr 2 R. = 2 .. As 2 2 sin 2 sin 6 = 3 π sin sin Therefore Therefore R = 4 sin 22 sin 22 sin 22 4 sin 66 = 22 . As a summary, r 22 R. R R 2 2 2 6 2 2 2 2 3 4.95 Solve the equation cos2 θ−cos2 φ = 2 cos3 θ(cos θ−cos φ)−2 sin33 θ(sin θ− 4.95 Solve the equation cos2 θ−cos2 φ = 2 cos3 θ(cos θ−cos φ)−2 sin3 θ(sin θ− 2 θ−cos2 4.95 Solve the equation cos φ= 2 cos33 θ(cos θ−cos φ)−2 sin33 θ(sin θ− sin φ). 4.95 sin 4.95φ). Solve Solve the the equation equation cos cos2 θ−cos θ−cos2 φ φ= = 22 cos cos θ(cos θ(cos θ−cos θ−cos φ)−2 φ)−2 sin sin θ(sin θ(sin θ− θ− sin φ). sin φ). sin φ). Solution: cos22 θ − cos22 φ = 2 cos33 θ(cos θ − cos φ) − 2 sin33 θ(sin θ − sin φ) ⇒ cos22 θ − Solution: cos cos23θθ + − 3cos 2 cos3 θ(cos θ −3 cos 2 sin θ − sin φ) ⇒ cos2 θ − sin θφ) −− sin 3θ 333 θ(sin cos222 φ θ = 2 3 2θ − Solution: cos θθ + − cos φ = 22 cos θθ − cos φ) − 22 sin θ(sin θθ − sin φ) ⇒ cos 2 23θ 3 θ(cos Solution: cos − cos φ = cos θ(cos − cos φ) − sin θ(sin cos − 3 sin θ − sin 3θ cos 3 cos θ θ− sinsin φ) φ) ⇒⇒ − cos φ) θ−− cos φ) − 2 sin(sin cos2 φ = cos θ − cos φ (cos Solution: = 2θθcos θ(cos θ(sin θ− − sin φ) ⇒2(cos cos22 θθθ − 3 sin θ − sin 3θ 3θ + 3 cos θ (sin θ − sin φ) ⇒ 2(cos (cos − cos φ) − cos2 φ = cos − 2 2 2 3 sin θ − sin 3θ cos 3θ + 3 cos θ θ − sin 3θ (sin θθ − 3θ + 3 cos θ (cos cos (cos θθ sin − cos φ) − − 3 sin φ) == cos 2φ 2 ⇒ 2(cos sin φ) 2(cos − cos φ − 3θ 22cos θ + sin θ −φ) cos φ 22− sin 3θ(sin sin θφ − + sin 3 cos 3 sin222 θθθ − (sin − sin φ)2 θθ⇒ ⇒− 2(cos (cos3θ θ sin − cos cos φ)cos −3θ cos22 φ) φ = == cos φ) cos 3θ cos θ + sin 3θ θ − cos 3θ cos φ − sin 3θ sin φ + 3 cos − 3 sin2 θ − − cos 2 2 2 2 2 2 2 2 2 2 = cos sin cos 3θ cos φ − 3θ sin − sin cos 3 cos θ cos φcos + 3θ 3 sin θ θθsin+ φsin ⇒3θ cos 2θθθ − cos(3θ − φ) − sin 3 cos(θ +φ φ)+ =33 cos 3 sin −33 cos − 2θ 2 φ) 2θ 2θ φ) = cos 3θ cos + sin 3θ sin − cos 3θ cos φ − sin 3θ sin φ + cos θ − sin θ − cos 3 cosφ) + 3θ 3 sin ⇒3θcos − φ) − sin 32cos(θ +φφ) 3 sinθ2 θ−−3 cos = φcos cosθ θsin+φsin sin2θθ − cos(3θ cos 3θ cos φ− 3θ sin 3 cos sin 2 θ − cos 2θ cos 2+= 2 2 θ cos φ + 3 sin θ sin φ ⇒ cos 2θ − cos(3θ − φ) − 3 cos(θ + φ) = 3 sin θ − cos θ − 233 cos φ ⇒ cos 2θ −cos(3θ −φ)−3 cos(θ +φ) = 3−4 cos θ −2 cos φ = 3−2(1+cos 2θ)− 2 2 2 2 2 cos θθφcos cos φcos + 332θsin sin sin φ φ −φ)−3 ⇒ cos cos 2θ 2θ − cos(3θ cos(3θ −3−4 φ) − − cos(θ + φ) φ)φ= == 333−2(1+cos sin θθ − − cos cos2θ)− − 23 cos ⇒φ −cos(3θ cos(θ +φ) =− cos332cos(θ θ −2 cos + θθ sin ⇒ − φ) + sin θθ − 2 2 2 2 22(1cos ⇒ cos(θ +φ) = cos θθ −2 cos = 3−2(1+cos + cos 2φ)cos = 2θ −2−cos(3θ cos 2θ −−φ)−3 cos 2φ ⇒ 3 cos 2θ − 33−4 cos(θ +22φ) − cos(3θ − φ) + cos 2φ =2θ)− 0⇒ 2φ 2φ cos φ ⇒ cos 2θ −cos(3θ −φ)−3 cos(θ +φ) = 3−4 cos −2 cos φ = 3−2(1+cos 2θ)− + cos 2φ)cos = 2θ −2−cos(3θ cos 2θ −−φ)−3 cos 2φ ⇒ 3 cos 2θ − cos(θ + φ) − cos(3θ + cos 2φ =2θ)− 0⇒ 2(1cos φ⇒ cos(θ +φ) = 33−4 cos θ −2 cos φ − =φ) 3−2(1+cos 3(θ −= 3θ− +φ) φ+ 3(θ − φ) + 3θ⇒ +33φcos θcos − 2θ φ − 3θ2φ) +φ φ) (1 + cos = −2 cos 2φ 2θ − 33 cos(θ φ) − cos(3θ cos 2φ 0⇒ (1 + cos 2φ) = −2 cos 2θ − cos 2φ ⇒ cos 2θ − cos(θ + φ) − cos(3θ − φ) + cos 2φ = 3(θ − 3θ + φ 3(θ − φ) 3θ + φ θ − φ 3θ + φ φ) (1 +sin cos 2φ) = sin −2 cos 2θ + − cos − cos(3θ + cos 2φ = 00 ⇒ ⇒ sin − φ) (sin sin2θ − 3 cos(θ=+ 0φ)⇒ 2 sin2φ ⇒ 3 cos − −6 + φ θ − φ 3θ + φ 3(θ − φ) 3θ + φ 3(θ − φ) (sin = 0 ⇒ sin sin + 2 sin sin − −6 sin 3θ 2+ 2− 2+ 2φ 2+ 2− 3(θ φ) 3θ φ 3(θ φ) 3θ φ θθ − 3θ φ 3(θ − φ) 3θ + φ 3(θ − φ) 3θ + φ − φ 3θ + φ −6 sin sin + 2 sin sin = 0 ⇒ sin (sin − 2 2 2 2 2 2 3θ + φ θ − φ θ − φ 3θ + φ θ − φ (sin = 0 ⇒ sin sin + 2 sin sin −6 sin − = 0 sin ⇒ 3θ sin+ φ 22= 0(sin ++ 2 sin −6 22 33 θsin 22= 0. Thus 22− φ − φ (−4 sin θ θ − φ222) =sin − φ = 0 ⇒ 222sin 3θ or sin 3 sinsin ) 2= 0 or sin θ2− 2 3θ− φ + φ + φ φ φ (−4 sin = ) = 0 ⇒ sin 3θ 3 sin θθ − ) 2= 0. Thus sin 3θ 2 2 2 2 2 3θ + φ − φ θ − φ 3θ + φ θ − φ 3 3θ + φ θ − φ θ − φ 3θ + φ θ − φ 33 sin 00 ⇒ sin sin 0. Thus sin = 00 or sin 2+ φ (−4 2 = 2 )) = 2 2 )) = 3 3θ θ − φ (−4 sin = ⇒ sin = or sin = 0. Thus sin (−4 sin ) = 0 ⇒ sin 30.sin sin = = 0 or sin ) = 0. Thus sin 22 22 = Therefore, we have 3θ 222+ φ = nπ (n 222∈ N ) or θ − φ = nπ 222(n ∈ N ). As a conclu2 2 + φ θ − φ 0. Therefore, we have 3θ = nπ (n ∈ N ) or = nπ (n ∈ N ). As a conclu2 2 3θ 2+ θθ − + ∈φ φN − φ= 0. Therefore, = (n ∈ N )) or 2φ 0. Therefore, we have = nπ (n ∈ N or = nπ nπ (n (n ∈ ∈N N ). ). As As aa concluconclusion, θ = nπ, φwe = have −nπ 3θ(n ).nπ 0. Therefore, we have = nπ (n ∈ N ) or 2 2 sion, θ = nπ, φ = −nπ (n 22 ∈ N ). 22 = nπ (n ∈ N ). As a conclusion, θ = nπ, φ= −nπ (n ∈N N ). sion, sion, θθ = = nπ, nπ, φ φ= = −nπ −nπ (n (n ∈ ∈ N ). ). 4.96 The interior angles A, B, C satisfy sin A cos B − sin B = sin C − 4.96 The interior angles A, B, C satisfy sin A cos B − sin B = sin C − 4.96A The interior angles is A,12, B, C satisfy sin A cos cosarea. B − sin sin B = = sin C C− sin cos C. If the perimeter of ABC find the maximal 4.96 The angles A, B, satisfy sin B 4.96 The interior interior angles is A,12, B, C C satisfy sin A A cosarea. B− − sin B B = sin sin C − − sin A cos C. If the perimeter of ABC find the maximal sin A A cos cos C. C. If If the the perimeter perimeter of of ABC ABC is is 12, 12, find find the the maximal maximal area. area. sin sin A cos C. If the perimeter of ABC is 12, find the maximal area. Solution: The given equation can be written as sin A(cos B + cos C) = sin B + sin C Solution: The given equation can be written as sin A(cos B + cos C) = sin B + sin C Solution: The given equation can beifwritten written as sin A(cos B+ coscos C) B = =sincos(π B +− sinC). C where cos B + cos C =equation 0. Otherwise, cos B +as cossin C A(cos = 0, we have Solution: The given can B cos C) = B sin C Solution: The given can be beifwritten B+ + coscos C) B = =sin sincos(π B+ +− sinC). C where cos B + cos C =equation 0. Otherwise, cos B +as cossin C A(cos = 0, we have where 0cos BB+ < cosπ,C0= =<0.πOtherwise, Otherwise, ifthen cos BB+ +=cosπ C C− = 0,that we have cos B= cos(π −conC). Since <B − C < π, if C,0, is Bcos +C = cos(π π. It− where cos cos we B C). where cos +< cosπ,C C0=<0. 0.πOtherwise, cos B BB+=cos cosπ C− = = we have have B= = C). Since 0cos <BB+ − C < π, ifthen C,0,that is Bcos +C = cos(π π. It−conSince 0 < B < π, 0 < π − C < π, then B = π − C, that is B + C = π. It contradicts to the condition A + B + C = π. Hence cos B + cos C = 0. Therefore Since 00 < π, < π, B π B C = It Since < Bthe< < condition π, 00 < < π π− − C <+ π, Cthen then B = =Hence π− − C, C, that iscos BC+ + = C 0. = π. π.Therefore It concontradicts to B AC +B = costhat B +is B+C B−Cπ. 2 sin cos sin B + sin C B + C π A B + C tradicts to the condition A + B + C = π. Hence cos B + cos C = 0. Therefore B+C B−C 2 2 tradicts to the condition A + B + C = π. Hence cos B + cos C = 0. Therefore 2 sin cos sinthe B +condition sin C = A + B + CB. +Since π − A, tradicts C = cos CB= +C 0. =Therefore sin A = to tan B cos B+C B−C 2 + cos 2 π. =Hence B+C B−C 2 sin sin B B+ + sin sin C C + + sin A = cos = tan B . Since B = π cos C = 22 cos 2C 2C 2 −A 2, sin B+C cos 2 2 sin B B π A B+C B−C sin cos B−C B+ + C .. Since B+ + C = π A sin B+ sin 2 2 sin A A = = cos = 2 cos = tan cos C C = 2C 2C 2 − 2 ,, 22 cos B−C 22 sin = tan Since = − B+C A = tan . Since = − sin A = = B+C cos B−C cos BC+ + cos cos C C π 22Acos 22 A 22 , A B−C A222 A 222 cos B +B cos A cos B+C cos 2 2 2 22 . C π cos cos 2 2 A A A A B +B C+=cos tan( −2Acos ) = cot . Then sin A = cot ⇒ 2 sin cos = then tan cos A 2 2 A cos A A A A A π + C = cos ⇒ 2 sin . Then sin A = cot ) = cot − = tan( . then tan B A 2 2 2 2 2 2 2 cos sin A A A A A π B A cos A A A A A π B+ + C = 22 22 sin sin A = cot cot tan( then tan 2 = 2 cos 2 ⇒ 2 .. Then 2 )) = 2− 2C sin 2 .. = cos ⇒ sin Then sin A = cot = cot − = tan( then tan A 2 = = tan( 22 − 22 ) = cot 22 . Then sin A = cot 22 ⇒ 2 sin 22 cos 22 then tan 22 A. sin sin 2 2 2 2 2 2 2 sin A222 Download free eBooks at bookboon.com 44.

(45) ELEMENTARY ALGEBRA EXERCISE BOOK II. Trigonometric Functions. √ π A 1 A A 2 2 A < , then cos = 0. Hence sin = ⇒ sin = . Therefore Since 0 < 2 2 2 2 2 2 2 π A = . After all, ABC is a right triangle. Let a, b, c are the side lengths correspond2 √ 2 ing to the angles A, B, C, and a is√ the√hypotenuse.√ We have b +√c + √ b2 + √c = 12. Since b > 0, c > 0, then b + c 2 bc, b2 + c2 2bc. Hence 2 bc + 2 bc 12. √ √ √ √ 1 12 1 √ = 6(2 − 2). SABC = bc 36(2 − 2)2 = 36(3 − 2 2). That is bc 2 2 2+ 2 √ When b = c, we have the the maximal area (SABC )max = 36(3 − 2 2). 4.97 Given a sin x + b cos x = 0, A sin 2x + B cos 2x = C where a and b are not zero at the same time, and 0 x 1800 , show 2abA+(b2 −a2 )B+(a2 +b2 )C = 0. Solution: (1) When a = 0, b = 0, we have b cos x = 0, then cos x = 0, 0 x 1800 . Hence x = 900 . That is A sin 1800 + B cos 1800 = C. Solving the equation, we have −B = C. That is 2abA + (b2 − a2 )B + (a2 + b2 )C = b2 (B + C) = 0. (2) When b = 0, a = 0, we have a sin x = 0, then sin x = 0, 0 x 1800 . Hence x = 00 or x = 1800 . That is A sin 0 + B cos 0 = C or A sin 3600 + B cos 3600 = C. Solving the equation, we have B = C. That is 2abA + (b2 − a2 )B + (a2 + b2 )C = a2 (C − B) = 0. (3) When a = 0, b = 0, the equation system is 1 a sin x + b cos x = 0, 2 A sin 2x + B cos 2x = C. b 1 results in tan x = − . We have sin 2x = 2 sin x cos x = Since cos x = 0, equation a b 2(− ) 2 tan x 2ab a 3 and cos 2x = 2 cos2 x − 1 == 2 tan x cos2 x = = =− 2 1 + tan2 x a + b2 1 + (− ab )2 2 a2 − b2 2 4 Substituting 3 and 4 into , 2 we have . −1 = −1 = 1 + tan2 x a2 + b 2 1 + (− ab )2 Top programmes Join 2ab the best a2 − b2 at 2abA (a2 − b2master’s )B 2 • 33 place= Financial worldwide ranking: A(− 2 )+B( ) = C ⇒ − + C ⇒Times −2abA+(a −b2 MSc )B = 2 2 + b2 2 + b2 2 + b2 a a a + b a International Business the Maastricht University (a2 + b2 )C ⇒ 2abA + (b2 − a2 )B + (a2 + b2 )C = 0.• 1 place: MSc International Business rd. School of Business and Economics! 4.98 Let a, b, c be the side lengths. st. • 1st place: MSc Financial Economics • 2nd place: MSc Management of Learning • 2nd place: MSc Economics • 2nd place: MSc Econometrics and Operations Research of• 2ABC to the annd place: MSc corresponding Global Supply Chain Management and CChange. B A A ) = (b + c − a) : 2a. − csc ) : (cot + cot Sources: Master ranking 2013; Elsevier ‘Beste Studies’ ranking 2012; 2 Keuzegids 2 2 4 Financial Times Global Masters in Management ranking 2012 2 sin B+C cos B−C − 2 sin A2 cos A2 sin B + sin C − sin A b+c−a 2 2 = = Proof: We have 2 sin A 2a 4 sin A2 cos A2 Maastricht cos A4 University cos B−C sin B2 sin C2 − cos B+C A A 1 is 2 2 the−best specialist 1 = − csc . Since cot = = = the 2 4 sin Ain 2 sin A2 sin A2 sin A4 university 2 gles A, B, C, show (cot. Visit us and find out why we are the best! Master’s Open Day: 22 February 2014. Netherlands (Elsevier). www.mastersopenday.nl. Download free eBooks at bookboon.com 45. Click on the ad to read more.

(46) or x = 1800 . That is A sin 0 + B cos 0 = C or A sin 3600 + B cos 3600 = C. Solving the equation, we have B = C. That is 2abA + (b2 − a2 )B + (a2 + b2 )C = a2 (C − B) = 0. (3) When a = 0, b = 0, the equation system is ELEMENTARY ALGEBRA EXERCISE BOOK IIa sin x + b cos x = 0, Trigonometric Functions 1 2 A sin 2x + B cos 2x = C. b 1 results in tan x = − . We have sin 2x = 2 sin x cos x = Since cos x = 0, equation a b 2(− ) 2 tan x 2ab a 3 and cos 2x = 2 cos2 x − 1 == = =− 2 2 tan x cos2 x = 1 + tan2 x a + b2 1 + (− ab )2 2 a2 − b2 2 4 Substituting 3 and 4 into , 2 we have . −1 = −1 = 1 + tan2 x a2 + b 2 1 + (− ab )2 2ab a2 − b 2 2abA (a2 − b2 )B A(− 2 )+B( ) = C ⇒ − + 2 = C ⇒ −2abA+(a2 −b2 )B = a + b2 a2 + b 2 a2 + b2 a + b2 (a2 + b2 )C ⇒ 2abA + (b2 − a2 )B + (a2 + b2 )C = 0. 4.98 . Let a, b, c be the side lengths of ABC corresponding to the anA B C A gles A, B, C, show (cot − csc ) : (cot + cot ) = (b + c − a) : 2a. 4 2 2 2 2 sin B+C cos B−C − 2 sin A2 cos A2 sin B + sin C − sin A b+c−a 2 2 = = Proof: We have 2a 2 sin A 4 sin A2 cos A2 cos A4 cos B−C sin B2 sin C2 − cos B+C A A 1 2 2 1 . Since cot − csc = = = − = A A A 4 2 2 sin 2 sin 2 sin 4 sin A2 cos B2 2 cos2 A4 − 1 cos A2 cos C2 cos A4 C 1 B + cot = − = = and cot + = 2 2 sin A4 2 sin A4 cos A4 2 sin A4 cos A4 sin A2 sin B2 sin C2 cos A2 cot A4 − csc A2 cos A2 sin B2 sin C2 sin B2 sin C2 sin B+C 2 2 . = . Then = = sin B2 sin C2 sin B2 sin C2 cot B2 + cot C2 sin A2 cos A2 sin A2 A A B C 1 and , 2 we have that (cot − csc ) : (cot + cot ) = (b + c − a) : 2a. From 4 2 2 2 4.99 . If the equation a sin x + b cos x + c = 0 has two distinct solutions a α+β = when b = 0. α, β which are in the interval [0, 2π], show that tan 2 b a sin x + Proof: The equation implies that a sin x + b cos x = −c. Then √ 2 a + b2 c a b b √ cos x = − √ . Let cos ϕ = √ , sin ϕ = √ . Then sin(x + 2 2 2 2 2 2 2 a +b a +b a +b a + b2 c ϕ) = − √ . Since α, β are two distinct solutions which are between 0 and 2π, a2 + b 2 c c 1 sin(β + ϕ) = − √ 2 Checking 1 − , 2 then sin(α + ϕ) = − √ , . a2 + b 2 a2 + b2 α+β α−β we have sin(α + ϕ) − sin(β + ϕ) = 0. That is 2 cos( + ϕ) sin = 0. S2 2 α−β α+β ince α = β and α, β ∈ [0, π], then = 0. Hence cos( + ϕ) = 0. That is 2 2 α+β α+β α+β α+β a b √ √ cos cos ϕ − sin sin ϕ = 0 ⇒ cos − sin = 0. 2 2 2 2 a2 + b 2 a2 + b 2 α+β α+β Since b = 0, then cos = 0. Dividing the equation by cos and multiplying 2 2 √ α+β a α+β it by a2 + b2 , we have a − b tan = 0. Therefore, tan = . 2 2 b 4.100 Proof: Let tan−1 √. b3 a3 1 1 a b csc2 ( tan−1 ) + sec2 ( tan−1 ) = (a + b)(a2 + b2 ). b a 2 2 2 2 Download free eBooks at bookboon.com a 1 2 2 a 2 θ = θ, then tan θ = , csc = = = = 46 2 θ b b 2 1 − cos θ sin 2 1 − √a2b+b2 √ √ √ Show.

(47) √ √ sin ϕ = 0 ⇒ cos − sin = 0. 2 2 2 a2 + b2 a2 + b2 α+β α+β and multiplying = 0. Dividing the equation by cos Since b = 0, then cos 2 2 √ α+β α + βTrigonometric a ELEMENTARY II Functions it by a2 ALGEBRA + b2 , weEXERCISE have aBOOK − b tan = 0. Therefore, tan = . 2 2 b cos. 2. cos ϕ − sin. b3 a3 1 1 a b csc2 ( tan−1 ) + sec2 ( tan−1 ) = (a + b)(a2 + b2 ). 2 2 b 2 2 a a 1 2 2 a θ = = = Proof: Let tan−1 = θ, then tan θ = , csc2 = 2 θ b b 2 1 − cos θ sin 2 1 − √a2b+b2 √ √ √ √ 2 a2 + b 2 2(a2 + b2 ) + 2b a2 + b2 a2 + b 2 + b 2 a2 + b 2 √ √ . We have the = √ = a2 a2 + b2 − b a2 + b 2 − b a2 + b 2 + b √ a3 a3 1 a θ equation csc2 ( tan−1 ) = csc2 = a(a2 + b2 ) + ab a2 + b2 . 2 2 b 2 2 b 1 2 2 b φ Let tan−1 = φ, then tan φ = , sec2 = = = = φ 2 √ a a 2 1 + cos φ 1 + a2a+b2 cos 2 √ √ √ √ 2 a2 + b 2 2(a2 + b2 ) − 2a a2 + b2 a2 + b 2 − a 2 a2 + b 2 √ √ . We have the = √ = b2 a2 + b2 + a a2 + b 2 + a a2 + b 2 − a √ b3 b3 φ 2 1 −1 b sec ( tan ) = sec2 = b(a2 + b2 ) − ab a2 + b2 . equation 2 2 a 2 2 √ a3 b3 1 b 2 1 −1 a Consequently, csc ( tan )+ sec2 ( tan−1 ) = a(a2 +b2 )+ab a2 + b2 +b(a2 + 2 2 b 2 2 a √ b2 ) − ab a2 + b2 = (a + b)(a2 + b2 ).. 4.100 . Show. 4.101 The side lengths a, b, c of ABC form a harmonic series, show B sin C sin A that cos = . 2 cos C + cos A Proof: From the given condition that side lengths a, b, c form a harmonic series, then 2ac 2 1 1 . Let the area of ABC is S, and the half of its perimeter is + = . Hence b = b a c a + c 1 + cos B a2 + c2 − b2 + 2ac B (a + c + b)(a + c − b) = p. We have cos = = = 2 4ac 4ac 2 p(p − b) S2 bc sin Aab sin C 2ab2 c sin A sin C = = = = b(a + c)[b2 − (c − a)2 ] ac ac(p − a)(p − c) 4ac(p − a)(p − c) 2abc sin A sin C sin A sin C 2abc sin A sin C = b2 +c2 −a2 a2 +b2 −c2 = = 2 2 2 2 3 3 2 2 2 2 2 2 ab + b c + ac + a c − c − a a(b + c − a ) + c(a + b − c ) + 2ab 2bc sin A sin C . cos A + cos C. > Apply now. redefine your future. π π globAl < . Find 4.102 If y = sin10 x + 10 sin2 x cos2 x +AxA cos10 x, − < x grAduAte 2 2 the maximum value and minimum value of y. progrAm 2015. - © Photononstop. Solution1: Applying the double angle formula and half angle formula, we obtain 5 y = sin10 x + 10 sin2 x cos2 x + cos10 x = (sin2 x)5 + (2 sin x cos x)2 + (cos2 x)5 = 2 1 − cos 2x 5 5 1 + cos 2x 5 1 5 ( ) + sin 2x + ( ) = [(1 − cos 2x)5 + (1 + cos 2x)5 ] + sin2 2x = 2 2 2 32 2 2 + 20 cos2 2x + 10 cos4 2x 5 1 + 10 cos2 2x + 5 cos4 2x + 40(1 − cos2 2x) 2 + (1−cos 2x) = = 32 2 16 5 5 41 5 cos4 2x − 30 cos2 2x + 41 1 = (cos4 2x − 6 cos2 2x + ) = (cos2 2x − 3)2 − . 19/12/13 16:36 axa_ad_grad_prog_170x115.indd 1 16 16 5 16 4 5 1 9 π 2 When cos 2x = 0 (i.e. x = ± ×9− = 2 . ), y has maximum value ymax = Download free the eBooks at bookboon.com Click16 on the ad4to read 16 more 4 5 1 47 2 When cos 2x = 1 (i.e. x = 0), y has the minimum value ymin = × 4 − = 1. 4 16.

(48) b ) − ab a + b = (a + b)(a + b ). 4.101 The side lengths a, b, c of ABC form a harmonic series, show Trigonometric Functions B sin C sin A . that cos = cos C + cos A 2. EXERCISE BOOK II ELEMENTARY ALGEBRA. Proof: From the given condition that side lengths a, b, c form a harmonic series, then 2 2ac 1 1 + = . Hence b = . Let the area of ABC is S, and the half of its perimeter is a c b a + c B 1 + cos B a2 + c2 − b2 + 2ac (a + c + b)(a + c − b) = = = p. We have cos = 2 4ac 4ac 2 p(p − b) S2 bc sin Aab sin C 2ab2 c sin A sin C = = = = ac ac(p − a)(p − c) 4ac(p − a)(p − c) b(a + c)[b2 − (c − a)2 ] 2abc sin A sin C 2abc sin A sin C sin A sin C = b2 +c2 −a2 a2 +b2 −c2 = = 2 2 2 2 3 3 2 2 2 2 2 2 ab + b c + ac + a c − c − a a(b + c − a ) + c(a + b − c ) + 2ab 2bc sin A sin C . cos A + cos C π π 4.102 If y = sin10 x + 10 sin2 x cos2 x + cos10 x, − < x < . Find 2 2 the maximum value and minimum value of y. Solution1: Applying the double angle formula and half angle formula, we obtain 5 y = sin10 x + 10 sin2 x cos2 x + cos10 x = (sin2 x)5 + (2 sin x cos x)2 + (cos2 x)5 = 2 1 − cos 2x 5 5 1 + cos 2x 5 1 5 ( ) + sin 2x + ( ) = [(1 − cos 2x)5 + (1 + cos 2x)5 ] + sin2 2x = 2 2 2 32 2 2 4 2x + 5 cos 2x + 40(1 − cos2 2x) 2 + 20 cos2 2x + 10 cos4 2x 5 1 + 10 cos + (1−cos2 2x) = = 32 2 16 4 2 5 5 41 1 5 cos 2x − 30 cos 2x + 41 = (cos4 2x − 6 cos2 2x + ) = (cos2 2x − 3)2 − . 16 16 5 16 4 1 9 π 5 2 When cos 2x = 0 (i.e. x = ± ), y has the maximum value ymax = ×9− = 2 . 4 16 4 16 1 5 2 When cos 2x = 1 (i.e. x = 0), y has the minimum value ymin = × 4 − = 1. 16 4 dy 9 Solution2: We can check the derivative of y, = 10 sin x cos x + 10(2 sin x cos3 x − dx 2 sin3 x cos x) − 10 cos9 x sin x = 10 sin x cos x[sin8 x + 2(cos2 x − sin2 x) − cos8 x] = 1 5 sin 2x[(sin4 x + cos4 x)(sin4 x − cos4 x) + 2 cos 2x] = 5 sin 2x[(1 − sin2 2x)(sin2 x − 2 1 2 2 cos x) + 2 cos 2x] = 5 sin 2x[(1 − sin 2x)(− cos 2x) + 2 cos 2x] = 5 sin 2x cos 2x(1 + 2 1 2 dy sin 2x). Let = 0, we obtain the stationary point: 2 dx dy (1) x1 = 0 when sin 2x = 0. We find that the sign of changes from negative to dx positive, then ymin (0) = 1 is a local minimum. dy π (2) x2,3 = ± when cos 2x = 0. We find that the sign of changes from positive to 4 dx 9 π negative, then ymax (± ) = 2 are local maxima. 4 16 1 (3)If 1 + sin2 2x = 0, then sin2 2x = −2. The equation has no solution. 2 4.103 . If n is an arbitrary natural number, show that sin α + sin 2α + sin nα sin (n+1)α 2 2 sin 3α + · · · + sin nα = . sin α2 Download free eBooks at bookboon.com. Proof: prove the conclusion by the method 48of mathematical induction. (1) When n = 1, the left side of the equation is sin α, the right side of the equation is.

(49) (2) x2,3 = ±. changes from positive to when cos 2x = 0. We find that the sign of dx 4 9 π negative, then ymax (± ) = 2 are local maxima. 16 4 1 2 2 sin 2xEXERCISE (3)If 1 + ALGEBRA = 0, then sin ELEMENTARY BOOK II 2x = −2. The equation has no solution. Trigonometric Functions 2 4.103 . If n is an arbitrary natural number, show that sin α + sin 2α + sin nα sin (n+1)α 2 2 sin 3α + · · · + sin nα = . sin α2 Proof: prove the conclusion by the method of mathematical induction. (1) When n = 1, the left side of the equation is sin α, the right side of the equation is sin α2 sin α which equals sin α. The left side equals the right side. The equation holds. sin α2 (2) Assume the equation holds when n = k. Then sin α+sin 2α+sin 3α+· · ·+sin kα = sin kα sin (k+1)α 2 2 . We add sin(k + 1)α to both sides, then sin α + sin 2α + · · · + sin kα + sin α2 sin (k+1)α sin (k+1)α + sin α2 sin(k + 1)α [sin kα + 2 sin α2 cos (k+1)α ] sin kα 2 2 2 2 2 = = sin(k+1)α = α α sin 2 sin 2 sin (k+1)α [sin kα + sin (k+2)α − sin kα ] sin [(k+1)+1]α sin (k+1)α 2 2 2 2 2 2 = . Hence the equation α α sin 2 sin 2 holds when n = k + 1. According to (1) and (2), the equation holds for all natural numbers n. 4.104 . If the maximum value of F (x) = | cos2 x + 2 sin x cos x − sin2 x + 3 Ax + B|, denoted M , is considered with parameters A and B for 0 x π, find the 2 values of A and B such that M has the minimum value. √ π Solution: F (x) = | cos 2x + sin 2x + Ax + B| = | 2 sin(2x + ) + Ax + B|. 4 √ √ 3 π Let f1 (x) = 2 sin(2x + ), (0 x π). Then f1 (x) has the maximum value 2 at 4 2 √ 9π π 5π . f1 (x) has the minimum value − 2 at x = . x = or x = 8 8 8 Let f2 (x) = Ax + B which is a monotone function. If A and B are not both zero, then 5π 9π π the sign of f2 (x) can not change twice. Hence when x = , x = , or x = , there 8 8 8 is at least √ point of f2 (x) which sign is same as the sign of f1 (x), and their sum is large than 2. Otherwise, for a pair of A√and B with at least one of them nonzero, the maximum value of F (x) is less than 2. Then √ √  πA + B| < 2 √  F ( π8 ) = | 2 + 8 √ 5π 5πA − 2 + 8 + B| √ < 2 F ( 8 ) = |√  9π 9πA F ( 8 ) = | 2 + 8 + B| < 2 ⇒.   . πA + B < 0, 8 5πA + B > 0, 8 9πA + B < 0, 8. 1 2 3 . 1 and , 2 we have A > 0. According to 2 and , 3 we have A < 0. It is According to √ a contradiction. Therefore, M has the minimum value 2 occurring at A = B = 0.. Download free eBooks at bookboon.com 49.

(50) ELEMENTARY ALGEBRA EXERCISE BOOK II. Sequences. 5 SEQUENCES 5. Sequences. 2an , a1 = 2. (1) Show sequence an + 2 (2) Find the explicit formula for an .. 5.1. Given an+1 =. 1 an. is an arithmetic sequence.. 2 2an an + 2 2 1 1 1 ⇒ = = 1+ ⇒ − = . Hence an + 2 an+1 an an an+1 an 2 1 1 sequence { } is an arithmetic sequence and its common difference is . an 2 1 1 1 1 1 1 1 1 (2)From the conclusion of (1), we have that − = , − = , · · · , − = a2 a1 2 a3 a2 2 an an−1 1 1 1 1 . We sum up these equations to obtain − = (n − 1). Since a1 = 2, then 2 an a1 2 2 an = (n ∈ N ∗ ). n Solution: (1) an+1 =. 1 Let an be the number of the integer roots of the equation f (x) = x2 + x + , x ∈ 2 1 ∗ [n, n + 1], where n ∈ N . (1) Find the general term of {an }. (2) Let bn = , an an+1 compute the nth partial sum of {bn } which is denoted by S.. 5.2. Solution: (1) Since f (x) is increasing on [n, n + 1], then the range is [f (n), f (n + 1)]. 1 1 Hence an = f (n + 1) − f (n) = [(n + 1)2 + (n + 1) + ] − (n2 + n + ) = 2n + 2, (n ∈ N ∗ ). 2 2 1 1 1 1 1 1 1 (2) bn = = = = ( − ). an an+1 n+2 (2n + 2)[2(n + 1) + 2] 4 (n + 1)(n + 2) 4 n+1 n 1 1 1 1 1 1 1 1 1 1 (n ∈ )= )] = ( − − Hence, S = [( − )+( − )+· · ·+( 8n + 16 4 2 n+2 n+1 n+2 3 4 3 2 4 N ∗ ). 5.3 Let sequence {an } satisfy a1 = a2 = 1 and an + an−1 + an−2 = n2 for n 3, compute a1996 . Solution: From the given condition, we have a1 = a2 = 1 and a1 + a2 + a3 = 9. Hence a3 = 7. Since an + an−1 + an−2 = n2 and an−1 + an−2 + an−3 = (n − 1)2 , adding them to obtain an −an−3 = 2n−1, which implies that that an−3 −an−6 = 2(n−3)−1, · · · . Thus a1996 −a1993 = 2×1996−1, a1993 −a1990 = 2×1993−1, · · · ,a4 −a1 = 2×4−1. Adding the 2(1996 + 4) × 665 above equations to obtain a1996 − a1 = − 665 = 2000 × 665 − 665 = 2 1329335. Therefore, a1996 = 1329335 + 1 = 1329336.. Download free eBooks at bookboon.com 50. Click on the ad to read more.

(51) 1 Let an be the number of the integer roots of the equation f (x) = x2 + x + , x ∈ 2 1 ∗ [n, n + 1], where n ∈ N . (1) Find the general term of {an }. (2) Let bn = , aSequences ELEMENTARY ALGEBRA EXERCISE BOOK II n an+1 compute the nth partial sum of {bn } which is denoted by S. 5.2. Solution: (1) Since f (x) is increasing on [n, n + 1], then the range is [f (n), f (n + 1)]. 1 1 Hence an = f (n + 1) − f (n) = [(n + 1)2 + (n + 1) + ] − (n2 + n + ) = 2n + 2, (n ∈ N ∗ ). 2 2 1 1 1 1 1 1 1 (2) bn = = = ( − ). = an an+1 (2n + 2)[2(n + 1) + 2] 4 (n + 1)(n + 2) 4 n+1 n+2 1 1 1 1 1 1 1 1 1 1 n Hence, S = [( − )+( − )+· · ·+( − )] = ( − )= (n ∈ 4 2 3 3 4 n+1 n+2 4 2 n+2 8n + 16 ∗ N ). 5.3 Let sequence {an } satisfy a1 = a2 = 1 and an + an−1 + an−2 = n2 for n 3, compute a1996 . Solution: From the given condition, we have a1 = a2 = 1 and a1 + a2 + a3 = 9. Hence a3 = 7. Since an + an−1 + an−2 = n2 and an−1 + an−2 + an−3 = (n − 1)2 , adding them to obtain an −an−3 = 2n−1, which implies that that an−3 −an−6 = 2(n−3)−1, · · · . Thus a1996 −a1993 = 2×1996−1, a1993 −a1990 = 2×1993−1, · · · ,a4 −a1 = 2×4−1. Adding the 2(1996 + 4) × 665 above equations to obtain a1996 − a1 = − 665 = 2000 × 665 − 665 = 2 1329335. Therefore, a1996 = 1329335 + 1 = 1329336. 5.4 If sequence {an } satisfies a1 = 3, and an+1 = 2an + 1 (n ∈ N ∗ ). Find the general term an of the sequence. Solution: Adding 1 to both sides of the equation an+1 = 2an + 1 to obtain an+1 + 1 = 2(an + 1). We apply the recurrent relation to obtain an + 1 = 2(an−1 + 1), an−1 + 1 = 2(an−2 + 1), · · · , a2 + 1 = 2(a1 + 1). Multiplying the above equations and applying a1 = 3 to generate an = 2n+1 − 1 (n ∈ N ∗ ). 5.5 Let the function f (x) = log2 x − logx 2 (0 < x < 1), and the sequence {an } satisfies f (2an ) = 2n. Find an . Solution: Since f (x) = log2 x−logx 2 = log2 x− log1 x , then f (2an ) = log2 2an − log 12an = 2 2 √ an − a1n = 2n. It leads to a2n − 2nan − 1 = 0. Hence an = n ± n2 + 1. Since 0 < x < 1, √ then 0 < 2an < 1 which means an < 0. Therefore an = n − n2 + 1 (n ∈ N ∗ ). 5.6 Given the general term of sequence {an } as an = 2n2 − n, do there exist nonzero an constants p, q such that the sequence { } is an arithmetic sequence? pn + q Solution: Assume that there exist nonzero constants p, q such that the sequence a1 a2 a3 an } is an arithmetic sequence. Then , , form an arithmetic { pn + q p + q 2p + q 3p + q 6 1 15 )×2 = + . sequence. From a1 = 1, a2 = 6, a3 = 15, we have ( 2p + q p+q 3p + q n 2n2 − n an = − when = Then pq + 2q 2 = 0. Since q = 0, then p = −2q, then −2qn + q q pn + q an p = −2q. We show that { } is an arithmetic sequence and the common difference pn + q 1 is − . q Download free eBooks at bookboon.com 51. ∗.

(52) pn + q Solution: Assume that there exist nonzero constants p, q such that the sequence a1 a2 a3 an } is an arithmetic sequence. Then , , form an arithmetic { pn + q ALGEBRA EXERCISE BOOK II p + q 2p + q 3p + q ELEMENTARY Sequences 6 1 15 )×2 = + . sequence. From a1 = 1, a2 = 6, a3 = 15, we have ( 2p + q p+q 3p + q 2n2 − n n an = = − when Then pq + 2q 2 = 0. Since q = 0, then p = −2q, then pn + q −2qn + q q an } is an arithmetic sequence and the common difference p = −2q. We show that { pn + q 1 is − . q 5.7. Given the sequence {an }, a1 = 1, an+1 = an + 3n (n ∈ N ∗ ), find a10 .. Solution: an+1 = an + 3n ⇒ an+1 − an = 3n. Then an − an−1 = 3(n − 1), an−1 − an−2 = 3(n − 2), an−2 − an−3 = 3(n − 3), · · · , a3 − a2 = 3 × 2, a2 − a1 = 3 × 1. Adding the n(n − 1) . Hence above equations to obtain an − a1 = 3[1 + 2 + 3 + · · · + (n − 1)] = 3 × 2 3 3n(n − 1) 3 = n2 − n + 1. Therefore, a10 = 150 − 15 + 1 = 136. an = a1 + 2 2 2 5.8 Given two arithmetic sequences {am } : 1, 5, 9, · · · and {bn } : 3, 10, 17, · · · . Consider their first 200th terms and find out the number of the terms with same values. Solution: Since am = 4m − 3, (m = 1, 2, · · · ), bn = 7n + 3, (0 = 0, 1, 2, · · · ), am < bn , 4m − 6 then n = (m, n ∈ N ∗ ) when 4m − 3 = 7n + 3. Thus m1 = 5, m2 = 12, · · · . 7 Assume mk = 7k − 2 200, (k ∈ N ∗ ). Hence k = 1, 2, 3, · · · , 28. Hence, there are 28 terms with same values.. Assume the four roots of the equation x2 −x+a = 0 and the equation x2 −x+b = 0 1 form an arithmetic sequence with the first term . If a < b, determine the values of a, b. 4 1 1 1 1 Solution: Let the four roots of the two equations are , + d, + 2d, + 3d. Applying 4 4 4 4 1 1 1 1 the relation between roots and coefficients, we have + + 3d = + d + + 2d = 1. 4 4 4 4 1 1 3 2 Then d = . Thus the two roots of the equation x − x + a = 0 are and . Hence 6 4 4 3 1 1 5 1 1 7 2 a = Get . And the feedback two roots of x from − x experts + b = in 0 your are + = and + 2 × = . in-depth & advice 16 4 6 12 4 6 12 topic area. 5 Find 7 out what 35 you can do to improve Hence,the b =quality×of your =dissertation! . 12 12 144 5.9. Need help with your dissertation?. √ √ 5.10 Let f (x) = ( x + 2)2 (x 0) and for {an }, a1 = 2, n 2, an > 0, Get Now Sn = f (Sn−1 ). Help Find the general term of {an }.. √ √ √ f (Sn−1 ), we Solution: According (x) = ( x√+ 2)2 and √ 2 √ have √ to f√ √ Sn = √ √ Sn = (√ Sn−1 + √2) . It√means√ Sn −√ Sn−1 √ Sn − Sn−1 = 2, Sn−1 − Sn−2 = √ = √2. Thus 2,· · · , S − S = 2, S − S = 2. Adding 2 2 1 √ √ equations √ to obtain √ √ 3 √ the above √ Sn − S1 = (n − 1) 2. Since S1 = a1 = 2, then Sn = 2 + (n − 1) 2 = n 2. more Hence Go Sn to =www.helpmyassignment.co.uk 2n2 , (n ∈ N ∗ ). an = Sn −for Sn−1 =info 2n2 − 2(n − 1)2 = 4n − 2 when n 2. And a1 = 2 when n = 1. Therefore, an = 4n − 2 (n ∈ N ∗ ). 5.11 . 1 1 Download free eBooks at bookboon.com Click on the ad that + to read+more Let each term of the sequence {an } is nonzero, show a 1 a2 a2 a3 52 1 n.

(53) 4m − 6 (m, n ∈ N ∗ ) when 4m − 3 = 7n + 3. Thus m1 = 5, m2 = 12, · · · . 7 Assume mk = 7k − 2 200, (k ∈ N ∗ ). Hence k = 1, 2, 3, · · · , 28. Hence, there are 28 terms with same values. ELEMENTARY ALGEBRA EXERCISE BOOK II Sequences then n =. Assume the four roots of the equation x2 −x+a = 0 and the equation x2 −x+b = 0 1 form an arithmetic sequence with the first term . If a < b, determine the values of a, b. 4 1 1 1 1 Solution: Let the four roots of the two equations are , + d, + 2d, + 3d. Applying 4 4 4 4 1 1 1 1 the relation between roots and coefficients, we have + + 3d = + d + + 2d = 1. 4 4 4 4 1 3 1 2 Then d = . Thus the two roots of the equation x − x + a = 0 are and . Hence 6 4 4 3 5 1 1 7 1 1 2 a= . And the two roots of x − x + b = 0 are + = and + 2 × = . 16 4 6 12 4 6 12 5 7 35 Hence, b = × = . 12 12 144 5.9. √ √ 5.10 Let f (x) = ( x + 2)2 (x 0) and for {an }, a1 = 2, n 2, an > 0, Sn = f (Sn−1 ). Find the general term of {an }.. √ 2 √ √ x + f (Sn−1 ), we Solution: According to f (x) = ( √ 2 √ 2) and √ have √ √ Sn = √ √ √ Sn = (√ Sn−1 + √2) . It√means√ Sn −√ Sn−1 √ Sn − Sn−1 = 2, Sn−1 − Sn−2 = √ = √2. Thus √ equations √ to obtain √ √ the above √2,· · · ,√ S3 − S2 = √ 2, S2 − S1 = 2. Adding Sn − S1 = (n − 1) 2. Since S1 = a1 = 2, then Sn = 2 + (n − 1) 2 = n 2. Hence Sn = 2n2 , (n ∈ N ∗ ). an = Sn − Sn−1 = 2n2 − 2(n − 1)2 = 4n − 2 when n 2. And a1 = 2 when n = 1. Therefore, an = 4n − 2 (n ∈ N ∗ ). 5.11 ··· +. Let each term of the sequence {an } is nonzero, show that. 1 n = . an an+1 a1 an+1. 1 1 + + a1 a2 a2 a3. Solution: Let the common difference of the sequence {an } is d. 1 an+1 − an d 1 1 1 1 1 1 1 − = = ⇒ = ( − ). Thus + +· · ·+ an an+1 an an+1 an an+1 an an+1 d an an+1 a1 a2 a2 a3 1 1 1 1 an+1 − a1 1 1 1 1 1 1 1 1 = ( − = ( = ( − + − +···+ − )= an an+1 d a1 a2 a2 a3 an an+1 ) d a1 an+1 ) d a1 an+1 n 1 nd = . d a1 an+1 a1 an+1 5.12 . Compute the sum of the sequence. 1 3 7 15 (1) Given , 2 , 4 , 6 , · · · , compute the nth partial sum Sn . 2 4 8 16 (2) S = an + an−1 b + an−2 b2 + · · · + an−r br + · · · + abn−1 + bn , where a = 0, b = 0, n ∈ N ∗ , evaluate S. (3) Given 1, 1 + 2, 1 + 2 + 22 , · · · , 1 + 2 + 22 + · · · + 2n−1 , compute the nth partial sum Sn of the sequence. 2n − 1 1 1 1 3 7 15 1 Solution: (1) Let M = + + + +· · ·+ n = (1− )+(1− )+(1− )+· · ·+(1− 8 4 2 4 8 16 2 2 1 1 n [1 − ( ) ] 1 1 1 1 1 1 1 2 ) = n−( + + +· · ·+ n ) = n− 2 = n−(1− n ) = n +n−1 (n ∈ N ∗ ). 2 2n 2 4 8 2 2 1 − 12 (n − 1)(n − 2) Let N = 2 + 4 + 6 + · · · + 2(n − 1) = (n − 1)2 + 2 = n2 − n (n ∈ N ∗ ). 2 Download free eBooks at bookboon.com 1 1 53 Thus Sn = M + N = n + n − 1 + n2 − n = + n2 − 1 (n ∈ N ∗ ). 2n 2.

(54) , 2 , 4 , 6 , · · · , compute the nth partial sum Sn . 2 4 8 16 (2) S = an + an−1 b + an−2 b2 + · · · + an−r br + · · · + abn−1 + bn , where a = 0, b = 0, n ∈ N ∗ , evaluate S. ELEMENTARY Sequences (3) Given ALGEBRA 1, 1 + 2,EXERCISE 1 + 2 +BOOK 22 , ·II· · , 1 + 2 + 22 + · · · + 2n−1 , compute the nth partial sum Sn of the sequence. (1) Given. 2n − 1 1 1 1 3 7 15 1 Solution: (1) Let M = + + + +· · ·+ n = (1− )+(1− )+(1− )+· · ·+(1− 2 4 8 16 2 2 4 8 1 [1 − ( 12 )n ] 1 1 1 1 1 1 1 2 ) = n−( + + +· · ·+ n ) = n− = n−(1− n ) = n +n−1 (n ∈ N ∗ ). 1 n 2 2 4 8 2 2 2 1− 2 (n − 1)(n − 2) 2 = n2 − n (n ∈ N ∗ ). Let N = 2 + 4 + 6 + · · · + 2(n − 1) = (n − 1)2 + 2 1 1 Thus Sn = M + N = n + n − 1 + n2 − n = n + n2 − 1 (n ∈ N ∗ ). 2 2 (2) Multiplying the equation by a or by b to obtain aS = an+1 + an b + an−1 b2 + · · · + an−r+1 br + · · · + a2 bn−1 + abn , bS = an b + an−1 b2 + an−2 b3 + · · · + an−r br+1 + · · · + abn + bn+1 . The first equation minus the second equation, we have (a − b)S = an+1 − bn+1 . Hence,   an+1 − bn+1 , a = b S= a − b  (n + 1)an , a = b. (3) Since an = (1 + 2 + 22 + · · · + 2n−1 )(2 − 1) = 2n − 1, then Sn = n . n . 2(1 − 2n ) − n = 2n+1 − n − 2 (n ∈ N ∗ ). 2k − 1= 1 − 2 k=1 k=1. n k=1. (2k − 1) =. 5.13 . Given sequence {an }, a1 = 1, and sequence {bn }, b1 = 0, with the re1 1 lationships an = (2an−1 + bn−1 ) and bn = (an−1 + 2bn−1 ) for n 2. Find an , bn . 3 3 1 1 Solution: an + bn = (2an−1 + bn−1 ) + (an−1 + 2bn−1 ) = an−1 + bn−1 = an−2 + bn−2 = 3 3 1 · · · = a1 + b1 = 1 . 1 1 1 1 And an −bn = (2an−1 +bn−1 )− (an−1 +2bn−1 ) = (an−1 −bn−1 ) = ( )2 (an−2 +bn−2 ) = 3 3 3 3 1 n−1 1 n−1 2 · · · = ( ) (a1 − b1 ) = ( ) . 3 3 1 1 1 1 1 and , 2 we have an = (1 + n−1 ), bn = (1 − n−1 ). According to 2 3 2 3 5.14 If the nth partial sum of the arithmetic sequence is 30 and the 2nth partial sum is 100, compute the 3nth partial sum. Solution 1: Let the first term be a1 and the common difference be d. From the given conditions, we have    na1 + n(n − 1) d = 30, 2   2na1 + 2n(2n − 1) d = 100. 2 40 10 20 Solving this equation system to obtain d = 2 , a1 = + 2 . Thus S3n = 3na1 + n n n 3n(3n − 1) 10(n + 2) 3n(3n − 1) 40 + = 210. d = 3n n2 2 n2 2 Download free eBooks at bookboon.com. Solution 2: According to the properties of 54 an arithmetic sequence, Sn , S2n − Sn , S3n −S2n form an arithmetic sequence. Hence 2(S2n −Sn ) = Sn +(S3n −S2n ). Therefore.

(55) Let and the the common common difference difference be be d. d. From From the the given given Let the the first first term term be be a a11 and we have  we have  n(n  n(n − − 1) 1) d = 30,   +  na 1 na + d = 30, 1 22 ELEMENTARY ALGEBRA EXERCISE BOOK II Sequences 2n(2n − 1)  2n(2n − 1)   d  2na d= = 100. 100. 2na11 + + 22 40 10 20 10 + 20 Solving this this equation equation system system to to obtain obtain d d= = 402 ,, a = = 3na1 + 1 + n22 .. Thus Solving a = Thus S S3n + 1 3n = 3na1 n n 2 n n n 40 3n(3n 10(n + + 2) 2) 3n(3n 3n(3n − − 1) 1) 40 3n(3n − − 1) 1) d = 3n 10(n + = 210. d = 3n + 2 n22 = 210. n2 22 n 22 n − Sn ,, Solution S2n Solution 2: 2: According According to to the the properties properties of of an an arithmetic arithmetic sequence, sequence, S Snn ,, S 2n − Sn S −S2n form −Sn )) = −S2n ). S3n form an an arithmetic arithmetic sequence. sequence. Hence Hence 2(S 2(S2n =S Snn +(S +(S3n ). Therefore Therefore 3n −S2n 2n −Sn 3n −S2n S = 3(S − S ) = 3(100 − 30) = 210. S3n 3n = 3(S2n 2n − Sn n ) = 3(100 − 30) = 210. Solution Solution 1: 1: conditions, conditions,. Solution Solution 3: 3: is that S n that Sn is aa. The The formula formula of of the the nth nth partial partial sum sum of of an an arithmetic arithmetic sequence sequence implies implies quadratic function. We have quadratic function. We have An22 + Bn = 30, An + Bn = 30, 2 A(2n) A(2n)2 + + B2n B2n = = 100, 100,. 10 20 10 20 where B = = n .. where A, A, B B are are constants. constants. Solving Solving the the equation equation system system to to obtain obtain A A= = n22 ,, B n n 20 2 20 (3n)22 + 10 10 3n = 210. 2 Therefore, S = A(3n) + B3n = Therefore, S3n (3n) + n 3n = 210. 3n = A(3n) + B3n = n2 n2 n 2 5.15 5.15 The The vertex vertex coordinates coordinates of of the the quadratic quadratic function function ff (x) (x) = = ax ax2 + + bx bx + + cc 33 11 is ), and and ff (3) (3) = = 2. 2. For For an an arbitrary arbitrary real real number number x, x, the the sequences sequences {a {ann }} and − ), is (( 2 ,, − and {b {bnn } } 2 44 n+1 ∗ satisfy x+ + bbnn = =x xn+1 (n (n ∈ ∈N N ∗ ), ), where where g(x) g(x) is is defined defined on on the the set set of of real real satisfy ff (x)g(x) (x)g(x) + +a ann x numbers numbers R. R. Find Find the the general general terms terms of of the the sequences sequences {a {ann }} and and {b {bnn }. }.. 11 33 2 Solution: − 4 Solution: From From the the given given condition, condition, we we have have ff (x) (x) = = a(x a(x − − 2 ))2 − 2 4. Brain power. (a = (a = 0). 0). SS-. By 2020, wind could provide one-tenth of our planet’s electricity needs. Already today, SKF’s innovative knowhow is crucial to running a large proportion of the world’s wind turbines. Up to 25 % of the generating costs relate to maintenance. These can be reduced dramatically thanks to our systems for on-line condition monitoring and automatic lubrication. We help make it more economical to create cleaner, cheaper energy out of thin air. By sharing our experience, expertise, and creativity, industries can boost performance beyond expectations. Therefore we need the best employees who can meet this challenge!. The Power of Knowledge Engineering. Plug into The Power of Knowledge Engineering. Visit us at www.skf.com/knowledge. Download free eBooks at bookboon.com 55. Click on the ad to read more.

(56) 5.15 The vertex coordinates of the quadratic function f (x) = ax2 + bx + c 3 1 is ( , − ), and f (3) = 2. For an arbitrary real number x, the sequences {an } and {bn } 2 4 ELEMENTARY ALGEBRA+EXERCISE II n+1 (n ∈ N ∗ ), where g(x) is defined on the set Sequences of real satisfy f (x)g(x) an x + bBOOK n = x numbers R. Find the general terms of the sequences {an } and {bn }. 3 1 Solution: From the given condition, we have f (x) = a(x − )2 − (a = 0). S2 4 33 2 11 3 )2 − 1 = ince ff (3) = 2, then a(3 − 2. Solving the equation to obtain a = 1. Hence ince (3) = 2, then a(3 − = 2. 2. Solving Solving the the equation equation to to obtain obtain a a= = 1. 1. Hence Hence ince f (3) = 2, then a(3 − 22 ))2 − − 44 = 2 2 4 0. f (x) = x − 3x 3x + + 2, 2, x x∈ ∈ R, R, fff (1) (1) = = 0, 0, fff (2) (2) = = (x) = =x x22 − − 3x + 2, x ∈ R, (1) = 0, (2) = 0. 0. ff (x) 11 On Applying f (1)g(1) + a + b = 1 to obtain + b = 11 . the other hand, n + bn = 1 to obtain a Applying f (1)g(1) + a + 1 On Applying f (1)g(1) + ann n+1 + bnn = 1 to obtain a annnn+1 + bbnnn = = 1 . . On the the other other hand, hand, 22 According 11 and 22 we ff (2)g(2) + 2a . to , bbn = 22n+1 2a bbn = 22n+1 n + n + n+1 ,, then n+1 (2)g(2) + 2a . According to and , + = then 2a + = n n n n 2 According to 1 and , 2 we f (2)g(2) + 2a . we , then n + bn = 2 n+1 n+1 2an + b∗n = 2 have a = 2 − 1, b = 2 − 2 , n ∈ N . n+1 n+1 ∗ n n n+1 n+1 ∗ have − 1, bbnn = have a ann = = 22 − 1, = 22 − − 22 ,, n n∈ ∈N N .. 21 1 21 21 , bb1 + bb2 + bb3 = let bbn = (( 11 ))aaannn ,, and 5.16 Given the arithmetic sequence {a n }, 5.16 Given the arithmetic sequence {a }, let = and + + = 5.16 Given the arithmetic sequence {ann }, let bnn = ( 22 ) , and b11 + b22 + b33 = 88 ,, 8 2 11 1 .. Find bb1 bb2 bb3 = the general term of the sequence {a n }. Find the general term of the sequence {a = }. b11 b22 b33 = 88 . Find the general term of the sequence {ann }. 8 1 1 1 1 +a (( 11 ))aaa111 (( 11 ))aaa222 (( 11 ))aaa333 = (( 11 ))aaa111 +a Solution: From the given condition, we have bb1 bb2 bb3 = +a22 +a33 = = = Solution: From the given condition, we have = Solution: From the given condition, we have b11 b22 b33 = ( 22 ) ( 22 ) ( 22 ) = ( 22 ) +a2 +a3 = 2 2 2 2 11 11 3 1= a1 +a +a3 = 3. Since {an }} is an arithmetic sequence, we assume a = = ((( 12 )))33 ... Hence Hence Hence a a11 +a +a222 +a +a33 = = 3. 3. Since Since {a {ann } is is an an arithmetic arithmetic sequence, sequence, we we assume assume a a111 = = 88 = 2 82 − d 2and a3 = a2 + d, where d is the common difference. Thus a2 − d + a2 + a2 + d = 3, a a a −d d and and a a33 = =a a22 + + d, d, where where1d d is is the the1common common difference. Thus a22 − − d+ +a a22 + +21 a22 + +d d= = 3, 3, a22 − 11 difference. 11 1−dThus 11 a 11 d a1 a2 a3 1+d 1 1 21 1 1 1 1 1 1 21 = 1. b + b + b = ( ) + ( ) + ( ) = ( ) + + ( ) = . Solving then a a a a 1−d 1+d + ( )1+d = Solving = 1. 1. bb111 + + bb222 + + bb333 = = (( 2 ))a11 + + (( 2 ))a22 + + (( 2 ))a33 = = (( 2 ))1−d + +2+ = 8 .. Solving then a a222 = then 2 22 22 2 2 ( 222 ) 2 2 2 88 17 d −d 17 . That means d = 2 or d = −2. the 2−d = 17 the equation, equation, we we have have 222dd + That means means d d= = 22 or or d d= = −2. −2. + the equation, we have + 22−d = = 44 .. That 4−1 + 2(n − 1) = 2n − 3 (n ∈ N ∗∗∗ ). When d = 2,a = 1 − d = −1, a = 1 n When d d= = 2,a 2,a11 = When = 11 − −d d= = −1, −1, a ann = = −1 −1 + + 2(n 2(n − − 1) 1) = = 2n 2n − − 33 (n (n ∈ ∈N N ∗ ). ). When d = −2,a = 1 − d = 3, a = 3 − 2(n − 1) = −2n + 5 (n ∈ N ∗∗ ). 1 n When d = −2,a = 1 − d = 3, a = 3 − 2(n − 1) = −2n + 5 (n ∈ N 1 n When d = −2,a1 = 1 − d = 3, an = 3 − 2(n − 1) = −2n + 5 (n ∈ N ). ). 5.17 5.17 5.17 tial sum, tial sum, tial sum,. Given {a aa sequence with positive terms, S the nth parn denotes Given {annn }}} as as sequence with positive terms, S denotes the nth par√ n as a sequence with positive terms, S denotes the nth parGiven {a n ∗ √ and 22√S Sn = a 11 (n ∈ N ). Find the general term of the sequence {a }. ∗ n+ ∗ and = a + (n ∈ N ). Find the general term of the sequence {a and 2 Snn = ann + 1 (n ∈ N ). Find the general term of the sequence {annn }. }. √ √ √Sn = an + 1, then 2√ √a1 = a1 + 1 for n Solution: Since 22√ n = = 1. 1. It It means means that that Solution: Since √ = a ann + + 1, 1, then then 22 a a11 = = a a11 + + 11 for for Solution: Since 2 S Snn = 2 2 n = 1. It means 2that √ ((√a a1 − 1) = 0. Thus a 1. We have 4S (a 1) ,, 4S (a 1) for 2 2 2 1 = n = n + n−1 = n−1 + 2 2 2 − 1) = 0. Thus a = 1. We have 4S = (a + 1) 4S = (a + 1) for ( a11 − 1) = 0. Thus a11 = 1. We have 4Snn = (ann + 1) , 4Sn−1 n−1 = (an−1 n−1 + 1) for n 2. Subtracting the second equation from the first equation to obtain 4a n n 2. 2. 2 Subtracting Subtracting2 the the second second 2 equation equation from from the first first equation equation to to obtain obtain 4a 4ann = = n the = 2 (a + 1) − (a + 1) ⇒ (a − 1) − (a + 1) = 0 ⇒ (a + a )(a − a − 2) 0. 2 2 2 2 n + 1)2 − (an−1 + 1)2 ⇒ (an − 1)2 − (an−1 + 1)2 = 0 ⇒ (an + an−1 )(an − an−1 − 2) = (a = 0. (ann + 1) − (an−1 n−1 + 1) ⇒ (an n − 1) − (an−1 n−1 + 1) = 0 ⇒ (an n + an−1 n−1 )(an n − an−1 n−1 − 2) = 0. Since a a 0, then a a 22 = 00 which means a a 2. Hence n + n+1 > n − n−1 − n − n−1 = Since a + a > 0, then a − a − = which means a − a n n+1 n n−1 n n−1 Since an + an+1 > 0, then an − an−1 − 2 = 0 which means an − an−1 = = 2. 2. Hence Hence {a } is an arithmetic sequence with the first term 1 and the common difference 2, n {ann }} is is an an arithmetic arithmetic sequence sequence with with the the first term 1 and the common difference 2, {a first term 1 and the common difference 2, ∗ a 11 + (n − 1) × 22 = 2n − 11 (n ∈ N ). ∗ n = ∗ a = + (n − 1) × = 2n − (n ∈ N ). ann = 1 + (n − 1) × 2 = 2n − 1 (n ∈ N ). n + 22 n + n + } be S , a = 1, a = (n = 5.18 Let the nth partial sum of {a n n 1 n+1 n S = n 2S (n = 5.18 n S be S Snn ,, a a11 = = 1, 1, a an+1 (n = 5.18 Let Let the the nth nth partial partial sum sum of of {a {ann }} be n+1 = n n n S n Snn }} is 1, 2, 3, · · · ). Show that (1) the sequence {S a geometric sequence; (2) S = 1, = 4a 4ann ... n+1 1, 2, 2, 3, 3, ·· ·· ·· ). ). Show Show that that (1) (1) the the sequence sequence {{ n } is is aa geometric geometric sequence; sequence; (2) (2) S Sn+1 n+1 = 4an n n n + 22 n + n + 2 and a = − = 1, 2, 3, · · · ), then Proof: (1) Since an+1 = S = n S = S Sn+1 −S Snn (n (n Proof: n+1 Snnn and and a an+1 (n = = 1, 1, 2, 2, 3, 3, ·· ·· ·· ), ), then then Proof: (1) (1) Since Since a an+1 n+1 = n+1 = Sn+1 n+1 − Sn n n S S n+1 n S Snn .. Therefore Sn+1 (n + 2)S = n(S − ⇒ nSn+1 = 22 S Therefore the n+1 = (n = −S Snn ))) ⇒ = 2(n 2(n + + 1)S 1)Snn ⇒ ⇒ n+1 (n + + 2)S 2)Snnn = = 2n . Therefore the the = n(S n(Sn+1 ⇒ nS nSn+1 ⇒ n + 1 n+1 − Sn n+1 = 2(n + 1)Sn n + 1 n n+1 n Sn sequence { } is a geometric sequence. n Sn−1 Sn+1 Sn−1 =4 , (n 2). Hence Sn+1 = 4(n + 1) = (2) We apply (1) to obtain n+1 n−1 n−1 4an , (n 2). Since a2 = 3S1 = 3, then S2 = a1 + a2 = 4 = 4a1 . Therefore Sn+1 = 4an holds for an arbitrary positive integer n 1. Download free eBooks at bookboon.com. 5.19 . 56. Let the common difference of arithmetic sequence {an } and the common.

(57) ELEMENTARY ALGEBRA EXERCISE BOOK II. Sequences. Sn sequence { } is a geometric sequence. n Sn+1 Sn−1 Sn−1 =4 , (n 2). Hence Sn+1 = 4(n + 1) = (2) We apply (1) to obtain n+1 n−1 n−1 4an , (n 2). Since a2 = 3S1 = 3, then S2 = a1 + a2 = 4 = 4a1 . Therefore Sn+1 = 4an holds for an arbitrary positive integer n 1. 5.19 Let the common difference of arithmetic sequence {an } and the common ratio of geometric sequence {bn } be both d (where d = 1 and d = 0), a1 = b1 , a4 = b4 , a10 = b10 . (1) Find the values of a1 and d. (2) Is b16 a term of {an }? If it is a term of {an }, which term is it? If it is not a term of {an }, explain the reason. Solution: (1) Since an = a1 + (n − 1)d, bn = b1 dn−1 = a1 dn−1 , and a4 = a4 , a10 = a10 , we have a1 + 3d = a1 d3 a1 + 9d = a1 d9 ⇒ 3d = a1 (d3 − 1) 9d = a1 (d9 − 1). Dividing the first equation by the second equation leads to d6 + d3 −√2 = 0 which means that d3 = 1 or d3 = −2. Since d √ = 1, then d3 = −2. √ Hence d =√− 3 2. Substituting it into the equation, we have a1 = 3 2. Therefore a1 = 3 2, d = − 3 2. √ 3 (2) We apply (1) to obtain that the general terms of {a } and {b } are a = (2 − n) 2, n n n √ √ √ √ √ √ bn = 3 2(− 3 2)n−1 = −(− 3 2)n . Hence b16 = −32 3 2. Since (2 − n) 3 2 = −32 3 2, then n = 34. Therefore, b16 is the 34th term of {an }.. 5.20 Given the sequence {an }, a1 = 1, an+1 = Sn + (n + 1), (n ∈ N ∗ ). (1) Show the sequence {an + 1} is a geometric sequence. (2) Find the general term an and the nth partial sum Sn . (1) Proof: We apply an+1 = Sn + (n + 1) to obtain Sn = an+1 − (n + 1) and Sn−1 = an − n. Then an = Sn − Sn−1 = [an+1 − (n + 1)] − (an − n). Thus an+1 + 1 an+1 = 2an + 1 ⇔ an+1 + 1 = 2(an + 1) ⇔ = 2. Therefore {an + 1} is an + 1 a geometric sequence with common ratio 2. (2) Solution: Since a1 + 1 = 2, then an + 1 = 2 · 2n−1 = 2n which is an = 2n − 1. 2(1 − 2n ) −n = Sn = (2 − 1) + (22 − 1) + · · · + (2n − 1) = (2 + 22 + · · · + 2n ) − n = 1−2 2n+1 − n − 2 (n ∈ N ∗ ).. Download free eBooks at bookboon.com 57.

(58) ELEMENTARY ALGEBRA EXERCISE BOOK II. Sequences. 5.21 Let P1 (x1 , y1 ), P2 (x2 , y2 ),· · · ,Pn (xn , yn ) (n 3) be the points on quadratic curve C, and a1 = |OP1 |2 , a2 = |OP2 |2 ,· · · ,an = |OPn |2 form an arithmetic sequence with common difference d (d = 0), and Sn = a1 + a2 + · · · + an . x2 y 2 (1) If the curve C is 2 + 2 = 1 (a = 10, b = 5), the point P1 (10, 0), and S3 = 255. a b Determine the point P3 . x2 y 2 (2) If the curve C is 2 + 2 = 1 (a > b > 0), the point P1 (a, 0), Find the minimum a b value of Sn as d varies. Solution: (1) Applying the point P1 (10, 0), we have a1 = |OP1 |2 = 102 = 100, 3 a1 + a3 S3 = a1 + + a3 = (a1 + a3 ) = 255. Then a3 = |OP3 |2 = 70. Thus 2 2 2 2 √ y x 3 2 Applying 1 and 2 to obtain x3 = ±2 15, 1 x23 + y32 = 70 , + 3 = 1 . 100 25 √ √ √ √ √ y3 =√± 10. Therefore coordinates of the point P3 are (2 15, 10), (2 15, − 10), √ √ the √ (−2 15, 10), (−2 15, − 10). (2) Since a1 = |OP1 |2 = a2 , then d < 0 and an = |OPn |2 = a2 + (n − 1)d b2 . That b2 − a2 n(n − 1) means d < 0. Since n 3, then Sn = na2 + d is increasing in n−1 2 n(a2 + b2 ) n(n − 1) b2 − a2 b2 − a2 , 0). Therefore (Sn )min = na2 + = . [ n−1 2 n−1 2 5.22 Let an arithmetic sequence has twelve terms where Seven : Sodd = 32 : 27 and the sum of sequence is 354. Find the common difference d. 1 Solution1: From the given condition, we have Seven − Sodd = nd = 6d. Since 2 Seven : Sodd = 32 : 27, let Seven = 32t, Sodd = 27t, then 32t + 27t = 354. Thus t = 6. Hence Seven = 32 × 6 = 192, Sodd = 27 × 6 = 162. Therefore Seven − Sodd = 30 = 6d, then d = 5. Seven Seven 32 32 32 ⇒ ⇒ Seven = × 354 = 192, Sodd = = = Sodd 27 Seven + Sodd 32 + 27 59 1 354 − 192 = 162. Since Seven − Sodd = nd = 6d, then 6d = 192 − 162 = 30. Therefore 2 d = 5. Solution2:. 5.23 Let {an }, a1 = 1, nan+1 = (n + 1)an + 1 (n 2). Compute the the nth partial sum Sn . Solution: nan+1 = (n + 1)an + 1 (n 2) ⇒ n(an+1 + 1) = (n + 1)(an + 1) (n 2). n+1 bn . Thus b1 = 2, b2 = 2×2, b3 = 3×2, b4 = 4×2,· · · , n bn = n×2. Therefore Sn = a1 +a2 +· · ·+an = b1 +b2 +· · ·+bn −n = 2(1+2+· · ·+n)−n = n2 (n 2). Let bn = an +1, then bn+1 =. 5.24 Given the quadratic function f (x) = n(n + 1)x2 − (2n + 1)x + 1, and n is chosen as all natural numbers, compute the sum of lengths of all line segments on the x-axis intercepted by theDownload graph. free eBooks at bookboon.com 58. 2.

(59) n+1 bn . Thus b1 = 2, b2 = 2×2, b3 = 3×2, b4 = 4×2,· · · , n bn = n×2. Therefore Sn = a1 +a2 +· · ·+an = b1 +b2 +· · ·+bn −n = 2(1+2+· · ·+n)−n = n2 (n ALGEBRA 2). ELEMENTARY EXERCISE BOOK II Sequences Let bn = an +1, then bn+1 =. 5.24 Given the quadratic function f (x) = n(n + 1)x2 − (2n + 1)x + 1, and n is chosen as all natural numbers, compute the sum of lengths of all line segments on the x-axis intercepted by the graph. Solution: n(n+1)x2 −(2n+1)x+1 = 0when f (x) = 0. Thus (nx−1)[(n+1)x−1] = 0. 1 1 Then x1 = , x2 = . Let the parabola intersects x-axis at the point An n n+1 and the point Bn , then the sum of lengths of the intercepted line segments Sn = 1 1 1 1 1 1 ) = 1− . |A1 B1 | + |A2 B2 | + · · · + |An Bn | = (1 − ) + ( − ) + · · · + ( − 2 2 3 n n+1 n+1 1 ) = 1. lim Sn = lim (1 − n→∞ n→∞ n+1 5.25 (1) Given an arithmetic sequence {an } that satisfies a1 = −60, a17 = −12. Let bn = |an |. Evaluate the 30th partial sum of {bn }. (2) If the general term of the arithmetic sequence {an } is an = 10 − 3n, compute |a1 | + |a2 | + · · · + |an |. Solution: (1) Let the common difference of the arithmetic sequence {an } is d. Since a1 = −60, a17 = −12, then −60 + 16d = −12. Thus d = 3. Hence an = −60 + (n − 1)3. It means an = 3n − 63. 3n − 63 = 0 if an = 0. Then n = 21. a21 = 0, a22 = 3. (30 − 21 + 1)(30 − 21) −60 + 0 Method 1: S21 = ×20 = −600. S30 −S21 = (30−21+1)×3+ × 2 2 3 = 165. Therefore the 30th partial sum of {bn } is S30 = |S21 | + |S30 − S21 | = 765. Method 2: Since an = 3n − 63, S30 = a1 + a2 + · · · + a30 − 2(a1 + a2 + · · · + a20 ) = −60 − 3 −60 + 27 × 30 − 2 × 20 = 765. 2 2 (2) Since an = 10 − 3n, then a1 > 0, a2 > 0, a3 > 0, a4 , a5 , · · · , an < 0. Hence a1 + a2 + · · · + an , (n 3) |a1 | + |a2 | + · · · + |an | = a1 + a2 + a3 − a4 − · · · − an , (n 4) =. . a1 + an n, (n 3) 2 2(a1 + a2 + a3 ) − (a1 + a2 + · · · + an ), (n 4). =.     . =.   .  . −3n2 + 17n , (n 3) 2 2 −3n + 17n 24 − , (n 4) 2 −3n2 + 17n , (n 3) 2 2 3n − 17n + 48 , (n 4) 2. 5.26 . Given f (x) is a linear function, and f (8) = 15. f (2), f (5), f (4) form a Sn geometric sequence. Denote Sn = f (1) + f (2) + · · · + f (n). Compute lim ( 2 ). n→∞ n Solution: Let f (x) = kx + b.Download From the given condition, we have free eBooks at bookboon.com 8k + b = 15 59 2.

(60) =.    . ELEMENTARY ALGEBRA EXERCISE BOOK II. , (n 3) 2 3n2 − 17n + 48 , (n 4) 2. Sequences. 5.26 . Given f (x) is a linear function, and f (8) = 15. f (2), f (5), f (4) form a Sn geometric sequence. Denote Sn = f (1) + f (2) + · · · + f (n). Compute lim ( 2 ). n→∞ n Solution: Let f (x) = kx + b. From the given condition, we have 8k + b = 15 (5k + b)2 = (2k + b)(4k + b) ⇒. . k=4 b = −17. Then f (x) = 4x − 17. Consist the sequence −13, −9, −5, · · · , (4n − 17) when x = (−13 + 4n − 17)n 2n2 − 15n Sn 1, 2, · · · , n. Sn = = 2n2 − 15n. lim ( 2 ) = lim = n→∞ n n→∞ 2 n2 15 2 − lim = 2. n→∞ n 5.27 . Let all terms of the arithmetic sequence {an } are positive. Show √. 1 √ + a1 + a2. 1 n−1 1 √ + ··· + √ √ =√ √ . a2 + a3 an−1 + an a1 + an 1 1 1 Proof: Let M = √ √ +√ √ + ··· + √ √ , the common difa 1 + a2 a2 + a3 an−1 + an √ √ an−1 − an √ 1 1 √ ference is d. We have √ = − ( an−1 − an ) ⇒ M = √ = an−1 + an an−1 − an d √ √ √ √ √ 1 √ 1 a1 − an 1 √ √ − ( a1 − a2 + a2 − a3 +· · ·+ an−1 − an ) = − ( a1 − an ) = − √ √ = d d a 1 + an d 1 a1 − [a1 + (n − 1)d] 1 −(n − 1)d n−1 − =− √ √ √ √ =√ √ . d a1 + an d a 1 + an a 1 + an √. 5.28 . Solve the nth partial sum of the sequence 1, 3a, 5a2 , 7a3 , · · · , (2n − 1)an−1 .. Download free eBooks at bookboon.com 60. Click on the ad to read more.

(61) 5.27 . Let all terms of the arithmetic sequence {an } are positive. Show √. ELEMENTARY ALGEBRA EXERCISE BOOK II. 1 √ + a2 a1 +Sequences. 1 n−1 1 √ + ··· + √ √ =√ √ . a 1 + an a2 + a3 an−1 + an 1 1 1 Proof: Let M = √ √ +√ √ + ··· + √ √ , the common difa1 + a2 a2 + a3 an−1 + an √ √ an−1 − an √ 1 1 √ ference is d. We have √ = − ( an−1 − an ) ⇒ M = √ = an−1 + an an−1 − an d √ √ √ √ √ 1 √ 1 a1 − an 1 √ √ − ( a1 − a2 + a2 − a3 +· · ·+ an−1 − an ) = − ( a1 − an ) = − √ √ = d d d a1 + an 1 a1 − [a1 + (n − 1)d] 1 −(n − 1)d n−1 − =− √ √ √ √ =√ √ . d a1 + an d a1 + an a1 + an √. 5.28 . Solve the nth partial sum of the sequence 1, 3a, 5a2 , 7a3 , · · · , (2n − 1)an−1 .. Solution: When a = 1, the sequence is 1, 3, 5, 7, · · · , (2n − 1). Sn =. [1 + (2n − 1)]n = 2. n2 (n ∈ N ∗ ). 1 Multiplying the equation When a = 1, Sn = 1+3a+5a2 +7a3 +· · ·+(2n−1)an−1 . 2 3 4 1 by a to obtain aSn = a + 3a + 5a + 7a + · · · + (2n − 1)an . 2 Using 1 − , 2 we have (1 − a)Sn = 1 + 2a + 2a2 + 2a3 + · · · + 2an−1 − (2n − 1)an = 1 − (2n − 1)an + 2(a + 2(a − an ) a(1 − an−1 ) a2 + a3 + · · · + an−1 ) = 1 − (2n − 1)an + 2 = 1 − (2n − 1)an + . 1−a 1−a n n 2(a − a ) 1 − (2n − 1)a While 1 − a = 0, then Sn = + (n ∈ N ∗ ). 2 1−a (1 − a) 5.29 Let the nth partial sum of the sequence {an } is Sn = 2n2 , {bn } is a geometric sequence, and a1 = b1 , b2 (a2 − a1 ) = b1 . (1) Find the general term of {an } an and {bn }. (2) Let cn = , evaluate the nth partial sum Tn of the sequence {cn }. bn Solution: (1) S1 = a1 = 2 when n = 1. an = Sn − Sn−1 = 2n2 − 2(n − 1)2 = 4n − 2 when n 2. 4n − 2 = 2 = a1 when n = 1. Hence the general term of {an } is an = 4n − 2 = 2 + 4(n − 1). Therefore {an } is an arithmetic sequence with the fist term 2 and the common difference 4. Let common ratio of {bn } is q. Since b2 (a2 − a1 ) = b1 and b2 = b1 q, then b1 qd = b1 . 1 1 2 Thus q = = . Otherwise, b1 = a1 , then bn = b1 q n−1 = n−1 . d 4 4 an = (2n − 1)4n−1 . Tn = c1 + c2 + · · · + cn = 1 + 3 × 4 + 5 × 42 + · · · + (2n − (2) cn = bn 1 Multiplying the equation 1 by 4 to obtain 4Tn = 1×4+3×42 +5×43 +· · ·+ 1)4n−1 . 2 Using − 1 , 2 we have 3Tn = −1−2×(4+42 +· · ·+4n−1 )+(2n−1)4n = (2n−1)4n . n−1 5 1 1 4(1 − 4 ) −1 − 2 + (2n − 1)4n = + (6n − 5)4n = [(6n − 5)4n + 5]. Therefore 1−4 3 3 3 1 n ∗ Tn = [(6n − 5)4 + 5] (n ∈ N ). 9 5.30 The sequence {an } is a geometric sequence, a1 = 8, bn = log2 an . If the first 7th partial sum S7 of {bn } is the maximum value, and S7 = S8 . Find the range of the common ratio q of the sequence {an }. an+1 = log2 q. Then {bn } is an an arithmetic sequence, and its first term is b1 = log2 a1 = 3, its common difference is log2 q. From the given condition, we have Download free eBooks at bookboon.com b761 0 b <0 Solution: bn+1 − bn = log2 an+1 − log2 an = log2.

(62) 9 5.30 . The sequence {an } is a geometric sequence, a1 = 8, bn = log2 an . If the Sequences is the maximum value, and S7 = S8 . Find the range of the common ratio q of the sequence {an }.. ELEMENTARY ALGEBRA EXERCISE first 7th partial sum S7 ofBOOK {bn }II. an+1 = log2 q. Then {bn } is an an arithmetic sequence, and its first term is b1 = log2 a1 = 3, its common difference is log2 q. From the given condition, we have b7 0 b8 < 0 Solution: bn+1 − bn = log2 an+1 − log2 an = log2. Then. . 3 + 6 log2 q 0 3 + 7 log2 q < 0 √ 1 3 2 −3 Thus − log2 q < − . Hence q ∈ [ , 2 7 ). 2 7 2 5.31 For a positive number, its decimal part, integer part and itself form a geometric sequence. Find this number. Solution: Let this number is x, its integer part is [x], its decimal part is x − [x]. 2 From the given condition, we have x(x − [x]) = [x]2 which means x2 − [x]x √ − [x] = 0 1+ 5 [x]. Since where [x] > 0, 0 < x − [x] < 1. Solve the equation, then x = 2 √ √ 1+ 5 5−1 0 < x − [x] < 1, then 0 < [x] − [x] < 1. Thus 0 < [x] < 1. Hence 2 2 √ √ 1+ 5 1+ 5 < 2. Therefore [x] = 1 ⇒ x = . 0 < [x] < 2 2 5.32 The sequence {an } has k terms (k is a fixed number). Its nth partial sum Sn = 2n2 + n (n k, n ∈ N ∗ ). If we remove one term (neither the first term nor the last term) from the k terms, the average value of the remaining (k − 1) terms is 79. (1) Find the general term for {an }. (2) Determine k and which term is removed. Solution: (1) From the given condition, we have S1 = a1 = 3, an = Sn − Sn−1 = (2n2 + n) − [2(n − 1)2 + (n − 1)] = 4n − 1, (n 2). Since a1 satisfies the above formula, then an = 4n − 1, (n k, n ∈ N ∗ ). (2) Let the removed term be the tth term, then 1 < t < k. From the given condition, we have Sk − at = 79(k − 1). Thus 2k 2 + k − (4t − 1) = 79k − 79 ⇒ 4t = 2k 2 − 78k + 80 ⇒ 4 < 2k 2 − 78k + 80 < 4k ⇒ 38 < k < 40. Since k ∈ N ∗ , then k = 39. k 2 − 39k + 40 = 20. Therefore the removed term is the 20th term. Hence t = 2 5.33 Given f (x) = x2 − (2n + 1)x + n2 + 5n − 7. (1) If the y-ordinate of the vertex of the graph of f (x) form a sequence {an }, show {an } is an arithmetic sequence. (2) If the distance from the vertex of the graph of f (x) to x-axis form a sequence {bn }, evaluate the nth partial sum of {bn }. Solution: (1) f (x) = x2 − (2n + 1)x + n2 + 5n − 7 = [x − (n + 1)]2 + 3n − 8, then Download free eBooks at bookboon.com 62.

(63) 5.33 Given f (x) = x2 − (2n + 1)x + n2 + 5n − 7. (1) If the y-ordinate of the vertex of the graph of f (x) form a sequence {an }, show {an } is an arithmetic sequence. (2) If the distance from the vertex of the graph of f (x) to x-axis form a sequenceSequences {bn }, ELEMENTARY ALGEBRA EXERCISE BOOK II evaluate the nth partial sum of {bn }. Solution: (1) f (x) = x2 − (2n + 1)x + n2 + 5n − 7 = [x − (n + 1)]2 + 3n − 8, then. an = 3n − 8. Since an+1 − an = [3(n + 1) − 8] − (3n − 8) = 3, then {an } is an arithmetic sequence which common difference is 3. (2) Applying (1), we have bn = |3n − 8|. When 1 n 2, then bn = 8 − 3n, 13n − 3n2 (5 + 8 − 3n)n b1 = 5, Sn = = . When n 3, then bn = 3n − 8, 2 2 3n2 − 13n + 28 (1 + 3n − 8)(n − 2) Sn = 5 + 2 + 1 + 4 + 7 + · · · + (3n − 8) = 7 + = . 2 2 Thus  2   13n − 3n , (1 n 2) 2 Sn = 2   3n − 13n + 28 , (n 3) 2 5.34 If we insert a number a between two positive numbers, the three numbers form an arithmetic sequence. If we insert two numbers b and c, the four numbers form a geometric sequence. Show (1) 2a > b + c (2) (a + 1)2 (b + 1)(c + 1). 1 Proof: (1) Let the two positive numbers be m and n (m, n > 0). Then m+n = 2a , 2 2 2 nb = c . 3 Applying 1 to obtain a > 0. Applying 2 and 3 to obtain mc = b , b2 c2 + = m + n = 2a. Then 2abc = b3 + c3 = (b + c)(b2 + c2 − bc) b, c > 0. Thus c b (b + c)(2bc − bc) = (b + c)bc. Hence 2a > b + c. √ √ m+n mn = bc. Thus a2 bc. Applying 2 (1) again, we have 2a > b + c. Thus a2 bc 2a b + c (2) Applying (1) to obtain a =. ⇒ a2 + 2a bc +Challenge b + c ⇒ (a 1)2we run bc + b + c + 1 = (b + 1)(c + 1). the+way 5.35 For an arbitrary real number x, [x] denotes the integer part of x. It means [x] is the biggest integer number which satisfies [x] x. (1) evaluate [log2 1] + [log2 2] + · · · + [log2 1024]. (2) Deduce the formula of [log2 1] + [log2 2] + · · · + [log2 (2n − 1)].. EXPERIENCE THE POWER OF FULL(1)ENGAGEMENT… Solution: [log2 1] = 0, [log2 2] = [log2 3]. = 1, [log2 4] = [log2 5] = [log2 6] = [log2 7] = 2,· · · , [log2 512] = [log2 513] = · · · = [log2 1023] = 9, [log2 1024] = 10. Thus [log2 1] + [log2 2] + · · · + [log2 1024] = 2 + 2 × 22 + 3 × 23 + · · · + 9 × 29 + 10 = 8204. [log2 2] + · · · + [log2 (2n − 1)] = Sn . Applying (1) to obtain Sn = (2) Let [log 2 1] + RUN FASTER.. RUN LONGER.. RUN EASIER…. READ MORE & PRE-ORDER TODAY WWW.GAITEYE.COM. 1349906_A6_4+0.indd 1. 22-08-2014 12:56:57. Download free eBooks at bookboon.com 63. Click on the ad to read more.

(64) 1 Proof: (1) Let the two positive numbers be m and n (m, n > 0). Then m+n = 2a , 3 Applying 1 to obtain a > 0. Applying 2 and 3 to obtain 2 nb = c2 . mc = b2 , b2 c2 + = m + n = 2a. Then 2abc = b3 + c3 = (b + c)(b2 + c2 −Sequences bc) b, c > 0. Thus ELEMENTARY ALGEBRA EXERCISE c b BOOK II (b + c)(2bc − bc) = (b + c)bc. Hence 2a > b + c. √ √ m+n mn = bc. Thus a2 bc. Applying 2 (1) again, we have 2a > b + c. Thus a2 bc 2a b + c (2) Applying (1) to obtain a =. ⇒ a2 + 2a bc + b + c ⇒ (a + 1)2 bc + b + c + 1 = (b + 1)(c + 1). 5.35 For an arbitrary real number x, [x] denotes the integer part of x. It means [x] is the biggest integer number which satisfies [x] x. (1) evaluate [log2 1] + [log2 2] + · · · + [log2 1024]. (2) Deduce the formula of [log2 1] + [log2 2] + · · · + [log2 (2n − 1)]. Solution: (1) [log2 1] = 0, [log2 2] = [log2 3] = 1, [log2 4] = [log2 5] = [log2 6] = [log2 7] = 2,· · · , [log2 512] = [log2 513] = · · · = [log2 1023] = 9, [log2 1024] = 10. Thus [log2 1] + [log2 2] + · · · + [log2 1024] = 2 + 2 × 22 + 3 × 23 + · · · + 9 × 29 + 10 = 8204. (2) Let [log2 1] + [log2 2] + · · · + [log2 (2n − 1)] = Sn . Applying (1) to obtain Sn =. 2 + 2 × 22 + 3 × 23 + · · · + (n − 1) × 2n−1 , 2Sn = 22 + 2 × 23 + 3 × 24 + · · · + (n − 1) × 2n . The second equation minus the first equation, we have Sn = (n − 1) × 2n − (2 + 22 + 23 + · · · + 2n−1 ) = (n − 1)2n − (2n − 1) = n2n − 2n+1 + 1 (n ∈ N ∗ ).. 5.36 How many terms are same in the first 100the terms of the arithmetic sequence 5, 8, 11, · · · and the arithmetic sequence 3, 7, 11, · · · ? Evaluate the sum of these same terms. Solution: The general term of the arithmetic sequence 5, 8, 11, · · · is an = 3n + 2, the general term of the geometric sequence 3, 7, 11, · · · is bn = 4m − 1, (m, n ∈ N ∗ ). 4 Let 3n + 2 = 4m − 1, then n = m − 1. Let m = 3k, (k ∈ N ∗ ), then n = 4k − 1. 3 Hence the general term of the same terms of the two sequences is ck = 12k − 1. Let 1 1 5 12k − 1 302, then k 25 . Thus k = 1, 2, · · · , 25. Therefore there are 25 2 4 terms which are same in the first 100th terms of the two sequences. Thus {ck } is an arithmetic sequence whose first term is c1 = 11 and common difference [11 + 11 + (25 − 1) × 12] = 3875. is d = c2 − c1 = 12. Hence S25 = 2 5.37 (1) Consider a geometric sequence {an }, a1 = 1, and it has even number of terms. The sum of all odd terms is 85. The sum of all even terms is 170. Evaluate the common ratio q and the number of terms n. (2) All terms of the geometric sequence {an } are positive, and it has even number of terms. The sum of all terms is four times of the sum of all even terms. The product of the 2th term and 4th term is nine times of the sum of the 3th term and 4th term. Compute a1 , the common ratio q, and the term number n when the nth partial sum of sequence {lg an } reaches the maximum value. free eBooks at bookboon.com (3) The nth partial sum ofDownload the geometric sequence {an } whose terms are all posi64 partial sum is 6560. Find the common tive is 80. The largest term is 54. The 2nth ratio q..

(65) 5.37 (1) Consider a geometric sequence {an }, a1 = 1, and it has even number of terms. ALGEBRA The sum of all BOOK odd terms is 85. The sum of all even terms is 170. Evaluate ELEMENTARY EXERCISE II Sequences the common ratio q and the number of terms n. (2) All terms of the geometric sequence {an } are positive, and it has even number of terms. The sum of all terms is four times of the sum of all even terms. The product of the 2th term and 4th term is nine times of the sum of the 3th term and 4th term. Compute a1 , the common ratio q, and the term number n when the nth partial sum of sequence {lg an } reaches the maximum value. (3) The nth partial sum of the geometric sequence {an } whose terms are all positive is 80. The largest term is 54. The 2nth partial sum is 6560. Find the common ratio q. Solution: (1) Since the term number is even, then n. 1 × [1 − (q 2 ) 2 ] = 85. Thus 2n = 256 = 28 ⇒ n = 8. 1 − q2. Seven 170 = 2, Seven = = q = Sodd 85. Seven 1 = = q. From the Sodd 3 condition a2 a4 = 9(a3 + a4 ), we have a21 q 4 = 9(a1 q 2 + a1 q 3 ). Thus a21 − 108a1 = 0. Then. (2) Sn = 4Seven ⇒ Seven + Sodd = 4Seven ⇒ Sodd = 3Seven ⇒. a1 = 108. Such that the nth partial sum of sequence {lg an } reaches the maximum 1 1 1 value, we have lg an = lg(a1 q n−1 ) > 0, then 108( )n−1 > 1. Thus ( )n−1 > . 3 3 108 11 1 n − 1 < log 1 = log 1 ( 3 ) = 3 + log3 4 ⇒ n < 4 + log3 4 4 + 1 = 5. Therefore 3 108 3 3 4 the nth partial sum of sequence {lg an } reaches the maximum value when n = 5.. (3) From the given condition, we know that the last nth partial sum of the positive sequence {lg an } is larger than the first nth partial sum, then q > 0, an = 54, a (1 − q n ) S − Sn 6560 − 80 2 q n = 2n 1 Sn = 1 = 80 , = 81. = an = a1 q n−1 = 54 , 1−q Sn 80 54 2 2 a 2 we have 1 to obtain 1 = = . Thus a1 = q. Substituting it into , Applying q 81 3 3 2 q(1 − 81) 3 = 80. Thus q = 3. 1−q 1 The function is defined on (−1, 1), f ( ) = −1 and satisfied that 2 x+y f (x) + f (y) = f ( ) for x, y ∈ (−1, 1) 1 + xy 2xn 1 . Compute f (xn ). (1) If the sequence {f (xn )} satisfies that x1 = and xn+1 = 2 1 + x2n 1 1 1 2n + 5 (2) Show + + ··· + >− . f (x1 ) f (x2 ) f (xn ) n+2 2xn x n + xn 1 ) = f( ) = f (xn ) + (1) Solution: f (x1 ) = f ( ) = −1. f (xn+1 ) = f ( 2 2 1 + xn 1 + xn xn f (xn+1 ) = 2. Thus {f (xn )} is a geometric sequence and its first f (xn ) = 2f (xn ) ⇒ f (xn ) term is −1, the common ratio is 2. Therefore f (xn ) = −2n−1 , (n ∈ N ∗ ). 5.38 . 1 − 21n 1 1 1 1 1 1 + + ··· + = −(1 + + 2 + · · · + n−1 ) = − = f (xn ) 2 2 f (x1 ) f (x2 ) 2 1 − 12 1 1 2n + 5 2(n + 2) + 1 −(2 − n−1 ) = −2 + n−1 > −2. On the other hand, − =− = 2 2 n+2 Download free eBooks at bookboon.comn + 2 1 1 1 1 2n + 5 1 < −2. Thus 65 + . ) = −2 − + ··· + >− −(2 + f (x1 ) f (x2 ) n+2 f (xn ) n+2 n+2 (2) Proof:.

(66) 1 2xn x n + xn ) = f( ) = f (xn ) + (1) Solution: f (x1 ) = f ( ) = −1. f (xn+1 ) = f ( 2 2 1 + xn 1 + xn xn f (xn+1 ) (xn ) ⇒ first f (xn ) = 2f ELEMENTARY ALGEBRA EXERCISE BOOK=II 2. Thus {f (xn )} is a geometric sequence and its Sequences f (xn ) term is −1, the common ratio is 2. Therefore f (xn ) = −2n−1 , (n ∈ N ∗ ). 1 − 21n 1 1 1 1 1 1 + + ··· + = −(1 + + 2 + · · · + n−1 ) = − (2) Proof: = f (x1 ) f (x2 ) f (xn ) 2 2 2 1 − 12 2(n + 2) + 1 1 1 2n + 5 =− = −(2 − n−1 ) = −2 + n−1 > −2. On the other hand, − 2 2 n+2 n+2 1 1 1 1 2n + 5 1 ) = −2 − < −2. Thus + + ··· + >− . −(2 + n+2 n+2 f (x1 ) f (x2 ) f (xn ) n+2 a1 + 2a2 + 3a3 + · · · + nan , 1 + 2 + 3 + ··· + n and {bn } is a geometric sequence. Show {an } is also a geometric sequence.. 5.39 . If the sequence {an } and {bn } satisfy bn =. Proof: Since (1+2+3+· · ·+n)bn = a1 +2a2 +3a3 +· · ·+nan , then. n(n + 1) bn = a1 +2a2 + 2. (n − 1)n 2 bn−1 = a1 +2a2 +3a3 +· · ·+(n−1)an−1 . 2 (n − 1)n n(n + 1) 1 − 2 to obtain bn − bn−1 = nan , (n = 2, 3, · · · ) ⇒ an = Checking 2 2 1 [(n + 1)bn − (n − 1)bn−1 ], (n = 2, 3, 4, · · · ) (∗). Since {bn } is a geometric sequence, 2 let the common difference be d. Substituting bn = b1 + (n − 1)d into (∗), we obtain 1 3 an = {(n + 1)[b1 + (n − 1)d] − (n − 1)[b1 + (n − 2)d]} = b1 + (n − 1)d, (n = 2, 3, · · · ). 2 2 3 a1 + 2a2 + 3a3 + · · · + nan For bn = , let n = 1, then b1 = a1 . Thus an = a1 + (n − 1 + 2 + 3 + ··· + n 2 3 3 3 1)d, (n = 2, 3, · · · ). Hence an − an−1 = [a1 + (n − 1)d] − [a1 + (n − 2)d] = d 2 2 2 (constant). Therefore {an } is also a geometric sequence. 3a3 +· · ·+nan. 1 We also have .. 3.40 Let the first nth partial sum Sn of sequence {an } satisfy Sn = 1 − 2 an (n ∈ N ∗ ). 3 (1) Calculate Sn and an . (2) If we let Tn denote the first nth partial sum of sequence {an Sn }, compute lim Tn . n→∞. 2 2 5 2 Solution: (1) Sn = 1 − an ⇒ Sn = 1 − (Sn − Sn−1 ) ⇒ Sn = 1 + Sn−1 ⇒ 3 3 3 3 2 2 Sn − 1 2 3 5 (Sn − 1) = (Sn−1 − 1) ⇒ = . Since S1 = 1 − S1 , then S1 = , 3 3 Sn−1 − 1 5 3 5 2 2 3 S1 − 1 = − 1 = − . Then {Sn − 1} is a geometric sequence and its first term is − , 5 5 5 2 2 2 n−1 2 n 2 n the common ratio is . Sn − 1 = (− )( ) = −( ) . Therefore Sn = 1 − ( ) , an = 5 5 5 5 5 2 n 2 n−1 2 n−1 2 n 2 n 5 3 2 n Sn −Sn−1 = [1−( ) ]−[1−( ) ] = ( ) −( ) = ( ) ( −1) = ( ) (n ∈ N ∗ ). 5 5 5 5 5 2 2 5 3 2 3 2 2 n−1 3 2 (2) Since an = ( )n = ( ) = ( )n−1 , then {an } is a geometric sequence 2 5 25 5 5 5 3 2 3 2 2 3 2 and its first term is , the common ratio is . an Sn = ( )n−1 [1 − ( )n ] = ( )n−1 − 5 5 5 5 5 5 5 3 6 5 2 6 2 2 n−1 25 [( ) ] . Thus lim Tn = lim an Sn = 5 2 − =1− = . n→∞ n→∞ 7 7 25 5 1− 5 1 − ( 25 )2 5.41 Let the common ratio of the sequence {an } is q > 1. The square Download free geometric eBooks at bookboon.com of the 17th term is equal to the 24th term. 66Compute the range of the integer number 1 1 1 1.

(67) n−1. 2 3 2 S1 − 1 = − 1 = − . Then {Sn − 1} is a geometric sequence and its first term is − , 5 5 5 2 2 2 n−1 2 n 2 n the common ratio is . Sn − 1 = (− )( ) = −( ) . Therefore Sn = 1 − ( ) , an = ELEMENTARY ALGEBRA EXERCISE 5 BOOK II 5 5 5 5 Sequences 2 n 2 n−1 2 n−1 2 n 2 n 5 3 2 n Sn −Sn−1 = [1−( ) ]−[1−( ) ] = ( ) −( ) = ( ) ( −1) = ( ) (n ∈ N ∗ ). 5 5 5 5 5 2 2 5 3 2 3 2 2 n−1 3 2 ( ) (2) Since an = ( )n = = ( )n−1 , then {an } is a geometric sequence 2 5 25 5 5 5 3 2 3 2 2 3 2 and its first term is , the common ratio is . an Sn = ( )n−1 [1 − ( )n ] = ( )n−1 − 5 5 5 5 5 5 5 3 6 5 2 6 2 2 n−1 25 [( ) ] . Thus lim Tn = lim an Sn = 5 2 − =1− = . n→∞ n→∞ 25 5 7 7 1− 5 1 − ( 25 )2 5.41 Let the common ratio of the geometric sequence {an } is q > 1. The square of the 17th term is equal to the 24th term. Compute the range of the integer number 1 1 1 1 n which satisfies a1 + a2 + a3 + · · · + an > + + + ··· + . a1 a2 a3 an Solution: a217 = a24 ⇒ (a1 q 16 )2 = a1 q 23 . Since q > 1 and a1 = 0, then a1 = q −9 . Since a1 + a2 + a3 + · · · + an >. 1 1 1 1 a1 (1 − q n ) > + + + · · · + , then a1 a2 a3 an 1−q. 1 (1 a1. −. 1−. 1 ) qn , 1 q. 1 1 Substituting a1 = q −9 into , 1 we have q −18 > q 1−n . . a1 q n−1 Since q > 1, then −18 > 1 − n. Thus n > 19. On the other hand, (n ∈ N ∗ ), then n 20. Hence the range of the integer number n is [20, +∞). which means a1 >. 5.42 Given the arithmetic sequence {an } and the x-dependent equations ai x2 + 2ai+1 x + ai+2 = 0, (i = 1, 2, · · · , n), and a1 and the common difference d are both nonzero real numbers. (1) Show these equations have same solutions. (2) If another 1 1 1 solution is βi , then , ,··· , form a geometric sequence. βn + 1 β1 + 1 β2 + 1 Proof: (1) Since ai x2 +2ai+1 x+ai+2 = 0 and {an } is a geometric sequence which means 2ai+1 = ai + ai+2 , then ai x2 + (ai + ai+2 )x + ai+2 = 0. Then (x2 + x)ai + (x + 1)ai+2 = 0. Since a1 and the common difference d are both nonzero real numbers, then ai = 0 and ai+2 = 0. Thus x2 + x = 0 and x + 1 = 0. Hence x = −1. Therefore these equations have same solutions x = −1. 2ai+1 = (2) Applying the relation of roots and coefficient to obtain βi + (−1) = − ai 1 1 2d 2d 1 2(ai + d) − − = = −2 − ⇒ βi = −1 − ⇒ − βi + 1 ai ai βi+1 + 1 ai −1 − a2d + 1 i+1 1 ai − ai+1 1 −d 1 1 1 = − (constant). Then β1 +1 , β2 +1 , · · · , βn +1 form a = = 2 2d 2d −1 − 2d +1 ai geometric sequence.. This e-book is made with. SetaPDF. SETASIGN. √ x2 − 4 (x −2). (1) Find the inverse function 1 f −1 (x). (2) Let a1 = 1, an = −f −1PDF (an−1 ),components evaluate an . (3)for If PHP b1 = developers , b2 = a1 + a2 1 1 , · · · , bn = , · · · , compute the first nth partial sum of {bn }. a2 + a3 an + an+1 √ Solution: (1) Since√ y = f (x) = x2 − 4 (x −2), then x = − y 2 + 4, which means f −1 (x) = − x2 + 4, (x 0). (2) From the given condition and (1), we have an = a2n−1 + 4. Squaring both sides of the equation, then a2n = a2n−1 + 4, that is a2n − a2n−1 = 4. Hence a22 − a21 = 4, Download free eBooks at bookboon.com Clickapplying on the ad a23 − a22 = 4, · · · , a2n − a2n−1 = 4.√Adding the above equations and a1to=read 1 tomore 67 obtain a2n = 4n − 3. Thus an = 4n − 3, (n ∈ N ∗ ). 5.43 . Given f (x) =. www.setasign.com.

(68) 1 1 Substituting a1 = q −9 into , 1 we have q −18 > q 1−n . . a1 q n−1 Since q > 1, then −18 > 1 − n. Thus n > 19. On the other hand, (n ∈ N ∗ ), then n 20. Hence theEXERCISE range BOOK of the ELEMENTARY ALGEBRA II integer number n is [20, +∞). Sequences which means a1 >. 5.42 Given the arithmetic sequence {an } and the x-dependent equations ai x2 + 2ai+1 x + ai+2 = 0, (i = 1, 2, · · · , n), and a1 and the common difference d are both nonzero real numbers. (1) Show these equations have same solutions. (2) If another 1 1 1 solution is βi , then , ,··· , form a geometric sequence. β1 + 1 β2 + 1 βn + 1 Proof: (1) Since ai x2 +2ai+1 x+ai+2 = 0 and {an } is a geometric sequence which means 2ai+1 = ai + ai+2 , then ai x2 + (ai + ai+2 )x + ai+2 = 0. Then (x2 + x)ai + (x + 1)ai+2 = 0. Since a1 and the common difference d are both nonzero real numbers, then ai = 0 and ai+2 = 0. Thus x2 + x = 0 and x + 1 = 0. Hence x = −1. Therefore these equations have same solutions x = −1. 2ai+1 = (2) Applying the relation of roots and coefficient to obtain βi + (−1) = − ai 1 1 2d 2d 1 2(ai + d) − − = = −2 − ⇒ βi = −1 − ⇒ − ai ai ai βi+1 + 1 βi + 1 −1 − a2d + 1 i+1 ai − ai+1 1 −d 1 1 1 1 = = = − (constant). Then β1 +1 , β2 +1 , · · · , βn +1 form a 2d 2d 2 −1 − 2d +1 ai geometric sequence. √ x2 − 4 (x −2). (1) Find the inverse function 1 f −1 (x). (2) Let a1 = 1, an = −f −1 (an−1 ), evaluate an . (3) If b1 = , b2 = a1 + a2 1 1 , · · · , bn = , · · · , compute the first nth partial sum of {bn }. a2 + a3 an + an+1 √ Solution: (1) Since√ y = f (x) = x2 − 4 (x −2), then x = − y 2 + 4, which means f −1 (x) = − x2 + 4, (x 0). (2) From the given condition and (1), we have an = a2n−1 + 4. Squaring both sides of the equation, then a2n = a2n−1 + 4, that is a2n − a2n−1 = 4. Hence a22 − a21 = 4, a23 − a22 = 4, · · · , a2n − a2n−1 = 4.√Adding the above equations and applying a1 = 1 to obtain a2n = 4n − 3. Thus an = 4n − 3, (n ∈ N ∗ ). 1 1 1 a2 − a1 a3 − a2 (3) Sn = b1 + b2 + · · · + bn = + + + +···+ = a1 + a2 a2 + a3 an + an+1 4 4 √ 4n + 1 − 1 an+1 − a1 an+1 − an = = , (n ∈ N ∗ ). + 4 4 4 5.43 . Given f (x) =. 5.44 For the arithmetic sequence {an }, a1 = 1, the common difference is d, the first nth partial sum is An . For the geometric sequence {bn }, b1 = 1, the common ratio is q (|q| < 1), the first nth partial sum is Bn . Let Sn = B1 + B2 + · · · + Bn . An If lim ( − Sn )=1, evaluate d and q. n→∞ n n(n − 1) 1(1 − q n ) , Solution: From the given condition, we have An = n + d, Bn = 1−q 2 n ) n − q(1−q (1 − q)n − q(1 − q n ) (1 − q) + (1 − q 2 ) + · · · + (1 − q n ) 1−q = Sn = = . S1−q (1 − q)2 1−q lim q n =0 An n−1 (1 − q)n − q(1 − q n ) n→∞ − Sn ) = 1, then lim [1 + ] = 1 ⇒ d− ince lim ( n→∞ n n→∞ (1 − q)2 2 q d (1 − q)n − q n−1 d 1 )n − + d− ]] = 0 ⇒ lim [( − ]=0⇒ lim [ 2 n→∞ n→∞ 2 2 (1 − q)2 (1 − q) 1−q 2 Download free eBooks at bookboon.com  168 d  =0 − .

(69) 5.44 For the arithmetic sequence {an }, a1 = 1, the common difference is d, the first nth partial sum is An . For the geometric sequence {bn }, b1 = 1, the common ratio ALGEBRA is q (|q|EXERCISE < 1), the first + Bn . ELEMENTARY BOOK II nth partial sum is Bn . Let Sn = B1 + B2 + · · ·Sequences An If lim ( − Sn )=1, evaluate d and q. n→∞ n n(n − 1) 1(1 − q n ) d, Bn = , Solution: From the given condition, we have An = n + 2 1−q n ) n − q(1−q (1 − q)n − q(1 − q n ) (1 − q) + (1 − q 2 ) + · · · + (1 − q n ) 1−q = = . SSn = 1−q 1−q (1 − q)2 lim q n =0 (1 − q)n − q(1 − q n ) An n−1 n→∞ − Sn ) = 1, then lim [1 + d− ] = 1 ⇒ ince lim ( n→∞ n n→∞ 2 (1 − q)2 (1 − q)n − q 1 d q n−1 d d− )n − + ]] = 0 ⇒ lim [( − ]=0⇒ lim [ 2 n→∞ n→∞ 2 2 (1 − q) 1−q 2 (1 − q)2  1 d  − =0  2 1−q d q   − + =0 2 (1 − q)2 ⇒. . d=4 1 q= 2. 5.45 If the product of the first 3th terms of an increasing geometric sequence {an } is 512, and subtracting 1,3,9 from these three terms respectively form an arith1 2 3 n metic sequence. Show + + + ··· + < 1. a1 a2 a3 an a Proof: Let the first 3th terms of the increasing geometric sequence be , a, aq. From q the given condition, we have a3 = 512, then a = 8. Similarly from the given condia 8 www.sylvania.com tion, we have ( − 1) + (aq − 9) = 2(a − 3), then ( − 1) + (8q − 9) = 10. Solving q q 1 or q = 2. Since {an } is an increasing sequence, then the equation, we have q = 2 8 1 not 2 reinvent 3 n We do = 4, an = 4 × 2n−1 = 2n+1 . Let S = q = 2, a1 = + + + ··· + = 2 a1 a2 a3 an wheel 1 2 3 n 1 1 2 the 3 n we reinvent 1 then S = + + + · · · + n+2 , . 2 ( 1 − ) 2 ×2 + + + · · · + n+1 , , 4 8 16 2 2 2 8 16 light. 32 Fascinating lighting offers an infinite spectrum of possibilities: Innovative technologies and new markets provide both opportunities and challenges. An environment in which your expertise is in high demand. Enjoy the supportive working atmosphere within our global group and benefit from international career paths. Implement sustainable ideas in close cooperation with other specialists and contribute to influencing our future. Come and join us in reinventing light every day.. Light is OSRAM. Download free eBooks at bookboon.com 69. Click on the ad to read more.

(70) 5.45 If the product of the first 3th terms of an increasing geometric sequence {an } is 512, and EXERCISE subtracting from these three terms respectively form anSequences arithELEMENTARY ALGEBRA BOOK 1,3,9 II 1 2 3 n metic sequence. Show + + + ··· + < 1. a1 a2 a3 an a Proof: Let the first 3th terms of the increasing geometric sequence be , a, aq. From q the given condition, we have a3 = 512, then a = 8. Similarly from the given condi8 a tion, we have ( − 1) + (aq − 9) = 2(a − 3), then ( − 1) + (8q − 9) = 10. Solving q q 1 or q = 2. Since {an } is an increasing sequence, then the equation, we have q = 2 8 1 2 3 n q = 2, a1 = = 4, an = 4 × 2n−1 = 2n+1 . Let S = + + + ··· + = 2 a1 a2 a3 an 1 2 3 n 1 2 3 n 1 1 then S = + + + · · · + n+2 , . 2 ( 1 − ) 2 ×2 + + + · · · + n+1 , , 4 8 16 2 2 8 16 32 2 1 1 1 n 1 n leads to S = + + · · · + n − n+1 = 1 − n − n+1 < 1. 2 4 2 2 2 2 5.46 Let the common difference of arithmetic sequence {an } is nonzero, {bn } is a geometric sequence. If a1 = 3, b1 = 1, a2 = b2 , 3a5 = b3 . For an arbitrary positive number n, there are constants α and β such that an = logα bn + β always holds. Evaluate α + β. Solution: Let the common difference of {an } is d, the common ratio of {bn } is q. Since a1 = 3, b1 = 1, a2 = b2 , 3a5 = b3 , then 3+d=q 3(3 + 4d) = q 2 Thus. . d=6 q=9. Hence an = 3 + (n − 1)6 = 6n − 3, bn = 9n−1 . Then 6n − 3 = logα 9n−1 + β = n logα 9 − logα 9 + β always holds for an arbitrary positive number n. Thus logα 9 = 6, √ 1 2 3 6 2 6 3 β − logα 9 = −3 ⇒ α = 9 = 3 . Since α > 0, then α = 3 = 3 = 3, √ log3 9 2 β = −3 + = −3 + 1 = 3. Therefore α + β = 3 3 + 3. log3 α 3 5.47 of a10 .. The third-order arithmetic sequence {an } is 1,2,8,22,47,· · · , find the value. Solution: Since {an } is a third-order arithmetic sequence, then let an = An3 + Bn2 + Cn + D where A, B, C, D are undetermined coefficients. From the given condition, we 1 When n = 2, have the following relationships: When n = 1, A + B + C + D = 1 ; 2 When n = 3, 27A + 9B + 3C + D = 8 ; 3 When n = 4, 8A + 4B + 2C + D = 2 ; 1 4 According to the , 1 , 2 3 and , 4 we have A = , 64A + 16B + 4C + D = 22 . 2 1 1 3 1 2 B = − , C = −1, D = 2. Thus the general term of {an } is an = n − n − n + 2. 2 2 2 1 1 3 2 Therefore a10 = × 10 − × 10 − 10 + 2 = 442. 2 2 5.48 . For the sequence {an }, fn (x) = a1 x + a2 x2 + · · · + an xn , and a1 = 3, 1 fn (1) = p(1 − n ). (1) Evaluate p. (2) Find the general term of {an }. (3) Find the Download free eBooks at bookboon.com 2 minimum value of positive integer n such that 3fn (2) 2005fn (1). 70.

(71) 2 1 1 3 1 2 B = − , C = −1, D = 2. Thus the general term of {an } is an = n − n − n + 2. 2 2 2 1 1 3 2 × 10 − 10 − 10 + 2 = 442. Therefore ALGEBRA a10 = EXERCISE ELEMENTARY BOOK× Sequences 2 2 II 5.48 . For the sequence {an }, fn (x) = a1 x + a2 x2 + · · · + an xn , and a1 = 3, 1 fn (1) = p(1 − n ). (1) Evaluate p. (2) Find the general term of {an }. (3) Find the 2 minimum value of positive integer n such that 3fn (2) 2005fn (1). 1 Solution: (1) From the given condition, we have fn (1) = a1 + a2 + · · · + an = p(1 − n ). 2 p Since a1 = 3, then = 3. Thus p = 6. 2 1 1 1 (2) when n 2, an = 6(1 − n ) − 6(1 − n−1 ) = 3( )n−1 , (n ∈ N ∗ ). 2 2 2 3 3 n (3) Since fn (2) = 2a1 + 4a2 + · · · + 2 an = 2 × 3 + 4 × + · · · + 2n n−1 = 6n and 2 2 1 3fn (2) 2005fn (1), then 3 × 6n 2005 × 6(1 − n ). Then 3n 2005(1 − 21n ). When 2 2005 n 10, it does not hold. Whenn > 10, < 1. Then 3n > 2004 which means n 2 n > 668. Since n ∈ N ∗ , then the minimum value of n is 669. 5.49 Given a0 = a1 = 1, and a0 an + a1 an−1 + · · · + an a0 = 2n an . Show 1 holds for all n ∈ N ∗ . an = n! 1 = 1. p(0) and p(1) hold. (2) Assume Proof: (1) When n = 0 or 1, a0 = a1 = 1! 1 when n = k, p(k) holds which means ak = . Applying the recurrence relation, we k! 1 1 1 have ak+1 + + + ··· + + ak+1 = 2k+1 ak+1 when n = k + 1. Then 1!k! 2!(k − 1)! k!1! 1 1 k+1 1 2 k (2 (Ck+1 + Ck+1 + · · · + Ck+1 (2k+1 − 2). Thus − 2)ak+1 = ) = (k + 1)! (k + 1)! 1 ak+1 = . When n = k + 1, p(k + 1) also holds. Therefore for all integers n 0, (k + 1)! 1 an = holds. n! 5.50 Let A, B, C be the three interior angles of ABC. lg A, lg B, lg C form an arithmetic sequence. Find the range of B. Solution: From the given condition, we have lg A + lg C = 2 lg B, then B 2 = AC. A+C 2 π ) . Since Thus C > B > A and B < . Hence [π − (A + C)]2 = AC ( 2 2 π A+C 2π 2π π A+C B < ⇒ π − (A + C) ⇒ A+C ⇒B π− = . 2 2 2 3 3 3 π π Therefore B < . 3 2 5.51 . For {an }, a1 = 1, 8an+1 an − 16an+1 + 2an + 5 = 0,. Download free eBooks at bookboon.com 71. (n ∈ N ∗ ). And.

(72) Deloitte & Touche LLP and affiliated entities.. ) . Since Thus C > B > A and B < . Hence [π − (A + C)] = AC ( 2 2 π 2π A+C 2π π A+C = . ⇒B π− ⇒ π − (A + C) ⇒ A+C B < 3 3 3 2 2 2 π π B EXERCISE < . BOOK II Therefore ALGEBRA ELEMENTARY Sequences 2 3 5.51 . For {an }, a1 = 1, 8an+1 an − 16an+1 + 2an + 5 = 0,. (n ∈ N ∗ ). And. 1 (n ∈ N ∗ ). 1, an − 2 (1) Find the value of b1 , b2 , b3 , b4 . (2) Find the general term of {bn }, and the nth partial sum of {an bn }, denoted as {Sn }.. bn =. 1 1 1 + . Substituting it into the equa1 , then an = bn 2 an − 2 4 6 3 tion 8an+1 an − 16an+1 + 2an + 5 = 0, we have − + = 0. It means bn+1 bn bn+1 bn 8 4 1 1 4 bn+1 = 2bn − . Since a1 = 1, then 1 = + . Thus b1 = 2. Hence b2 = 2b1 − = , 3 b1 2 3 3 4 20 4 b3 = 2b2 − = 4, b4 = 2b3 − = . 3 3 3 2 4 4 4 4 (2) From (1), we have bn+1 = 2bn − . Then bn+1 − = 2(bn − ). Since b1 − = = 0, 3 3 3 3 3 2 4 then {bn − } is a geometric sequence and its first term is , the common ratio q is 2. 3 3 2 n−1 4 1 n 1 n 4 1 Hence bn − = 2 = 2 . Thus bn = 2 + , (n ∈ N ∗ ). Since bn = , 3 3 3 3 3 an − 12 1 then an bn = 12 bn + 1. Therefore Sn = a1 b1 + a2 b2 + · · · + an bn = (b1 + b2 + +bn ) + n = 2 2 (1 − 2n ) 5 5 1 n 4 4 4 1 3 [(b1 − )+(b2 − )+· · ·+(bn − )]+ n = + n = (2 +5n−1), (n ∈ N ∗ ). 2 3 3 3 3 2(1 − 2) 3 3 Solution: (1) Since bn =. 360° thinking. .. 5.52 If the increasing sequence {an } satisfies an+2 = an+1 + an when n 1, a7 = 120. Find the value of a8 . Solution: Let a1 = x, a2 = y, x, y ∈ N ∗ . From the given condition, we have x < y, and a3 = x + y, a4 = x + 2y, a5 = 2x + 3y, a6 = 3x + 5y, a7 = 5x + 8y, a8 = 8x + 13y. Since a7 = 120, then 5x + 8y = 120. We have x = 8t y = 15 − 5t. 360° thinking. .. 360° thinking. .. 15 where t is a integer number. Since y > x > 0, then 15t−5t > 8t > 0. Thus 0 < t < . 13 Then t = 1. Hence x = 8, y = 10. Therefore a8 = 8 × 8 + 13 × 10 = 194.. 5.53 If the nth partial sum of an arithmetic sequence {an } with positive common difference is Sn , a3 a4 = 117, a2 + a5 = 22. (1) Find the general term an . (2) If Sn Discover at www.deloitte.ca/careers the arithmetic sequence {bn } satisfies , evaluate the nonzero constant c. (3) bn =the truth n+c bn (n ∈ N ∗ ). Calculate the maximum value of f (n) = (n + 36)bn+1 © Deloitte & Touche LLP and affiliated entities.. Discover the truth at www.deloitte.ca/careers. © Deloitte & Touche LLP and affiliated entities.. Discoverfree theeBooks truth atatbookboon.com www.deloitte.ca/careers Download. Click on the ad to read more. 72. Dis.

(73) n. 2. 1 then an bn = + 1. Therefore Sn = a1 b1 + a2 b2 + · · · + an bn = (b1 + b2 + +bn ) + n = 2 2 n (1 − 2 ) 5 1 4 4 4 5 1 3 [(b1 − )+(b ∈ N ∗ ). + n = (2n +5n−1), (n Sequences 2 − )+· · ·+(bn − )]+ n = ELEMENTARY 3 3 3 ALGEBRA3EXERCISE BOOK II3 2 2(1 − 2) 3 1 b 2 n. 5.52 If the increasing sequence {an } satisfies an+2 = an+1 + an when n 1, a7 = 120. Find the value of a8 . Solution: Let a1 = x, a2 = y, x, y ∈ N ∗ . From the given condition, we have x < y, and a3 = x + y, a4 = x + 2y, a5 = 2x + 3y, a6 = 3x + 5y, a7 = 5x + 8y, a8 = 8x + 13y. Since a7 = 120, then 5x + 8y = 120. We have x = 8t y = 15 − 5t 15 where t is a integer number. Since y > x > 0, then 15t−5t > 8t > 0. Thus 0 < t < . 13 Then t = 1. Hence x = 8, y = 10. Therefore a8 = 8 × 8 + 13 × 10 = 194.. 5.53 If the nth partial sum of an arithmetic sequence {an } with positive common difference is Sn , a3 a4 = 117, a2 + a5 = 22. (1) Find the general term an . (2) If Sn the arithmetic sequence {bn } satisfies bn = , evaluate the nonzero constant c. (3) n+c bn (n ∈ N ∗ ). Calculate the maximum value of f (n) = (n + 36)bn+1 Solution: (1) Since {an } is an arithmetic sequence, then a3 + a4 = a2 + a5 = 22. On the other hand, a3 a4 = 117, then a3 , a4 are the two roots of the equation x2 −22x+117 = 0. Since the common difference d > 0, then a3 < a4 . Solving the equation to obtain a3 = 9, a4 = 13. Then a1 + 2d = 9 a1 + 3d = 13 ⇒ a1 = 1 d=4. Thus an = 1 + (n − 1)4 = 4n − 3 (n ∈ N ∗ ). n(n − 1) Sn 4 = 2n2 − n. Then bn = = (2) From (1), we have Sn = n × 1 + 2 n+c 2n2 − n 1 6 15 (n ∈ N ∗ ) ⇒ b1 = , b2 = , b3 = . Since {bn } is an arithmetic n+c 1+c 2+c 3+c 1 15 12 = + ⇒ 2c2 + c = 0. Therefore sequence, then 2b2 = b1 + b3 . Thus 2+c 1+c 3+c 1 1 c = − or c = 0. Since c is nonzero, then c = − . 2 2 n 2n2 − n 2n (3) From (2), we have bn = = 2 = 1 = 2n, then f (n) = 2(n + 1)(n + 36) n + 37n + 36 n− 2 1 1 1 1 . f (n)max = when n = 6. Thus the maximum = 36 49 49 n + n + 37 2 n 36 + 37 n. 1 value of f (n) is . 49 5.54 . 1. Given the function f (x) = √ 1. x2. −4. , (x < −2).. = −f −1 (an ), (n ∈ N ∗ ). Evaluate an . an+1 (2) Let Sn = a21 + a22 + · · · + a2n , bn = Sn+1 − Sn . Determine whether there exists the Download free eBooks at bookboon.com m minimum value of positive integer m such that b < holds for n ∈ N ∗ . If yes, find n 73 25 the value of m. Otherwise, explain the reason. (1) Let a1 = 1,.

(74) 1 1 1 when n = 6. Thus the maximum . f (n)max = = 49 49 n + + 37 + 37 2 n 36 n 1 ELEMENTARY BOOK II Sequences . value of f ALGEBRA (n) is EXERCISE 49 1. 36 n. 5.54 . Given the function f (x) = √ 1. 1 x2 − 4. , (x < −2).. = −f −1 (an ), (n ∈ N ∗ ). Evaluate an . an+1 (2) Let Sn = a21 + a22 + · · · + a2n , bn = Sn+1 − Sn . Determine whether there exists the m minimum value of positive integer m such that bn < holds for n ∈ N ∗ . If yes, find 25 the value of m. Otherwise, explain the reason. 1 1 , (x < −2), then x = − 4 + 2 . Thus Solution: (1) Since y = f (x) = √ 2 y x −4 1 1 1 1 1 f −1 (x) = − 4 + 2 , (x > 0). Since = −f −1 (an ) = 4 + 2 ⇒ 2 − 2 = 4. x an+1 an an+1 an 1 Hence { 2 } is an arithmetic sequence and a1 = 1, the common difference d is 4. Then an (1) Let a1 = 1,. 1 1 1 , (n ∈ N ∗ ). = 2 + (4n − 1) = 4n − 3. Since an > 0, then an = √ 2 an a1 4n − 3. (2) From the given condition and (1), we have bn = Sn+1 −Sn = a2n+1 =. 1 = 4(n + 1) − 3. 1 m 25 m 1 . Since bn < , then < . Thus m > holds for n ∈ N ∗ . 4n + 1 25 4n + 1 25 4n + 1 25 5 ⇒ m > 5. Therefore there exists the minimum value of positive integer 4n + 1 m holds for n ∈ N ∗ . m = 6 such that bn < 25. 5.55 . The nth partial sum of a geometric sequence {an } is S, the product S is p, the sum of the reciprocal of every term is T . Show p2 = ( )n , (n ∈ N ∗ ). T n , p = an1 , p2 = Proof: (1) When the common ratio q = 1, then S = na1 , T = a1 S n S n 2 n 2n 2 a2n 1 ⇒ ( ) = (a1 ) = a1 . That means p = ( ) holds. T T 1 (1 − q1n ) a1 (1 − q n ) qn − 1 a1 (2) when the common ratio q = 1, then S = ,T = ,p = = 1−q a1 q n−1 (q − 1) 1 − 1q n(n−1) S n a1 (1 − q n ) a1 q n−1 (q − 1) n n(n−1) an1 q 1+2+···+(n−1) = an1 q 2 , p2 = a2n ) ] = q , then ( = [ 1 T 1−q qn − 1 S n(n−1) = p2 . As a conclusion, p2 = ( )n , (n ∈ N ∗ ) holds. a2n 1 q T 5.56 There are n numbers which form a sequence. Their numerators form an arithmetic sequence, the first term is a, the common difference is d. The denominators form a geometric sequence, the first term is b, the common ratio is q. Show the (a − aq − d)(1 − q n ) + nd(1 − q) first nth sum Sn satisfies Sn = . bq n−1 (1 − q)2 a a + d a + 2d a + (n − 1)d Proof: From the given condition, we have Sn = + + +· · ·+ = 2 b bq bq bq n−1 1 {(1 + q + q 2 + · · · + q n−1 )a + [q n−2 + 2q n−3 + 3q n−4 + · · · + (n − 3)q 2 + (n − 2)q + (n − bq n−1 1 − qn at bookboon.com 1 1)]d} · · · (∗). For the above Download equation,free weeBooks find that the coefficient for a is . 1−q 74 The coefficient for d can be solved by the following method:.

(75) 5.56 There are n numbers which form a sequence. Their numerators form an arithmetic sequence, the first term is a, the common difference is d. The denominators form a geometric sequence, the first term is b, the common ratio is q. Show the ELEMENTARY ALGEBRA EXERCISE BOOK II Sequences (a − aq − d)(1 − q n ) + nd(1 − q) first nth sum Sn satisfies Sn = . bq n−1 (1 − q)2 a a + d a + 2d a + (n − 1)d + Proof: From the given condition, we have Sn = + +· · ·+ = b bq bq 2 bq n−1 1 {(1 + q + q 2 + · · · + q n−1 )a + [q n−2 + 2q n−3 + 3q n−4 + · · · + (n − 3)q 2 + (n − 2)q + (n − n−1 bq 1 − qn 1 1)]d} · · · (∗). For the above equation, we find that the coefficient for a is . 1−q The coefficient for d can be solved by the following method: Let A = q n−2 + 2q n−3 + 3q n−4 + · · · + (n − 3)q 2 + (n − 2)q + (n − 1). Dividing the both (n − 1) A = q n−3 + 2q n−4 + · · · + (n − 3)q + (n − 2) + . Subtracting sides by q to obtain q q n−1 A . the above two equations, we have A − = q n−2 + q n−3 + q n−4 + · · · + q + 1 − q q 1 − q n−1 n−1 q (1 − q n−1 )q − (n − 1)(1 − q) 1 ) = − ⇒ A = = q 1−q q q−1 (1 − q)q 1 1 − qn n(1 − q) − (1 − q n ) 2 Substituting 1 and 2 into (∗), we have Sn = n−1 [ . a+ (1 − q)2 bq 1−q n(1 − q) − (1 − q n ) (1 − q n )(1 − q)a + [n(1 − q) − (1 − q n )]d d] = (1 − q)2 bq n−1 (1 − q)2 n (a − aq − d)(1 − q ) + nd(1 − q) = . bq n−1 (1 − q)2 Then A(1 −. 5.57 . We put n2 (n 4) positive numbers as n rows and n columns: a11 a21 a31 a41 an1. a12 a22 a32 a42. a13 a23 a33 a43. a14 a24 a34 a44. an2. an3. an4. · · · a1n · · · a2n · · · a3n · · · a4n ··· ··· · · · ann. Wethewill turninyour CVform into where numbers each row an arithmetic sequence and the numbers in each column form a geometric of sequence which common ratios are all equal. Given a24 = 1, an opportunity a lifetime 1 3 a42 = , a43 = . Evaluate a11 + a22 + a33 + a44 + · · · + ann . 8 16 Solution: Let the common difference of the sequence from the first row is d, the common ratio of the sequences from columns is q. then the common difference of the 4th term is dq 3 . Then   a24 = (a11 + 3d)q = 1    1 a42 = (a11 + d)q 3 = 8   1 3   a43 = + dq 3 = 8 16 1 Do you like Would yousystem, like to be we a part of a successful Send us your on Solving thecars? equations have a11 = d brand? = q = ± . Since these n2CVnumberWe will appreciate and reward both your enthusiasm and talent. 2 www.employerforlife.com 1 you. Send us your CV. You will be surprised where it can take s are all positive, then a11 = d = q = . For any 1 k n, akk = a1k q k−1 = 2 1 1 1 1 1 k−1 [a11 + (k − 1)d]q = k k , S = a11 + a22 + a33 + · · · + ann = + 2 2 + 3 3 + · · · + n n 2 2 2 2 2 1 1 1 1 1 Since S = 2 + 2 · 3 + 3 ·Download + · · free · + eBooks n · n+1 . Subtracting these two equations, we 2 2 2 24 2 at bookboon.com Click on the ad to read more 1 (1 − 21n ) 1 1 1 1 1 75 1 1 1 − n n+1 . Therefore have S = + 2 + 3 + 4 + · · · + n − n n+1 = 2 1.

(76) (1 − q) n(1 − q) − (1 − q n ) (1 − q n )(1 − q)a + [n(1 − q) − (1 − q n )]d d] = (1 − q)2 bq n−1 (1 − q)2 n (a − aq − d)(1 − q ) + nd(1 − q) = . ELEMENTARY ALGEBRA EXERCISE BOOK II bq n−1 (1 − q)2 5.57 . bq. 1−q. Sequences. We put n2 (n 4) positive numbers as n rows and n columns: a11 a21 a31 a41 an1. a12 a22 a32 a42. a13 a23 a33 a43. a14 a24 a34 a44. an2. an3. an4. · · · a1n · · · a2n · · · a3n · · · a4n ··· ··· · · · ann. where the numbers in each row form an arithmetic sequence and the numbers in each column form a geometric sequence which common ratios are all equal. Given a24 = 1, 1 3 a42 = , a43 = . Evaluate a11 + a22 + a33 + a44 + · · · + ann . 8 16 Solution: Let the common difference of the sequence from the first row is d, the common ratio of the sequences from columns is q. then the common difference of the 4th term is dq 3 . Then   a24 = (a11 + 3d)q = 1    1 a42 = (a11 + d)q 3 = 8   1 3   a43 = + dq 3 = 8 16 1 Solving the equations system, we have a11 = d = q = ± . Since these n2 number2 1 s are all positive, then a11 = d = q = . For any 1 k n, akk = a1k q k−1 = 2 1 1 1 1 1 k−1 [a11 + (k − 1)d]q = k k , S = a11 + a22 + a33 + · · · + ann = + 2 2 + 3 3 + · · · + n n 2 2 2 2 2 1 1 1 1 1 Since S = 2 + 2 · 3 + 3 · 4 + · · · + n · n+1 . Subtracting these two equations, we 2 2 2 2 2 1 (1 − 21n ) 1 1 1 1 1 1 1 1 − n n+1 . Therefore have S = + 2 + 3 + 4 + · · · + n − n n+1 = 2 1 2 2 2 2 2 2 2 2 1− 2 1 n S = 2 − n−1 − n , (n 4). 2 2. 5.58 Let {an }, {bn }, {cn } satisfy bn = an − an+2 , cn = an + 2an+1 + 3an+2 , (n ∈ N ∗ ). Show that {an } is an arithmetic sequence if and only if {cn } is an arithmetic sequence and bn bn+1 (n ∈ N ∗ ). Proof: “⇒”: Let the common difference of the arithmetic sequence {an } be d1 , then bn+1 − bn = (an+1 − an+3 ) − (an − an+2 ) = (an+1 − an ) − (an+3 − an+2 ) = d1 − d1 = 0. Thus bn bn+1 , (n ∈ N ∗ ). On the other hand, cn+1 − cn = (an+1 − an ) + 2(an+2 − an+1 ) + 3(an+3 − an+2 ) = d1 + 2d1 + 3d1 = 6d1 (constant), then {cn } is an arithmetic sequence. “⇐”: Let the common difference of the arithmetic sequence {cn } is d2 , and bn 1 then cn+2 = an+2 + 2an+3 + bn+1 , (n ∈ N ∗ ). Since cn = an + 2an+1 + 3an+2 , 2 Using 1 − , 2 we have cn − cn+2 = (an − an+2 ) + 2(an+1 − an+3 ) + 3(an+2 − 3an+4 . an+4 ) = bn +2bn+1 +3bn+2 . On the other hand, cn −cn+2 = (cn −cn+1 )+(cn+1 −cn+2 ) = 3 and bn+1 +2bn+2 +3bn+3 = −2d2 . 4 Using −2d2 , then bn +2bn+1 +3bn+2 = −2d2 , 4 3 5 − , we have (bn+1 −bn )+2(bn+2 −bn+1 )+3(bn+3 −bn+2 ) = 0 . Since bn+1 −bn 0, 5 Assume bn = d3 , bn+2 −bn+1 0, bn+3 −bn+2 Download 0, we have = 0, (n ∈ N ∗ ) by . n+1 −bat n bookboon.com free b eBooks then an − an+2 = d3 (constant). Then cn =76an + 2an+1 + 3an+2 = 4an + 2an+1 − 3d3 , cn+1 = 4an+1 + 2an+2 − 3d3 = 4an+1 + 2an − 3d3 = 4an+1 + 2an − 5d3 . Subtract-.

(77) sequence and bn bn+1 (n ∈ N ). Proof: “⇒”: Let the common difference of the arithmetic sequence {an } be d1 , then bn+1 − bn = (an+1 − an+3 ) − (an − an+2 ) = (an+1 − an ) − (an+3 − an+2 ) = d1 − d1 = 0. Thus bn ALGEBRA bn+1 , (n ∈ N ∗BOOK ). On n+2 − ELEMENTARY EXERCISE II the other hand, cn+1 − cn = (an+1 − an ) + 2(a Sequences an+1 ) + 3(an+3 − an+2 ) = d1 + 2d1 + 3d1 = 6d1 (constant), then {cn } is an arithmetic sequence. “⇐”: Let the common difference of the arithmetic sequence {cn } is d2 , and bn 1 then cn+2 = an+2 + 2an+3 + bn+1 , (n ∈ N ∗ ). Since cn = an + 2an+1 + 3an+2 , 2 Using 1 − , 2 we have cn − cn+2 = (an − an+2 ) + 2(an+1 − an+3 ) + 3(an+2 − 3an+4 . an+4 ) = bn +2bn+1 +3bn+2 . On the other hand, cn −cn+2 = (cn −cn+1 )+(cn+1 −cn+2 ) = 3 and bn+1 +2bn+2 +3bn+3 = −2d2 . 4 Using −2d2 , then bn +2bn+1 +3bn+2 = −2d2 , 4 , 3 we have (bn+1 −bn )+2(bn+2 −bn+1 )+3(bn+3 −bn+2 ) = 0 . 5 Since bn+1 −bn 0, − 5 Assume bn = d3 , bn+2 −bn+1 0, bn+3 −bn+2 0, we have bn+1 −bn = 0, (n ∈ N ∗ ) by . then an − an+2 = d3 (constant). Then cn = an + 2an+1 + 3an+2 = 4an + 2an+1 − 3d3 , cn+1 = 4an+1 + 2an+2 − 3d3 = 4an+1 + 2an − 3d3 = 4an+1 + 2an − 5d3 . Subtracting the above two equations, we have cn+1 − cn = 2(an+1 − an ) − 2d3 which means 1 1 an+1 − an = (cn+1 − cn ) + d3 = d2 + d3 (constant) (n ∈ N ∗ ). 2 2 Therefore {an } is an arithmetic sequence.. 5.59 See Figure 1, let the radius of sector AOB be R, ∠AOB = θ(0 < θ < π ). AB1 is perpendicular to OB, B1 A1 is parallel to AB, A1 B2 is perpendicular to OB, 2 B2 A2 is parallel to AB, and keep going, then we obtain two sequences of points {An } and {Bn } on OA an OB. Let the areas of ABB1 , A1 B1 B2 ,· · · ,An Bn Bn+1 · · · be S1 , S2 ,· · · ,Sn+1 ,· · · . Evaluate the sum S of all these areas. π−θ π θ = − . ConsiderSolution: From the given condition, we have ∠ABO = 2 2 2 ing the line perpendicular to AB through the point O, we have AB = 2R cos ∠ABO = θ π θ 2R cos( − ) = 2R sin . 2 2 2 π θ θ θ In right triangle AB1 B, BB1 = AB cos ∠ABO = 2R sin cos( − ) = 2R sin2 = 2 2 2 2 π θ θ (1 − cos θ)R, AB1 = 2R sin sin( − ) = R sin θ, OB1 = R − BB1 = R cos θ, OB2 = 2 2 2 1 1 2 3 R cos θ, OB3 = R cos θ,· · · . Thus SABB1 = BB1 · AB1 = (1 − cos θ)R · R sin θ = 2 2 1 2 (1−cos θ)R sin θ = S1 . Since ABB1 A1 B1 B2 A2 B2 B3 · · · An Bn Bn+1 . 2 A1 B12 OB12 S2 On the other hand, AB||A1 B1 ||A2 B2 , we have q = = ··· = = = S1 AB 2 OB 2 n 1 (1 − cos θ)R2 sin θ 1 2 sin θ R2 cos2 θ 2 2 R = = = cos θ. Hence S = lim S = k 2θ n→∞ 2 R2 1 + cos θ 1 − cos k=1 Download free eBooks at bookboon.com. 1 2 2 sin 2θ cos 2θ 1 2 θ R · = R tan . θ 2 2 2 2 cos2 2. 77.

(78) 2 2 2 2 π θ θ (1 − cos θ)R, AB1 = 2R sin sin( − ) = R sin θ, OB1 = R − BB1 = R cos θ, OB2 = 2 2 2 1 1 2 3 R cos θ, OB3 = REXERCISE cos θ,·BOOK · · . Thus SABB1 = BB1 · AB1 = (1 − cos θ)R · R Sequences sin θ = ELEMENTARY ALGEBRA II 2 2 1 (1−cos θ)R2 sin θ = S1 . Since ABB1 A1 B1 B2 A2 B2 B3 · · · An Bn Bn+1 . 2 A1 B12 OB12 S2 On the other hand, AB||A1 B1 ||A2 B2 , we have q = = ··· = = = S1 AB 2 OB 2 n 1 (1 − cos θ)R2 sin θ 1 sin θ R2 cos2 θ 2 2 = R2 = = cos θ. Hence S = lim Sk = 2 2 n→∞ R 1 − cos θ 2 1 + cos θ k=1 1 2 2 sin 2θ cos 2θ 1 θ R · = R2 tan . θ 2 2 2 2 2 cos 2. 5.60 Given a > 0 and a = 1. The sequence {an } is a geometric sequence. The first term is a, the common ratio is also a. If bn = an lg an (n ∈ N ∗ ). Does there exist a such that every term of {bn } is less than its next term? If yes, find the range of a. If no, please explain the reason. Solution: Assume there exists a real number a such that bn < bn+1 for all n ∈ N ∗ . From the given condition, we have an = a·an−1 = an , then bn = an lg an = an lg an = nan lg a. Thus nan lg a < (n + 1)an+1 lg a for all n ∈ N ∗ . n (1) When a > 1, since lg a > 0, then n < (n+1)a for all n ∈ N ∗ which means a > n+1 n < 1 < a always holds. Therefore bn < bn+1 always holds for all n ∈ N ∗ . Thus n+1 for a > 1 and n ∈ N ∗ . (2) When 0 < a < 1, since lg a < 0, then n > (n + 1)a for all n ∈ N ∗ which means n n 1 for all n ∈ N ∗ . Since a< =1− is increasing when n is increasing. n+1 n+1 n+1 n 1 n 1 Thus the minimum value of is when n = 1. That means that n+1 2 n+1 2 n 1 Programme always holds when 0 < a < . Therefore�e bnGraduate < bn+1 always holds. Thus because a< I joined MITAS for Engineers and Geoscientists n+1 2 1 I wanted real responsibili� Maersk.com/Mitas www.discovermitas.com and nMITAS ∈ N ∗ . because always holds when 0 < a 1. 2. �e G for Engine. Ma. 1 5.61 Given a sequence {an } with a1 = 0, an = (an−1 + 3) (n = 2, 3, · · · ). 4 Find the general term an .. Month 16 I was a construction Mo supervisor ina const I was the North Sea super advising and the No he helping foremen advis ssolve problems he helping fo �ree wo work or placements ssolve pr. 1 1 Solution: From the given condition an = (an−1 + 3), we have an−1 = (an−2 + 3). 4 4 1 1 1 Then an −an−1 = (an−1 −an−2 ), an−1 −an−2 = (an−2 −an−3 ),· · · ,a3 −a2 = (a2 −a1 ). 4 4 4 1 n−2 Multiplying all the above equations together to obtain an − an−1 = ( ) (a2 − a1 ) = 4 3 1 n−2 3 1 n−3 3 3 1 n−2 ( ) (n 2). Then an − an−1 = ( ) , an−1 − an−2 = ( ) ,· · · ,a2 − a1 = . 4 4 4 4 4 4 4 3 1 1 n−3 1 n−2 Adding all the above equations to obtain an − a1 = [1 + + · · · + ( ) + ( ) ]. Real work4 4 4 4 1 n−1 International al opportunities ) 3 1 − (Internationa 1 n−1 ∗ 4 �ree= work wo or Since a1 = 0, then an = · 1 −placements ( ) (n ∈ N ). Real work 4 4 1 − 14 International Internationa al opportunities. 5.62 Consider a sequence {an } with a1 = a2 = 1, a3 = 2, and an an+1 an+2 = 1 an for arbitrary natural number n, and an an+1 an+2 an+3 = an + an+1 + an+2 + an+3 . Download free eBooks at bookboon.com 78. Click on the ad to read more.

(79) the given condition, we have an = a·a = a , then bn = an lg an = a lg a = na lg a. Thus nan lg a < (n + 1)an+1 lg a for all n ∈ N ∗ . n (1) When a > 1, since lg a > 0, then n < (n+1)a for all n ∈ N ∗ which means a > n+1 ELEMENTARY ALGEBRA EXERCISEnBOOK II Sequences ∗ < 1 < a always holds. Therefore bn < bn+1 always holds for all n ∈ N . Thus n + 1 for a > 1 and n ∈ N ∗ . (2) When 0 < a < 1, since lg a < 0, then n > (n + 1)a for all n ∈ N ∗ which means n 1 n a< for all n ∈ N ∗ . Since =1− is increasing when n is increasing. n+1 n+1 n+1 1 n 1 n is when n = 1. That means that Thus the minimum value of n+1 2 n+1 2 n 1 always holds when 0 < a < . Therefore bn < bn+1 always holds. Thus a < n+1 2 1 always holds when 0 < a < and n ∈ N ∗ . 2 1 According to (1) and (2), the range of a is 0 < a < or a > 1. 2 1 5.61 Given a sequence {an } with a1 = 0, an = (an−1 + 3) (n = 2, 3, · · · ). 4 Find the general term an . 1 1 Solution: From the given condition an = (an−1 + 3), we have an−1 = (an−2 + 3). 4 4 1 1 1 Then an −an−1 = (an−1 −an−2 ), an−1 −an−2 = (an−2 −an−3 ),· · · ,a3 −a2 = (a2 −a1 ). 4 4 4 1 n−2 Multiplying all the above equations together to obtain an − an−1 = ( ) (a2 − a1 ) = 4 3 1 n−2 3 1 n−3 3 3 1 n−2 ( ) (n 2). Then an − an−1 = ( ) , an−1 − an−2 = ( ) ,· · · ,a2 − a1 = . 4 4 4 4 4 4 4 1 1 n−3 3 1 n−2 Adding all the above equations to obtain an − a1 = [1 + + · · · + ( ) + ( ) ]. 4 4 4 4 1 n−1 3 1 − (4) 1 n−1 ∗ Since a1 = 0, then an = · =1−( ) (n ∈ N ). 4 4 1 − 14 5.62 Consider a sequence {an } with a1 = a2 = 1, a3 = 2, and an an+1 an+2 = 1 an for arbitrary natural number n, and an an+1 an+2 an+3 = an + an+1 + an+2 + an+3 . Evaluate a1 + a2 + · · · + a100 . Solution: Since an an+1 an+2 an+3 = an + an+1 + an+2 + an+3 , then a1 a2 a3 a4 = a1 + a2 + a3 + a4 , and we have a1 = a2 = 1, a3 = 2, thus a4 = 4. From the given condition, we have an an+1 an+2 an+3 = an + an+1 + an+2 + an+3 , an+1 an+2 an+3 an+4 = an+1 + an+2 + an+3 + an+4 . Subtracting the second equation from the first equation to obtain an+1 an+2 an+3 (an − an+4 ) = an − an+4 which means (an − an+4 )(an+1 an+2 an+3 − 1) = 0. 100 100 (a1 +a2 +a3 +a4 ) = 200. ai = Since an an+1 an+2 = 1, then an+4 = an . Therefore 4 i=1 5.63 If two sequences {an } and {bn } satisfy a1 = 1, a2 = r (r > 0), bn = an an+1 , and {bn } is a geometric sequence with common ratio q (q > 0). Let lg cn+1 cn = a2n−1 + a2n (n ∈ N ∗ ). (1) Find the general term of {cn }. (2) If dn = , lg cn 1 r = 219.2 − 1, q = , find the maximum and minimum terms. 2 bn+1 Solution: (1) Since {bn } is a geometric sequence with common ratio q, then = q. bn an+1 an+2 an+2 Since bn = an an+1 , then = q. It means = q, (n ∈ N ∗ ). Thus the an an+1 an sequence a1 , a3 , a5 ,· · · ,a2n−1Download ,· · · andfree theeBooks sequence a2 , a4 , a6 ,· · · ,a2n ,· · · are both geoat bookboon.com metric sequences with common ratio q. Then a2n−1 = a1 q n−1 = q n−1 , a2n = a2 q n−1 = 79 r · q n−1 , (n ∈ N ∗ ). Therefore cn = a2n−1 + a2n = q n−1 + rq n−1 = (1 + r)q n−1 , (n ∈ N ∗ )..

(80) 5.63 If two sequences {an } and {bn } satisfy a1 = 1, a2 = r (r > 0), bn = an an+1 , and {bn } is a geometric sequence with common ratio q (q > 0). Let lg cn+1 cn = a2n−1 + a2n (n ∈ N ∗ ). (1) Find the general term of {cn }. (2) If dn = , lg cn ELEMENTARY ALGEBRA EXERCISE BOOK II Sequences 1 r = 219.2 − 1, q = , find the maximum and minimum terms. 2 bn+1 Solution: (1) Since {bn } is a geometric sequence with common ratio q, then = q. bn an+1 an+2 an+2 Since bn = an an+1 , then = q. It means = q, (n ∈ N ∗ ). Thus the an an+1 an sequence a1 , a3 , a5 ,· · · ,a2n−1 ,· · · and the sequence a2 , a4 , a6 ,· · · ,a2n ,· · · are both geometric sequences with common ratio q. Then a2n−1 = a1 q n−1 = q n−1 , a2n = a2 q n−1 = r · q n−1 , (n ∈ N ∗ ). Therefore cn = a2n−1 + a2n = q n−1 + rq n−1 = (1 + r)q n−1 , (n ∈ N ∗ ). 1 lg cn+1 = (2) Since cn = (1 + r)q n−1 = (1 + 219.2 − 1)( )n−1 = 220.2−n , then dn = 2 lg cn 1 lg 220.2−(n+1) (n ∈ N ∗ ). =1+ 20.2−n lg 2 n − 20.2 When n − 20.2 > 0 which means n 21 (n ∈ N ∗ ), then dn is decreasing as n is 1 1 increasing. Thus 1 < dn d21 = 1 + = 2.25 . 21 − 20.2 When n − 20.2 < 0 which means n 20 (n ∈ N ∗ ), then dn is decreasing as n is 1 2 increasing. Thus 1 > dn d20 = 1 + = −4 . 20 − 20.2 1 and , 2 we have d20 dn d21 , (n ∈ N ∗ ). Therefore the maximum By applying term of {dn } is d21 = 2.25, and minimum term is d20 = −4. 5.64 . Given c1 = 1, c2 = 1, cn+2 = cn+1 + cn , find the general term cn .. Solution: From the given condition, we have that the roots of the characteristic equa√ √ √ √ 1 − 1 + 1 + 1 − 5 5 5 5 n and . Assume cn = A( )n + B( ) . tion x2 − x − 1 = 0 are 2 2 √2 √2 1− 5 1+ 5 +B = 1. When n = 1, then A 2√ 2 √ 1+ 5 2 1− 5 2 When n = 2, then A( ) + B( ) = 1. 2 2 √ 1 1 1+ 5 n 1 ) − By solving the above equations, we have A = √ , B = − √ . Thus cn = √ [( 2 5 5 5 √ 1− 5 n ( ) ]. 2. Download free eBooks at bookboon.com 80.

(81) ELEMENTARY ALGEBRA EXERCISE BOOK II. Sequences. 5.65 See Figure 2, in RtABC (Rt represents right triangle), there are infinitely many squares S1 , S2 , S3 , S4 ,· · · , and the leg BC = a. The sum of areas of all these squares is half of the area of RtABC. Compute the length of the other leg AC. Solution: Let the side length of the first square is b1 , from left to right, the others are b2 , b3 ,· · · ,bn ,· · · . Let AC = x. b1 a ax a − b1 = ⇒ b1 = = a = By applying the similarity of triangles, we have a x a+x +1 x tan B a tan B tan B 2 = a. Similarly, we have b2 = b1 = ( ) a, b3 = cot B + 1 1 + tan B 1 + tan B 1 + tan B tan B 3 tan B n tan B tan B 2 2 ( ) a,· · · ,bn = ( ) a. Then b21 = ( a)2 , b22 = [( ) a] , 1 + tan B 1 + tan B 1 + tan B 1 + tan B tan B 3 2 tan B n 2 ) a] ,· · · , b2n = [( ) a] . Thus {b2n } is a geometric sequence b23 = [( 1 + tan B 1 + tan B tan B 2 tan B a)2 , the common ratio is ( ) . From the given with the first term ( 1 + tan B 1 + tan B ( tan BB )2 a2 1 1 condition, we have lim {b2n } = ax, which means 1+tantan ax. Since x = = B n→∞ 4 4 1 − ( 1+tan B )2 a2 · tan2 B 1 1 1 = a2 tan B. Then tan B = . Therefore AC = a · tan B = a. a tan B ⇒ 1 + 2 tan B 4 2 2 5.66 If the sequence {an } satisfies a1 = 1, a2 = r (r > 0). {an an+1 } is a geometric sequence with the common ratio q (q > 0). Let bn = a2n−1 + a2n (n ∈ N ∗ ). (1) Find the range of q such that an an+1 + an+1 an+2 > an+2 an+3 . 1 (2) Find bn and lim where Sn = b1 + b2 + · · · + bn . n→∞ Sn Solution: (1) a1 a2 = r when n = 1. Since {an an+1 } is a geometric sequence with the common ratio q. Then an an+1 = r · q n−1 (n ∈ N ∗ ). a2 a3 = rq when n = 2. Thus a3 = q. a3 a4 = rq 2 when n = 3. Thus a4 = rq. a4 a5 = rq 3 when n = 4. Thus a5 = q 2 · · · . Hence the sequence {an } is 1, r, q, rq, q 2 , rq 2 , · · · , q n−1 , rq n−1 , · · · . Since √ 1+ 5 n−1 n n+1 . an an+1 + an+1 an+2 > an+2 an+3 , then rq + rq > rq . Therefore 0 < q < 2 (2) By applying bn = a2n−1 + a2n (n ∈ N ∗ ) and (1), we have b1 = 1 + r, b2 = (1 + r)q, b3 = (1 + r)q 2 ,· · · ,bn = (1 + r)q n−1 . Thus Sn = b1 + b2 + · · · + bn = (1 + r)(1 + q + q 2 + 1 − qn . · · · + q n−1 ) = (1 + r) 1−q  (1 + r)n, q = 1, n ∈ N ∗ ;   1 1−q 1−q lim = = lim , 0 < q < 1; n n→∞ Sn n→∞ (1 + r)(1 − q )   1+r 0, q > 1. a2n + 3 5.67 Given the sequence {an } with a1 = 2, an+1 = . The sequence 2an bn+1 1 {bn } satisfies bn = 3 − a2n , (n ∈ N ). Show (1) bn < 0. (2) | | < . (3) bn 2 1 n−1 |bn | < ( ) , (n 2). 2 Download free eBooks at bookboon.com positive sequence, andmore Proof: (1) From the given condition, we on the ad to read know that {an } is a Click 81 √ a2 + 3 3 3 an−1 an−1 + · = > 2 = 3. Thus bn = 3 − a2n < an = n−1.

(82) ( tan BB )2 a2 1 1 ax, which means 1+tantan = ax. Since x = B n→∞ 4 4 1 − ( 1+tan B )2 1 2 1 1 a2 · tan2 B = BOOK a tan B. Then tan B = . Therefore AC = a · tan BSequences = a. a tan B ⇒ALGEBRA EXERCISE ELEMENTARY II 1 + 2 tan B 4 2 2 condition, we have lim {b2n } =. 5.66 If the sequence {an } satisfies a1 = 1, a2 = r (r > 0). {an an+1 } is a geometric sequence with the common ratio q (q > 0). Let bn = a2n−1 + a2n (n ∈ N ∗ ). (1) Find the range of q such that an an+1 + an+1 an+2 > an+2 an+3 . 1 (2) Find bn and lim where Sn = b1 + b2 + · · · + bn . n→∞ Sn Solution: (1) a1 a2 = r when n = 1. Since {an an+1 } is a geometric sequence with the common ratio q. Then an an+1 = r · q n−1 (n ∈ N ∗ ). a2 a3 = rq when n = 2. Thus a3 = q. a3 a4 = rq 2 when n = 3. Thus a4 = rq. a4 a5 = rq 3 when n = 4. Thus a5 = q 2 · · · . Hence the sequence {an } is 1, r, q, rq, q 2 , rq 2 , · · · , q n−1 , rq n−1 , · · · . Since √ 1+ 5 n−1 n n+1 . an an+1 + an+1 an+2 > an+2 an+3 , then rq + rq > rq . Therefore 0 < q < 2 (2) By applying bn = a2n−1 + a2n (n ∈ N ∗ ) and (1), we have b1 = 1 + r, b2 = (1 + r)q, b3 = (1 + r)q 2 ,· · · ,bn = (1 + r)q n−1 . Thus Sn = b1 + b2 + · · · + bn = (1 + r)(1 + q + q 2 + 1 − qn . · · · + q n−1 ) = (1 + r) 1−q  (1 + r)n, q = 1, n ∈ N ∗ ;   1 1−q 1−q lim = = lim , 0 < q < 1; n n→∞ Sn n→∞ (1 + r)(1 − q )   1+r 0, q > 1. 5.67 . Given the sequence {an } with a1 = 2, an+1 =. {bn } satisfies bn = 3 − a2n , (n ∈ N ).. Show (1) bn < 0.. 1 |bn | < ( )n−1 , (n 2). 2. a2n + 3 . The sequence 2an bn+1 1 (2) | | < . (3) bn 2. Proof: (1) From the given condition, we know that {an } is a positive sequence, and 2 √ a +3 3 3 an−1 an−1 + · = > 2 = 3. Thus bn = 3 − a2n < an = n−1 2an−1 2 2an−1 2 2an−1 √ 2 3 − ( 3) = 0. This means bn < 0. 2. n +3 2 ) 3 − ( a2a 3 − a2n+1 a2n + 3 bn+1 1 1 1 bn+1 1 n = = − (2) | |− =| | − − < 0. Thus | |< . 2 2 2 bn 2 3 − an 2 3 − an 2 4an bn 2 1 1 n−1 b2 b3 bn 1 1 | < | − 1| · · · · · = ( ) , (n 2). (3) |bn | = |b1 | · | | · | | · · · | b1 b2 bn−1 2 2 2 2. 5.68 Let the first nth partial sum of {an } be Sn , and a1 = 1, Sn+1 = 4an + 2, (n ∈ N ∗ ). (1) Let bn = an+1 − 2an , show {bn } is a geometric sequence. an (2) Let cn = n , show {cn } is a geometric sequence. 2 (3) Calculate Sn = a1 + a2 + · · · + an . (1) Proof: Since an+1 = Sn+1 − Sn = 4an − 4an−1 , then an+1 − 2an = 2(an − 2an−1 ). Since bn = an+1 − 2an , then bn = 2bn−1 , (n 2). On the other hand, b1 = a2 − 2a1 = bn S2 − 3a1 = (4a1 + 2) − 3a1 = a1 + 2 = 3, q = = 2. Thus {bn } is a geometric bn−1 freecommon eBooks at bookboon.com sequence with the first termDownload 3 and the ratio 2. n−1 (2) Proof: By Applying (1), we have bn = 82 3 · 2 , then an+1 − 2an = 3 · 2n−1 . Since 1 1 1 3 a1 an n−1.

(83) n. 4an + 2, (n ∈ N ∗ ). (1) Let bn = an+1 − 2an , show {bn } is a geometric sequence. an (2) Let c = n , show {cn } is a geometric sequence. ELEMENTARYn ALGEBRA 2 EXERCISE BOOK II (3) Calculate Sn = a1 + a2 + · · · + an .. n. 1. n+1. Sequences. (1) Proof: Since an+1 = Sn+1 − Sn = 4an − 4an−1 , then an+1 − 2an = 2(an − 2an−1 ). Since bn = an+1 − 2an , then bn = 2bn−1 , (n 2). On the other hand, b1 = a2 − 2a1 = bn S2 − 3a1 = (4a1 + 2) − 3a1 = a1 + 2 = 3, q = = 2. Thus {bn } is a geometric bn−1 sequence with the first term 3 and the common ratio 2. (2) Proof: By Applying (1), we have bn = 3 · 2n−1 , then an+1 − 2an = 3 · 2n−1 . Since 1 1 1 3 an a1 cn = n , then cn+1 − cn = n+1 (an+1 − 2an ) = n+1 · 3 · 2n−1 = , and c1 = = . 2 2 2 4 2 2 3 1 Thus {cn } is a geometric sequence with the first term , the common ratio is . 2 4 3 3 1 1 (3) Solution: By Applying (2), we have cn = + (n − 1) · = n − . Then 2 4 4 4 n n−2 an = 2 · cn = 2 (3n − 1). When n 2, then Sn = 4an−1 + 2 = 4 × 2n−3 (3n − 4) + 2 = 2n−1 (3n − 4) + 2. When n = 1, then S1 = a1 = 1 which also satisfies the above equation. Therefore Sn = 2n−1 (3n − 4) + 2 always holds for all n ∈ N ∗ . 5.69 The adjacent terms of {an }, an and an+1 , are the roots of the equa1 tion x2 −cn x+( )n = 0, and a1 = 2. Find the sum of infinite sequence c1 , c2 , · · · , cn , · · · . 3 Solution: By applying the Vieta’s theorem, we have cn = an + an+1 , an · an+1 = 1 1 1 Then an+1 · an+2 = ( )n+1 . 2 Dividing the equation 2 by the equation ( )n . 3 3 ∞ ∞ ∞ ∞ ∞ ∞ an+2 1 1 we have , = ⇒ cn = an + an+1 = a2k+1 +2 a2k + a2k+1 = an 3 n=1 n=1 n=1 k=0 k=1 k=1 ∞ ∞ 2 a2k+1 + 2 a2k − a1 . k=0. k=1. an+2 1 1 1 , then a2 = . By applying = and the formula for the 3 6 an 3 ∞ 1 2 6 cn = 2· sum of infinite decreasing geometric sequence, we have +2· −2 = 1 − 13 1 − 31 n=1 9 . 2 Since a1 = 2, a1 a2 =. 5.70 Given {an } is an arithmetic sequence, {bn } is a geometric sequence, and a1 = b2 , a2 = b2 , a1 = a2 , an > 0, n = 1, 2, · · · . Show an < bn when n > 2. Proof: Let the common difference be d and the common ratio be q. Since a2 = a1 + d, d b2 = b1 q = a1 q, a2 = b2 , we have a1 + d = a1 q. Then q = 1 + . a1 On the other hand, an > 0, n = 1, 2, · · · , then d > 0. Thus q > 1. Hence an − bn = a1 +(n−1)d−a1 q n−1 = a1 (1−q n−1 )+(n−1)d = a1 (1−q)(1+q +· · ·+q n−2 )+(n−1)d < a1 (1 − q)(n − 1) + (n − 1)d = (n − 1)[a1 (1 − q) + d] = (n − 1)(a2 − b2 ) = 0. Therefore an < bn when n > 2. 5.71 For an arbitrary real number sequence A = (a1 , a2 , · · · ), we define ∆A as the sequence A = (a2 − a1 , a3 − a2 , · · · ). Its nth term is an+1 − an . Assume the Download eBooks bookboon.com differences between the adjacent twofree terms areatall 1, and a19 = a92 = 0. Evaluate a1 . 83. Solution 1: Let an+1 −an = bn , then bn −bn−1 = 1. Thus {bn } is an arithmetic sequence.

(84) On the other hand, an > 0, n = 1, 2, · · · , then d > 0. Thus q > 1. Hence an − bn = a1 +(n−1)d−a1 q n−1 = a1 (1−q n−1 )+(n−1)d = a1 (1−q)(1+q +· · ·+q n−2 )+(n−1)d < a1 (1 − q)(n − 1) + (n − 1)d = (n − 1)[a1 (1 − q) + d] = (n − 1)(a2 − b2 ) = 0. Therefore ALGEBRA an < bnEXERCISE when nBOOK > 2.II ELEMENTARY Sequences 5.71 For an arbitrary real number sequence A = (a1 , a2 , · · · ), we define ∆A as the sequence A = (a2 − a1 , a3 − a2 , · · · ). Its nth term is an+1 − an . Assume the differences between the adjacent two terms are all 1, and a19 = a92 = 0. Evaluate a1 . Solution 1: Let an+1 −an = bn , then bn −bn−1 = 1. Thus {bn } is an arithmetic sequence and its common difference is 1. Hence bn = b1 +n−1. This means an+1 −an = b1 +n−1. n−1 (n − 1)(n − 2) We obtain an = a1 + . Since b1 = a2 −a1 , (b1 +k−1) = a1 +(n−1)b1 + 2 k=1 (n − 1)(n − 2) then an = (n − 1)a2 − (n − 2)a1 + . 2 18 × 17 1 a19 = 0 when n = 19. Then 18a2 − 17a1 = − . 2 91 × 90 2 a92 = 0 when n = 92. Then 91a2 − 90a1 = − . 2 1 and the equation 2 to generate a1 = 819. Solving the equation Solution 2: Let the first term of the sequence ∆A be d. Then the sequence ∆A is (d, d + 1, d + 2, · · · ). Its nth term is d + (n − 1). Thus the nth term of sequence A is n−1 1 an = a1 + (ak+1 −ak ) = a1 +d+(d+1)+· · ·+(d+n−2) = a1 +(n−1)d+ (n−1)(n−2). 2 k=2 This shows that an is a quadratic polynomial with respect to n, and the coefficient of 1 1 its first term is . Since a19 = a92 = 0, then an = (n − 19)(n − 92). Therefore 2 2 1 a1 = (1 − 19)(1 − 92) = 819. 2 Solution 3: From the given condition, we obtain that {an } is a second order arithmetic sequence. Thus its general term is a quadratic polynomial whit respect to n. Since a19 = a92 = 0, we let an = A(n − 19)(n − 92) where A is an undetermined coefficient. Since a3 −2a2 +a1 = 1, then A[(3−19)(3−92)−2(2−19)(2−92)+(1−19)(1−92)] = 1. 1 1 By solving the equation, we have A = . Therefore a1 = (1 − 19)(1 − 92) = 819. 2 2 5.72 Given sequence {an } with a1 = 1, a1 + a2 + · · · + an = n2 an , (n ∈ N ∗ ). Find the general term of the sequence {an } and S100 . Solution: From the given condition a1 +a2 +· · ·+an = n2 an , we have a1 +a2 +· · ·+an−1 = (n − 1)2 an−1 by substituting n for n − 1. Subtracting the second equation from n−1 the first one, we have an + (n − 1)2 an−1 = n2 an . This means an = an−1 = n+1 1 n−1n−2n−3 n−1n−2n−3 2 n−1n−2 an−2 = an−3 = · · · = · · · a1 = , n+1 n n+1 n n−1 n+1 n n−1 3 n(n + 1) N ∗ ). 200 2 S100 = a1 + a2 + · · · + a100 = 1002 = . 100(100 + 1) 101 5.73 Given the nth partial sum of sequence {an } as Sn = 1 + kan , k is a constant and k = 1. Find an and Sn . 1 , 1−k k 1 . = − a2 = S2 − S1 = 1 + ka2 − a1 . This means (1 − k)a2 = 1 − 1−k 1−k Download free eBooks at2bookboon.com k n−2 k k n−2 84 Thus a2 = − . Similarly, a = , · · · , a = (−1) , 3 n−1 (1 − k)3 (1 − k)n−1 (1 − k)2. Solution: Since Sn = 1 + kan , we have S1 = a1 = 1 + ka1 , then a1 =. (n ∈.

(85) n+1 N ∗ ).. n. an−2 =. n+1. n. n−1. an−3 = · · · =. 2 S100 = a1 + a2 + ·EXERCISE · · + a100 = 1002 ELEMENTARY ALGEBRA BOOK II 100(100. + 1). =. n+1. n. n−1. · · · a1 = , 3 n(n + 1). 200 . 101. (n ∈. Sequences. 5.73 Given the nth partial sum of sequence {an } as Sn = 1 + kan , k is a constant and k = 1. Find an and Sn . 1 , 1−k k 1 = − . a2 = S2 − S1 = 1 + ka2 − a1 . This means (1 − k)a2 = 1 − 1−k 1−k k n−2 k k2 n−2 . Similarly, a = , · · · , a = (−1) , Thus a2 = − 3 n−1 (1 − k)2 (1 − k)3 (1 − k)n−1 k a3 a2 k 1 k2 k n−1 an = (−1)n−1 = − , . Thus = − / = /− (1 − k)n a1 (1 − k)2 1 − k 1 − k a2 (1 − k)3 k an k k ,· · · , . Hence {an } is a geometric sequence with = − = − 2 (1 − k) 1−k an−1 1−k k 1 k n−1 a1 = and q = − . Therefore an = (−1)n−1 , (n ∈ N ∗ ), Sn = 1−k 1−k (1 − k)n k n k n−1 n−1 1 + kan = 1 + k(−1)n−1 ) , (k = 1, n ∈ N ∗ ). = 1 + (−1) ( (1 − k)n 1−k Solution: Since Sn = 1 + kan , we have S1 = a1 = 1 + ka1 , then a1 =. 5.74 The three sides of ABC form an arithmetic sequence, and the dif0 ference between √ the largest √ angle √ and the smallest angle is 90 . Show the ratio of the three sides is ( 7 + 1) : 7 : ( 7 − 1).. no.1. Sw. ed. en. nine years in a row. STUDY AT A TOP RANKED INTERNATIONAL BUSINESS SCHOOL Reach your full potential at the Stockholm School of Economics, in one of the most innovative cities in the world. The School is ranked by the Financial Times as the number one business school in the Nordic and Baltic countries.. Stockholm. Visit us at www.hhs.se. Download free eBooks at bookboon.com 85. Click on the ad to read more.

(86) ELEMENTARY ALGEBRA EXERCISE BOOK II. Sequences. Proof: Let the side lengths of ABC be a − d, a, a + d, the smallest angle is α, the a+d a−d = = largest angle is 900 + α. From the given condition, we have sin α sin(900 + α) a−d+a+d a 2a a a ⇒ = ⇒ = ⇒ 0 0 0 sin(90 − 2α) sin α + sin(90 + α) sin(90 − 2α) sin α + cos α cos 2α 1 2 cos 2α = sin α + cos α ⇒ 2(cos2 α − sin2 α) = sin α + cos α ⇒ cos α − sin α = ⇒ 2 √ √ √ √ 1 2 2 2 14 . Then cos(450 − α) = 1 − ( ) = . 2 sin(450 − α) = ⇒ sin(450 − α) = 2 4 4 √ √4 2 14 Thus sin α = sin[450 −(450 −α)] = sin 450 cos(450 −α)−cos 450 sin(450 −α) = − 2 4 √ √ √ 2 2 7−1 = , sin(900 + α) = cos α = cos[450 − (450 − α)] = cos 450 cos(450 − 2 4 4 √ √ √ √ √ 2 14 2 2 7+1 0 0 α) + sin 45 sin(45 − α) = + = . sin(900 − 2α) = cos 2α = 2 4 2 4 4 √ √ √ 7+1 2 7−1 2 7 2 2 ) −( ) = . cos α − sin α = ( 4 4 4 √ √ √ As a conclusion, the ratio of the three sides is ( 7 + 1) : 7 : ( 7 − 1). 5.75 term an .. Given the sequence {an }, a1 = 5, an+1 =. 5an + 6 . Find the general an + 4. 5x + 6 which means x2 − x − 6 = 0, then the two fixed points x+4 5x + 6 2(an − 3) 5an + 6 1 of f (x) = are x = 3, x = −2. Hence an+1 − 3 = −3= , x+4 an + 4 an + 4 7(an + 2) a 2 an − 3 5an + 6 −3 2 By 1 ÷ , 2 then n+1 +2 = . = ( ). an+1 + 2 = an + 4 an + 4 an+1 + 2 7 an + 2 a1 − 3 2 an − 3 } is a geometric sequence with the first term = and the comHence { an + 2 a1 + 2 7 2 an − 3 2 2 n−1 2 n 3 · 7n + 2n+1 mon ratio ⇒ = ( ) = ( ) . Therefore an = . 7 an + 2 7 7 7 7n − 2n Solution: Let x =. 1 an + (n ∈ N ∗ ). Show 5.76 Given sequence {an }, a1 = 2, an+1 = 2 an √ √ 1 2 < an < 2 + . n √ √ 1 √ Proof:(1) 2 < a1 = 2 < 2 + = 2 + 1 when n = 1. Thus p(1) holds. 1 √ √ 1 (2) Assume p(k) holds when n = k. This means 2 < ak < 2 + . k. Download free eBooks at bookboon.com 86.

(87) ELEMENTARY ALGEBRA EXERCISE BOOK II. Sequences. √ 1 ak ak 1 + When n = k + 1, then ak+1 = 2 = 2, and the equation holds if and 2 ak 2 ak √ √ 1 ak = which means ak = 2. Since ak > 2, then equal sign cannot hold only if 2 ak √ √ 2 + k1 1 1 ak for the above formula. Then ak+1 > 2. Since ak+1 = + +√ = < 2 ak 2 2 √ √ 1 1 2+ . Hence p(k + 1) holds when n = k + 1. 2+ 2k k+1 √ √ 1 2 < an < 2 + always holds for all n ∈ N ∗ . n Given sequence {an }, 3an+1 + an = 4 (n 1), a1 = 9, the nth 1 partial sum is Sn and |Sn − n − 6| < . Find the smallest positive integer number n. 125. 5.77 . Solution: By applying the recurrence relation, we have 3(an+1 − 1) = −(an − 1) where n = 1 is a root of equation 3n2 + n = 4. Let bn = an − 1, bn+1 = an+1 − 1, 1 then bn+1 = − bn , b1 = a1 − 1 = 8. Hence {bn } is a geometric sequence with 3 1 1 the first term 8 and the common ratio − . Therefore bn = 8(− )n−1 . Sn − n = 3 3 8 · [1 − (− 13 )n ] 1 (a1 − 1) + (a2 − 1) + · · · + (an − 1) = b1 + b2 + · · · + bn = = 6 − 6 · (− )n . 1 3 1 − (− 3 ) 1 1 ⇒ 2 · 31−n < ⇒ 3n−1 > 250. Therefore the smallest Since |Sn − n − 6| < 125 125 positive integer number such that the inequality holds is n = 7. 5.78 Given sequence {an }, a1 = 1, a2 = 2, an+2 = 4an+1 − 3an + 2, find the general term of {an }. Solution: By applying the recurrence relation, we have an+2 − an+1 = 3(an+1 − an ) + 2. cn+1 cn 1 cn 1 Let an+1 − an = cn , then c1 = 1, cn+1 = 3cn + 2, n = n−1 + 2( )n = n−1 + 2 × × 3 3 3 3 3 n−1 cn 1 k 1 1 − ( 13 )n−1 1 n−1 1 n−1 2×( ) = 1+2×( ) = 2 − ( ) . Then ( ) . Thus n−1 = c1 + 1 3 3 3 3 3 1 − 3 k=1 cn = 2 × 3. n−1. n−1. − 1, an+1 = an + 2 · 3. − 1. Therefore an = a1 +. 3n−1 − 1 − (n − 1) = 3n−1 − n + 1 (n ∈ N ∗ ). 1+2× 2. Download free eBooks at bookboon.com 87. n−1 k=1. (2 · 3k−1 − 1) =.

(88) ELEMENTARY ALGEBRA EXERCISE BOOK II. 5.79 lim an = 1.. Sequences. Let the sequence {an } satisfy (2 − an ) · an+1 = 1, n 1. Show. n→∞. 1 1 , then an = . 2 − an 2 − an−1 an−1 − 1 1 = −1 + . Thus Subtracting both sides by 1 to obtain an − 1 = 2 − an−1 an − 1 1 1 1 − (n − 1)(a1 − 1) 1 = −2 + = · · · = −(n − 1) + = . Hence an−1 − 1 an−2 − 1 a1 − 1 a1 − 1 a1 − 1 a1 − 1 + 1. Therefore lim an = lim [ + 1] = 1. an = n→∞ n→∞ 1 − (n − 1)(a1 − 1) 1 − (n − 1)(a1 − 1) Proof: From the given condition, we have an+1 =. 5.80 If {an } and {bn } are both positive infinite sequences, a1 = √ a, b1 = b, and an , bn , an+1 are arithmetic, bn , an+1 , bn+1 are geometric. (1) Show { bn } is an arithmetic sequence. (2) Find the general term of {bn }. (3) Compare an and bn . (1) Proof: From condition, we have a2n+1 = bn bn+1 , a2n = bn−1 bn . Thus the given √ bn · bn−1 + bn · bn+1 = 2bn . Divide the both sides by bn , then an + an+1= √ √ bn−1 + bn+1 = 2 bn . Hence { bn } is an arithmetic sequence. √ √ (2b − a)2 . On (2) Solution: Since b1 = b, a22 = b1 b2 and a1 + a2 = 2b1 , we have b2 = b 2b − a the other hand, since {bn } is a positive sequence and an < bn , then b2 = √ . Thus b b−a 3b − 2a 2b − a √ b−a d = b2 − b1 = √ − b = √ > 0, b3 = b1 + (3 − 1) √ = √ (all b b b b [nb − (n − 1)a]2 nb − (n − 1)a √ terms are positive)· · · , bn = . Thus bn = . b b √ (3) Solution: Since { bn } is an increasing sequence and a2n = bn−1 bn , then bn−1 < an < bn which means an < bn . . 3 + an . (1)Find the range of 2 a1 such that an+1 > an for any positive number n. (2) If a1 = 4, bn = |an+1 − an | (n ∈ 5 N ∗ ), and let the nth partial sum of {bn } be Sn , show Sn < . 2 an − an−1 3 + an−1 3 + an (1) Solution: an+1 − an = − = when n 2. 2 2 3+an 2( + 3+an−1 ) 5.81 . Given sequence {an }, a1 > 0, an+1 =. 2. 2. Since the denominator is positive, then an+1 − an > 0 ⇔ an − an−1 > 0 ⇔ · · · ⇔ 3 + a1 3 a2 − a1 = − a1 > 0, and a1 > 0. Thus 0 < a1 < . Hence the range of a1 is 2 2. Download free eBooks at bookboon.com 88. Click on the ad to read more.

(89) and an , bn , an+1 are arithmetic, bn , an+1 , bn+1 are geometric. (1) Show { bn } is an arithmetic sequence. (2) Find the general term of {bn }. (3) Compare an and bn . (1) Proof: From the given condition, we have a2n+1 = bn bn+1 , a2n = bn−1 bn . Thus √ Sequences · bn+1 = 2bn . Divide the both sides by bn , then √ √ bn−1 + bn+1 = 2 bn . Hence { bn } is an arithmetic sequence. √ √ (2b − a)2 . On (2) Solution: Since b1 = b, a22 = b1 b2 and a1 + a2 = 2b1 , we have b2 = b 2b − a the other hand, since {bn } is a positive sequence and an < bn , then b2 = √ . Thus b 2b − a √ b−a b−a 3b − 2a d = b2 − b1 = √ − b = √ > 0, b3 = b1 + (3 − 1) √ = √ (all b b b b nb − (n − 1)a [nb − (n − 1)a]2 √ terms are positive)· · · , bn = . . Thus bn = b b √ (3) Solution: Since { bn } is an increasing sequence and a2n = bn−1 bn , then bn−1 < an < bn which means an < bn .. EXERCISE BOOK II ELEMENTARY ALGEBRA + a = b · b + bn a n n+1 n n−1 . . 3 + an . (1)Find the range of 2 a1 such that an+1 > an for any positive number n. (2) If a1 = 4, bn = |an+1 − an | (n ∈ 5 N ∗ ), and let the nth partial sum of {bn } be Sn , show Sn < . 2 an − an−1 3 + an 3 + an−1 − = (1) Solution: an+1 − an = when n 2. 3+an−1 2 2 3+an 2( + ) 5.81 . Given sequence {an }, a1 > 0, an+1 =. 2. 2. Since the denominator is positive, then an+1 − an > 0 ⇔ an − an−1 > 0 ⇔ · · · ⇔ 3 + a1 3 a 2 − a1 = − a1 > 0, and a1 > 0. Thus 0 < a1 < . Hence the range of a1 is 2 2 3 a1 ∈ (0, ). 2 (2) Proof: By applying the method of (1), we obtain that an+1 − an < 0 holds for any 3 positive n when a1 > . Since a1 = 4, an+1 − an < 0, then bn = |an+1 − an | = an − an+1 . 2 Hence Sn = b1 + b2 + · · · + bn = (a1 − a2 ) + (a2 − a3 ) + · · · + (an − an+1 ) = 4 − an+1 . Addi3 3 + an+1 < an+1 , then an+1 > . Therefore tionally, since an+2 < an+1 which means 2 2 5 3 Sn < 4 − = . 2 2 √ 5.82 Let the positive sequence a0 , a1 , a2 , · · · , an , · · · satisfy an an−2 − √ an−1 an−2 = 2an−1 (n 2), and a0 = a1 = 1. Find the general term of {an }. √ √ Solution: The given equation leads to an an−2 − 2an−1 = an−1 an−2 . Dividing both an an−1 an−1 an−2 √ 1 Thus sides by an−1 an−2 to obtain −2 = 1 . −2 = a a a a n−1 n−2 n−2 n−3 a2 a1 2 2 ···, 1 2 3 1 , −2 = 1 n-1 . By applying ×1+ ×2+ ×2 +· · ·+n-1 ×2n−2 , a a 1 0 an a1 an n−1 2 n−2 we have −2 = 1 + 2 + 2 + · · · + 2 . This means =1+2+ an−1 a0 an−1 22 + · · · + 2n−2 + 2n−1 = 2n − 1. Hence an = (2n − 1)2 an−1 = (2n − 1)2 (2n−1 − 1)2 an−2 = n ··· = (2k − 1)2 . k=1. As a conclusion,.  . 1, n = 0; n 2 (2k − 1) , n ∈ N ∗. Download at bookboon.com  free eBooks. an =. k=1. 89.

(90) an an a1 n−1 2 n−2 we have −2 = 1 + 2 + 2 + · · · + 2 . This means =1+2+ an−1 a0 an−1 22 + · · · + 2n−2 + 2n−1 = 2n − 1. Hence an = (2n − 1)2 an−1 = (2n − 1)2 (2n−1 − 1)2 an−2 = n ALGEBRA EXERCISE BOOK II ELEMENTARY Sequences k 2 ··· = (2 − 1) . . k=1. As a conclusion,. an =. 5.83 .   . 1, n = 0; n (2k − 1)2 , n ∈ N ∗ . k=1. Given sequence {an }, a1 = 2, the nth partial sum is Sn . an is the 5 arithmetic mean of 3Sn −4 and 2− Sn−1 for any n ∈ N ∗ . (1) Show {an } is a geometric 2 1 sequence, and find the general term an . (2) Show (log2 Sn + log2 Sn+2 ) < log2 Sn+1 . 2 4 2 4 (3) If bn = − 1, cn = log2 ( ) . Let Tn is the nth partial sum of {bn }, and Rn is an an the nth partial sum of {cn }. Does there exist a positive integer n such that Tn > Rn . If yes, find its range. If no, please explain the reason. 5 (1) Proof: 2an = 3Sn − 4 + 2 − Sn−1 when n 2 which leads to 2(Sn − Sn−1 ) = 3Sn − 2 5 1 1 1 2− Sn−1 . Then Sn = Sn−1 +2, Sn+1 = Sn +2. Since a1 = 2, then 2+a2 = ×2+2. 2 2 2 2 ( 12 Sn + 2) − ( 12 Sn−1 + 2) an+1 Sn+1 − Sn 1 Thus a2 = 1. Hence = = = . Therefore an Sn − Sn−1 Sn − Sn−1 2 a2 1 1 = . After all, {an } is a geometric sequence with the common ratio . a1 2 2 2(1 − 21n ) 1 n−2 1 = 4 − ( ) . To show (log2 Sn + (2) Proof: From (1), we have Sn = 1 2 2 1− 2 2 log2 Sn+2 ) < log2 Sn+1 , we only need to show Sn Sn+2 < Sn+1 . Since Sn Sn+2 = [4 − 1 1 1 1 1 1 1 2 = [4−( )n−1 ]2 = 16−4( )n−2 +( )2n−2 , ( )n−2 ][4−( )n ] = 16−5( )n−2 +( )2n−2 , Sn+1 2 2 2 2 2 2 2 1 2 then Sn Sn+2 < Sn+1 . Therefore (log2 Sn + log2 Sn+2 ) < log2 Sn+1 . 2 4 4 (3) Proof: From the given condition and (1), we have bn = − 1 = 1 − 1 = 2n − 1, an 2n−2 4 2 cn = log2 ( ) = log2 (2n )2 = 2n. Tn = 2(1 + 2 + 22 + · · · + 2n−1 ) − n = 2(2n − 1) − n = an n(n + 1) = n2 + n. Tn < Rn when 2n+1 − n − 2. Rn = 2(1 + 2 + 3 + · · · + n) = 2 × 2 n = 1, 2, 3. Tn > Rn when n = 4, 5 which means 2n+1 > n2 + 2n + 2. When n 6, 0 1 2 then 2n+1 = (1 + 1)n+1 = c0n+1 + c1n+1 + c2n+1 + · · · + cn+1 n+1 > 2(cn+1 + cn+1 + cn+1 ) = n2 + 3n + 4 > n2 + 2n + 2. Thus Tn > Rn when n 4. 5.84 Find an and Sn .. 1 1 Given sequence {an }, ak > 0 (k = 1, 2, · · · , n), and Sn = (an + ). an 2. 1 1 1 1 Solution: Since a1 = S1 = (a1 + ), then a1 = 1. Since a2 = S2 −S1 = (a2 + )−1, 2 a1 2 a2 √ √ √ 1 1 then a2 = −1 + 2, S2 = a2 + S1 = 2. Since a3 = S3 − S2 = (a3 + ) − 2, then 2 a3√ √ √ √ √ a3 = − 2+ 3, S 3, · · · . We have the conjecture: a 3 = a3 +S2 = n = − n − 1+ n, √ Sn = n, (n ∈ N ∗ ). Now we show the conjecture using √ mathematical induction. Proof: (1) When n = 1, thenDownload a1 = free 1 =eBooks S1 . p(1) is correct.√ Click on at bookboon.com √ the ad to read √ more (2) Suppose p(k) is correct when n = k. This means ak = − k − 1 + k and Sk = k 90 √ 1 1.

(91) ( 1 Sn + 2) − ( 12 Sn−1 + 2) an+1 Sn+1 − Sn 1 = = 2 = . Therefore 2 anBOOK IISn − Sn−1 Sn − Sn−1 ELEMENTARY ALGEBRA EXERCISE Sequences a2 1 1 = . After all, {an } is a geometric sequence with the common ratio . a1 2 2 2(1 − 21n ) 1 n−2 1 = 4 − ( ) . To show (log2 Sn + (2) Proof: From (1), we have Sn = 1 2 2 1− 2 2 log2 Sn+2 ) < log2 Sn+1 , we only need to show Sn Sn+2 < Sn+1 . Since Sn Sn+2 = [4 − 1 1 1 1 1 1 1 2 = [4−( )n−1 ]2 = 16−4( )n−2 +( )2n−2 , ( )n−2 ][4−( )n ] = 16−5( )n−2 +( )2n−2 , Sn+1 2 2 2 2 2 2 2 1 2 then Sn Sn+2 < Sn+1 . Therefore (log2 Sn + log2 Sn+2 ) < log2 Sn+1 . 2 4 4 (3) Proof: From the given condition and (1), we have bn = − 1 = 1 − 1 = 2n − 1, an 2n−2 4 2 cn = log2 ( ) = log2 (2n )2 = 2n. Tn = 2(1 + 2 + 22 + · · · + 2n−1 ) − n = 2(2n − 1) − n = an n(n + 1) = n2 + n. Tn < Rn when 2n+1 − n − 2. Rn = 2(1 + 2 + 3 + · · · + n) = 2 × 2 n = 1, 2, 3. Tn > Rn when n = 4, 5 which means 2n+1 > n2 + 2n + 2. When n 6, 0 1 2 then 2n+1 = (1 + 1)n+1 = c0n+1 + c1n+1 + c2n+1 + · · · + cn+1 n+1 > 2(cn+1 + cn+1 + cn+1 ) = n2 + 3n + 4 > n2 + 2n + 2. Thus Tn > Rn when n 4. Thus a2 = 1. Hence. 5.84 Find an and Sn .. 1 1 Given sequence {an }, ak > 0 (k = 1, 2, · · · , n), and Sn = (an + ). 2 an. 1 1 1 1 Solution: Since a1 = S1 = (a1 + ), then a1 = 1. Since a2 = S2 −S1 = (a2 + )−1, 2 a1 2 a2 √ √ √ 1 1 then a2 = −1 + 2, S2 = a2 + S1 = 2. Since a3 = S3 − S2 = (a3 + ) − 2, then 2 a3√ √ √ √ √ a3 = − 2+ 3, S = a +S = 3, · · · . We have the conjecture: a = − n − 1+ n, 3 3 2 n √ ∗ Sn = n, (n ∈ N ). Now we show the conjecture using √ mathematical induction. Proof: (1) When n = 1, then a1 = 1 = S1 . p(1) is correct.√ √ √ (2) Suppose p(k) is correct when n = k. This means ak = − k − 1 + k and Sk = k √ 1 1 both hold. Then when n = k + 1, we have ak+1 = Sk+1 − Sk = (ak+1 + )− k ⇒ 2 ak+1 √ √ √ 2 1 2 − 2 k ⇒ ak+1 + 2 kak+1 − 1 = 0 ⇒ (ak+1 + k) = k + 1. Since 2ak+1 = ak+1 + ak+1 √ √ √ √ √ a √k+1 > 0, then ak+1 = − k+ k + 1. Thus Sk+1 = ak+1 +Sk = (− k+ k + 1)+ k = k + 1. Hence p(k when n√ = k + 1. √+ 1) is correct √ Therefore an = − n − 1 + n and Sn = n both hold for any n ∈ N ∗ . 5.85 Given sequence {an }, a1 = 0, an+1 = 2an + n2 nth partial sum Sn of {an }.. (n ∈ N ∗ ). Find the. Solution: an+1 = 2an + n2 implies an+1 − 2an = n2 . Since Sn = a1 + a2 + · · · + an , then 2Sn = 2a1 + 2a2 + · · · + 2an . Subtracting the second equation from the first equation, we have −Sn = a1 + (a2 − 2a1 ) + (a3 − 2a2 ) + · · · + (an − 2an−1 ) − 2an . This means −Sn = 0 + 12 + 22 + · · · + (n − 1)2 − 2an . n(n − 1)(2n − 1) (n ∈ N ∗ )(∗). Let an = an2 + bn + c, then Thus Sn = 2an − 6 a(n + 1)2 + b(n + 1) + c = 2(an2 + bn + c) + n2 . Simplifying the equation to generate (a + 1)n2 + (b − 2a)n − [(a + b) − c] = 0. Thus a = −1, b = −2, c = −3. Hence an+1 + [(n + 1)2 + 2(n + 1) + 3] 2 an+1 +[(n+1)2 +2(n+1)+3] Download = = 2(an +n +2n+3)⇒ free eBooks at bookboon.com an + n2 + 2n + 3 91 {an + n2 + 2n + 3} is a geometric 2. a1 + 12 + 2 × 1 + 3 = 6 when n = 1. Therefore sequence with the first term 6 and the common ratio 2. Then a + n2 + 2n + 3 =.

(92) nth partial sum Sn of {an }. Solution: an+1 = 2an + n2 implies an+1 − 2an = n2 . Since Sn = a1 + EXERCISE a2 + · · BOOK · + aIIn , then 2Sn = 2a1 + 2a2 + · · · + 2an . Subtracting the ELEMENTARY ALGEBRA Sequences second equation from the first equation, we have −Sn = a1 + (a2 − 2a1 ) + (a3 − 2a2 ) + · · · + (an − 2an−1 ) − 2an . This means −Sn = 0 + 12 + 22 + · · · + (n − 1)2 − 2an . n(n − 1)(2n − 1) (n ∈ N ∗ )(∗). Let an = an2 + bn + c, then Thus Sn = 2an − 6 a(n + 1)2 + b(n + 1) + c = 2(an2 + bn + c) + n2 . Simplifying the equation to generate (a + 1)n2 + (b − 2a)n − [(a + b) − c] = 0. Thus a = −1, b = −2, c = −3. Hence an+1 + [(n + 1)2 + 2(n + 1) + 3] 2 2 an+1 +[(n+1) +2(n+1)+3] = 2(an +n +2n+3)⇒ = an + n2 + 2n + 3 2. a1 + 12 + 2 × 1 + 3 = 6 when n = 1. Therefore {an + n2 + 2n + 3} is a geometric sequence with the first term 6 and the common ratio 2. Then an + n2 + 2n + 3 = 6 × 2n−1 , 2an = 6 × 2n − 2 × n2 − 4n − 6. By substituting it into (∗), we have n(n − 1)(2n − 1) (n ∈ N ∗ ). Sn = 6 × 2n − 2n2 − 4n − 6 − 6 5.86 . Let the function f1 (x) =. 2 . 1+x. Define fn+1 (x) = f1 [fn (x)],. fn (0) − 1 , n ∈ N ∗ . (1) Find the general term of {an }. (2) If T2n = fn (0) + 2 2 +n ∗ a1 + 2a2 + · · · + 2na2n , Qn = 4n4n 2 +4n+1 , n ∈ N . Compare 9T2n and Qn . and an =. Solution: (1) From the given condition, we have f1 (0) = 2, a1 = 2. 1 2−1 = , fn+1 (0) = 2+2 4 −1 1 − fn (0) = = 4 + 2fn (0) +2. 2 fn+1 (0) − 1 1+fn (0) . Thus an+1 = = f1 [fn (0)] = 2 1 + fn (0) fn+1 (0) + 2 1+fn (0) 1 an+1 1 1 fn (0) − 1 = − an . This means = − . Thus {an } is a geometric sequence − 2 fn (0) + 2 2 an 2 1 1 1 1 with the first term and the common ratio − , and an = (− )n−1 , n ∈ N ∗ . 4 2 4 2 1 Subtracting both sides of 1 by (2) T2n = a1 + 2a2 + · · · + (2n − 1)a2n−1 + 2na2n . 1 1 1 1 1 1 − to obtain − T2n = (− )a1 + (− )2a2 + · · · + (− )(2n − 1)a2n−1 + (− )2na2n = 2 2 2 2 2 2 3 2 Using 1 − , 2 we have T2n = a1 + a2 + a3 + a2 + 2a3 + · · · + (2n − 1)a2n − na2n . 2 1 1 2n [1 − (− 2 ) ] 1 1 2n−1 1 1 1 2n n 1 2n−1 4 + n (− ) = − (− ) + (− ) . Thus · · · + a2n + na2n = 4 2 6 6 2 4 2 1 + 12 1 1 1 1 1 3n + 1 1 n 1 1 n 1 T2n = − (− )2n + (− )2n−1 = − − · 2 · 2n = (1 − ). This 2n 9 9 2 6 2 9 92 6 2 9 22n 3n + 1 3n + 1 4n2 + n =1− . Q = . n 2n 2 2 4n + 4n + 1 (2n + 1)2 When n = 1, then 22n = 4, (2n + 1)2 = 9. Thus 9T2n < Qn . When n = 2, then 22n = 16, (2n + 1)2 = 25. Thus 9T2n < Qn . When n 3, then 22n = [(1 + 1)n ]2 = (c0n + c1n + · · · + cnn )2 > (2n + 1)2 . Thus 9T2n > Qn . means 9T2n = 1 −. 5.87 Let x1 and x2 be the two real roots of equation x2 − 6x + 1 = 0. Show xn1 + xn2 is always an integer but not a multiple of 5, for any natural number n. Proof: According to the relation of roots and coefficients, we have x1 +x2 = 6, x1 x2 = 1. Let an = xn1 + xn2 , then a1 = x1 + x2 = 6, a2 = x21 + x22 = (x1 + x2 )2 − 2x1 x2 = 34. Since xn1 + xn2 = (x1 + x2 )(xn−1 + xn−1 ) − x1 x2 (xn−2 + xn−2 ), then an = 6an−1 − an−2 (n 3). 1 2 1 2 Let bn be the remainder of an divided by 5. By applying the above recursive formula, we have bn = bn−1 − bn−2 , bn+2 =Download bn+1 − free bn , beBooks − bn+1 = (bn+1 − bn ) − bn+1 = −bn . n+3 =atbn+2 bookboon.com Then bn+6 = −bn+3 = bn . Thus {bn } is a sequence whose period is 6. 92 Since a1 = 6, then b1 = 1. Since a2 = 34, then b2 = 4. Since a3 = x31 + x32 = (x1 +.

(93) When n = 1, then 22n = 4, (2n + 1)2 = 9. Thus 9T2n < Qn . When n = 2, then 22n = 16, (2n + 1)2 = 25. Thus 9T2n < Qn . When n 3, then 22n = [(1 + 1)n ]2 = (c0n + c1n + · · · + cnn )2 > (2n + 1)2 . Thus 9T2n > Qn .. ELEMENTARY ALGEBRA EXERCISE BOOK II. Sequences. 5.87 Let x1 and x2 be the two real roots of equation x2 − 6x + 1 = 0. Show xn1 + xn2 is always an integer but not a multiple of 5, for any natural number n. Proof: According to the relation of roots and coefficients, we have x1 +x2 = 6, x1 x2 = 1. Let an = xn1 + xn2 , then a1 = x1 + x2 = 6, a2 = x21 + x22 = (x1 + x2 )2 − 2x1 x2 = 34. Since xn1 + xn2 = (x1 + x2 )(xn−1 + xn−1 ) − x1 x2 (xn−2 + xn−2 ), then an = 6an−1 − an−2 (n 3). 1 2 1 2 Let bn be the remainder of an divided by 5. By applying the above recursive formula, we have bn = bn−1 − bn−2 , bn+2 = bn+1 − bn , bn+3 = bn+2 − bn+1 = (bn+1 − bn ) − bn+1 = −bn . Then bn+6 = −bn+3 = bn . Thus {bn } is a sequence whose period is 6. Since a1 = 6, then b1 = 1. Since a2 = 34, then b2 = 4. Since a3 = x31 + x32 = (x1 + x2 )[(x1 + x2 )2 − 3x1 x2 ] = 198. Thus b3 = 3. Since a4 = x41 + x42 = [(x1 + x2 )2 − 2x1 x2 ]2 − 2(x1 x2 )2 = 1154, then b4 = −1. Since a5 = x51 + x52 = (x31 + x32 )(x21 + x22 ) − x31 x22 − x21 x32 = (x1 +x2 )[(x1 +x2 )2 −3x1 x2 ][(x1 +x2 )2 −2x1 x2 ]−(x1 x2 )2 (x1 +x2 ) = 6726, then b5 = −4. Since a6 = x61 +x62 = (x21 )3 +(x22 )3 = (x21 +x22 )(x41 −x21 x22 +x42 ) = [(x1 +x2 )2 −2x1 x2 ]{[(x1 + x2 )2 − 2x1 x2 ]2 − 3(x1 x2 )2 } = 39202, then b6 = −3. Therefore bn = 0 for any natural number n and an is not a multiple of 5. 5.88 . Show an. Let sequence {an } and sequence {bn } satisfy a0 = 1, b0 = 0, and 1 an+1 = 7an + 6bn − 3 (n = 0, 1, 2, · · · ) 2 bn+1 = 8an + 7bn − 4 . (n = 0, 1, 2, · · · ) are complete squares.. 1 1 we get bn = (an+1 − 7an + 3). Substituting Proof: By applying the equation , 6 Excellent Economics and Business programmes at: 1 2 we get bn+1 = (7an+1 − an − 3) . 3 From the equation , 1 we get it into , 6 1 4 From the equation 3 and equation , 4 we get bn+1 = (an+2 − 7an+1 + 3) . 6 1 1 1 1 an+2 = 14an+1 − an − 6. This means an+2 − = 14(an+1 − ) − (an − ) where is 2 2 2 2 the root of the equation x = 14x − x − 6. 1 7 1 1 1 1 Let dn = an − , then d0 = 1 − = , d1 = a1 − = 7a0 − 6b0 − 3 − = , dn+2 = 2 2 2 2 2 2 14dn+1 − dn . The characteristic equation is x2 = 14x − 1, and the characteristic roots. “The perfect start of a successful, international career.” CLICK HERE. to discover why both socially and academically the University of Groningen is one of the best places for a student to be. www.rug.nl/feb/education. Download free eBooks at bookboon.com 93. Click on the ad to read more.

(94) 5.88 . Let sequence {an } and sequence {bn } satisfy a0 = 1, b0 = 0, and 1 an+1 = 7an + 6bn − 3 (n = 0, 1, 2, · · · ) 2 b = 8a + 7b − 4 n n ELEMENTARY ALGEBRA EXERCISE n+1 BOOK II Sequences Show an. (n = 0, 1, 2, · · · ) are complete squares.. 1 1 we get bn = (an+1 − 7an + 3). Substituting Proof: By applying the equation , 6 1 2 we get bn+1 = (7an+1 − an − 3) . 3 From the equation , 1 we get it into , 6 1 4 From the equation 3 and equation , 4 we get bn+1 = (an+2 − 7an+1 + 3) . 6 1 1 1 1 an+2 = 14an+1 − an − 6. This means an+2 − = 14(an+1 − ) − (an − ) where is 2 2 2 2 the root of the equation x = 14x − x − 6. 1 7 1 1 1 1 Let dn = an − , then d0 = 1 − = , d1 = a1 − = 7a0 − 6b0 − 3 − = , dn+2 = 2 2 2 2 2 2 14dn+1 − dn . The characteristic equation is x2 = 14x − 1, and the characteristic roots √ √ √ 1 7 are x1,2 = 7 ± 4 3. Then dn = c1 (7 + 4 3)n + c2 (7 − 4 3)n . Since d0 = , d1 = , 2 2 then    c1 + c2 = 1 2 √ 7   7(c1 + c2 ) + 4 3(c1 − c2 ) = 2 √ √ 1 1 Solving the equations to obtain c1 = c2 = . Thus dn = [(7 + 4 3)n + (7 − 4 3)n ]. 4 4 √ 2n √ √ n √ n √ 1 1 1 an = dn + = [(2 + 3) + 2(2 + 3) (2 − 3) + (2 − 3)2n ] = [(2 + 3)n + (2 − 2 4√ 4 √ n2 √ 3) ] . Let (2 + 3)n = An + Bn 3 (An , Bn are all positive integer number). Then √ √ √ √ 1 1 (2 − 3)n = An − Bn 3. Hence an = (An + Bn 3 + An − Bn 3)2 = (2An )2 = A2n . 4 4 Therefore an (n = 0, 1, 2, · · · ) are complete squares. 5.89 Given sequences {an } and {bn }, a1 = 1, a2 = −1, b1 = 2, b2 = −3, and an+1 = 3an − 2bn , bn+1 = 5an − 4bn . Find the general terms of sequence {an } and {bn }. Solution: From the given condition, we have an+2 = 3an+1 − 2bn+1 = 3an+1 − 2(5an − 4bn ) = 3an+1 −2(5an +2an+1 −6an ) = −an+1 +2an . let an+2 −r1 an+1 = r2 (an+1 −r1 an ). Comparing the coefficients to obtain r1 + r2 = −1, r1 r2 = −2. Then r1 , r2 are the two roots of the characteristic equation x2 + x − 2 = 0. Thus r1 = 1, r2 = −2. This an − an−1 = means an+2 − an+1 = −2(an+1 − an ) ⇒ an − an−1 = −(an−1 − an−2 ) ⇒ an−1 − an−2 a3 − a2 an−1 − an−2 = −2, · · · , = −2. Multiplying the above equations to ob−2, an−2 − an−3 a2 − a1 an − an−1 = (−2)n−2 . Thus an − an−1 = (−2)n−1 , an−1 − an−2 = (−2)n−2 , · · · , tain a2 − a1 a2 − a1 = (−2). Adding the above equations to obtain an − a1 = −2 + (−2)2 + 1 − (−2)n −2[1 − (−2)n−1 ] · · · + (−2)n−1 . Hence an = 1 + = (n ∈ N ∗ ). similarly, 1 − (−2) 3 1 + 5(−2)n−1 bn = (n ∈ N ∗ ). 3. Download free eBooks at bookboon.com 94.

(95) ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions. 6 FUNCTIONS 6. Functions. 1 − 2x , and the graphs of g(x) and y = f −1 (x + 1) are 1+x symmetric about y = x. Evaluate g(2).. 6.1. Let the function f (x) =. 1−y 1−x 1 − 2x , then x = . Thus f −1 (x) = , 1+x y+2 x+2 −x −x . Hence g(x) and f (x) = are inverse functions for each other. f −1 (x + 1) = x+3 x+3 x , then g(2) = −2. Since 2 = − x+3 Solution 2: Since y = f −1 (x + 1), then x = f (y) − 1. Thus g(x) = f (x) − 1. Then g(2) = f (2) − 1 = −2. Solution 1: Since y = f (x) =. 1 log 1. 1 . log 1 13 2 5 1 1 1 1 1 1 + log 1 = log 1 = log3 10, and 2 = Solution: x = 1 + 1 = log 3 3 3 2 5 10 log 1 3 log 1 3 2 5 log3 9 < log3 10 < log3 27 = 3. Hence x ∈ (2, 3). 6.2. Compute the range of x =. 1 3. +. 6.3 Let x1 and x2 be the two real roots of the equation x2 − (k − 2)x + (k 2 + 3k + 5) = 0 (k ∈ R). Find the maximum value of x21 + x22 . Solution: According to the Vieta’s theorem, we have x1 +x2 = k −2, x1 x2 = k 2 +3k +5. Then x21 + x22 = (x1 + x2 )2 − 2x1 x2 = (k − 2)2 − 2(k 2 + 3k + 5) = −(k + 5)2 + 19. Since the equation has two real roots, then ∆ = (k − 2)2 − 4(k 2 + 3k + 5) 0. This 4 means 3k 2 + 16k + 16 0. The range is −4 k − . To find the maximum value of 3 4 2 2 x1 + x2 , we need to find the maximum value of y = −(k + 5)2 + 19 when −4 k − . 3 4 Since the symmetric axis k = −5 is not in [−4, − ], then the function is decreasing in 3 4 [−4, − ]. Therefore the the maximum value is ymax = −(−4 + 5)2 + 19 = 18. 3 6.4 If the function f (x) is defined for all real numbers R, and f (10 + x) = f (10 − x), f (20 − x) = −f (20 + x). Is f (x) a periodic function? And determine f (x) is odd or even. Solution: From the first given equation, we have f [10 + (10 − x]) = f [10 − (10 − x)]. 1 Combining the given second equation, we have f (x) = Thus f (x) = f (20 − x) . 2 −f (20 + x) .. Download free eBooks at bookboon.com 95.

(96) ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions. Then f (40 + x) = f [20 + (20 + x)] = −f (20 + x) = f (x). Hence f (x) is a periodic 1 and , 2 we have f (−x) = f (20 + x) = −f (x). Therefore function. By applying f (x) is an odd function. The function F (x) is an odd function, and a > 0, a = 1. Determine the function 1 1 G(x) = F (x)( x + ) is odd or even. a −1 2 1 1 + = Proof: Since F (x) is an odd function, then F (−x) = −F (x). Let g(x) = x a −1 2 x 1 +1 a−x + 1 ax + 1 a + 1 x , then g(−x) = = ax1 = −g(x). This means = 2(ax − 1) 2(a−x − 1) 2(1 − ax ) 2( axx − 1) g(x) is also an odd function. G(x) = F (x)g(x) holds in R when a > 0 and a = 1. Since G(−x) = F (−x)g(−x) = F (x)g(x), then G(x) is an even function. 6.5. Given the set M 1 valuate (x + ) + (x2 + y. = {x, xy, lg(xy)}, the set N = {0, |x|, y}, and M = N , e1 1 1 3 2011 ) + (x + ) + · · · + (x + ). y2 y3 y 2011. 6.6. Proof: Since M = N , we get that at least one element of M is zero. From the definition of logarithmic function, we have xy = 0. This means x and y are both nonzero. Thus 1 lg(xy) = 0. Then xy = 1. Hence M = {x, 1, 0}, N = {0, |x|, }. Additionally, by x applying the equal of sets, we have x = |x| 1 1= x In the past four years we have drilled or x = x1 1 = |x| But it is contradicting to the element distinction in a set when x = 1. Thus x = −1, That’s more than twice around the world. 1 1 y = −1. Then x2k+1 + 2k+1 = −2, x2k + 2k = 2, (k = 0, 1, 2, · · · ). y y 1 1 1 we? 1 2 3 2011 Who are +are the ) = −2. Therefore (x + ) + (x + 2 ) + (x + 3 ) + · · · + (x We 2011 y y y world’s largest oilfield services company . y Working globally—often in remote and challenging locations—. 89,000 km 1. Given f (x) = √ 3 3. 6.7. f (5) + · · · + f (2011).. x2 + 2x + 1 +. Solution: f (x) = √ 3 3. x2 + 2x + 1 +. √ 3 3. x2 − 1 +. 1. √ 3 3. we invent, design, engineer, and apply technology to help our customers find and produce oil and gas safely.. 1. x2 − 1 +. √ 3 3. √ 3 3. , solve f (1) + f (3) + 2x +are 1 we looking for? x2 − Who Every year, we need thousands of graduates to begin dynamic careers in the following domains: n Engineering, Research and Operations n Geoscience and Petrotechnical n Commercial and Business. x2 − 2x + 1. What will you be?. careers.slb.com Based on Fortune 500 ranking 2011. Copyright © 2015 Schlumberger. All rights reserved.. 1. Download free eBooks at bookboon.com 96. Click on the ad to read more.

(97) y. y. y. y. Proof: Since M = N , we get that at least one element of M is zero. From the definition of logarithmic function, we have xy = 0. This means x and y are both nonzero. Thus 1 ELEMENTARY ALGEBRA EXERCISE BOOK II Functions lg(xy) = 0. Then xy = 1. Hence M = {x, 1, 0}, N = {0, |x|, }. Additionally, by x applying the equal of sets, we have x = |x| 1 1= x or x = x1 1 = |x| But it is contradicting to the element distinction in a set when x = 1. Thus x = −1, 1 1 y = −1. Then x2k+1 + 2k+1 = −2, x2k + 2k = 2, (k = 0, 1, 2, · · · ). y y 1 1 1 1 Therefore (x + ) + (x2 + 2 ) + (x3 + 3 ) + · · · + (x2011 + 2011 ) = −2. y y y y 6.7. Given f (x) = √ 3. f (5) + · · · + f (2011). Solution: f (x) = √ 3 = =. x2 + 2x + 1 +. x2 + 2x + 1 +. 1 x2 − 1 +. 1 x2 − 1 +. √ 3. √ 3. x2 − 2x + 1. , solve f (1) + f (3) +. x2 − 2x + 1. √ 3 x+1− √ x−1 √ 2 2 (x−1)− (x+1) (x−1)+ (x −1)(x+1)− 3 (x2 −1)(x−1)+ 3 (x−1)2 (x+1)−(x−1) √ √ 3 x+1− 3 √ x−1 √ √ √ 3 3 2 2 2− (x+1)(x −1)+ (x −1)(x+1)− 3 (x2 −1)(x−1)+ 3 (x2 −1)(x−1). √ 3. √ 3. √ 3. √ 3. √ 3. √ 1 √ = ( 3 x + 1 − 3 x − 1). 2 √ √ √ √ 1 √ 3 3 3 3 3 Thus f (1) + f (3) + f (5) + · · · + f (2011) = ( 2 − 0 + 4 − 2 + 6 − 4 + · · · + 2 √ 3 √ √ √ √ 2012 3 3 3 3 . 2010 − 2008 + 2012 − 2010) = 2 √ 6.8 Given f (x) = a sin x + b 3 x + 4, (a, b ∈ R), and f (lg log3 10) = 5. Find the value of f (lg lg 3). √ Solution: Since f (x) − 4 = a sin x + b 3 x, then f (x) − 4 is an odd function. Thus f (−x) − 4 = −(f (x) − 4). This means f (−x) = −f (x) + 8. Additionally, since lg lg 3 = − lg log3 10, then f (lg lg 3) = f (− lg log3 10) = −f (lg log3 10)+8 = −5+8 = 3. 6.9 Given f (x) is an odd function, g(x) is an even function, and f (x)−g(x) = x2 −x. Find f (x) and g(x). Solution: Since f (x) is an odd function , then f (−x) = −f (x). Since g(x) is an even function , then g(−x) = g(x). Thus f (x) − g(x) = x2 − x ⇒ f (−x) − g(−x) = x2 + x ⇒ −f (x) − g(x) = x2 + x ⇒ f (x) + g(x) = −x2 − x. Then f (x) − g(x) = x2 − x f (x) + g(x) = −x2 − x ⇒ f (x) = −x, g(x) = −x2 .. 1 If the domain of the function y = f (x2 ) is [− , 1], find the domain of Download free eBooks at bookboon.com 4 g(x) = f (x + a) + f (x − a). 97. 6.10 .

(98) . 2 ELEMENTARY EXERCISE ⇒ f (x) =ALGEBRA −x, g(x) = −xBOOK . II. f (x) − g(x) = x2 − x f (x) + g(x) = −x2 − x. Functions. 1 If the domain of the function y = f (x2 ) is [− , 1], find the domain of 4 g(x) = f (x + a) + f (x − a).. 6.10 . 1 1 Solution: Since the domain of the function y = f (x2 ) is [− , 1], then − x 1. 4 4 Thus 0 x2 1. Hence the domain of the function y = f (x) is {x|0 x 1}. Then the domain of g(x) is the solution set of the following system: 0x+a1 0x−a1 Then. . −a x 1 − a ax1+a. 1 The domain of g(x) = f (x + a) + f (x − a) is {x| − a x 1 + a} when − < a < 0. 2 1 The domain of g(x) = f (x + a) + f (x − a) is {x|a x 1 − a} when 0 a . Note 2 1 1 that x ∈ ∅ when a < − or a > . 2 2 6.11 . Given the range of y =. of a and b.. ax2 + 8x + b as {y|1 y 9}, find the value x2 + 1. ax2 + 8x + b , then (y − a)x2 − 8x + y − b = 0. Since x ∈ R, x2 + 1 then ∆ = 64 − 4(y − a)(y − b) 0 when y = a. Simplifying the formula to generate y 2 − (a + b)y + ab − 16 0. Since the range of y is {y|1 y 9}, then 1 and 9 are the two roots of the equation with respect to y. According to the relationship between roots and coefficients, we have a + b = 10 ab − 16 = 9. Solution: Since y =. Then a = b = 5. a−b x= ∈ R when y = a. Therefore a = b = 5. 8 6.12 For arbitrary x, y ∈ R, the function y = f (x) always satisfies f (x + y) = f (x) + f (y) − 1. And f (x) > 1 when x > 0 and f (3) = 4. (1) Show y = f (x) is an increasing function. (2) Find the maximum value and the minimum value of f (x) in [1, 2]. (1) Proof: Let x1 , x2 ∈ R, and x1 < x2 , then f (x2 ) = f (x2 − x1 + x1 ) = f (x2 − x1 ) + f (x1 ) − 1. Thus f (x2 ) − f (x1 ) = f (x2 − x1 ) − 1. Since x1 < x2 , x2 − x1 > 0, we have f (x2 − x1 ) > 1 which means f (x2 − x1 ) − 1 > 0. Hence f (x2 ) − f (x1 ) > 0. Then f (x2 ) > f (x1 ). Therefore y = f (x) is an increasing function. (2) Solution: From (1), we know that f (x) is an increasing function in R. Then f (x) is an increasing function in [1, 2]. The minimum value of f (x) is f (1) = 2 when x = 1. The maximum value of f (x) is f (2) = 2f (1) − 1 = 3 when x = 2. Therefore the maximum value of f (x) is 3 and the minimum value of f (x) is 2 when x ∈ [1, 2]. Download free eBooks at bookboon.com 98 6.13 For all ordered pairs of positive integers (x, y), f (x, 1) = 1 holds. f (x, y) = 0 and f (x + 1, y) = y[f (x, y) + f (x, y − 1)] both hold when y > x. Evaluate f (5, 5)..

(99) creasing function. (2) Find the maximum value and the minimum value of f (x) in [1, 2]. (1) Proof: Let x1 , x2 ∈ R, and x1 < x2 , then f (x2 ) = f (x2 − x1 + x1 ) = f (x2 − x1 ) + f (x1ALGEBRA ) − 1. Thus f (x2 ) −II f (x1 ) = f (x2 − x1 ) − 1. Since x1 < x2 , x2 − x1 >Functions 0, we ELEMENTARY EXERCISE BOOK have f (x2 − x1 ) > 1 which means f (x2 − x1 ) − 1 > 0. Hence f (x2 ) − f (x1 ) > 0. Then f (x2 ) > f (x1 ). Therefore y = f (x) is an increasing function. (2) Solution: From (1), we know that f (x) is an increasing function in R. Then f (x) is an increasing function in [1, 2]. The minimum value of f (x) is f (1) = 2 when x = 1. The maximum value of f (x) is f (2) = 2f (1) − 1 = 3 when x = 2. Therefore the maximum value of f (x) is 3 and the minimum value of f (x) is 2 when x ∈ [1, 2]. 6.13 For all ordered pairs of positive integers (x, y), f (x, 1) = 1 holds. f (x, y) = 0 and f (x + 1, y) = y[f (x, y) + f (x, y − 1)] both hold when y > x. Evaluate f (5, 5). Solution: Since f (x, 1) = 1, then f (1, 1) = 1, f (2, 2) = f (1+1, 2) = 2[f (1, 2)+f (1, 1)] = 2[0 + f (1, 1)] = 2f (1, 1) = 2 = 2 × 1. f (3, 3) = f (2 + 1, 3) = 3[f (2, 3) + f (2, 2)] = 3f (2, 2) = 3 × 2 × 1, · · · . Thus f (5, 5) = 5 × 4 × 3 × 2 × 1 = 120.. 6.14 If the real numbers x and θ satisfy log3 (x + 7) + 2 cos(θ + 2012) = 4. Compute |x − 2| + |x − 722|. Solution: log3 (x + 7) + 2 cos(θ + 2012) = 4 ⇒ 2 log3 (x + 7) = 4 − 2 cos(θ + 2012) 6 ⇒ 9 x + 7 729 ⇒ 2 x 722. Thus |x − 2| + |x − 722| = x − 2 + 722 − x = 720. 6.15 . Let A = [1, b] (b > 1). The function f (x) =. of f (x) is A when x ∈ A. Find the value of b.. 1 2 3 x − x + . The range 2 2. 3 1 1 2 x − x + = (x − 1)2 + 1 is a parabolic curve, and its sym2 2 2 metric axis is x = 1, the vertex is (1, 1). Thus f (x) is an increasing function when x ∈ [1, b] (b > 1). Then f (x) reaches the maximum value f (b) when x = b. Since 1 f (b) ∈ [1, b], then f (b) = b. This means (b − 1)2 + 1 = b. Then b2 − 4b + 3 = 0. Hence 2 b = 1 orisb currently = 3. Sinceenrolling b > 1, then b = 3. in the Solution: f (x) =. American online LIGS University. Interactive Online BBA, MBA, MSc, DBA the andfunction PhD programs: 6.16 Given f (x) = ax2 + bx + 1 where a, b are real numbers, x ∈ R. . f (x), (x > 0) −fSeptember (x), (x < 0) 30th, 2014 and ▶▶ enroll by (1) If f (−1) = 0, and the range of f (x) is [0, +∞), Find the analytic formula of ▶ save to 16% the tuition! F▶(x). (2)up Under theon condition of (1), g(x) = f (x) − kx is a monotone function when pay in installments x▶∈▶ [−2, 2].10Find the range/ 2ofyears k. ▶▶ Interactive Online education a−b+1=0 ▶▶ visit www.ligsuniversity.com to . By solving the Solution: (1) From the given condition, we have b − = −1 2 2a find out more! a=1 x + 2x + 1, (x > 0) equation system, we have . Thus F (x) = . b=2 −x2 − 2x − 1, (x < 0) Note: LIGS University is not accredited by any (2) g(x) = recognized x2 + (2 − k)x + 1 is a agency monotone function when x ∈ [−2, 2] if and only nationally accrediting listed 2 − k 2 − k Secretary −2. Then k 6 or k −2. Thus the range of k is 2 or − of Education. ifby−the US More 2info here. 2 (−∞, −2] ∪ [6, +∞).. F (x) =. Download at bookboon.com 6.17 The graph of f (x) = kx free + beBooks intersects x axis at AClick andon intersects axismore the ad toyread −→ at B. AB = 2i + 2j, (i is the unit vector99of positive x axis, j is the unit vector of 2.

(100) 3. |x − 2| + |x − 722|. 2012) = 4 ⇒ 2 log3 (x + 7) = 4 − 2 cos(θ + 2012) 6 ⇒ 9 x + 7 729 ⇒ 2 x 722. Thus |x − 2| + |x − 722| = x − 2 + 722 − x Functions = 720.. Solution: log3 (x + 7) + 2 cos(θ + ELEMENTARY ALGEBRA EXERCISE BOOK II. 6.15 . Let A = [1, b] (b > 1). The function f (x) =. of f (x) is A when x ∈ A. Find the value of b.. 1 2 3 x − x + . The range 2 2. 1 3 1 2 x − x + = (x − 1)2 + 1 is a parabolic curve, and its sym2 2 2 metric axis is x = 1, the vertex is (1, 1). Thus f (x) is an increasing function when x ∈ [1, b] (b > 1). Then f (x) reaches the maximum value f (b) when x = b. Since 1 f (b) ∈ [1, b], then f (b) = b. This means (b − 1)2 + 1 = b. Then b2 − 4b + 3 = 0. Hence 2 b = 1 or b = 3. Since b > 1, then b = 3.. Solution: f (x) =. Given the function f (x) = ax2 + bx + 1 where a, b are real numbers, x ∈ R. f (x), (x > 0) F (x) = −f (x), (x < 0) (1) If f (−1) = 0, and the range of f (x) is [0, +∞), Find the analytic formula of F (x). (2) Under the condition of (1), g(x) = f (x) − kx is a monotone function when x ∈ [−2, 2]. Find the range of k. a−b+1=0 Solution: (1) From the given condition, we have . By solving the b − = −1 2 2a a=1 x + 2x + 1, (x > 0) equation system, we have . Thus F (x) = . b=2 −x2 − 2x − 1, (x < 0) (2) g(x) = x2 + (2 − k)x + 1 is a monotone function when x ∈ [−2, 2] if and only 2−k 2−k if − 2 or − −2. Then k 6 or k −2. Thus the range of k is 2 2 (−∞, −2] ∪ [6, +∞). 6.16. 6.17 The graph of f (x) = kx + b intersects x axis at A and intersects y axis −→ at B. AB = 2i + 2j, (i is the unit vector of positive x axis, j is the unit vector of positive y axis), and g(x) = x2 − x − 6. g(x) + 1 when f (x) > g(x). f (x) −→ b b Solution: (1) From the given condition, we have A(− , 0), B(0, b). Then AB = { , b}. k k b Thus = 2, b = 2, k = 1. k g(x) + 1 = (2) Since f (x) > g(x), then x + 2 > x2 − x − 6. thus −2 < x < 4. f (x) x2 − x − 5 1 g(x) + 1 1 = x+2+ −5. Since x+2 > 0, then 2 (x + 2) −5 = x+2 x+2 f (x) x+2 −3, and the equation holds if and only if x + 2 = 1 i.e. x = −1. Hence the minimum g(x) + 1 value of is −3. f (x) (1) Evaluate k and b. (2) Find the minimum value of. 6.18 If the equation (2 − 2−|x−3| )2 = 3 + a with respect to x has real roots, find the range of real number a. Download free eBooks at bookboon.com. Solution: We simply the given equation to get a = (2 − 2−|x−3| )2 − 3. Let t = 2−|x−3| , 100 then 0 < t 1, a = f (t) = (t − 2)2 − 3. Since a = f (t) is decreasing on (0, 1], then.

(101) 1 g(x) + 1 1 x −x−5 = x+2+ −5. Since x+2 > 0, then 2 (x + 2) −5 = x+2 x+2 f (x) x+2 −3, and the equation holds if and only if x + 2 = 1 i.e. x = −1. Hence the minimum g(x) + 1 is −3. BOOK II value of ALGEBRA EXERCISE ELEMENTARY Functions f (x) 6.18 If the equation (2 − 2−|x−3| )2 = 3 + a with respect to x has real roots, find the range of real number a. Solution: We simply the given equation to get a = (2 − 2−|x−3| )2 − 3. Let t = 2−|x−3| , then 0 < t 1, a = f (t) = (t − 2)2 − 3. Since a = f (t) is decreasing on (0, 1], then f (1) f (t) < f (0). This means −2 f (t) < 1. Thus the range of real number a is a ∈ [−2, 1). 6.19. If the maximum value of the function f (x) = −3x2 − 3x + 4b2 +. [−b, 1 − b] is 25, find the value of b.. 9 4. (b > 0) on. 1 Solution: From the given condition, we have f (x) = −3(x + )2 + 4b2 + 3. 2 1 3 1 2 (1) The maximum value of f (x) is 4b + 3 = 25 when −b − 1 − b i.e. b . 2 2 2 1 3 11 2 Then b = . It is contradicting to b . 2 2 2 1 1 (2) f (x) is decreasing on the interval [−b, 1 − b] when − −b i.e. 0 1 − b i.e. b > . 2 2 5 15 2 Hence f (1 − b) = b + 9b − = 25. Then b = . 4 2 6.20 . If for all x, y ∈ R, f (x + y) = f (x) + f (y) holds. (1) Show f (x) is an. odd function. (2) if f (−3) = a, express f (12) as a function a.. (1) Proof: Obviously, the domain of f (x) is R. Let y = −x where f (x+y) = f (x)+f (y), then f (0) = f (x) + f (−x). Let x = y = 0 where f (0) = f (0) + f (0), then f (0) = 0. Thus f (x) + f (−x) = 0 which means f (x) = −f (−x). Hence f (x is an odd function. (2) Solution: Since f (−3) = a, f (x + y) = f (x) + f (y), and f (x) is an odd function, we have f (12) = 2f (6) = 4f (3) = −4f (−3) = −4a. 6.21 If M is a set of functions that satisfy the following conditions: (1) the domain of f (x) is [−1, 1]. (2) If x1 , x2 ∈ [−1, 1], then |f (x1 ) − f (x2 )| 4|x1 − x2 |. Determine whether the function g(x) = x2 + 2x − 1 defined on the interval [−1, 1] belongs to the set M . Proof: From the given condition, we know that g(x) satisfies the condition (1) obviously. Let x1 , x2 ∈ [−1, 1]. Then |x1 | 1, |x2 | 1. Since |g(x1 ) − g(x2 )| = |(x21 + 2x1 − 1) − (x22 + 2x2 − 1)| = |(x1 − x2 )(x1 + x2 + 2)| |x1 − x2 ||x1 + x2 + 2| (|x1 | + |x2 | + 2)|x1 − x2 | 4|x1 − x2 |. Thus g(x) satisfies the condition (2). Hence g(x) ∈ M . (x − 2a) < 0}. x − (a2 + 1) Download free eBooks at bookboon.com (1) Find A ∩ B when a = 2. (2) Find the range of the real number a such that B ⊆ A. 101 x−4 < 0} = (4, 5). Thus Solution: (1) A = {x|(x − 2)(x − 7) < 0} = (2, 7), B = {x|. 6.22 . Given the set A = {x|(x − 2)[x − (3a + 1)] < 0}, B = {x|.

(102) viously. Let x1 , x2 ∈ [−1, 1]. Then |x1 | 1, |x2 | 1. Since |g(x1 ) − g(x2 )| = |(x21 + 2x1 − 1) − (x22 + 2x2 − 1)| = |(x1 − x2 )(x1 + x2 + 2)| |x1 − x2 ||x1 + x2 + 2| (|x1 | + |x2 | + 2)|x1 − x2 | 4|x1 − x2 |. Thus g(x) satisfies the condition (2). Hence ELEMENTARY Functions g(x) ∈ M .ALGEBRA EXERCISE BOOK II (x − 2a) < 0}. x − (a2 + 1) (1) Find A ∩ B when a = 2. (2) Find the range of the real number a such that B ⊆ A. x−4 Solution: (1) A = {x|(x − 2)(x − 7) < 0} = (2, 7), B = {x| < 0} = (4, 5). Thus x−5 A ∩ B = (4, 5). 1 (2) Since B = (2a, a2 + 1), then A = (3a + 1, 2) when a < . In order to have B ⊆ A, 3 1 2a 3a + 1 for which a = −1. A = φ When a = . There is no a we must have 2 a +12 3 1 such that B ⊆ A. Then A = (2, 3a + 1) when a > . In order to have B ⊆ A, we 3 2a 2 . Thus 1 a 3. As a conclusion, the range of the real must have a2 + 1 3a + 1 number a such that B ⊆ A is [1, 3] ∪ {−1}. 6.22 . Given the set A = {x|(x − 2)[x − (3a + 1)] < 0}, B = {x|. 6.23 The statement p is that the equation a2 x2 + ax − 2 = 0 has solutions on the interval [−1, 1], and the statement q is that there is only one real number x such that the inequality x2 + 2ax + 2a 0 holds. If the statement “p or q” is a false statement, find the range of a.. .. Download free eBooks at bookboon.com 102. Click on the ad to read more.

(103) ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions. Solution: a2 x2 + ax − 2 = 0 ⇒ (ax + 2)(ax − 1) = 0. Obviously, a = 0, then 2 1 2 1 x = − or x = . Since x ∈ [−1, 1], then | | 1 or | | 1. Thus |a| 1. Since a a a a there is only one real number x such that the inequality x2 + 2ax + 2a 0 holds, then there is only one intersection of the parabolic curve y = x2 + 2ax + 2a and x-axis. Thus ∆ = 4a2 − 8a = 0. Hence a = 0 or a = 2. Then |a| 1 or a = 0 when the statement “p or q” is a true statement. Therefore the range of a is {a| − 1 < a < 1 or 0 < a < 1} when the statement “p or q” is a false statement. 6.24 . The function f (x) is defined on the interval [0, 1], and f (0) = f (1). If for arbi1 trary distinct x1 , x2 ∈ [0, 1], |f (x2 )−f (x1 )| < |x2 −x1 | holds. Show |f (x2 )−f (x1 )| < . 2 1 1 Proof: Let 0 x1 x2 1. (1) If x2 −x1 , then |f (x2 )−f (x1 )| < |x2 −x1 | . (2) 2 2 1 If x2 − x1 > , since f (0) = f (1), then |f (x2 ) − f (x1 )| = |f (x2 ) − f (1) + f (0) − f (x1 )| 2 1 |f (x2 ) − f (1)| + |f (0) − f (x1 )| < (1 − x2 ) + (x1 − 0) = 1 − (x2 − x1 ) < . 2 1 As a conclusion, |f (x2 ) − f (x1 )| < . 2 6.25 Given f (x) is an increasing function on the interval (0, +∞), and f (1) = 0, f (x) + f (y) = f (x, y), show |f (x)| > |f (y)| when 0 < x < y < 1. Proof: Since f (x) is an increasing function on the interval (0, +∞), and 0 < x < y, 1 Since 0 < x < y < 1, we have that f (x) < f (y). This means f (x) − f (y) < 0 . 2 By applying then f (x) + f (y) = f (xy) < f (1) = 0. This means f (x) + f (y) < 0 . 1 × , 2 we have [f (x)]2 − [f (y)]2 > 0. Thus |f (x)| > |f (y)|. 6.26 Let f (x) is a function defined on R → R. Show f (x) can be expressed as the sum of an odd function and an even function. Proof: Assume f ((x) = g(x) + h(x) where g(x) is an even function and h(x) is odd f (x) = g(x) + h(x) function. Then f (−x) = g(−x)+h(−x) = g(x)−h(x). Thus . f (−x) = g(x) − h(x) 1 1 Solving the equation system, we have g(x) = [f (x) + f (−x)], h(x) = [f (x) − f (−x)]. 2 2 1 Conversely, since g(−x) = [f (−x)+f (x)] = g(x), then g(x) is an even function. Since 2 1 1 h(−x) = [f (−x) − f (x)] = − [f (x) − f (−x)] = −h(x), then h(x) is an odd function. 2 2 Therefore f ((x) can be expressed as the sum of an odd function and an even function.. Download free eBooks at bookboon.com 103.

(104) ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions. 6.27 As shown in Figure 3, the circle D intersects y-axis at the points A√and B, and intersects x-axis at the point C on the left. The straight line y = −2 2x − 8 intersects y-axis at the point P . The coordinates of the center D is (0, 1). (1) Show P C is a tangent line of the circle D. (2) Determine whether there exists a point E on the straight line CP such that SEOP = 4SCOD . If yes, find the coordinates of E. If no, please explain the reason. (3) When the straight line CP turns around the point at the point F (here F does not coincide with A or P , it intersects the inferior arc AC C). We connect OF . Let P F = m, OF = n, find the relation between m and n, and determine the range of the variable n. √ (1) Proof: The straight line y = −2 2x − 8 passing through C intersects x-axis at √ √ CO C(−2 2, 0) and y-axis at P (0, −8). Then cot ∠OCD = = 2 2, cot ∠OP C = |OD| √ |OP | = 2 2. Since ∠OP C + ∠P CO = 900 , then ∠OCD + ∠P CO = 900 . Hence, P C |OC| is a tangent line of the circle D. (2) Let the point E(x, y) on the straight line CP such that SEOP = 4SCOD . Then √ √ √ 1 1 × 8 × |x| = 4 × × 1 × 2 2. Thus x = ± 2. Since y = −2 2x − 8, then y = −12 2 2 √ √ √ when√x = 2 and y = −4 when x = − 2. Thus there exists a point E( 2, −12) or E(− 2, −4) on the straight line CP such that SEOP = 4SCOD . at the point F , and intersect(3) Let the straight line P F intersects the arc AC at the point Q. We connect DQ. By applying the cutting theorem, s the arc BC 1 In CP D and OP C, ∠P CD = ∠P OC = 900 , we have P C 2 = P F · P Q . PC PD = which means P C 2 = ∠CP D = ∠OP C. Thus CP D OP C, PO PC 2 According to 1 and , 2 we have P O · P D = P F · P Q. AdditionP O · P D . PF PD m ally, since ∠F P O = ∠DP Q,Download then F O DP Q. Hence = = . Since freePeBooks at bookboon.com FO DQ n √ m 9 104 = = 3, OA = 3 − 1 = 2. P D = 9, DQ = CD = (2 2)2 + 12 = 3. Thus n 3.

(105) is a tangent line of the circle D. (2) Let the point E(x, y) on the straight line CP such that SEOP = 4SCOD . Then √ √ √ 1 1 × 8 × |x| = 4 × × 1 × 2 2. Thus x = ± 2. Since y = −2 2x − 8, then y = −12 2 2 ELEMENTARY ALGEBRA EXERCISE BOOK II √ √ √ Functions when√x = 2 and y = −4 when x = − 2. Thus there exists a point E( 2, −12) or E(− 2, −4) on the straight line CP such that SEOP = 4SCOD . at the point F , and intersect(3) Let the straight line P F intersects the arc AC at the point Q. We connect DQ. By applying the cutting theorem, s the arc BC we have P C 2 = P F · P Q . 1 In CP D and OP C, ∠P CD = ∠P OC = 900 , PC PD = which means P C 2 = ∠CP D = ∠OP C. Thus CP D OP C, PO PC 2 According to 1 and , 2 we have P O · P D = P F · P Q. AdditionP O · P D . PD m PF = = . Since ally, since ∠F P O = ∠DP Q, then F P O DP Q. Hence FO DQ n √ 9 m P D = 9, DQ = CD = (2 2)2 + 12 = 3. Thus = 3, OA = 3 − 1 = 2. = 3 n √ Therefore m = 3n (2 < n < 2 2). 6.28 Given the function f (x) = log2 (x + 1), and the point (x, y) moves on the x y graph of f (x), the point ( , ) moves on the graph of y = g(x). Find the maximum 3 2 value of the function p(x) = g(x) − f (x). 1 Solution: From the given condition, we have g(x) = log2 (3x + 1). Then P (x) = 2 1 3x + 1 3x + 1 log2 (3x + 1) − log2 (x + 1) = log2 . Let u = , then ux2 + (2u − 2 2 2 (x + 1) (x + 1) 2 3)x + u − 1 = 0. Since u has meaning, then ∆ = (2u − 3) − 4u(u − 1) = −8u + 9 0. 9 3 9 = log2 3 − . This means u . Thus pmax (x) = log2 8 8 2 . x+3 has the domain A, and the funcx+1 tion g(x) = lg[(x − a − 1)(2a − x)](a < 1) has the domain B. (1) Find A and B. (2) If B ⊆ A, find the range of the real number a. Top master’s programmes Join the best at 2x +Times 2 −worldwide x − 3 ranking:xMSc −1 • 33 place Financial Solution: (1) From the given condition, we have f (x) = = . International Business the Maastricht University x + 1 x + 1 • 1 place: MSc International Business x−1 • 1 place: MSc Financial Economics Since 0, then x 1 or x −1. Thus A = (−∞, −1] ∪ [1, +∞). Since School of Business and • 2 place: MSc Management of Learning x + 1 place: MSc Economics [x − (a + 1)](x • • −22 2a) <0 (xEconomics! − a − 1)(2a − x) > 0 place: MSc Econometrics , then 2aand<Operations x < aResearch + 1. ⇒ • 2 place: MSc Global Supply Chain Management and a<1 a<1 6.29 . Suppose the function f (x) =. 2−. rd. st st. nd. nd nd nd. Change. Sources: Keuzegids Master ranking 2013; Elsevier ‘Beste Studies’ ranking 2012; Financial Times Global Masters in Management ranking 2012. Maastricht University is the best specialist university in the Netherlands (Elsevier). Visit us and find out why we are the best! Master’s Open Day: 22 February 2014. www.mastersopenday.nl. Download free eBooks at bookboon.com 105. Click on the ad to read more.

(106) log2 (3x + 1) − log2 (x + 1) = log2 . Let u = , then ux + (2u − 2 (x + 1)2 (x + 1)2 2 3)x + u − 1 = 0. Since u has meaning, then ∆ = (2u − 3) − 4u(u − 1) = −8u + 9 0. 3 9 9 ELEMENTARY ALGEBRA Functions This means u EXERCISE . ThusBOOK pmaxII(x) = log2 = log2 3 − . 2 8 8 . x+3 has the domain A, and the funcx+1 tion g(x) = lg[(x − a − 1)(2a − x)](a < 1) has the domain B. (1) Find A and B. (2) If B ⊆ A, find the range of the real number a. 2x + 2 − x − 3 x−1 Solution: (1) From the given condition, we have f (x) = = . x+1 x+1 x−1 Since 0, then x 1 or x −1. Thus A = (−∞, −1] ∪ [1, +∞). Since x+1 (x − a − 1)(2a − x) > 0 [x − (a + 1)](x − 2a) < 0 ⇒ , then 2a < x < a + 1. a<1 a<1 6.29 . Suppose the function f (x) =. 2−. Thus B = (2a, a + 1).. 1 (2) Since B ⊆ A, then 2a 1 or a + 1 −1. This means a , or a −2. Since 2 1 a < 1, then a < 1, or a −2. For B ⊆ A, the range of real number a is 2 1 (−∞, −2] ∪ [ , 1). 2 6.30 The graph of the linear function f (x) = ax + b passes through the point (10, 13), and its x-intercept is (p, 0) and its y-intercept is (0, q), where p is a prime number and q is a positive integer number. Find all linear functions that satisfy the above conditions.. Solution: Since the x- and y-intercepts of the linear function f (x) = ax + b are ap + b = 0 (p, 0) and (0, q) respectively, we have . Solving these equations to obtain b=q q x y p a = − , b = q. Thus y = − x + q. This means + = 1. Since the linear function q p p q passes through (10, 13), we have 10q + 13p = pq. Then (p − 10)(q − 13) = 130. Since p is a prime number, then p is only 11 or 23. Hence, q = 143 when p = 11; q = 23 when p = 23. The linear functions which satisfy the conditions are y = −13x + 143; y = −x + 23. Given a quadratic function f (x) = ax2 + bx + c (a > 0). The two roots 1 of the equation f (x) − x = 0 are x1 ,x2 , which satisfy 0 < x1 < x2 < . In addition, a x1 the graph of f (x) is symmetric about the straight line x = x0 . Show x0 < . 2 6.31. Proof: From the given condition, we have f (x) − x = ax2 + (b − 1)x + c. Since 1 the two roots of the equation f (x) − x = 0 are x1 ,x2 , which satisfy 0 < x1 < x2 < , a b−1 1 b−1 b−1 1 b−1 we have 0 < x1 < < x2 < , and − x1 = x 2 − < − . This −2a a −2a −2a a −2a x1 b b means − < x1 . Thus x0 = − < . a 2a 2 6.32 . Given f (x) = x4 + ax3 + bx2 + cx + d (a,b are constants), and f (1) = 1 2009f (2) = 4018f (3) = 6027.Download Evaluate [f (11)at + f ((−7)]. free eBooks bookboon.com 4 106. Solution: Let n = 2009, F (x) = f (x) − nx, then F (1) = F (2) = F (3) = 0. Thus.

(107) a b−1 1 b−1 b−1 1 b−1 . This − x1 = x2 − we have 0 < x1 < < x2 < , and < − −2a −2a a −2a a −2a b x1 b < x1 . EXERCISE means − ALGEBRA Thus x0BOOK = −II < . ELEMENTARY Functions a 2a 2 6.32 . Given f (x) = x4 + ax3 + bx2 + cx + d (a,b are constants), and f (1) = 1 2009f (2) = 4018f (3) = 6027. Evaluate [f (11) + f ((−7)]. 4 Solution: Let n = 2009, F (x) = f (x) − nx, then F (1) = F (2) = F (3) = 0. Thus F (x) = (x − 1)(x − 2)(x − 3)(x − r). 1 1 1 1 [f (11)+f (−7)] = [F (11)+F ((−7)+11n−7n] = [F (11)+F (−7)]+n = [10×9× 4 4 4 4 1 8×(11−r)+(−8)·(−9)·(−10)·(−7−r)]+2009 = [10×9×8×(11−r+7+r)]+2009 = 4 1 × 10 × 9 × 8 × 18 + 2009 = 5249. 4 6.33 If a > b > c, show that the equation 3x2 −2(a+b+c)x+ab+bc+ca = 0 has two real roots with one located in the interval (c, b) and the other one in the interval (b, a). Proof: Let f (x) = 3x2 − 2(a + b + c)x + ab + bc + ca. Since ∆ = [−2(a + b + c)]2 − 4 × 3 · (ab + bc + ca) = 4(a + b + c)2 − 12(ab + bc + ca) = 2(2a2 + 2b2 + 2c2 − 2ab − 2bc − 2ca) = 2[(a − b)2 + (b − c)2 + (c − a)2 ]. Since a > b > c, then ∆ > 0. Thus the equation has two distinct real roots. f (a) = 3a2 − 2(a + b + c)a + ab + bc + ca = a2 − ca + bc − ab = (a − c)(a − b) > 0. f (b) = 3b2 − 2(a + b + c)b + ab + bc + ca = b2 − ab − bc + ca = (b − a)(b − c) < 0. f (c) = 3c2 − 2(a + b + c)c + ab + bc + ca = c2 − ca − bc + ab = (c − a)(c − b) > 0. The two x-intercepts of the graph are in the interval (c, b) and (b, a). Thus the equation has two real roots with one located in the interval (c, b) and the other one in the interval (b, a). 6.34 Let p be a real number, and the graph of the quadratic function y = x2 −2px−p has two distinct x-intercepts A(x1 , 0), B(x2 , 0). (1) Show 2px1 +x22 +3p > 0. (2) If the distance between the two points A and B is not larger than |2p − 3|. Find the maximum value of p. (1) Proof: According to the relationship between roots and coefficients, we have x1 + x2 = 2p, x1 x2 = −p. From the above equations, we have x22 = 2px2 + p. Since ∆ = (−2p)2 − 4(−p) = 4p2 + 4p > 0, then 2px1 + x22 + 3p = 2px1 + (2px2 + p) + 3p = 2p(x1 + x2 ) + 4p = 4p2 + 4p > 0. (2) Solution: AB = |x2 − x1 | = (x2 + x1 )2 − 4x2 x1 = 4p2 + 4p |2p − 3|. Squaring both sides to generate 4p2 + 4p 4p2 − 12p + 9. Solving the inequality to obtain 9 9 . p . Thus the maximum value of p is 16 16. Download free eBooks at bookboon.com 107.

(108) ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions. 6.35 As shown in Figure 4, A, B, C are three points on the graph of the function y = log 1 x. Their x-coordinates are t, t + 2, t + 4 (t 1), respectively. (1) Let the 3 area SABC be S, find S = f (t). (2) Determine the monotonicity of S = f (t). (3) Find the maximum value of S = f (t). Solution: (1) Starting from the points A, B, C, we draw lines AA , BB ,CC perpendicular to x-axis with the foot points A , B , C . Then S = Strapezoid AA B B + 1 Strapezoid BB C C − Strapezoid AA C C = −[log 1 t + log 1 (t + 2)](t + 2 − t) × − [log 1 (t + 3 3 3 2 2 1 1 t + 4t 2)+log 1 (t+4)]·[(t+4)−(t+2)] +[log 1 t+log 1 (t+4)](t+4−t)× = log 1 = 3 3 3 3 2 2 (t + 2)2 4 log3 (1 + 2 ), (t 1). t + 4t 2 (2) Let u = t + 4t be an increasing function on the interval [1, +∞), and u 5. Let 4 9 v = 1 + be a decreasing function on the interval [5, +∞), and 1 < v , which 5 u 9 means S = log3 v is an increasing function on the interval (1, ]. Hence, the composite redefine your future 5 4 function S = f (t) = log3 (1 + 2 ) is a decreasing function on the interval [1, +∞). AxA globAl grAduAte t + 4t. > Apply now. - © Photononstop. progrAm 2015. axa_ad_grad_prog_170x115.indd 1. 19/12/13 16:36. Download free eBooks at bookboon.com 108. Click on the ad to read more.

(109) 6.35 As shown in Figure 4, A, B, C are three points on the graph of the function y = log 1 x. Their x-coordinates are t, t + 2, t + 4 (t 1), respectively. (1) Let the 3 area SABC be S,EXERCISE find SBOOK = fII(t). (2) Determine the monotonicity of S = f (t). (3) ELEMENTARY ALGEBRA Functions Find the maximum value of S = f (t). Solution: (1) Starting from the points A, B, C, we draw lines AA , BB ,CC perpendicular to x-axis with the foot points A , B , C . Then S = Strapezoid AA B B + 1 Strapezoid BB C C − Strapezoid AA C C = −[log 1 t + log 1 (t + 2)](t + 2 − t) × − [log 1 (t + 3 3 3 2 1 1 t2 + 4t 2)+log 1 (t+4)]·[(t+4)−(t+2)] +[log 1 t+log 1 (t+4)](t+4−t)× = log 1 = 3 3 3 3 (t + 2)2 2 2 4 log3 (1 + 2 ), (t 1). t + 4t (2) Let u = t2 + 4t be an increasing function on the interval [1, +∞), and u 5. Let 4 9 v = 1 + be a decreasing function on the interval [5, +∞), and 1 < v , which u 5 9 means S = log3 v is an increasing function on the interval (1, ]. Hence, the composite 5 4 function S = f (t) = log3 (1 + 2 ) is a decreasing function on the interval [1, +∞). t + 4t (3) According to (2), S reaches the maximum value when t = 1, and its maximum 9 value is Smax = f (1) = log3 = 2 − log3 5. 5 6.36 . Given f (2x − 1) = x2. (x ∈ R), find the range of the function f [f (x)].. Solution 1 : Let A(2x − 1)2 + B(2x − 1) + C ≡ x2 . Comparing the coefficients to 1 1 1 1 1 1 1 obtain A = , B = , C = . Thus f (x) = x2 + x + = (x + 1)2 . Hence 4 2 4 4 2 4 4 1 1 1 1 2 2 2 2 f [f (x)] = [ (x + 1) + 1] = [(x + 1) + 4] . Then f [f (x)] (The equation 4 4 64 4 1 holds when x = −1). This means the minimum value of f [f (x)] is . The range of the 4 function f [f (x)] is [1/4, +∞). t+1 Solution 2 : Let 2x − 1 = t, then x = . From the given condition, we have 2 t+1 2 x+1 2 1 f (t) = ( ) . Notice that t ∈ R and x ∈ R, then f (x) = ( ) = (x + 1)2 , 2 2 4 1 1 1 1 2 2 2 2 f [f (x)] = [ (x + 1) + 1] = [(x + 1) + 4] . Thus f [f (x)] (The equation holds 4 4 64 4 when x = −1). The range of the function f [f (x)] is [1/4, +∞). 6.37 Let A = [−1, 0] ∪ (1, 2] and B = [0, 2]. The map f : A → B maps x to y = |x|. Show that f : A → B is a one-to-one map and find its inverse map. Proof: Let x1 , x2 ∈ A, and x1 = x2 , with f (x1 ) = |x1 |, f (x2 ) = |x2 |. If |x1 | = |x2 |, according the given condition x1 = x2 , then x1 = −x2 . This means x1 , x2 are opposite numbers. Without loss of generality, let x1 > 0, x2 < 0, according to the given A, we have x1 ∈ (1, 2], x2 ∈ [−1, 0). Obviously, |x1 | > 1, |x2 | 1 for this case. Thus |x1 | = |x2 |, which is contradicting to the given condition. On the other hand, let y1 ∈ B, then 0 y1 2. Notice B = [0, 1] ∪ (1, 2]. Then y1 ∈ [0, 1], or y1 ∈ (1, 2]. When y1 ∈ [0, 1], then there is a unique element in A, x1 = −y1 , corresponding to it. When y1 ∈ (1, 2], then there is a unique element in A, x2 = y1 , corresponding to it. According to the property of a one-to-one map, we know y, (y ∈ (1, 2]) that f : A → B is a one-to-one map, and its inverse map is x = . −y, (y ∈ [0, 1]) Download free eBooks at bookboon.com. 6.38 . 109. Given the set A = {(x, y)|x2 +mx−y+2 = 0, x ∈ R}, B = {(x, y)|x−y+1 =.

(110) Proof: Let x1 , x2 ∈ A, and x1 = x2 , with f (x1 ) = |x1 |, f (x2 ) = |x2 |. If |x1 | = |x2 |, according the given condition x1 = x2 , then x1 = −x2 . This means x1 , x2 are opposite numbers. Without loss of generality, let x1 > 0, x2 < 0, according to the given A, ELEMENTARY ALGEBRA EXERCISE BOOK II Functions we have x1 ∈ (1, 2], x2 ∈ [−1, 0). Obviously, |x1 | > 1, |x2 | 1 for this case. Thus |x1 | = |x2 |, which is contradicting to the given condition. On the other hand, let y1 ∈ B, then 0 y1 2. Notice B = [0, 1] ∪ (1, 2]. Then y1 ∈ [0, 1], or y1 ∈ (1, 2]. When y1 ∈ [0, 1], then there is a unique element in A, x1 = −y1 , corresponding to it. When y1 ∈ (1, 2], then there is a unique element in A, x2 = y1 , corresponding to it. According to the property of a one-to-one map, we know y, (y ∈ (1, 2]) that f : A → B is a one-to-one map, and its inverse map is x = . −y, (y ∈ [0, 1]) 6.38 Given the set A = {(x, y)|x2 +mx−y+2 = 0, x ∈ R}, B = {(x, y)|x−y+1 = 0, 0 x 2}. If A ∩ B = φ, find the range of the real number m. y = x2 + mx + 2 Solution: The problem is equivalent to the following: the equation system y =x+1 has solutions on the interval [0, 2]. This means x2 + (m − 1)x + 1 = 0 has solutions on the interval [0, 2]. Let f (x) = x2 + (m − 1)x + 1, since f (0) = 1, then the parabolic curve y = f (x) passes through the point (0, 1). Thus the parabolic curve y = f (x) has 1 an  x-intercept in the interval [0, 2]. It is equivalent to f (2) = 22 + 2(m − 1) + 1 0 2 ∆ = (m − 1) − 4 0   1−m 2 or . 0< <2  22  f (2) = 2 + 2(m − 1) + 1 > 0 3 3 2 we have − < m −1. Thus the range of m is 1 we have m − . From , From , 2 2 (−∞, −1]. 6.39 If the functions f (x) and g(x) defined for the real numbers satisfy f (0) = 0, and for arbitrary x, y ∈ R, g(x − y) = f (x)f (y) + g(x)g(y) holds. Show [f (x)]2012 + [g(x)]2012 1. Proof : Let x = y, then [f (x)]2 + [g(x)]2 = g(0)(∗). In (∗), let x = 0, then [f (0)]2 + [g(0)]2 = g(0). Since f (0) = 0, then [g(0)]2 = g(0). Thus g(0) = 0 or g(0) = 1. (1) If we substitute g(0) = 0 into (∗), then [f (x)]2 + [g(x)]2 = g(0). Thus f (x) = g(x) = 0. Hence [f (x)]2012 + [g(x)]2012 1. (2) If we substitute g(0) = 1 into (∗), then f (x)2 + g(x)2 = 1. Thus |f (x)| 1, |g(x)| 1. Hence [f (x)]2012 [f (x)]2 , [g(x)]2012 [g(x)]2 . Therefore [f (x)]2012 + [g(x)]2012 [f (x)]2 + [g(x)]2 1. 6.40 Let f (x) = x2 + px + q, A = {x|x = f (x)}, B = {x|f [f (x)] = x}. (1) Show A ⊆ B. (2) If A = {−1, 3}, find B. (1) Proof: Let x0 is an arbitrary element in the set A which means x0 ∈ A. Since A = {x|x = f (x)}, then x0 = f (x0 ). Thus f [f (x0 )] = f (x0 ) = x0 . Thus x0 ∈ B. Therefore A ⊆ B. (2) Solution: Since A = {−1, 3} = {x|x2 + px + q = x}, then the equation x2 + (p −1 and 3. By applying the Vieta’s theorem, we have − 1)x + q = 0 has two roots p = −1 −1 + 3 = −(p − 1) ⇒ . Then f (x) = x2 − x − 3. Thus the elements q = −3 (−1) × 3 = q 2 2 free eBooks at bookboon.com in the set B is the roots ofDownload the equation f [f (x)] = x. This means x− −more Click on(xthe−ad to 3) read 2 2 2 2 110 (x − x − 3) − 3 = x. Simplifying the equation to generate (x − x − 3) − x = 0√⇒ (x2 − x − 3 + x)(x2 − x − 3 − x) = 0 ⇒ (x2 − 3)(x2 − 2x − 3) = 0. Thus x1 = − 3,.

(111) . f (2) = 22 + 2(m − 1) + 1 > 0 3 3 2 we have − < m −1. Thus the range of m is 1 we have m − . From , From , 2 2 (−∞, −1].ALGEBRA EXERCISE BOOK II ELEMENTARY Functions 6.39 If the functions f (x) and g(x) defined for the real numbers satisfy f (0) = 0, and for arbitrary x, y ∈ R, g(x − y) = f (x)f (y) + g(x)g(y) holds. Show [f (x)]2012 + [g(x)]2012 1. Proof : Let x = y, then [f (x)]2 + [g(x)]2 = g(0)(∗). In (∗), let x = 0, then [f (0)]2 + [g(0)]2 = g(0). Since f (0) = 0, then [g(0)]2 = g(0). Thus g(0) = 0 or g(0) = 1. (1) If we substitute g(0) = 0 into (∗), then [f (x)]2 + [g(x)]2 = g(0). Thus f (x) = g(x) = 0. Hence [f (x)]2012 + [g(x)]2012 1. (2) If we substitute g(0) = 1 into (∗), then f (x)2 + g(x)2 = 1. Thus |f (x)| 1, |g(x)| 1. Hence [f (x)]2012 [f (x)]2 , [g(x)]2012 [g(x)]2 . Therefore [f (x)]2012 + [g(x)]2012 [f (x)]2 + [g(x)]2 1. 6.40 Let f (x) = x2 + px + q, A = {x|x = f (x)}, B = {x|f [f (x)] = x}. (1) Show A ⊆ B. (2) If A = {−1, 3}, find B. (1) Proof: Let x0 is an arbitrary element in the set A which means x0 ∈ A. Since A = {x|x = f (x)}, then x0 = f (x0 ). Thus f [f (x0 )] = f (x0 ) = x0 . Thus x0 ∈ B. Therefore A ⊆ B. (2) Solution: Since A = {−1, 3} = {x|x2 + px + q = x}, then the equation x2 + (p −1 and 3. By applying the Vieta’s theorem, we have − 1)x + q = 0 has two roots p = −1 −1 + 3 = −(p − 1) . Then f (x) = x2 − x − 3. Thus the elements ⇒ q = −3 (−1) × 3 = q in the set B is the roots of the equation f [f (x)] = x. This means (x2 − x − 3)2 − (x2 − x − 3) − 3 = x. Simplifying the equation to generate (x2 − x − 3)2 − x2 = 0√⇒ (x2 −√ x − 3 + x)(x2 − x − 3 − x) = 0 ⇒ (x2 −√3)(x2 −√2x − 3) = 0. Thus x1 = − 3, x2 = 3, x3 = −1, x4 = 3. Therefore B = {− 3, −1, 3, 3}. 1 2 1000 4x , compute f ( ) + f( ) + · · · + f( ). x 4 +2 1001 1001 1001 4 2 41−x = , then f (x) + f (1 − x) = = x Solution: Since f (1 − x) = 1−x x 4 + 2 4 + 2 × 4 4 + 2 2 1 2 1000 1 1000 4x + x = 1. Thus f ( ) + f( ) + · · · + f( ) = f( ) + f( )+ x 4 +2 4 +2 1001 1001 1001 1001 1001 999 500 501 2 ) + f( ) + · · · + f( ) + f( ) = 1 + 1 +· · · + 1 = 500. f( 1001 1001 1001 1001 6.41 . Let f (x) =. 500. 3 3 x x π Given vectors a = (cos x, sin x), b = (cos , − sin ), and x ∈ [0, ]. 2 2 2 2 2 3 If the minimum value of f (x) = ab − 2λ|a + b| is − . Find the value of λ. 2 x 3 3 x x 3x x 3x Solution: ab = (cos x, sin x)(cos , − sin ) = cos cos − sin sin = cos 2x, 2 2 2 2 2 2 2 2 3x x 2 3x x 2 − sin ) = 2(1 + cos 2x) = 2| cos x|. Since |a + b| = (cos + cos ) + (sin 2 2 2 2 π x ∈ [0, ], then cos x > 0. Thus |a + b| = 2 cos x. Hence f (x) = ab − 2λ|a + b| = 2 π cos 2x−4λ cos x+2λ2 −2λ2 = 2(cos x−λ)2 −1−2λ2 . Since x ∈ [0, ], then 0 cos x 1. 2 Download free eBooks at bookboon.com (1) If λ < 0, f (x) reaches the minimum value −1 if and only if cos x = 0. It is contra111 dicting to the given condition. 6.42 .

(112) 1001. 1001. 1001. 1001. 500. π x x 3 3 Given vectors a = (cos x, sin x), b = (cos , − sin ), and x ∈ [0, ]. ELEMENTARY ALGEBRA EXERCISE BOOK II Functions 2 2 2 2 2 3 If the minimum value of f (x) = ab − 2λ|a + b| is − . Find the value of λ. 2 3 x x 3x x 3x x 3 cos − sin sin = cos 2x, Solution: ab = (cos x, sin x)(cos , − sin ) = cos 2 2 2 2 2 2 2 2 x x 3x 3x + cos )2 + (sin − sin )2 = 2(1 + cos 2x) = 2| cos x|. Since |a + b| = (cos 2 2 2 2 π x ∈ [0, ], then cos x > 0. Thus |a + b| = 2 cos x. Hence f (x) = ab − 2λ|a + b| = 2 π cos 2x−4λ cos x+2λ2 −2λ2 = 2(cos x−λ)2 −1−2λ2 . Since x ∈ [0, ], then 0 cos x 1. 2 (1) If λ < 0, f (x) reaches the minimum value −1 if and only if cos x = 0. It is contradicting to the given condition. (2) If 0 < λ 1, f (x) reaches the minimum value −1 − 2λ2 if and only if cos x = λ. 1 3 Since −1 − 2λ2 = − , then λ = . 2 2 (3) If λ > 1, f (x) reaches the minimum value 1 − 4λ if and only if cos x = 1. Since 3 5 1 − 4λ = − , then λ = . It is contradicting to λ > 1. 2 8 As a conclusion, λ = 12 . 6.42 . 6.43 Given f (x) = x2 + (lg a + 2)x + lg b, and f (−1) = −2. f (x) 2x holds for all x ∈ R. Evaluate a + b. Solution: Since f (−1) = −2 ⇒ 1−(lg a+2)x+lg b = −2 ⇒ lg a = lg 10b ⇒ a = 10b. Since f (x) 2x holds for all x ∈ R, then x2 +(lg a+2)x+lg b 2x ⇒ x2 +(lg a)x+lg b 0. Since the efficient of x2 is 1 > 0, then ∆ = lg2 a − 4 lg b 0 ⇒ lg2 a − 4(lg a − 1) 0 ⇒ (lg a − 2)2 0. Since (lg a − 2)2 0, then (lg a − 2)2 = 0. Thus a = 100, b = 10. Therefore a + b = 110.. 6.44 Let the equation of the curve C is y = x3 − x. The graph of C shifts t (t = 0) units to the positive x direction and then shifts s units to the positive y direction. Then we obtain the curve C1 . (1) Write the equation of the curve C1 . (2) t s Show the curves C and C1 are symmetric about the point A( , ). (3) If there is only 2 2 t3 one intersection between the curves C and C1 , show s = − t. 4 (1) Solution: The equation of the curve C1 is y = (x − t)3 − (x − t) + s (t = 0). (2) Proof: We choose an arbitrary point B1 (x1 , y1 ) on the curve C. Assume B2 (x2 , y2 ) x1 + x2 t y1 + y2 s t s = , = . Thus is the symmetric point of B1 about A( , ), then 2 2 2 2 2 2 x1 = t − x2 , y1 = s − y2 . By substituting them into the equation of C, we have s − y2 = (t − x2 )3 − (t − x2 ). This means y2 = (x2 − t)3 − (x2 − t) + s. Then we show that the point B2 (x2 , y2 ) is on the curve C1 . Similarly, we can show that the t s symmetric point of C1 about A( , ) is on the curve C. Therefore the curves C and 2 2 t s C1 are symmetric about the point A( , ). 2 2 (3) Proof: Since the curves C and C have a unique intersection point, then the equa1 3 y =x −x has only one solution. By eliminating y, we tion system y = (x − t)3 − (x − t) + s obtain 3tx2 − 3t2 x + (t3 − t − s) = 0. Then ∆ = 9t4 − 12t(t3 − t − s) = 0. This means t3 3 t(t3 − 4t − 4s) = 0. Since t = − t (t = 0). 0, then t − 4t − 4s = 0. Thus s = Download free eBooks at bookboon.com 4 112. √.

(113) s 1− y2 = (t2− x12 ) − (t −2 x2 ). This means y2 = (x2 − t) − (x2 − t) + s. Then we 3 sshow − y2that = (tthe − xpoint (t2 (x −2x, 2y). This means y2 = − t)3 − (x t) + s. Then we 2 ) −B 2 − C1(x . 2Similarly, we can show that the 2 ) is on the curve show that the point B2 (x2 , y2 ) ist on the curve C . Similarly, we can show that the s 1 symmetric point of C1 about A( t , s ) is on the curve C. Therefore the curves C and 2 , 2 ) is on the curve C. Therefore the curves Functions C and symmetricALGEBRA point of C1 about ELEMENTARY EXERCISE BOOK A( II t s 2 2 C1 are symmetric about the point A( t , s ). C1 are symmetric about the point A( 2 , 2 ). 2 a unique intersection point, then the equa(3) Proof: Since the 3curves C and C1 2have (3) Proof: Since the curves C and C have a unique intersection point, then the equa1 y =x −x has only one solution. By eliminating y, we tion system y = x3 − x 3 tion system y = (x − t)3 − (x − t) + s has only one solution. By eliminating y, we (x(t−3 − t) t − s ∆ = 9t4 − 12t(t3 − t − s) = 0. This means obtain 3tx2 − 3ty2= x+ − (x s) − = t) 0. +Then 2 2 3 obtain 3tx − 3t x + (t − t − s) = 0. Then ∆ = 9t4 − 12t(t3 − t −t3s) = 0. This means t(t3 − 4t − 4s) = 0. Since t = 0, then t3 − 4t − 4s = 0. Thus s = t3 − t (t = 0). t(t3 − 4t − 4s) = 0. Since t = 0, then t3 − 4t − 4s = 0. Thus s = 4 − t (t = 0). 4 √ 6.45 Given f (x) = lg(x + √x2 + 1), show f (x) and f −1 (x) are both odd func−1 2 6.45 tions. Given f (x) = lg(x + x + 1), show f (x) and f (x) are both odd functions. √ √ Proof: Let y = f (x) = lg(x + √x2 + 1) ⇒ x + √x2 + 1 = 10yy ⇒ x22 + 1 = x22 + x2 + 1) ⇒ x + x2 +10 Proof: Let y = f (x) = lg(x +10−y 1 x=− 10 10y − 10−x⇒ x + 1 = x + 2y y y −y x −x , (x ∈ R). Since (x) = 102y − 2x · 10y ⇒ x = 10 − 10 . Thus f −1 10 − −1 2 2 10 , (x ∈ R). Since . Thus f x = (x) = 10 − 2x · 10 ⇒ √ √ 2 1 f (−x) = lg(−x + √x2 + 1) = lg2 √ 1) = −f (x), then f (x) = − lg(x + √x2 + 1 f (−x) = lg(−x + x2 + 1) = lg √x2 + 1 + x = − lg(x + x2 + 1) = −f (x), then f (x) x2 + 1 + x is an odd function. −x x is an odd function. 10x − 10−x 10 − 10 −1 −x x Since f −1 (−x) = 10 − 10 = − 10x − 10−x = −f (x), then f −1 (x) is an odd function 2 2 =− = −f (x), then f −1 (x) is an odd function Since f (−x) = on R. 2 2 on R. x+1−u 6.46 Given f (x) = x + 1 − u (u ∈ R). (1) Is the grapy of 6.46 Given f (x) = u − x (u ∈ R). (1) Is the grapy of u− y = f (x) centrally symmetric? If it xis centrally symmetric, please point y = f (x) centrally symmetric? If of it fis(x) centrally please point metric center. (2) Find the range for x ∈symmetric, [u + 1, u + 2]. metric center. (2) Find the range of f (x) for x ∈ [u + 1, u + 2].. the the out out. function function its symits sym-. Need help with your dissertation? Get in-depth feedback & advice from experts in your topic area. Find out what you can do to improve the quality of your dissertation!. Get Help Now. Go to www.helpmyassignment.co.uk for more info. Download free eBooks at bookboon.com 113. Click on the ad to read more.

(114) ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions. 1 x+1−u = −1 − , u−x x−u 1 then the graph of y = f (x) can be obtained by shifting the graph of g(x) = − in the x 1 horizontal direction. Since the graph of g(x) = − is centrally symmetric about the x original point, then the grapy of the function y = f (x) is centrally symmetric about (u, −1), the symmetric center is (u, −1). x+1−u u+1−x 3 x+1−u 3 u+2−x (2) Since f (x)+2 = +2 = , f (x)+ = + = . u−x u−x 2 u−x 2 2(u − x) u+1−x u+2−x (x − u − 1)(x − u − 2) 3 · = . On the Thus [f (x) + 2][f (x) + ] = 2 u−x 2(u − x) 2(u − x)2 other hand, since x ∈ [u + 1, u + 2], (u − x)2 > 0, then (x − u − 1) 0, (x − u − 2) 0. 3 3 Thus [f (x) + 2][f (x) + ] 0. Hence −2 f (x) − . 2 2 Solution: (1) From the given condition, we have f (x) =. 1+t t +i , t ∈ R, t = −1, t = 0}, N = {z|z = 6.47 If M = {z|z = 1+t t √ 2[cos(arcsin t) + i · cos(arccos t)], t ∈ R, |t| 1}. Determine how many elements in M ∩ N.  t   x= 1+t Solution: The points of the set M are on the curve M : 1 + t (t ∈ R, t =   y= t −1, t = 0). x = √ 2(1 − t2 ) (t ∈ R, |t| 1). The points of the set N are on the curve N : y = 2t The √ general equations of the curves M and N are xy = 1(x = 0, 1), x2 + y 2 = 2, (0 x 2), respectively. Thus the x-coordinates of the intersection points of M and N 1 satisfy x2 + 2 = 2. This means x = ±1. Obviously, M ∩ N = φ. Hence, M ∩ N has x zero elements.  < a)   0,x(x −a 2 1 ) , (a x b) . (1) Show f (x) ( always 6.48 Given f (x) =  4  a−b 1, (x > b) a+b a+b holds for arbitrary x . (2) Is there a real number c such that f (c) ? If 2 2 c exists, find its range. If c does not exist, please explain the reason.. Download free eBooks at bookboon.com 114.

(115) ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions. a+b 1 a+b 1 if , x b, then f (x) = (x − a)2 is 2 2 (a − b)2 1 a+b 1 2 2 if x > b, then an increasing function. Thus f (x) − a) . ( = (a − b)2 2 4 1 1 a+b f (x) = 1 > . Thus f (x) holds when x . 4 4 2 a+b (2) Solution: When a + b 0, since f (x) 0, then f (c) always holds for 2 a+b arbitrary real number c. When > 1, since f (x) 1, then c does not exist. When 2 a+b (c − a)2 a+b 0< 1, if c > b, then f (x) = 1. If a < c b, then f (c) = . We 2 2 (a − b) 2 a+b a+b a+b + a c b. Hence f (c) when c (b − a) + a. get that (b − a) 2 2 2 a+b , does not After all, when a + b > 2, the real number c, which satisfies f (c) 2 a+b exist. When a + b 0, the real number c, which satisfies f (c) , exists. When 2 a+b a+b + a, +∞] makes f (c) hold. 0 < a + b 2, c ∈ [(b − a) 2 2 (1) Proof: When x . 6.49 lg x + lg y}.. Determine how many elements in the set {(x, y)| lg(x3 +. 1 3 1 y + ) = 3 9. Solution: From the properties of the logarithmic function, we get x > 0, y > 0. Then 1 1 1 1 1 1 1 1 3 lg(x3 + y 3 + ) = lg x+lg y ⇒ x3 + y 3 + = xy ⇒ x3 + y 3 + 3 x3 ( y 3 ) = xy, 3 9 9 3 9 3 9 3 1   x3 = 9 . Thus x = 3 1 , y = 3 1 . Hence there the equation holds if and only if 1 1  9 3  y3 = 3 9 3 1 3 1 , ) in the set. Therefore the set has one element. is a unique point ( 9 3 6.50 . Given the functions f (x) = 3x − 1 and g(x) = log9 (3x + 1). (1) If 1 f −1 (x) g(x), find the range D of x. (2) Let the function H(x) = g(x) − f −1 (x). 2 What is the range of H(x) when x ∈ D?. −1 g(x) ⇒ Solution: (1) Since f (x) = 3x − 1, then f −1 (x) = log3 (x + 1). Since f (x) 2 (x + 1) 3x + 1 . log3 (x + 1) log9 (3x + 1) ⇒ log9 (x + 1)2 log9 (3x + 1) ⇒ x+1>0. Download free eBooks at bookboon.com 115.

(116) ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions. Then 0 x 1. Thus x ∈ D = [0, 1]. 1 1 (2) H(x) = g(x) − f −1 (x) = log9 (3x + 1) − log3 (x + 1) = log9 (3x + 1) − log9 (x + 1) = 2 2 3x + 1 2 3x + 1 log9 , x ∈ [0, 1]. Let t = = 3− . Obviously, t is increasing on the x+1 x+1 x+1 interval [0, 1], then 1 t 2. Hence 0 H(x) log9 2. Therefore the range of H(x) is {y|0 y log9 2}. 6.51 The function f (x) = log2 (x + m), and the numbers f (0), f (2), f (6) form an arithmetic sequence. (1) Find the value of the real number m. (2) If a, b, c are distinct positive numbers, and they form a geometric sequence, determine the order of f (a) + f (c) and 2f (b). Solution: (1) Since f (0), f (2), f (6) form an arithmetic sequence, then 2 log2 (2 + m) = log2 m + log2 (m + 6) ⇒ (m + 2)2 = m(m + 6), and m > 0. Thus m = 2. (2) Since f (x) = log2 (x + 2), then 2f (b) = 2 log2 (b + 2) = log2 (b + 2)2 , f (a) + f (c) = log2 (a √ + 2) + log2 (c + 2) = log2 [(a + 2)(c + 2)]. Thus (a + 2)(c + 2) = ac + 2(a + c) + 4 > ac + 4 ac + 4 = b2 + 4b + 4 = (b + 2)2 . Hence log2 [(a + 2)(c + 2) > log2 (b + 2)2 . This means log2 (a + 2) + log2 (c + 2) > 2 log2 (b + 2). Therefore f (a) + f (c) > 2f (b). 6.52 Given the function f (x) = x−k +k+2 (k ∈ Z) and f (2) < f (3). (1) Evaluate k. (2) Is there a positive number p such that the range of the function g(x) = 1 − pf (x) + (2p − 1)x is [−4, 17 ] when x ∈ [−1, 2]. If p exists, find its range. If 8 p does not exist, please explain the reason. 2. Solution: (1) Since f (2) < f (3), then −k 2 + k + 2 > 0. Thus k 2 − k − 2 < 0. Since k ∈ Z, then k = 0 or k = 1. By 2020, wind could provide one-tenth of our planet’s 2 electricity needs. Already today, SKF’s)2innovative know(2) From (1), we have f (x) = x2 , g(x) = 1 − px2 + (2p + 4p4p+1 . − 1)x = −p(x − 2p−1 2p how is crucial to running a large proportion of the 2 1 17 2p − 1 1 world’s4p wind+ turbines. = generating . Wecosts have 2 When ∈ [−1, 2] which means p ∈ [ , +∞), thenUp to 25 % of the relatepto= mainte4p 8 2p 4 nance. These can be reduced dramatically thanks to our 1 1 automatic g(−1)for=on-line −4,condition g(2) monitoring = −1. and When or p = . Since p ∈ [ , +∞), then p = 2. Thus systems lubrication. We help make it more economical to create 4 8 cleaner, cheaper energy 2pout−of1thin air. 2p − 1 ∈ (−∞, −1) ∈ (2, +∞) Since p > 0, then such p does not exist. When By sharing our experience, expertise, and creativity, 2p 2p industries can boost performance beyond expectations. 17 1 Therefore we need the best employees who can , g(2) = meet −4.thisThus such p does not exist. which means p ∈ (0, ), then g(−1) = challenge! 8 4 After all, p = 2.. Brain power. The Power of Knowledge Engineering. 6.53 . x Suppose f (x) is an increasing function defined on (0, +∞), and f ( ) = y. Plug into The Power of Knowledge Engineering. Visit us at www.skf.com/knowledge. Download free eBooks at bookboon.com 116. Click on the ad to read more.

(117) (2) Since f (x) = log2 (x + 2), then 2f (b) = 2 log2 (b + 2) = log2 (b + 2)2 , f (a) + f (c) = log2 (a √ + 2) + log2 (c + 2) = log2 [(a + 2)(c + 2)]. Thus (a + 2)(c + 2) = ac + 2(a + c) + 4 > ac + 4 ac + 4 = b2 + 4b + 4 = (b + 2)2 . Hence log2 [(a + 2)(c + 2) > log2 (b + 2)2 . This means logALGEBRA + log2 (c + 2) ELEMENTARY BOOK II > 2 log2 (b + 2). Therefore f (a) + f (c) > 2f (b). Functions 2 (a + 2)EXERCISE 6.52 Given the function f (x) = x−k +k+2 (k ∈ Z) and f (2) < f (3). (1) Evaluate k. (2) Is there a positive number p such that the range of the function g(x) = 1 − pf (x) + (2p − 1)x is [−4, 17 ] when x ∈ [−1, 2]. If p exists, find its range. If 8 p does not exist, please explain the reason. 2. Solution: (1) Since f (2) < f (3), then −k 2 + k + 2 > 0. Thus k 2 − k − 2 < 0. Since k ∈ Z, then k = 0 or k = 1. 2 (2) From (1), we have f (x) = x2 , g(x) = 1 − px2 + (2p − 1)x = −p(x − 2p−1 )2 + 4p4p+1 . 2p 1 4p2 + 1 17 2p − 1 ∈ [−1, 2] which means p ∈ [ , +∞), then = . We have p = 2 When 2p 4 4p 8 1 1 or p = . Since p ∈ [ , +∞), then p = 2. Thus g(−1) = −4, g(2) = −1. When 8 4 2p − 1 2p − 1 ∈ (2, +∞) Since p > 0, then such p does not exist. When ∈ (−∞, −1) 2p 2p 17 1 , g(2) = −4. Thus such p does not exist. which means p ∈ (0, ), then g(−1) = 4 8 After all, p = 2. x Suppose f (x) is an increasing function defined on (0, +∞), and f ( ) = y 1 ) 2. f (x) − f (y), f (2) = 1. Solve the inequality f (x) − f ( x−3 x x Proof: Since 1 = f (2), then 2 = f (2)+f (2). And f ( ) = f (x)−f (y) ⇒ f (y)+f ( ) = y y x f (x). Let y = 2, = 2, we have x = 2y = 4. Then f (2) + f (2) = f (4). Thus f (4) = 2. y 1 ) 2 ⇒ f [x(x − 3)] f (4). Since f (x) is an increasing funcHence f (x) − f ( x−3   x(x − 3) 4 x>0 . Solving the equation system to obtain tion on (0, +∞), we have  x−3>0 3 < x 4. Therefore the solution set of the inequality is {x|3 < x 4}. 6.53 . 6.54 Let f (x) = x2 + ax + b cos x, {x|f (x) = 0, x ∈ R} = {x|f (f (x)) = 0, x ∈ R} = φ. Find all values of a and b which satisfy the conditions. Solution: Let x0 ∈ {x|f (x) = 0, x ∈ R}, then b = f (0) = f (f (x0 )) = 0. Thus f (x) = x(x + a), f (f (x)) = f (x)(f (x) + a) = x(x + a)(x2 + ax + a). Obviously, a = 0 satisfies the problem. If a = 0, since the roots of x2 + ax + a = 0 are neither 0 nor −a. Hence x2 + ax + a = 0 does not have real roots. Therefore ∆ = a2 − 4a < 0. We have 0 < a < 4. After all, all a and b which satisfy the conditions are 0 a < 4, b = 0. 6.55 Suppose a, b, c ∈ R, and their absolute values are no more than 1. Show ab + bc + ca + 1 0. Proof: We introduce the function f (a) = ab + bc + ca + 1. Since a ∈ [−1, 1], then we only need to show f (−1) 0, f (1) 0. Then we show f (a) 0. When a = −1, then f (−1) =Download −b + bcfree − ceBooks + 1 =at(bbookboon.com − 1)(c − 1). Since b, c ∈ [−1, 1], then 117 (b − 1)(c − 1) 0. Thus f (−1) 0. When a = 1, then f (1) = b + bc + c + 1 = (b + 1)(c + 1). Since b, c ∈ [−1, 1], then.

(118) satisfies the problem. If a = 0, since the roots of x2 + ax + a = 0 are neither 0 nor −a. Hence x2 + ax + a = 0 does not have real roots. Therefore ∆ = a2 − 4a < 0. We have 0 < a < 4. After all, all a and b which satisfy the conditions are 0 a < 4, b = 0. ELEMENTARY ALGEBRA EXERCISE BOOK II Functions 6.55 Suppose a, b, c ∈ R, and their absolute values are no more than 1. Show ab + bc + ca + 1 0. Proof: We introduce the function f (a) = ab + bc + ca + 1. Since a ∈ [−1, 1], then we only need to show f (−1) 0, f (1) 0. Then we show f (a) 0. When a = −1, then f (−1) = −b + bc − c + 1 = (b − 1)(c − 1). Since b, c ∈ [−1, 1], then (b − 1)(c − 1) 0. Thus f (−1) 0. When a = 1, then f (1) = b + bc + c + 1 = (b + 1)(c + 1). Since b, c ∈ [−1, 1], then (b + 1)(c + 1) 0. Thus f (1) 0. Therefore ab + bc + ca + 1 0. a · 2x − 1 (a ∈ R) is an odd function. (1) Evalu2x + 1 ate a. (2) Find the inverse function of f (x). (3) For arbitrary k ∈ (0, +∞), solve the 1+x . inequality of f −1 (x) > log2 k a−1 = 0. Solution: (1) Since the f (x) is an odd function, we have f (0) = 0. Then 2 6.56 . The function f (x) =. Thus a = 1. While f (x) + f (−x) =. 2x − 1 1 − 2x 2x − 1 2−x − 1 + = + = 0 satisfies 2x + 1 2−x + 1 2 x + 1 1 + 2x. that f (x) is an odd function. 2 2x − 1 = 1− x , then (2) Since y = f (x) = x 2 +1 2 +1 1+x , (−1 < x < 1). f −1 (x) = log2 1−x 1 + x 1 + x > log2 ⇒⇒ (3) Since f −1 (x) = log2 1−x k. 2x =. 1+y , (−1 < y < 1). Thus 1−y. 1+x 1+x x>1−k > ⇒ . 1−x k −1 < x < 1 −1 < x < 1 When 0 < k < 2, the solution set of the inequality is {x|1 − k < x < 1}. When k 2, the solution set of the inequality is {x| − 1 < x < 1}.. 6.57 The function f (x) is defined for real numbers, and f (x + 2)f (1 √ − f (x)) = 1 + f (x). (1) Show that f (x) is a periodic function. (2) If f (1) = 2 + 3, find the value of f (2013). (1) Proof: Since f (x + 2)(1 − f (x)) = 1 + f (x) holds on R, then f (1) = 1. Thus (x) 1 + 1+f 1 + f (x + 2) 1 + f (x) 1−f (x) . Hence f (x + 4) = f [(x + 2) + 2] = = = f (x + 2) = (x) 1 − f (x) 1 − f (x + 2) 1 − 1+f 1−f (x) 1 1 2 =− . On the other hand, f (x + 8) = f [(x + 4) + 4] = − = f (x), −2f (x) f (x) f (x + 4) then f (x) is a periodic function √ with the period 8. (2)Solution: Since f (1)√= 2 + 3 and f (x) is a periodic function with the period 8,√we have f (8k + 1) = 2 + 3, (k ∈ Z). Thus f (2009) = f (251 × 8 + 1) = f (1) = 2 + 3, √ 1 1 √ = 3 − 2. f (2013) = f (2009 + 4) = − = f (2009) 2+ 3 6.58 . Find the range of the function y = x +. √. x2 − 3x + 2.. √ Download free Solution: From the given condition, weeBooks have at bookboon.com x2 − 3x + 2 = y − x 0. Then we 2 square both sides of the equation to obtain118x − 3x + 2 = y 2 − 2xy + x2 . This means 3 y2 − 2.

(119) then f (x) is a periodic function √ with the period 8. (2)Solution: Since f (1)√= 2 + 3 and f (x) is a periodic function with the period 8,√we have f (8k + 1) = 2 + 3, (k ∈ Z). Thus f (2009) = f (251 × 8 + 1) = f (1) = 2 + 3, √ 1 1 √ = 3 − 2. = f (2013) =ALGEBRA f (2009EXERCISE + 4) =BOOK − II ELEMENTARY Functions f (2009) 2+ 3 6.58 . Find the range of the function y = x +. √. x2 − 3x + 2.. √ Solution: From the given condition, we have x2 − 3x + 2 = y − x 0. Then we square both sides of the equation to obtain x2 − 3x + 2 = y 2 − 2xy + x2 . This means y2 − 2 3 . Ad(2y − 3)x = y 2 − 2. From the above equation, we have y = and x = 2 2y − 3 y 2 − 3y + 2 (y − 1)(y − 2) y2 − 2 ⇒ 0⇒ 0. ditionally, since y x, then y 2y − 3 2y − 3 y − 32 3 Then 1 y < or y 2. 2 y2 − 2 y2 − 2 Now we have y0 ∈ [2, +∞) arbitrarily. Let x0 = 0 , then x0 − 2 = 0 −2 = 2y0 − 3 2y0 − 3 (y0 − 2)2 2 0. Thus x0 2. Hence x0 − 3x0 + 2 0, and y0 = x0 + x20 − 3x0 + 2. 2y0 − 3 y2 − 2 y2 − 2 3 , then x0 − 1 = 0 −1 = We have y ∈ [1, ) arbitrarily. Let x0 = 0 2 2y0 − 3 2y0 − 3 (y0 − 1)2 0. Thus x0 1. Hence x20 − 3x0 + 2 0, and y0 = x0 + x20 − 3x0 + 2. 2y0 − 3 √ 3 As a conclusion, the range of the function y = x + x2 − 3x + 2 is [1, ) ∪ [2, +∞). 2 6.59 Given the coefficient of the quadratic term of the quadratic function f (x) is a, and the solution of the inequality f (x) > −2x is (1, 3). (1) If the function f (x) + 6a = 0 has two equal roots, find the analytic form of f (x). (2) If the maximum value of f (x) is a positive number, find the range of a. Solution: (1) Since the solution of the inequality f (x) > −2x is (1, 3), we let f (x)+2x = 1 a(x−1)(x−3) and a < 0. Then f (x) = a(x−1)(x−3)−2x = ax2 −(2+4a)x+3a . 2 1 we have ax −(2+4a)x+9a = 0 . 2 By substituting the equation f (x)+6a = 0 into , 2 2 2 has two equal roots, then ∆ = [−(2 + 4a)] − 36a = 0. Thus Since the equation 1 1 1 1 we have a = 1 or a = − . Since a < 0, then a = − . By substituting a = − into , 5 5 5 1 6 3 f (x) = − x2 − x − . 5 5 5 1 + 2a 2 a2 + 4a + 1 2 and a < 0, (2) Since f (x) = ax − 2(1 + 2a)x + 3a = a(x − ) − a a   a2 + 4a + 1 a2 + 4a + 1 − > 0 , then then the maximum value of f (x) is − . Since  a<0 a a √ √ √ a < −2√− 3 or −2 + 3 < a < 0. Therefore the range of a is (−∞, −2 − 3) or (−2 + 3, 0) when the maximum value of f (x) is positive. 6.60 Given f (x) = (xn + c)m , g(x) = (axm + 1), h(x) = (bxn + 1), and f (x) ≡ g(x)h(x) where m, n are both positive integers. Compute |a + b + c|. Solution: Since f (x) ≡ g(x)h(x), then (xn + c)m ≡ (axm + 1)(bxn + 1). Comparing the leading terms, we have mn = m + n ⇒ (m − 1)(n − 1) = 1. Since m, n are both positive integers, then m = 2, n =  2. Thus (x2 + c)2 ≡ (ax2 + 1)(bx2 + 1) ⇒  ab = 1 Download eBooks at bookboon.com a + b = 2c . ThenClick a = on 1, bthe = ad 1, cto=read 1 ormore b)x2 +free 1⇒ x4 + 2cx2 + c2 ≡ abx4 + (a + 119 2 c =1.

(120) 2 2y0 − 3 2y − 3 0 (y0 − 1)2 0. Thus x0 1. Hence x20 − 3x0 + 2 0, and y0 = x0 + x20 − 3x0 + 2. 2y0 − 3 √ 3 As a conclusion, range of the function y = x + x2 − 3x + 2 is [1, ) ∪ [2, +∞). ELEMENTARY ALGEBRA the EXERCISE BOOK II Functions 2 6.59 Given the coefficient of the quadratic term of the quadratic function f (x) is a, and the solution of the inequality f (x) > −2x is (1, 3). (1) If the function f (x) + 6a = 0 has two equal roots, find the analytic form of f (x). (2) If the maximum value of f (x) is a positive number, find the range of a. Solution: (1) Since the solution of the inequality f (x) > −2x is (1, 3), we let f (x)+2x = 1 a(x−1)(x−3) and a < 0. Then f (x) = a(x−1)(x−3)−2x = ax2 −(2+4a)x+3a . 2 1 we have ax −(2+4a)x+9a = 0 . 2 By substituting the equation f (x)+6a = 0 into , 2 2 2 Since the equation has two equal roots, then ∆ = [−(2 + 4a)] − 36a = 0. Thus 1 1 1 1 we have a = 1 or a = − . Since a < 0, then a = − . By substituting a = − into , 5 5 5 1 3 6 f (x) = − x2 − x − . 5 5 5 1 + 2a 2 a2 + 4a + 1 2 (2) Since f (x) = ax − 2(1 + 2a)x + 3a = a(x − ) − and a < 0, a  a  a2 + 4a + 1 a2 + 4a + 1 − > 0 , then then the maximum value of f (x) is − . Since  a<0 a a √ √ √ a < −2√− 3 or −2 + 3 < a < 0. Therefore the range of a is (−∞, −2 − 3) or (−2 + 3, 0) when the maximum value of f (x) is positive. 6.60 Given f (x) = (xn + c)m , g(x) = (axm + 1), h(x) = (bxn + 1), and f (x) ≡ g(x)h(x) where m, n are both positive integers. Compute |a + b + c|. Solution: Since f (x) ≡ g(x)h(x), then (xn + c)m ≡ (axm + 1)(bxn + 1). Comparing the leading terms, we have mn = m + n ⇒ (m − 1)(n − 1) = 1. Since m, n are both positive integers, then m = 2, n =  2. Thus (x2 + c)2 ≡ (ax2 + 1)(bx2 + 1) ⇒  ab = 1 a + b = 2c . Then a = 1, b = 1, c = 1 or x4 + 2cx2 + c2 ≡ abx4 + (a + b)x2 + 1 ⇒  2 c =1 a = −1, b = −1, c = −1. Hence |a + b + c| = 3. 6.61 Given the function f (x) = ax2 + 4x + b (a < 0, a, b ∈ R), the two real roots of the equation f (x) = 0 with respect to x are x1 , x2 , and the two real roots of the equation f (x) = x with respect to x are α, β. (1) If |α − β| = 1, Find the relation formula between a and b. (2) If a and b are both negative integers, and |α − β| = 1, find the analytic expression of f (x). (3) If α < 1 < β < 2, show (x1 + 1)(x2 + 1) < 7. Solution: (1) Since f (x) = ax2 +4x + b and f (x) = x, then ax2 + 3x + b = 0. 3   α + β = −   a b . Then a2 + 4ab = 9. From the given condition, we have αβ =     |α − β|a = 1 (2) Since a, b are both negative integers, then a + 4b is also a negative integer, and a + 4b −5. Since a2 + 4ab = 9, then a(a + 4b) = 9. Thus a = −1, a + 4b = −9. Then b = −2. Hence f (x) = −x2 + 4x − 2. (3)Let g(x) = ax2 +3x+b, the sufficient and necessary condition of α < 1 < β < 2 then Download free eBooks at bookboon.com b 4 g(1) > 0 a+b+3>0 120 is , that is . Since x1 + x2 = − , x1 x2 = , then g(2) < 0 4a + b + 6 < 0 a a.

(121) Solution: (1) Since f (x) = ax2 +4x + b and f (x) = x, then ax2 + 3x + b = 0. 3   α + β = −   a b . Then a2 + 4ab = 9. From the ALGEBRA given condition, we IIhave ELEMENTARY EXERCISE BOOK Functions αβ =     |α − β|a = 1 (2) Since a, b are both negative integers, then a + 4b is also a negative integer, and a + 4b −5. Since a2 + 4ab = 9, then a(a + 4b) = 9. Thus a = −1, a + 4b = −9. Then b = −2. Hence f (x) = −x2 + 4x − 2. (3)Let g(x) = ax2 +3x+b, then the sufficient and necessary condition of α < 1 < β < 2 b 4 g(1) > 0 a+b+3>0 is , that is . Since x1 + x2 = − , x1 x2 = , then g(2) < 0 4a + b + 6 < 0 a a 10 g(1) − 73 g(2) −6a + b − 4 b 4 3 = . (x1 +1)(x2 +1)−7 = x1 x2 +(x1 +x2 )−6 = − −6 = a a a a Since g(1) > 0, g(2) < 0, a < 0, then (x1 + 1)(x2 + 1) − 7 < 0. Thus we show that (x1 + 1)(x2 + 1) < 7. 6.62 Let f (x) = ax2 + bx + c (a > b > c), f (1) = 0, g(x) = ax + b. (1) Show the graphs of y = f (x) and y = g(x) have two intersection points. (2) Let the two intersection points of the graphs of y = f (x) and y = g(x) are A, B, their projections on x-axis are A1 , B1 . Find the range of |A1 B1 |. a+b+c=0 (1) Proof: From the given condition, we have ⇒ a > 0, c < 0. Let a>b>c ax2 +bx+c = ax+b, then ax2 +(b−a)x+c−b = 0, ∆ = (b−a)2 −4a(c−b) = (b+a)2 −4ac. Since a > 0, c < 0, then ∆ > 0. Thus the graphs of y = f (x) and y = g(x) have two intersection points. (2) Solution: Let the two roots of ax2 + (b − a)x + c − b = 0 are x1 , x2 , then a−b c−b x1 + x2 = , x1 x2 = . Thus |A1 B1 | = |x1 − x2 | = (x1 + x2 )2 − 4x1 x2 = a a √ (b + a)2 − 4ac c2 − 4ac c c . Since b = −(a + c), then |A1 B1 | = = ( )2 − 4( ) = a a a a c a > −(a + c) . ( − 2)2 − 4. Since b = −(a + c) and a > b > c, a > 0, c < 0, then −(a + c) > c a √ 1 3 c Solve the equation system, then −2 < < − . Thus < |A1 B1 | < 2 3. Hence the a 2 2 3 √ range of |A1 B1 | is ( , 2 3). 2 6.63 The function f (x) is defined for real numbers, and for arbitrary x, y ∈ R, f (x) + f (y) = f (x + y) − xy − 1. Since f (1) = 1, Find the integer number n such that f (n) = n. Solution: For the given equation, let y = 1, then f (x + 1) = f (x) + x + 2. Thus f (x + 1) > x + 1 when x ∈ Z. For the given equation, let x = y = 0, then 2f 0) = f (0) − 1. Thus f (0) = −1. Let x = −1, y = 1, then f (−1) + f (1) = f (0). Thus f (−1) = f (0) − f (1) = −2. Let x = −2, y = 1, then f (−2) + f (1) = f (−1) + 1. Thus f (−2) = f (−1) − f (1) + 1 = −2. When x ∈ Z, x −3, then f (x) > x. Therefore the integer number n such that f (n) = n is n = 1 or n = −2. 6.64 Given the odd function f (x) defined for real numbers, and f (x) > 0 when x 0. Does there exist a real number λ such that f (cos 2θ−3)+f (4λ−2λ cos θ) > f (0) π hold for all θ ∈ [0, ]? If yes, find its range. If no, please explain the reason. 2 Download free eBooks at bookboon.com 121 Solution: Since f (x) is an odd function and is an increasing function on [0, +∞),.

(122) 2f 0) = f (0) − 1. Thus f (0) = −1. Let x = −1, y = 1, then f (−1) + f (1) = f (0). Thus f (−1) = f (0) − f (1) = −2. Let x = −2, y = 1, then f (−2) + f (1) = f (−1) + 1. Thus f (−2) = f (−1) − f (1) + 1 = −2. When x ∈ Z, x −3, then f (x) > x. Therefore the ELEMENTARY ALGEBRA BOOK II Functions integer number n EXERCISE such that f (n) = n is n = 1 or n = −2. 6.64 Given the odd function f (x) defined for real numbers, and f (x) > 0 when x 0. Does there exist a real number λ such that f (cos 2θ−3)+f (4λ−2λ cos θ) > f (0) π hold for all θ ∈ [0, ]? If yes, find its range. If no, please explain the reason. 2 Solution: Since f (x) is an odd function and is an increasing function on [0, +∞), then f (x) is an increasing function on R, and f (0) = 0. Since f (cos 2θ − 3) + f (4λ − 2λ cos θ) > f (0) = 0 ⇒ f (cos 2θ − 3) > −f (4λ − 2λ cos θ) = f (2λ cos θ − 4λ) ⇒ cos 2θ − 3 > 2λ cos θ − 4λ ⇒ cos2 θ − λ cos θ + 2λ − 2 > 0. If the inequality hold for ar2 − cos2 θ π . bitrary θ ∈ [0, ], we should have λ is lager than the maximum value of y = 2 2 − cos θ 2 − t2 ⇒ t2 − yt + 2y − 2 = 0, ∆ = y 2 − 8y + 8 0. Let t = cos θ ∈ [0, 1], then y = 2 − t √ √ Thus ymax = 4 − 2 2. Hence λ ∈ (4 − 2 2, +∞). 6.65 If the quadratic function f (x) = ax2 + bx + 1 (a, b ∈ R, a > 0), and the function f (x) = x has two real roots x1 , x2 . (1) If x1 < 2 < x2 < 4, let the symmetric axis of f (x) is x = x0 . Show x0 > −1. (2) If |x1 | < 2, |x2 − x1 | = 2, find the range of b. 1, then the two roots of g(x) = 0 are (1) Proof: Let g(x) = f (x) − x = ax2 + (b − 1)x + g(2) < 0 4a + 2b − 1 < 0 ⇒ x1 , x2 . Since a > 0 and x1 < 2 < x2 < 4, we have g(4) > 0 16a + 4b − 3 > 0. Challenge the way we run. EXPERIENCE THE POWER OF FULL ENGAGEMENT… RUN FASTER. RUN LONGER.. RUN EASIER…. READ MORE & PRE-ORDER TODAY WWW.GAITEYE.COM. 1349906_A6_4+0.indd 1. 22-08-2014 12:56:57. Download free eBooks at bookboon.com 122. Click on the ad to read more.

(123) ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions.    3+3· b − 3 <0 1 b b 2a 4a 1 + , 2 we have By < 1. Thus x0 = − > −1. ⇒ 3 b  2a 2a  −4 − 2 · 2 + <0 2a 4a 4 b−1 2 (2) Solution: (x1 − x2 )2 = (x1 + x2 )2 − 4x1 x2 = ( ) − = 4 ⇒ 2a + 1 = a a 1 (b − 1)2 + 1. Since x1 x2 = > 0, then the signs of x1 , x2 are same. Thus |x1 | < a 2 < x2 0 < x1 < x2 < −2 < x1 < 0 2, |x2 −x1 | = 2 are equivalent to or 2a + 1 = (b − 1)2 + 1 2a + 1 = (b − 1)2 + 1    g(−2) < 0  g(2) < 0 1 g(0) > 0 g(0) > 0 or . Then we have b < ⇒   4 2a + 1 = (b − 1)2 + 1 2a + 1 = (b − 1)2 + 1 7 or b > . 4 6.66 Given a linear function f (x) = kx + h (k = 0). f (m) > 0 and f (n) > 0 when m < n. (1) Show f (x) > 0 holds for arbitrary x ∈ (m, n). (2) By applying the condition of (1), show if a, b, c ∈ R and |a| < 1, |b| < 1, |c| < 1, then ab + bc + ca > −1. Proof: (1) When k > 0, then the function f (x) = kx + h is an increasing function on x ∈ R, m < x < n, f (x) > f (m) > 0. When k < 0, then the function f (x) = kx + h is a decreasing function on x ∈ R, m < x < n, f (x) > f (n) > 0. Thus f (x) > 0 holds for arbitrary x ∈ (m, n). (2) Rewrite ab + bc + ca + 1 as (b + c)a + bc + 1. Introduce the function f (x) = (b + c)x + bc + 1. Then f (a) = (b + c)a + bc + 1. f (a) = bc + 1 = −c2 + 1 when b + c = 0 which means b = −c. Since |c| < 1, then f (a) = −c2 + 1 > 0. When b + c = 0, then f (x) = (b + c)x + bc + 1 is a linear function. Since |b| < 1, |c| < 1, then f (1) = b + c + bc + 1 = (1 + b)(1 + c) > 0, f (−1) = −b − c + bc + 1 = (1 − b)(1 − c) > 0. From (1), we obtain all f (a) > 0 for |a| < 1. Thus (b+c)a+bc+1 = ab+bc+ca+1 > 0. Hence we show that ab + bc + ca > −1. 6.67 Let P (x + a, y1 ), Q(x, y2 ), R(2 + a, y3 ) be the three distinct points on the graph of the inverse function of the function f (x) = 2x + a. If there is one but the only one real number x such that y1 , y2 , y3 form an arithmetic sequence, find the range of a. And Compute the area of P QR when the distance from the origin to the point R is smallest. Solution: From the given condition, P (x + a, y1 ), Q(x, y2 ), R(2 + a, y3 ) are on the graph of the function f −1 (x) = log2 (x − a). Then y1 = log2 x, y2 = log2 (x − a), x>a . y3 = 1. Since y1 + y3 = 2y2 , then log2 x + 1 = 2 log2 (x − a) ⇔ (x − a)2 = 2x From the given condition, the equation x2 − (2a + 2)x + a2 = 0 has an unique 1 root. Then ∆ = (2a + 2)2 − 4a2 = 0. Thus a = − or a 0. Additionally, 2 1 2 2 |OR| = (2 + a) + 1. When a = − , then |OR| reaches the minimum value, for which 2 √ 1 3 1 5 1 2 2 , 1). free Since |P Q| = ( − 0) + (0 + 1) = , the x = , P (0, −1), Q( , 0), R( Download eBooks at bookboon.com 2 2 2 2 2 1 123 distance from the point R to the straight line P Q : 2x − y − 1 = 0 is d = √ . Hence.

(124) . ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions. x>a . y3 = 1. Since y1 + y3 = 2y2 , then log2 x + 1 = 2 log2 (x − a) ⇔ (x − a)2 = 2x From the given condition, the equation x2 − (2a + 2)x + a2 = 0 has an unique 1 root. Then ∆ = (2a + 2)2 − 4a2 = 0. Thus a = − or a 0. Additionally, 2 1 2 2 |OR| = (2 + a) + 1. When a = − , then |OR| reaches the minimum value, for which 2 √ 1 3 1 1 5 , the x = , P (0, −1), Q( , 0), R( , 1). Since |P Q| = ( − 0)2 + (0 + 1)2 = 2 2 2 2 2 1 distance from the point R to the straight line P Q : 2x − y − 1 = 0 is d = √ . Hence 5 1 1 SP QR = |P Q| · d = . 2 4. a . (1) The graph of y = g(x) is the 2x graph of y = f (x) shifted to the right by 2 units. Find the analytic expression of g(x). (2) The graph of y = h(x) and the graph of y = g(x) are symmetric about the 1 line y = 1. Find the analytic expression of h(x). (3) Let F (x) = f (x) + h(x), the a √ minimum value of F (x) is m, and m > 2 + 7, find the range of the real number a. 6.68 . Given the function f (x) = 2x −. Solution: (1) g(x) = f (x − 2) = 2x−2 −. a 2x−2. (2) An arbitrary point on y = h(x) is P (x, y). Then the symmetric point of P about the linear line y = 1 is Q(x, 2 − y) and the point Q is on the graph y = g(x). Thus a h(x) = 2 − 2x−2 + x−2 . 2 1 x a 1 1 a 1 (3) F (x) = (2 − x ) + 2 − 2x−2 + x−2 = ( − )2x + (4a − 1) x + 2. a 2 2 a 4 2 1 1 1 When a < 0, − < 0, 4a − 1 < 0, then F (x) < 2. It is contradictory with a 4 √ m > 2 + 7. 1 1 1 2 When 0 < a , − > 0, 4a − 1 0, then F (x) is an increasing function on R. 4 a 4 Thus F (x) has no minimum value. 1 1 1 (4 − a)(4a − 1) 3 When < a < 4, − > 0, 4a − 1 > 0, then F (x) 2 + 2 = m. 4 a  4 4a 1   <a<4 √ 1 4 . Then < a < 2. Since m > 2 + 7, then (4 − a)(4a − 1)  2  >7 a 6.69 . If an odd function f (x) is defined on R, and f (x) = 2x − x2 when x 0. 1 1 Let the range of the function y = f (x), x ∈ [a, b] be [ , ], (a = b). Find the values of b a a, b.. 2 Solution: Since y = f2(x) is an odd function, then f (x) = x + 2x when x < 0. 1 1 2x − x , (x 0) . Since [a, b] and [ , ] exist at the same time, and Thus f (x) = 2 x + 2x, (x < 0) b a 1 1 a = b, then a < b, < . Thus the signs of a, b are same. b a    2a − a2 = 1 (a − 1)(a2 − a − 1) = 0 a 1 When 1 a < b, since We have ⇒ 1 (b − 1)(b2 − b − 1) = 0   2b − b2 = Download free eBooks b at bookboon.com √ 1+ 5 124 a = 1, b = . 2 .

(125) 2 Solution: Since y = f2(x) is an odd function, then f (x) = x + 2x when x < 0. 1 1 2x − x , (x 0) . Since [a, b] and [ , ] exist at the same time, and Thus f (x) = 2 + 2x, BOOK (x < II0) ELEMENTARY ALGEBRAxEXERCISE Functions b a 1 1 a = b, then a < b, < . Thus the signs of a, b are same. a b    2a − a2 = 1 (a − 1)(a2 − a − 1) = 0 a ⇒ 1 When 1 a < b, since We have (b − 1)(b2 − b − 1) = 0  2b − b2 = 1  b √ 1+ 5 a = 1, b = . 2    2a + a2 = 1 2 a ⇒ (a + 1)(a2 + a − 1) = 0 We 2 When −∞ < a < b −1, since 1 (b + 1)(b + b − 1) = 0   2b + b2 = b √ −1 − 5 have a = , b = −1. 2    2a − a2 = 1 b 3 When 0 < a < b < 1, since 1 . The equation system has no solution.   2b − b2 = a   2  2a + a = 1 b 4 When −1 < a < b < 0, since 1 . The equation system has no solu 2  2b + b = a tion. 1 5 When 0 < a < 1 −1.   √  a=1 √  −1 − 5 a= . As a conclusion, we have 1 + 5 or  b=  b = −1 2 2. 6.70 Given a function f (x), and f (x + y) − f (y) = (x + 2y + 1)x holds for all x, y, and f (1) = 0. 1 1 (1) Evaluate f (0). (2) If f (x1 ) + 2 < loga x2 holds for arbitrary x1 ∈ (0, ), x2 ∈ (0, ), 2 2 find the range of a.. This e-book is made with. SETASIGN. Solution: (1) Let x = 1 and y = 0 in the formula f (x + y) − f (y) = (x + 2y + 1)x, then f (1) − f (0) = 2. Since f (1) = 0, then f (0) = −2.. SetaPDF. PDF components for PHP developers. www.setasign.com Download free eBooks at bookboon.com 125. Click on the ad to read more.

(126) 2b + b =. a tion. 1 5 When 0 < a < 1 −1.   √  a=1 √  −1 − 5 a= . As a conclusion, we have 1 + 5 or  b=  b = −1 2 2 6.70 Given a function f (x), and f (x + y) − f (y) = (x + 2y + 1)x holds for all x, y, and f (1) = 0. 1 1 (1) Evaluate f (0). (2) If f (x1 ) + 2 < loga x2 holds for arbitrary x1 ∈ (0, ), x2 ∈ (0, ), 2 2 find the range of a. Solution: (1) Let x = 1 and y = 0 in the formula f (x + y) − f (y) = (x + 2y + 1)x, then f (1) − f (0) = 2. Since f (1) = 0, then f (0) = −2.. = 0 in the formula f (x + y) − f (y) = (x + 2y + 1)x, then f (x) − f (0) = = 0 in the formula f (x + y) − f (y) = (x + 2y + 1)x, then f (x) − f (0)1= From (1), we have f (0) = −2. Then f (x) + 2 = x22 + x. Since x1 ∈ (0, 1 ), From (1), we have f (0) = −2. Then f (x) + 2 = x + x. Since x1 ∈ (0, 2 ), 2 1 1 1 then f (x1 ) + 2 = x221 + x1 = (x1 + 1 )22 − 1 is increasing when x1 ∈ (0, 1 ). Thus then f (x1 ) + 2 = x1 + x1 = (x1 + 2 ) − 4 is increasing when x1 ∈ (0, 2 ). Thus 2 4 2 3 1 f (x1 ) + 2 ∈ (0, 3 ). To make f (x1 ) + 2 < loga x2 hold for arbitrary x1 ∈ (0, 1 ), f (x1 ) + 2 ∈ (0, 4 ). To make f (x1 ) + 2 < loga x2 hold for arbitrary x1 ∈ (0, 2 ), 4 2 1 1 x2 ∈ (0, 1 ), when a > 1, loga x2 < loga 1 , obviously the inequality does not hold. when x2 ∈ (0, 2 ), when a > 1, loga x2 < log a 22 , obviously the inequality does not hold. when 2 0<a<1 1 0 < a1 < 13 . Solving the inequality system to 0 < a < 1, loga x2 > loga 1 , then 0 < a < 1, loga x2 > loga 2 , then loga 1 3 . Solving the inequality system to loga 2 4 2 √ 3 2 4 √ 4 3 generate 4 a < 1. generate 4 a < 1. 4. (2) Let y (2) Let y (x + 1)x. (x + 1)x.. 6.71 If on the interval [1, 2], the function f (x) = x22 + px + q (p ∈ [−4, −2]) 6.71 If 1on the interval [1, 2], the function f (x) = x + px + q (p ∈ [−4, −2]) and g(x) = x + 12 reach the same minimum value on the same point. Find the maxiand g(x) = x + x2 reach the same minimum value on the same point. Find the maxix on the interval. mum value of f (x) mum value of f (x) on the interval. 1 1 x 1 x x 3√ 3 3 x 1 x 1 x Solution: When x ∈ [1, 2], g(x) = x + 2 = + + 2 3 · 3 x · x · 12 = 3 √ 3 2, and Solution: When x ∈ [1, 2], g(x) = x + x2 = 2 + 2 + x2 3 · 2 · 2 · x2 = 2 2, and x x 2 2 x 2 2 2 √ 1 3√ x 3 3 √ the equation holds if and only if x = 12 i.e. x = √ 3 2 ∈ [1, 2]. Thus g(x)min = 3 3 2. the equation holds if and only if 2 = x2 i.e. x = 2 ∈ [1, 2]. Thus g(x)min = 2 2. x 2 2 p 4q − p2 On the other hand, f (x) = x22 + px + q = (x + p )22 + 4q − p2 . Since f (x) and g(x) On the other hand, f (x) = x + px + q = (x + 2 ) + 4 . Since f (x) and g(x) 4p 2 √ 3√ 4q − p22 3 3 4q − p = 3√ reach the same minimum value on the same point, then − p = √ 2, 3 3 2. reach the same minimum value on the same point, then − 2 = 2, = 2 2. 4 √ √ 2 2 4 3√ 3 3 3 √ √ Thus p = −2√ 3 2, q = 3 3 2 + 3 4. Thus p = −2 2, q = 2 2 + 4. 3 2 of f (x) = x2 + px + q is x = √ √ 2 and the coefficient of the Since the symmetric axis Since the symmetric axis of f (x) = x2 + px + q is x = 3 2 and the coefficient of√ 3the quadratic term is√ positive, then the function f (x) is decreasing on the interval [1, √ 3 2]. quadratic term is√ 2]. 3positive, then the function f (x) is decreasing on the interval [1, √ 3 2 3 √ Thus f (1) = 1 − 2 + 3 4 when x = 1. The function f (x) is increasing on the interval Thus f (1) = 1 − 2 + 4 when x = 1. The function f (x) is increasing on the interval √ 2 √ 5√ 3 3 3 √ √ [√ 3 2, 2]. Thus f (2) = 4 − 5 3 2 + 3 4 when x = 2. Since f (2) > f (1). Therefore the [ 2, 2]. Thus f (2) = 4 − 2 2 + 4 when x = 2. Since f (2) > f (1). Therefore the √ 2 5√ 3 3 √ maximum value of f (x) on the interval [1, 2] is fmax (x) = f (2) = 4 − 5 √ 3 2 + 3 4. maximum value of f (x) on the interval [1, 2] is fmax (x) = f (2) = 4 − 2 2 + 4. 2 6.72 Suppose the function y = f (x) is a periodic function defined on R at bookboon.com 6.72 Suppose the Download functionfree y eBooks = f (x) is a periodic function defined on R with the period 5. The function y = f (x) is an odd function on the interval [−1, 1]. with the period 5. The function y = f (x) 126 is an odd function on the interval [−1, 1]. y = f (x) is a linear function on the [0, 1], and it is a quadratic function on the interval y = f (x) is a linear function on the [0, 1], and it is a quadratic function on the interval.

(127) 2 √ 5√ 3 3 [ 2, 2]. Thus f (2) = 4 − 2 + 4 when x = 2. Since f (2) > f (1). Therefore the 2 √ 5√ 3 3 maximumALGEBRA value ofEXERCISE f (x) on theII interval [1, 2] is fmax (x) = f (2) = 4 − 2 + 4. ELEMENTARY BOOK Functions 2 √ 3. 6.72 Suppose the function y = f (x) is a periodic function defined on R with the period 5. The function y = f (x) is an odd function on the interval [−1, 1]. y = f (x) is a linear function on the [0, 1], and it is a quadratic function on the interval [1, 4]. The function reaches its minimum value -5 occurring at x = 2. (1) Show f (1) + f (4) = 0. (2) Find the analytic expression of y = f (x) when x ∈ [1, 4]. (3) Find the analytic expression of y = f (x) when x ∈ [4, 9]. (1) Proof: Since f (x) is a periodic function with the period 5, then f (4) = f (4 − 5) = f (−1). Since y = f (x) is an odd function on the interval [−1, 1], then f (1) = −f (−1) = −f (4). Thus f (1) + f (4) = 0. (2) Solution: According to the given condition, we assume f (x) = a(x − 2)2 − 5(a > 0) when x ∈ [1, 4]. From (1), then f (1) + f (4) = 0, that is a(1 − 2)2 − 5 + a(4 − 2)2 − 5 = 0. Thus a = 2. Hence f (x) = 2(x − 2)2 − 5, (1 x 4). (3) Solution: Since y = f (x) is an odd function on the interval [−1, 1], then f (0) = 0. Since y = f (x) is a linear function on the [0, 1], we choose f (x) = kx (0 x 1). Since f (1) = 2(1 − 2)2 − 5 = −3, then k = −3. Thus, when 0 x 1, then f (x) = −3x. When −1 x < 0, then f (x) = −f (−x) = −3x. when −1 x 1, then f (x) = −3x. Hence, when 4 x 6, which is equivalent to −1 x − 5 1, then f (x) = f (x − 5) = −3(x − 5) = −3x + 15. when 6 < x 9, which is equivalent 2 2 to 1 < x − 5 4, then f (x) = f (x − 5) = 2[(x − 5) − 2] − 5 = 2(x − 7) − 5. −3x + 15, (4 x 6) . Thus f (x) = 2(x − 7)2 − 5, (6 < x 9) 6.73 Given the function f (x) = |x − a|, g(x) = x2 + 2ax + 1 (a is a positive constant). The y-intercepts of the graphs of f (x) and g(x) are equal. (1) Evaluate a. (2) What is the monotone increasing interval of f (x) + g(x). (3) If 4 n ∈ N ∗ , show 10f (n) · ( )g(n) < 4. 5 (1) Solution: From the given condition, we have f (0) = g(0). Then |a| = 1. Since a > 0, then a = 1. (2) Solution: f (x) + g(x) = |x − 1| + x2 + 2x + 1 . When x 1, then f (x) + g(x) = x2 + 3x. It is monotonous increasing on the interval [1 + ∞). When x < 1, then 1 f (x) + g(x) = x2 + x + 2. It is monotonous decreasing on the interval [− , 1). 2 4 g(n) Tn+1 f (n) (3) Proof: Let Tn = 10 · ( ) . Solving the inequality < 1 with Tn > 0 to 5 Tn 4 3 1 obtain 10( )2n+3 < 1. By solving the inequality, we have n > − ≈ 3.7. Since 5 2 lg 0.8 2 n ∈ N ∗ , then n 4. Thus T1 T2 T3 T4 , while T4 > T5 > T6 > · · · . 4 4 4 Therefore 10f (n) · ( )g(n) 10f (4) · ( )g(4) = 103 · ( )25 < 4. 5 5 5 6.74 If the symmetric axis of the quadratic function f (x) = x2 + bx + c is on the right of y-axis. Its y-intercept is P (0, −3), and its x-intercepts are A, B. Its vertex is Q. If the area of QAB is 8, find the analytic expression of the quadratic function.. Download free eBooks at bookboon.com 127.

(128) ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions. Solution: Let A(x1 , 0), B(x2 , 0), then x1 , x2 are the two roots of the equation x2 + bx + c = 0. Since x1 + x2 = −b, x1 · x2 = c, then AB = |x1 − x2 | = (x1 + x2 )2 − 4x1 x2 = √ 4c − b2 1√ 2 2 . Thus SQAB = b − 4c. The vertex of the parabolic curve is b − 4c · 4 2 2 √ 4c − b 1√ 2 1 Since | | which is equivalent to 8 = b − 4c·(b2 −4c). Then b2 − 4c = 4 . 4 8 the intersection of the parabolic curve and y-axis is (0, −3), then substituting c = −3 2 1 to generate b = ±2. When b = 2, then the symmetric axis is x = − = −1 into 2 which is on the left of y-axis. It is contradicting to the given conditions. Thus b = −2. When b = −2, the analytic expression of the quadratic function is y = x2 − 2x − 3. x4 + kx2 + 1 . f (a), f (b), f (c) form three sides of a x 4 + x2 + 1 triangle for arbitrary real numbers a, b, c. Find the range of k. 6.75 . If k ∈ R, f (x) =. Solution: To have f (x) > 0, we only need x4 + kx2 + 1 > 0. x4 + kx2 + 1 > 0 holds when k 0. We need ∆ = k 2 − 4 < 0 when k < 0. This means −2 < k < 0. Thus f (x) > 0 when k > −2. 1 When k = 1, then f (x) = 1 which satisfies the given conditions. (k − 1)x2 2 When k > 1, then f (x) = 1 + 4 1 and the equation holds when x = 0. x + x2 + 1 2 (k − 1)x2 (k − 1)x k+2 1+ Additionally, f (x) = 1 + 4 = and the equation holds 2 2 x +x +1 3x 3 k+2 . when x = 1. Thus fmin (x) = 1, fmax (x) = 3 According to the property that the sum of two sides is larger than the third side, we k+2 . Then k < 4. Thus 1 < k < 4 satisfies the given condition. have 2www.sylvania.com ×1> 3 k+2 3 When −2 < k < 1, similarly with , 2 we have fmax (x) = 1, fmin (x) = . Since 3 1 1 k+2 do not reinvent 1. > 1, then k > − . Therefore − < k < We 2× 2 2 3 1 the wheel we reinvent As a conclusion, the range of k is − < k < 4. 2. light.. Fascinating lighting offers an infinite spectrum of possibilities: Innovative technologies and new markets provide both opportunities and challenges. An environment in which your expertise is in high demand. Enjoy the supportive working atmosphere within our global group and benefit from international career paths. Implement sustainable ideas in close cooperation with other specialists and contribute to influencing our future. Come and join us in reinventing light every day.. Light is OSRAM. Download free eBooks at bookboon.com 128. Click on the ad to read more.

(129) Solution: To have f (x) > 0, we only need x4 + kx2 + 1 > 0. x4 + kx2 + 1 > 0 holds when k 0. We need ∆ = k 2 − 4 < 0 when k < 0. This means −2 < k < 0. Thus f (x) > 0 when k > −2. 1 When kALGEBRA = 1, then f (x) = 1II which satisfies the given conditions. ELEMENTARY EXERCISE BOOK Functions (k − 1)x2 2 When k > 1, then f (x) = 1 + 4 1 and the equation holds when x = 0. x + x2 + 1 (k − 1)x2 (k − 1)x2 k+2 1 + and the equation holds Additionally, f (x) = 1 + 4 = 2 2 x +x +1 3x 3 k+2 . when x = 1. Thus fmin (x) = 1, fmax (x) = 3 According to the property that the sum of two sides is larger than the third side, we k+2 . Then k < 4. Thus 1 < k < 4 satisfies the given condition. have 2 × 1 > 3 k+2 3 When −2 < k < 1, similarly with , 2 we have fmax (x) = 1, fmin (x) = . Since 3 k+2 1 1 2× > 1, then k > − . Therefore − < k < 1. 3 2 2 1 As a conclusion, the range of k is − < k < 4. 2. 6.76 As shown in Figure 5, in the Cartesian coordinates, the points B, C are on the negative x-axis, the points A is on the negative y-axis. The circle with diameter = AO. If AB = 10, and AC intersects the extended line of AB at the point D. CD the lengths of AO and BO are the two roots of the quadratic function x2 + kx + 48 = 0, AO > BO > 0. (1) Find the coordinates of D. (2) If the point P is on the straight line AC, and 1 AP = AC. Is the point (−2, −10) on the straight line DP ? Please explain the rea4 son. Solution: (1) Since the lengths of AO and BO are the two roots of the quadratic 1 In RtABO, AB 2 = AO2 + BO2 . function x2 + kx + 48 = 0, then AO · BO = 48 . 2 From 1 and , 2 we Since AO > BO > 0, AB = 10, then AO2 + BO2 = 100 . have AO = 8, BO = 6. = AO, then ∠BAC = ∠BCA which means CDO = AOD. Thus AB = Since CD BC = 10, AD = CO = CB + BO = 16. Then DB = AD − AB = 6. We draw DE⊥BC at the point E through the point D as shown in Figure 5, then RtDEB DE DB EB AO · DB 8×6 24 = Download = free . eBooks This means DE = = = , RtAOB. Thus at bookboon.com AO AB OB AB 10 5 48 18 18 6×6 DB · OB , then D is +6 = . Since EO129= EB + BO = = = EB = 5 5 5 10 AB.

(130) son. Solution: (1) Since the lengths of AO and BO are the two roots of the quadratic 2 1 In RtABO, AB 2 = AO2 +Functions function xALGEBRA + kx +EXERCISE 48 = 0,BOOK thenII AO · BO = 48 . BO2 . ELEMENTARY 2 From 1 and , 2 we Since AO > BO > 0, AB = 10, then AO2 + BO2 = 100 . have AO = 8, BO = 6. = AO, then ∠BAC = ∠BCA which means CDO = AOD. Thus AB = Since CD BC = 10, AD = CO = CB + BO = 16. Then DB = AD − AB = 6. We draw DE⊥BC at the point E through the point D as shown in Figure 5, then RtDEB DE DB EB AO · DB 8×6 24 RtAOB. Thus = = . This means DE = = = , AO AB OB AB 10 5 DB · OB 6×6 18 18 48 EB = = = . Since EO = EB + BO = +6 = , then D is AB 10 5 5 5 48 24 (− , ). 5 5 1 (2) Since the point P is on AC, AP = AC, A(0, −8), C(−16, 0), P (−4, −6). Let 4 the straight line passing through the points D and P is y =kx + b. Substituting the 48 24 − k+b= coordinates of D and P into the linear function to generate 5 5 . Then −4k + b = −6 96 27 k = − , b = − . Hence the straight line passing through the points D and P is 14 7 27 96 y = − x − . By substituting −2, the x-coordinate of the point (−2, −10), into the 14 7 69 equation of the straight line DP , we have y = − = −10. Thus the point (−2, −10) 7 is not on the straight line DP . 6.77 Let M = {(x, y)||xy| = 1, x > 0}, N = {(x, y)| arctan x + arccot y = π}. Show M ∪ N = M . Proof: In the set M , |xy| = 1. This means xy = 1 or xy = −1. Since x > 0, then the graph of the reciprocal function is in the first and fourth quadrants. In the set N , arctan x + arccot y = π, then arctan x = π − arccot y. Thus x = 1 1 tan(π − arccot y) = − tan(arccot y) = − = − . Then xy = −1. Thus cot(π − arccot y) y π the reciprocal function is in the second and fourth quadrants. Since − < arctan x < 2 π π π , 0 < arccot y < π, if x < 0, then − < arctan x < 0, y > 0, 0 < arccot y < . At 2 2 2 π π this time − < arctan x + arccot y < . It is contradicting to arctan x + arccot y = π. 2 2 Thus x 0. Then N = {(x, y)|xy = −1, x > 0} is the reciprocal function xy = −1 in the fourth quadrant. Hence N ⊂ M . therefore M ∪ N = M . c b a = 0 where m + + m m+1 m+2 m is a positive integer. For f (x) = ax2 + bx + c, (1) if a = 0, show af ( ) < 0. (2) m+1 If a = 0, show the equation f (x) = 0 has real roots on the interval (0, 1). 6.78 . The real numbers a, b, c satisfy. m 2 m m ) +b ) = a[a( + m+1 m+1 m+1 a b c am bm 1 Since c] . + + = 0, then c = −( + ). By substituting m+2 m+1 m m+1 m+1 m 2 m 1 1 m 1 we have af ( ]. ) − ] = a2 m 2 [ − 2 it into , ) = a2 [( 2 m+1 m+1 (m + 1) m + 2m m+1. Proof: (1) From the given condition, we have af (. Download free eBooks at bookboon.com 130.

(131) Deloitte & Touche LLP and affiliated entities.. 2 2 2 π π this time − < arctan x + arccot y < . It is contradicting to arctan x + arccot y = π. 2 2 Thus x 0. Then N = {(x, y)|xy = −1, x > 0} is the reciprocal function xy = −1 in ELEMENTARY EXERCISE BOOK Functions the fourthALGEBRA quadrant. Hence NII ⊂ M . therefore M ∪ N = M . b c a + + = 0 where m m+2 m+1 m m ) < 0. (2) is a positive integer. For f (x) = ax2 + bx + c, (1) if a = 0, show af ( m+1 If a = 0, show the equation f (x) = 0 has real roots on the interval (0, 1). 6.78 . The real numbers a, b, c satisfy. m 2 m m ) = a[a( ) +b + m+1 m+1 m+1 b c am bm a 1 Since + + = 0, then c = −( + ). By substituting c] . m+2 m+1 m m+1 m+1 1 m 2 m 1 m 1 we have af ( ) = a2 [( ) − ] = a2 m 2 [ ]. − 2 it into , 2 m+1 m+1 m+1 (m + 1) m + 2m. Proof: (1) From the given condition, we have af (. m 1 1 < 0. Thus af ( ) < 0. − 2 2 (m + 1) m + 2m m+1 m (2) If a > 0, from the conclusion of (1), we have f ( ) < 0. We only need to prove m+1 one of f (0) and f (1) is larger than 0. Since f (0) = c, f (1) = a+b+c, then f (0) > 0 holds if c > 0. We prove the conclusion. If c 0, we only need to prove f (1) = a + b + c > 0. b c a(m + 1) c(m + 1) a + + = 0, we have b = −[ + ]. Thus By applying m+2 m+1 m m+2 m a(m + 1) c(m + 1) a c f (1) = a − − +c = − . Since a > 0, m > 0, c 0, then m+2 m m+2 m f (1) > 0. We prove the conclusion. Hence when a > 0, the equation f (x) = 0 has real roots on the interval (0, 1). Similarly, when a < 0, the equation f (x) = 0 also has real roots on the interval (0, 1). Therefore, the equation f (x) = 0 has real roots on the interval (0, 1) when a = 0. Since (m + 1)2 > m2 + 2m > 0, then. 360° thinking. .. 2 − 1 (x ∈ R), If the function f (x) is the inverse function of y = x 10 + 1 1 the graph of g(x) and the graph of y = − are symmetric about the straight line x+2 x = −2. Let F (x) = f (x) + g(x). (1) Find the analytic expression and the domain of F (x). (2) Determine whether there exist two distinct points A, B on the graph of the function F (x) such that the straight line AB is perpendicular to y-axis. If yes, find the coordinates of A, B. If no, please explain the reason. 6.79 . 360° thinking. .. 360° 1−x thinking . − 1, then its inverse function is f (x) = lg. .. 2 1+x +1 1 are symmetric about the Additionally, the graph of g(x) and the graph of y = − x+2 1 1−x 1 , x ∈ (−1, 1). + . Thus F (x) = lg straight line x = −2, then g(x) = 1+x x+2 x+2 Discover thegraph truth at www.deloitte.ca/careers (2) Let the two distinct points A, B on the of the function F (x) be A(x1 , y1 ), 1 − x1 1 B(x2 , y2 ), −1 < x1 < x2 < 1, then y1 − y2 = F (x1 ) − F (x2 ) = lg − + 1 + x1 x1 + 2 1 − x2 1 − x1 1 + x2 1 1 − x1 1 + x2 1 1 lg = lg( − ) = lg( − )+( )+ 1© Deloitte + x&2Touche LLPxand2 affiliated + 2entities. 1 + x1 1 − x2 x2 + 2 1 + x1 1 − x2 x1 + 2 1 + x2 1 − x1 x2 − x1 . Since −1 < x1 < x2 < 1, then > 1, © Deloitte> 1, xLLP2 and −affiliated x1 > & Touche entities. Discover the truth at www.deloitte.ca/careers (x1 + 2)(x2 + 2) 1 + x1 1 − x2 1 − x1 1 + x2 x2 − x1 0, (x1 + 2)(x2 + 2) > 0. Thus lg( > 0. Hence ) > 0, 1 + x1 1 − x2 (x1 + 2)(x2 + 2) y1 > y2 , which means F (x) is monotone decreasing on (−1, 1). Therefore there do not Discoverfree theeBooks truth atatbookboon.com www.deloitte.ca/careers Download thethe ad straight to read more exist two distinct points A, B on the graph of the function F (x)Click suchon that 131 line AB is perpendicular to y-axis. Solution: (1) Since y =. 10x. Dis.

(132) f (1) > 0. We prove the conclusion. Hence when a > 0, the equation f (x) = 0 has real roots on the interval (0, 1). Similarly, when a < 0, the equation f (x) = 0 also has real roots on the interval (0, 1). Therefore, the equation f (x) = 0 has real roots on the ELEMENTARY EXERCISE BOOK II Functions interval (0,ALGEBRA 1) when a = 0. 2 − 1 (x ∈ R), If the function f (x) is the inverse function of y = x 10 + 1 1 are symmetric about the straight line the graph of g(x) and the graph of y = − x+2 x = −2. Let F (x) = f (x) + g(x). (1) Find the analytic expression and the domain of F (x). (2) Determine whether there exist two distinct points A, B on the graph of the function F (x) such that the straight line AB is perpendicular to y-axis. If yes, find the coordinates of A, B. If no, please explain the reason. 6.79 . 2 1−x − 1, then its inverse function is f (x) = lg . +1 1+x 1 are symmetric about the Additionally, the graph of g(x) and the graph of y = − x+2 1−x 1 1 . Thus F (x) = lg + , x ∈ (−1, 1). straight line x = −2, then g(x) = x+2 1+x x+2 (2) Let the two distinct points A, B on the graph of the function F (x) be A(x1 , y1 ), 1 − x1 1 B(x2 , y2 ), −1 < x1 < x2 < 1, then y1 − y2 = F (x1 ) − F (x2 ) = lg − + 1 + x1 x1 + 2 1 − x2 1 − x1 1 + x2 1 1 − x1 1 + x2 1 1 lg = lg( − ) = lg( − )+( )+ 1 + x2 x2 + 2 1 + x1 1 − x2 x1 + 2 x2 + 2 1 + x1 1 − x2 1 + x2 1 − x1 x2 − x1 . Since −1 < x1 < x2 < 1, then > 1, > 1, x2 − x1 > (x1 + 2)(x2 + 2) 1 + x1 1 − x2 1 − x1 1 + x2 x2 − x1 0, (x1 + 2)(x2 + 2) > 0. Thus lg( > 0. Hence ) > 0, 1 + x1 1 − x2 (x1 + 2)(x2 + 2) y1 > y2 , which means F (x) is monotone decreasing on (−1, 1). Therefore there do not exist two distinct points A, B on the graph of the function F (x) such that the straight line AB is perpendicular to y-axis. Solution: (1) Since y =. 10x. 6.80 Given the function f (x) = ax2 + (b + 1)x + b − 2 (a = 0). If there is a real number x0 such that f (x0 ) = x0 holds, then x0 is called the fixed point of f (x). (1) Find the fixed point of f (x) when a = 2, b = −2. (2) If for an arbitrary real number b, the function f (x) has two distinct fixed points. Find the range of the real number a. (3) On the condition of (2), if the x-coordinates of the points A, B on 1 the curve of f (x) are the fixed points of f (x), and the straight line y = kx + 2 2a + 1 is the perpendicular bisector of the line segment AB, find the range of the real number b. Solution: (1) Since f (x) = ax2 + (b + 1)x + b − 2, (a = 0), a = 2, b = −2, then f (x) = 2x2 − x − 4. Assume x is a fixed point, then 2x2 − x − 4 = x. Thus x1 = −1, x2 = 2. Hence the fixed points of f (x) are 2x2 − x − 4 = x are x1 = −1 and x2 = 2. (2) Since f (x) = x, then ax2 + bx + b − 2 = 0. Since the function f (x) has two distinct fixed points, then ∆ > 0. Thus b2 − 4a(b − 2) > 0 ⇒ b2 − 4ab + 8a > 0 always holds for arbitrary b ∈ R. This means ∆b < 0, 16a2 − 32a < 0. Hence 0 < a < 2. 1 is the per(3) Let A(x1 , x1 ), B(x2 , x2 ). Since the straight line y = kx + 2 2a + 1 pendicular bisector of the line segment AB, then k = −1. Let the midpoint of b AB is M (x0 , x0 ). From (2), we have x0 = − . Since M is on the straight line 2a 1 b 1 b + = y = kx + 2 , then − . By simplifying the equation, we have 2a 2a2 + √ 2a 2a + 1 1 a 1 1 2 Download free eBooks and the equation holds if and only if b=− 2 =− = − at ,bookboon.com 1 − 2a + 1 2a + a 1 132 4 2 2a · a √ √.

(133) Solution: (1) Since f (x) = ax2 + (b + 1)x + b − 2, (a = 0), a = 2, b = −2, then f (x) = 2x2 − x − 4. Assume x is a fixed point, then 2x2 − x − 4 = x. Thus x1 = −1, x2 = 2. Hence the fixed points of f (x) are 2x2 − x − 4 = x are x1 = −1 and x2 = 2. 2 (2) Since fALGEBRA (x) = x, then ax +IIbx + b − 2 = 0. Since the function f (x) has two distinct ELEMENTARY EXERCISE BOOK Functions fixed points, then ∆ > 0. Thus b2 − 4a(b − 2) > 0 ⇒ b2 − 4ab + 8a > 0 always holds for arbitrary b ∈ R. This means ∆b < 0, 16a2 − 32a < 0. Hence 0 < a < 2. 1 is the per(3) Let A(x1 , x1 ), B(x2 , x2 ). Since the straight line y = kx + 2 2a + 1 pendicular bisector of the line segment AB, then k = −1. Let the midpoint of b AB is M (x0 , x0 ). From (2), we have x0 = − . Since M is on the straight line 2a 1 b b 1 y = kx + 2 , then − = + . By simplifying the equation, we have 2a + 1 2a 2a 2a2 + √ 1 a 1 1 2 b=− 2 =− , and the equation holds if and only if =− 1 − 2a + 1 4 2a + a 2 2a · a1 √ √ 2 2 a= . Hence b ∈ [− , +∞). 2 4 6.81 The curve of the quadratic function y = x2 − (2m + 4)x + m2 − 4 passes through y-axis, and its y-intercept is below the origin. The curve of the quadratic function y passes through x-axis, and its two x-intercepts are A, B. A is on the left of B. The distances from A, B to the origin are |AO|, |OB|. |AO| and |BO| satisfy 3(|OB| − |AO|) = 2|AO| · |OB|. The intersection of the straight line y = kx + k and the curve of the quadratic function is P . the tangent function of acute angle ∠P OB is 4. (1) Find the analytic expression of the quadratic function. (2) Find the analytic expression of the linear function y = kx + k. (3) Find the coordinates of P . Solution: (1) Let the coordinates of A, B be A(x1 , 0), B(x2 , 0), and x1 < x2 , then x1 , x2 are the two real roots of the equation x2 − (2m + 4)x + m2 − 4 = 0. ∆ = [−(2m + 4)]2 − 4(m2 − 4) > 0, then m > −2. On the other hand, since the quadratic function has y-intercept below the origin, and x1 < x2 , then x1 < 0, x2 > 0. Since 3(|OB| − |AO|) = 2|AO||OB|, then 3[x2 − (−x1 )] = 2(−x1 )x2 . 3(2m + 4) = −2(m2 − 4) ⇒ m2 + 3m + 2 = 0. Thus m1 = −1, m2 = −2. Since m > −2, m1 = −1. Hence the analytic expression of the quadratic function is y = x2 − 2x − 3. (2) Since y = x2 − 2x − 3, then A(−1, 0),B(3, 0). Since the straight line y = x1 = −1 y = x2 − 2x − 3 . Thus kx + k intersects the quadratic curve, then y1 = 0 y = kx + k x2 = k + 3 . Since ∠P OB is an acute angle, then the point is on the right of yor y2 = k 2 + 4k |k 2 + 4k| = 4. axis. Thus, P is (k+3, k 2 +4k), and k+3 > 0. Since tan ∠P OB = 4, then k+3 √ √ |k 2 + 4k| = 4, we solve the equation to get k1 = 2 3, k2 = −2 3. We can When k+3 √ √ validate that k1 = 2 3 and k2 = −2 3 are both the solutions of the equation. Addi√ k 2 + 4k = −4, we solve the equation to get tionally, k2 + 3 < 0, then k1 = 2 3. When k+3 k3 = −2, k4 = −6. We can validate that k3 = −2 and k4 = −6 are both the solutions of the equation. Additionally,√k4 + 3 <√0, then k3 = −2. Thus the analytic expression of the linear function is y = 2 3x + 2 3 or y = −2x − 2. √ √ (3) From (2), the analytic expression of the linear function is y = 2 3x + 2 3 or y = −2x − 2. √ √ √ √ √ √ 2 3x + 2 3 = 4. Thus x = 2 3 + 3, y = 12 + 8 3. When y = 2 3x + 2 3, then x √ √ Hence, P is (2 3 + 3, 12 + 8 3). −2x − 2 at = bookboon.com When y = −2x − 2, then Download=free 4. eBooks Thus x 1, y = −4. Hence, P is (1, −4). −x 133.

(134) = −4, we solve the equation to get k+3 k3 = −2, k4 = −6. We can validate that k3 = −2 and k4 = −6 are both the solutions of the equation. Additionally,√k4 + 3 <√0, then k3 = −2. Thus the analytic expression of the linear function is yBOOK = 2 II 3x + 2 3 or y = −2x − 2. ELEMENTARY ALGEBRA EXERCISE Functions √ √ (3) From (2), the analytic expression of the linear function is y = 2 3x + 2 3 or y = −2x − 2. √ √ √ √ √ √ 2 3x + 2 3 = 4. Thus x = 2 3 + 3, y = 12 + 8 3. When y = 2 3x + 2 3, then x √ √ Hence, P is (2 3 + 3, 12 + 8 3). −2x − 2 When y = −2x − 2, then = 4. Thus x = 1, y = −4. Hence, P is (1, −4). −x tionally, k2 + 3 < 0, then k1 = 2 3. When. 6.82 If the quadratic curve y = f1 (x) has the origin as its vertex and passes through the point (1, 1). The distance between the two intersection points of the reciprocal curve y = f2 (x) and the diagonal line y = x is 8. Let f (x) = f1 (x) + f2 (x). (1) Find the analytic expression of f (x). (2) Show the function f (x) = f (a) with respect to x has three real roots when a > 3. (1) Solution: From the given condition, we let f1 (x) = ax2 . Since f (1) = 1, then k a = 1. Thus f1 (x) = x2 . Let f2 (x) = (k > 0). The intersection points of its graph √ √ √ x√ k, k) and B(− k, − k). Since |AB| = 8, then and the diagonal line y = x are A( √ √ √ √ 8 8 ( k + k)2 + ( k + k)2 = 8. Thus k = 8. Then f2 (x) = . Hence f (x) = x2 + . x x 8 8 8 2 2 (2) Proof: Since f (x) = f (a), then x + = a + which is equivalent to = x a x 8 8 8 −x2 + a2 + . The curve f2 (x) = and the curve f3 (x) = −x2 + a2 + are sketched a x a in Figure 6. From the figure, we observe that f2 (x) and f3 (x) have one intersection point in the third quadrant. Then f (x) = f (a) has a negative solution. 8 Since f2 (2) = 4, f3 (2) = −4 + a2 + . When a > 3, the point (2, f (2)) on the curve a Download free eBooks at bookboon.com 134.

(135) x x 8 8 8 2 = (2) Proof: Since f (x) = f (a), then x + = a + which is equivalent to x x a 8 8 8 −x2 + a2 + . The curve f2 (x) = and the curve f3 (x) = −x2 + a2 + are sketched ELEMENTARY ALGEBRA EXERCISE BOOK II Functions a x a in Figure 6. From the figure, we observe that f2 (x) and f3 (x) have one intersection point in the third quadrant. Then f (x) = f (a) has a negative solution. 8 Since f2 (2) = 4, f3 (2) = −4 + a2 + . When a > 3, the point (2, f (2)) on the curve a f3 (x) in the first quadrant is above the curve f2 (x). Thus the curves f2 (x) and f3 (x) have two intersection points in the first quadrant. Hence f (x) = f (a) has two positive solutions. Therefore f (x) = f (a) has three real solutions. 2. 6.83 If the domain of the function f (x) is symmetric about the origin but does not include zero. For an arbitrary real number x in the domain, there exist x1 , x2 in the domain such that x = x1 − x2 , f (x1 ) = f (x2 ), and the following conditions (x2 )+1 . (B) f (a) = 1 (a is a hold: (A) If 0 < |x1 − x2 | < 2a, then f (x1 − x2 ) = ff(x(x12)f)−f (x1 ) positive constant). (C) f (x) > 0 when 0 < x < 2a. Show (1) f (x) is an odd function. (2) f (x) is a periodic function. And evaluate the period. (3) f (x) is a decreasing function on (0, 4a). (x2 )+1 Proof: (1) From the given condition, we have f (x) = f (x1 − x2 ) = ff(x(x12)f)−f = (x1 ) −f (x2 − x1 ) = −f (−x). Thus f (x) is an odd function. (2) Since f (a) = 1, then f (−a) = −f (a) = −1. Thus f (−2a) = f (−a − a) = f (−a)f (a) + 1 f (x)f (−2a) + 1 = 0. If f (x) = 0, then f (x + 2a) = f [x − (−2a)] = = f (a) − f (−a) f (−2a) − f (x) 1 1 − , f (x + 4a) = f [(x + 2a) + 2a] = − = f (x). f (x) f (x + 2a) f (x)f (−a) + 1 If f (x) = 0, then f (x + a) = f [x − (−a)] = = −1, f (x + 3a) = f [(x + f (−a) − f (x) f (x + 3a)f (−a) + 1 1 = 0. = 1. f (x + 4a) = f [(x + 3a) − (−a)] = a) + 2a] = − f (−a) − f (x + 3a) f (x + a) Thus f (x + 4a) = f (x). Hence f (x) is a periodic function, and the period is 4a. (3) When 0 < x1 < x2 2a, then 0 < x2 − x1 < 2a. Thus f (x1 ) > 0, f (x2 ) 0 f (x2 )f (x1 ) + 1 = f (x2 − x1 ) > 0. Hence (f (x2 ) = −f (−2a) = 0 when x2 = 2a). and f (x1 ) − f (x2 ) f (x1 ) > f (x2 ). We will turn your CV into When 2a < x1 < x2 < 4a, then 0 < x1 − 2a < x2 − 2a < 2a. Thus f (x1 − 2a) > 1 an opportunity of a lifetime f (x2 − 2a) > 0. Hence f (x) = f [(x − 2a) + 2a] = − . Then f (x1 ) − f (x2 ) = f (x − 2a) 1 1 − + > 0. Hence f (x) is a decreasing function on (2a, 4a). f (x1 − 2a) f (x2 − 2a) Therefore, f (x) is a decreasing function on (0, 4a).. √ If the inverse function of the function f (x) = log (x + x2 − 2) (a > a √ n −n √ 3 +3 2 −1 0, a = 1) is f −1 (x) and let g(n) = (n ∈ N ∗ ). f (n + loga 2). If g(n) < 2 2 6.84 . Do you like cars? Would you like to be a part of a successful brand? We will appreciate and reward both your enthusiasm and talent. Send us your CV. You will be surprised where it can take you.. Download free eBooks at bookboon.com 135. Send us your CV on www.employerforlife.com. Click on the ad to read more.

(136) −f (x2 − x1 ) = −f (−x). Thus f (x) is an odd function. (2) Since f (a) = 1, then f (−a) = −f (a) = −1. Thus f (−2a) = f (−a − a) = f (−a)f (a) + 1 f (x)f (−2a) + 1 = 0. If f (x) = 0, then f (x + 2a) = f [x − (−2a)] = = f (a) − f (−a) f (−2a) − fFunctions (x) ELEMENTARY ALGEBRA EXERCISE BOOK II 1 1 − , f (x + 4a) = f [(x + 2a) + 2a] = − = f (x). f (x) f (x + 2a) f (x)f (−a) + 1 = −1, f (x + 3a) = f [(x + If f (x) = 0, then f (x + a) = f [x − (−a)] = f (−a) − f (x) f (x + 3a)f (−a) + 1 1 = 1. f (x + 4a) = f [(x + 3a) − (−a)] = = 0. a) + 2a] = − f (x + a) f (−a) − f (x + 3a) Thus f (x + 4a) = f (x). Hence f (x) is a periodic function, and the period is 4a. (3) When 0 < x1 < x2 2a, then 0 < x2 − x1 < 2a. Thus f (x1 ) > 0, f (x2 ) 0 f (x2 )f (x1 ) + 1 = f (x2 − x1 ) > 0. Hence (f (x2 ) = −f (−2a) = 0 when x2 = 2a). and f (x1 ) − f (x2 ) f (x1 ) > f (x2 ). When 2a < x1 < x2 < 4a, then 0 < x1 − 2a < x2 − 2a < 2a. Thus f (x1 − 2a) > 1 f (x2 − 2a) > 0. Hence f (x) = f [(x − 2a) + 2a] = − . Then f (x1 ) − f (x2 ) = f (x − 2a) 1 1 − + > 0. Hence f (x) is a decreasing function on (2a, 4a). f (x1 − 2a) f (x2 − 2a) Therefore, f (x) is a decreasing function on (0, 4a).. √ If the inverse function of the function f (x) = log (x + x2 − 2) (a > a √ n −n √ 3 +3 2 −1 0, a = 1) is f −1 (x) and let g(n) = f (n + loga 2). If g(n) < (n ∈ N ∗ ). 2 2 Find the range of a. 6.84 . √ √ 2 − 2 > 0, then the domain of f (x) is [ 2, +∞). x Thus Solution: Since x + √ √ x + x2 − 2 2. √ −1 When a > 1, the domain of f (x) is [log 2, +∞). When 0 < a < of a √ √ √ 1, the domain −1 y 2 2 1 f (x) is (−∞, loga 2]. Since y = loga (x + x − 2), √ we have x + x − 2 = a . 1 to obtain x − x2 − 2 = 2a−y . 2 From 1 Rationalizing the numerator of √ √ ay + 2a−y 2 we have x = and , . Since n + loga 2 ∈ [loga 2, +∞), then a > 1, 2 √ √ √ 2 −1 ax + 2a−x −1 (x loga 2). Hence g(n) = f (n + loga 2) = and then f (x) = 2 2 √ √ √ n √ √ a 2 1 n+loga √2 2 2 + a−n [a [ 2an + 2a−n ] = . Since g(n) < + 2a−(n+loga 2) ] = 2n 2 −n 4 2 2 3 +3 , then an + a−n < 3n + 3−n ⇒ 3n a2n + 3n < 32n an + an ⇒ (3n an − 1)(an − 3n ) < 2 1 0 ⇒ < a < 3. Since a > 1, then 1 < a < 3. 3 6.85 The straight line l with dip angle 450 passes through √ the point A(1−2) and the point B where B is in the first quadrant and |AB| = 3 2. (1) Fine the coordinates of the point B. (2) If the straight line l passes through the x2 hyperbolic curve C : 2 − y 2 = 1(a > 0) at the two points E and F , and the middle a point of the line segment EF is (4, 1). Evaluate a. (3) For an arbitrary point P in the plane, when Q is moving on the line segment AB, we denote the minimum value of |P Q| as the distance from the point P to the line segment AB. If the point P moves on the x-axis, find the function for the distance h(t) from the point P (t, 0) to the line segment AB. Download free eBooks at bookboon.com. Solution: (1) Let the equation of the straight line l is y = tan 450 x + b = x + b. Since 136 the straight line passes through the point A(1 −2), then −2 = 1 + b. Thus b = −3..

(137) 4 2 2 2 2 3n + 3−n , then an + a−n < 3n + 3−n ⇒ 3n a2n + 3n < 32n an + an ⇒ (3n an − 1)(an − 3n ) < 2 1 0 ⇒ < aALGEBRA < 3. Since a> 1, then 1 < a < 3. ELEMENTARY EXERCISE BOOK II Functions 3 6.85 The straight line l with dip angle 450 passes through √ the point A(1−2) and the point B where B is in the first quadrant and |AB| = 3 2. (1) Fine the coordinates of the point B. (2) If the straight line l passes through the x2 hyperbolic curve C : 2 − y 2 = 1(a > 0) at the two points E and F , and the middle a point of the line segment EF is (4, 1). Evaluate a. (3) For an arbitrary point P in the plane, when Q is moving on the line segment AB, we denote the minimum value of |P Q| as the distance from the point P to the line segment AB. If the point P moves on the x-axis, find the function for the distance h(t) from the point P (t, 0) to the line segment AB. Solution: (1) Let the equation of the straight line l is y = tan 450 x + b = x + b. Since the straight line passes through the point A(1 −2), then −2 = 1 + b. Thus b = −3. y =x−3 √ And y = x − 3. Let the point B = (x, y). Since and (x − 1)2 + (y + 2)2 = (3 2)2 x > 0, y > 0, then x = 4, y = 1. Hence, the coordinate of B is (4, 1).  y =x−3 1 (2) Since , then ( − 1)x2 + 6x − 10 = 0. Let E(x1 , y1 ), F (x2 , y2 ). x2 2  2 − y2 = 1 a a 6a2 Since the middle point of EF is (4, 1), then x1 + x2 = − = 8. Thus a = 2. 1 − a2 (3) Let the coordinates of an arbitrary point Q on the line segment AB is (x, x − 3). |P Q| = (t − x)2 + (x − 3)2 .. I joined MITAS because I wanted real responsibili� I joined MITAS because I wanted real responsibili�. Real work International Internationa al opportunities �ree wo work or placements. Maersk.com/Mitas www.discovermitas.com. �e G for Engine. Ma. Month 16 I was a construction Mo supervisor ina const I was the North Sea super advising and the No he helping foremen advis ssolve problems Real work he helping fo International Internationa al opportunities �ree wo work or placements ssolve pr. Download free eBooks at bookboon.com 137. �e Graduate Programme for Engineers and Geoscientists. Click on the ad to read more.

(138) ELEMENTARY ALGEBRA EXERCISE BOOK II. . Functions. . t + 3 2 (t − 3)2 ) + (1 t 4). When 2 2 |t − 3| t+3 t+3 t+3 4, i.e. −1 t 5, then |P Q|min = f ( ) = √ . When > 4, 1 2 2 2 2 i.e. t > 5, then f (x) is monotonous decreasing on [1, 4]. Thus |P Q|min = f (4) = t+3 < 1, i.e. t < −1, then f (x) is increasing on [1, 4). Thus (t − 4)2 + 1. When 2 |P Q|min = f (4) = (t −1)2+ 4. (t − 1)2 + 4 (t < −1)    |t − 3| √ (−1 t 5) As a conclusion, h(t) =  2   (t − 4)2 + 1 (t > 5) Denote f (x) =. (t − x)2 + (x − 3)2 =. 2(x −. x−2 (a > 1). Show (1) The funcx+1 tion f (x) is increasing on (−1, +∞). (2) The equation f (x) = 0 has no negative roots.. 6.86 . Let the function f (x) = ax +. Proof: (1) Let −1 < x1 < x2 , then f (x1 ) − f (x2 ) = ax1 +. x2 − 2 x1 − 2 − ax2 − = x1 + 1 x2 + 1. 3(x1 − x2 ) . Since −1 < x1 < x2 , then x1 +1 > 0, x2 +1 > 0, x1 −x2 < 0, (x1 + 1)(x2 + 1) 3(x1 − x2 ) < 0. then (x1 + 1)(x2 + 1) Since −1 < x1 < x2 and a > 1, then ax1 < ax2 , ax1 − ax2 < 0. Thus f (x1 ) − f (x2 ) < 0 which is equivalent to f (x1 ) < f (x2 ). Hence f (x) is increasing on (−1, +∞). (2) Assume x0 is a negative root of the equation f (x) = 0, and x0 = −1, then 3 − (x0 + 1) 3 x0 − 2 2 − x0 1 When −1 < = 0 ⇒ ax0 = = = − 1 . ax0 + x0 + 1 x0 + 1 x0 + 1 x0 + 1 3 x0 < 0, i.e. 0 < x0 + 1 < 1, then − 1 > 2. Since a > 1, then ax0 < 1, then x0 + 1 3 1 does not hold. When x0 < −1, i.e. x0 + 1 < 0, then the formula − 1 < −1. x0 + 1 1 does not hold. Since ax0 > 0, then ax0 < 1, then the formula As a conclusion, the equation f (x) = 0 has no negative roots. ax1 −ax2 +. 6.87 . Let f (x) = ax2 +bx+c (a = 0). If |f (0)| 1, |f (1)| 1, |f (−1)| 1. 5 Show |f (x)| holds for any x ∈ [−1, 1]. 4 1 Proof: Since f (−1) = a − b + c, f (1) = a + b + c, f (0) = c, then a = (f (1) + f (−1) + 2 1 2f (0)), b = (f (1) − f (−1)), c = f (0). Substituting a, b, c into f (x) = ax2 + bx + c and 2 x2 + x x2 − x simplifying the equation to obtain f (x) = f (1)( )+f (−1)( )+f (0)(1−x2 ). 2 2 x2 + x x2 − x | + |f (0)||1 − x2 | When −1 x < 0, then |f (x)| |f (1)|| | + |f (−1)|| 2 2 x2 − x x2 − x x2 + x x2 + x 2 |+| | + |1 − x | = −( )+( ) + (1 − x2 ) = −x2 − x + 1 = | 2 2 2 2 5 1 2 5 −(x + ) + . 4 2 4 x2 + x x2 − x When 0 x 1, then |f (x)| |f free (1)||eBooks at| bookboon.com | + |f (0)||1 − x2 | + |f (−1)|| Download 2 2 x2 − x x2 + x x2 + x −x2 138 1 5 +x 2 2 2 |+| |+|1−x | = + +1−x = −x +x+1 = −(x− )2 + |.

(139) ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions. x2 − x x2 + x )+f (−1)( )+f (0)(1−x2 ). simplifying the equation to obtain f (x) = f (1)( 2 2 x2 − x x2 + x | + |f (−1)|| | + |f (0)||1 − x2 | When −1 x < 0, then |f (x)| |f (1)|| 2 2 x2 + x x2 − x x2 − x x2 + x | |+| | + |1 − x2 | = −( )+( ) + (1 − x2 ) = −x2 − x + 1 = 2 2 2 2 1 5 5 −(x + )2 + . 2 4 4 x2 + x x2 − x When 0 x 1, then |f (x)| |f (1)|| | + |f (−1)|| | + |f (0)||1 − x2 | 2 2 x2 + x x2 − x x2 + x −x2 + x 1 5 2 2 2 | |+| |+|1−x | = + +1−x = −x +x+1 = −(x− )2 + 2 2 2 2 2 4 5 . 4 5 As a conclusion, |f (x)| holds for any x ∈ [−1, 1]. 4. 6.88 (1) If x is an arbitrary positive integer, and the values of the quadratic function f (x) = ax2 + bx + c are all integers. Show 2a, a − b, c are all integers. (2) Write the inverse statement of the above statement. Judge the inverse statement is true or false and provide your reason. Proof (1): From the given condition, the values of the quadratic function f (x) = ax2 + bx + c are all integers when x is an arbitrary positive integer, we have f (0) = c is an integer when x = 0. Similarly, when x = −1, then f (−1) = a − b + c is an integer. Thus a − b = f (−1) − c is an integer. When x = −2, then f (−2) = (−2)2 a + (−2)b + c is an integer. Thus 2a = f (−2) − 2f (−1) + c is an integer. Hence 2a, a − b, c are all integers. (2) The inverse statement is that if 2a, a − b, c are all integers, then the values of the quadratic function f (x) = ax2 + bx + c are all integers when x is an arbitrary positive integer. This inverse statement is true and the proof is provided below. If 2a, a − b, c are all integers, then f (x) = ax2 + bx + c = ax2 + ax − ax + bx + c = ax(x + 1) − (a − b)x + c. When x is an integer, then x(x + 1) is an even function. Thus 1 1 x(x + 1) is an integer. Additionally, 2a is an integer, then 2a · x(x + 1) is an integer. 2 2 Since a − b, c are integers, then −(a − b)x + c is an integer. Hence the values of the quadratic function f (x) = ax2 + bx + c are all integers when x is an arbitrary positive integer. An alternative proof: If 2a, a − b, c are all integers, then when x is an even number (let x = 2k), we have f (2k) = a(2k)2 + b(2k) + c = 2a · 2k 2 + [2a − 2(a − b)]k + c is an integer. When x is an odd number, let x = 2k−1, we have f (2k−1) = a(2k−1)2 +b(2k−1)+c = (4k 2 − 4k)a + +a + 2kb − b + c = 2a(2k 2 − 2k) + [2a − 2(a − b)] + (a − b) + c is an integer. Therefore, the inverse statement is true. 6.89 Let the function fn (x)(n ∈ N ∗ ) satisfy f1 (x) = 2, fn+1 (x) = xfn (x) + 1. Find the analytic expression of fn (x) and prove the conclusion. Solution: Since f1 (x) = 2, fn+1 (x) = xfn (x) + 1, then f2 (x) = xf1 (x) + 1 = 2x + 1, f3 (x) = xf2 (x) + 1 = 2x2 + x + 1, f4 (x) = xf3 (x) + 1 = 2x3 + x2 + x + 1, · · · . Thus we have fn (x) = 2xn−1 + xn−2 + · · · + x + 1 and the proof is provided below. (1) When n = 1, then f1 (x) = 2x = 2. Thus p(1) holds. free eBooks at bookboon.com (2) Assume p(k) holds whenDownload n = k, i.e. fk (x) = 2xk−1 + xk−2 + · · · + x + 1. When n = k + 1, we have fk+1 (x) = xfk (x)139+ 1 = x(2xk−1 + xk−2 + · · · + x + 1) + 1 = k. k−1. k. k−1.

(140) integer. Therefore, the inverse ELEMENTARY ALGEBRA EXERCISE BOOK II statement. is true.. Functions. 6.89 Let the function fn (x)(n ∈ N ∗ ) satisfy f1 (x) = 2, fn+1 (x) = xfn (x) + 1. Find the analytic expression of fn (x) and prove the conclusion. Solution: Since f1 (x) = 2, fn+1 (x) = xfn (x) + 1, then f2 (x) = xf1 (x) + 1 = 2x + 1, f3 (x) = xf2 (x) + 1 = 2x2 + x + 1, f4 (x) = xf3 (x) + 1 = 2x3 + x2 + x + 1, · · · . Thus we have fn (x) = 2xn−1 + xn−2 + · · · + x + 1 and the proof is provided below. (1) When n = 1, then f1 (x) = 2x = 2. Thus p(1) holds. (2) Assume p(k) holds when n = k, i.e. fk (x) = 2xk−1 + xk−2 + · · · + x + 1. When n = k + 1, we have fk+1 (x) = xfk (x) + 1 = x(2xk−1 + xk−2 + · · · + x + 1) + 1 = 2xk + xk−1 + · · · + x + 1. Thus fk+1 (x) = 2xk + xk−1 + · · · + x + 1. p(k + 1) holds. fn (x) = 2xn−1 + xn−2 + · · · + x + 1 holds for all n ∈ N ∗ . 6.90 If the function f (x) is defined on R, f (0) = 2008, and for any x ∈ R, f (x + 2) − f (x) 3 · 2x and f (x + 6) − f (x) 63 · 2x both hold. Compute f (2008). Solution: From the given condition, we have f (x + 2) − f (x) = −[f (x + 4) − f (x + 2)] − [f (x + 6) − f (x + 4)] + [f (x + 6) − f (x)] −3 · 2x+2 − 3 · 2x+4 + 63 · 2x = −12·2x −48·2x +63·2x = 3·2x . Since f (x+2)−f (x) 3·2x , then f (x+2)−f (x) = 3·2x . Thus f (2008) = f (2008) − f (2006) + f (2006) − f (2004) + · · · + f (2) − f (0) + f (0) = 41004 − 1 3(22006 + 22004 + · · · + 22 + 1) + f (0) = 3 + 2008 = 22008 + 2007. 4−1 6.91 If the function f (x) is defined on (0, +∞) and satisfies f (x)+f (y) = √ f (xy), and f (x) < 0 when x > 1. If the inequality f ( x2 + y 2 ) f ( xy) + f (a) always holds for any x, y ∈ (0, +∞), find the range of the real number a. x2 Solution: Let x1 , x2 ∈ (0, +∞), and x1 < x2 , then > 1. We get f (x1 ) − f (x2 ) = x1 x2 x2 x2 f (x1 ) − f (x1 ) = f (x1 ) − [f (x1 ) + f ( )] = −f ( ) > 0. Since x > 1, then f (x) < 0. x1 x1 x1 Thus f (x1 ) − f (x2 ) > 0. Hence f (x ) > f (x ). We obtain that the function f (x) 2 1 √ 2 2 2 2 is decreasing on (0, +∞). Then f ( x + y ) f ( xy) + f (a) ⇒ f( x + y ) x2 + y 2 ) x2 + y 2 √ √ √ , then . Since xy f (a xy) ⇒ x2 + y 2 a xy. Thus a √ 2 xy √ √ √ x2 + y 2 2. Hence a 2. Additionally, since a > 0, we have 0 < a 2. √ xy. Download free eBooks at bookboon.com 140. Click on the ad to read more.

(141) 2)] − [f (x + 6) − f (x + 4)] + [f (x + 6) − f (x)] −3 · 2 −3·2 + 63 · 2 = −12·2x −48·2x +63·2x = 3·2x . Since f (x+2)−f (x) 3·2x , then f (x+2)−f (x) = 3·2x . Thus f (2008) = f (2008) − f (2006) + f (2006) − f (2004) + · · · + f (2) − f (0) + f (0) = 41004 − 1 2004 ELEMENTARY II f (0) = 3 Functions 3(22006 + 2ALGEBRA + 2008 = 22008 + 2007. + ·EXERCISE · · + 22 BOOK + 1) + 4−1 6.91 If the function f (x) is defined on (0, +∞) and satisfies f (x)+f (y) = √ f (xy), and f (x) < 0 when x > 1. If the inequality f ( x2 + y 2 ) f ( xy) + f (a) always holds for any x, y ∈ (0, +∞), find the range of the real number a. x2 Solution: Let x1 , x2 ∈ (0, +∞), and x1 < x2 , then > 1. We get f (x1 ) − f (x2 ) = x1 x2 x2 x2 f (x1 ) − f (x1 ) = f (x1 ) − [f (x1 ) + f ( )] = −f ( ) > 0. Since x > 1, then f (x) < 0. x1 x1 x1 Thus f (x1 ) − f (x2 ) > 0. Hence f (x ) > f (x ). We obtain that the function f (x) 1 2 √ 2 2 2 2 is decreasing on (0, +∞). Then f ( x + y ) f ( xy) + f (a) ⇒ f( x + y ) 2 2 x +y ) x2 + y 2 √ √ √ . Since xy , then f (a xy) ⇒ x2 + y 2 a xy. Thus a √ xy 2 √ √ √ x2 + y 2 2. Hence a 2. Additionally, since a > 0, we have 0 < a 2. √ xy √ After all, the range of the real number a is (0, 2]. x (x = −1). x+1 (1) Find the intervals on which f (x) is monotone. (2) If a > b > 0, c = 6.92 . Given f (x) =. 3 f (a) + f (c) > . 4. 1 , show (a − b)b. 1 x = 1− , then f (x) is an monotone function x+1 x+1 on the interval (−∞, −1) ∪ (−1, +∞). Let −∞ < x1 < x2 < −1 ∪ −1 < x1 < 1 x 2 − x1 1 x2 < +∞, we have f (x2 ) − f (x1 ) = 1 − −1+ = > 0. x2 + 1 x1 + 1 (x1 + 1)(x2 + 1) Thus f (x2 ) > f (x1 ). Hence f (x) is a monotone increasing function on the intervals (−∞, −1) ∪ (−1, +∞). x y xy + xy + x + y xy + x + y (2) If x > y > 0, since f (x)+f (y) = + = > = x+1 y+1 xy + x + y + 1 xy + x + y + 1 f (xy + x + y). And xy + x + y > x + y. From (1), we have f (xy + x + y) > f (x + y). 1 1 4 a−b+b 2 = 2 > 0, Thus f (x) + f (y) > f (x + y). On the other hand, c = (a − b)b a ( 2 ) 4 a 4 a 4 a 3 a then a + c a + 2 = + + 2 3 · · 2 = 3. Therefore f (a) + f (c) > a 2 2 a 2 2 a 3 f (a + c) f (3) = . 4 (1) Solution: Since f (x) =. 6.93 If the monotone function f (x) defined on R satisfies f (3) = log2 3, and for any x, y ∈ R, f (x + y) = f (x) + f (y). (1) Determine f (x) is odd or even. (2) If f (k3x ) + f (3x − 9x − 2) < 0 holds for any x ∈ R, find the range of the real number k. Solution: (1) Since f (x + y) = f (x) + f (y) (x, y ∈ R). Let x = y = 0, then f (0) = f (0) + f (0). Thus f (0) = 0. Let y = −x, then f (0) = f (x) + f (−x), i.e. f (x) + f (−x) = 0. Hence f (−x) = −f (x) holds for any x, y ∈ R. Therefore f (x) is an odd function. (2) Since f (3) = log2 3 > 0, f (3) > f (0). Since f (x) is a monotone function, then f (x) eBooks at bookboon.com is an increasing function onDownload R. Andfree Since f (x) is an odd function according to (1), x x x x141 we get f (k3 ) < −f (3 − 9 − 2) = f (−3 + 9x + 2). Hence k3x < −3x + 9x + 2 ⇒ 32x − (k + 1)3x + 2 > 0 holds for any x ∈ R. Let t = 3x , the question is equivalent to.

(142) 6.93 If the monotone function f (x) defined on R satisfies f (3) = log2 3, and for any x, y ∈ R, f (x + y) = f (x) + f (y). (1) Determine f (x) is odd or even. (2) If f (k3x ) + f (3x − 9x − 2) < 0II holds for any x ∈ R, find the range of the real number ELEMENTARY ALGEBRA EXERCISE BOOK Functions k. Solution: (1) Since f (x + y) = f (x) + f (y) (x, y ∈ R). Let x = y = 0, then f (0) = f (0) + f (0). Thus f (0) = 0. Let y = −x, then f (0) = f (x) + f (−x), i.e. f (x) + f (−x) = 0. Hence f (−x) = −f (x) holds for any x, y ∈ R. Therefore f (x) is an odd function. (2) Since f (3) = log2 3 > 0, f (3) > f (0). Since f (x) is a monotone function, then f (x) is an increasing function on R. And Since f (x) is an odd function according to (1), we get f (k3x ) < −f (3x − 9x − 2) = f (−3x + 9x + 2). Hence k3x < −3x + 9x + 2 ⇒ 32x − (k + 1)3x + 2 > 0 holds for any x ∈ R. Let t = 3x , the question is equivalent to the following: t2 − (k + 1)t + 2 > 0 holds for any t > 0. Let f (t) = t2 − (k + 1)t + 2, k+1 k+1 the symmetric axis is x = . When < 0, i.e. k < −1, then f (0) = 2 > 0 2 2 k+1 0, i.e. k −1, for any t > 0, f (t) > 0 satisfies the given problem. When 2 k+1 √ 0 always holds ⇔ ⇔ −1 k −1 + 2 2. 2 ∆ = (k + 1)2 − 8 < 0 √ As a conclusion, when k < −1 + 2 2, f (k3x ) + f (3x − 9x − √2) < 0 holds for any x ∈ R. Therefore the range of the real number k is (−∞, −1 + 2 2).  (x = 0)   0 1 6.94 Let the function f (x) = − x (4k−1 |x| < 2 · 4k−1 , k ∈ Z)   2 2x (2 · 4k−1 |x| 4k , k ∈ Z) (1) What is the domain of f (x)? (2) We rotate the curve of y = f (x) around the origin π by to obtain the curve of y = g(x). Find the analytic expression of g(x). (3) For the 2 π function f (x) defined on R, if we rotate the curve of y = f (x) around the origin by 2 to obtain the same curve, show that the function f (x) = x has a unique solution. (1) Solution: Let the domain of the function f (x) is D. For any x ∈ R, x ∈ D when x = 0; when x = 0, then |x| > 0. There exists an integer k such that 4k−1 |x| 4k , then x ∈ D, which means R ⊆ D. Hence D = R. Therefore the domain of f (x) is x ∈ R. (2) Solution: We rotate an arbitrary point (x0 , y0 ) on y = f (x) around the origin by π , then the coordinates of the new point is (−y0 , x0 ). f (0) = 0 when x0 = 0, then 2 1 1 g(0) = 0. When 4k−1 |x| < 2 · 4k−1 , then f (x0 ) = − x0 . Thus g( x0 ) = x0 . Let 2 2 1 x0 = x1 , then g(x1 ) = 2x1 . Thus 2 × 4k−2 |x1 | < 4k−1 . When 2 × 4k−1 |x0 | 4k , 2 1 f (x0 ) = 2x0 . Thus g(−2x0 ) = x0 . Let −2x0 = x1 , then g(x1 ) = − x1 . Thus 2 4k |x1 | 2 × 4k .  (x = 0)   0 2x (2 · 4k−2 |x| < 4k−1 , k ∈ Z) As a conclusion, g(x) =   − 1 x (4k |x| 2 · 4k , k ∈ Z) 2 (3) Proof: Let f (0) = y0 , then (0, y0 ) is on the curve of the function f (x). We rotate π the point twice (by each time) in the same direction around the origin to obtain the 2 point (0, −y0 ) which is still on the curve of y = f (x). Since y0 = f (0) = −y0 , then y0 = 0, f (0) = 0. Hence x = 0 is a solution of the equation f (x) = x. Download bookboon.com Assume f (x0 ) = x0 , then the point free (x0 ,eBooks x0 ) is aton the curve of y = f (x). If it rotates π three around the origin to generate the 142 point (x0 , −x0 ). And the point is also on 2.

(143) g(0) = 0. When 4k−1 |x| < 2 · 4k−1 , then f (x0 ) = − x0 . Thus g( x0 ) = x0 . Let 2 2 1 x0 = x1 , then g(x1 ) = 2x1 . Thus 2 × 4k−2 |x1 | < 4k−1 . When 2 × 4k−1 |x0 | 4k , 2 1 ELEMENTARY ALGEBRA EXERCISE BOOK II Functions f (x0 ) = 2x0 . Thus g(−2x0 ) = x0 . Let −2x0 = x1 , then g(x1 ) = − x1 . Thus 2 4k |x1 | 2 × 4k .  (x = 0)   0 2x (2 · 4k−2 |x| < 4k−1 , k ∈ Z) As a conclusion, g(x) =   − 1 x (4k |x| 2 · 4k , k ∈ Z) 2 (3) Proof: Let f (0) = y0 , then (0, y0 ) is on the curve of the function f (x). We rotate π the point twice (by each time) in the same direction around the origin to obtain the 2 point (0, −y0 ) which is still on the curve of y = f (x). Since y0 = f (0) = −y0 , then y0 = 0, f (0) = 0. Hence x = 0 is a solution of the equation f (x) = x. Assume f (x0 ) = x0 , then the point (x0 , x0 ) is on the curve of y = f (x). If it rotates π three around the origin to generate the point (x0 , −x0 ). And the point is also on 2 the curve of y = f (x). Hence x0 = f (x0 ) = −x0 . Then x0 = 0. After all, the function f (x) = x has a unique solution x = 0. 6.95 Let N be the set of natural numbers, and k ∈ N . If the function f : N → N is strictly increasing, and for every n ∈ N , f (f (n)) = kn. Show for 2k k+1 an arbitrary n ∈ N , n f (n) n. k+1 2 Proof: Let a, b ∈ N , and a < b. Since f : N → N is a strictly increasing, we have f (a + 1) − f (a) > 0. Thus f (a + 1) − f (a) 1. Then f (b) − f (a) = [f (b) − f (b − 1)] + [f (b − 1) − f (b − 2)] + · · · + [f (a + 1) − f (a)] 1 + 1 + · · · + 1 = b − a. From the above conclusion, we have f (f (f (n)))−f (f (n)) f (f (n))−f (n) f (n)−n, which is equivalent to kf (n)−kn kn−f (n) f (n)−n. Since kf (n)−kn kn−f (n), k+1 2k n. n. Since kn − f (n) f (n) − n, then f (n) then f (n) 2 k+1 k+1 2k n holds. n f (n) Therefore for any n ∈ N , 2 k+1. no.1. 6.96 . STUDY AT A TOP RANKED INTERNATIONALb BUSINESS SCHOOL. (x = 0). (1) If the function |x| f (x) is increasing on (0, +∞), findyour the full range of b. (2) When b = 2, School if f (x)of< Economics, x holds Reach potential at the Stockholm on (1, +∞), find the range of (3)ofIf the the most rangeinnovative of f (x) is cities also [m, x ∈The [m,School n], in a. one in n] thewhen world. is ranked by n]. theFind Financial Times as that the a, number onesatisfy. business we call f (x) as the closed function on [m, the conditions b should. Given the function f (x) = a −. Sw. ed. en. nine years in a row. school in the Nordic and Baltic countries.. b b = a − is an increasing function. |x| x Visit us at www.hhs.se b b b(x2 − x1 ) Let 0 < x1 < x2 , then f (x1 ) < f (x2 ). Thus f (x2 )−f (x1 ) = − + = > 0. x 1 x2 x2 x 1 Since 0 < x1 < x2 , then x2 − x1 > 0, x2 x1 > 0. Thus b > 0 which is equivalent to b ∈ (0, +∞). b 2 < x holds on (1, +∞), i.e. a < x + . Since (2) When b = 2, f (x) = a − |x| x √ 2 2 2 x + 2 x = 2 2 and the equation holds if and only if x = which is equivax x x √ √ √ 2 lent to x = 2 and 2 ∈ (1, +∞). Thus the minimum value of x + is 2 2 when √ √ x x ∈ (1, +∞). Hence a 2 2. Therefore the range of a is (−∞, 2 2]. b is {x|x = 0}. (3) From the given condition, we know that the domain of f (x) = a − |x| Let f (x) be a closed function on [m,free n], eBooks then mn > 0, and b = 0. Download at bookboon.com Click on the ad to read more b 143 = a − (i) If 0 < m < n, when b > 0, then f (x) is an increasing function on |x| Solution:(1)Stockholm When x ∈ (0, +∞), then f (x) = a −.

(144) an arbitrary n ∈ N ,. k+1 2k n. n f (n) 2 k+1. Proof: Let a, b ∈EXERCISE N , and a II< ELEMENTARY ALGEBRA BOOK. b. Since f : N → N is a strictly increasing, we Functions have f (a + 1) − f (a) > 0. Thus f (a + 1) − f (a) 1. Then f (b) − f (a) = [f (b) − f (b − 1)] + [f (b − 1) − f (b − 2)] + · · · + [f (a + 1) − f (a)] 1 + 1 + · · · + 1 = b − a. From the above conclusion, we have f (f (f (n)))−f (f (n)) f (f (n))−f (n) f (n)−n, which is equivalent to kf (n)−kn kn−f (n) f (n)−n. Since kf (n)−kn kn−f (n), 2k k+1 then f (n) n. Since kn − f (n) f (n) − n, then f (n) n. k+1 2 k+1 2k n f (n) n holds. Therefore for any n ∈ N , k+1 2. b (x = 0). (1) If the function |x| f (x) is increasing on (0, +∞), find the range of b. (2) When b = 2, if f (x) < x holds on (1, +∞), find the range of a. (3) If the range of f (x) is also [m, n] when x ∈ [m, n], we call f (x) as the closed function on [m, n]. Find the conditions that a, b should satisfy. 6.96 . Given the function f (x) = a −. b b = a − is an increasing function. |x| x b b b(x2 − x1 ) > 0. Let 0 < x1 < x2 , then f (x1 ) < f (x2 ). Thus f (x2 )−f (x1 ) = − + = x2 x1 x1 x2 Since 0 < x1 < x2 , then x2 − x1 > 0, x2 x1 > 0. Thus b > 0 which is equivalent to b ∈ (0, +∞). b 2 (2) When b = 2, f (x) = a − < x holds on (1, +∞), i.e. a < x + . Since |x| x √ 2 2 2 x + 2 x = 2 2 and the equation holds if and only if x = which is equivax x x √ √ √ 2 lent to x = 2 and 2 ∈ (1, +∞). Thus the minimum value of x + is 2 2 when √ x √ x ∈ (1, +∞). Hence a 2 2. Therefore the range of a is (−∞, 2 2]. b (3) From the given condition, we know that the domain of f (x) = a − is {x|x = 0}. |x| Let f (x) be a closed function on [m, n], then mn > 0, and b = 0. b (i) If 0 < m < n, when b > 0, then f (x) = a − is an increasing function on |x| b f (m) = m . Thus the equation a − = x has two distinct roots (0, +∞). We have f (n) = n x on (0, +∞). This means x2 − ax + b = 0 has two distinct roots on (0, +∞). Hence ∆ = a2 − 4b > 0, x1 + x2 = a > 0, x1 x2 = b > 0. Then a > 0, b > 0 and a2 − 4b > 0. b −b =a+ is an decreasing function on (0, +∞). We When b < 0, then f (x) = a − |x| x  b   a− =n f (m) = n a=0 m have ⇒ ⇒ . Thus a = 0, b < 0. b f (n) = m mn = −b   a− =m n b b (ii) If m < n < 0, when b > 0, then f (x) = a − = a + is a decreasing function on |x| x  b   a+ =n a=0 f (m) = n m ⇒ . Thus a = 0, b > 0. (−∞, 0). We have ⇒ b mn = b f (n) = m   a+ =m n b b = a + is an increasing function on (−∞, 0). We When b < 0, then f (x) = a − x |x| b f (m) = m have . Thus the equation a + = x has two distinct roots on (−∞, 0). f (n) = n x 2 This means x − ax − b = 0 has two distinct roots on (−∞, 0). Hence ∆ = a2 + 4b > 0, Download free eBooks at bookboon.com x1 + x2 = a < 0, x1 x2 = −b > 0. Then a < 0, b < 0 and a2 + 4b > 0. 144 After all, a = 0, b = 0 or a < 0, b < 0 and a2 + 4b > 0 or a > 0, b > 0 and a2 − 4b > 0. Solution:(1) When x ∈ (0, +∞), then f (x) = a −.

(145)    a− b =n a=0 f (m) = n m ⇒ . Thus a = 0, b < 0. have ⇒ b mn = −b f (n) = m   a− =m ELEMENTARY ALGEBRA EXERCISE BOOK n II Functions b b = a + is a decreasing function on (ii) If m < n < 0, when b > 0, then f (x) = a − x |x|  b   a+ =n a=0 f (m) = n m ⇒ . Thus a = 0, b > 0. (−∞, 0). We have ⇒ b mn = b f (n) = m   a+ =m n b b When b < 0, then f (x) = a − = a + is an increasing function on (−∞, 0). We |x| x b f (m) = m have . Thus the equation a + = x has two distinct roots on (−∞, 0). f (n) = n x This means x2 − ax − b = 0 has two distinct roots on (−∞, 0). Hence ∆ = a2 + 4b > 0, x1 + x2 = a < 0, x1 x2 = −b > 0. Then a < 0, b < 0 and a2 + 4b > 0. After all, a = 0, b = 0 or a < 0, b < 0 and a2 + 4b > 0 or a > 0, b > 0 and a2 − 4b > 0. Thus a, b should satisfy the conditions: a = 0, b = 0 or ab > 0 and a2 − 4|b| > 0. . 6.97 The function f (t) satisfies f (x + y) = f (x) + f (y) + xy + 1 and f (−2) = −2. (1) Evaluate f (1). (2) Show f (t) > t always holds for any positive integer t larger than 1. (3) Compute the number of integers which satisfy f (t) = t, and explain the reason. (1) Solution: Let x = y = 0, then f (0) = −1. Let x = y = −1, since f (−2) = −2, then f (−2) = 2f (−1) + 2. Thus f (−1) = −2. Let x = 1, y = −1, then f (0) = f (1) + f (−1). Thus f (1) = f (0) − f (−1) = 1. (2) Solution: Let x = 1, then f (y +1) = f (y)+y +2. Thus f (y +1)−f (y) = y +2 (∗). When y ∈ N , then f (y + 1) − f (y) > 0. Since f (y + 1) > f (y) and f (1) = 1, then f (y) > 0 holds for any integer y. Thus when y ∈ N , f (y + 1) = f (y) + 1 + y + 1 > y + 1. Then f (t) > t always holds for any positive integer t larger than 1. (3) From (∗) and (1), we have f (−3) = −1, f (−4) = 1. Now we can show that f (t) > t when t −4. Since t −4, then −(t + 2) 2 > 0. From (∗), we have f (t) − f (t + 1) = −(t + 2) > 0 which is equivalent to f (−5)−f (−4) > 0, f (−6)−f (−5) > 0,· · · , f (t+1)−f (t+2) > 0, f (t) − f (t + 1) > 0. Adding the above inequalities to generate f (t) − f (−4) > 0. Thus. f (t) > f (−4) = 1. Hence t −4. Therefore, the number of integers t which satisfy f (t) = t is two, and t = 1 or t = −2.. 1 The function f (x) is defined on (−1, 1), and f ( ) = 1. f (x)−f (y) = 2 1 2xn x−y f ( 1−xy ) for x, y ∈ (−1, 1). The sequence {xn }, x1 = , xn+1 = . (1) Show f (x) 2 1 + x2n is an odd function on (−1, 1). (2) Find the analytic expression of f (xn ). (3) Is there a 1 1 1 m−8 natural number m such that + +···+ < for any n ∈ N ∗ . If f (x1 ) f (x2 ) f (xn ) 4 m exists, find its minimum value. If m does not exist, please explain the reason. 6.98 . (1) Proof: Let x = y = 0, then f (0) = 0. Let x = 0, then f (0) − f (y) = f (−y). Thus f (−y) + f (y) = 0. Hence for arbitrary x ∈ (−1, 1), f (−x) + f (x) = 0 holds. Therefore f (x) is an odd function on (−1, 1). 1 2xn (2) Solution: Since the sequence {xn }, x1 = , xn+1 = , then 0 < xn < 1. 1 + x2n 2 2xn xn − (−xn ) ] = f( Since f (xn ) − f (−xn ) = f [ ). Additionally, since f (x) is an 1 − xn (−xn ) 1 + x2n Download free eBooks at bookboon.com 1 f (xn+1 ) = 2. Since f ( ) = 1, odd function on (−1, 1), then f (xn+1 ) = 2f145(xn ). Thus f (xn ) 2 1.

(146) x−y f ( 1−xy ) for x, y ∈ (−1, 1). The sequence {xn }, x1 = , xn+1 = . (1) Show f (x) 1 + x2n 2 is an odd function on (−1, 1). (2) Find the analytic expression of f (xn ). (3) Is there a 1 1 1 m−8 for any n ∈ Functions N ∗ . If natural number m such that + +···+ < ELEMENTARY ALGEBRA EXERCISE BOOKfII(x ) f (x f (x 4 ) ) 1 2 n m exists, find its minimum value. If m does not exist, please explain the reason.. (1) Proof: Let x = y = 0, then f (0) = 0. Let x = 0, then f (0) − f (y) = f (−y). Thus f (−y) + f (y) = 0. Hence for arbitrary x ∈ (−1, 1), f (−x) + f (x) = 0 holds. Therefore f (x) is an odd function on (−1, 1). 1 2xn (2) Solution: Since the sequence {xn }, x1 = , xn+1 = , then 0 < xn < 1. 2 1 + x2n 2xn xn − (−xn ) ] = f( Since f (xn ) − f (−xn ) = f [ ). Additionally, since f (x) is an 1 − xn (−xn ) 1 + x2n 1 f (xn+1 ) = 2. Since f ( ) = 1, odd function on (−1, 1), then f (xn+1 ) = 2f (xn ). Thus f (xn ) 2 1 x1 = , then f (x1 ) = 1. Hence the sequence {f (xn )} is a geometric sequence with the 2 first term 1 and the common ratio 2. Thus f (xn ) = 2n−1 . 1 − 21n 1 1 1 1 1 1 + + ··· + = 1 + + 2 + · · · + n−1 = = (3) Solution: f (x1 ) f (x2 ) f (xn ) 2 2 2 1 − 12 1 2 − n−1 (n ∈ N ∗ ). 2 1 1 1 m−8 Assume there is a natural number m such that + +· · ·+ < for f (x1 ) f (x2 ) f (xn ) 4 m−8 1 m−8 arbitrary n ∈ N ∗ . Then 2 − n−1 < holds. Thus 2. We have m 16. 2 4 4 1 1 1 m−8 Therefore there is a natural number m, and + + ··· + < f (x1 ) f (x2 ) f (xn ) 4 holds when m 16 for arbitrary n ∈ N ∗ . The minimum value of m is 16. 6.99 . The domain of the function f (x) is R+ , for arbitrary x, y ∈ R+ , 1 f (xy) = f (x) + f (y) holds. (1) Show f ( ) = −f (x) when x ∈ R+ . (2) If f (x) < 0 x holds when x > 1, show f (x) has an inverse function. (3) Let f −1 (x) is the inverse function of f (x). Show that in the domain of f −1 (x), f −1 (x1 + x2 ) = f −1 (x1 ) · f −1 (x2 ).. Download free eBooks at bookboon.com 146. Click on the ad to read more.

(147) < holds. Thus 2. We have m 16. 2n−1 4 4 1 1 1 m−8 Therefore there is a natural number m, and + + ··· + < f (x1 ) f (x2 ) f (xn ) 4 ELEMENTARY ALGEBRA EXERCISE BOOK II Functions holds when m 16 for arbitrary n ∈ N ∗ . The minimum value of m is 16. arbitrary n ∈ N ∗ . Then 2 −. 6.99 . The domain of the function f (x) is R+ , for arbitrary x, y ∈ R+ , 1 f (xy) = f (x) + f (y) holds. (1) Show f ( ) = −f (x) when x ∈ R+ . (2) If f (x) < 0 x holds when x > 1, show f (x) has an inverse function. (3) Let f −1 (x) is the inverse function of f (x). Show that in the domain of f −1 (x), f −1 (x1 + x2 ) = f −1 (x1 ) · f −1 (x2 ). 1 1 1 in the given equation, then f (x) + f ( ) = f (x · ) = f (1). x x x 1 Let x = y = 1, then f (1) = f (1) + f (1). Thus f (1) = 0. Hence f (x) + f ( ) = 0. x 1 + Therefore f ( ) = −f (x) when x ∈ R . x x2 (2) Proof: Let x1 , x2 ∈ R+ , and x1 < x2 , then > 1. Thus f (x2 ) − f (x1 ) = x1 1 x2 f (x2 ) + f ( ) = f ( ) < 0. Hence the function f (x) is decreasing in R+ . Therefore x1 x2 f (x) has an inverse function. (3) Proof: Since x1 , x2 , x1 + x2 are in the domain of f −1 (x), then f −1 (x1 ), f −1 (x2 ), f −1 (x1 + x2 ) ∈ R+ . Thus f [f −1 (x1 ) · f −1 (x2 )] = f [f −1 (x1 )] + f [f −1 (x2 )] = x1 + x2 = f [f −1 (x1 + x2 )]. Hence f −1 (x1 + x2 ) = f −1 (x1 )f −1 (x2 ). (1) Proof: Let y =. 6.100 The function f (x) = 2x3 + (m − x)3 (m ∈ N ∗ ). x1 + x2 (1) If x1 , x2 ∈ (0, m), show f (x1 ) + f (x2 ) 2f ( ). (2) If an = f (n) (n = 2 m 2 1, 2, · · · , m − 1), show a1 + am−1 a2 + am−2 . (3) For arbitrary a, b, c ∈ [ , m], can 2 3 the values of f (a), f (b), f (c) form the three side lengths of a triangle? Please explain the reason. (1) Proof: From the given condition, we have x1 , x2 ∈ (0, m), f (x1 ) = 2x31 + (m − x1 )3 , 3 x 1 + x2 3 f (x2 ) = 2x32 + (m − x2 )3 . Since x31 + x32 − 2( ) = (x1 + x2 )(x1 − x2 )2 , 2 4 3 x 1 + x2 3 2 3 3 x1 , x2 ∈ (0, m), then (x1 + x2 )(x1 − x2 ) 0. Thus x1 + x2 2( ) which 4 2 x 1 + x2 3 is equivalent to 2x31 + 2x32 2 × 2( ) . Similarly, (m − x1 )3 + (m − x2 )3 2 m − x1 + m − x2 3 x1 + x2 3 x1 + x2 2( ) = 2(m − ) . Therefore f (x1 ) + f (x2 ) 2f ( ). 2 2 2 (2) Proof: From (1), we have a1 + a3 2a2 , a2 + a4 2a3 , a3 + a5 2a4 ,· · · , am−3 + am−1 2am−2 . Adding the above (m − 3) inequalities to generate a1 + am−1 a2 + am−2 . (3) Solution: Since f (x) = 2x3 + (m − x)3 , then f (x) = 6x2 − 3(m − x)2 = 3x2 + m 2 6mx − 3m2 . Obviously, f (x) > 0 when x ∈ [ , m]. This means f (x) is an increasing 2 3 m 2 3 m3 m3 function on [ , m]. The minimum value of f (x) is f (x)min = 2 × + = m3 2 3 8 8 8 m 8 3 1 3 17 3 m when x = . The maximum value of f (x) is f (x)max = 2 × m + m = 2 27 27 27 2 when x = m. 3. Download free eBooks at bookboon.com 147.

(148) ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions. 3 17 3 m . Thus f (a) + f (b) We let a b c, then m3 f (a) f (b) f (c) 8 27 3 17 3 3 m · 2 = m3 > m3 f (c). 8 4 27 Therefore f (a), f (b), f (c) can be the three side lengths of a triangle. 6.101 Given f (x) = √. x , and fn (x) = f (f · · · (f (x))), n ∈ N ∗ . Find 1 − x2 nf. the analytic expression of fn (x) and prove it.. Solution: From the given condition, we have f1 (x) = f (x) = √. x , f2 (x) = 1 − x2. x f (x) = ··· = √ , f3 (x) = f (f (f (x))) = f (f2 (x)) = · · · = f (f (x)) = 1 − 2x2 1 − [f (x)]2 x x √ , · · · . Then we generalize fn (x) = √ (n ∈ N ∗ ). 2 1 − 3x 1 − nx2 Now we prove the conclusion by mathematical induction. x = f (x). p(1) holds. (1) When n = 1, f1 (x) = √ 1 − x2 x (2) Assume p(k) holds when n = k. This means that fk (x) = √ holds. 1 − kx2 √ x fk (x) 2 1−kx = = When n = k +1, fk+1 (x) = f (f · · · (f (x))) = f (fk (x)) = x2 1 − fk2 (x) 1 − 1−kx2 (k+1)f x x . Then fk+1 (x) = . Hence p(k + 1) also holds. 1 − (k + 1)x2 1 − (k + 1)x2 x Therefore, for all n ∈ N ∗ , fn (x) = f (x) = √ always holds. 1 − nx2 π 6.102 Let the function fn (θ) = sinn θ + (−1)n cosn θ, 0 θ , where n 4 is a positive integer. (1) Determine the monotonicity of f1 (θ) and f3 (θ). Prove your conclusions. (2) Show 2f6 (θ) − f4 (θ) = (cos4 θ − sin4 θ)(cos2 θ − sin2 θ). (3) For an arbitrary given positive integer n, find the maximum value and minimum value of the function fn (θ). π (1) Solution: We can show that f1 (θ) and f3 (θ) are both increasing functions on [0, ]. 4 Now we provide the proof for the monotonicity of f1 (θ). π Since f1 (θ) = sin θ−cos θ, let θ1 , θ2 ∈ [0, ], and θ1 < θ2 , then f1 (θ1 )−f1 (θ2 ) = (sin θ1 − 4 cos θ1 ) − (sin θ2 − cos θ2 ) = (sin θ1 − sin θ2 ) + (cos θ2 − cos θ1 ). Since sin θ1 < sin θ2 ), cos θ2 < cos θ1 ), then f1 (θ1 ) − f1 (θ2 ) < 0. Thus f1 (θ1 ) < f1 (θ2 ). Hence f1 (θ) is increasπ ing on [0, ]. 4 π Similarly, f3 (θ) is increasing on [0, ] 4 (2) Proof: The left-hand side of the equation 2f6 (θ) − f4 (θ) = 2(sin6 θ + cos6 θ) − (sin4 θ + cos4 θ) = 2(sin2 θ + cos2 θ)(sin4 θ − sin2 θ cos2 θ + cos4 θ) − (sin4 θ + cos4 θ) = sin4 θ − 2 sin2 θ cos2 θ + cos4 θ = (sin2 θ + cos2 θ)2 − 4 sin2 θ cos2 θ = 1 − sin2 2θ = cos2 2θ. The right-hand side of the equation = (cos2 θ + sin2 θ)(cos2 θ − sin2 θ)2 = cos2 2θ. Thus, The left-hand side of the equation equals the right-hand side. π π Download free eBooks at bookboon.com (3) When n = 1, the function f1 (θ) is increasing on [0, ], then f1 (θ)max = f1 ( ) = 0, 4 4 148 f1 (θ)min = f1 (0) = −1. When n = 2, f2 (θ)max = f2 (θ)min = 1. When n = 3, the funcπ π.

(149) ELEMENTARY ALGEBRA EXERCISE BOOK II. Functions. π ing on [0, ]. 4 π Similarly, f3 (θ) is increasing on [0, ] 4 (2) Proof: The left-hand side of the equation 2f6 (θ) − f4 (θ) = 2(sin6 θ + cos6 θ) − (sin4 θ + cos4 θ) = 2(sin2 θ + cos2 θ)(sin4 θ − sin2 θ cos2 θ + cos4 θ) − (sin4 θ + cos4 θ) = sin4 θ − 2 sin2 θ cos2 θ + cos4 θ = (sin2 θ + cos2 θ)2 − 4 sin2 θ cos2 θ = 1 − sin2 2θ = cos2 2θ. The right-hand side of the equation = (cos2 θ + sin2 θ)(cos2 θ − sin2 θ)2 = cos2 2θ. Thus, The left-hand side of the equation equals the right-hand side. π π (3) When n = 1, the function f1 (θ) is increasing on [0, ], then f1 (θ)max = f1 ( ) = 0, 4 4 f1 (θ)min = f1 (0) = −1. When n = 2, f2 (θ)max = f2 (θ)min = 1. When n = 3, the funcπ π tion f3 (θ) is increasing on [0, ], then f3 (θ)max = f3 ( ) = 0, f3 (θ)min = f3 (0) = −1. 4 4 1 2 π When n = 4, the function f4 (θ) = 1 − sin 2θ is decreasing on [0, ], then f4 (θ)max = 2 4 1 π f4 (0) = 1, f4 (θ)min = f4 ( ) = . Now we discuss the case n 5. 4 2 π When n is an odd number, for arbitrary θ1 , θ2 ∈ [0, ], and θ1 < θ2 , since fn (θ1 ) − 4 fn (θ2 ) = (sinn θ1 −sinn θ2 )+(cosn θ2 −cosn θ1 ), and 0 sin θ1 < sin θ2 < 1, 0 < cos θ2 < cos θ1 1. Thus sinn θ1 < sinn θ2 , cosn θ2 < cosn θ1 . Hence fn (θ1 ) < fn (θ2 ). Then fn (θ) π π is increasing on [0, ]. We have fn (θ)max = fn ( ) = 0, fn (θ)min = fn (0) = −1. 4 4 When n is an even number, on one hand, fn (θ) = sinn θ + cosn θ sin2 θ + cos2 θ 1 = fn (0), and on the other hand, for an arbitrary positive integer l 2, we have 1 2f2l (θ) − f2l−2 (θ) = (cos2l−2 θ − sin2l−2 θ)(cos2 θ − sin2 θ) 0, then fn (θ) fn−2 (θ) 2 1 1 π π · · · n −1 f2 (θ) = n −1 = fn ( ). Thus fn (θ)max = fn (0) = 1, fn (θ)min = f ( ) = 4 4 22 22 1 2 ( )n . 2 As a conclusion, when n is an odd number, the maximum value of fn (θ) is 0, the minimum value of fn (θ) is −1. When nis an even number, the maximum value of fn (θ) is 1 1, the minimum value of fn (θ) is 2 ( )n . 2. Download free eBooks at bookboon.com 149.

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