Calculus I: Practice Problems, Methods, and Solutions [2 ed.] 3031450272, 9783031450273, 9783031450280, 9783031450303

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Mehdi Rahmani-Andebili

Calculus I

Practice Problems, Methods, and Solutions Second Edition

Calculus I

Mehdi Rahmani-Andebili

Calculus I Practice Problems, Methods, and Solutions Second Edition

Mehdi Rahmani-Andebili Electrical Engineering Department Arkansas Tech University Russellville, AR, USA

ISBN 978-3-031-45027-3 ISBN 978-3-031-45028-0 https://doi.org/10.1007/978-3-031-45028-0

(eBook)

# The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021, 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland Paper in this product is recyclable.

Preface

Calculus is one of the most important courses of many majors, including engineering and science, and even some non-engineering majors like economics and business, which is taught in three successive courses at universities and colleges worldwide. Moreover, in many universities and colleges, a precalculus course is mandatory for under-prepared students as the prerequisite course of Calculus 1. Unfortunately, some students do not have a solid background and knowledge in math and calculus when they start their education in universities or colleges. This issue prevents them from learning calculus-based courses such as physics and engineering courses. Sometimes, the problem escalates, so they give up and leave the university. Based on my real professorship experience, students do not have a serious issue comprehending physics and engineering courses. In fact, it is the lack of enough knowledge of calculus that hinders them from understanding those courses. Therefore, a series of calculus textbooks covering Precalculus, Calculus 1, Calculus 2, and Calculus 3 have been prepared to help students succeed in their major. This book, Calculus 1: Practice Problems, Methods, and Solutions, is the second edition of the book Calculus: Practice Problems, Methods, and Solutions, which was published in 2021. In the new version of the book, many new problems have been added to each chapter. The subjects of the calculus series books are as follows.

Precalculus: Practice Problems, Methods, and Solution . . . . . . .

Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities Systems of Equations Quadratic Equations Functions, Algebra of Functions, and Inverse Functions Factorization of Polynomials Trigonometric and Inverse Trigonometric Functions Arithmetic and Geometric Sequences

Calculus 1: Practice Problems, Methods, and Solution . . . . .

Characteristics of Functions Trigonometric Equations and Identities Limits and Continuities Derivatives and Their Applications Definite and Indefinite Integrals

v

vi

Preface

Calculus 2: Practice Problems, Methods, and Solution . . . .

Applications of Integration Sequences and Series and Their Applications Polar Coordinate System Complex Numbers

Calculus 3: Practice Problems, Methods, and Solution . . . . . .

Linear Algebra and Analytical Geometry Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System Multivariable Functions Double Integrals and their Applications Triple Integrals and their Applications Line Integrals and Their Applications

The textbooks include basic and advanced calculus problems with very detailed problem solutions. They can be used as practicing study guides by students and as supplementary teaching sources by instructors. Since the problems have very detailed solutions, the textbooks are helpful for under-prepared students. In addition, they are beneficial for knowledgeable students because they include advanced problems. In preparing the problems and solutions, care has been taken to use methods typically found in the primary instructor-recommended textbooks. By considering this key point, the textbooks are in the direction of instructors’ lectures, and the instructors will not see any untaught and unusual problem solutions in their students’ answer sheets. To help students study in the most efficient way, the problems have been categorized into nine different levels. In this regard, for each problem, a difficulty level (easy, normal, or hard) and a calculation amount (small, normal, or large) have been assigned. Moreover, problems have been ordered in each chapter from the easiest problem with the smallest calculations to the most difficult problems with the largest ones. Therefore, students are suggested to start studying the textbooks from the easiest problems and continue practicing until they reach the normal and then the hardest ones. This classification can also help instructors choose their desirable problems to conduct a quiz or a test. Moreover, the classification of computation amount can help students manage their time during future exams, and instructors assign appropriate problems based on the exam duration. Russellville, AR, USA

Mehdi Rahmani-Andebili

The Other Works Published by the Author

The author has already published the books and textbooks below with Springer Nature. Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2023. Calculus III – Practice Problems, Methods, and Solutions, Springer Nature, 2023. Calculus II – Practice Problems, Methods, and Solutions, Springer Nature, 2023. Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2023. Planning and Operation of Electric Vehicles in Smart Grid, Springer Nature, 2023. Applications of Artificial Intelligence in Planning and Operation of Smart Grid, Springer Nature, 2022. AC Electric Machines – Practice Problems, Methods, and Solutions, Springer Nature, 2022. DC Electric Machines, Electromechanical Energy Conversion Principles, and Magnetic Circuit Analysis- Practice Problems, Methods, and Solutions, Springer Nature, 2022. Applications of Fuzzy Logic in Planning and Operation of Smart Grids, Springer Nature, 2021. Differential Equations – Practice Problems, Methods, and Solutions, Springer Nature, 2022. Feedback Control Systems Analysis and Design – Practice Problems, Methods, and Solutions, Springer Nature, 2022. Power System Analysis – Practice Problems, Methods, and Solutions, Springer Nature, 2022. Advanced Electrical Circuit Analysis – Practice Problems, Methods, and Solutions, Springer Nature, 2022. Design, Control, and Operation of Microgrids in Smart Grids, Springer Nature, 2021. Applications of Fuzzy Logic in Planning and Operation of Smart Grids, Springer Nature, 2021. Operation of Smart Homes, Springer Nature, 2021. AC Electrical Circuit Analysis – Practice Problems, Methods, and Solutions, Springer Nature, 2021. Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. DC Electrical Circuit Analysis – Practice Problems, Methods, and Solutions, Springer Nature, 2020. Planning and Operation of Plug-in Electric Vehicles: Technical, Geographical, and Social Aspects, Springer Nature, 2019.

vii

Contents

1

Problems: Characteristics of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

2

Solutions of Problems: Characteristics of Functions . . . . . . . . . . . . . . . . . . . .

13

3

Problems: Trigonometric Equations and Identities . . . . . . . . . . . . . . . . . . . . .

39

4

Solutions of Problems: Trigonometric Equations and Identities . . . . . . . . . . .

61

5

Problems: Limits and Continuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

6

Solutions of Problems: Limits and Continuities . . . . . . . . . . . . . . . . . . . . . . . . 119

7

Problems: Derivatives and Their Applications . . . . . . . . . . . . . . . . . . . . . . . . 139

8

Solutions of Problems: Derivatives and Their Applications . . . . . . . . . . . . . . . 151

9

Problems: Definite and Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

10

Solutions of Problems: Definite and Indefinite Integrals . . . . . . . . . . . . . . . . . 195

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225

1

Problems: Characteristics of Functions

Abstract

In this chapter, the basic and advanced problems of functions and inverse functions, algebra of functions, characteristics of functions such as domain of functions, range of functions, axis of symmetry of functions, and types of functions in terms of being odd or even are presented. To help students study the chapter in the most efficient way, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Moreover, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations. 1.1 Determine the reflection of the graph of the function below with respect to the origin [1, 2]. y ¼ log

x-1 xþ1

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large x-1 1) y ¼ log xþ1 1-x 2) y ¼ log 1þx 1þx 3) y ¼ log 1-x xþ1 4) y ¼ log x-1 1.2 Determine the relation of f (g(h(x))) for the information below. f ðxÞ ¼ ln x9 gð x Þ ¼

p 3

x2

h ð x Þ ¼ ex Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 6x 2) 9x 3) 10x 4) e9x # The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_1

1

2

1

1.3 If f

1 x

¼

2x - 1 x2

and g(x) ¼ 2cos2(x), calculate the value of fog

π 3

:

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 1 2) 2p 3 3) 2 4) 2 p 1.4 Determine the value of fog(3) if f ðxÞ ¼ x - 2 and g(x) ¼ x + 1. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 1 3) 2 4) 3 1.5 If f (x) ¼ 2x - 2 and g(x) ¼ x2 - 1, solve the equation of fog(x) ¼ 0. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large p 1) ± 2 2) ±2 p 3) ± 3 p 4) 2 1.6 In the function below, calculate the value of f (-2) + f (2). f ð xÞ ¼

2x2 þ 4

x≥2

½ x - 4

x0

1

x≤0

As can be noticed from f (x), the value of function is always positive. Therefore, the value of -f (x) is always negative. Hence: f ð-f ðxÞÞ ¼ 1 Choice (1) is the answer. 2.46 A function in the form of f (x, y) ¼ 0 is symmetric with respect to the line of y ¼ x if f (x, y) ¼ f(y, x). Based on the information given in the problem, we have: 3x2 þ 4xy þ ð2a - 1Þy2 þ a2 - 4 x ¼ 7 ) f ðx, yÞ ¼ 3x2 þ 4xy þ ð2a - 1Þy2 þ a2 - 4 x - 7 Moreover, from f (x, y) ¼ f(y, x), we have: ) 3x2 þ 4xy þ ð2a - 1Þy2 þ a2 - 4 x - 7 ¼ 3y2 þ 4yx þ ð2a - 1Þx2 þ a2 - 4 y - 7 ) ð4 - 2aÞx2 þ ð2a - 4Þy2 þ a2 - 4 x - a2 - 4 y ¼ 0 4 - 2a ¼ 0 )

2a - 4 ¼ 0 a2 - 4 ¼ 0

) a ¼ f2g \ f2g \ f-2, 2g \ f-2, 2g ) a ¼ 2

a2 - 4 ¼ 0 Choice (4) is the answer. 2.47 Based on the information given in the problem, we have: p

2-1

ð1Þ

f ðxÞ ¼ cos x

ð2Þ

gof ðxÞ ¼ 1 þ tan 2 x

ð3Þ

gof ðxÞ ¼ gðf ðxÞÞ ¼ gðcos xÞ ¼ 1 þ tan 2 x

ð4Þ

x¼

Solving (2) and (3):

32

2

Solutions of Problems: Characteristics of Functions

From trigonometry, we know that: 1 cos 2 x

1 þ tan 2 x ¼

ð5Þ

Solving (4) and (5): gðcos xÞ ¼

1 1 ) gð x Þ ¼ 2 x cos 2 x

) gogðxÞ ¼ gðgðxÞÞ ¼ g

ð6Þ

1 1 ¼ x4 ¼ 1 2 x2 2

ð7Þ

x

Solving (1) and (7): ðgogÞ

p

2-1 ¼

p

2-1

4

p ¼ 2-2 2þ1

) ðgogÞ

p

2

p ¼ 3-2 2

2

p ¼ 9 - 12 2 þ 8

p 2 - 1 ¼ 17 - 12 2

Choice (4) is the answer. 2.48 Based on the information given in the problem, we have: f ðxÞ ¼

log

5x - x2 4

The domain of a function in radical form, including even root, is determined by considering the radicand equal and greater than zero. Moreover, the domain of a logarithmic function with the base of 10 can be determined by considering its argument greater than zero. Therefore: log

5x - x2 ≥0 4

5x - x2 >0 4

)

)

log

5x - x2 ≥ logð1Þ 4 ) 2 x - 5x < 0

ð x - 4Þ ð x - 1 Þ ≤ 0 xð x - 5Þ < 0

)

5x - x2 ≥1 ) 4 xð x - 5Þ < 0

x2 - 5x þ 4 ≤ 0 x ð x - 5Þ < 0

1≤x≤4 \ ¼) 1 ≤ x ≤ 4 0 0, x ≠ 1

log x x2 þ 9 ≥ log x ð1Þ )

x2 þ 9 > 0

x2 þ 9 ≥ 1 )

x > 0, x ≠ 1

x2 þ 9 > 0 x > 0, x ≠ 1

Note that x2 + 9 > 0 and x2 + 8 ≥ 0 are true for any x. Hence: \ ¼) x > 0, x ≠ 1 ) Df ¼ ð0, 1Þ - f1g Choice (4) is the answer. 2.52 Based on the information given in the problem, we have: f ðxÞ ¼ 2x - 2½x] þ 1 Based on the definition, we know that: -½x] þ1 x2 ¼ ¼ ¼ ¼ ¼ ¼ ¼) 0 ≤ x - ½x] < 1 ¼ ½x] ≤ x < ½x] þ 1 ¼ ¼ ¼ ¼) 0 ≤ 2x - 2½x] < 2 ¼ ¼ ¼ ¼) 1 ≤ 2x - 2½x] þ 1 < 3 ) 1 ≤ f ðxÞ < 3 ) Rf ¼ ½1, 3Þ Choice (2) is the answer.

2

Solutions of Problems: Characteristics of Functions

35

2.53 Based on the information given in the problem, we have: f ð x Þ ¼ x2 þ 1 gð x Þ ¼ ) fogðxÞ ¼ f ðgðxÞÞ ¼ f

p

p

x-1

x-1 ¼

p x-1

2

þ1¼x-1þ1¼x

Next, we need to determine the domain of the function. As we know, the domain of a function in radical form, including even root, is determined by considering the radicand equal and greater than zero. x - 1 ≥ 0 ) x ≥ 1 ) Dfog ¼ ½1, 1Þ Now, we can determine the range of the function based its domain as follows: Dfog ¼ ½1, 1Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) Rfog ¼ ½1, 1Þ fogðxÞ ¼ x ¼ Choice (2) is the answer. 2.54 Based on the information given in the problem, we have: f ðxÞ ¼

x2 - 2x þ 3

The problem can be solved as follows: ) f ðxÞ ¼

ð x - 1Þ 2 þ 2

As we know: p þ2 2 ðx - 1Þ ≥ 0 ¼ ¼ ¼) ðx - 1Þ þ 2 ≥ 2 ¼ ¼ ¼) 2

p p p ðx - 1Þ2 þ 2 ≥ 2 ) f ðxÞ ≥ 2 ) Rf ðxÞ ¼ 2, 1

Choice (1) is the answer. 2.55 Based on the information given in the problem, we have: f ð x Þ ¼ j x - 2j ) D f ¼ ℝ Based on the definition, two functions are equivalent if they are equal, and their domains are the same. Choice (1): g1 ð xÞ ¼

ð x - 1Þ ð x - 2Þ x2 - 3x þ 2 ¼ ¼ j x - 2j x-1 x-1 ) D g 1 ¼ ℝ - f 1g

Therefore, the functions are not equivalent, since their domains are different. Note that x ¼ 1 makes the denominator zero; thus, it is not in the domain.

36

2

Solutions of Problems: Characteristics of Functions

Choice (2): g2 ð x Þ ¼

ðx - 2Þðx þ 2Þ x2 - 4 ¼ ¼ j x - 2j xþ2 xþ2 ) Dg2 ¼ ℝ - f-2g

Therefore, the functions are not equivalent because their domains are not the same. Note that x ¼ - 2 makes the denominator zero; thus, it must be excluded from the domain. Choice (3): g3 ð xÞ ¼

ð x - 2Þ 2 j x - 2j 2 ¼ ¼ j x - 2j j x - 2j jx - 2j ) Dg3 ¼ ℝ - f-2g

Therefore, the functions are not equivalent because their domains are not the same. Note that x ¼ - 2 makes the denominator zero; thus, it must be excluded from the domain. Choice (4): g 4 ð xÞ ¼

j6x - 12j 6jx - 2j ¼ ¼ jx - 2j 6 6 ) D g4 ¼ ℝ

Therefore, the functions are equivalent because their functions and domains are the same. Choice (4) is the answer. 2.56 Based on the information given in the problem, we have: f ðxÞ ¼

p 3

x-

p 3 xþ2

The line of x ¼ a is the axis of symmetry of the function of f (x, y) if f(a + x, y) ¼ f(a -x, y). For Choice (1), we have x ¼ -2 ) a ¼ -2. f ð-2 þ x, yÞ ¼

3

ð-2 þ xÞ -

3

ð-2 þ xÞ þ 2

f ð-2 - x, yÞ ¼

3

ð-2 - xÞ -

3

ð-2 - xÞ þ 2

)

p 3

x-2-

p 3

p p 3 x ≠ -x - 2 þ 3 x

) f ð-2 þ x, yÞ ≠ f ð-2 - x, yÞ For Choice (2), we have x ¼ -1 ) a ¼ -1. f ð-1 þ x, yÞ ¼

3

ð-1 þ xÞ -

3

ð-1 þ xÞ þ 2

f ð-1 - x, yÞ ¼

3

ð-1 - xÞ -

3

ð-1 - xÞ þ 2

References

37

)

p 3

x-1-

p p p 3 xþ1¼-3 xþ1þ 3 x-1

) f ð-1 þ x, yÞ ¼ f ð-1 - x, yÞ For Choice (3), we have x ¼ 1 ) a ¼ 1. f ð1 þ x, yÞ ¼

3

ð 1 þ xÞ -

3

ð 1 þ xÞ þ 2

f ð1 - x, yÞ ¼

3

ð 1 - xÞ -

3

ð 1 - xÞ þ 2

)

p 3

xþ1-

p 3

x þ 3≠ -

p 3

x-1þ

p 3

x-3

) f ð1 þ x, yÞ ≠ f ð1 - x, yÞ For Choice (4), we have x ¼ 2 ) a ¼ 2. f ð2 þ x, yÞ ¼

3

ð 2 þ xÞ -

3

ð 2 þ xÞ þ 2

f ð2 - x, yÞ ¼

3

ð 2 - xÞ -

3

ð 2 - xÞ þ 2

)

p 3

xþ2-

p 3

x þ 4≠ -

p 3

x-2þ

p 3

x-4

) f ð2 þ x, yÞ ≠ f ð2 - x, yÞ In the calculations, the rule below was used because n was an odd number: n

-f ðxÞ ¼ - n f ðxÞ

Choice (2) is the answer.

References 1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.

3

Problems: Trigonometric Equations and Identities

Abstract

In this chapter, the basic and advanced problems of trigonometric equations and trigonometric identities are presented. The subjects include trigonometric equations, trigonometric identities, domain, range, period, half angle formulas, reciprocal identities, Pythagorean identities, expressing sum of sine and cosine as a product, expressing product of sine and cosine as a sum, even and odd functions, periodic functions, degrees to radians formula, cofunction formulas, unit circle, inverse trigonometric functions, and domain and range of inverse trigonometric functions. To help students study the chapter in the most efficient way, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Moreover, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations. 3.1. Calculate the value of sin x cos x(1 - 2sin2x) for x ¼ 7.5° [1, 2]. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large p 3 1) 8 1 2) 8 p 3 3) 4 1 4) 4 3.2. Calculate the value of tan3x + cot3x if tan x + cot x ¼ 3. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 18 2) 9 3) 27 4) 3 3.3. Which one of the following choices is correct about the extension of hyperbolic functions? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large coshða þ bÞ ¼ cosh a cosh b - sinh a sinh b 1) sinhða þ bÞ ¼ sinh a cosh b þ cosh a sinh b coshða þ bÞ ¼ cosh a cosh b - sinh a sinh b 2) sinhða þ bÞ ¼ sinh a cosh b - cosh a sinh b

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_3

39

40

3

3) 4)

coshða þ bÞ ¼ cosh a cosh b þ sinh a sinh b sinhða þ bÞ ¼ sinh a cosh b - cosh a sinh b coshða þ bÞ ¼ cosh a cosh b þ sinh a sinh b sinhða þ bÞ ¼ sinh a cosh b þ cosh a sinh b

3.4. Calculate the value of tan(2θ) if cot(θ) ¼ 5. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 5 1) 12 5 2) 13 5 3) 12 5 4) 13 3.5. Determine the value of tan(-2100°). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large p 1) 3 p 3 2) 3p 3) - 3 p 3 4) 3 3.6. Simplify and calculate the final value of the following term: 1 þ cos ð40 ° Þ sin ð40 ° Þ Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) sin (20°) 2) cos (20°) 3) tan (20°) 4) cot (20°) 3m - 1 π 2π and ≤ α ≤ . 4 6 3 Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large p 2 3-1 1) 1, 3 p 2 3þ1 2) 1, 3 3) [1, 2] 5 4) 1, 3

3.7. Determine the range of m if sinðαÞ ¼

Problems: Trigonometric Equations and Identities

3

Problems: Trigonometric Equations and Identities

2m - 1 π π and - ≤ x ≤ . 6 3 3 Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 7 1) 2, 2 3 7 2) , 2 2

3.8. Determine the range of m if cosðxÞ ¼

3) 2, 4)

5 2

3 5 , 2 2

2x ? 3 Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π 2) 2π 3) 3π 4) 4π

3.9. What is the main period of cos 2 ðxÞ - 5 cos

3x 2x ? þ cos 3 5 3 Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 3π 2) 5π 3) 15π 4) 30π

3.10. What is the main period of sin 4

πx þ cosðπxÞ þ 5. 3 Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 2 3) 3 4) 6

3.11. Determine the main period of sin 4

3.12. Figure 3.1 illustrates part of the function of y ¼ sin (kx). Determine the value of k. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 2 1) 3 3 2) 4 3 3) 2 4 4) 3

41

42

3

Problems: Trigonometric Equations and Identities

Figure 3.1 The graph of problem 3.12

3.13. Figure 3.2 illustrates part of the function of y ¼ cos

ax þ

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1 1) 2 3 2) 2 2 3) 3 7 4) 4

1 π . Determine the value of a. 2

Figure 3.2 The graph of problem 3.13

3.14. Which one of the following choices is correct if α + β ¼ 19π? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) sin(α) ¼ sin (β) 2) cos(α) ¼ cos (β) 3) tan(α) ¼ tan (β) 4) cot(α) ¼ cot (β) 3.15. Calculate the final value of the following equation. sinð5π þ xÞ þ sin x Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

π 7π þ sin x þ 3 3

3

Problems: Trigonometric Equations and Identities

43

1) 0

π 3 π 3) 2 sin 3 π 4) - sin 3 2) sin

3.16. Calculate the value of the term below. sin arccos

p -1 - 3 þ arcsin 2 2

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large p 3 1) 2 2) 1 1 3) 2 4) -1 1 . 2 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 4 1) arctan 3 3 2) arctan 4 2 3) arctan 3 3 4) arctan 2

3.17. Calculate the value of 2 arctan

3.18. What is the value of cos(π sinh ln 3)? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) -1 1 2) p2 2 3) 2 p 3 4) 2 3.19. Calculate the value of cos(20°) if sin(50°) + sin (10°) ¼ m. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large m 1) 2 2) m 3) 2m 2m 4) 3

44

3

Problems: Trigonometric Equations and Identities

3.20. Simplify and calculate the final value of the following term: ð1 þ tan 2 ð5 ° ÞÞ sinð10 ° Þ ð1 - tan 2 ð5 ° ÞÞ tanð10 ° Þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) tan(15°) 2) tan(25°) 3) tan(35°) 4) tan(45°) 3.21. Which one of the following relations is correct if cot(α) ¼ m and cos(α) ¼ n? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) m2(1 + n2) ¼ n2 2) m2(1 - n2) ¼ n2 3) m2(2 + n2) ¼ 1 4) m2(2 - n2) ¼ 1 3.22. Determine the main period of sin(3x) cos (5x) + 11. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π 2) 2π 2π 3) 3 2π 4) 5 4π . 3 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large π 1) 6 5π 2) 6 π 3) 3 π 4) 6 17π 3.24. Calculate the value of arc sin sin . 5 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 2π 1) 5 3π 2) 5 2π 3) 5 3π 4) 5 3.23. Calculate the value of arc cos sin

3

Problems: Trigonometric Equations and Identities

45

19π . 5 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large π 1) 5 4π 2) 5 π 3) 5 4π 4) 5

3.25. Calculate the value of arc cos cos

1 . 2 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 3 2) 4 4 3) 3 3 4) 5

3.26. Calculate the value of tan 2arc tan

3 3 þ arc tan 5 4 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 10 1) 13 9 2) 13 12 3) 35 24 4) 25

3.27. Calculate the final value of sin arc sin

4 3 - arc cot 3 4 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π 2π 2) 3 π 3) 2 π 4) 3

3.28. Calculate the final value of arc cot -

3 . 2 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large π 1) 4 π 2) 4

3.29. Calculate the final value of arcðtanð5ÞÞ þ arc tan

.

.

46

3

Problems: Trigonometric Equations and Identities

3π 4 5π 4) 4 3)

3 4 þ cos arc sin 5 5 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 7 1) 5 1 2) 5 1 3) 5 7 4) 5

3.30. Calculate the final value of sin arc cos

.

3.31. Figure 3.3 shows a unit circle. Which one of the choices shows the value of tan(θ) and cot(θ), respectively? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) OA, OB 2) HA, HB 3) OA, AB 4) OB, BH

Figure 3.3 The graph of problem 3.31

3.32. Figure 3.4 illustrates a unit circle. Which one of the choices shows the value of sec(θ)? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) HA 2) MB 3) OB 4) OM

3

Problems: Trigonometric Equations and Identities

47

Figure 3.4 The graph of problem 3.32

3.33. Figure 3.5 illustrates a unit circle. Which one of the choices shows the value of csc(θ)? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) MB 2) OB 3) HC 4) OM

Figure 3.5 The graph of problem 3.33

2 1 . þ arc tan 3 5 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large π 1) 6 π 2) 4 π 3) 3 π 4) 2

3.34. Calculate the value of arc tan

3.35. Calculate the final value of the term below. arcðtanðmÞÞ þ arc tan

1 m

þ arcðcotðmÞÞ þ arcðcotð- mÞÞ

48

3

Problems: Trigonometric Equations and Identities

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π or 2π π 3π 2) or 2 2 3π 3) 2 π 4) 2 3.36. Determine the range of x in the inequality below. Herein, x is an acute angle. - 1 ≤ cosð4xÞ cosð2xÞ þ sinð4xÞ sinð2xÞ ≤ 0 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large π 3π , 1) 6 8 π π 2) , 8 4 π π 3) , 6 3 π π 4) , 4 2 3.37. Calculate the value of tan(2y) if tan(x + y) ¼ 5 and tan(x - y) ¼ 7. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1 1) 18 1 2) 18 1 3) 36 1 4) 36 3.38. Simplify and calculate the value of the following term: 5π 5π þ cos 12 12 5π 5π sin - cos 12 12

sin

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large p 1) 3 p 3 2) 3 p 3) - 2 3 p 3 4) 3

3

Problems: Trigonometric Equations and Identities

49

3.39. Figure 3.6 illustrates part of the function of y ¼ a sin (bπx). Determine the value of a + b. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 4 1) 3 5 2) 3 7 3) 3 8 4) 3

Figure 3.6 The graph of problem 3.39

3.40. Figure 3.7 illustrates part of the function of y ¼ a sin (bπx). Determine the value of a  b. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) -6 2) -3 9 3) 2 4) 6

Figure 3.7 The graph of problem 3.40

3.41. Figure 3.8 illustrates part of the function of y ¼ a sin Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2 5 2) 2

bx þ

1 π . Determine the value of a  b. 2

50

3

3) 3 7 4) 2

Figure 3.8 The graph of problem 3.41

3.42. Simplify and calculate the value of the following term: cosð5 ° Þ cosð10 ° Þ cosð20 ° Þ cosð50 ° Þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 4 cosð85 ° Þ 1 2) 8 cosð85 ° Þ 1 3) 8 sinð85 ° Þ 1 4) 4 sinð85 ° Þ x 7 if sinðxÞ þ cosðxÞ ¼ . 2 5 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 or 3 1 1 2) or 3 2 3 3) 2 or 5 2 4) 3 or 5

3.43. Calculate the value of tan

3.44. Simplify and calculate the value of the following term: sin 4 ðαÞ - cos 4 ðαÞ sinðαÞ cosðαÞ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

Problems: Trigonometric Equations and Identities

3

Problems: Trigonometric Equations and Identities

1) 2) 3) 4)

51

2 cot (2α) -2 cot (2α) 2 tan (3α) -2 tan (3α)

1 3.45. Calculate the value of cot2(2α) if sin 4 ðαÞ þ cos 4 ðαÞ ¼ . 2 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 2 3) 0 4) 3 3.46. Calculate the value of the following relation for x ¼

3π : 8

sin 3 ðxÞ cosðxÞ - cos 3 ðxÞ sinðxÞ þ 3 sin 2 ðxÞ cos 2 ðxÞ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 3 1) 8 5 2) 8 5 3) 8 3 4) 8 3.47. Calculate the value of the following relation for α ¼

π : 15

sinð2αÞ þ sinð5αÞ þ sinð8αÞ cosð2αÞ þ cosð5αÞ þ cosð8αÞ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large p 3 1) 3 p 2) - 3 p 3) 3 p 3 4) 3 3.48. Calculate the value of the following relation for x ¼

π : 12

ðsinðxÞ - cosðxÞ þ 2ÞðsinðxÞ - cosðxÞ - 2Þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

52

3

Problems: Trigonometric Equations and Identities

7 2 5 2) 2 1)

5 2 7 4) 2

3) -

3.49. Calculate the value of 4sin2(α)cos2(α)(tan(α) + cot (α))2. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 2 3) 3 4) 4 3.50. Determine the number of roots of the equation below. sinðπxÞ cos 2 ðπxÞ þ sin 2 ðπxÞ cosðπxÞ ¼ 0: Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 11 2) 12 3) 13 4) 14 1 16π . þ 2 cosðmxÞ. Determine the value of the function for x ¼ 2 3 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 2 1 2) 2 3) 1 4) 0

3.51. Figure 3.9 illustrates part of the function of y ¼

Figure 3.9 The graph of problem 3.51

3

Problems: Trigonometric Equations and Identities

3.52. Figure 3.10 shows part of the function of y ¼ 1 + sin (mx). Determine the value of the function for x ¼ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 0 1 2) 2 3) 1 4) 2

53

7π . 6

Figure 3.10 The graph of problem 3.52

3.53. Figure 3.11 shows part of the function of y ¼ a - sin (bπx). Determine the value of the function for x ¼ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 5 2) 2 3) 3 7 4) 2

Figure 3.11 The graph of problem 3.53

25 . 3

54

3

Problems: Trigonometric Equations and Identities

π x for 0 < x < 4. Determine the value of b. 2 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) -2 2) -1 3) 1 4) 2

3.54. Figure 3.12 shows the function of y ¼ a þ b cos

Figure 3.12 The graph of problem 3.54

4 3.55. Figure 3.13 shows the function of y ¼ 1 + a sin (bπx) for 0 < x < . Determine the value of a + b. 3 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 3 2) 4 3) 5 4) 6

Figure 3.13 The graph of problem 3.55

3.56. Figure 3.14 shows part the function of y ¼ a - 2 cos bx þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 2 2) 1 3 3) 2 4) 2

π . Determine the value of a + b. 2

3

Problems: Trigonometric Equations and Identities

55

Figure 3.14 The graph of problem 3.56

3 3.57. Calculate the value of cos(25 ° - α) if tanðα þ 20 ° Þ ¼ . 4 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 5 2) 6 3) 7 4) 8 π 3 þ α assuming that α is an acute angle and sinðαÞ ¼ . 4 5 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) -7 1 2) 7 1 3) 7 4) 7

3.58. Calculate the value of tan

2 π π - α if tan - α ¼ . 4 2 3 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 3 1 2) 5 1 3) 5 1 4) 3

3.59. Calculate the value of tan

3.60. Calculate the value of tan(2a) while we know that tanða þ bÞ ¼ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 3 1 2) 2 3) 3 4) 1

2 3 and tanða - bÞ ¼ . 5 7

56

3

3.61. Calculate the value of tan(x) if we have: π 4 ¼2 π cos x 4 sin x -

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) -3 1 2) 3 2 3) 3 4) 3 π 3.62. Calculate the value of (1 + tan (α))(1 + tan (β)) if α þ β ¼ . 4 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) -2 2) 2 1 3) 3 1 4) 2 π π þ α - tan - α . 4 4 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 tan (2α) 2) 2 cos (2α) 3) 0 4) 2 sin (2α)

3.63. Calculate the value of tan

π 1 -α ¼ . 4 5 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 1.5 2) 1.8 3) 2.4 4) 2.5

3.64. Calculate the value of tan(2α) if tan

1 3.65. Calculate the value of tan(2α - β) if tan(α) ¼ 2 and tanðβÞ ¼ . 3 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) -3 2) -2 3) 0.5 4) 3

Problems: Trigonometric Equations and Identities

3

Problems: Trigonometric Equations and Identities

57

3.66. Determine the common solution of the equation of cos(3x) + cos (x) ¼ 0 assuming cos(x) ≠ 0. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large kπ π 1) þ 2 4 kπ π 2) þ 2 8 π 3) kπ 4 π 4) kπ þ 4 3.67. Calculate the sum of the positive acute roots of the equation of tan(4x) ¼ cot (x). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 2π 1) 5 4π 2) 5 3π 3) 5 π 4) 5 3.68. Determine the common solution of the equation of 2sin2(x) + 3 cos (x) ¼ 0. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 2π 1) 2kπ ± 3 π 2) 2kπ ± 3 5π 3) 2kπ ± 6 π 4) kπ 3 3.69. Determine the common solution of the equation of 2sin2(x) ¼ 3 cos (x). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large π 1) kπ ± 6 π 2) kπ ± 3 π 3) 2kπ ± 6 π 4) 2kπ ± 3 3.70. Two lines with the equations of x tan (α) + y cot (α) ¼ 2 and x tan (α) - y cot (α) ¼ 1 are intersecting each other at point M. By changing the value of α, what is the position equation of the point? Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1 1) y ¼ x 3 2) y ¼ x 1 3) y ¼ 4x 3 4) y ¼ 4x

58

3

Problems: Trigonometric Equations and Identities

3.71. What is the position equation of the point of (2 - 3 sin (α), 1 + 4 cos (α)) if the value of α changes? Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) Circle 2) Ellipse 3) Parabola 4) Hyperbola 3.72. What is the position equation of the point of (2 - 5 cos (α), 4) if the value of α changes? Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) A horizontal line 2) A vertical line 3) A horizontal line segment 4) A vertical line segment 3.73. Calculate the value of y if 2 cos (x - y) + 3 sin (x + y) ¼ 5 and 0 < x, y < 2π. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large π 2π 1) or 3 3 π 5π 2) or 4 4 π 5π 3) or 6 6 π 3π 4) or 2 2 π 3.74. Calculate the value of m if tan(α) ≠ tan (β), α þ β ¼ , and α and β are the two roots of the equation below. 4 tan 2 ðxÞ þ ðm þ 2Þ tanðxÞ þ 2m - 2 ¼ 0 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 3 3) 5 4) 7 3.75. Calculate the final value of the following relation: sin 6 ðαÞ þ cos 6 ðαÞ þ 3 sin 2 ðαÞ cos 2 ðαÞ sin 4 ðαÞ þ cos 4 ðαÞ þ 2 sin 2 ðαÞ cos 2 ðαÞ Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) sin2(α) 2) cos2(α) 3) sin2(α) - cos2(α) 4) 1

3

Problems: Trigonometric Equations and Identities

59

3.76. Calculate the final value of the relation below. sinð135 ° Þ cosð210 ° Þ þ cosð135 ° Þ sinð420 ° Þ tanð210 ° Þ cotð420 ° Þ þ cotð120 ° Þ tanð330 ° Þ Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large p 6 1) 4 p 3 6 2) 4 p 6 3) 2 p 3 6 4) 2 π 3.77. Calculate the final value of (1 + cot (x))(1 + cot ( y)) if x þ y ¼ kπ þ . 4 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) tan(x) tan ( y) 2) 2 tan (x) tan ( y) 3) cot(x) cot ( y) 4) 2 cot (x) cot ( y) 3.78. Determine the common solution of the equation below. ðsinðxÞ - tanðxÞÞ tan

3π 4π - x ¼ cos 2 3

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large π 1) kπ 6 π 2) kπ þ 3 π 3) 2kπ ± 3 π 4) 2kπ ± 6 3.79. Calculate the sum of the roots of the equation below for x 2 [0, π]. sinð2xÞðsinðxÞ þ cosðxÞÞ ¼ cosð2xÞðcosðxÞ - sinðxÞÞ Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 3π 1) 4 5π 2) 4 3π 3) 2 7π 4) 4

60

3

3.80. Determine the common solution of the equation of

p

2 sin

Problems: Trigonometric Equations and Identities

π 5π þx . - x ¼ 1 þ sin 4 2

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large π 1) kπ þ 2 π 2) 2kπ 4 π 3) 2kπ 2 π 4) 2kπ þ 2 3.81. Which one of the following choices shows one of the common solutions of the equation of cosð2xÞ þ Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large π 1) kπ 6 π 2) kπ 3 π 3) kπ þ 6 π 4) kπ þ 3 3.82. Calculate the value of the following term: cos 3α þ sin α sin 2α sin α  sin 3α - sin 2α cos α cos α Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) tanα 2) cotα 3) 1 4) -1

References 1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.

p

3 sinð2xÞ ¼ 1?

4

Solutions of Problems: Trigonometric Equations and Identities

Abstract

In this chapter, the problems of the third chapter are fully solved, in detail, step-by-step, and with different methods. 4.1. From trigonometry, we know that [1, 2]: sin 2x = 2 sinðxÞ cosðxÞ 1 -cos 2x = 2 sin 2 x sin 30 ° =

1 2

Therefore: sin x cos x 1 - 2 sin 2 x =

1 1 sin 2x ðcos 2xÞ = sin 4x 4 2

For x = 7.5°, we have: 1 1 1 × sinð4 × 7:5 ° Þ = sin 30 ° = 4 4 8 Choice (2) is the answer. 4.2. Based on the information given in the problem, we have: tan x þ cot x = 3 From algebra, we know that: a3 þ b3 = ða þ bÞ3 - 3abða þ bÞ In addition, from trigonometry, we know that: tan x cot x = 1

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_4

61

62

4 Solutions of Problems: Trigonometric Equations and Identities

Therefore: tan 3 x þ cot 3 x = ðtan x þ cot xÞ3 - 3 tan x cot xðtan x þ cot xÞ = ð3Þ3 - 3ð1Þð3Þ ) tan 3 x þ cot 3 x = 18 Choice (1) is the answer. 4.3. From trigonometry of hyperbolic functions, we know that: coshða ± bÞ = cosh a cosh b ± sinh a sinh b sinhða ± bÞ = sinh a cosh b ± cosh a sinh b Choice (4) is the answer. 4.4. From trigonometry, we know that: tanðθÞ = tanð2θÞ =

1 cotðθÞ

2 tanðθÞ 1 - tan 2 ðθÞ

Based on the information given in the problem: cotðθÞ = 5 ) tanðθÞ =

1 5

Therefore: 2 tanðθÞ tanð2θÞ = = 1 - tan 2 ðθÞ

) tanð2θÞ =

1 5 1 15 2×

2

2 = 5 24 25

5 12

Choice (1) is the answer. 4.5. From trigonometry, we know that: tanðα þ nπ Þ = tanðαÞ, 8n 2 ℤ tanð- αÞ = - tanðαÞ p tan 60 ° = 3 Therefore: tan - 2100 ° = -tan 2100 ° = -tan 12 × 180 - 60 ° = -tan - 60 ° = tan 60 ° p ) tan - 2100 ° = 3 Choice (1) is the answer.

4

Solutions of Problems: Trigonometric Equations and Identities

63

4.6. From trigonometry, we know that: 1 þ cosðθÞ = 2 cos 2

sinðθÞ = 2 sin cotðθÞ =

θ 2

θ θ cos 2 2 cosðθÞ sinðθÞ

Therefore: 1 þ cos 40 ° 2 cos 2 20 ° cos 20 ° = = sin 40 ° 2 sin 20 ° cos 20 ° sin 20 ° )

1 þ cos 40 ° = cot 20 ° sin 40 °

Choice (4) is the answer. 4.7. For the given range of α, we can conclude that: π 2π 1 ≤α≤ ) ≤ sinðαÞ ≤ 1 6 3 2 Therefore, based on the given information, i.e., sinðαÞ =

3m - 1 4 ,

we can write:

1 3m - 1 ≤ ≤ 1 ) 2 ≤ 3m - 1 ≤ 4 ) 3 ≤ 3m ≤ 5 2 4 ) 1≤m≤

5 3

Choice (4) is the answer. 4.8. For the given range of x, we can conclude that: -

π π 1 ≤ x ≤ ) ≤ cosðxÞ ≤ 1 3 3 2

Therefore, based on the given information, i.e., cosðxÞ =

2m - 1 6 ,

we can write:

1 2m - 1 ≤ ≤ 1 ) 3 ≤ 2m - 1 ≤ 6 ) 4 ≤ 2m ≤ 7 2 6 ) 2≤m≤ Choice (1) is the answer.

7 2

64

4 Solutions of Problems: Trigonometric Equations and Identities

4.9. From trigonometry, we know that: f 1 ðxÞ = cos 2n ðaxÞ, 8n 2 ℤ ) T 1 =

π j aj

f 2 ðxÞ = cos 2nþ1 ðaxÞ, 8n 2 ℤ ) T 2 =

2π jaj

Therefore: f 1 ðxÞ = cos 2 ðxÞ ) T 1 = f 2 ðxÞ = - 5 cos

π =π 1

2x 2π ) T 2 = 2 = 3π 3 3

The main period of the given expression is the least common multiple (LCM) of the main periods of the terms, as can be seen in the following: T = LCMðπ, 3π Þ ) T = 3π Choice (3) is the answer. 4.10. From trigonometry, we know that: f 1 ðxÞ = sin 2n ðaxÞ, 8n 2 ℤ ) T 1 =

π jaj

f 2 ðxÞ = cos 2nþ1 ðaxÞ, 8n 2 ℤ ) T 2 =

2π jaj

Therefore: f 1 ðxÞ = sin 4

π 5π 3x ) T1 = 3 = 3 5 5

f 2 ðxÞ = cos 3

2π 2x ) T 2 = 2 = 3π 3 3

The main period of the given expression is: the least common multiple (LCM) of the main periods of the terms as follows: T = LCM

5π , 3π 3

) T = 15π Choice (3) is the answer.

4

Solutions of Problems: Trigonometric Equations and Identities

65

4.11. From trigonometry, we know that: f 1 ðxÞ = sin 2n ðaxÞ, 8n 2 ℤ ) T 1 =

π jaj

f 2 ðxÞ = cos 2nþ1 ðaxÞ, 8n 2 ℤ ) T 2 =

2π jaj

Therefore: f 1 ðxÞ = sin 4

π πx ) T1 = π = 3 3 3

f 2 ðxÞ = cosðπxÞ ) T 2 =

2π =2 π

The main period of the given expression is: the least common multiple (LCM) of the main periods of the terms, as can be seen in the following: T = LCMð3, 2Þ ) T =6 Choice (4) is the answer. 4.12. From trigonometry, we know that: y = sinðkxÞ ) T =

2π jk j

Therefore: )

3π 1 2π = 4 2 jk j

) jk j =

4 4 ) k= ± 3 3

Based on the graph and the function, the positive value of k is acceptable. ) k=

4 3

Choice (4) is the answer (Fig. 4.1).

Figure 4.1 The graph of solution of problem 4.12

66

4 Solutions of Problems: Trigonometric Equations and Identities

4.13. From trigonometry, we know that: y = cos πax þ

π = -sinðπaxÞ 2

y = sinðmxÞ ) T =

2π jmj

Therefore: ) 1- -

1 2π 4 2 3 = ) = ) a= ± 3 3 2 jπaj j aj

Based on the graph and y = - sin (πax), the positive value of a is acceptable. ) a=

3 2

Choice (2) is the answer (Fig. 4.2).

Figure 4.2 The graph of solution of problem 4.13

4.14. From trigonometry, we know that: sinðα þ 2nπ Þ = sinðαÞ, 8n 2 ℤ cosðα þ 2nπ Þ = cosðαÞ, 8n 2 ℤ tanðα þ nπ Þ = tanðαÞ, 8n 2 ℤ cotðα þ nπ Þ = cotðαÞ, 8n 2 ℤ sinðπ - αÞ = sinðαÞ cosðπ - αÞ = -cosðαÞ tanð- αÞ = -tanðαÞ cotð- αÞ = -cotðαÞ

4

Solutions of Problems: Trigonometric Equations and Identities

67

Based on the information given in the problem, we have: α þ β = 19π ) α = 19π - β Therefore: sinðαÞ = sinð19π - βÞ = sinðπ - βÞ = sinðβÞ cosðαÞ = cosð19π - βÞ = cosðπ - βÞ = -cosðβÞ tanðαÞ = tanð19π - βÞ = tanð- βÞ = -tanðβÞ cotðαÞ = cotð19π - βÞ = cotð- βÞ = -cotðβÞ Choice (1) is the answer. 4.15. From trigonometry, we know that: sinðα þ 2nπ Þ = sinðαÞ, 8n 2 ℤ cosðα þ 2nπ Þ = cosðαÞ, 8n 2 ℤ sinðα þ π Þ = -sinðαÞ sinðαÞ þ sinðβÞ = 2 sin

αþβ α-β cos 2 2

Therefore: sinð5π þ xÞ þ sin x -

π 7π π π þ sin x þ = sinðx þ π Þ þ sin x þ sin x þ 3 3 3 3 π = -sinðxÞ þ 2 sinðxÞ cos = -sinðxÞ þ sinðxÞ 3

) sinð5π þ xÞ þ sin x -

π 7π þ sin x þ =0 3 3

Choice (1) is the answer. 4.16. Let us assume: cos

2π 1 1 ≜ - ) arc cos 3 2 2

p -π 1 - 3 sin ≜ - ) arc sin 3 2 2

=

2π 3

=

-π 3

Therefore: 1 sin arc cos 2

p - 3 þ arc sin 2

= sin

2π -π þ 3 3

= sin

π 3

68

4 Solutions of Problems: Trigonometric Equations and Identities

p

p - 3 2

) sin arc cos -

1 2

þ arc sin

arc tan

1 2

≜ α ) tanðαÞ =

=

3 2

Choice (1) is the answer. 4.17. Let us assume:

We need to find the value of 2 arctan

1 2

1 which is equal to 2α. 2

From trigonometry, we know that: tanð2αÞ =

2 tan α 1 - tan 2 α

Hence: tanð2αÞ =

2× 1-

1 2 1 2 2

) 2α = arc tan

=

4 3

4 3

Choice (1) is the answer. 4.18. As we know: e ln a = a Moreover, from trigonometry of hyperbolic functions, we know that: cos

π 1 = 3 2

sinh x =

ex - e - x 2

Thus, for x = ln 3, we can write: sinh ln 3 =

e ln 3 - e - ln 3 3 = 2 2

1 3

=

4 3

Therefore, cosðπ sinh ln 3Þ = cos π ×

4 π π = cos π þ = -cos 3 3 3

) cosðπ sinh ln 3Þ = Choice (2) is the answer.

1 2

4

Solutions of Problems: Trigonometric Equations and Identities

69

4.19. From trigonometry, we know that: sin 30 ° =

1 2

αþβ α-β cos 2 2

sinðαÞ þ sinðβÞ = 2 sin Therefore:

sin 50 ° þ sin 10 ° = m ) 2 sin 30 ° cos 20 ° = m ) 2 ×

1 cos 20 ° = m ) cos 20 ° = m 2

Choice (2) is the answer. 4.20. From trigonometry, we know that: cosðθÞ =

1 - tan 2 1 þ tan 2

tanðθÞ =

θ 2 θ 2

sinðθÞ cosðθÞ

tan 45 ° = 1 Therefore: 1 þ tan 2 5 °

sin 10 °

°

tan 10 °

1 - tan 2 5

)

=

sin 10 ° 1 × =1 ° sinð10 ° Þ cos 10 ° cosð10 Þ

1 þ tan 2 5 °

sin 10 °

1 - tan 2 5 °

tan 10 °

= tan 45 °

Choice (4) is the answer. 4.21. From trigonometry, we know that: 1 = 1 þ tan 2 ðαÞ cos 2 ðαÞ tanðαÞ =

1 cotðαÞ

Based on the information given in the problem, we have: cotðαÞ = m cosðαÞ = n

70

4 Solutions of Problems: Trigonometric Equations and Identities

Therefore: 1 1 =1 þ cos 2 ðαÞ cot 2 ðαÞ 1 1 × m 2 n2 2 =1 þ 2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) m = m2 n2 þ n2 2 n m

)

) m2 1 - n2 = n2 Choice (2) is the answer. 4.22. From trigonometry, we can know that: sin 2nþ1 ðaxÞ, 8n 2 ℤ ) T = sinðαÞ cosðβÞ =

2π jaj

1 ðsinðα þ βÞ þ sinðα - βÞÞ 2

We need to change the product expression to the summation one, as follows: y = sinð3xÞ cosð5xÞ þ 11 ) y =

1 1 sinð8xÞ - sinð2xÞ þ 11 2 2

Then: 2π π 1 = sinð8xÞ ) T 1 = 8 4 2 -

1 2π sinð2xÞ ) T 2 = =π 2 2

The main period of the given expression is: the least common multiple (LCM) of the main periods of the terms, as is presented in the following: ) T = LCM

π ,π 4

) T =π Choice (1) is the answer. 4.23. From trigonometry, we know that: sin

p 4π π π 3 = sin π þ = -sin =3 3 2 3 arcðcosð- αÞÞ = π - arcðcosðαÞÞ p arc cos

3 2

=

π 6

4

Solutions of Problems: Trigonometric Equations and Identities

71

Therefore: arc cos sin

p

4π 3

= arc cos -

p

3 2

) arc cos sin

= π - arc cos 4π 3

=

3 2

=π-

π 6

5π 6

Choice (2) is the answer. 4.24. From trigonometry, we know that: 17π 3π 3π = sin 4π = sin 5 5 5

sin

sin -

3π 3π ≜ α ) arcðsinðαÞÞ = 5 5

Therefore: arc sin sin

17π 5

= arc sin sin -

) arc sin sin

17π 5

3π 5 =-

= arcðsinðαÞÞ 3π 5

Choice (4) is the answer. 4.25. From trigonometry, we know that: cos

19π π π = cos 4π = cos 5 5 5 cos

π π ≜ α ) arcðcosðαÞÞ = 5 5

Therefore: arc cos cos

19π 5

= arc cos cos -

π 5

) arc cos cos

= arc cos cos 19π 5

=

π 5

Choice (1) is the answer. 4.26. From trigonometry, we know that: tanð2αÞ =

arc tan

1 2

2 tanðαÞ 1 - tan 2 ðαÞ

≜ α ) tanðαÞ =

1 2

π 5

= arcðcosðαÞÞ

72

4 Solutions of Problems: Trigonometric Equations and Identities

Therefore: 1 2

tan 2arc tan

= tanð2αÞ =

2 × 12 2 tanðαÞ 1 = 3 = 2 1 - tan ðαÞ 1 - 1 2 4 2 1 2

) tan 2arc tan

=

4 3

Choice (3) is the answer. 4.27. From trigonometry, we know that: sin 2 ðαÞ þ cos 2 ðαÞ = 1 1 þ tan 2 ðαÞ =

1 cos 2 ðαÞ

Therefore: arc sin arc tan ) sin arc sin

3 5

3 4

þ arc tan

3 5

≜ α ) sinðαÞ =

≜ β ) tanðβÞ = 3 4

3 4 ) cosðαÞ = 5 5

3 4 3 ) cosðβÞ = ) sinðβÞ = 4 5 5

= sinðα þ βÞ = sinðαÞ cosðβÞ þ cosðαÞ sinðβÞ = 3 5

) sin arc sin

þ arc tan

3 4

=

3 4 4 3 × þ × 5 5 5 5

24 25

Choice (4) is the answer. 4.28. From trigonometry, we know that: arcðcotð- αÞÞ = π - arcðcotðαÞÞ arcðcotðαÞÞ þ arc cot

1 α

=

π 2

Therefore: arc cot -

4 3

- arc cot

3 4

= π - arc cot

4 3

- arc cot

) arc cot Choice (3) is the answer.

4 3

3 4

- arc cot

= π - arc cot 3 4

=

π 2

4 3

þ arc cot

3 4

=π-

π 2

4

Solutions of Problems: Trigonometric Equations and Identities

73

4.29. From trigonometry, we know that: tanðα þ βÞ =

tanðαÞ þ tanðβÞ 1 -tanðαÞ tanðβÞ

arcðtanð5ÞÞ≜ α ) tanðαÞ = 5 arc tan

3 2

≜ β ) tanðβÞ =

arcðtanð- 1ÞÞ = tan - 1 ð- 1Þ =

3 2 3π 4

Therefore: tan arcðtanð5ÞÞ þ arc tan

3 2

= tanðα þ βÞ =

) arcðtanð5ÞÞ þ arc tan

13 5 þ 32 tanðαÞ þ tanðβÞ 2 = = = -1 1 -tanðαÞ tanðβÞ 1 - 15 - 13 2 2

3 2

) arcðtanð5ÞÞ þ arc tan

= tan - 1 ð- 1Þ 3 2

=

3π 4

Choice (3) is the answer. 4.30. From trigonometry, we know that: sin 2 ðαÞ þ cos 2 ðαÞ = 1 3 5

arc cos

3 5

≜ α ) cosðαÞ =

arc sin -

4 5

≜ β ) sinðβÞ = -

4 5

Therefore:

sin arc cos

3 5

þ cos arc sin -

4 5

= sinðαÞ þ cosðβÞ =

) sin arc cos

3 5

þ cos arc sin -

1-

3 5

2

4 5

=

7 5

Choice (1) is the answer. 4.31. From trigonometry, we know that: cotðθÞ = tan

π -θ 2

Based on the definition of tan(θ) and cot(θ), we can write: tanðθÞ =

Opposite for θ HA HA = = = HA Adjacent for θ OH 1

þ

1- -

4 5

2

=

4 3 þ 5 5

74

4 Solutions of Problems: Trigonometric Equations and Identities

cotðθÞ = tan HOB =

Opposite for HOB HB HB = = = HB OH 1 Adjacent for HOB

Choice (2) is the answer (Fig. 4.3).

Figure 4.3 The graph of solution of problem 4.31

4.32. Based on the definition of sec(θ) and cos(θ), we can write: secðθÞ =

1 = cosðθÞ

1 Adjacent for θ Hypotenuse for θ

=

1 OA OB

=

1 1 OB

= OB

Choice (3) is the answer (Fig. 4.4).

Figure 4.4 The graph of solution of problem 4.32

4.33. From trigonometry, we know that: sinðθÞ = cos

π -θ 2

Based on the definition of csc(θ) and sin(θ), we can write: cscðθÞ =

1 1 = = sinðθÞ cos COB

1 Adjacent for COB Hypotenuse for COB

Choice (2) is the answer (Fig. 4.5).

=

1 OC OB

=

1 1 OB

= OB

4

Solutions of Problems: Trigonometric Equations and Identities

75

Figure 4.5 The graph of solution of problem 4.33

4.34. From trigonometry, we know that: tanðα þ βÞ =

tanðαÞ þ tanðβÞ 1 -tanðαÞ tanðβÞ

arc tan

2 3

≜ α ) tanðαÞ =

2 3

arc tan

1 5

≜ β ) tanðβÞ =

1 5

π 4

arcðtanð1ÞÞ = Therefore: tanðα þ βÞ =

2 þ1 tanðαÞ þ tanðβÞ = 3 52 = 1 -tanðαÞ tanðβÞ 1 - 15

13 15 13 15

=1

) α þ β = arcðtanð1ÞÞ ) αþβ=

π 4

Choice (2) is the answer. 4.35. From trigonometry, we know that: 1 arcðtanðαÞÞ þ arc tan α

=

π 2 π 2

if α > 0 if α < 0

arcðcotðαÞÞ þ arcðcotð- αÞÞ = π Therefore: 1 arcðtanðmÞÞ þ arc tan m

þ arcðcotðmÞÞ þ arcðcotð- mÞÞ = π þ

π 2 π 2

if m > 0 if m < 0

76

4 Solutions of Problems: Trigonometric Equations and Identities

1 ) arcðtanðmÞÞ þ arc tan m

þ arcðcotðmÞÞ þ arcðcotð- mÞÞ =

3π 2 π 2

if m > 0 if m < 0

Choice (2) is the answer. 4.36. From trigonometry, we know that: cosðα - βÞ = cosðαÞ cosðβÞ þ sinðαÞ sinðβÞ Therefore: - 1 ≤ cosð4xÞ cosð2xÞ þ sinð4xÞ sinð2xÞ ≤ 0 ) - 1 ≤ cosð4x - 2xÞ ≤ 0 ) - 1 ≤ cosð2xÞ ≤ 0 Since x is an acute angle: )

π π π ≤ 2x ≤ π ) ≤ x ≤ 2 4 2

Choice (4) is the answer. 4.37. From trigonometry, we know that: tanðα - βÞ =

tanðαÞ -tanðβÞ 1 þ tanðαÞ tanðβÞ

Therefore: tanð2yÞ = tanððx þ yÞ - ðx - yÞÞ =

tanðx þ yÞ -tanðx - yÞ 5-7 -2 = = 1 þ tanðx þ yÞ tanðx - yÞ 1 þ 5 × 7 1 þ 35

) tanð2yÞ =

-1 18

Choice (2) is the answer. 4.38. From trigonometry, we know that: p π sinðαÞ þ cosðαÞ = 2 sin α þ 4 p π sinðαÞ -cosðαÞ = 2 sin α 4 p 2π 3 sin = 3 2 sin

π 1 = 6 2

4

Solutions of Problems: Trigonometric Equations and Identities

77

Therefore: p 5π 2 sin sin 5π 12 þ cos 12 =p 5π 5π sin 12 -cos 12 2 sin )

5π 12 5π 12

þ π4 -

π 4

sin 2π 3 = = sin π6

p

3 2 1 2

5π p sin 5π 12 þ cos 12 = 3 5π 5π sin 12 -cos 12

Choice (1) is the answer. 4.39. From trigonometry, we know that: y = a sinðmxÞ ) T =

2π jmj

Therefore: ) 6=

2π 1 1 ) jbj = ) b = ± 3 3 jbπ j

Based on the graph and the function, the positive value of b is acceptable. ) b=

1 3

Moreover, based on y = a sin (bπx) and the given graph, it is concluded that a = 2. Therefore: ) a þ b=2 þ ) a þ b=

1 3

7 3

Choice (3) is the answer (Fig. 4.6).

Figure 4.6 The graph of solution of problem 4.39

78

4 Solutions of Problems: Trigonometric Equations and Identities

4.40. From trigonometry, we know that: y = a sinðmxÞ ) T =

2π jmj

Therefore: ) 3=3×

2π = 1 ) jbj = 2 ) b = ± 2 jbπ j

Based on the graph and the given function, the negative value of b is accepted. ) b= -2 In addition, based on y = a sin (bπx) and the given graph, it is clear that a = 3. Therefore: ) a × b = 3 × ð- 2Þ ) a×b= -6 Choice (1) is the answer (Fig. 4.7).

Figure 4.7 The graph of solution of problem 4.40

4.41. From trigonometry, we know that: y = a sin

π þ bπx = a cosðbπxÞ 2

y = a cosðmxÞ ) T =

2π jmj

Therefore: ) 3:5 - ð- 2:5Þ = 3 ×

2π 6 ) 6= ) j bj = 1 ) b = ± 1 jbπ j j bj

Based on the graph and y = a cos (bπx), the positive value of b is accepted. ) b=1

4

Solutions of Problems: Trigonometric Equations and Identities

79

In addition, based on y = a cos (bπx) and the given graph, it is clear that a = 2. Therefore: ) a×b=2×1 ) a×b=2 Choice (1) is the answer (Fig. 4.8).

Figure 4.8 The graph of solution of problem 4.41

4.42. From trigonometry, we know that: cosðαÞ = sin

π -α 2

sinðαÞ = cos

π -α 2

sinð2αÞ = 2 sinðαÞ cosðαÞ Therefore: cos 5 ° cos 10 ° cos 20 ° cos 50 ° =

=

cos 5 ° cos 10 ° cos 20 ° cos 5 ° cos 10 ° cos 20 ° = sin 40 ° 2 sin 20 ° cos 20 °

cos 5 ° cos 10 ° cos 5 ° 1 = = 2 × 2 sin 10 ° cos 10 ° 4 × 2 sin 5 ° cos 5 ° 8 sin 5 ° )

cos 5 ° cos 10 ° cos 20 ° cos 50

°

=

Choice (2) is the answer. 4.43. From trigonometry, we know that: sinðxÞ =

2 tan 2x 1 þ tan 2

x 2

cosðxÞ =

1 - tan 2 1 þ tan 2

x 2 x 2

1 8 cos 85 °

80

4 Solutions of Problems: Trigonometric Equations and Identities

Therefore: sinðxÞ þ cosðxÞ =

) 12 tan

2

2 tan 2x 7 ) 5 1 þ tan 2

x 2

þ

1 - tan 2 1 þ tan 2

x 2 x 2

2 tan 2x þ 1 - tan 2 7 ) 5 1 þ tan 2 2x

=

x 2

=

p x x x 10 ± 102 - 4 × 12 × 2 10 ± 2 - 10 tan þ 2 = 0 ) tan = = 24 2 2 2 24 ) tan

x 1 1 = or 2 2 3

Choice (2) is the answer. 4.44. From trigonometry, we know that: sinð2αÞ = 2 sinðαÞ cosðαÞ sin 2 ðαÞ þ cos 2 ðαÞ = 1 cos 2 ðαÞ - sin 2 ðαÞ = cosð2αÞ cotð2αÞ =

cosð2αÞ sinð2αÞ

In addition, from factoring rule, we know that: a4 - b 4 = a 2 - b 2 a 2 þ b 2 Therefore: sin 2 ðαÞ - cos 2 ðαÞ sin 2 ðαÞ þ cos 2 ðαÞ sin 4 ðαÞ - cos 4 ðαÞ -cosð2αÞ × 1 = 1 = sinðαÞ cosðαÞ sinðαÞ cosðαÞ 2 sinð2αÞ )

sin 4 ðαÞ - cos 4 ðαÞ = - 2 cotð2αÞ sinðαÞ cosðαÞ

Choice (2) is the answer. 4.45. Based on the information given in the problem, we have: sin 4 ðαÞ þ cos 4 ðαÞ =

1 2

From trigonometry, we know that: sinð2αÞ = 2 sinðαÞ cosðαÞ sin 2 ðαÞ þ sin 2 ðαÞ = 1 ) sin 2 ðαÞ þ sin 2 ðαÞ

2

=1

) sin 4 ðαÞ þ cos 4 ðαÞ þ 2 sin 2 ðαÞ cos 2 ðαÞ = 1

7 5

4

Solutions of Problems: Trigonometric Equations and Identities

81

Therefore: 1 þ 2 sin 2 ðαÞ cos 2 ðαÞ = 1 ) 4 sin 2 ðαÞ cos 2 ðαÞ = 1 ) ð2sinðαÞ cosðαÞÞ2 = 1 2 ) sin 2 ð2αÞ = ± 1 ) cos 2 ð2αÞ = 0 )

cos 2 ð2αÞ =0 sin 2 ð2αÞ

) cot 2 ð2αÞ = 0 Choice (3) is the answer. 4.46. From trigonometry, we know that: sinð2xÞ = 2 sinðxÞ cosðxÞ cosð2xÞ = cos 2 ðxÞ - sin 2 ðxÞ Therefore: sin 3 ðxÞ cosðxÞ - cos 3 ðxÞ sinðxÞ þ 3 sin 2 ðxÞ cos 2 ðxÞ = sinðxÞ cosðxÞ sin 2 ðxÞ - cos 2 ðxÞ þ =

1 3 sinð2xÞð -cosð2xÞÞ þ sin 2 ð2xÞ 4 2 =-

For x =

3 × 4 sin 2 ðxÞ cos 2 ðxÞ 4

1 3 sinð4xÞ þ sin 2 ð2xÞ 4 4

3π , we have: 8 3π 1 3 1 3π 3 = - ð- 1Þ þ þ sin 2 2 × - sin 4 × 8 4 4 8 4 4

p

2 2

2

=

1 3 5 þ = 4 8 8

Choice (2) is the answer. 4.47. From trigonometry, we know that: sinðαÞ þ sinðβÞ = 2 sin

αþβ α-β cos 2 2

cosðαÞ þ cosðβÞ = 2 cos

α-β αþβ cos 2 2

Therefore: sinð2αÞ þ sinð5αÞ þ sinð8αÞ sinð8αÞ þ sinð2αÞ þ sinð5αÞ = cosð2αÞ þ cosð5αÞ þ cosð8αÞ cosð8αÞ þ cosð2αÞ þ cosð5αÞ =

sinð5αÞð2cosð3αÞ þ 1Þ 2 sinð5αÞ cosð3αÞ þ sinð5αÞ = = tanð5αÞ 2 cosð5αÞ cosð3αÞ þ cosð5αÞ cosð5αÞð2cosð3αÞ þ 1Þ

82

4 Solutions of Problems: Trigonometric Equations and Identities

π α= p 15 π π = tan = 3 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) tan 5 × ¼ 15 3 Choice (3) is the answer. 4.48. From trigonometry, we know that: sin 2 ðxÞ þ cos 2 ðxÞ = 1 2 sinðxÞ cosðxÞ = sinð2xÞ In addition, from factoring rule, we know that: ð a þ bÞ ð a - b Þ = a 2 - b 2 Therefore: ðsinðxÞ -cosðxÞ þ 2ÞðsinðxÞ -cosðxÞ - 2Þ = ðsinðxÞ -cosðxÞÞ2 - 4 = sin 2 ðxÞ þ cos 2 ðxÞ - 2 sinðxÞ cosðxÞ - 4 = 1 -sinð2xÞ - 4 = - 3 -sinð2xÞ π x= 12 π 1 7 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) - 3 -sin = -3- = 6 2 2 Choice (4) is the answer. 4.49. From trigonometry, we know that: tanðαÞ cotðαÞ = 1 1 þ tan 2 ðαÞ =

1 cos 2 ðαÞ

1 þ cot 2 ðαÞ =

1 sin 2 ðαÞ

sin 2 ðαÞ þ cos 2 ðαÞ = 1 Therefore: 4 sin 2 ðαÞ cos 2 ðαÞðtanðαÞ þ cotðαÞÞ2 = 4 sin 2 ðαÞ cos 2 ðαÞ tan 2 ðαÞ þ cot 2 ðαÞ þ 2 tanðαÞ cotðαÞ = 4 sin 2 ðαÞ cos 2 ðαÞ 1 þ tan 2 ðαÞ þ 1 þ cot 2 ðαÞ

4

Solutions of Problems: Trigonometric Equations and Identities

83

= 4 sin 2 ðαÞ cos 2 ðαÞ

1 1 þ 2 cos ðαÞ sin 2 ðαÞ

= 4 sin 2 ðαÞ cos 2 ðαÞ

sin 2 ðαÞ þ cos 2 ðαÞ sin 2 ðαÞ cos 2 ðαÞ

= 4 sin 2 ðαÞ cos 2 ðαÞ

1 =4 sin ðαÞ cos 2 ðαÞ 2

Choice (4) is the answer. 4.50. From trigonometry, we know the common solution of the equations below. sinðαÞ = 0 ) α = kπ, 8k 2 ℤ π cosðαÞ = 0 ) α = kπ þ , 8k 2 ℤ 2 tanðαÞ = - 1 ) α = kπ -

π , 8k 2 ℤ 4

Hence: sinðπxÞ cos 2 ðπxÞ þ sin 2 ðπxÞ cosðπxÞ = 0 ) sinðπxÞ cosðπxÞðcosðπxÞ þ sinðπxÞÞ = 0

)

-2≤x≤2 sinðπxÞ = 0 ) πx = kπ ) x = k ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) x = - 2, - 1, 0, 1, 2 3 1 1 3 π 1 -2≤x≤2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) x = - , - , , cosðπxÞ = 0 ) πx = kπ þ ) x = k þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ 2 2 2 2 2 2 5 1 3 7 π 1 -2≤x≤2 sinðπxÞ þ cosðπxÞ = 0 ) tanðπxÞ = - 1 ) πx = kπ - ) x = k - ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) x = - , - , , ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ 4 4 4 4 4 4

Therefore, the number of roots of the equation is: 5 + 4 + 4 = 13. Choice (3) is the answer. 4.51. From trigonometry, we know that: y = a þ b cosðmxÞ ) T =

2π jmj

Therefore: ) 4π =

2π 1 1 ) jm j = ) m = ± 2 2 jmj

Based on the graph and the function, the positive value of m is accepted. ) m= The value of function for x = y

16π 3

1 1 1 ) y = þ 2 cos x 2 2 2

is:

1 16π 1 8π 1 2π 1 2π 16π 1 = þ 2 cos = þ 2 cos 2π þ = þ 2 cos = þ 2 cos × 2 3 2 3 2 3 2 3 3 2

84

4 Solutions of Problems: Trigonometric Equations and Identities

=

1 π 1 π 1 1 1 = - 2 cos = -2 =þ 2 cos π 2 3 2 3 2 2 2

Choice (1) is the answer (Fig. 4.9).

Figure 4.9 The graph of solution of problem 4.51

4.52. From trigonometry, we know that: y = a þ b sinðmxÞ ) T =

2π jmj

Therefore: )

2π 2π = ) jmj = 3 ) m = ± 3 3 jmj

Based on the graph and the given function, the positive value of m is accepted. ) m = 3 ) y = 1 -sinð3xÞ The value of function for x = y

7π 6

is:

7π 7π 7π 3π 3π = 1 -sin 3 × = 1 -sin = 1 -sin 2π þ = 1 -sin = 1 - ð- 1Þ = 2 6 6 2 2 2

Choice (4) is the answer (Fig. 4.10).

Figure 4.10 The graph of solution of problem 4.52

4

Solutions of Problems: Trigonometric Equations and Identities

85

4.53. From trigonometry, we know that: y = a þ b sinðmxÞ ) T =

2π jmj

Therefore: ) 5-1=

2π 1 ) b= ± 2 jbπ j

Based on the graph and the given function, the positive value of m is accepted. ) b= -

1 π ) y = a þ sin - x 2 2

By testing the point of (0, 3) in the function, we have: 3 = a þ sin The value of function for x = y

25 3

π π × 0 ) a = 3 ) y = 3 þ sin - x 2 2

is: 25 π 25 π = 3 þ sin - × = 3 þ sin - 4π 3 2 3 6 = 3 þ sin -

π 1 = 3 - = 2:5 6 2

Choice (2) is the answer (Fig. 4.11).

Figure 4.11 The graph of solution of problem 4.53

86

4 Solutions of Problems: Trigonometric Equations and Identities

4.54. By testing the point of (0, 0) in the function, we have: 0 = a þ b cos

π ×0 ) a þ b=0 2

ð1Þ

Based on the function and the graph given in the problem, we can write: ymax = a þ jbj ) a þ jbj = 4

ð2Þ

The assumption of b < 0 is not acceptable because it results in the equations with an impossible solution, as can be seen in the following: Using ð1Þ, ð2Þ a þ b = 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) ) Impossible a þ b=4 However, for the assumption of b > 0, we have: Using ð1Þ, ð2Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼

a þ b=0 a-b=4

) 2b = - 4 ) b = - 2

Choice (1) is the answer (Fig. 4.12).

Figure 4.12 The graph of solution of problem 4.54

4.55. From trigonometry, we know that: y = 1 þ a sinðmxÞ ) T =

2π jmj

Therefore: )

4 2π =2× ) jbj = 3 ) b = ± 3 3 jbπ j

Based on the function and the graph given in the problem, we can write: ymin = 1 - jaj ) - 1 = 1 - jaj ) jaj = 2 ) a = ± 2 Based on the graph and the given function, both of a and b must be either positive or negative. Hence:

4

Solutions of Problems: Trigonometric Equations and Identities

87

a=2 ) a þ b=5 b=3 a= -2 ) a þ b= -5 b= -3 Only a + b = 5 exists in the choices. Choice (3) is the answer (Fig. 4.13).

Figure 4.13 The graph of solution of problem 4.55

4.56. From trigonometry, we know that: cos α þ

π = -sinðαÞ 2

Therefore: y = a - 2 cos bx þ

π = a þ 2 sinðbxÞ 2

In addition, from trigonometry, we know that: y = a þ 2 sinðbxÞ ) T =

2π 13π π 2π ) = ) jbj = 3 ) b = ± 3 18 18 jbj jbj

Based on the graph and the simplified function, i.e., y = a + 2 sin (bx), the positive value of b is acceptable. ) b=3 Based on the simplified function and the given graph, we can write: ymax = a þ 2 ) 1 = a þ 2 ) a = - 1 Hence: a þ b= -1 þ 3=2 Choice (4) is the answer (Fig. 4.14).

88

4 Solutions of Problems: Trigonometric Equations and Identities

Figure 4.14 The graph of solution of problem 4.56

4.57. From trigonometry, we know that: tan 45 ° = 1 tanðα - βÞ =

tanðαÞ -tanðβÞ 1 þ tanðαÞ tanðβÞ

cotðαÞ =

1 tanðαÞ

In addition, based on the information given in the problem, we have: tan α þ 20 ° =

3 4

Therefore: cot 25 ° - α =

=

1 1 = tan 45 ° - α þ 20 ° tan 25 ° - α

1 þ tan 45 ° tan α þ 20 ° tan 45

°

-tan α þ 20 °

=

1 þ tan α þ 20 ° 1 þ 34 = ° 1 - 34 1 -tan α þ 20

) cot 25 ° - α = 7 Choice (3) is the answer. 4.58. From trigonometry, we know that: cosðαÞ = -

1 - sin 2 ðαÞ for an obtuse angle tanðαÞ =

sinðαÞ cosðαÞ

4

Solutions of Problems: Trigonometric Equations and Identities

89

tan tanðα þ βÞ =

π =1 4 tanðαÞ þ tanðβÞ 1 -tanðαÞ tanðβÞ

In addition, based on the information given in the problem, we have: sinðαÞ =

3 5

Therefore: cosðαÞ = -

1 - sin 2 ðαÞ = -

tanðαÞ =

tan

1-

3 5

2

=-

4 5

3 sinðαÞ 3 = 54=4 cosðαÞ -5

tan π4 þ tanðaÞ 1 þ tanðαÞ 1 þ - 34 π = = = þα = π 4 1 -tanðαÞ 1 -tan 4 tanðaÞ 1 - - 34 ) tan

π 1 þα = 4 7

Choice (3) is the answer. 4.59. From trigonometry, we know that: tan

π - α = cotðαÞ 2

cotðαÞ = tan tanðα - βÞ =

1 tanðαÞ

π =1 4 tanðαÞ -tanðβÞ 1 þ tanðαÞ tanðβÞ

In addition, based on the information given in the problem, we have: tan

π 2 -α = 2 3

Therefore: 2 π 1 3 ) tanðαÞ = = tan - α = cotðαÞ = 3 2 2 tanðαÞ

1 4 7 4

90

4 Solutions of Problems: Trigonometric Equations and Identities

tan

tan π4 -tanðaÞ -1 1 - 32 1 -tanðαÞ π = 52 = = -α = 3 π 4 1 þ tan 4 tanðaÞ 1 þ tanðαÞ 1 þ 2 2 ) tan

π 1 -α = 4 5

Choice (2) is the answer. 4.60. From trigonometry, we know that: tanðα þ βÞ =

tanðαÞ þ tanðβÞ 1 -tanðαÞ tanðβÞ

In addition, based on the information given in the problem, we have: tanða þ bÞ =

2 5

tanða - bÞ =

3 7

Therefore: tanð2aÞ = tanðða þ bÞ þ ða - bÞÞ =

2 þ3 tanða þ bÞ þ tanða - bÞ = 5 27 1 -tanða þ bÞ tanða - bÞ 1 - 5 ×

) tanð2aÞ = 1 Choice (4) is the answer. 4.61. From trigonometry, we know that: sinðαÞ = cos

π -α 2

cosðαÞ = cosð- αÞ tan tanðα - βÞ =

π =1 4 tanðαÞ -tanðβÞ 1 þ tanðαÞ tanðβÞ

Therefore: sin

π π π -x þ x = cos 2 4 4 ) 2=

= cos

π π - x = cos x 4 4

sin x - π4 sin x - π4 = sin x þ π4 cos x - π4

3 7

=

29 35 29 35

4

Solutions of Problems: Trigonometric Equations and Identities

) tan x -

)

91 π

tanðxÞ -tan 4 π =2 ) =2 4 1 þ tanðxÞ tan π4

tanðxÞ - 1 = 2 ) tanðxÞ - 1 = 2 þ 2 tanðxÞ 1 þ tanðxÞ ) tanðxÞ = - 3

Choice (1) is the answer. 4.62. From trigonometry, we know that: π =1 4

tan

tanðαÞ þ tanðβÞ 1 -tanðαÞ tanðβÞ

tanðα þ βÞ =

Moreover, based on the information given in the problem, we have: αþβ=

π 4

If we calculate the tangent value of each side of the abovementioned relation, we will have: tanðα þ βÞ = tan

tanðαÞ þ tanðβÞ π ) = 1 ) tanðαÞ þ tanðβÞ = 1 -tanðαÞ tanðβÞ 4 1 -tanðαÞ tanðβÞ

Therefore: ð1 þ tanðαÞÞð1 þ tanðβÞÞ = 1 þ tanðαÞ þ tanðβÞ þ tanðαÞ tanðβÞ = 1 þ ð1 -tanðαÞ tanðβÞÞ þ tanðαÞ tanðβÞ ) ð1 þ tanðαÞÞð1 þ tanðβÞÞ = 2 Choice (2) is the answer. 4.63. From trigonometry, we know that: π =1 4

tan tanðα þ βÞ =

tanðαÞ þ tanðβÞ 1 -tanðαÞ tanðβÞ

tanðα - βÞ =

tanðαÞ -tanðβÞ 1 þ tanðαÞ tanðβÞ

tanð2αÞ =

2 tanðαÞ 1 - tan 2 ðαÞ

92

4 Solutions of Problems: Trigonometric Equations and Identities

Therefore: tan

tan π4 -tanðαÞ tan π4 þ tanðαÞ π π þ α -tan - α = π 4 4 1 -tan 4 tanðαÞ 1 þ tan π4 tanðαÞ =

=

1 þ tanðαÞ 1 -tanðαÞ 1 -tanðαÞ 1 þ tanðαÞ

ð1 þ tanðαÞÞ2 - ð1 -tanðαÞÞ2 1 - tan 2 ðαÞ =

) tan

4 tanðαÞ 1 - tan 2 ðαÞ

π π þ α -tan - α = 2 tanð2αÞ 4 4

Choice (1) is the answer. 4.64. From trigonometry, we know that: π =1 4

tan tanðα - βÞ =

tanðαÞ -tanðβÞ 1 þ tanðαÞ tanðβÞ

tanð2αÞ =

2 tanðαÞ 1 - tan 2 ðαÞ

Moreover, based on the information given in the problem, we have: tan

tan π4 -tanðαÞ 1 -tanðαÞ π 1 1 = = -α = ) 4 5 1 þ tan π4 tanðαÞ 1 þ tanðαÞ 5 ) 5 - 5 tanðαÞ = 1 þ tanðαÞ ) tanðαÞ = ) tanð2αÞ =

2 3

2 × 23 2 tanðαÞ 12 = = 5 1 - tan 2 ðαÞ 1 - 2 2 3 ) tanð2αÞ = 2:4

Choice (3) is the answer. 4.65. From trigonometry, we know that: tanð2αÞ = tanðα - βÞ =

2 tanðαÞ 1 - tan 2 ðαÞ

tanðαÞ -tanðβÞ 1 þ tanðαÞ tanðβÞ

4

Solutions of Problems: Trigonometric Equations and Identities

93

Moreover, based on the information given in the problem, we have: tanðαÞ = 2 tanðβÞ =

1 3

Therefore: tanð2αÞ =

2 tanðαÞ 2×2 4 =) tanð2αÞ = 3 1 - tan 2 ðαÞ 1 - 22

tanð2α - βÞ =

tanð2αÞ -tanðβÞ 1 þ tanð2αÞ tanðβÞ

) tanð2α - βÞ =

- 43 1 þ - 43

1 3 1 3

=

-

5 3

5 9

) tanð2α - βÞ = - 3 Choice (1) is the answer. 4.66. From trigonometry, we know that: cosðπ - xÞ = -cosðxÞ cosðαÞ = cosðα0 Þ ) α = 2kπ ± α0 , 8k 2 ℤ Moreover, based on the information given in the problem, we have: cosðxÞ ≠ 0 Therefore: cosð3xÞ þ cosðxÞ = 0 ) cosð3xÞ = -cosðxÞ ) cosð3xÞ = cosðπ - xÞ

) 3x = 2kπ ± ðπ - xÞ )

3x = 2kπ þ π - x ) 4x = 2kπ þ π ) 3x = 2kπ - π þ x ) 2x = 2kπ - π

However: cosðxÞ ≠ 0 ) x = Choice (1) is the answer.

kπ π þ 2 4

kπ π þ 2 4 π x = kπ 2

x=

94

4 Solutions of Problems: Trigonometric Equations and Identities

4.67. From trigonometry, we know that: cotðαÞ = tan

π -α 2

tanðαÞ = tanðα0 Þ ) α = kπ þ α0 , 8k 2 ℤ Therefore: tanð4xÞ = cotðxÞ ) tanð4xÞ = tan ) 4x = kπ þ

)

π -x 2

π π kπ π - x ) 5x = kπ þ ) x = þ 2 2 5 10

π is not a positive angle k = - 1 ) x1 = 10 π k = 0 ) x2 = is an acute angle 2π 10 ) x2 þ x 3 = 3π 5 k = 1 ) x3 = is an acute angle 10 π k = 2 ) x4 = is not an acute angle 2

Choice (1) is the answer. 4.68. From trigonometry, we know that: sin 2 ðxÞ þ cos 2 ðxÞ = 1 cosðxÞ = cosðx0 Þ ) x = 2kπ ± x0 Therefore: 2 sin 2 ðxÞ þ 3 cosðxÞ = 0 ) 2 1 - cos 2 ðxÞ þ 3 cosðxÞ = 0 ) 2 cos 2 ðxÞ - 3 cosðxÞ - 2 = 0 ) cos 2 ðxÞ -

3 1 cosðxÞ - 1 = 0 ) cosðxÞ þ ðcosðxÞ - 2Þ = 0 2 2 )

1 2π ) x = 2kπ ± 2 3 cosðxÞ = 2 ) not acceptable

cosðxÞ = -

Choice (1) is the answer. 4.69. From trigonometry, we know that: sin 2 ðxÞ þ cos 2 ðxÞ = 1 cosðxÞ = cosðx0 Þ ) x = 2kπ ± x0 Therefore: 2 sin 2 ðxÞ = 3 cosðxÞ ) 2 1 - cos 2 ðxÞ - 3 cosðxÞ = 0 ) 2 cos 2 ðxÞ þ 3 cosðxÞ - 2 = 0

4

Solutions of Problems: Trigonometric Equations and Identities

) cos 2 ðxÞ þ

95

3 1 cosðxÞ - 1 = 0 ) cosðxÞ - ðcosðxÞ þ 2Þ = 0 2 2 )

1 π ) x = 2kπ ± 2 3 cosðxÞ = - 2 ) not acceptable

cosðxÞ =

Choice (4) is the answer. 4.70. From trigonometry, we know that: tanðαÞ: cotðαÞ = 1 Now, let us find the intersection point of the lines, as follows: x tanðαÞ - y cotðαÞ = 1 ) x tanðαÞ þ y cotðαÞ = 2

) xy =

3 2 tanðαÞ 1 2y cotðαÞ = 1 ) y = 2 cotðαÞ

2x tanðαÞ = 3 ) x =

3 1 3 3 × = ) y= 4x 2 tanðαÞ 2 cotðαÞ 4

Choice (4) is the answer. 4.71. From trigonometry, we know that: sin 2 ðαÞ þ cos 2 ðαÞ = 1 Based on the information given in the problem, we have: x-2 -3 y-1 y = 1 þ 4 cosðαÞ ) cosðαÞ = 4

x = 2 - 3 sinðαÞ ) sinðαÞ =

Therefore: )

ðx - 2Þ2 ðy - 1Þ2 þ =1 9 16

which is the equation of an ellipse. Choice (2) is the answer. 4.72. Based on the information given in the problem, we have: x = 2 - 5 cosðαÞ y=4 From trigonometry, we know that: - 1 ≤ cosðαÞ ≤ 1 ) - 1 ≤

2-x ≤1 5

96

4 Solutions of Problems: Trigonometric Equations and Identities

) -5≤2-x≤5 ) -7≤ -x≤3 ) -3≤x≤7 Therefore: )

-3≤x≤7 y=4

which is the equation of a horizontal line segment. Choice (3) is the answer. 4.73. From trigonometry, we know that the maximum value of cos(.) and sin(.) is one. Therefore, the only solution of the given equation is: cosðx - yÞ = 1 sinðx þ yÞ = 1 The common solution of the equations can be calculated as follows: )

x - y = 2kπ

0 < x, y < 2π ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) π¼ x þ y = 2kπ þ 2

x-y=0 π 5π π 5π ) y = 4 or 4 x þ y = or 2 2

Choice (2) is the answer. 4.74. From trigonometry, we know that: tanðα þ βÞ =

tanðαÞ þ tanðβÞ 1 -tanðαÞ tanðβÞ

In addition, we know that the sum and the product of the roots of a quadratic equation in the form of ax2 + bx + c = 0 are - ba and ac, respectively. Based on the information given in the problem, we have: tan 2 ðxÞ þ ðm þ 2Þ tanðxÞ þ 2m - 2 = 0 αþβ=

π 4

Therefore: tanð:Þ tanðαÞ þ tanðβÞ =1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) tanðα þ βÞ = 1 ) ¼ ¼ 1 -tanðαÞ tanðβÞ - ðmþ2Þ

)

sum of the roots of the quadratic equation -m-2 1 = = =1 1 - product of the roots of the quadratic equation 1 - 2m1- 2 3 - 2m ) - m - 2 = 3 - 2m ) m = 5

Choice (3) is the answer.

4

Solutions of Problems: Trigonometric Equations and Identities

97

4.75. From trigonometry, we know that: sin 6 ðαÞ þ cos 6 ðαÞ = 1 - 3 sin 2 ðαÞ cos 2 ðαÞ sin 4 ðαÞ þ cos 4 ðαÞ = 1 - 2 sin 2 ðαÞ cos 2 ðαÞ Therefore: sin 6 ðαÞ þ cos 6 ðαÞ þ 3 sin 2 ðαÞ cos 2 ðαÞ 1 - 3 sin 2 ðαÞ cos 2 ðαÞ þ 3 sin 2 ðαÞ cos 2 ðαÞ = sin 4 ðαÞ þ cos 4 ðαÞ þ 2 sin 2 ðαÞ cos 2 ðαÞ 1 - 2 sin 2 ðαÞ cos 2 ðαÞ þ 2 sin 2 ðαÞ cos 2 ðαÞ )

sin 6 ðαÞ þ cos 6 ðαÞ þ 3 sin 2 ðαÞ cos 2 ðαÞ =1 sin 4 ðαÞ þ cos 4 ðαÞ þ 2 sin 2 ðαÞ cos 2 ðαÞ

Choice (4) is the answer. 4.76. From trigonometry, we know that: sin 135 ° = sin 180 ° - 45 ° = sin 45 ° cos 210 ° = cos 180 ° þ 30 ° = -cos 30 ° cos 135 ° = cos 180 ° - 45 ° = -cos 45 ° sin 420 ° = sin 360 ° þ 60 ° = sin 60 ° tan 210 ° = tan 180 ° þ 30 ° = tan 30 ° cot 420 ° = cot 360 ° þ 60 ° = cot 60 ° cot 120 ° = cot 180 ° - 60 ° = -cot 60 ° tan 330 ° = tan 360 ° - 30 ° = -tan 30 ° Therefore: sin 45 ° -cos 30 ° þ -cos 45 ° sin 60 ° = -tan 30 ° tan 30 ° cot 60 ° þ -cot 60 °

p

p - 3 2 2 × 2 p p 3 3 × 3 þ 3

þ -

p 3 3

p sin 45 ° -cos 30 ° þ -cos 45 ° sin 60 ° -3 6 = ) 4 -tan 30 ° tan 30 ° cot 60 ° þ -cot 60 ° Choice (2) is the answer.

p

2 2

-

p p

3 2

3 3

98

4 Solutions of Problems: Trigonometric Equations and Identities

4.77. From trigonometry, we know that: cotðx þ yÞ =

cotðxÞ cotðyÞ - 1 cotðxÞ þ cotðyÞ

Based on the information given in the problem, we have: π k=0 π cotð:Þ cotðxÞ cotðyÞ - 1 x þ y = kπ þ ¼ =1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) x þ y = ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 4 4 cotðxÞ þ cotðyÞ ) 1 þ cotðxÞ þ cotðyÞ = cotðxÞ cotðyÞ

ð1Þ

ð1 þ cotðxÞÞð1 þ cotðyÞÞ = ð1 þ cotðxÞ þ cotðyÞÞ þ cotðxÞ cotðyÞ

ð2Þ

On the other hand, we can write:

Solving (1) and (2): ð1 þ cotðxÞÞð1 þ cotðyÞÞ = cotðxÞ cotðyÞ þ cotðxÞ cotðyÞ ) ð1 þ cotðxÞÞð1 þ cotðyÞÞ = 2 cotðxÞ cotðyÞ Choice (4) is the answer. 4.78. From trigonometry, we know that: tan cos

3π - x = cotðxÞ 2

4π π 1 = -cos =3 3 2 tanðxÞ cotðxÞ = 1 cotðxÞ =

cosðxÞ sinðxÞ

cosðxÞ = cosðx0 Þ ) x = 2kπ ± x0 Therefore: ðsinðxÞ -tanðxÞÞ tan

3π 4π 1 ) ðsinðxÞ -tanðxÞÞ cotðxÞ = - x = cos 2 3 2

) sinðxÞ cotðxÞ -tanðxÞ cotðxÞ = -

1 1 1 ) cosðxÞ - 1 = - ) cosðxÞ = 2 2 2

) x = 2kπ ± Choice (3) is the answer.

π 3

4

Solutions of Problems: Trigonometric Equations and Identities

99

4.79. From trigonometry, we know that: sinðα þ βÞ = sinðαÞ cosðβÞ þ sinðαÞ cosðβÞ cosðα þ βÞ = cosðαÞ cosðβÞ -sinðαÞ sinðβÞ tanðxÞ =

sinðxÞ cosðxÞ

tanðxÞ = tanðx0 Þ ) x = kπ þ x0 Therefore: sinð2xÞðsinðxÞ þ cosðxÞÞ = cosð2xÞðcosðxÞ -sinðxÞÞ ) sinð2xÞ sinðxÞ þ sinð2xÞ cosðxÞ = cosð2xÞ cosðxÞ -cosð2xÞ sinðxÞ ) sinð2xÞ cosðxÞ þ cosð2xÞ sinðxÞ = cosð2xÞ cosðxÞ -sinð2xÞ sinðxÞ 1 × & cosð3xÞ ≠ 0 cosð3xÞ ¼ ¼ ¼ ¼ ¼ ) sinð2x þ xÞ = cosð2x þ xÞ ) sinð3xÞ = cosð3xÞ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) tanð3xÞ = 1 ) 3x = kπ þ

π kπ π k = 0, 1, 2&x 2 ½0, π ] π 5π 9π ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) x1 = þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) x= ,x = ,x = 4 3 12 12 2 12 3 12 ) x 1 þ x2 þ x3 =

5π 4

Choice (2) is the answer. 4.80. From trigonometry, we know that: sin

5π π þ x = sin þ x = cosðxÞ 2 2

sinðα þ βÞ = sinðαÞ cosðβÞ þ sinðαÞ cosðβÞ sin

p π π 2 = cos = 4 4 2

sinðxÞ = sinðx0 Þ )

x = 2kπ þ x0 x = 2kπ þ π - x0

Therefore: )

p

2 sin

p π π π cosðxÞ -cos sinðxÞ = 1 þ cosðxÞ - x = 1 þ cosðxÞ ) 2 sin 4 4 4

100

4 Solutions of Problems: Trigonometric Equations and Identities

) cosðxÞ -sinðxÞ = 1 þ cosðxÞ ) sinðxÞ = - 1 )

x = 2kπ þ -

π 2

x = 2kπ þ π - -

π 2

π 2 3π x = 2kπ þ 2 x = 2kπ -

)

) x = 2kπ -

π 2

Choice (3) is the answer. 4.81. From trigonometry, we know that: tan

p π = 3 3

tanðxÞ =

cos

sinðxÞ cosðxÞ

π 1 = 3 2

cosðα - βÞ = cosðαÞ cosðβÞ þ sinðαÞ sinðβÞ cosðxÞ = cosðx0 Þ ) x = 2kπ ± x0 Therefore: cosð2xÞ þ

p

3 sinð2xÞ = 1 ) cosð2xÞ þ tan

) cosð2xÞ cos

sin π sinð2xÞ = 1 ) cosð2xÞ þ 3 cos

π 3 π 3

sinð2xÞ = 1

π π π π π þ sin sinð2xÞ = cos ) cos 2x = cos 3 3 3 3 3

π π ) 2x - = 2kπ ± ) 3 3

π π π = 2kπ þ ) x = kπ þ 3 3 3 π π 2x - = 2kπ - ) x = kπ 3 3 2x -

Choice (4) is the answer. 4.82. From trigonometry, we know that: sin a sin b =

1 ½cosða - bÞ -cosða þ bÞ] 2

sin a cos b =

1 ½sinða þ bÞ þ sinða - bÞ] 2

cos a þ cos b = 2 cos

aþb a-b cos 2 2

References

101

sin a -sin b = 2 cos

aþb a-b sin 2 2

Therefore: cos 3α þ 12 ½cos α -cos 3α] sin α cos 3α þ sin α sin 2α sin α × × = sin 3α -sin 2α cos α cos α sin 3α - 12 ½sin 3α þ sin α] cos α =

1 2 cos 3α þ 1 2 sin 3α -

1 2 1 2

cos α sin α cos 3α þ cos α sin α 2 cos 2α cos α sin α × = × = × cos α sin 3α -sin α cos α 2 cos 2α sin α cos α sin α )

cos 3α þ sin α sin 2α sin α × =1 sin 3α -sin 2α cos α cos α

Choice (3) is the answer.

References 1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.

5

Problems: Limits and Continuities

Abstract

In this chapter, the basic and advanced problems of limits and continuities are presented. The subjects include limits by direct substitution, limits by factoring, limits by rationalization, limits at infinity, trigonometric limits, limits of absolute value functions, limits involving Euler’s number, limits by L’Hopital’s rule, application of Taylor series in limits, and limits and continuity. To help students study the chapter in the most efficient way, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Moreover, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations. 5.1. Determine the continuity status of the following function [1, 2]: f ð xÞ ¼

10jxj x ≠ 0 0

x¼0

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) It is continuous everywhere except from the right-hand side of x ¼ 0. 2) It is continuous everywhere except from the left-hand side of x ¼ 0. 3) It is continuous everywhere. 4) It is continuous everywhere except at x ¼ 0. 5.2. What is the continuity status of the function below? f ðxÞ ¼

j xj x ≠ 0 1 x¼0

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) It is continuous everywhere except from the right-hand side of x ¼ 0. 2) It is continuous everywhere except from the left-hand side of x ¼ 0. 3) It is continuous everywhere. 4) It is continuous everywhere except at x ¼ 0.

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_5

103

104

5

5.3. Calculate the value of k if the function below is continuous at x ¼ 2. f ðxÞ ¼

ð x þ 2Þ ½ - x  x < 2 xþk

x≥2

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 10 2) -10 3) 6 4) -8 5.4. For which value of the parameter of “a” the function below is continuous at ¼ -2 ? f ðxÞ ¼

jxj½x þ a j xj þ ½ x

x < -2 x ≥ -2

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 2 3) 3 4) 6 5.5. Calculate the value of the following limit: lim

þ

x → ð- 1Þ

½ x þ 1 x2 - 1

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1 1) 2 2) 0 1 3) 2 4) 1 5.6. Calculate the limit of the following function if x → 2+. f ðxÞ ¼ Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) -1 3) 2 4) -2

xþ4 ½ -x -3

Problems: Limits and Continuities

5

Problems: Limits and Continuities

105

5.7. Determine the value of the following limit: lim

x → 0-

xþ2 ½ x

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2 2) -2 3) 1 4) -1 5.8. Determine the value of the limit below. lim

x→ -1

½x þ 3x ½x - 3x

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2 2) -2 3) 4 4) -4 5.9. Calculate the limit of the function below if x → 0. f ðxÞ ¼

p xþ 3 x p x- 3 x

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 1 3) -1 4) 1 5.10. Calculate the value of the following limit: lim-

x→0

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) -1 3) 0 4) 1

½ x x

106

5

5.11. Determine the limit of the function below if x → 0+. f ðxÞ ¼

p ðx2 - 1Þ x p ðx x þ 1Þx

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) -1 3) 0 4) -1 5.12. Determine the value of the following limit: lim

x → 0þ

1 1 x x3

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) -1 3) 0 4) 1 5.13. For the function below, calculate the value of limþ f ðxÞ - lim- f ðxÞ. x→1

x→1

f ðxÞ ¼

2x ½2x þ 2

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) -1 1 2) 6 2 3) 3 4) 1 5.14. Calculate the value of limþ ð½x - 2Þ½x. x→2

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) -2 2) -1 3) 0 4) 1 5.15. Calculate the limit of the following function if x → 4-: f ðxÞ ¼

½ x - 4 x2 - 16

Problems: Limits and Continuities

5

Problems: Limits and Continuities

107

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 1 2) 8 3) 1 4) -1 5.16. Determine the value of the limit below. lim-

x→1

1 - x3 arcðcosðxÞÞ

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) -1 3) 0 4) -3 5.17. Calculate the value of the limit below. lim

x→0

tanðxÞ - tanð3xÞ þ tanð2xÞ x3

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) -6 2) 6 3) 10 4) -10 5.18. Calculate the value of the following limit: 9 - x2 p x→3 2 - x þ 1 lim

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 6 2) 12 3) 18 4) 24 5.19. Determine the limit of the function below if x → + 1. f ð xÞ ¼ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large

sinðxÞ x

108

5

1) 2) 3) 4)

Undefined 0 1 1

5.20. Determine the value of the limit below. ½ x2  - x2 x tanðxÞ

lim

x→0

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) -1 3) 2 4) -2 5.21. Determine the value of the following limit: lim x sin

x → þ1

1 x

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) -1 3) 0 4) Undefined 5.22. Calculate the value of the following limit: lim

x→ -1

x2 þ x - 1 p - 3x þ 4 - x

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1 1) 3 1 2) 3 3) 1 4) -1 5.23. Calculate the value of the limit below. limþ

x→0

p ðx þ 1Þ x x2 - x

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0

Problems: Limits and Continuities

5

Problems: Limits and Continuities

109

2) -1 3) 1 4) -1 5.24. Determine the limit of the following function if x → + 1: f ðxÞ ¼

x-1 þ

x p x2 þ x - 1

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 0 1 3) 2 1 4) 2 5.25. Calculate the value of lim x cotðxÞ. x→0

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 1 3) 1 4) 2 5.26. For what value of “a” the following function has a definite limit at x ¼ 1? f ðxÞ ¼

x2 þ ax x-3

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 3 3) -3 4) -2 5.27. Determine the value of the limit below. lim-

x→2

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) -24 2) -16 3) 16 4) 24

x3 - 8 p x - 2x

x>1 x n

Therefore: lim

x→ -1

x2 þ x - 1 p - 3x þ 4 - x

- lim

x→ -1

x x2 ¼ þ1 ¼ lim 3 - 3x x → - 1

Choice (3) is the answer. 6.23. From calculus, we know that: lim am xm þ am - 1 xm - 1 þ . . . þ am - n xm - n þ am - n - 1 xm - n - 1 - am - n - 1 xm - n - 1

x→0

or: lim ðam xm þ an xn Þ - an xn

x→0

if m > n

The problem can be solved as follows: p p p p ð x þ 1Þ x x x xþ x -1 ¼ limþ - limþ limþ ¼ limþ p ¼ - 1 2-x 2-x x x x x x→0 x→0 x→0 x→0 Choice (4) is the answer.

6

Solutions of Problems: Limits and Continuities

125

6.24. From calculus, we know that: lim

x → þ1

x2 þ ax þ b - x þ

a 2

The problem can be solved as follows: lim

x → þ1 x - 1

þ

x x x x 1 1 p ¼ lim - lim - lim ¼ lim ¼ x → þ1 2x - 1 x → þ1 2x x → þ1 2 2 x2 þ x - 1 x → þ1 x - 1 þ x þ 12 2

Choice (4) is the answer. 6.25. From trigonometry, we know that: cotðxÞ ¼

1 tanðxÞ

From application of Taylor series in limit, we know that: lim tanðxÞ - x

x→0

The problem can be solved as follows: lim x cotðxÞ ¼ lim

x→0

x→0

x x - lim ¼ 1 tanðxÞ x → 0 x

Choice (3) is the answer. 6.26. As we know, the limit of a function at the point of x0 exits if: lim f ðxÞ ¼ lim þ f ðxÞ ) lim- f ðxÞ ¼ limþ f ðxÞ

ð1Þ

lim f ðxÞ ¼ lim- ðx - 3Þ ¼ 1 - 3 ¼ - 2

ð2Þ

lim f ðxÞ ¼ limþ x2 þ ax ¼ 1 þ a

ð3Þ

x → x0 -

x→1

x → x0

x→1

Therefore: x → 1-

x→1

x → 1þ

x→1

Using ð1Þ, ð2Þ, ð3Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) - 2 ¼ 1 þ a ) a ¼ -3 ¼ ¼ ¼ ¼ ¼ Choice (3) is the answer. 6.27. The problem can be solved as follows: lim-

x→2

x3 - 8 - ðx3 - 8Þ 0þ p p ¼ lim¼ 0 x - 2x x → 2 x - 2x

d ð- ðx3 - 8ÞÞ H - 3x2 - 3 x 22 - 12 p ¼ ¼ ¼ ¼ ¼ ¼) lim- dxd ¼ ¼ 1 ¼ -24 ¼ lim1 1 p p x→2 x → 2 1 1 2 dx x - 2x 2x 2x2

Choice (1) is the answer.

126

6

Solutions of Problems: Limits and Continuities

6.28. The problem can be solved as follows: lim-

x→0

½x] þ x - 1 þ x - 1 þ 0-1 ¼ lim¼ - ¼ þ1 ¼ 0 þ 00þx 0 ½ - x] þ x x → 0

Choice (1) is the answer. 6.29. The problem can be solved as follows: sinð3xÞ þ sinð7xÞ 0 ¼ 0 3x þ tanð2xÞ

lim

x→0 d dx ðsinð3xÞ d x→0 dx ð3x þ

H ¼ ¼ ¼ ¼ ¼ ¼) lim

þ sinð7xÞÞ 3 cosð3xÞ þ 7 cosð7xÞ 3 þ 7 ¼ ¼2 ¼ lim 3þ2 x → 0 3 þ 2ð1 þ tan 2 ð2xÞÞ tanð2xÞÞ

Choice (2) is the answer. 6.30. The problem can be solved as follows: p lim

x→0

p p p p p x þ 3-3 x þ 3- 3 xþ3- 3 xþ3þ 3 p p ¼ lim p ¼ lim xp x x x→0 x → 0 x xþ3þ 3 xþ3þ 3 ¼ lim p x→0

p 1 1 1 3 p ¼p p ¼ p ¼ 6 xþ3þ 3 3þ 3 2 3

Choice (2) is the answer. 6.31. The problem can be solved as follows: lim

x→0

1 - cosðxÞ 0 ¼ 0 sinðxÞ

d dx ð1 - cosðxÞÞ d x→0 dx sinðxÞ

H ¼ ¼ ¼ ¼ ¼ ¼) lim

¼ lim

x→0

sinðxÞ 0 ¼ ¼0 cosðxÞ 1

Choice (1) is the answer. 6.32. The problem can be solved as follows: lim

x→0

5x - sinðxÞ 0 ¼ 2x þ cosðxÞ - 1 0

d dx ð5x - sinðxÞÞ d x→0 dx ð2x þ cosðxÞ - 1Þ

H ¼ ¼ ¼ ¼ ¼ ¼) lim Choice (2) is the answer.

¼ lim

x→0

5 - cosðxÞ 5 - 1 ¼ ¼2 2 - sinðxÞ 2 - 0

6

Solutions of Problems: Limits and Continuities

127

6.33. The problem can be solved as follows: lim-

x→2

ðx - 2Þðx2 þ 2x þ 4Þ x3 - 8 þ 5x þ 5x ¼ limx→2 - ð x - 2Þ j x - 2j

¼ lim- - x2 - 2x - 4 þ 5x x→2

lim - x2 þ 3x - 4 ¼ - 4 þ 6 - 4 ¼ - 2

x → 2-

Choice (2) is the answer. 6.34. The problem can be solved as follows: sinðxÞ þ cosðxÞ 1 - þ 0 - 1 ¼ ¼ - ¼ -1 00 cosðxÞ

lim þ x → ðπ2Þ Choice (2) is the answer.

6.35. The problem can be solved as follows: lim arcðsinðxÞÞ - x

x→0

lim 3x4 þ 2x3 - 2x3

x→0

lim

x→0

3x4 þ 2x3 2x3 - lim 3 ¼ lim 2 ¼ 2 3 x→0 x x→0 ðarcðsinðxÞÞÞ

Choice (2) is the answer. 6.36. The problem can be solved as follows: lim

x→ -1

2 x ¼ lim ½0 - ]x ¼ ð- 1Þð- 1Þ ¼ þ1 x→ -1 xþ1

Choice (1) is the answer. 6.37. The problem can be solved as follows: lim p

x → 3þ

x-4 ¼ lim x2 - 4x þ 3 x → 3þ

x-4 -1 ¼ þ ¼ -1 0 ð x - 3Þ ð x - 1Þ

Choice (2) is the answer. 6.38. From calculus, we know that: lim

x→ ±1

x2 þ ax þ b - x þ

a 2

The problem can be solved as follows: lim

x→ -1

xþ

x2 þ 4x - 10 - lim ðx þ jx þ 2jÞ ¼ lim ðx - x - 2Þ ¼ lim ð- 2Þ ¼ - 2

Choice (2) is the answer.

x→ -1

x→ -1

x→ -1

128

6

Solutions of Problems: Limits and Continuities

6.39. The problem can be solved as follows: 0 4 - x2 p ¼ 2 0 x→2 6 - 2 x þ 5 lim

H ¼ ¼ ¼ ¼ ¼ ¼) lim

x→2 d dx

2 d dx ð4 - x Þ p 6 - 2 x2 þ

5

¼ lim

x→2

- 2x ¼ lim x→2 - 2 x 2p2x x2 þ5

x2 þ 5 ¼ 3

Choice (3) is the answer. 6.40. The problem can be solved as follows: lim p

x→0

d dx ðsinð2xÞÞ

H ¼ ¼ ¼ ¼ ¼ ¼) lim

p

x→0 d dx

x þ 1-1

sinð2xÞ 0 ¼ x þ 1-1 0 ¼ lim

2 cosð2xÞ

x→0

p1 2 xþ1

¼

2x1 1 2

¼4

Choice (2) is the answer. 6.41. From calculus, we know that: lim

x→ -1

lim

x→ ±1

x4 þ 2x2 þ x - x2

am x m þ am - 1 x m - 1 þ . . . þ a2 x 2 þ a1 x þ a0 - am x m

or: lim ðam xm þ an xn Þ - am xm

x→ ±1

if m > n

The problem can be solved as follows: p lim x→ -1

x4 þ 2x2 þ x - x ¼ x →lim -1 2

x4 þ 2x2 þ x - x

2

x p

x4 þ 2x2 þ x þ x2 x4 þ 2x2 þ x þ x2

x4 þ 2x2 þ x - x4 2x2 þ x 2x2 p lim ¼ x →lim 1¼1 ¼ x →lim ¼ lim 2 2 -1 x → - 1 ðx þ x Þ x → - 1 2x2 -1 x4 þ 2x2 þ x þ x2 Choice (1) is the answer. 6.42. From calculus, we know that the limit of a function at a specific point (x0) exits if: lim f ðxÞ ¼ lim þ f ðxÞ

x → x0 -

x → x0

Therefore, we must have: lim

x → ð- 3Þ

-

x2 - 9 x2 - 9 ¼ lim þ xþ3 xþ3 x → ð- 3Þ

ð1Þ

6

Solutions of Problems: Limits and Continuities

lim

x → ð- 3Þ

lim

x → ð- 3Þþ

129

x2 - 9 x2 - 9 ¼ lim ¼ lim - ðx - 3Þ ¼ - 6 xþ3 x → ð- 3Þ x þ 3 x → ð- 3Þ

ð2Þ

x2 - 9 - ð x2 - 9Þ ¼ lim þ ¼ lim þ - ðx - 3Þ ¼ 6 xþ3 xþ3 x → ð- 3Þ x → ð- 3Þ

ð3Þ

-

ð1Þ, ð2Þ, ð3Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) - 6 ≠ 6 ) lim

x→ -3

x2 - 9 ¼ Undefined xþ3

Choice (4) is the answer. 6.43. The problem can be solved as follows: lim

x → 12

H ¼ ¼ ¼ ¼ ¼ ¼) lim

d dx

x → 12

tan πx 0 2 -1 ¼ 0 cosðπxÞ

tan πx 2 -1 ¼ lim d x → 12 dx ðcosðπxÞÞ

π 2

1 þ tan 2 πx 2 - π sinðπxÞ

π ð 1 þ 1Þ ¼2 ¼ -1 -π x 1

Choice (2) is the answer. 6.44. The problem can be solved as follows: lim

p

x → þ1

p x þ 5 - x þ 1 ¼ lim

x → þ1

¼ lim

x → þ1

p

p

p

x þ 5- x þ 1 x p

xþ5þ xþ5þ

p p

xþ1 xþ1

x þ 5 - ð x þ 1Þ 4 p p p ¼0 ¼ lim p x → þ1 xþ5þ xþ1 xþ5þ xþ1

Choice (3) is the answer. 6.45. From trigonometry, we know that: 1 þ cosð2xÞ ¼ 2 cos 2 ðxÞ tanðxÞ ¼

sinðxÞ cosðxÞ

sinð2xÞ ¼ 2 sinðxÞ cosðxÞ The problem can be solved as follows: limπ

x→2

tanð2xÞ cosðxÞ sinð2xÞ cosðxÞ 2 sinðxÞ cos 2 ðxÞ sinðxÞ 1 ¼ limπ ¼ ¼ limπ ¼ limπ ¼ -1 2 -1 x → 2 cosð2xÞ x 2 cos ðxÞ x → 2 cosð2xÞ x 2 cos 2 ðxÞ x → 2 cosð2xÞ 1 þ cosð2xÞ

Choice (3) is the answer.

130

6

Solutions of Problems: Limits and Continuities

6.46. From calculus and trigonometry, we know that: 1 - cos 2 ðxÞ ¼ sin 2 ðxÞ Moreover, from application of Taylor series in limit, we know that: lim sinðxÞ - x

x → 0-

lim tanðxÞ - x

x → 0-

The problem can be solved as follows: tanð2xÞ ¼ lim 1 - cosðxÞ x → 0 -

lim

x → 0-

¼ lim-

tanð2xÞ x

x→0

p

2

sin 2 ðxÞ

tanð2xÞ x 1 - cosðxÞ

1 þ cosðxÞ 1 þ cosðxÞ

¼ lim-

tanð2xÞ

1 þ cosðxÞ

1 - cos 2 ðxÞ

x→0

p p p tanð2xÞ x 2 tanð2xÞ x 2 2x x 2 ¼ lim¼ lim¼ lim-x x→0 x→0 x→0 - sinðxÞ j sinðxÞj p p ¼ lim- - 2 2 ¼ - 2 2 x→0

Choice (1) is the answer. 6.47. The problem can be solved as follows: lim

p 3

p n þ 1000 - 3 n - 20

3

ðn þ 1000Þ2 þ

3

ðn þ 1000Þðn - 20Þ þ

3

ðn - 20Þ2

3

ðn þ 1000Þ2 þ

3

ðn þ 1000Þðn - 20Þ þ

3

ðn - 20Þ2

x→ -1

¼ lim

x→ -1

p 3

p 3

n þ 1000 - n - 20 x

¼ lim

x→ -1

n þ 1000 - ðn - 20Þ 3

ðn þ 1000Þ2 þ

3

ðn þ 1000Þðn - 20Þ þ

Choice (2) is the answer. 6.48. From application of Taylor series in limit, we know that: lim sinðxÞ - x

x→0

In addition, from trigonometry, we know that: 1 þ cosðxÞ ¼ 2 cos 2 sinðxÞ ¼ 2 sin

x 2

x x cos 2 2

3

ðn - 20Þ2

¼

1020 ¼0 þ1

6

Solutions of Problems: Limits and Continuities

131

The problem can be solved as follows: limþ

sinðπ sinðxÞÞ sin

x 2

1 þ cosðxÞ

x→π

¼ limþ x→π

- limþ

π sinðxÞ sin

x→π

π x 2 sin 2 2x cos p 2 - cos 2x

x 2

2 cos 2

x 2

¼ limþ x→π

x 2

π x 2 sin 2x cos 2x sin p 2 cos 2x

x 2

p p ¼ -π 2 x 1 ¼ -π 2

p x ¼ limþ - π 2 sin 2 2 x→π

Choice (1) is the answer. 6.49. From application of Taylor series in limit, we know that: lim

α→0

p n

1 þ α - lim 1 þ α→0

α n

lim am xm þ am - 1 xm - 1 þ . . . þ am - n xm - n þ am - n - 1 xm - n - 1 - am - n - 1 xm - n - 1

x→0

or: lim ðam xm þ an xn Þ - an xn

x→0

if m > n

The problem can be solved as follows: p 3 lim

x→0

p 2 1 þ x3 - 1 1 þ x2 - 4 1 - 2x - lim x→0 2x2 þ 2x 2x2 þ 2x

2x 4

¼

x2 þx lim 32 2 x → 0 2x þ 2x

- lim

x 2

x → 0 2x

¼

1 4

Choice (1) is the answer. 6.50. From trigonometry, we know that: sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1 The problem can be solved as follows: lim

x→0

sin 2 ðxÞ þ sinðxÞ þ cos 2 ðxÞ - cosðxÞ 1 þ sinðxÞ - cosðxÞ 0 ¼ lim ¼ sin 2 ðxÞ - sinðxÞ þ cos 2 ðxÞ - cosðxÞ x → 0 1 - sinðxÞ - cosðxÞ 0

d dx ð1 þ x → 0 d ð1 dx

H ¼ ¼ ¼ ¼ ¼ ¼) lim Choice (2) is the answer.

sinðxÞ - cosðxÞÞ cosðxÞ þ sinðxÞ 1þ0 ¼ ¼ lim ¼ -1 sinðxÞ - cosðxÞÞ x → 0 - cosðxÞ þ sinðxÞ - 1 þ 0

132

6

Solutions of Problems: Limits and Continuities

6.51. From application of Taylor series in limit, we know that: lim sinðuðxÞÞ - uðxÞ

uðxÞ → 0

The problem can be solved as follows: lim

x→0 d dx ðcosðmxÞ - cosðnxÞÞ d 2 x→0 dx ðx Þ

H ¼ ¼ ¼ ¼ ¼ ¼) lim

cosðmxÞ - cosðnxÞ 0 ¼ 0 x2 - m sinðmxÞ þ n sinðnxÞ - mðmxÞ þ nðnxÞ - lim 2x 2x x→0

¼ lim

x→0

- m2 þ n2 n2 - m2 ¼ 2 2

¼ lim

x→0

Choice (3) is the answer. 6.52. From application of Taylor series in limit, we know that: lim sinðxÞ - x

x→0

lim tanðxÞ - x

x→0

The problem can be solved as follows: lim

x→0 d dx ðsinðxÞ - xÞ d x → 0 ðtanðxÞ - xÞ dx

H ¼ ¼ ¼ ¼ ¼ ¼) lim

d dx ðcosðxÞ - 1Þ x → 0 d ðtan 2 ðxÞÞ dx

H ¼ ¼ ¼ ¼ ¼ ¼) lim

¼ lim

x→0

sinðxÞ - x 0 ¼ tanðxÞ - x 0

¼ lim

x→0 1

cosðxÞ - 1 cosðxÞ - 1 0 ¼ lim ¼ 0 þ tan 2 ðxÞ - 1 x → 0 tan 2 ðxÞ

- sinðxÞ -x -1 ¼ lim - lim 2 tanðxÞð1 þ tan 2 ðxÞÞ x → 0 2xð1 þ x2 Þ x → 0 2ð1 þ x2 Þ ¼

-1 -1 ¼ 2 2ð 1 þ 0 Þ

Choice (1) is the answer. 6.53. From trigonometry, we know that: 1 þ cos 3 ðxÞ ¼ ð1 þ cosðxÞÞ 1 - cosðxÞ þ cos 2 ðxÞ 1 - cos 2 ðxÞ ¼ ð1 þ cosðxÞÞð1 - cosðxÞÞ The problem can be solved as follows: 1 þ cos 3 ðxÞ 0 ¼ x → π 1 - cos 2 ðxÞ 0 lim

6

Solutions of Problems: Limits and Continuities

) lim

x→π

133

1 þ cos 3 ðxÞ ð1 þ cosðxÞÞð1 - cosðxÞ þ cos 2 ðxÞÞ 1 - cosðxÞ þ cos 2 ðxÞ ¼ lim ¼ lim 2 x → π x → π ð1 þ cosðxÞÞð1 - cosðxÞÞ 1 - cosðxÞ 1 - cos ðxÞ ¼

1 - ð- 1Þ þ ð- 1Þ2 3 ¼ 2 1 - ð- 1Þ

Choice (1) is the answer. 6.54. Based on the information given in the problem, we have: f ð 0Þ ¼ 0 1 f ðxÞ ¼ xð- 1Þ½x] ,

x 2 ℝ - f 0g

A function is right continuous at this given point of x0 if: lim f ðxÞ ¼ f ðx0 Þ

x → x0 þ

Moreover, a function is left continuous at this given point of x0 if: lim f ðxÞ ¼ f ðx0 Þ

x → x0 -

In addition, a function is continuous at this given point of x0 if: lim f ðxÞ ¼ lim - f ðxÞ ¼ f ðx0 Þ

x → x0 þ

x → x0

For the given function, we have: 1 lim- f ðxÞ ¼ lim- xð- 1Þ½x] ¼ 0 x ðfinite quantityÞ ¼ 0

x→0

x→0

1 limþ f ðxÞ ¼ limþ xð- 1Þ½x] ¼ 0 x ðfinite quantityÞ ¼ 0

x→0

x→0

f ð 0Þ ¼ 0 ) limþ f ðxÞ ¼ lim- f ðxÞ ¼ f ð0Þ x→0

x→0

Thus, the function is continuous at x ¼ 0. Choice (3) is the answer. 6.55. As we know from calculus: nk ¼0 n → þ1 an

If a > 1, k 2 N ) lim Hence: 3n2 n2 lim p n ¼ lim 3 p n → þ1 5 n → þ1 5 Choice (1) is the answer.

n

¼3x0¼0

134

6

Solutions of Problems: Limits and Continuities

6.56. From trigonometry, we know that: x 2

1 - cosðxÞ ¼ 2 sin 2

Moreover, from application of Taylor series in limit, we know that: lim sin n ðxÞ - xn

x→0

Thus:

lim

x→0

x3 - sinðxÞ 2sin 2 x3 - sinðxÞð1 - cosðxÞÞ ¼ lim x→0 x3 x3 ¼ lim

x→0

x3 x3

x3 2

x3 2 x → 0 x3

¼ lim

x 2

- lim

x→0

x3 - x x 2 x3

x 2 2

1 1 ¼ 2 2

¼ lim

x→0

Choice (2) is the answer. 6.57. From trigonometry and calculus, we know that: arcðcosð1 - ÞÞ ¼ 0þ d -1 ðarcðcos xÞÞ ¼ p dx 1 - x2 -1 d p 1-x ¼ p dx 2 1-x The problem can be solved as follows: lim-

x→1

H ¼ ¼ ¼ ¼ ¼ ¼) limx→1

d dx ðarcðcos xÞÞ p d 1-x dx

¼ limx→1

arcðcos xÞ 0þ p ¼ þ 0 1-x

p-1 1 - x2 p- 1 2 1-x

p ¼ limx→1

1 ð1 - xÞð1þxÞ p1 2 1-x

¼ lim- p x→1

p 2 ¼ 2 1þx

Choice (1) is the answer. 6.58. Based on the information given in the problem, we have: lim x2 - 1 cotðxn - 1Þ ¼

x→1

From calculus, we know that: cotðxÞ ¼

1 tanðxÞ

From application of Taylor series in limit, we know that: lim

uðxÞ → 0

tanðuðxÞÞ - uðxÞ

1 2

ð1Þ

6

Solutions of Problems: Limits and Continuities

135

The problem can be solved as follows: x2 - 1 x2 - 1 0 - lim n ¼ n tanðx - 1Þ x → 1 x - 1 0

lim x2 - 1 cotðxn - 1Þ ¼ lim

x→1

x→1

H ¼ ¼ ¼ ¼ ¼ ¼) lim

x→1

2 2x ¼ n 1 n nx

ð2Þ

Solving (1) and (2): 2 1 ¼ )n¼4 n 2 Choice (2) is the answer. 6.59. From trigonometry, we know that: sinð4xÞ ¼ 2 sinð2xÞ cosð2xÞ cosðxÞ sinðxÞ

cotðxÞ ¼

sinðx - yÞ ¼ sinðxÞ cosðyÞ - cosðxÞ sinðyÞ The problem can be solved as follows: lim sinð4xÞðcotð2xÞ - cotðxÞÞ ¼ lim 2 sinð2xÞ cosð2xÞ

x→0

x→0

¼ lim 2 sinð2xÞ cosð2xÞ x→0

sinðxÞ cosð2xÞ - cosðxÞ sinð2xÞ sinð2xÞ sinðxÞ

sinðx - 2xÞ sinð2xÞ sinðxÞ

¼ lim 2 sinð2xÞ cosð2xÞ x→0

cosð2xÞ cosðxÞ sinð2xÞ sinðxÞ

¼ lim ð- 2 cosð2xÞÞ ¼ - 2 x→0

Choice (3) is the answer. 6.60. From application of Taylor series in limit, we know that: lim sinðxÞ - x -

x3 6

lim cosðxÞ - 1 -

x2 2

x→0

x→0

The problem can be solved as follows:

lim

x → 0- 1 2

sinðxÞ - x - lim sinð2xÞ - x cosðxÞ x → 0 -

Choice (3) is the answer.

1 2

2x -

ð2xÞ3 6

x3 6

-x 1-

x2 2

¼ limx→0

-

x3 6 x3 6

¼ lim- 1 ¼ 1 x→0

136

6

Solutions of Problems: Limits and Continuities

6.61. The problem can be solved as follows: limπ

x→4

1 - 3 tanðxÞ 0 ¼ 1 - 2 sin 2 ðxÞ 0 tan ðxÞ p - 1þ 3 2 2

H ¼ ¼ ¼ ¼ ¼ ¼) limπ

x→4

d dx d dx

1-

3

tanðxÞ

1 - 2 sin ðxÞ 2

¼ limπ

x→4

3

tan ðxÞ

- 4 sinðxÞ cosðxÞ

¼

4x

1þ1 3x1 p p 2 2 2 x 2

¼

1 3

Choice (1) is the answer. 6.62. From calculus and trigonometry, we know that: sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1 1 - cos 3 ðxÞ ¼ ð1 - cosðxÞÞ 1 þ cosðxÞ þ cos 2 ðxÞ From application of Taylor series in limit, we know that: lim sinðxÞ - x

x→0

lim tanðuðxÞÞ - uðxÞ

u ð xÞ → 0

The problem can be solved as follows: lim

x→0

) lim

x→0

¼ lim

1 - cos 3 ðxÞ 0 ¼ sinðxÞ tanð2xÞ 0

1 - cos 3 ðxÞ ð1 - cosðxÞÞð1 þ cosðxÞ þ cos 2 ðxÞÞ ð1 - cosðxÞÞ ¼ lim x sinðxÞ tanð2xÞ x → 0 sinðxÞ tanð2xÞ ð1 þ cosðxÞÞ

x→0

ð1 - cos 2 ðxÞÞð1 þ cosðxÞ þ cos 2 ðxÞÞ sin 2 ðxÞ x ð1 þ 1 þ 1Þ ¼ lim x → 0 sinðxÞ tanð2xÞ x ð1 þ 1Þ sinðxÞ tanð2xÞð1 þ cosðxÞÞ 3 sin 2 ðxÞ 3x2 3 3 - lim ¼ lim ¼ 4 x → 0 2 sinðxÞ tanð2xÞ x → 0 2x x 2x x→0 4 lim

Choice (3) is the answer. 6.63. From trigonometry, we know that: 1 - cosðxÞ ¼ 2 sin 2

x 2

sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1 From application of Taylor series in limit, we know that: lim sin n ðuðxÞÞ -

uðxÞ → 0þ

lim ðuðxÞÞn

uðxÞ → 0þ

References

137

The problem can be solved as follows: limþ

x→0

1 - cosðxÞ 1 - cosðxÞ 1 þ p ¼ lim p x 1 - cosð xÞ x → 0þ 1 - cosð xÞ 1 þ

cosðxÞ cosðxÞ

x

p 1 þ cosð xÞ p 1 þ cosð xÞ

¼ limþ

p ð1 - cosðxÞÞð1 þ cosð xÞÞ ð1 - cosðxÞÞð1 þ 1Þ p ¼ limþ p 2 ð 1 - cos 2 ð xÞÞð1 þ 1Þ x → 0 ð1 - cos ð xÞÞ 1 þ cosðxÞ

¼ limþ

2 sin 2 2x 2 x 1 - cosðxÞ x p ¼ limþ p - limþ p2 2 ¼ limþ ¼ 0 2 2 1 - cos ð xÞ x → 0 sin ð xÞ x → 0 ð xÞ x→0 2

x→0

2

x→0

Choice (1) is the answer.

References 1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.

7

Problems: Derivatives and Their Applications

Abstract

In this chapter, the basic and advanced problems of derivatives and their applications are presented. The subjects include the definition of derivative, differentiation formulas, product rule, quotient rule, chain rule, derivatives of trigonometric functions, derivatives of exponential functions, derivatives of logarithm functions, derivatives of inverse trigonometric functions, derivatives of hyperbolic functions, implicit differentiation, higher-order derivatives, logarithmic differentiation, applications of derivatives, rates of change, critical points, minimum and maximum values, and absolute extrema. To help students study the chapter in the most efficient way, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Moreover, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations. 7.1. The current population of a specific animal in a jungle is about 820. How long will it take for the population to be 3280 if the growth constant is about 0.2 [1, 2]? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 4 ln 2 2) 10 ln 2 3) 2 ln 2 4) 2 ln 10 7.2. Which one of the following choices presents the nondifferentiable point(s) of the function below? f ð x Þ ¼ x ð x þ 2Þ 2 ð x - 3Þ 3 Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) x ¼ -2 2) x ¼ 3 3) x ¼ 0 4) x ¼ -2, 0, 3 7.3. Calculate the value of f ′(x ¼ 1) if f (x) ¼ xex - ex. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 0 3) -e 4) e

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_7

139

140

7

Problems: Derivatives and Their Applications

7.4. If f(x) + g(x3) ¼ 5x - 1 and f ′(1) ¼ 2, calculate the value of g′(1). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) -1 3) 2 4) -2 7.5. Determine the range of x where the function of y(x) ¼ 1 - 4x2 is ascending. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) x < 0 2) x > 0 3) -2 < x < 2 4) -4 < x < 4 1 7.6. Determine the derivative of the function below at x ¼ . 4 p x- x p f ð xÞ ¼ 1- x Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) -1 1 2) 2 1 3) 2 4) 1 7.7. Determine the first derivative of the function of (x100 + x50 + 50x2 + 50x + 1)10 at x ¼ 0. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 100 2) 200 3) 400 4) 500 7.8. Calculate the derivative of the function of f (x) ¼ tan3(2x) at Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 4 1) 3 4 2) 9 8 3) 3 8 4) 9

π . 12

7

Problems: Derivatives and Their Applications

141

7.9. What is the first derivative of the inverse function of f (x) ¼ x3 + x - 2 at a point with the length of zero on the inverse function? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) -1 2) 0 3) 1 1 4) 4 7.10. If f(x) ¼ x5 + 3x3 + x + 1, then calculate the first derivative of the inverse function at a point with the length of six on the inverse function. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1 1) 5 1 2) 15 1 3) 6 1 4) 16 7.11. For the following function, calculate the value of ( f-1(x))′ for x ¼ 2. f ðxÞ ¼

4x3 þ1

x2

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 112 1) 25 25 2) 112 1 3) 4 4) 4 7.12. Calculate the value of the limit below if f(x) ¼ x tan x.

lim

f ðxÞ - f

x⟶π4

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 π 2) 1 4 π 3) 1 þ 4 π 4) 1 þ 2

x-

π 4

π 4

142

7

Problems: Derivatives and Their Applications

7.13. If the function of f (x) ¼ jx3 - 3x + aj does not have a derivate at x ¼ 2, calculate the value of a. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2 2) -2 3) 1 4) -1 7.14. Calculate the value of f ′(2) + f ′(4) if f (x) ¼ jx2 - 6j. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) -8 2) 8 3) -4 4) 4 7.15. If f 0 ðxÞ ¼ 5x, calculate the first derivative of f (x5). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 5 1) x 25 2) x 25 3) x 5 4) 5 x 7.16. If the first derivative of f (sin (x)) is equal to cos3(x), determine the value of f ′(x). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 + x2 2) 1 - x2 3) x3 4) -x3 1 7.17. Calculate the derivative of the function of f (x) ¼ arc(tan(3x)) at x ¼ . 3 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 3 1) 2 4 2) 3 2 3) 3 3 4) 4 p 1 þ g t ¼ t 2 þ 1 and g′(1) ¼ 5, calculate the value of f ′(1). t Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

7.18. If f

7

Problems: Derivatives and Their Applications

1) 1 2) 2 1 3) 2 4) -

1 2

7.19. If 2 cos ( y) - sin (x + y) + 2 ¼ 0, calculate the value of y0x at (0, π). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 2 1 2) 2 3) -1 4) 1 7.20. The equation of a curve is given by x3 + y3 ¼ 16. Calculate the second derivate of y with respect to x. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 16y 1) - 5 x 16x 2) 5 y 32y 3) - 5 x 32x 4) - 5 y π 7.21. If x ¼ 2 + 3 sin (t) and y ¼ 3 - 2 cos (t), calculate the value of y0x for t ¼ . 6 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large p 2 3 1) 9 p 2 3 2) 3 p 2 2 3) 3 p 4 2 4) 3 7.22. If x ¼ t2 + t and y ¼ t2 - 2t, calculate the value of x0y þ y0x for t ¼ -1. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 11 1) 4 13 2) 4 15 3) 4 17 4) 4

143

144

7

Problems: Derivatives and Their Applications

7.23. Calculate the value of f ′(4) if we know that: lim

h→0

p f ð x þ hÞ - f ð x - hÞ ¼2 x h

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 2 1) 3 4 2) 3 3) 4 4) 2 7.24. Which one of the choices is true about the function of f (x) ¼ x2jxj at x ¼ 0? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) The first derivative exists, but the second derivative does not. 2) The second derivative exists, but the first derivative does not. 3) The first and second derivatives do not exist. 4) The first and second derivatives exist. π 7.25. The function below is differentiable at x ¼ . Determine the value of b. 4 f ðxÞ ¼

sin 2 ðxÞ - cosð2xÞ a tanðxÞ þ b sinð2xÞ

π 0 0 ) x < 0 Choice (1) is the answer. 8.6. From list of derivative rules, we know that: f ðxÞ ¼

p

1 x ) f 0 ð xÞ ¼ p 2 x

First, we should simplify the function as follows: f ðxÞ ¼

p p p p x ð x - 1Þ x- x p ¼ p ¼ - x 1- x 1- x

Therefore: 1 1 1 ¼ ) f 0 ð xÞ ¼ - p ) f 0 ¼ -1 4 2 x 2 14 Choice (1) is the answer. 8.7. From list of derivative rules, we know that: f ðxÞ ¼ un ðxÞ ) f 0 ðxÞ ¼ nu0 ðxÞun - 1 ðxÞ Therefore: f ðxÞ ¼ x100 þ x50 þ 50x2 þ 50x þ 1

10

) f 0 ðxÞ ¼ 10 100x99 þ 50x49 þ 100x þ 50 x100 þ x50 þ 50x2 þ 50x þ 1 ) f 0 ð0Þ ¼ 10ð0 þ 0 þ 0 þ 50Þð0 þ 0 þ 0 þ 0 þ 1Þ9 ¼ 500 Choice (4) is the answer. 8.8. From list of derivative rules, we know that: f ðxÞ ¼ tan n ðuðxÞÞ ) f 0 ðxÞ ¼ nu0 ðxÞ 1 þ tan 2 ðuðxÞÞ tan n - 1 ðuðxÞÞ Therefore: f ðxÞ ¼ tan 3 ð2xÞ ) f 0 ðxÞ ¼ 3  2 1 þ tan 2 ð2xÞ tan 2 ð2xÞ ) f0 Choice (3) is the answer.

π π ¼ 3  2 1 þ tan 2 12 6

tan 2

π 1 1 8 ¼6 1þ  ¼ 6 3 3 3

9

154

8 Solutions of Problems: Derivatives and Their Applications

8.9. The derivative of the inverse function of f(x) at a point with the length of “b” on the inverse function can be calculated as follows: 0

f - 1 ðbÞ ¼

1 f 0 ð aÞ

On the other hand, we know that: f ð a Þ ¼ b , f - 1 ð bÞ ¼ a Therefore, for the function of f(x) ¼ x3 + x - 2 and the point with the length of zero (b ¼ 0) on the inverse function, we can calculate “a” as follows: 0 ¼ a3 þ a - 2 ) a ¼ 1 Also, we have: f ðxÞ ¼ x3 þ x - 2 ) f 0 ðxÞ ¼ 3x2 þ 1 Hence: 0

f - 1 ð 0Þ ¼

1 1 1 ¼ ¼ f 0 ð 1Þ 3  1 2 þ 1 4

Choice (4) is the answer. 8.10. As we know, the derivative of the inverse function of f(x) at a point with the length of “b” on the inverse function can be calculated as follows: 0

f - 1 ðbÞ ¼

1 f 0 ð aÞ

Moreover: f ð a Þ ¼ b , f - 1 ð bÞ ¼ a Thus, for the function of f(x) ¼ x5 + 3x3 + x + 1 and the point with the length of six (b ¼ 6) on the inverse function, we can calculate “a” as follows: 6 ¼ a5 þ 3a3 þ a þ 1 ) a ¼ 1 On the other hand, we have: f ðxÞ ¼ x5 þ 3x3 þ x þ 1 ) f 0 ðxÞ ¼ 5x4 þ 9x2 þ 1

Hence: 0

f - 1 ð 6Þ ¼ Choice (2) is the answer.

1 1 1 ¼ ¼ f 0 ð1Þ 5  14 þ 9  12 þ 1 15

8

Solutions of Problems: Derivatives and Their Applications

155

8.11. As we know, the derivative of the inverse function of f(x) at a point with the length of “b” on the inverse function can be calculated as follows: 0

f - 1 ðbÞ ¼

1 f 0 ð aÞ

In addition, we know that: f ð a Þ ¼ b , f - 1 ð bÞ ¼ a 3

Hence, for the function of f ðxÞ ¼ x4x 2 þ1 and the point with the length of two (b ¼ 2) on the inverse function, we can calculate “a” as follows: 2¼

4a3 )a¼1 a2 þ 1

On the other hand, we can write: f ðxÞ ¼

12x2 ðx2 þ 1Þ - ð2xÞð4x3 Þ 4x3 ) f 0 ð xÞ ¼ þ1 ðx2 þ 1Þ2

x2

Thus: 0

f - 1 ð 2Þ ¼

2

12 þ 1 1 4 1 ¼ ¼ ¼ 0 2 2 24 - 8 4 f ð1Þ 12  1 1 þ 1 - ð2  1Þ 4  13

Choice (3) is the answer. 8.12. We know that: d ðuðxÞvðxÞÞ ¼ u0 ðxÞvðxÞ þ uðxÞv0 ðxÞ dx

ð1Þ

d ðtan xÞ ¼ 1 þ tan 2 x dx tan

π ¼1 4

The first derivative of a function is defined as follows: lim

x⟶x0

f ð xÞ - f ð x0 Þ x - x0

Therefore, the value of the following limit for the function of f(x) ¼ x tan x is equal to the first derivative of the function at x0 ¼ π4. f ðxÞ - f x - π4 x⟶4 limπ

π 4

¼ f 0 ð x0 Þ

156

8 Solutions of Problems: Derivatives and Their Applications

The first derivative of the function can be calculated as follows: f 0 ðxÞ ¼ tan x þ 1 þ tan 2 x x For x0 ¼ π4, we have: f0

π π π π π ¼ tan þ 1 þ tan 2 ¼1þ 4 4 4 4 2

Choice (4) is the answer. 8.13. Based on the information given in the problem, we have: f ðxÞ ¼ x3 - 3x þ a A derivative of an absolute function does not exist at its simple roots. Therefore, we need to solve the equation below: f ð 2Þ ¼ 0 ) 23 - 3  2 þ a ¼ 0 ) 8 - 6 þ a ¼ 0 ) a ¼ - 2 Choice (2) is the answer. 8.14. From list of derivative rules, we know that: f ð x Þ ¼ j uð x Þ j ) f 0 ð xÞ ¼

u0 ðxÞuðxÞ j uð x Þ j

Based on the information given in the problem, we have: f ð x Þ ¼ x2 - 6 Therefore: ) f 0 ð xÞ ¼ ) f 0 ð 2Þ ¼

2x  ðx2 - 6Þ j x 2 - 6j

4  ð 4 - 6Þ 8  ð16 - 6Þ ¼ - 4 and f 0 ð4Þ ¼ ¼8 j 4 - 6j j16 - 6j ) f 0 ð 2Þ þ f 0 ð 4Þ ¼ - 4 þ 8 ¼ 4

Choice (4) is the answer. 8.15. From list of derivative rules, we know that: f ðxÞ ¼ gðhðxÞÞ ) f 0 ðxÞ ¼ h0 ðxÞg0 ðhðxÞÞ Based on the information given in the problem, we have: f 0 ðxÞ ¼

5 x

8

Solutions of Problems: Derivatives and Their Applications

157

The problem can be solved as follows: f x5

0

¼ 5x4  f 0 x5 ¼ 5x4 

5 ) f x5 x5

0

¼

25 x

Choice (3) is the answer. 8.16. From list of derivative rules, we have: ðf ðgðxÞÞÞ0 ¼ g0 ðxÞf 0 ðgðxÞÞ Based on the information given in the problem, we have: ðf ðsinðxÞÞÞ0 ¼ cos 3 ðxÞ The problem can be solved as follows: ðf ðsinðxÞÞÞ0 ¼ cos 3 ðxÞ ) cosðxÞ  f 0 ðsinðxÞÞ ¼ cos 3 ðxÞ ) f 0 ðsinðxÞÞ ¼ cos 2 ðxÞ ) f 0 ðsinðxÞÞ ¼ 1 - sin 2 ðxÞ ) f 0 ðxÞ ¼ 1 - x2 Choice (2) is the answer. 8.17. From list of derivative rules, we know that: f ðxÞ ¼ arcðtanðuðxÞÞÞ ) f 0 ðxÞ ¼

u0 ð x Þ 1 þ u2 ð xÞ

Therefore: f ðxÞ ¼ arcðtanð3xÞÞ ) f 0 ðxÞ ¼ f0

1 3 ¼ 3 1þ9

1 2 3

¼

3 1 þ 9x2

3 2

Choice (1) is the answer. 8.18. From list of derivative rules, we know that: f ðxÞ ¼ axn ) f 0 ðxÞ ¼ anxn - 1 f ðxÞ ¼ gðhðxÞÞ ) f 0 ðxÞ ¼ h0 ðxÞg0 ðhðxÞÞ Based on the information given in the problem, we have: f

p 1 þ g t ¼ t2 þ 1 t g0 ð1Þ ¼ 5

158

8 Solutions of Problems: Derivatives and Their Applications

The problem can be solved as follows: d p p dx 1 1 1 1 þ p  g0 t ¼ 2t f þ g t ¼ t 2 þ 1¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) - 2  f 0 t t t 2 t g0 ð 1Þ ¼ 5 1 5 t ¼ 1 ) - f 0 ð1Þ þ g0 ð1Þ ¼ 2¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) - f 0 ð 1Þ þ ¼ 2 2 2 ) f 0 ð 1Þ ¼

1 2

Choice (3) is the answer. 8.19. From derivative rules, we know that: f ðx, yÞ ¼ 0 ) y0x ¼ -

f 0x ðx, yÞ ¼ f 0y ðx, yÞ

d dx f ðx, yÞ d dy f ðx, yÞ

Based on the information given in the problem, we have: 2 cosðyÞ - sinðx þ yÞ þ 2 ¼ 0 The problem can be solved as follows: y0x ¼ -

d dx ð2cosðyÞ d dy ð2cosðyÞ -

sinðx þ yÞ þ 2Þ - cosðx þ yÞ cosðx þ yÞ ¼ ¼ 2 sin ð y Þ cos ð x þ y Þ 2 sin ð y Þ þ cosðx þ yÞ sinðx þ yÞ þ 2Þ ðx, yÞ ¼ ð0, π Þ 0 -1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) yx ¼ 0-1 ) y0x ¼ - 1

Choice (3) is the answer. 8.20. From derivative rules, we know that: f ðx, yÞ ¼ 0 ) y0x ¼ -

f 0x ðx, yÞ ¼ f 0y ðx, yÞ

Based on the information given in the problem, we have: x3 þ y3 ¼ 16 The problem can be solved as follows: y0x ¼ y0 ¼ ) y00 ¼ -

3x2 x2 ¼ - 2 2 3y y

2xy2 - 2yy0 x2 y4

d dx f ðx, yÞ d dy f ðx, yÞ

8

Solutions of Problems: Derivatives and Their Applications

159

x2 y0 ¼ - 2 2xy2 - 2y y 00 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) y ¼ y4

x2 y2

x2

¼ -

2xðy3 þ x3 Þ 2xy3 þ 2x4 ¼ 5 y5 y

x3 þ y3 ¼ 16 00 2x  16 32x ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) y ¼ ) y00 ¼ - 5 y5 y Choice (4) is the answer. 8.21. From derivative rules, we know that: y ¼ yð t Þ y0 d yð t Þ ) y0x ¼ t0 ¼ dtd xt dt xðt Þ x ¼ xð t Þ Based on the information given in the problem, we have: x ¼ 2 þ 3 sinðt Þ y ¼ 3 - 2 cosðt Þ Hence: y0x ¼

y0t 2 sinðt Þ 2 ¼ ¼ tanðt Þ x0t 3 cosðt Þ 3

p p 2 3 2 3 π 0 0 ) yx ¼ t ¼ ) yx ¼  3 3 9 6 Choice (1) is the answer. 8.22. From derivative rules, we know that: y ¼ yð t Þ y0 d yð t Þ x0 d xð t Þ ) y0x ¼ t0 ¼ dtd , x0y ¼ t0 ¼ dtd xt dt xðt Þ yt dt yðt Þ x ¼ xð t Þ Based on the information given in the problem, we have: x ¼ t2 þ t y ¼ t 2 - 2t Therefore:

)

y0t 2t - 2 ¼ x0t 2t þ 1 t ¼ -1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 0 x 2t þ 1 t 0 xy ¼ 0 ¼ yt 2t - 2 y0x ¼

Choice (4) is the answer.

-2-2 ¼4 1 17 -2 þ 1 ) y0x þ x0x ¼ 4 þ ) y0x þ x0x ¼ 4 4 2 þ 1 1 ¼ x0y ¼ -2-2 4

y0x ¼

160

8 Solutions of Problems: Derivatives and Their Applications

8.23. Based on the information given in the problem, we have: lim

h→0

p f ð x þ hÞ - f ð x - hÞ ¼2 x h

ð1Þ

From definition of derivative, we know: f 0 ðxÞ ¼ lim

h→0

f ð x þ hÞ - f ð x Þ f ð x Þ - f ð x - hÞ ¼ lim h h h→0

The problem can be solved as follows: lim

h→0

f ð x þ hÞ - f ð x - hÞ f ð x þ h Þ - f ð x Þ þ f ð x Þ - f ð x - hÞ ¼ lim h h h→0 ¼ lim

h→0

f ðx þ h Þ - f ðxÞ f ð x Þ - f ð x - hÞ þ lim ¼ 2f 0 ðxÞ h h h→0

ð2Þ

Solving (1) and (2): p p p 2f 0 ðxÞ ¼ 2 x ) f 0 ðxÞ ¼ x ) f 0 ð4Þ ¼ 4 ¼ 2 Choice (4) is the answer. 8.24. The function can be simplified as follows: f ðxÞ ¼ x2 jxj ¼

) f ð 0Þ ¼

0 0

x≥0

x3 -x

3

x