Calculus: Practice Problems, Methods, and Solutions 3030649792, 9783030649791

Table of contents :
Preface
Contents
Chapter 1: Problems: Trigonometric Equations and Identities
Reference
Chapter 2: Solutions of Problems: Trigonometric Equations and Identities
Reference
Chapter 3: Problems: Limits
Reference
Chapter 4: Solutions of Problems: Limits
Reference
Chapter 5: Problems: Derivatives and Its Applications
Reference
Chapter 6: Solutions of Problems: Derivatives and its Applications
Reference
Chapter 7: Problems: Definite and Indefinite Integrals
Reference
Chapter 8: Solutions of Problems: Definite and Indefinite Integrals
Reference
Index

Citation preview

Mehdi Rahmani-Andebili

Calculus

Practice Problems, Methods, and Solutions

Calculus

Mehdi Rahmani-Andebili

Calculus Practice Problems, Methods, and Solutions

Mehdi Rahmani-Andebili State University of New York Buffalo State, NY, USA

ISBN 978-3-030-64979-1 ISBN 978-3-030-64980-7 https://doi.org/10.1007/978-3-030-64980-7

(eBook)

# Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are reserved 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 AGThe registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

Calculus is one of the most important courses of many majors, including engineering and science and even some of the non-engineering majors like economics and business, which is taught in the first semester at universities and colleges all over the world. Moreover, in many universities and colleges, precalculus is a mandatory course for the under-prepared students as the prerequisite course of calculus. Unfortunately, some students do not have a solid background and knowledge in math and calculus when they begin their education in universities or colleges. This issue prevents them from learning other important courses like physics and calculus-based courses. Sometimes, the problem is so escalated that they give up and leave university or college. Based on my real professorship experience, students do not have a serious issue to comprehend physics and engineering courses. In fact, it is the lack of enough knowledge of calculus that hinders their understanding of calculus-based courses. Therefore, this textbook along with the other one (Precalculus: Practice Problems, Methods, and Solutions, Springer, 2020) have been prepared to help students succeed in their major. The textbooks include basic and advanced problems of calculus with very detailed problem solutions. They can be used as a practicing textbooks by students and as a supplementary teaching source by instructors. Since the problems have very detailed solutions, the textbooks are useful for under-prepared students. In addition, the textbooks are beneficial for knowledgeable students because they include advanced problems. In the preparation of problem solutions, effort has been made to use typical methods to present the textbooks as an instructor-recommended one. By considering this key point, both 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 the textbooks in the most efficient way, the problems have been categorized in 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, in each chapter, problems have been ordered from the easiest one with the smallest amount of calculations to the most difficult one with the largest amount of calculations. Therefore, students are advised to study the textbooks from the easiest problem and continue practicing until they reach the normal and then the hardest one. On the other hand, this classification can 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. Buffalo, NY, USA

Mehdi Rahmani-Andebili

Contents

1

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

1

2

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

19

3

Problems: Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

53

4

Solutions of Problems: Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

67

5

Problems: Derivatives and Its Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83

6

Solutions of Problems: Derivatives and its Applications . . . . . . . . . . . . . . . . . .

91

7

Problems: Definite and Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

8

Solutions of Problems: Definite and Indefinite Integrals . . . . . . . . . . . . . . . . . . 125

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

1

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, sine and cosine identities, tangent and cotangent identities, half angle formulas, reciprocal identities, Pythagorean identities, sum and difference to product formulas, product to sum formulas, even and odd formulas, periodic formulas, sum to product formulas, double angle formulas, degrees to radians formulas, cofunction formulas, unit circle, inverse trigonometric functions, inverse properties, alternate notation, and domain and range of inverse trigonometric functions. To help students study the chapter in the most efficient way, the problems are categorized 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. Calculate the value of tan(2θ) if cot(θ) ¼ 5 [1]. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 5 1) 12 5 2) 13 5 3)  12 5 4)  13 

1.2. Determine the value of tan(2100 ). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large pffiffiffi 1) 3 pffiffi 2) 33 pffiffiffi 3)  3 pffiffi 4)  33 1.3. Simplify and calculate the final value of the following term.   1 þ cos 40  sin ð40 Þ Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Calculus, https://doi.org/10.1007/978-3-030-64980-7_1

1

2

1

1) 2) 3) 4)



sin(20 )  cos(20 )  tan(20 )  cot(20 )

π 2π 1.4. Determine the range of m if sin ðαÞ ¼ 3m1 4 and 6  α  3 . Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large h pffiffi i 1) 1, 2 331 h pffiffi i 2) 1, 2 33þ1

3) [1, 2]   4) 1, 53 π π 1.5. Determine the range of m if cos ðαÞ ¼ 2m1 6 and  3  α  3. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large   1) 2, 72   2) 32 , 72   3) 2, 52   4) 32 , 52

  1.6. What is the main period of cos 2 ðxÞ  5 cos 2x3 ? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π 2) 2π 3) 3π 4) 4π     3 2x 1.7. What is the main period of sin 4 3x 5 þ cos 3 ? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 3π 2) 5π 3) 15π 4) 30π   1.8. Determine the main period of sin 4 πx 3 þ cos ðπxÞ þ 5. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 2 3) 3 4) 6

Problems: Trigonometric Equations and Identities

1

Problems: Trigonometric Equations and Identities

3

1.9. Figure 1.1 illustrates a part of the function of y ¼ sin (kx). Determine the value of k. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)

2 3 3 4 3 2 4 3

Figure 1.1 The graph of problem 1.9

1.10. Figure 1.2 illustrates a part of the function of y ¼ cos Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)



  ax þ 12 π . Determine the value of a.

1 2 3 2 2 3 7 4

Figure 1.2 The graph of problem 1.10

1.11. 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) 1.12. Calculate the final value of the following equation.     π 7π sin ð5π þ xÞ þ sin x  þ sin x þ 3 3 Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

4

1

1) 2) 3) 4)

Problems: Trigonometric Equations and Identities

0  sin π3  2 sin π3   sin π3 





1.13. Calculate the value of cos(20 ) if sin(50 ) + sin (10 ) ¼ m. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) m2 2) m 3) 2m 4) 2m 3 1.14. Simplify and calculate the final value of the following term.        1 þ tan 2 5 sin 10      1  tan 2 5 tan ð10 Þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large  1) tan(15 )  2) tan(25 )  3) tan(35 )  4) tan(45 ) 1.15. 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 1.16. Determine the main period of sin(3x) cos (5x) + 11. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π 2) 2π 3) 2π 3 4) 2π 5     1.17. Calculate the value of arc cos sin 4π . 3 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π6 2) 5π 6 3) π3 4)  π6

1

Problems: Trigonometric Equations and Identities

    1.18. Calculate the value of arc sin sin 17π . 5 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2π 5 2) 3π 5 3)  2π 5 4)  3π 5     1.19. Calculate the value of arc cos cos 19π . 5 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π5 2) 4π 5 3)  π5 4)  4π 5     1.20. Calculate the value of tan 2arc tan 12 . Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 34 3) 43 4) 35        1.21. Calculate the final value of sin arc sin 35 þ arc tan 34 . Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)

10 13 9 13 12 35 24 25

      1.22. Calculate the final value of arc cot  43  arc cot 34 . Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π 2) 2π 3 3) π2 4) π3    1.23. Calculate the final value of arcð tan ð5ÞÞ þ arc tan 32 . Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π4 2)  π4 3) 3π 4 4) 5π 4

5

6

1

Problems: Trigonometric Equations and Identities

        1.24. Calculate the final value of sin arc cos 35 þ cos arc sin  45 . Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 75 2)  15 3) 15 4)  75 1.25. Figure 1.3 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 1.3 The graph of problem 1.25

1.26. Figure 1.4 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

Figure 1.4 The graph of problem 1.26

1

Problems: Trigonometric Equations and Identities

1.27. Figure 1.5 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 1.5 The graph of problem 1.27

      1.28. Calculate the value of arc tan 23 þ arc tan 15 . Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π6 2) π4 3) π3 4) π2 1.29. Calculate the final value of the term below.    1 arcð tan ðmÞÞ þ arc tan þ arcð cot ðmÞÞ þ arcð cot ðmÞÞ m Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π or 2π 2) π2 or 3π 2 3) 3π 2 4) π2 1.30. Determine the range of x in the inequality below. Herein, x is an acute angle. 1  cos ð4xÞ cos ð2xÞ þ sin ð4xÞ sin ð2xÞ  0 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large   1) π6 , 3π π π8 2) 8 , 4   3) π6 , π3 π π  4) 4 , 2

7

8

1

Problems: Trigonometric Equations and Identities

1.31. 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

1.32. Simplify and calculate the value of the following term.     sin 5π þ cos 5π 12 12   5π sin 5π  cos 12 12 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large pffiffiffi 1) 3 pffiffi 2) 33 pffiffiffi 3) 2 3 pffiffi 4)  33 1.33. Figure 1.6 illustrates a 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) 2) 3) 4)

4 3 5 3 7 3 8 3

Figure 1.6 The graph of problem 1.33

1.34. Figure 1.7 illustrates a 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 3) 92 4) 6

1

Problems: Trigonometric Equations and Identities

9

Figure 1.7 The graph of problem 1.34

1.35. Figure 1.8 illustrates a part of the function of y ¼ a sin Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2 2) 52 3) 3 4) 72



  bx þ 12 π . Determine the value of a  b.

Figure 1.8 The graph of problem 1.35

1.36. Simplify and calculate the value of the following term.       cos 5 cos 10 cos 20   cos 50 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2) 3) 4)

1  4 cos ð85 Þ 1  8 cos ð85 Þ 1  8 sin ð85 Þ 1  4 sin ð85 Þ

10

1

Problems: Trigonometric Equations and Identities

 1.37. Calculate the value of tan 2x if sin ðxÞ þ cos ðxÞ ¼ 75. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 or 3 2) 12 or 13 3) 2 or 35 4) 3 or 25 1.38. Simplify and calculate the value of the following term. sin 4 ðαÞ  cos 4 ðαÞ sin ðαÞ cos ðαÞ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 cot (2α) 2) 2 cot (2α) 3) 2 tan (3α) 4) 2 tan (3α) 1.39. Calculate the value of cot2(2α) if sin 4 ðαÞ þ cos 4 ðαÞ ¼ 12. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 2 3) 0 4) 3 1.40. Calculate the value of the following relation for x ¼ 3π 8. sin 3 ðxÞ cos ðxÞ  cos 3 ðxÞ sin ðxÞ þ 3 sin 2 ðxÞ cos 2 ðxÞ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 38 2) 58 3)  58 4)  38 π 1.41. Calculate the value of the following relation for α ¼ 15 .

sin ð2αÞ þ sin ð5αÞ þ sin ð8αÞ cos ð2αÞ þ cos ð5αÞ þ cos ð8αÞ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large pffiffi 1)  33 pffiffiffi 2)  3 pffiffiffi 3) 3 pffiffi 4) 33

1

Problems: Trigonometric Equations and Identities π 1.42. Calculate the value of the following relation for x ¼ 12 .

ð sin ðxÞ  cos ðxÞ þ 2Þð sin ðxÞ  cos ðxÞ  2Þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 72 2) 52 3)  52 4)  72 1.43. 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 1.44. Determine the number of roots of the equation below. sin(πx) cos2(πx) + sin2(πx) cos(πx) ¼ 0 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 11 2) 12 3) 13 4) 14 1.45. Figure 1.9 illustrates a part of the function of y ¼ 12 þ 2 cos ðmxÞ. Determine the value of the function for x ¼ 16π 3 . Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1)  12 2) 12 3) 1 4) 0

Figure 1.9 The graph of problem 1.45

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12

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Problems: Trigonometric Equations and Identities

1.46. Figure 1.10 shows a part of the function of y ¼ 1 + sin (mx). Determine the value of the function for x ¼ 7π 6. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 0 2) 12 3) 1 4) 2

Figure 1.10 The graph of problem 1.46

1.47. Figure 1.11 shows a part of the function of y ¼ a  sin (bπx). Determine the value of the function for x ¼ 25 3. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 2) 52 3) 3 4) 72

Figure 1.11 The graph of problem 1.47

  1.48. Figure 1.12 shows the function of y ¼ a þ b cos π2 x for 0 < x < 4. Determine the value of b. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 2) 1 3) 1 4) 2

1

Problems: Trigonometric Equations and Identities

Figure 1.12 The graph of problem 1.48

1.49. Figure 1.13 shows the function of y ¼ 1 + a sin (bπx) for 0 < x < 43. Determine the value of a + b. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 3 2) 4 3) 5 4) 6

Figure 1.13 The graph of problem 1.49

  1.50. Figure 1.14 shows a part the function of y ¼ a  2 cos bx þ π2 . Determine the value of a + b. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 12 2) 1 3) 32 4) 2

Figure 1.14 The graph of problem 1.50

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14

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Problems: Trigonometric Equations and Identities

   1.51. Calculate the value of cos(25  α) if tan α þ 20 ¼ 34. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 5 2) 6 3) 7 4) 8   1.52. Calculate the value of tan π4 þ α assuming that α is an acute angle and sin ðαÞ ¼ 35. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 7 2)  17 3) 17 4) 7     1.53. Calculate the value of tan π4  α if tan π2  α ¼ 23. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1)  13 2)  15 3) 15 4) 13 1.54. Calculate the value of tan(2a) while we know that tan ða þ bÞ ¼ 25 and tan ða  bÞ ¼ 37. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2) 3) 4)

 13  12 3 1

1.55. Calculate the value of tan(x) if we have:   sin x  π4  ¼2 cos x  π4 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 3 2) 13 3) 23 4) 3

1

Problems: Trigonometric Equations and Identities

1.56. Calculate the value of (1 + tan (α))(1 + tan (β)) if α þ β ¼ π4. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 2) 2 3) 13 4)  12     1.57. Calculate the value of tan π4 þ α  tan π4  α . Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 tan (2α) 2) 2 cos (2α) 3) 0 4) 2 sin (2α)   1.58. Calculate the value of tan(2α) if tan π4  α ¼ 15. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 1.5 2) 1.8 3) 2.4 4) 2.5 1.59. Calculate the value of tan(2α  β) if tan(α) ¼ 2 and tan ðβÞ ¼ 13. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 3 2) 2 3) 0.5 4) 3 1.60. Determine the common solution of the equation of cos(3x) + cos (x) ¼ 0 assuming cos(x) 6¼ 0. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) kπ2 þ π4 2) kπ2 þ π8 3) kπ  π4 4) kπ þ π4 1.61. 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 1) 2) 3) 4)

2π 5 4π 5 3π 5 π 5

15

16

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Problems: Trigonometric Equations and Identities

1.62. 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 1) 2) 3) 4)

2kπ  2π 3 2kπ  π3 2kπ  5π 6 kπ  π3

1.63. Determine the common solution of the equation of 2sin2(x) ¼ 3 cos (x). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) kπ  π6 2) kπ  π3 3) 2kπ  π6 4) 2kπ  π3 1.64. 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) 2) 3) 4)

y ¼ 1x y ¼ 3x y ¼ 4x1 y ¼ 4x3

1.65. 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 1.66. 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 1.67. 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 1) 2) 3) 4)

π 3 π 4 π 6 π 2

or or or or

2π 3 5π 4 5π 6 3π 2

1

Problems: Trigonometric Equations and Identities

17

1.68. Calculate the value of m if tan(α) 6¼ tan (β), α þ β ¼ π4 and α and β are the two roots of the equation below. tan 2 ðxÞ þ ðm þ 2Þ tan ðxÞ þ 2m  2 ¼ 0 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 3 3) 5 4) 7 1.69. 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 1.70. Calculate the final value of the relation below.         sin 135 cos 210 þ cos 135 sin 420     tan ð210 Þ cot ð420 Þ þ cot ð120 Þ tan ð330 Þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large pffiffi 1)  46 pffiffi 2)  3 4 6 pffiffi 3)  26 pffiffi 4)  3 2 6 1.71. Calculate the final value of (1 + cot (x))(1 + cot ( y)) if x þ y ¼ kπ þ π4. 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) 1.72. Determine the common solution of the equation below. ð sin ðxÞ  tan ðxÞÞ tan Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large

    3π 4π  x ¼ cos 2 3

18

1

1) 2) 3) 4)

Problems: Trigonometric Equations and Identities

kπ  π6 kπ þ π3 2kπ  π3 2kπ  π6

1.73. Calculate the sum of the roots of the equation below for x 2 [0, π]. sin ð2xÞð sin ðxÞ þ cos ðxÞÞ ¼ cos ð2xÞð cos ðxÞ  sin ðxÞÞ Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 2) 3) 4)

3π 4 5π 4 3π 2 7π 4

pffiffiffi     1.74. Determine the common solution of the equation of 2 sin π4  x ¼ 1 þ sin 5π 2 þx . Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) kπ þ π2 2) 2kπ  π4 3) 2kπ  π2 4) 2kπ þ π2 1.75. Which one of the following choices shows one of the common solutions of the equation of cos ð2xÞ þ Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) kπ  π6 2) kπ  π3 3) kπ þ π6 4) kπ þ π3

Reference 1. Rahmani-Andebili, M. (2020). Precalculus: Practice problems, methods, and solutions, Springer Nature, 2020.

pffiffiffi 3 sin ð2xÞ ¼ 1.

2

Solutions of Problems: Trigonometric Equations and Identities

Abstract

In this chapter, the problems of the first chapter are fully solved, in detail, step-by-step, and with different methods. The subjects include trigonometric equations, trigonometric identities, domain, range, period, sine and cosine identities, tangent and cotangent identities, half angle formulas, reciprocal identities, Pythagorean identities, sum and difference to product formulas, product to sum formulas, even and odd formulas, periodic formulas, sum to product formulas, double angle formulas, degrees to radians formulas, cofunction formulas, unit circle, inverse trigonometric functions, inverse properties, alternate notation, and domain and range of inverse trigonometric functions. 2.1. From trigonometry, we know that [1] tan ðθÞ ¼ tan ð2θÞ ¼

1 cot ðθÞ

2 tan ðθÞ 1  tan 2 ðθÞ

Based on the information given in the problem: cot ðθÞ ¼ 5 ) tan ðθÞ ¼

1 5

Therefore, 2 tan ðθÞ tan ð2θÞ ¼ ¼ 1  tan 2 ðθÞ

1 2 2 5 5 5
2 ¼ 24 ¼ 12 1 25 1 5

Choice (1) is the answer. 2.2. From trigonometry, we know that: tan ðα þ nπ Þ ¼ tan ðαÞ, 8n 2 ℤ tan ðαÞ ¼  tan ðαÞ Therefore,

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Calculus, https://doi.org/10.1007/978-3-030-64980-7_2

19

20

2 Solutions of Problems: Trigonometric Equations and Identities

       pffiffiffi     tan 2100 ¼  tan 2100 ¼  tan 12  180  60 ¼  tan 60 ¼ tan 60 ¼ 3 Choice (1) is the answer. 2.3. From trigonometry, we know that: 1 þ cos ðθÞ ¼ 2 cos 2

  θ 2

    θ θ sin ðθÞ ¼ 2 sin cos 2 2 cot ðθÞ ¼

cos ðθÞ sin ðθÞ

Therefore,         1 þ cos 40 2 cos 2 20 cos 20 ¼ ¼ ¼ cot 20     sin ð40 Þ 2 sin ð20 Þ cos ð20 Þ sin ð20 Þ Choice (4) is the answer. 2.4. For the given range of α, we can conclude that: π 2π 1 α ⟹  sin ðαÞ  1 6 3 2 Therefore, based on the given information, i.e., sin ðαÞ ¼ 3m1 4 , we can write: 1 3m  1 5   1 ⟹1  m  2 4 3 Choice (4) is the answer. 2.5. For the given range of x, we can conclude that: π π 1   x  ⟹  cos ðxÞ  1 3 3 2 Therefore, based on the given information, i.e., cos ðαÞ ¼ 2m1 6 , we can write: 1 2m  1 7   1 ⟹2  m  2 6 2 Choice (1) is the answer. 2.6. From trigonometry, we know that: f 1 ðxÞ ¼ cos 2n ðaxÞ, 8n 2 ℤ ⟹ T 1 ¼

π jaj

f 2 ðxÞ ¼ cos 2nþ1 ðaxÞ, 8n 2 ℤ ⟹ T 2 ¼

2π j aj

2

Solutions of Problems: Trigonometric Equations and Identities

21

Therefore, f 1 ðxÞ ¼ cos 2 ðxÞ ⟹ T 1 ¼ f 2 ðxÞ ¼ 5 cos

π ¼π 1


 2x 2π ⟹ T 2 ¼ 2 ¼ 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 ¼ LCMðπ, 3π Þ ¼ 3π Choice (3) is the answer. 2.7. From trigonometry, we know that: f 1 ðxÞ ¼ sin 2n ðaxÞ, 8n 2 ℤ ⟹ T 1 ¼

π j aj

f 2 ðxÞ ¼ cos 2nþ1 ðaxÞ, 8n 2 ℤ ⟹ T 2 ¼

2π j aj

Therefore, f 1 ðxÞ ¼ sin 4 f 2 ðxÞ ¼ cos 3


 3x π 5π ⟹ T1 ¼ 3 ¼ 5 3 5


 2x 2π ⟹ T 2 ¼ 2 ¼ 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 ¼ LCM

5π , 3π 3

 ¼ 15π

Choice (3) is the answer. 2.8. From trigonometry, we can know that: f 1 ðxÞ ¼ sin 2n ðaxÞ, 8n 2 ℤ ⟹ T 1 ¼

π j aj

f 2 ðxÞ ¼ cos 2nþ1 ðaxÞ, 8n 2 ℤ ⟹ T 2 ¼ Therefore, f 1 ðxÞ ¼ sin 4


 πx π ⟹T1 ¼ π ¼ 3 3 3

f 2 ðxÞ ¼ cos ðπxÞ ⟹ T 2 ¼

2π ¼2 π

2π j aj

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2 Solutions of Problems: Trigonometric Equations and Identities

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 ¼ LCMð3, 2Þ ¼ 6 Choice (4) is the answer. 2.9. From trigonometry, we know that: y ¼ sin ðkxÞ ⟹ T ¼

2π jk j

Therefore,   3π 1 2π 4 4 ⟹ ¼ ⟹ jk j ¼ ⟹ k ¼  4 2 jk j 3 3 Based on the graph and the function, the positive value of k is acceptable. ⟹k ¼

4 3

Choice (4) is the answer.

Figure 2.1 The graph of the solution of problem 2.9

2.10. From trigonometry, we know that:
 π y ¼ cos πax þ ¼  sin ðπaxÞ 2 y ¼ sin ðmxÞ ⟹ T ¼

2π jm j

Therefore,
 1 2π 4 2 3 ¼ ⟹ ¼ ⟹a ¼  ⟹1   3 3 j aj 2 jπaj

2

Solutions of Problems: Trigonometric Equations and Identities

23

Based on the graph and y ¼  sin (πax), the positive value of a is acceptable. ⟹a ¼

3 2

Choice (2) is the answer.

Figure 2.2 The graph of the solution of problem 2.10

2.11. From trigonometry, we know that: sin ðα þ 2nπ Þ ¼ sin ðαÞ, 8n 2 ℤ cos ðα þ 2nπ Þ ¼ cos ðαÞ, 8n 2 ℤ tan ðα þ nπ Þ ¼ tan ðαÞ, 8n 2 ℤ cot ðα þ nπ Þ ¼ cot ðαÞ, 8n 2 ℤ sin ðπ  αÞ ¼ sin ðαÞ cos ðπ  αÞ ¼  cos ðαÞ tan ðαÞ ¼  tan ðαÞ cot ðαÞ ¼  cot ðαÞ Based on the information given in the problem, we have: α þ β ¼ 19π ⟹ α ¼ 19π  β Therefore, sin ðαÞ ¼ sin ð19π  βÞ ¼ sin ðπ  βÞ ¼ sin ðβÞ cos ðαÞ ¼ cos ð19π  βÞ ¼ cos ðπ  βÞ ¼  cos ðβÞ tan ðαÞ ¼ tan ð19π  βÞ ¼ tan ðβÞ ¼  tan ðβÞ cot ðαÞ ¼ cot ð19π  βÞ ¼ cot ðβÞ ¼  cot ðβÞ Choice (1) is the answer.

24

2 Solutions of Problems: Trigonometric Equations and Identities

2.12. From trigonometry, we know that: sin ðα þ 2nπ Þ ¼ sin ðαÞ, 8n 2 ℤ cos ðα þ 2nπ Þ ¼ cos ðαÞ, 8n 2 ℤ sin ðα þ π Þ ¼  sin ðαÞ sin ðαÞ þ sin ðβÞ ¼ 2 sin

    αþβ αβ cos 2 2

Therefore,



 π 7π π π sin ð5π þ xÞ þ sin x  þ sin x þ ¼ sin ðx þ π Þ þ sin x  þ sin x þ 3 3 3 3 ¼  sin ðxÞ þ 2 sin ðxÞ cos


 π ¼  sin ðxÞ þ sin ðxÞ ¼ 0 3

Choice (1) is the answer. 2.13. From trigonometry, we know that:     αþβ αβ sin ðαÞ þ sin ðβÞ ¼ 2 sin cos 2 2 Therefore,             1 sin 50 þ sin 10 ¼ m ) 2 sin 30 cos 20 ¼ m ) 2  cos 20 ¼ m ) cos 20 ¼ m 2 Choice (2) is the answer. 2.14. From trigonometry, we know that:  1  tan 2 θ2  cos ðθÞ ¼ 1 þ tan 2 θ2 tan ðθÞ ¼

sin ðθÞ cos ðθÞ

  tan 45 ¼ 1 Therefore,            1 þ tan 2 5 sin 10 sin 10 1     ¼ ¼ 1 ¼ tan 45     2 sinð10 Þ cos ð10 Þ 1  tan 5 tan ð10 Þ  cos ð10 Þ Choice (4) is the answer.

2

Solutions of Problems: Trigonometric Equations and Identities

25

2.15. From trigonometry, we know that: 1 þ tan 2 ðαÞ ¼

1 cos 2 ðαÞ

1 cot ðαÞ

tan ðαÞ ¼ Based on the information given in the problem, we have:

cot ðαÞ ¼ m cos ðαÞ ¼ n Therefore, 1 1 1 1 m2 n2 2 2 ð α Þ ¼ 1 þ ⟹ ¼ 1 þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) m ¼ m2 n2 þ n2 ¼ 1 þ tan cos 2 ðαÞ n2 m2 cot 2 ðαÞ   ⟹ m2 1  n2 ¼ n2 Choice (2) is the answer. 2.16. From trigonometry, we can know that: sin 2nþ1 ðaxÞ, 8n 2 ℤ ⟹ T ¼

2π j aj

1 sin ðαÞ cos ðβÞ ¼ ð sin ðα þ βÞ þ sin ðα  βÞÞ 2 We need to change the product expression to the summation one, as follows. y ¼ sin ð3xÞ cos ð5xÞ þ 11 ⟹ y ¼

1 1 sin ð8xÞ  sin ð2xÞ þ 11 2 2

Then, 1 2π π sin ð8xÞ ⟹ T 1 ¼ ¼ 2 8 4 1 2π  sin ð2xÞ ⟹ T 2 ¼ ¼π 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 ¼ LCM Choice (1) is the answer.




 π ,π ¼ π 4

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2 Solutions of Problems: Trigonometric Equations and Identities

2.17. From trigonometry, we know that:
sin

pffiffiffi 

 3 4π π π ¼ sin π þ ¼  sin ¼ 2 3 3 3 arcð cos ðαÞÞ ¼ π  arcð cos ðαÞÞ  pffiffiffi 3 π arc cos ¼ 2 6

Therefore,   pffiffiffi  pffiffiffi


 3 3 4π π 5π ¼ π  arc cos ¼π ¼ arc cos sin ¼ arc cos  2 2 3 6 6 Choice (2) is the answer. 2.18. From trigonometry, we know that:
sin



 17π 3π 3π ¼ sin 4π  ¼ sin  5 5 5


 3π 3π sin  ≜ α ) arcð sin ðαÞÞ ¼  5 5 Therefore,





 17π 3π 3π arc sin sin ¼ arc sin sin  ¼ arcð sin ðαÞÞ ¼  5 5 5 Choice (4) is the answer. 2.19. From trigonometry, we know that: cos




 19π π π ¼ cos 4π  ¼ cos  5 5 5 cos


 π π ≜ α ) arcð cos ðαÞÞ ¼ 5 5

Therefore,








 19π π π π arc cos cos ¼ arc cos cos  ¼ arc cos cos ¼ arcð cos ðαÞÞ ¼ 5 5 5 5 Choice (1) is the answer. 2.20. From trigonometry, we know that: tan ð2αÞ ¼

2 tan ðαÞ 1  tan 2 ðαÞ



 1 1 ≜ α ) tan ðαÞ ¼ arc tan 2 2

2

Solutions of Problems: Trigonometric Equations and Identities

27

Therefore,


 2  12 2 tan ðαÞ 1 1 4 tan 2arc tan ¼ tan ð2αÞ ¼ ¼  ¼ ¼ 2 2 1  tan ðαÞ 1  1 2 34 3 2

Choice (3) is the answer. 2.21. From trigonometry, we know that: sin 2 ðαÞ þ cos 2 ðαÞ ¼ 1 1 þ tan 2 ðαÞ ¼

1 cos 2 ðαÞ

Therefore,

 3 3 4 ≜ α ⟹ sin ðαÞ ¼ ⟹ cos ðαÞ ¼ arc sin 5 5 5

 3 3 4 3 arc tan ≜ β ⟹ tan ðβÞ ¼ ⟹ cos ðβÞ ¼ ⟹ sin ðβÞ ¼ 4 4 5 5




 3 3 ⟹ sin arc sin þ arc tan ¼ sin ðα þ βÞ ¼ sin ðαÞ cos ðβÞ þ cos ðαÞ sin ðβÞ 5 4 ¼

3 4 4 3 24  þ  ¼ 5 5 5 5 25

Choice (4) is the answer. 2.22. From trigonometry, we know that: arcð cot ðαÞÞ ¼ π  arcð cot ðαÞÞ

 1 π arcð cot ðαÞÞ þ arc cot ¼ α 2 Therefore,







 4 3 4 3 arc cot   arc cot ¼ π  arc cot  arc cot 3 4 3 4




 4 3 π π ¼ π  arc cot þ arc cot ¼π ¼ 3 4 2 2 Choice (3) is the answer. 2.23. From trigonometry, we know that: tan ðα þ βÞ ¼

tan ðαÞ þ tan ðβÞ 1  tan ðαÞ tan ðβÞ

arcð tan ð5ÞÞ ≜ α ⟹ tan ðαÞ ¼ 5

28

2 Solutions of Problems: Trigonometric Equations and Identities



 3 3 arc tan ≜ β ⟹ tan ðβÞ ¼ 2 2 Therefore,


 13 5 þ 32 tan ðαÞ þ tan ðβÞ 3 2 tan arcð tan ð5ÞÞ þ arc tan ¼ tan ðα þ βÞ ¼ ¼ ¼ 1 ¼ 13 2 1  tan ðαÞ tan ðβÞ 1  15  2 2

 3 3π ¼ tan 1 ð1Þ ¼ arcð tan ð5ÞÞ þ arc tan 2 4 Choice (3) is the answer. 2.24. From trigonometry, we know that: sin 2 ðαÞ þ cos 2 ðαÞ ¼ 1

 3 3 ≜ α ⟹ cos ðαÞ ¼ arc cos 5 5

 4 4 arc sin  ≜ β ⟹ sin ðβÞ ¼  5 5 Therefore,





 3 4 sin arc cos þ cos arc sin  ¼ sin ðαÞ þ cos ðβÞ ¼ 5 5 ¼

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2 3 4 þ 1  1 5 5

4 3 7 þ ¼ 5 5 5

Choice (1) is the answer. 2.25. From trigonometry, we know that: cot ðθÞ ¼ tan


 π θ 2

Based on the definition of tan(θ) and cot(θ), we can write: tan ðθÞ ¼

Opposite for θ HA HA ¼ ¼ HA ¼ 1 Adjacent for θ OH


 d d ¼ Opposite for HOB ¼ HB ¼ HB ¼ HB cot ðθÞ ¼ tan HOB OH 1 d Adjacent for HOB Choice (2) is the answer.

2

Solutions of Problems: Trigonometric Equations and Identities

29

Figure 2.3 The graph of the solution of problem 2.25

2.26. Based on the definition of sec(θ) and cos(θ), we can write: sec ðθÞ ¼

1 ¼ cos ðθÞ

1 Adjacent for θ Hypotenuse for θ

1 1 ¼ OA ¼ 1 ¼ OB OB

OB

Choice (3) is the answer.

Figure 2.4 The graph of the solution of problem 2.26

2.27. From trigonometry, we know that: sin ðθÞ ¼ cos




π θ 2



Based on the definition of csc(θ) and sin(θ), we can write: cscðθÞ ¼

1 1
¼ ¼ sin ðθÞ cos COB d

1 c Adjacent for COB c Hypotenuse for COB

1 1 ¼ OC ¼ 1 ¼ OB OB

Choice (2) is the answer.

Figure 2.5 The graph of the solution of problem 2.27

OB

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2 Solutions of Problems: Trigonometric Equations and Identities

2.28. From trigonometry, we know that: tan ðα þ βÞ ¼

tan ðαÞ þ tan ðβÞ 1  tan ðαÞ tan ðβÞ



 2 2 ≜ α ⟹ tan ðαÞ ¼ arc tan 3 3

 1 1 arc tan ≜ β ⟹ tan ðβÞ ¼ 5 5 Therefore, tan ðα þ βÞ ¼

2 13 þ1 tan ðαÞ þ tan ðβÞ π ¼ 1 ⟹ α þ β ¼ arcð tan ð1ÞÞ ¼ ¼ 3 52 ¼ 15 4 1  tan ðαÞ tan ðβÞ 1  15 13 15

Choice (2) is the answer. 2.29. From trigonometry, we know that: 8 π <
 > 2 1 arcð tan ðαÞÞ þ arc tan ¼ α > : π 2


if α > 0 if α < 0

arcð cot ðαÞÞ þ arcð cot ðαÞÞ ¼ π Therefore, 8 π > <
 2 1 arcð tan ðmÞÞ þ arc tan þ arcð cot ðmÞÞ þ arcð cot ðmÞÞ ¼ π þ m > : π 2


8 3π > < 2 ¼ > :π 2

if m > 0 if m < 0

if m > 0 if m < 0

Choice (2) is the answer. 2.30. From trigonometry, we know that: cos ðα  βÞ ¼ cos ðαÞ cos ðβÞ þ sin ðαÞ sin ðβÞ Therefore, 1  cos ð4xÞ cos ð2xÞ þ sin ð4xÞ sin ð2xÞ  0 ⟹  1  cos ð4x  2xÞ  0 ⟹  1  cos ð2xÞ  0 Since x is an acute angle: ⟹ Choice (4) is the answer.

π π π  2x  π ⟹  x  2 4 2

2

Solutions of Problems: Trigonometric Equations and Identities

31

2.31. From trigonometry, we know that: tan ðα  βÞ ¼

tan ðαÞ  tan ðβÞ 1 þ tan ðαÞ tan ðβÞ

Therefore, tan ð2yÞ ¼ tan ððx þ yÞ  ðx  yÞÞ ¼

tan ðx þ yÞ  tan ðx  yÞ 57 2 1 ¼ ¼ ¼ 1 þ tan ðx þ yÞ tan ðx  yÞ 1 þ 5  7 1 þ 35 18

Choice (2) is the answer. 2.32. From trigonometry, we know that: sin ðαÞ þ cos ðαÞ ¼

 pffiffiffi
π 2 sin α þ 4

sin ðαÞ  cos ðαÞ ¼

 pffiffiffi
π 2 sin α  4

Therefore,   5π pffiffiffi 5π π 2π  pffiffi3 pffiffiffi sin 2 sin sin 5π þ þ cos 12 12 12 4 π3 ¼ 21 ¼ 3 5π 5π  ¼ ¼ pffiffiffi 5π sin 12  cos 12 sin 6 2 sin 12  π4 2 Choice (1) is the answer. 2.33. From trigonometry, we know that: y ¼ a sin ðmxÞ ⟹ T ¼

2π jm j

Therefore, ⟹6 ¼

2π 1 1 ⟹ j bj ¼ ⟹ b ¼  3 3 jbπ j

Based on the graph and the function, the positive value of b is acceptable. ⟹b ¼

1 3

Moreover, based on y ¼ a sin (bπx) and the given graph, it is concluded that a ¼ 2. Therefore, ⟹a þ b ¼ 2 þ Choice (3) is the answer.

1 7 ¼ 3 3

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2 Solutions of Problems: Trigonometric Equations and Identities

Figure 2.6 The graph of the solution of problem 2.33

2.34. From trigonometry, we know that: y ¼ a sin ðmxÞ ⟹ T ¼

2π jm j

Therefore, ⟹3 ¼ 3 

2π ¼ 1 ⟹ jbj ¼ 2 ⟹ b ¼ 2 jbπ j

Based on the graph and the given function, the negative value of b is accepted. ⟹ b ¼ 2 In addition, based on y ¼ a sin (bπx) and the given graph, it is clear that a ¼ 3. Therefore, ⟹ a  b ¼ 3  ð2Þ ¼ 6 Choice (1) is the answer.

Figure 2.7 The graph of the solution of problem 2.34

2.35. From trigonometry, we know that: y ¼ a sin


 π þ bπx ¼ a cos ðbπxÞ 2

y ¼ a cos ðmxÞ ⟹ T ¼

2π jmj

Therefore, ⟹ 3:5  ð2:5Þ ¼ 3 

2π 6 ⟹6 ¼ ⟹ jbj ¼ 1 ⟹ b ¼ 1 jbπ j j bj

2

Solutions of Problems: Trigonometric Equations and Identities

33

Based on the graph and y ¼ a cos (bπx), the positive value of b is accepted. ⟹b ¼ 1 In addition, based on y ¼ a cos (bπx) and the given graph, it is clear that a ¼ 2. Therefore, ⟹a  b ¼ 2  1 ¼ 2 Choice (1) is the answer.

Figure 2.8 The graph of the solution of problem 2.35

2.36. From trigonometry, we know that: cos ðαÞ ¼ sin sin ðαÞ ¼ cos




 π α 2




 π α 2

sin ð2αÞ ¼ 2 sin ðαÞ cos ðαÞ Therefore,                   cos 5 cos 10 cos 20 cos 5 cos 10 cos 20 cos 5 cos 10 cos 20   ¼ ¼    sin ð40 Þ 2 sin ð20 Þ cos ð20 Þ cos 50       cos 5 cos 10 cos 5 1 1     ¼   ¼   ¼ ¼   2  2 sin ð10 Þ cos ð10 Þ 4  2 sin 5 cos 5 8 sin 5 8 cos 85 Choice (2) is the answer. 2.37. From trigonometry, we know that: sin ðxÞ ¼

 2 tan 2x  1 þ tan 2 2x

 1  tan 2 2x  cos ðxÞ ¼ 1 þ tan 2 2x

34

2 Solutions of Problems: Trigonometric Equations and Identities

Therefore,     2 tan 2x 1  tan 2 2x 2 tan 2x þ 1  tan 2 2x 7 7 7  þ ¼ ⟹  sin ðxÞ þ cos ðxÞ ¼ ⟹ ¼ 5 5 5 1 þ tan 2 2x 1 þ tan 2 2x 1 þ tan 2 2x pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi


 x x x 10  102  4  12  2 10  2 ⟹ 12 tan  10 tan þ 2 ¼ 0 ⟹ tan ¼ ¼ 2 2 2 24 24 2

⟹ tan


 x 1 1 ¼ or 2 2 3

Choice (2) is the answer. 2.38. From trigonometry, we know that: sin ð2αÞ ¼ 2 sin ðαÞ cos ðαÞ sin 2 ðαÞ þ cos 2 ðαÞ ¼ 1 cos 2 ðαÞ  sin 2 ðαÞ ¼ cos ð2αÞ cot ð2αÞ ¼

cos ð2αÞ sin ð2αÞ

In addition, from the factoring rule, we know that:    a4  b4 ¼ a2  b2 a2 þ b2 Therefore,  2   sin ðαÞ  cos 2 ðαÞ sin 2 ðαÞ þ cos 2 ðαÞ sin 4 ðαÞ  cos 4 ðαÞ  cos ð2αÞ  1 ¼ 2 cot ð2αÞ ¼ ¼ 1 sin ðαÞ cos ðαÞ sin ðαÞ cos ðαÞ 2 sin ð2αÞ Choice (2) is the answer. 2.39. Based on the information given in the problem, we have: sin 4 ðαÞ þ cos 4 ðαÞ ¼

1 2

From trigonometry, we know that: sin ð2αÞ ¼ 2 sin ðαÞ cos ðαÞ  2 sin 2 ðαÞ þ sin 2 ðαÞ ¼ 1 ⟹ sin 2 ðαÞ þ sin 2 ðαÞ ¼ 1 ⟹ sin 4 ðαÞ þ cos 4 ðαÞ þ 2 sin 2 ðαÞ cos 2 ðαÞ ¼ 1

2

Solutions of Problems: Trigonometric Equations and Identities

35

Therefore, 1 þ 2 sin 2 ðαÞ cos 2 ðαÞ ¼ 1 ⟹ 4 sin 2 ðαÞ cos 2 ðαÞ ¼ 1 ⟹ ð2 sin ðαÞ cos ðαÞÞ2 ¼ 1 2 ⟹ sin 2 ð2αÞ ¼ 1 ⟹ cos 2 ð2αÞ ¼ 0 ⟹

cos 2 ð2αÞ ¼ 0 ⟹ cot 2 ð2αÞ ¼ 0 sin 2 ð2αÞ

Choice (3) is the answer. 2.40. From trigonometry, we know that: sin ð2xÞ ¼ 2 sin ðxÞ cos ðxÞ cos ð2xÞ ¼ cos 2 ðxÞ  sin 2 ðxÞ Therefore, sin 3 ðxÞ cos ðxÞ  cos 3 ðxÞ sin ðxÞ þ 3 sin 2 ðxÞ cos 2 ðxÞ   3 ¼ sin ðxÞ cos ðxÞ sin 2 ðxÞ  cos 2 ðxÞ þ  4 sin 2 ðxÞ cos 2 ðxÞ 4 ¼ x¼

1 3 1 3 sin ð2xÞð cos ð2xÞÞ þ sin 2 ð2xÞ ¼  sin ð4xÞ þ sin 2 ð2xÞ 2 4 4 4

pffiffiffi2

 2 3π 1 3π 3 3π 1 3 1 3 5 ⟹  sin 4  þ sin 2 2  ¼  ð1Þ þ ¼ þ ¼ 8 4 8 4 8 4 4 2 4 8 8

Choice (2) is the answer. 2.41. From trigonometry, we know that: sin ðαÞ þ sin ðβÞ ¼ 2 sin

    αþβ αβ cos 2 2

    αþβ αβ cos ðαÞ þ cos ðβÞ ¼ 2 cos cos 2 2 Therefore, sin ð2αÞ þ sin ð5αÞ þ sin ð8αÞ sin ð8αÞ þ sin ð2αÞ þ sin ð5αÞ ¼ cos ð2αÞ þ cos ð5αÞ þ cos ð8αÞ cos ð8αÞ þ cos ð2αÞ þ cos ð5αÞ ¼

2 sin ð5αÞ cos ð3αÞ þ sin ð5αÞ sin ð5αÞð2 cos ð3αÞ þ 1Þ ¼ ¼ tan ð5αÞ 2 cos ð5αÞ cos ð3αÞ þ cos ð5αÞ cos ð5αÞð2 cos ð3αÞ þ 1Þ α¼

Choice (3) is the answer.


 pffiffiffi π π ⟹ tan ð5αÞ ¼ tan ¼ 3 15 3

36

2 Solutions of Problems: Trigonometric Equations and Identities

2.42. From trigonometry, we know that: sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1 2 sin ðxÞ cos ðxÞ ¼ sin ð2xÞ In addition, from the factoring rule, we know that: ð a þ bÞ ð a  bÞ ¼ a2  b2 Therefore, ð sin ðxÞ  cos ðxÞ þ 2Þð sin ðxÞ  cos ðxÞ  2Þ ¼ ð sin ðxÞ  cos ðxÞÞ2  4 ¼ sin 2 ðxÞ þ cos 2 ðxÞ 2 sin ðxÞ cos ðxÞ  4 ¼ 1  sin ð2xÞ  4 ¼ 3  sin ð2xÞ x¼


 π π 1 7 ⟹  3  sin ¼ 3  ¼  12 6 2 2

Choice (4) is the answer. 2.43. From trigonometry, we know that: tan ðαÞ cot ðαÞ ¼ 1 1 þ tan 2 ðαÞ ¼

1 cos 2 ðαÞ

1 þ cot 2 ðαÞ ¼

1 sin 2 ðαÞ

sin 2 ðαÞ þ cos 2 ðαÞ ¼ 1 Therefore, 4 sin 2 ðαÞ cos 2 ðαÞð tan ðαÞ þ cot ðαÞÞ2   ¼ 4 sin 2 ðαÞ cos 2 ðαÞ tan 2 ðαÞ þ cot 2 ðαÞ þ 2 tan ðαÞ cot ðαÞ    2 2 2 2 ¼ 4 sin ðαÞ cos ðαÞ 1 þ tan ðαÞ þ 1 þ cot ðαÞ ¼ 4 sin ðαÞ cos ðαÞ 2

2



1 1 þ cos 2 ðαÞ sin 2 ðαÞ

   sin 2 ðαÞ þ cos 2 ðαÞ 1 2 2 ¼ 4 sin ðαÞ cos ðαÞ ¼ 4 sin ðαÞ cos ðαÞ ¼4 sin 2 ðαÞ cos 2 ðαÞ sin 2 ðαÞ cos 2 ðαÞ 2

2

Choice (4) is the answer. 2.44. From trigonometry, we know the common solution of the equations below. sin ðαÞ ¼ 0 ) α ¼ kπ, 8k 2 ℤ



2

Solutions of Problems: Trigonometric Equations and Identities

37

π cos ðαÞ ¼ 0 ) α ¼ kπ þ , 8k 2 ℤ 2 π tan ðαÞ ¼ 1 ) α ¼ kπ  , 8k 2 ℤ 4 Hence, sin ðπxÞ cos 2 ðπxÞ þ sin 2 ðπxÞ cos ðπxÞ ¼ 0 ) sin ðπxÞ cos ðπxÞð cos ðπxÞ þ sin ðπxÞÞ ¼ 0 8 2  x  2 > > sin ðπxÞ ¼ 0 ) πx ¼ kπ ) x ¼ 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 ¼ ¼ ¼ ¼ ¼ ¼) x ¼ 2,  1, 0, 1, 2 > > > < π 1 2  x  2 3 1 1 3 ) cos ðπxÞ ¼ 0 ) πx ¼ kπ þ ) x ¼ k þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ 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  ,  , , 2 2 2 2 2 2 > > > > 2  x  2 > π 5 1 3 7 : sin ðπxÞ þ cos ðπxÞ ¼ 0 ) tan ðπxÞ ¼ 1 ) πx ¼ kπ  ) x ¼ k  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) x ¼  ,  , , 4 4 4 4 4 4 Therefore, the number of roots of the equation is 5 + 4 + 4 ¼ 13. Choice (3) is the answer. 2.45. From trigonometry, we know that: y ¼ a þ b cos ðmxÞ ⟹ T ¼

2π jmj

Therefore, ⟹ 4π ¼

2π 1 1 ⟹ jmj ¼ ⟹ m ¼  2 2 jmj

Based on the graph and the function, the positive value of m is accepted. ⟹m ¼


 1 1 1 ⟹ y ¼ þ 2 cos x 2 2 2

The value of function for x ¼ 16π 3 is:  y

16π 3

 ¼

 


 1 1 16π 1 8π 1 2π 1 2π þ 2 cos  ¼ þ 2 cos ¼ þ 2 cos 2π þ ¼ þ 2 cos 2 2 3 2 3 2 3 2 3


 1 π 1 π 1 1 1 ¼ þ 2 cos π  ¼  2 cos ¼ 2 ¼ 2 3 2 3 2 2 2

Choice (1) is the answer.

Figure 2.9 The graph of the solution of problem 2.45

38

2 Solutions of Problems: Trigonometric Equations and Identities

2.46. From trigonometry, we know that: y ¼ a þ b sin ðmxÞ ⟹ T ¼

2π jm j

Therefore, ⟹

2π 2π ¼ ⟹ jmj ¼ 3 ⟹ m ¼ 3 3 jmj

Based on the graph and the given function, the positive value of m is accepted. ⟹ m ¼ 3 ⟹ y ¼ 1  sin ð3xÞ The value of function for x ¼ 7π 6 is: y






 7π 7π 7π 3π 3π ¼ 1  sin 3  ¼ 1  sin ¼ 1  sin 2π þ ¼ 1  sin ¼ 1  ð1Þ ¼ 2 6 6 2 2 2

Choice (4) is the answer.

Figure 2.10 The graph of the solution of problem 2.46

2.47. From trigonometry, we know that: y ¼ a þ b sin ðmxÞ ⟹ T ¼

2π jm j

Therefore, ⟹5  1 ¼

2π 1 ⟹b ¼  2 jbπ j

Based on the graph and the given function, the positive value of m is accepted. ⟹b ¼ 


 1 π ⟹ y ¼ a þ sin  x 2 2

By testing the point of (0, 3) in the function, we have:

 π π 3 ¼ a þ sin   0 ⟹ a ¼ 3 ⟹ y ¼ 3 þ sin  x 2 2

2

Solutions of Problems: Trigonometric Equations and Identities

39

The value of function for x ¼ 25 3 is: y

   
 25 π 25 π ¼ 3 þ sin   ¼ 3 þ sin 4π  3 2 3 6
 π 1 ¼ 3 þ sin  ¼ 3  ¼ 2:5 6 2

Choice (2) is the answer.

Figure 2.11 The graph of the solution of problem 2.47

2.48. By testing the point of (0, 0) in the function, we have: 0 ¼ a þ b cos


 π  0 ⟹a þ b ¼ 0 2

ð1Þ

Based on the function and the graph given in the problem, we can write: ymax ¼ a þ jbj ⟹ a þ jbj ¼ 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 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)

(

aþb¼0 aþb¼4

) Impossible

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) Choice (1) is the answer.

(

aþb¼0 ab¼4

⟹ 2b ¼ 4 ⟹ b ¼ 2

40

2 Solutions of Problems: Trigonometric Equations and Identities

Figure 2.12 The graph of the solution of problem 2.48

2.49. From trigonometry, we know that: y ¼ 1 þ a sin ðmxÞ ⟹ T ¼

2π jm j

Therefore, ⟹

4 2π ¼2 ⟹ jbj ¼ 3 ⟹ b ¼ 3 3 jbπ j

Based on the function and the graph given in the problem, we can write: ymin ¼ 1  jaj ⟹  1 ¼ 1  jaj ⟹ jaj ¼ 2 ⟹ a ¼ 2 Based on the graph and the given function, both a and b must be either positive or negative. Hence: (

a¼2 b¼3

(

a ¼ 2 b ¼ 3

⟹a þ b ¼ 5

⟹ a þ b ¼ 5

Only a + b ¼ 5 exists in the choices. Choice (3) is the answer.

Figure 2.13 The graph of the solution of problem 2.49

2.50. From trigonometry, we know that:
 π cos α þ ¼  sin ðαÞ 2

2

Solutions of Problems: Trigonometric Equations and Identities

41

Therefore,
 π y ¼ a  2 cos bx þ ¼ a þ 2 sin ðbxÞ 2 In addition, from trigonometry, we know that: y ¼ a þ 2 sin ðbxÞ ⟹ T ¼

2π 13π π 2π ⟹  ¼ ⟹ jbj ¼ 3 ⟹ b ¼ 3 18 18 j bj j bj

Based on the graph and the simplified function, i.e., y ¼ a + 2 sin (bx), the positive value of b is acceptable. ⟹b ¼ 3 Based on the simplified function and the given graph, we can write: ymax ¼ a þ 2 ⟹ 1 ¼ a þ 2 ⟹ a ¼ 1 Hence, a þ b ¼ 1 þ 3 ¼ 2 Choice (4) is the answer.

Figure 2.14 The graph of the solution of problem 2.50

2.51. From trigonometry, we know that:   tan 45 ¼ 1 tan ðα  βÞ ¼

tan ðαÞ  tan ðβÞ 1 þ tan ðαÞ tan ðβÞ

cot ðαÞ ¼

1 tan ðαÞ

In addition, based on the information given in the problem, we have:   3 tan α þ 20 ¼ 4

42

2 Solutions of Problems: Trigonometric Equations and Identities

Therefore,    cot 25  α ¼

    1 þ tan 45 tan α þ 20 1 1      ¼ ¼      tan 45  tan ðα þ 20 Þ tan 25  α tan 45  ðα þ 20 Þ 

  1 þ tan α þ 20 1 þ 34 ¼7 ¼ ¼  1  tan ðα þ 20 Þ 1  34 Choice (3) is the answer. 2.52. From trigonometry, we know that: cos ðαÞ ¼ 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  sin 2 ðαÞ for an obtuse angle tan ðαÞ ¼ tan

tan ðα þ βÞ ¼

sin ðαÞ cos ðαÞ


 π ¼1 4 tan ðαÞ þ tan ðβÞ 1  tan ðαÞ tan ðβÞ

In addition, based on the information given in the problem, we have: sin ðαÞ ¼

3 5

Therefore, rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 3 4 2 cos ðαÞ ¼  1  sin ðαÞ ¼  1  ¼ 5 5 tan ðαÞ ¼

3 sin ðαÞ 3 ¼ 54 ¼  4 cos ðαÞ  5

   1  tan π4 þ tan ðaÞ 1 þ tan ðαÞ 1 þ  34 π 1    ¼ 47 ¼ tan þα ¼ ¼ ¼ 4 7 1  tan π4 tan ðaÞ 1  tan ðαÞ 1   34 4


Choice (3) is the answer. 2.53. From trigonometry, we know that:
tan

 π  α ¼ cot ðαÞ 2

cot ðαÞ ¼ tan

1 tan ðαÞ


 π ¼1 4

2

Solutions of Problems: Trigonometric Equations and Identities

43

tan ðα  βÞ ¼

tan ðαÞ  tan ðβÞ 1 þ tan ðαÞ tan ðβÞ

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

 π 2 α ¼ 2 3

Therefore,
 2 π 1 3 ¼ tan  α ¼ cot ðαÞ ¼ ⟹ tan ðαÞ ¼ 3 2 2 tan ðαÞ   tan π4  tan ðaÞ 1  tan ðαÞ 1  32  12 π 1 π  ¼ tan α ¼ ¼ 5 ¼  ¼ 4 5 1 þ tan 4 tan ðaÞ 1 þ tan ðαÞ 1 þ 32 2


Choice (2) is the answer. 2.54. From trigonometry, we know that: tan ðα þ βÞ ¼

tan ðαÞ þ tan ðβÞ 1  tan ðαÞ tan ðβÞ

In addition, based on the information given in the problem, we have: tan ða þ bÞ ¼

2 5

tan ða  bÞ ¼

3 7

Therefore, tan ð2aÞ ¼ tan ðða þ bÞ þ ða  bÞÞ ¼

2 29 þ3 tan ða þ bÞ þ tan ða  bÞ ¼1 ¼ 5 2 7 3 ¼ 35 1  tan ða þ bÞ tan ða  bÞ 1  5  7 29 35

Choice (4) is the answer. 2.55. From trigonometry, we know that: sin ðαÞ ¼ cos




 π α 2

cos ðαÞ ¼ cos ðαÞ tan tan ðα  βÞ ¼


 π ¼1 4 tan ðαÞ  tan ðβÞ 1 þ tan ðαÞ tan ðβÞ

44

2 Solutions of Problems: Trigonometric Equations and Identities

Therefore, sin

⟹2 ¼






 π π π π π þ x ¼ cos   x ¼ cos  x ¼ cos x  4 2 4 4 4

    
 sin x  π4 sin x  π4 tan ðxÞ  tan π4 π      ¼2 ¼ 2 ⟹ ¼ ⟹ tan x  4 sin x þ π4 cos x  π4 1 þ tan ðxÞ tan π4

⟹

tan ðxÞ  1 ¼ 2 ⟹ tan ðxÞ  1 ¼ 2 þ 2 tan ðxÞ ⟹ tan ðxÞ ¼ 3 1 þ tan ðxÞ

Choice (1) is the answer. 2.56. From trigonometry, we know that: tan tan ðα þ βÞ ¼


 π ¼1 4 tan ðαÞ þ tan ðβÞ 1  tan ðαÞ tan ðβÞ

Moreover, based on the information given in the problem, we have: αþβ ¼

π 4

If we calculate the tangent value of each side of the abovementioned relation, we will have: tan ðα þ βÞ ¼ tan


 tan ðαÞ þ tan ðβÞ π ) ¼ 1 ) tan ðαÞ þ tan ðβÞ ¼ 1  tan ðαÞ tan ðβÞ 4 1  tan ðαÞ tan ðβÞ

Therefore, ð1 þ tan ðαÞÞð1 þ tan ðβÞÞ ¼ 1 þ tan ðαÞ þ tan ðβÞ þ tan ðαÞ tan ðβÞ ¼ 1 þ ð1  tan ðαÞ tan ðβÞÞ þ tan ðαÞ tan ðβÞ ¼ 2 Choice (2) is the answer. 2.57. From trigonometry, we know that: tan


 π ¼1 4

tan ðα þ βÞ ¼

tan ðαÞ þ tan ðβÞ 1  tan ðαÞ tan ðβÞ

tan ðα  βÞ ¼

tan ðαÞ  tan ðβÞ 1 þ tan ðαÞ tan ðβÞ

tan ð2αÞ ¼

2 tan ðαÞ 1  tan 2 ðαÞ

2

Solutions of Problems: Trigonometric Equations and Identities

45

Therefore,   
 tan π4 þ tan ðαÞ tan π4  tan ðαÞ π π   þ α  tan α ¼ tan  4 4 1  tan π4 tan ðαÞ 1 þ tan π4 tan ðαÞ


¼

1 þ tan ðαÞ 1  tan ðαÞ ð1 þ tan ðαÞÞ2  ð1  tan ðαÞÞ2 4 tan ðαÞ ¼ ¼ 2 tan ð2αÞ  ¼ 1  tan ðαÞ 1 þ tan ðαÞ 1  tan 2 ðαÞ 1  tan 2 ðαÞ

Choice (1) is the answer. 2.58. From trigonometry, we know that: tan tan ðα  βÞ ¼


 π ¼1 4 tan ðαÞ  tan ðβÞ 1 þ tan ðαÞ tan ðβÞ

tan ð2αÞ ¼

2 tan ðαÞ 1  tan 2 ðαÞ

Moreover, based on the information given in the problem, we have: 
 tan π4  tan ðαÞ 1  tan ðαÞ 1 π 1 2 π  ¼ tan α ¼ ⟹ ¼ ⟹ 5  5 tan ðαÞ ¼ 1 þ tan ðαÞ ⟹ tan ðαÞ ¼ 4 5 5 3 1 þ tan 4 tan ðαÞ 1 þ tan ðαÞ tan ð2αÞ ¼

2  23 2 tan ðαÞ 12 ¼   ¼ 5 ¼ 2:4 1  tan 2 ðαÞ 1  2 2 3

Choice (3) is the answer. 2.59. From trigonometry, we know that: tan ð2αÞ ¼ tan ðα  βÞ ¼

2 tan ðαÞ 1  tan 2 ðαÞ

tan ðαÞ  tan ðβÞ 1 þ tan ðαÞ tan ðβÞ

Moreover, based on the information given in the problem, we have: tan ðαÞ ¼ 2 tan ðβÞ ¼

1 3

Therefore, tan ð2αÞ ¼

2 tan ðαÞ 22 4 ¼ ⟹ tan ð2αÞ ¼ 2 3 1  tan 2 ðαÞ 12

46

2 Solutions of Problems: Trigonometric Equations and Identities

tan ð2α  βÞ ¼

 43  13 5 tan ð2αÞ  tan ðβÞ  41 ¼ 5 3 ¼ 3 ⟹ tan ð2α  βÞ ¼ 1 þ tan ð2αÞ tan ðβÞ 1 þ 3 3 9

Choice (1) is the answer. 2.60. From trigonometry, we know that: cos ðπ  xÞ ¼  cos ðxÞ cos ðαÞ ¼ cos ðα0 Þ ⟹ α ¼ 2kπ  α0 , 8k 2 ℤ Moreover, based on the information given in the problem, we have: cos ðxÞ 6¼ 0 Therefore, cos ð3xÞ þ cos ðxÞ ¼ 0 ⟹ cos ð3xÞ ¼  cos ðxÞ ⟹ cos ð3xÞ ¼ cos ðπ  xÞ ( ⟹ 3x ¼ 2kπ  ðπ  xÞ ⟹

8 kπ π >

3x ¼ 2kπ  π þ x ⟹ 2x ¼ 2kπ  π : x ¼ kπ  π 2

However, cos ðxÞ 6¼ 0 ⟹ x ¼

kπ π þ 2 4

Choice (1) is the answer. 2.61. From trigonometry, we know that: cot ðαÞ ¼ tan


 π α 2

tan ðαÞ ¼ tan ðα0 Þ ⟹ α ¼ kπ þ α0 , 8k 2 ℤ Therefore, tan ð4xÞ ¼ cot ðxÞ ⟹ tan ð4xÞ ¼ tan ⟹ 4x ¼ kπ þ





 π x 2

 π π kπ π  x ⟹ 5x ¼ kπ þ ⟹ x ¼ þ 2 2 5 10

8 π k ¼ 1 ⟹ x1 ¼  is not a positive angle > > 10 > > > π > > is an acute angle < k ¼ 0 ⟹ x2 ¼ 10 ⟹ > k ¼ 1 ⟹ x ¼ 3π is an acute angle > > 3 > 10 > > > : k ¼ 2 ⟹ x ¼ π is not an acute angle 4 2 Choice (1) is the answer.

⟹ x2 þ x3 ¼

2π 5

2

Solutions of Problems: Trigonometric Equations and Identities

47

2.62. From trigonometry, we know that: sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1 cos ðxÞ ¼ cos ðx0 Þ ⟹ x ¼ 2kπ  x0 Therefore,   2 sin 2 ðxÞ þ 3 cos ðxÞ ¼ 0 ⟹ 2 1  cos 2 ðxÞ þ 3 cos ðxÞ ¼ 0 ⟹ 2 cos 2 ðxÞ  3 cos ðxÞ  2 ¼ 0 ⟹ cos 2 ðxÞ 


 3 1 cos ðxÞ  1 ¼ 0 ⟹ cos ðxÞ þ ð cos ðxÞ  2Þ ¼ 0 2 2 8 < cos ðxÞ ¼  1 ⟹ x ¼ 2kπ  2π 2 3 ⟹ : cos ðxÞ ¼ 2 ⟹ not acceptable

Choice (1) is the answer. 2.63. From trigonometry, we know that: sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1 cos ðxÞ ¼ cos ðx0 Þ ⟹ x ¼ 2kπ  x0 Therefore,   2 sin 2 ðxÞ ¼ 3 cos ðxÞ ⟹ 2 1  cos 2 ðxÞ  3 cos ðxÞ ¼ 0 ⟹ 2 cos 2 ðxÞ þ 3 cos ðxÞ  2 ¼ 0 ⟹ cos 2 ðxÞ þ


 3 1 cos ðxÞ  1 ¼ 0 ⟹ cos ðxÞ  ð cos ðxÞ þ 2Þ ¼ 0 2 2

8 < cos ðxÞ ¼ 1 ⟹ x ¼ 2kπ  π 2 3 ⟹ : cos ðxÞ ¼ 2 ⟹ not acceptable Choice (4) is the answer. 2.64. From trigonometry, we know that: tan ðαÞ: cot ðαÞ ¼ 1 Now, let us find the intersection point of the lines, as follows. (

x tan ðαÞ  y cot ðαÞ ¼ 1 x tan ðαÞ þ y cot ðαÞ ¼ 2

⟹ xy ¼ Choice (4) is the answer.

⟹

8 3 > > < 2x tan ðαÞ ¼ 3 ⟹ x ¼ 2 tan ðαÞ > > : 2y cot ðαÞ ¼ 1 ⟹ y ¼

3 1 3 3  ¼ ⟹y ¼ 4x 2 tan ðαÞ 2 cot ðαÞ 4

1 2 cot ðαÞ

48

2 Solutions of Problems: Trigonometric Equations and Identities

2.65. From trigonometry, we know that: sin 2 ðαÞ þ cos 2 ðαÞ ¼ 1 Based on the information given in the problem, we have: 8 x2 > < x ¼ 2  3 sin ðαÞ ⟹ sin ðαÞ ¼ 3 > : y ¼ 1 þ 4 cos ðαÞ ⟹ cos ðαÞ ¼ y  1 4 Therefore, ⟹

ðx  2Þ2 ðy  1Þ2 þ ¼1 9 16

which is the equation of an ellipse. Choice (2) is the answer. 2.66. Based on the information given in the problem, we have: (

x ¼ 2  5 cos ðαÞ y¼4

From trigonometry, we know that: 1  cos ðαÞ  1 ⟹  1 

2x  1 ⟹  5  2  x  5 ⟹  7  x  3 ⟹  3  x  7 5

Therefore, ( ⟹

3x7 y¼4

which is the equation of a horizontal line segment. Choice (3) is the answer. 2.67. From trigonometry, we know that the maximum value of cos(.) and sin(.) is one. Therefore, the only solution of the given equation is: (

cos ðx  yÞ ¼ 1 sin ðx þ yÞ ¼ 1

The common solution of the equations can be calculated as follows: ( ⟹

8 0 < x, y < 2π < x  y ¼ 0 π 5π ⟹ y ¼ or ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) π ¼ 4 4 : x þ y ¼ π or 5π x þ y ¼ 2kπ þ 2 2 2 x  y ¼ 2kπ

Choice (2) is the answer.

2

Solutions of Problems: Trigonometric Equations and Identities

49

2.68. From trigonometry, we know that: tan ðα þ βÞ ¼

tan ðαÞ þ tan ðβÞ 1  tan ðαÞ tan ðβÞ

In addition, we know that the sum and the product of the roots of a quadratic equation in the form of ax2 + bx + c ¼ 0 are  ba and ac, respectively. Based on the information given in the problem, we have: tan 2 ðxÞ þ ðm þ 2Þ tan ðxÞ þ 2m  2 ¼ 0 αþβ ¼

π 4

Therefore, tan ð:Þ tan ðαÞ þ tan ðβÞ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) tan ðα þ βÞ ¼ 1 ⟹ ¼1 1  tan ðαÞ tan ðβÞ ðmþ2Þ

⟹

sum of the roots of the quadratic equation m  2 1 ¼ ¼1 ¼ 1  product of the roots of the quadratic equation 1  2m2 3  2m 1 ⟹  m  2 ¼ 3  2m ⟹ m ¼ 5

Choice (3) is the answer. 2.69. From trigonometry, we know that: sin 6 ðαÞ þ cos 6 ðαÞ ¼ 1  3 sin 2 ðαÞ cos 2 ðαÞ sin 4 ðαÞ þ cos 4 ðαÞ ¼ 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 ¼ ¼1 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 Choice (4) is the answer. 2.70. From trigonometry, we know that:        sin 135 ¼ sin 180  45 ¼ sin 45        cos 210 ¼ cos 180 þ 30 ¼  cos 30        cos 135 ¼ cos 180  45 ¼  cos 45        sin 420 ¼ sin 360 þ 60 ¼ sin 60        tan 210 ¼ tan 180 þ 30 ¼ tan 30

50

2 Solutions of Problems: Trigonometric Equations and Identities

       cot 420 ¼ cot 360 þ 60 ¼ cot 60        cot 120 ¼ cot 180  60 ¼  cot 60        tan 330 ¼ tan 360  30 ¼  tan 30 Therefore,
pffiffi
pffiffi pffiffi              pffiffi2  3 þ  22 23 3pffiffi6ffi sin 45  cos 30 þ  cos 45 sin 60 2 2 
¼ p p pffiffi
pffiffi ¼ ffiffi ffiffi     4 3 tan ð30 Þ cot ð60 Þ þ ð cot ð60 ÞÞð tan ð30 ÞÞ  3þ  3  3 3

3

3

3

Choice (2) is the answer. 2.71. From trigonometry, we know that: cot ðx þ yÞ ¼

cot ðxÞ cot ðyÞ  1 cot ðxÞ þ cot ðyÞ

Based on the information given in the problem, we have: πk¼0 π cot ð:Þ cot ðxÞ cot ðyÞ  1 x þ y ¼ kπ þ ¼ 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 ¼) ¼1 4 4 cot ðxÞ þ cot ðyÞ ⟹ 1 þ cot ðxÞ þ cot ðyÞ ¼ cot ðxÞ cot ðyÞ

ð1Þ

ð1 þ cot ðxÞÞð1 þ cot ðyÞÞ ¼ ð1 þ cot ðxÞ þ cot ðyÞÞ þ cot ðxÞ cot ðyÞ

ð2Þ

On the other hand, we can write:

Solving (1) and (2): ð1 þ cot ðxÞÞð1 þ cot ðyÞÞ ¼ cot ðxÞ cot ðyÞ þ cot ðxÞ cot ðyÞ ¼ 2 cot ðxÞ cot ðyÞ Choice (4) is the answer. 2.72. From trigonometry, we know that:
tan
cos

 3π  x ¼ cot ðxÞ 2


 4π π 1 ¼  cos ¼ 3 3 2 tan ðxÞ cot ðxÞ ¼ 1 cot ðxÞ ¼

cos ðxÞ sin ðxÞ

cos ðxÞ ¼ cos ðx0 Þ ⟹ x ¼ 2kπ  x0

2

Solutions of Problems: Trigonometric Equations and Identities

51

Therefore, ð sin ðxÞ  tan ðxÞÞ tan





 3π 4π 1  x ¼ cos ⟹ ð sin ðxÞ  tan ðxÞÞ cot ðxÞ ¼  2 3 2

⟹ sin ðxÞ cot ðxÞ  tan ðxÞ cot ðxÞ ¼ 

1 1 1 π ⟹ cos ðxÞ  1 ¼  ⟹ cos ðxÞ ¼ ⟹ x ¼ 2kπ  2 2 2 3

Choice (3) is the answer. 2.73. From trigonometry, we know that: sin ðα þ βÞ ¼ sin ðαÞ cos ðβÞ þ sin ðαÞ cos ðβÞ cos ðα þ βÞ ¼ cos ðαÞ cos ðβÞ  sin ðαÞ sin ðβÞ tan ðxÞ ¼

sin ðxÞ cos ðxÞ

tan ðxÞ ¼ tan ðx0 Þ ⟹ x ¼ kπ þ x0 Therefore, sin ð2xÞð sin ðxÞ þ cos ðxÞÞ ¼ cos ð2xÞð cos ðxÞ  sin ðxÞÞ ⟹ sin ð2xÞ sin ðxÞ þ sin ð2xÞ cos ðxÞ ¼ cos ð2xÞ cos ðxÞ  cos ð2xÞ sin ðxÞ ⟹ sin ð2xÞ cos ðxÞ þ cos ð2xÞ sin ðxÞ ¼ cos ð2xÞ cos ðxÞ  sin ð2xÞ sin ðxÞ 1  & cos ð3xÞ 6¼ 0 cos ð3xÞ ⟹ sin ð2x þ xÞ ¼ cos ð2x þ xÞ ⟹ sin ð3xÞ ¼ cos ð3xÞ¼ ¼ ¼ ¼) tan ð3xÞ 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 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ⟹ 3x ¼ kπ þ

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

5π 4

Choice (2) is the answer. 2.74. From trigonometry, we know that:  
 5π π sin þ x ¼ sin þ x ¼ cos ðxÞ 2 2 sin ðα þ βÞ ¼ sin ðαÞ cos ðβÞ þ sin ðαÞ cos ðβÞ sin



 pffiffiffi 2 π π ¼ cos ¼ 2 4 4 (

sin ðxÞ ¼ sin ðx0 Þ ⟹

x ¼ 2kπ þ x0 x ¼ 2kπ þ π  x0

52

2 Solutions of Problems: Trigonometric Equations and Identities

Therefore, 
  pffiffiffi
π pffiffiffi

π  π ⟹ 2 sin  x ¼ 1 þ cos ðxÞ ⟹ 2 sin cos ðxÞ  cos sin ðxÞ ¼ 1 þ cos ðxÞ 4 4 4 8
 π > < x ¼ 2kπ þ  2
 ⟹ cos ðxÞ  sin ðxÞ ¼ 1 þ cos ðxÞ ⟹ sin ðxÞ ¼ 1 ⟹ > : x ¼ 2kπ þ π   π 2

⟹

8 π > < x ¼ 2kπ  2 > : x ¼ 2kπ þ 3π 2

⟹ x ¼ 2kπ 

π 2

Choice (3) is the answer. 2.75. From trigonometry, we know that: tan


 pffiffiffi π ¼ 3 3

tan ðxÞ ¼

cos

sin ðxÞ cos ðxÞ


 π 1 ¼ 3 2

cos ðα  βÞ ¼ cos ðαÞ cos ðβÞ þ sin ðαÞ sin ðβÞ cos ðxÞ ¼ cos ðx0 Þ ⟹ x ¼ 2kπ  x0 Therefore, 
 pffiffiffi sin π3 π   sin ð2xÞ ¼ 1 sin ð2xÞ ¼ 1 ⟹ cos ð2xÞ þ cos ð2xÞ þ 3 sin ð2xÞ ¼ 1 ⟹ cos ð2xÞ þ tan 3 cos π3 ⟹ cos ð2xÞ cos






 π π π π π þ sin sin ð2xÞ ¼ cos ⟹ cos 2x  ¼ cos 3 3 3 3 3

8 π π π < 2x  ¼ 2kπ þ ⟹ x ¼ kπ þ π π 3 3 3 ⟹ 2x  ¼ 2kπ  ⟹ 3 3 : 2x  π ¼ 2kπ  π ⟹ x ¼ kπ 3 3 Choice (4) is the answer.

Reference 1. Rahmani-Andebili, M. (2020). Precalculus: Practice problems, methods, and solutions, Springer Nature, 2020.

3

Problems: Limits

Abstract

In this chapter, the basic and advanced problems of limits 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 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 the following limit [1].

lim

þ

x!ð1Þ

½ x þ 1 x2  1

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1)  12 2) 0 3) 12 4) 1 3.2. Calculate the limit of the following function if x ! 2+. f ð xÞ ¼

xþ4 ½x  3

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

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Calculus, https://doi.org/10.1007/978-3-030-64980-7_3

53

54

3

3.3. 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 3.4. Determine the value of the limit below. lim

½x þ 3x  3x

x!1 ½x

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2 2) 2 3) 4 4) 4 3.5. Calculate the limit of the function below if x ! 0. pffiffiffi xþ 3 x pffiffiffi f ð xÞ ¼ x 3 x Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 1 3) 1 4) 1 3.6. 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

Problems: Limits

3

Problems: Limits

55

3.7. Determine the limit of the function below if x ! 0+. pffiffiffi ð x 2  1Þ x pffiffiffi ðx x þ 1Þx

f ð xÞ ¼ Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 1 3) 0 4) 1 3.8. Determine the value of the following limit. limþ

x!0

  1 1  3 x x

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 1 3) 0 4) 1 3.9. For the function below, calculate the value of limþ f ðxÞ  lim f ðxÞ. x!1

f ð xÞ ¼

x!1

2x ½2x þ 2

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2)  16 3) 23 4) 1 3.10. 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 3.11. Calculate the limit of the following function if x ! 4. f ð xÞ ¼ Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large

½ x  4 x2  16

56

3

1) 0 2) 18 3) 1 4) 1 3.12. Determine the value of the limit below. lim

x!1

1  x3 arcð cos ðxÞÞ

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 1 3) 0 4) 3 3.13. Calculate the value of the limit below. lim

x!0

tan ðxÞ  tan ð3xÞ þ tan ð2xÞ x3

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 6 2) 6 3) 10 4) 10 3.14. Calculate the value of the following limit. lim

x!3

9  x2 pffiffiffiffiffiffiffiffiffiffiffi 2 xþ1

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 6 2) 12 3) 18 4) 24 3.15. Determine the limit of the function below if x ! + 1. f ð xÞ ¼ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) Undefined 2) 0 3) 1 4) 1

sin ðxÞ x

Problems: Limits

3

Problems: Limits

57

3.16. Determine the value of the limit below. lim

x!0

½ x2   x2 x tan ðxÞ

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 1 3) 2 4) 2 3.17. 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 3.18. Calculate the value of the following limit.  lim

x!1

x2 þ x  1 pffiffiffiffiffiffi 3x þ 4 x

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 13 2)  13 3) 1 4) 1 3.19. Calculate the value of the limit below. pffiffiffi ð x þ 1Þ x lim x2  x x!0þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 1 3) 1 4) 1



58

3

3.20. Determine the limit of the following function if x ! + 1. f ðxÞ ¼

x pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x  1 þ x2 þ x  1

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 0 3)  12 4) 12 3.21. Calculate the value of lim x cot ðxÞ. x!0

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 1 3) 1 4) 2 3.22. For what value of “a”, the following function has a definite limit at x ¼ 1? ( f ð xÞ ¼

x2 þ ax x3

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 3 3) 3 4) 2 3.23. Determine the value of the limit below.  3   x  8 pffiffiffiffiffi lim x!2 x  2x Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 24 2) 16 3) 16 4) 24 3.24. Calculate the value of the limit below. lim

x!0

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large

½x þ x ½x þ x

x>1 x n

x!1

Therefore,  lim

x!1

   x2 þ x  1 x2 x pffiffiffiffiffiffi  lim ¼ þ1 ¼ lim  x!1 3x x!1 3 3x þ 4 x

Choice (3) is the answer. 4.19. From calculus, we know that:   lim am xm þ am1 xm1 þ . . . þ amn xmn þ amn1 xmn1  amn1 xmn1

x!0

or lim ðam xm þ an xn Þ  an xn if m > n

x!0

The problem can be solved as follows. limþ

x!0

Choice (4) is the answer.

pffiffiffi pffiffiffi pffiffiffi pffiffiffi ð x þ 1Þ x x x xþ x 1 ¼ limþ pffiffiffi ¼ 1 ¼ lim  lim x2  x x2  x x x!0þ x!0þ x x!0

4

Solutions of Problems: Limits

71

4.20. From calculus, we know that: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a x2 þ ax þ b  x þ x!þ1 2 lim

The problem can be solved as follows. lim

x!þ1 x

1þ

x x x x 1 1 ¼ lim pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  lim ¼ lim ¼  lim x2 þ x  1 x!þ1 x  1 þ x þ 12 x!þ1 2x  12 x!þ1 2x x!þ1 2 2

Choice (4) is the answer. 4.21. From trigonometry, we know that:

cot ðxÞ ¼

1 tan ðxÞ

From the application of the Taylor series in limit, we know that: lim tan ðxÞ  x

x!0

The problem can be solved as follows. lim x cot ðxÞ ¼ lim

x!0

x!0

x x  lim ¼ 1 tan ðxÞ x!0 x

Choice (3) is the answer. 4.22. As we know, the limit of a function at the point of x0 exits if: lim f ðxÞ ¼ limþ f ðxÞ ) lim f ðxÞ ¼ limþ f ðxÞ

ð1Þ

lim f ðxÞ ¼ lim ðx  3Þ ¼ 1  3 ¼ 2

ð2Þ

  limþ 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!1

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 ¼ ¼ ¼)  2 ¼ 1 þ a ) a ¼ 3 Choice (3) is the answer. 4.23. The problem can be solved as follows. 3 x  8 ðx3  8Þ 0þ pffiffiffiffiffi ¼ lim pffiffiffiffiffi ¼  lim 0 x!2 x  2x x!2 x  2x

72

4

Solutions of Problems: Limits

d ððx3  8ÞÞ H 3x2 3  22 12 pffiffiffiffiffi ¼ lim ) lim dxd  ¼ ¼ 1 ¼ 24 1 1 x!2 x!2 1  pffiffiffiffi 1  pffiffiffiffiffiffi 2 dx x  2x 2x 22

Choice (1) is the answer. 4.24. The problem can be solved as follows. ½ x þ x 1 þ x 1 þ 0 1 ¼ ¼  ¼ þ1 ¼ lim 0 0 þ 0 ½x þ x x!0 0 þ x

lim

x!0

Choice (1) is the answer. 4.25. The problem can be solved as follows. sin ð3xÞ þ sin ð7xÞ 0 ¼ 0 3x þ tan ð2xÞ

lim

x!0

H

d dx ð sin ð3xÞ d x!0 dx ð3x þ

) lim

þ sin ð7xÞÞ 3 cos ð3xÞ þ 7 cos ð7xÞ 3 þ 7 ¼ lim ¼2 ¼ 3þ2 x!0 3 þ 2ð1 þ tan 2 ð2xÞÞ tan ð2xÞÞ

Choice (2) is the answer. 4.26. The problem can be solved as follows. pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi xþ33 xþ3 3 xþ3 3 xþ3þ 3 lim ¼ lim  pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ¼ lim pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi x x x!0 x!0 x!0 x xþ3þ 3 xþ3þ 3 pffiffiffi 3 1 1 1 ¼ lim pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ¼ pffiffiffi pffiffiffi ¼ pffiffiffi ¼ 6 x!0 xþ3þ 3 3þ 3 2 3 Choice (2) is the answer. 4.27. The problem can be solved as follows. lim

x!0

H

1  cos ðxÞ 0 ¼ 0 sin ðxÞ

d dx ð1  cos ðxÞÞ d x!0 dx sin ðxÞ

) lim

¼ lim

x!0

sin ðxÞ 0 ¼ ¼0 cos ðxÞ 1

Choice (1) is the answer. 4.28. The problem can be solved as follows. lim

x!0

H

) lim

x!0

Choice (2) is the answer.

d dx ð5x d dx ð2x

5x  sin ðxÞ 0 ¼ 2x þ cos ðxÞ  1 0

 sin ðxÞÞ 5  cos ðxÞ 5  1 ¼ lim ¼2 ¼ þ cos ðxÞ  1Þ x!0 2  sin ðxÞ 2  0

4

Solutions of Problems: Limits

73

4.29. The problem can be solved as follows. lim

x!2

    ðx  2Þðx2 þ 2x þ 4Þ x3  8 þ 5x ¼ lim þ 5x ¼ lim x2  2x  4 þ 5x x!2 x!2 jx  2j  ðx  2 Þ   lim x2 þ 3x  4 ¼ 4 þ 6  4 ¼ 2

x!2

Choice (2) is the answer. 4.30. The problem can be solved as follows. sin ðxÞ þ cos ðxÞ 1 þ 0 1 ¼  ¼ 1 ¼ 0 0 cos ðxÞ

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

4.31. The problem can be solved as follows. lim arcð sin ðxÞÞ  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 ðarcð sin ðxÞÞÞ

Choice (2) is the answer. 4.32. The problem can be solved as follows. h lim

x!1

i 2 x ¼ lim ½0 x ¼ ð1Þð1Þ ¼ þ1 x!1 xþ1

Choice (1) is the answer. 4.33. The problem can be solved as follows. x4 x4 1 lim pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ limþ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ þ ¼ 1 x2  4x þ 3 x!3 ð x  3Þ ð x  1Þ 0

x!3þ

Choice (2) is the answer. 4.34. From calculus, we know that: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a x2 þ ax þ b  x þ 2 x!1 lim

The problem can be solved as follows.  lim

x!1

xþ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 4x  10  lim ðx þ jx þ 2jÞ ¼ lim ðx  x  2Þ ¼ lim ð2Þ ¼ 2

Choice (2) is the answer.

x!1

x!1

x!1

74

4

Solutions of Problems: Limits

4.35. The problem can be solved as follows. lim

x!2

H

) lim

x!2 d dx



4  x2 0 pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 6  2 x2 þ 5 0

pffiffiffiffiffiffiffiffiffiffiffiffiffi  x2 Þ 2x pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ lim x2 þ 5 ¼ 3 ¼ lim ffiffiffiffiffiffiffi ffi x!2 x!2 2  p2x 6  2 x2 þ 5 2 2 x þ5 d dx ð4

Choice (3) is the answer. 4.36. The problem can be solved as follows. sin ð2xÞ 0 lim pffiffiffiffiffiffiffiffiffiffiffi ¼ xþ11 0

x!0

d dx ð sin ð2xÞÞ

H

) lim

x!0 d dx

pffiffiffiffiffiffiffiffiffiffiffi  ¼ lim x!0 xþ11

2 cos ð2xÞ 2  1 ¼ 1 ¼4 p1ffiffiffiffiffiffi 2 2 xþ1

Choice (2) is the answer. 4.37. From calculus, we know that: lim

x!1

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x4 þ 2x2 þ x  x2

  lim am xm þ am1 xm1 þ . . . þ a2 x2 þ a1 x þ a0  am xm

x!1

or, lim ðam xm þ an xn Þ  am xm if m > n

x!1

The problem can be solved as follows. ffi  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 þ 2x2 þ x þ x2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 2 x4 þ 2x2 þ x  x ¼ lim x4 þ 2x2 þ x  x  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lim  x!1 x!1 x4 þ 2x2 þ x þ x2 x4 þ 2x2 þ x  x4 2x2 þ x 2x2  lim ¼ lim pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ lim 1 ¼ 1 ¼ lim  2 2 x!1 x!1 ðx þ x Þ x!1 2x2 x!1 x4 þ 2x2 þ x þ x2 Choice (1) is the answer. 4.38. From calculus, we know that the limit of a function at a specific point (x0) exits if lim f ðxÞ ¼ limþ f ðxÞ

x!x0 

x!x0

Therefore, we must have: lim

x!ð3Þ

2 x  9 xþ3

¼

lim

x!ð3Þþ

2 x  9 xþ3

ð1Þ

4

Solutions of Problems: Limits

75

lim

2 x  9 

x!ð3Þ

lim

xþ3

2 x  9

x!ð3Þþ

xþ3

¼

¼

lim



x!ð3Þ

x2  9 ¼ lim  ðx  3Þ ¼ 6 xþ3 x!ð3Þ

 ð x 2  9Þ ¼ lim þ  ðx  3Þ ¼ 6 xþ3 x!ð3Þþ x!ð3Þ lim

2 x  9 ð1Þ, ð2Þ, ð3Þ ¼ Undefined ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ 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 6¼ 6 ) lim x!3 x þ 3 Choice (4) is the answer. 4.39. The problem can be solved as follows.   tan πx 0 2 1 lim ¼ 1 0 cos ðπxÞ x!2 H

) lim x!12

d dx



   tan πx 2 1 ¼ lim d x!12 dx ð cos ðπxÞÞ

π 2



  π 1 þ tan 2 πx ð1 þ 1 Þ 2 ¼ 1 ¼2 π  1 π sin ðπxÞ

Choice (2) is the answer. 4.40. The problem can be solved as follows. ffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi xþ5þ xþ1 x þ 5  x þ 1 ¼ lim x þ 5  x þ 1  pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi x!þ1 x!þ1 xþ5þ xþ1 lim

x þ 5  ð x þ 1Þ 4 ¼ lim pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi ¼ lim pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi ¼ 0 x!þ1 x!þ1 xþ5þ xþ1 xþ5þ xþ1 Choice (3) is the answer. 4.41. From trigonometry, we know that: 1 þ cos ð2xÞ ¼ 2 cos 2 ðxÞ tan ðxÞ ¼

sin ðxÞ cos ðxÞ

sin ð2xÞ ¼ 2 sin ðxÞ cos ðxÞ The problem can be solved as follows. limπ x!2

tan ð2xÞ cos ðxÞ sin ð2xÞ cos ðxÞ 2 sin ðxÞ cos 2 ðxÞ sin ðxÞ 1 ¼ 1 ¼ limπ ¼ limπ ¼ limπ ¼ 2 1 x!2 cos ð2xÞ  2 cos ðxÞ x!2 cos ð2xÞ  2 cos 2 ðxÞ x!2 cos ð2xÞ 1 þ cos ð2xÞ

Choice (3) is the answer.

ð2Þ

ð3Þ

76

4

Solutions of Problems: Limits

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

x!0

lim tan ðxÞ  x

x!0

The problem can be solved as follows. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi tan ð2xÞ 1 þ cos ðxÞ 1 þ cos ðxÞ tan ð2xÞ tan ð2xÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lim pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ lim pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ lim x!0 1 þ cos ðxÞ x!0 1  cos ðxÞ x!0 1  cos ðxÞ 1  cos 2 ðxÞ pffiffiffi pffiffiffi pffiffiffi pffiffiffi tan ð2xÞ  2 tan ð2xÞ  2 tan ð2xÞ  2 2x  2 ¼ lim qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ lim ¼ lim ¼ lim x x!0 x!0 x!0 x!0  sin ðxÞ j sin ðxÞj sin 2 ðxÞ  pffiffiffi pffiffiffi ¼ lim 2 2 ¼ 2 2 x!0

Choice (1) is the answer. 4.43. The problem can be solved as follows. lim x!1

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3 n þ 1000  n  20

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q p 3 3 2 3 ð n þ 1000 Þ þ ð n þ 1000 Þ ð n  20 Þ þ ðn  20Þ2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3 n þ 1000  n  20  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ lim x!1 3 3 ðn þ 1000Þ2 þ 3 ðn þ 1000Þðn  20Þ þ ðn  20Þ2 n þ 1000  ðn  20Þ 1020 ¼0 ¼ lim qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ x!1 3 þ1 3 2 2 3 ðn þ 1000Þ þ ðn þ 1000Þðn  20Þ þ ðn  20Þ Choice (2) is the answer. 4.44. From the application of the Taylor series in limit, we know that: lim sin ðxÞ  x

x!0

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

  x 2

    x x cos 2 2

4

Solutions of Problems: Limits

77

The problem can be solved as follows. sin ðπ sin ðxÞÞ sin pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lim x!π þ 1 þ cos ðxÞ

 x 2

    π sin ðxÞ sin 2x π  2 sin 2x cos 2x sin 2x pffiffiffi    limþ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi limþ  ffi ¼ x!π x!π 2 cos 2x 2 cos 2 2x

   pffiffiffi   pffiffiffi pffiffiffi π  2 sin 2 2x cos 2x x pffiffiffi x ¼ limþ π 2 sin 2 ¼ π 2  1 ¼ π 2 ¼ limþ 2 x!π x!π 2  cos 2 Choice (1) is the answer. 4.45. From the application of the Taylor series in limit, we know that: p ffiffiffiffiffiffiffiffiffiffiffi α n 1 þ α  lim 1 þ n α!0 α!0 lim

  lim am xm þ am1 xm1 þ . . . þ amn xmn þ amn1 xmn1  amn1 xmn1

x!0

or, lim ðam xm þ an xn Þ  an xn if m > n

x!0

The problem can be solved as follows.   p ffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffiffiffi 2 x2 x x 3 1 þ x3  1  2x 1 þ x2  4 1  2x 1 3 þ2 4 2 ¼ lim  lim ¼ lim  lim 2 2 2 4 2x x!0 x!0 x!0 2x þ 2x x!0 2x þ 2x 2x þ 2x Choice (1) is the answer. 4.46. From trigonometry, we know that: sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1 The problem can be solved as follows. lim

x!0

H

sin 2 ðxÞ þ sin ðxÞ þ cos 2 ðxÞ  cos ðxÞ 1 þ sin ðxÞ  cos ðxÞ 0 ¼ lim ¼ 2 2 x!0 1  sin ðxÞ  cos ðxÞ 0 sin ðxÞ  sin ðxÞ þ cos ðxÞ  cos ðxÞ d dx ð1 x!0 d ð1 dx

) lim

þ sin ðxÞ  cos ðxÞÞ cos ðxÞ þ sin ðxÞ 1þ0 ¼ lim ¼ 1 ¼ 1 þ0  sin ðxÞ  cos ðxÞÞ x!0  cos ðxÞ þ sin ðxÞ

Choice (2) is the answer. 4.47. From the application of the Taylor series in limit, we know that: lim sin ðuðxÞÞ  uðxÞ

uðxÞ!0

The problem can be solved as follows. lim

x!0

cos ðmxÞ  cos ðnxÞ 0 ¼ 0 x2

78

4 d dx ð cos ðmxÞ  cos ðnxÞÞ d 2 x!0 dx ðx Þ

H

) lim

¼ lim

x!0

¼ lim

x!0

Solutions of Problems: Limits

m sin ðmxÞ þ n sin ðnxÞ mðmxÞ þ nðnxÞ  lim 2x 2x x!0

m2 þ n2 n2  m2 ¼ 2 2

Choice (3) is the answer. 4.48. From the application of the Taylor series in limit, we know that: lim sin ðxÞ  x

x!0

lim tan ðxÞ  x

x!0

The problem can be solved as follows. lim

x!0

H

d dx ð sin ðxÞ  xÞ d x!0 dx ð tan ðxÞ  xÞ

) lim

H

d dx ð cos ðxÞ  1Þ d 2 x!0 dx ð tan ðxÞÞ

¼ lim

) lim

x!0

sin ðxÞ  x 0 ¼ tan ðxÞ  x 0

¼ lim

x!0

cos ðxÞ  1 cos ðxÞ  1 0 ¼ lim ¼ 0 1 þ tan 2 ðxÞ  1 x!0 tan 2 ðxÞ

 sin ðxÞ x 1  lim ¼ lim 2 tan ðxÞð1 þ tan 2 ðxÞÞ x!0 2xð1 þ x2 Þ x!0 2ð1 þ x2 Þ ¼

1 1 ¼ 2 2ð 1 þ 0Þ

Choice (1) is the answer. 4.49. From trigonometry, we know that:   1 þ cos 3 ðxÞ ¼ ð1 þ cos ðxÞÞ 1  cos ðxÞ þ cos 2 ðxÞ 1  cos 2 ðxÞ ¼ ð1 þ cos ðxÞÞð1  cos ðxÞÞ The problem can be solved as follows. lim x!π ⟹ lim

x!π

1 þ cos 3 ðxÞ 0 ¼ 1  cos 2 ðxÞ 0

1 þ cos 3 ðxÞ ð1 þ cos ðxÞÞð1  cos ðxÞ þ cos 2 ðxÞÞ 1  cos ðxÞ þ cos 2 ðxÞ ¼ lim ¼ lim 2 x!π x!π ð1 þ cos ðxÞÞð1  cos ðxÞÞ 1  cos ðxÞ 1  cos ðxÞ ¼

Choice (1) is the answer.

1  ð1Þ þ ð1Þ2 3 ¼ 2 1  ð1Þ

4

Solutions of Problems: Limits

79

4.50. As we know from calculus: nk ¼0 n!þ1 an

If a > 1, k 2 N ⟹ lim Hence,

3n2 n2 lim pffiffiffiffinffi ¼ lim 3 pffiffiffin ¼ 3  0 ¼ 0 n!þ1 5 n!þ1 5 Choice (1) is the answer. 4.51. From trigonometry, we know that: 1  cos ðxÞ ¼ 2 sin 2

  x 2

Moreover, from the application of the Taylor series in limit, we know that: lim sin n ðxÞ  xn x!0

Thus,     2 x3  sin ðxÞ 2 sin 2 2x x3  x  2 2x x3  sin ðxÞð1  cos ðxÞÞ lim ¼ lim  lim x!0 x!0 x!0 x3 x3 x3 3

¼ lim

x!0

3

x x3  x2 1 1 ¼ lim 23 ¼ lim ¼ 3 2 x!0 x x!0 2 x

Choice (2) is the answer. 4.52. From trigonometry and calculus, we know that: arcð cos ð1 ÞÞ ¼ 0þ d 1 ðarcð cos xÞÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi dx 1  x2   d pffiffiffiffiffiffiffiffiffiffiffi 1 1  x ¼ pffiffiffiffiffiffiffiffiffiffiffi dx 2 1x The problem can be solved as follows. lim

x!1

H

) lim x!1

Choice (1) is the answer.

d dx ðarcð cos xÞÞ pffiffiffiffiffiffiffiffiffiffiffi d 1x dx

¼ lim x!1

arcð cos xÞ 0þ pffiffiffiffiffiffiffiffiffiffiffi ¼ þ 0 1x

1 ffi pffiffiffiffiffiffiffi 1x2 ffiffiffiffiffiffi p1 2 1x

1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

¼ lim x!1

ð1xÞð1þxÞ p1ffiffiffiffiffiffi 2 1x

pffiffiffi 2 ¼ lim pffiffiffiffiffiffiffiffiffiffiffi ¼ 2 x!1 1þx

80

4

Solutions of Problems: Limits

4.53. Based on the information given in the problem, we have:   1 lim x2  1 cot ðxn  1Þ ¼ 2

ð1Þ

x!1

From calculus, we know that: cot ðxÞ ¼

1 tan ðxÞ

From the application of the Taylor series in limit, we know that: lim tan ðuðxÞÞ  uðxÞ

uðxÞ!0

The problem can be solved as follows.   lim x2  1 cot ðxn  1Þ ¼ lim

x!1

x!1

H

) lim

x!1

x2  1 x2  1 0 ¼  lim tan ðxn  1Þ x!1 xn  1 0

2x 2 ¼ nxn1 n

ð2Þ

Solving (1) and (2): 2 1 ¼ )n¼4 n 2 Choice (2) is the answer. 4.54. From trigonometry, we know that: sin ð4xÞ ¼ 2 sin ð2xÞ cos ð2xÞ cot ðxÞ ¼

cos ðxÞ sin ðxÞ

sin ðx  yÞ ¼ sin ðxÞ cos ðyÞ  cos ðxÞ sin ðyÞ The problem can be solved as follows.  lim sin ð4xÞð cot ð2xÞ  cot ðxÞÞ ¼ lim 2 sin ð2xÞ cos ð2xÞ

x!0

x!0



 cos ð2xÞ cos ðxÞ  sin ð2xÞ sin ðxÞ

sin ðxÞ cos ð2xÞ  cos ðxÞ sin ð2xÞ ¼ lim 2 sin ð2xÞ cos ð2xÞ x!0 sin ð2xÞ sin ðxÞ



  sin ðx  2xÞ 2 sin ð 2x Þ cos ð 2x Þ ¼ lim ¼ lim ð2 cos ð2xÞÞ ¼ 2 x!0 x!0 sin ð2xÞ sin ðxÞ Choice (3) is the answer.

4

Solutions of Problems: Limits

81

4.55. From the application of the Taylor series in limit, we know that: lim sin ðxÞ  x 

x3 6

lim cos ðxÞ  1 

x2 2

x!0

x!0

The problem can be solved as follows. 3

lim

x!0 1 2

3

 x6  x6 sin ðxÞ  x   lim  ¼ lim 3 3 ¼ lim 1 ¼ 1    x!0  x x!0 sin ð2xÞ  x cos ðxÞ x!0 1 2x  ð2xÞ  x 1  x2 6 2

6

2

Choice (3) is the answer. 4.56. The problem can be solved as follows. pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  3 tan ðxÞ 0 ¼ limπ 0 x!4 1  2 sin 2 ðxÞ 

H

) limπ x!4

d dx 1  d dx 1 

2 ðxÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ptan ffiffiffiffiffiffiffiffiffiffiffi  1þ 3 1þ1 tan ðxÞ 3 3 tan 2 ðxÞ 1 31 p ffiffi pffiffi ¼  ¼ lim ¼ x!π4 4 sin ðxÞ cos ðxÞ 2 sin 2 ðxÞ 4 2 2 3

2

2

Choice (1) is the answer. 4.57. From calculus and trigonometry, we know that: sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1   1  cos 3 ðxÞ ¼ ð1  cos ðxÞÞ 1 þ cos ðxÞ þ cos 2 ðxÞ From the application of the Taylor series in limit, we know that: lim sin ðxÞ  x

x!0

lim tan ðuðxÞÞ  uðxÞ

uðxÞ!0

The problem can be solved as follows. lim

x!0

) lim

x!0

1  cos 3 ðxÞ 0 ¼ sin ðxÞ tan ð2xÞ 0

1  cos 3 ðxÞ ð1  cos ðxÞÞð1 þ cos ðxÞ þ cos 2 ðxÞÞ ð1  cos ðxÞÞ ¼ lim  sin ðxÞ tan ð2xÞ x!0 sin ðxÞ tan ð2xÞ ð1 þ cos ðxÞÞ

¼ lim

x!0

ð1  cos 2 ðxÞÞð1 þ cos ðxÞ þ cos 2 ðxÞÞ sin 2 ðxÞ  ð1 þ 1 þ 1Þ ¼ lim x!0 sin ðxÞ tan ð2xÞ  ð1 þ 1Þ sin ðxÞ tan ð2xÞð1 þ cos ðxÞÞ

82

4

lim

x!0

Solutions of Problems: Limits

3 sin 2 ðxÞ 3x2 3 3 ¼ lim ¼  lim 2 sin ðxÞ tan ð2xÞ x!0 2x  2x x!0 4 4

Choice (3) is the answer. 4.58. From trigonometry, we know that: 1  cos ðxÞ ¼ 2 sin 2

  x 2

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

uðxÞ!0þ

uðxÞ!0

The problem can be solved as follows. pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 1  cos ðxÞ 1  cos ðxÞ 1 þ cos ðxÞ 1 þ cos ð xÞ pffiffiffi ¼ limþ pffiffiffi  pffiffiffiffiffiffiffiffiffiffiffiffiffiffi  pffiffiffi limþ x!0 1  cos ð xÞ x!0 1  cos ð xÞ 1 þ cos ðxÞ 1 þ cos ð xÞ ¼ limþ x!0

pffiffiffi ð1  cos ðxÞÞð1 þ cos ð xÞÞ ð1  cos ðxÞÞð1 þ 1Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ limþ pffiffiffi  p ffiffi ffi 2 x!0 ð1  cos 2 ð xÞÞð1 þ 1Þ ð1  cos ð xÞÞ 1 þ cos ðxÞ

  2 2 sin 2 2x 2 2x 1  cos ðxÞ x pffiffiffi ¼ limþ ¼ limþ  limþ pffiffiffi 2 ¼ limþ ¼ 0 2 pffiffiffi x!0 1  cos 2 ð xÞ x!0 sin ð xÞ x!0 ð xÞ x!0 2 Choice (1) is the answer.

Reference 1. Rahmani-Andebili, M. (2020). Precalculus: Practice problems, methods, and solutions, Springer Nature, 2020.

5

Problems: Derivatives and Its Applications

Abstract

In this chapter, the basic and advanced problems of derivatives and its applications are presented. The subjects include definition of derivative, differentiation formulas, product rule, quotient rule, chain rule, derivatives of trigonometric functions, derivatives of exponential, 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 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. Calculate the value of f 0(x ¼ 1) if f(x) ¼ xex  ex [1]. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 0 3) e 4) e 5.2. If f(x) + g(x3) ¼ 5x  1 and f 0(1) ¼ 2, calculate the value of g0(1). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 1 3) 2 4) 2 5.3. 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

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Calculus, https://doi.org/10.1007/978-3-030-64980-7_5

83

84

5

Problems: Derivatives and Its Applications

5.4. Determine the derivative of the function below at x ¼ 14. f ðxÞ ¼

pffiffiffi x x pffiffiffi 1 x

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2)  12 3) 12 4) 1 5.5. 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 π 5.6. Calculate the derivative of the function of f(x) ¼ tan3(2x) at 12 . Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 43 2) 49 3) 83 4) 89

5.7. If the function of f(x) ¼ |x3  3x + a| 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 5.8. Calculate the value of f 0(2) + f 0(4) if f(x) ¼ |x2  6|. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 8 2) 8 3) 4 4) 4 5.9. If f 0 ðxÞ ¼ 5x, calculate the first derivative of f(x5). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1)  5x 2)  25 x 3) 25 x 4) x55

5

Problems: Derivatives and Its Applications

5.10. If the first derivative of f(sin(x)) is equal to cos3(x), determine the value of f 0(x). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 + x2 2) 1  x2 3) x3 4) x3 5.11. Calculate the derivative of the function of f(x) ¼ arc(tan(3x)) at 13. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 32 2) 43 3) 23 4) 34  pffi  5.12. If f 1t þ g t ¼ t 2 þ 1 and g0(1) ¼ 5, calculate the value of f 0(1). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 2 3) 12 4)  12 5.13. 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) 12 2)  12 3) 1 4) 1 5.14. 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 1)  16y x5 2) 16x 5 y 3)  32y x5 4)  32x y5 5.15. If x ¼ 2 + 3 sin (t) and y ¼ 3  2 cos (t), calculate the value of y0x for t ¼ π6. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large pffiffi 1) 2 9 3 pffiffi 2) 2 3 3 pffiffi 3) 2 3 2 pffiffi 4) 4 3 2

85

86

5

Problems: Derivatives and Its Applications

5.16. 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 1) 11 4 2) 13 4 3) 15 4 4) 17 4 5.17. Calculate the value of f 0(4) if we know that: lim

h!0

pffiffiffi f ð x þ h Þ  f ð x  hÞ ¼2 x h

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 23 2) 43 3) 4 4) 2 5.18. Which one of the choices is true about the function of f(x) ¼ x2|x| 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. 5.19. The function below is differentiable at x ¼ π4. Determine the value of b. f ð xÞ ¼

8 < sin 2 ðxÞ  cos ð2xÞ : a tan ðxÞ þ b sin ð2xÞ

π 0 > > y0x ¼ t0 ¼ < xt 2t þ 1

8 2  2 0 > ¼4 < yx ¼ 2 þ 1 t ¼ 1 1 17 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) ⟹ y0x þ x0x ¼ 4 þ ⟹ y0x þ x0x ¼ ⟹ 0 4 4 > > 2 þ 1 1 x 2t þ 1 : > > x0y ¼ ¼ : x0y ¼ t0 ¼ 2  2 4 yt 2t  2 Choice (4) is the answer. 6.17. Based on the information given in the problem, we have: lim

h!0

pffiffiffi f ð x þ hÞ  f ð x  hÞ ¼2 x h

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

ð1Þ

6

Solutions of Problems: Derivatives and its Applications

97

The problem can be solved as follows. lim

h!0

f ð x þ hÞ  f ð x  hÞ f ðx þ hÞ  f ðxÞ þ f ðxÞ  f ðx  hÞ ¼ lim h h h!0 ¼ lim

h!0

f ð x þ hÞ  f ð x Þ f ð x Þ  f ð x  hÞ þ lim ¼ 2 f 0 ð xÞ h h h!0

ð2Þ

Solving (1) and (2): pffiffiffi pffiffiffi pffiffiffi 2 f 0 ð x Þ ¼ 2 x ⟹ f 0 ð x Þ ¼ x ⟹ f 0 ð 4Þ ¼ 4 ¼ 2 Choice (4) is the answer. 6.18. The function can be simplified as follows:  f ð xÞ ¼ x2 j xj ¼  ⟹ f ð 0Þ ¼

0

x0

0

x > > x5 , x < 0 : : 5x4 , x < 0 20x3 , x < 0 5

8 ,x > 0 60x2 > > < , x ¼ 0 ⟹ f 000 ðxÞ ¼ 60xjxj ⟹ f 000 ðxÞ ¼ 0 > > : 60x2 , x < 0 Choice (4) is the answer.

6

Solutions of Problems: Derivatives and its Applications

101

6.27. Based on the information given in the problem, we have: pffiffiffiffiffiffiffiffiffiffiffi 1 x þ a ⟹ f 0 ð 2Þ ¼ 4

ð1Þ

pffiffiffiffiffiffiffiffiffi u0 ð x Þ uðxÞ ⟹ f 0 ðxÞ ¼ pffiffiffiffiffiffiffiffiffi 2 uð x Þ

ð2Þ

f ðxÞ ¼ From the list of derivative rules, we know that: f ð xÞ ¼ Solving (1) and (2):

 pffiffiffiffiffiffiffiffiffiffiffi 1  1 1 1 pffiffiffiffiffiffiffiffiffiffiffi ¼ ⟹ pffiffiffiffiffiffiffiffiffiffiffi ¼ ⟹ 2 þ a ¼ 2 ⟹ a ¼ 2 4 4 2 x þ a x¼2 2 2þa Choice (4) is the answer. 6.28. First, we should simplify the function as follows: yðxÞ ¼ ln e

pffiffiffiffiffiffiffiffiffiffi sin ðxÞ

¼

pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi sin ðxÞ ln e ¼ sin ðxÞ

ð1Þ

From the list of derivative rules, we know that: f ð xÞ ¼

pffiffiffiffiffiffiffiffiffi u0 ð x Þ uðxÞ ⟹ f 0 ðxÞ ¼ pffiffiffiffiffiffiffiffiffi 2 uð x Þ

ð2Þ

Solving (1) and (2): pffiffi
π  pffiffiffi 3   cos cos ð x Þ 6 π 0 0 6 2 q ffiffi q p ffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ¼ ¼ ⟹y ⟹ y ð xÞ ¼
π  ¼ 4 6 1 2 sin ðxÞ 2 2 sin 6

2

Choice (4) is the answer. 6.29. From the list of derivative rules, we know that: f ðxÞ ¼ axn ⟹ f 0 ðxÞ ¼ anxn1 To determine the maximum value of a function for a given range, we need to calculate the value of the function at its critical points including the extremum points and the beginning and the end of the range. yðxÞ ¼ x3  3x2  9x þ 5 ⟹ y0 ðxÞ ¼ 3x2  6x  9 To find the extremum points of the function: ⟹ y0 ðxÞ ¼ 0 ⟹ 3x2  6x  9 ¼ 0 ⟹ x2  2x  3 ¼ 0 ⟹ x ¼ 3,  1 x ¼ 3 is not acceptable because it is out of the range. The value of the function at the critical points can be calculated as follows: yð2Þ ¼ ð2Þ3  3ð2Þ2  9ð2Þ þ 5 ¼ 3

102

6 Solutions of Problems: Derivatives and its Applications

yð1Þ ¼ ð1Þ3  3ð1Þ2  9ð1Þ þ 5 ¼ 10 yð2Þ ¼ ð2Þ3  3ð2Þ2  9ð2Þ þ 5 ¼ 17 Therefore, the maximum value of the function is 10. Choice (2) is the answer. 6.30. From the list of derivative rules, we know that: f ð xÞ ¼

gð x Þ g0 ðxÞhðxÞ  h0 ðxÞgðxÞ ⟹ f 0 ð xÞ ¼ hð x Þ h2 ð x Þ

ð1Þ

First, we need to take its derivative as follows: yð xÞ ¼

1  sin ðxÞ  cos ðxÞ cos ðxÞ  ð sin ðxÞÞð1  sin ðxÞÞ ⟹ y0 ð xÞ ¼ cos ðxÞ cos 2 ðxÞ y0 ðxÞ ¼

 cos 2 ðxÞ þ sin ðxÞ  sin 2 ðxÞ 1 þ sin ðxÞ ¼ cos 2 ðxÞ cos 2 ðxÞ

The y0(x) is always nonpositive because sin(x)  1. Hence, the function is a descending function. Choice (2) is the answer. 6.31. Based on the information given in the problem, we have: f ð xÞ ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffi 5 1 1 þ x2  x , gðxÞ ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffi 5 1 þ x2 þ x

ð1Þ

From the list of derivative rules, we know that:

f ð xÞ gð x Þ

0

¼

f 0 ðxÞgðxÞ  g0 ðxÞf ðxÞ ⟹ f 0 ðxÞgðxÞ  g0 ðxÞf ðxÞ ¼ ðgðxÞÞ2



0 f ð xÞ ðgðxÞÞ2 gð x Þ

Solving (1) and (2): 0
pffiffiffiffiffiffiffiffiffiffiffiffiffi !2  5 10 2x 1 þ x 1 0 0 A
pffiffiffiffiffiffiffiffiffiffiffiffiffi f ðxÞgðxÞ  g ðxÞf ðxÞ ¼ @ 5 1 pffiffiffiffiffiffiffiffi 5 1 þ x2 þ x 2 ð 1þx þxÞ ¼




1 þ x2  x2

5 0

1 1 1 0
pffiffiffiffiffiffiffiffiffiffiffiffiffi 10 ¼ ð1Þ 
pffiffiffiffiffiffiffiffiffiffiffiffiffi 10 ¼ 0 
pffiffiffiffiffiffiffiffiffiffiffiffiffi 10 ¼ 0 2 2 1þx þx 1þx þx 1 þ x2 þ x

Choice (2) is the answer. 6.32. Since the tangent lines are parallel to x-axis, their slope angles must be zero. Therefore, y ¼ x3  6x þ 12 ⟹ y0 ¼ 3x2  6 ¼ 0 ⟹ x ¼ x1 ¼

pffiffiffi pffiffiffi 2,  2

pffiffiffi3 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 2 ⟹ y1 ¼ 2  6 2 þ 12 ¼ 2 2  6 2 þ 12 ¼ 4 2 þ 12

ð2Þ

6

Solutions of Problems: Derivatives and its Applications

103

 pffiffiffi3  pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi x2 ¼  2 ⟹ y2 ¼  2  6  2 þ 12 ¼ 2 2 þ 6 2 þ 12 ¼ 4 2 þ 12  pffiffiffi   pffiffiffi  pffiffiffi y2  y1 ¼ 4 2 þ 12  4 2 þ 12 ⟹ y2  y1 ¼ 8 2 Choice (4) is the answer. 6.33. Based on the information given in the problem, we have: 8 5 < ð x þ 1Þ f ð xÞ ¼ j x þ 1j : 0

x 6¼ 1 x ¼ 1

This problem can be solved by using the definition of derivative of a function as follows. lim f 0 ðx0 Þ ¼ x!x

0

0

⟹ f ð1Þ ¼ lim

ðxþ1Þ5 jxþ1j

x!ð1Þ x

0

 ð1Þ

¼ lim

f ð xÞ  f ð x0 Þ x  x0

ðxþ1Þ5 jxþ1j

x!ð1Þ x

þ1

ðx þ 1Þ4 ¼ lim jx þ 1j3 ¼ 0 x!ð1Þ jx þ 1j x!ð1Þ

¼ lim

Choice (1) is the answer. 6.34. From the list of derivative rules, we know that: f ð xÞ ¼

pffiffiffiffiffiffiffiffiffi u0 ð x Þ uðxÞ ⟹ f 0 ðxÞ ¼ pffiffiffiffiffiffiffiffiffi 2 uð x Þ

ð1Þ

The distance of the point from the origin can be calculated as follows: DðxÞ ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffi 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x þ 8  0 ¼ x2 þ x þ 8 ð x  0Þ 2 þ

In addition, the changing rate of the distance can be determined as follows: D0 ðxÞ ¼

ffi d d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ x þ 8 DðxÞ ¼ dx dx

Solving (1) and (2): 2x þ 1 27þ1 15 D0 ðxÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ⟹ D0 ð7Þ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2 x2 þ x þ 8 2 72 þ 7 þ 8 16 Choice (1) is the answer. 6.35. From the list of derivative rules, we know that: f ðxÞ ¼ gðhðxÞÞ ⟹ f 0 ðxÞ ¼ h0 ðxÞg0 ðhðxÞÞ f ð xÞ ¼

pffiffiffiffiffiffiffiffiffi u0 ð x Þ uðxÞ ⟹ f 0 ðxÞ ¼ pffiffiffiffiffiffiffiffiffi 2 uð x Þ

ð2Þ

104

6 Solutions of Problems: Derivatives and its Applications

The problem should be solved by using the definition of derivative of a function as follows. lim f 0 ðx0 Þ ¼ x!x

0

f ð xÞ  f ð x0 Þ x  x0

ð1Þ

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

x!2

f ðxÞ  f ð2Þ 1 ¼ x2 3

ð2Þ

Solving (1) and (3): f 0 ð 2Þ ¼ 

1 3

ð3Þ

Therefore, d dx

 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  f jxj þ 3 

x¼1

¼


pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 f x þ 3 x¼1 ⟹ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f 0 x þ 3  ¼  f 0 ð 2Þ dx 4 2 x þ 3 x¼1

ð4Þ

Solving (3) and (4): d dx

 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  f jxj þ 3 

x¼1

  1 1 1 ¼ ¼   4 3 12

Choice (2) is the answer. 6.36. Based on the information given in the problem, we have: f ð xÞ ¼

ðx þ 1ÞhðxÞ , hð1Þ 6¼ 0 ð2x þ 1Þhð2x þ 1Þ

ð1Þ

The derivative of this function should be solved by using the definition of derivative of a function as follows: f 0 ðx0 Þ ¼ lim

x!x0

f ðxÞ  f ðx0 Þ f ðxÞ  f ð1Þ ⟹ f 0 ð1Þ ¼ lim x  x0 x!1 x  ð1Þ

ð2Þ

Solving (1) and (2): 0

f ð1Þ ¼ lim

ðxþ1ÞhðxÞ ð2xþ1Þhð2xþ1Þ

ð1þ1Þhð1Þ  ð2þ1 Þhð2þ1Þ

x  ð1Þ

x!1

f 0 ð1Þ ¼ lim

x!1 ð2x

¼ lim

x!1

ðxþ1ÞhðxÞ ð2xþ1Þhð2xþ1Þ

0

xþ1

hð x Þ hð1Þ ¼ ¼ 1 þ 1Þhð2x þ 1Þ ð1Þhð1Þ

Choice (2) is the answer. 6.37. Based on the information given in the problem, the width of the extremum point is as follows: yðxM Þ ¼

3 4

ð1Þ

To determine the extremum points of a function, we need to find the roots of the derivative of the function as follows.

Reference

105

f 0 ð xÞ ¼ 0 f ðxÞ ¼ cos 2 ðxÞ þ ⟹ f 0 ðxÞ ¼ 2 sin ðxÞ cos ðxÞ þ

pffiffiffi 3 sin ðxÞ þ a

 pffiffiffi pffiffiffi 3 cos ðxÞ ¼ cos ðxÞ 2 sin ðxÞ þ 3

ð2Þ ð3Þ ð4Þ

Solving (2) and (4): 8 < cos ðxÞ ¼ 0  pffiffiffi pffiffiffi cos ðxÞ 2 sin ðxÞ þ 3 ¼ 0 ⟹ : sin ðxÞ ¼ 3 2 There is no answer for equation (5) in the range of 0 < x < π2. However, x ¼ π3 is only answer for equation (6). Therefore, by using xM ¼ π3 and (1) in (3), we have:   pffiffiffi   3 π π 3 1 3 ¼ cos 2 þ 3 sin þ a ⟹ ¼ þ þ a ⟹ a ¼ 1 4 3 3 4 4 2 Choice (4) is the answer.

Reference 1. Rahmani-Andebili, M. (2020). Precalculus: Practice problems, methods, and solutions, Springer Nature, 2020.

ð 5Þ ð 6Þ

7

Problems: Definite and Indefinite Integrals

Abstract

In this chapter, the basic and advanced problems of definite and indefinite integrals are presented. The subjects include definite integrals, indefinite integrals, substitution rule for integrals, integration techniques, integration by parts, integrals involving trigonometric functions, trigonometric substitutions, integration using partial fractions, integrals involving roots, integrals involving quadratics, applications of integrals, average value, area between curves, and volume of solid of revolution. To help students study the chapter in the most efficient way, the problems are categorized 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. Calculate the value of the definite integral below [1]. ð2 1

xþ4 dx x3

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 2 3) 3 4) 4 7.2. Solve the following indefinite integral. ð

 2 ex þ 2xex dx

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

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Calculus, https://doi.org/10.1007/978-3-030-64980-7_7

107

108

7 Problems: Definite and Indefinite Integrals

7.3. Calculate the value of f 00(1) if we know that: ð f ð xÞ ¼



 x3 þ 5x dx

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2 2) 4 3) 6 4) 8 7.4. Calculate the value of f 00

π  2

if f(x) ¼

Ð

cos3(x)dx.

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 1 3) 1 pffiffi 4) 3 2 2 7.5. Calculate the value of the definite integral below. ð 1

x3 þ x2  1 dx x2 2

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 1 3) 12 4)  12 7.6. Solve the indefinite integral below. ð

x2 pffiffiffi dx x

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large pffiffiffi 1) 23 xðx þ 6Þ þ c pffiffiffi 2) 23 xðx  6Þ þ c pffiffiffi 3) 13 xðx þ 6Þ þ c pffiffiffi 4) 23 xðx  6Þ þ c

7

Problems: Definite and Indefinite Integrals

109

7.7. Calculate the value of the following definite integral. ð1

x2 dx 4 0 ðx3 þ 1Þ

Difficulty level ● Easy ○ Normal Calculation amount ○ Small ● Normal 5 1) 36 7 2) 36 5 3) 72 7 4) 72

○ Hard ○ Large

7.8. Calculate the integral of the function below for the range of 1 < x < + 1. f ð xÞ ¼

1 x2 þ 4

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π2 2) π 3) 3π 2 4) 2π 7.9. Calculate the value of the definite integral below. ð4

pffiffiffi x x pffiffiffi dx x 1

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 13 2) 23 3) 43 4) 53 3

7.10. If the primary function of f(x) is equal to x6 , determine the first derivate of f Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1)  x13 2)  6x 3)  12 4) x13

1 x

with respect to x.

110

7 Problems: Definite and Indefinite Integrals

7.11. Calculate the value of F0(λ ¼ 0) if: F ðλÞ ¼

ðλ

1 dx 4þ2 x 0

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 12 3) 13 4) 0 7.12. Calculate the value of the definite integral below. ð1

x2 arcð tan ðxÞÞdx 2 1 1 þ x

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 1 3) π4 4) π2 7.13. Calculate the integral of the function below for the range of  12 < x < 12. 1 f ðxÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi 1  x2 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) π6 2) π3 3)  π6 4)  π3 7.14. Calculate the value of the definite integral below. ð1

1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx 2x  x2 0

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) π4 3) π2 4) π

7

Problems: Definite and Indefinite Integrals

111

7.15. Calculate the value of the definite integral below. ð1



1

  x2 þ 1 x3 þ 3x dx

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 21 2) 0 3) 11 4) 2 7.16. Solve the following indefinite integral. ð

sin ðxÞ dx 1 þ cos ð cos ðxÞÞ

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large   1)  tan 12 cos ðxÞ þ c 2) tan(cos(x)) + c 3)  tan (cos(x)) + c   4) tan 12 cos ðxÞ þ c 7.17. Calculate the surface area enclosed between the curves of y ¼ 2x2  2x and y ¼ x2. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 13 2) 23 3) 43 4) 73 7.18. Determine the function of a curve that passes from the point of (3, 4) and its derivative is  xy. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2x2 + y2 ¼ 34 2) x2 + y2 ¼ 16 3) y2 ¼ 4x + 4 4) x2 + y2 ¼ 25 7.19. Solve the following indefinite integral. ð

x pffiffiffiffiffiffiffiffiffiffiffi dx x1

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

112

7 Problems: Definite and Indefinite Integrals 3

1

3

1

3

1

1) 23 ðx  1Þ2  2ðx  1Þ2 þ c 2) 23 ðx  1Þ2 þ 2ðx  1Þ2 þ c 3) 13 ðx  1Þ2  2ðx  1Þ2 þ c 3

1

4)  13 ðx  1Þ2  2ðx  1Þ2 þ c 7.20. Calculate the value of the definite integral below. ð5 2

jx  3jdx

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 25 2 2) 27 2 3) 29 2 4) 31 2 7.21. Calculate the value of the definite integral below. ð2

x ð x þ 1Þ 2 þ 2 dx ðx þ 1Þ2 1

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 12 5 2) 95 3) 11 6 4) 74 7.22. Calculate the surface area enclosed between the curves of y ¼ x2 and y ¼ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 23 2) 1 3) 13 4) 16 7.23. Solve the following indefinite integral. ð

1 dx 1 þ ex

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) x + ln (1 + ex) + c 2) x  ln (1 + ex) + c 3) 12 x2 þ ln ð1 þ ex Þ þ c 4) 12 x2  ln ð1 þ ex Þ þ c

pffiffiffi x.

7

Problems: Definite and Indefinite Integrals

113

7.24. Calculate the surface area enclosed between the curve of y ¼ x3 + 2x2 + x and x-axis. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 12 1 2) 10 1 3) 9 4) 17 7.25. Calculate the mean value of the function of y ¼ ax + b in the range of [2, 5]. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 52 a þ 3b 2) 72 a þ 3b 3) 52 a þ b 4) 72 a þ b 7.26. Determine the function of a curve that passes from the point of (1, 1) and its derivative is as follows. y0 ¼

xþ1 1y

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) x2 + y2 + 2x  2y  2 ¼ 0 2) x2  y2 + 4x  4y + 1 ¼ 0 3) x2 + y2  2x + 2y  2 ¼ 0 4) x2  y2 + 3x  2y  1 ¼ 0 7.27. Consider the functions of f(x) ¼ 2x and g(x) ¼ 3x2  2x. Calculate the value of λ if the mean value of the functions in the range of [1, λ] is the same. Difficulty level ○ Easy Calculation amount ○ Small pffiffi 1) 1þ2 5 pffiffi 2) 1þ2 3 pffiffi 3) 25 pffiffi 4) 3 2 3

● Normal ○ Hard ● Normal ○ Large

7.28. What is the function of a curve that passes from the point of (1, 1) and the relation below holds. y0 ¼ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2y2  3x2 + 1 ¼ 0 2) y2  2x2 + 1 ¼ 0 3) 2y2 + x2  3 ¼ 0 4) 2y2  x2  1 ¼ 0

3x 2y

114

7 Problems: Definite and Indefinite Integrals

7.29. In the equation below, determine the value of A. ð

pffiffiffiffiffiffiffiffiffiffiffiffiffi 3x pffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ A x2 þ 1 þ c x2 þ 1

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 12 2) 1 3) 32 4) 3 7.30. Calculate the value of the definite integral below. ð2 1

jxjdx

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 32 2) 52 3) 72 4) 92 7.31. Calculate the value of the following definite integral. ðπ

2

sin 2 ðxÞdx

0

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 0 2) π2 3) π4 4) π8 7.32. Calculate the value of the definite integral below. ð 3π 4



 tan 5 ðxÞ þ tan 7 ðxÞ dx

0

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 0 2) 12 3) 13 4) 16

7

Problems: Definite and Indefinite Integrals

115

7.33. Which one of the points below is on a curve that passes from the point of (π, 1) and y0 ¼ y2 cos (x) holds for that? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large   ,2 1) 3π 2   2) π2 , 1 π  3) 2 , 1 4) (0, 1) 7.34. Calculate the value of the definite integral of I1 if I2 ¼ m. I1 ¼

I2 ¼

ð5

3x dx x 3 2

ð5

1 dx x  2 3

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) m + 2 2) 4m  6 3) 6m + 6 4) 6m  4 7.35. Calculate the value of the definite integral below. ð3 2

x dx x2  1

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large  1) ln 83   pffiffi 2) arc sin 2 5 3    3) arc tan 32 qffiffi 8 4) ln 3 7.36. Calculate the volume resulted from the rotation of the surface area around x-axis enclosed between one period of the curve of y ¼ sin (x) and x-axis. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) π 2 2) 2π 2 2 3) π2 4)

π2 4

7.37. Calculate the value of the definite integral of

Ð4 0

f 0 ðxÞdx if we have f ðxÞ ¼

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

Ð x pffi a t dt.

116

7 Problems: Definite and Indefinite Integrals

1) 83 2) 16 3 3)  83 4)  16 3 7.38. Solve the following indefinite integral. ð

  8 tan 6 ðxÞ þ tan 8 ðxÞ dx

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) tan7(x) + c 2) 17 tan 8 ðxÞ þ c 3) 85 tan 5 ðxÞ þ c 4) 87 tan 7 ðxÞ þ c 7.39. Calculate the value of the definite integral below. ð1 1

  ðx þ 1Þ x2 þ 2x þ 3 dx

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 6 2) 4 3) 8 4) 10 7.40. Calculate the value of the definite integral below. ð1h i 1 1 dx 3 1 x x 2

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 2 3) 12 4) 32 7.41. Calculate the value of y0x if we have: y ¼ u þ v, u ¼

ð x2 1

sin ðt Þ dt, v ¼ t

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

ð1 x2

sin ðuÞ du u

7

Problems: Definite and Indefinite Integrals

1)

117

4 sin ðx2 Þ x 4 sin ðx2 Þ  x

2) 3) 0 4) 1

7.42. Calculate the surface area enclosed between the curve of y ¼ x2 + 1 and the line of y ¼ 2. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 13 2) 23 3) 1 4) 43 7.43. Calculate the value of f(3x + 2) if we have: ð hð x Þ ¼

f 0 ð3x þ 2Þdx, hð0Þ ¼ 1, f ð2Þ ¼ 3

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) h(x)  1 2) 2h(x) + 1 3) 3h(x) 4) 3h(x)  1  pffiffiffi pffiffiffi 7.44. Calculate the surface area enclosed between the curve with the function below and x-axis in the range of  2, 2 . y¼

1 2 þ x2

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large pffiffi 1) π 4 2 2) π4 pffiffi 3) π 2 2 4) π2 7.45. A curve is tangent to y ¼ x in the origin and its second derivative is 2x + 1. Which one of the points below is on the curve? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ○ Normal ● Large   1) 1, 11 6   2) 1, 13 6   3) 2, 11 6  13 4) 2, 6

118

7 Problems: Definite and Indefinite Integrals

7.46. Solve the indefinite integral of

Ð

sin (2x) cos (4x)dx.

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ○ Normal ● Large 1) 13 cos 3 ð2xÞ  12 cos ð2xÞ þ c 2)  13 cos 3 ð2xÞ þ 12 cos ð2xÞ þ c 3) 13 cos 3 ð2xÞ þ 12 cos ð2xÞ þ c 4)  13 cos 3 ð2xÞ  12 cos ð2xÞ þ c 7.47. Calculate the value of the definite integral below. ð1 h i x dx 3 1 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 1 3) 1 4) 3 7.48. What is the function of a curve that passes from the point of (1, 2) and the relation of xy0 + y ¼ 1 holds. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) y ¼ 1 þ 1x 2) y ¼ 2  1x 3) y ¼ 3x  1 4) y ¼ 3x þ 1 7.49. Solve the indefinite integral below. ð

pffiffiffi f 0 ð 3 xÞ p ffiffiffiffiffi dx 3 2 x

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large pffiffiffi 1) 13 f ð 3 xÞ þ c pffiffiffi 2) 23 f ð 3 xÞ þ c pffiffiffi 3) f ð 3 xÞ þ c pffiffiffi 4) 3f ð 3 xÞ þ c 7.50. Solve the indefinite integral below. ð ln xdx Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large

7

Problems: Definite and Indefinite Integrals

1) 2) 3) 4)

119

x ln x  x + c x ln x + x + c x ln x + x + c x ln x  x + c

7.51. Solve the following indefinite integral. ð

1 dx sin ðxÞ cos ðxÞ

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) ln|sin(2x)| + c 2) ln|tan(x)| + c 3) ln|cos(2x)| + c 4) ln|cot(x)| + c 7.52. Calculate the value of the definite integral below. ðπ

4

0

1 dx cos 4 ðxÞ

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 13 2) 23 3) 1 4) 43 7.53. Calculate the value of the definite integral below. ðπ

4

0

1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx 3 2 sin ðxÞ cos 4 ðxÞ

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 2 3) 3 4) 4 7.54. Calculate the value of f(x ¼ e) if the derivative of f(x2) with respect to x is 6x and f(x ¼ 1) ¼ 0. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 0 2) 1 3) 3 4) 6

120

7 Problems: Definite and Indefinite Integrals

7.55. Calculate the value of f(x ¼  1) if f 0(cos2(x)) ¼ cos (2x) and f(x ¼ 1) ¼ 1. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 2 3) 3 4) 4 7.56. Solve the following indefinite integral. ð

cos ð2xÞ dx sin ðxÞ cos 2 ðxÞ 2

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1)  sin2ð2xÞ þ c 2)  sin1ð2xÞ þ c 3) 4)

2 sin ð2xÞ 1 sin ð2xÞ

þc þc

7.57. Calculate the value of the definite integral below. ðe

ð2x þ ln ðxÞÞdx

1

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) e2 2) 1 3) 1 + e 4) e  1 7.58. Solve the indefinite integral below. ð

 50 sin ð2xÞ 2 þ cos 2 ðxÞ dx

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

51 1 2 51 ð2 þ cos ðxÞÞ þ c 51 1 ð2 þ cos 2 ðxÞÞ þ  51 50 1 2 51 ð2 þ cos ðxÞÞ þ c 50 1 ð2 þ cos 2 ðxÞÞ þ  51

c c

7

Problems: Definite and Indefinite Integrals

121

7.59. Calculate the value of the definite integral below. ðe 1

ln ðxÞ dx x

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 12 3) 2 4) 1e 7.60. Determine the function of a curve that passes from the point of (0, 1) and the relation below holds. y0 ¼ 

2x þ 2 4y þ 1

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 2x2 + y2 ¼ 34 + 3x 2) x2  y2 ¼  7y + 5 3) x2 + y2 ¼ 4x + 4y  1 4) x2 + 2y2 ¼  y  2x + 3 7.61. Calculate the value of the definite integral below. ðπ

2

π 6

cot ðxÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx 1  cos ð2xÞ

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large pffiffiffi 1) 2 pffiffi 2) 22 pffiffiffi 3) 3 pffiffi 4) 23 7.62. Calculate the value of the definite integral below. ð6 3

xþ2 pffiffiffiffiffiffiffiffiffiffiffi dx x2

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 25 3 2) 38 3 3) 23 3 4) 34 3

122

7 Problems: Definite and Indefinite Integrals

7.63. Calculate the value of the definite integral below. pffiffiffi ð 4 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1þ x pffiffiffi dx x 1 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large pffiffiffi  1) 2 3 þ 13 pffiffiffi 1 2) 2 3  3 pffiffiffi pffiffi 3) 4 3 þ 2 3 2 pffiffiffi pffiffi 4) 4 3  2 3 2 7.64. Solve the following indefinite integral if we know that 0 < x < π. ð cot ðxÞ

pffiffiffiffiffiffiffiffiffiffiffiffiffi sin ðxÞdx

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1) cot ðxÞ þ c pffiffiffiffiffiffiffiffiffiffiffiffiffi 2) 2 sin ðxÞ þ c pffiffiffiffiffiffiffiffiffiffiffiffiffi 3) sin ðxÞ sin ðxÞ þ c pffiffiffiffiffiffiffiffiffiffiffiffiffi 4) 12 sin ðxÞ þ c 7.65. Calculate the volume resulted from the rotation of the surface area around y-axis enclosed between the curve of y ¼ 1  14 x2 and x-axis. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) π 2) 2π 3) 3π 4) 4π 7.66. Calculate the value of the definite integral below. ðπ

3

sec ðxÞ tan ðxÞdx

0

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 2 3) 12 4) 32

7

Problems: Definite and Indefinite Integrals

123

7.67. Calculate the value of the definite integral below. ðπ

4

π 6

cscðxÞ cot ðxÞdx

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large pffiffiffi 1) 2 þ 2 pffiffiffi 2) 2  2 pffiffiffi pffiffiffi 3) 3  2 pffiffiffi pffiffiffi 4) 3 þ 2 7.68. Calculate the value of the definite integral below. ðπ

4

π 6

1 dx sin ðxÞ cos 2 ðxÞ 2

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large pffiffi 1) 33 pffiffi 2) 2 3 3 pffiffiffi 3) 3 pffiffiffi 4) 2 3 7.69. Solve the following indefinite integral. ð ð tan ðxÞ  cot ðxÞÞð tan ðxÞ þ cot ðxÞÞ5 dx Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) 14 ð tan ðxÞ þ cot ðxÞÞ4 þ c 2) 15 ð tan ðxÞ þ cot ðxÞÞ5 þ c 3) 13 ð tan ðxÞ þ cot ðxÞÞ3 þ c 4) 15 ð tan ðxÞ  cot ðxÞÞ5 þ c 7.70. Calculate the volume resulted from the rotation of the surface area around x-axis enclosed between the curve with the function below, x-axis, x ¼ π4, and x ¼ π2. y¼ Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) π 2) 2π 3 3) 2π 4) 4π 3

1 sin 2 ðxÞ

124

7 Problems: Definite and Indefinite Integrals

7.71. Which one of the choices is not an acceptable solution for the indefinite integral of

Ð

sin (x) cos (x)dx.

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1)  14 cos ð2xÞ þ c 2)  14 sin ð2xÞ þ c 3)  12 cos 2 ðxÞ þ c 4) 12 sin 2 ðxÞ þ c

Reference 1. Rahmani-Andebili, M. (2020). Precalculus: Practice problems, methods, and solutions, Springer Nature, 2020.

8

Solutions of Problems: Definite and Indefinite Integrals

Abstract

In this chapter, the problems of the seventh chapter are fully solved, in detail, step-by-step, and with different methods. The subjects include definite integrals, indefinite integrals, substitution rule for integrals, integration techniques, integration by parts, integrals involving trigonometric functions, trigonometric substitutions, integration using partial fractions, integrals involving roots, integrals involving quadratics, applications of integrals, average value, area between curves, and volume of solid of revolution. 8.1. From the list of integral of trigonometric functions, we know that [1]: ð xn dx ¼

1 nþ1 x þc nþ1

The problem can be solved as follows. ð2

xþ4 dx ¼ 3 1 x

ð2 1

   1 4 1 2 2 1 2  þ dx ¼   ¼    ð1  2Þ ¼ 2 x x2 1 2 4 x2 x3

Choice (2) is the answer. 8.2. From the list of integral of trigonometric functions, we know that: ð eu du ¼ eu The problem can be solved as follows. ð  2 2 ex þ 2xex dx ¼ ex þ ex þ c Choice (4) is the answer. 8.3. Based on the information given in the problem, we have: ð f ð xÞ ¼



 x3 þ 5x dx

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Calculus, https://doi.org/10.1007/978-3-030-64980-7_8

125

126

8 d dx

d dx

Solutions of Problems: Definite and Indefinite Integrals

x¼1

) f 0 ðxÞ ¼ x3 þ 5x ) f 00 ðxÞ ¼ 3x2 þ 5 ) f 00 ð1Þ ¼ 3 þ 5 ¼ 8 Choice (4) is the answer. 8.4. The problem can be solved as follows. ð f ðxÞ ¼

cos 3 ðxÞdx ⟹ f 0 ðxÞ ¼ cos 3 ðxÞ ⟹ f 00 ðxÞ ¼ 3 cos 2 ðxÞ sin ðxÞ f 00

  π ¼ 3  0  1 ¼ 0 2

Choice (1) is the answer. 8.5. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 þc x nþ1

The problem can be solved as follows. ð 1

x3 þ x2  1 dx ¼ x2 2 ¼



 2  ð 1   1 x 1 1 x þ 1  2 dx ¼ þxþ  x 2 2 x 2

   1 1 11  22 ¼ 1 2 2

Choice (2) is the answer. 8.6. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

The problem can be solved as follows. ð

x2 pffiffiffi dx ¼ x

ð

 1 1 1 2 3 2 pffiffiffi x ð x  6Þ þ c x2  2x2 dx ¼ x2  4x2 þ c ¼ 3 3

Choice (2) is the answer. 8.7. The problem can be solved by changing the variable of the integral as follows. d dx

x3 þ 1 ≜ u ) 3x2 dx ¼ du ) x2 dx ¼ ð1

x2 dx ¼ 4 3 0 ð x þ 1Þ

ð u2 u u1

4

du 3

 3 1 du 1 u3 u2 1 3 ¼  þ 1 ¼ x  3 3 3 u1 9 0

1 1 1 1 1 þ 8 7  ð1 þ 1Þ3 þ ð0 þ 1Þ3 ¼  þ ¼ ¼ 9 9 72 9 72 72 Choice (4) is the answer.

8

Solutions of Problems: Definite and Indefinite Integrals

127

8.8. From the list of integral of trigonometric functions, we know that: ð

  1 1 x arc tan þc dx ¼ a a x 2 þ a2

Therefore, ð þ1

      1 1 x þ1 1 π π π ¼ arc tan   ¼ dx ¼  2þ4 1 2 2 2 2 2 2 x 1

Choice (1) is the answer. 8.9. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

The problem can be solved as follows.    ð4 pffiffiffi    3  1 x x 2 2 16 2 5 4 pffiffiffi dx ¼ x x  ¼ 4  1 ¼ x2  1 dx ¼ 1 3 3 3 3 x 1 1

ð4

Choice (4) is the answer. 8.10. Based on the information given in the problem, we have: ð

d

f ðxÞdx ¼

x3 dx x2 ) f ð xÞ ¼ 6 2

Therefore,   12 1 1 )f ¼ x ¼ 2 x 2 2x )

d dx

     1 d 1 4x 1 f ¼ ¼ 4 ¼ 3 x dx 2x2 4x x

Choice (1) is the answer. 8.11. As we know: F ðxÞ ¼

ð uðxÞ vðxÞ

f ðxÞdx ⟹ F 0 ðxÞ ¼ u0 ðxÞF ðuðxÞÞ  v0 ðxÞF ðvðxÞÞ

The problem can be solved as follows. F ðλÞ ¼ Choice (2) is the answer.

ðλ

1 1 1 1 0¼ 4 ⟹ F 0 ð 0Þ ¼ dx ⟹ F 0 ðλÞ ¼ 1  4 4þ2 2 x λ þ2 λ þ2 0

128

8

Solutions of Problems: Definite and Indefinite Integrals

8.12. Since the function is an odd function and the range of the integral is symmetric, the final answer is zero. ð1

x2 arcð tan ðxÞÞdx ¼ 0 2 1 1 þ x

Choice (1) is the answer. 8.13. From the list of integral of trigonometric functions, we know that: ð

1 pffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ arcð sin ðxÞÞ 1  x2

Therefore:  1    1 π π π pffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ arcð sin ðxÞÞ 2 1 ¼   ¼ 2 6 6 3  1 2 1  x 2

ð1 2

Choice (2) is the answer. 8.14. From the list of integral of trigonometric functions, we know that: ð

1 pffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ arcð sin ðuÞÞ 1  u2

The problem can be solved by changing the variable of the integral as follows. d dx

x  1 ≜ u ) dx ¼ du ð1

1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ 2x  x2 0

ð1

1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ 0 1  ð x  1Þ 2

ð u2 u1

 u 1 pffiffiffiffiffiffiffiffiffiffiffiffiffi du ¼ ðarcð sin ðuÞÞÞ 2 2 u1 1u

   π π 1 ¼ ðarcð sin ðx  1ÞÞÞ ¼ 0   ¼ 0 2 2 Choice (3) is the answer. 8.15. The final answer is zero, since the function is an odd function and the range of the integral is symmetric. ð1



1

  x2 þ 1 x3 þ 3x dx ¼ 0

Choice (2) is the answer. 8.16. From trigonometry, we know that: 1 þ cos ðuÞ ¼ 2 cos 2

  u 2

  1 u u ¼ 1 þ tan 2 2 cos 2 2

8

Solutions of Problems: Definite and Indefinite Integrals

129

In addition, from the list of integral of trigonometric functions, we know that: ð

1 þ tan 2

    u u du ¼ a tan þc a a

The problem can be solved by changing the variable of the integral as follows. cos ðxÞ≜u ⟹

d d cos ðxÞ ¼ u ⟹  sin ðxÞdx ¼ du dx dx

ð ð sin ðxÞ 1 1   du du ¼  ⟹ dx ¼  1 þ cos ðuÞ 1 þ cos ð cos ðxÞÞ 2 cos 2 u2 ð

1 ¼ 2

  ð     cos ðxÞ u 2 u 1 þ tan þc du ¼  tan þ c ¼  tan 2 2 2

Choice (1) is the answer. 8.17. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 þc x nþ1

First, we need to find the intersection points of the curves as follows: 2x2  2x ¼ x2 ⟹ x2  2x ¼ 0 ⟹ x ¼ 0, 2 S¼

ð x2

ðy2  y1 Þdx ¼

x1

ð2





x  2x þ 2x dx ¼ 2

2

0

ð2 0



 3  x 2 2 x þ 2x dx ¼  þ x  3 0 2



 3  2 4 2 ¼  þ 2  ð 0 þ 0Þ ⟹ S ¼ 3 3 Choice (3) is the answer. 8.18. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

Based on the information given in the problem, we have: y ð 3Þ ¼ 4 y0 ¼ 

x y

The problem can be solved as follows. ð y0 ¼ 

dx x y2 x2 ) yy0 þ x ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) þ ¼ c0 ) y2 þ x2 ¼ c y 2 2

ð1Þ

130

8

Solutions of Problems: Definite and Indefinite Integrals

y ð 3Þ ¼ 4 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 ) 4 þ 32 ¼ c ) c ¼ 25 ð1Þ, ð2Þ 2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) y þ x2 ¼ 25 Choice (4) is the answer. 8.19. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 þc x nþ1

The problem can be solved as follows. ð

ð ð  1 1 3 1 x x1þ1 2 pffiffiffiffiffiffiffiffiffiffiffi dx ¼ pffiffiffiffiffiffiffiffiffiffiffi dx ¼ ðx  1Þ2 þ ðx  1Þ2 dx ¼ ðx  1Þ2 þ 2ðx  1Þ2 þ c 3 x1 x1

Choice (2) is the answer. 8.20. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

The problem can be solved as follows. ð5 2

jx  3jdx ¼

ð3 2

ð3  xÞdx þ

ð5

 ðx  3Þdx ¼

3x 

3

  2  5 x2  3 x  3x   þ 2 2 2 3

      9 25 9 9 5 9 9 þ 16  5 þ 9 29 ¼ 9  ð6  2Þ þ  15  9 ¼ þ8 þ ¼ ¼ 2 2 2 2 2 2 2 2 Choice (3) is the answer. 8.21. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 þc x nþ1

The problem can be solved as follows. ð2

x ð x þ 1Þ 2 þ 2 dx ¼ ð x þ 1Þ 2 1

ð2 1

 2       2 2 1 4 1 11 x 2  1 ¼ þ ¼  x þ 2ðx þ 1Þ dx ¼  ¼ 2 3 2 3 2 6 2 xþ1 1 2

Choice (3) is the answer. 8.22. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

First, we need to find the intersection points of the curves as follows: (

 pffiffiffi pffiffiffi pffiffiffi y1 ¼ x2 pffiffiffi ⟹ y2 ¼ y1 ⟹ x ¼ x2 ⟹ x x x  1 ¼ 0 ⟹ x ¼ 0, 1 y2 ¼ x

ð2Þ

8

Solutions of Problems: Definite and Indefinite Integrals

S¼

ð x2

ðy2  y1 Þdx ¼

ð1

131

pffiffiffi  x  x2 dx ¼

0

x1



 2 32 x3 1 2 1 1 x   ¼  ⟹S¼ 3 3 3 0 3 3

Choice (3) is the answer. 8.23. From the list of integral of trigonometric functions, we know that: ð xn dx ¼ ð

1 nþ1 x þc nþ1

1 du ¼ ln juj þ c u

The problem can be solved as follows. ð

 ð ð ð ð 1 1 þ ex  ex ex ex dx ¼ dx ¼ 1  dx dx ¼ 1dx  x x x 1þe 1þe 1þe 1 þ ex ¼ x þ c0 

ð

ex dx 1 þ ex

ð1Þ

Now, we should change the variable of the integral as follows. 1 þ ex ≜u ⟹ ex dx ¼ du Solving (1) and (2): x þ c0 

ð

1 du ¼ x þ c0  ln juj þ c00 ¼ x  ln j1 þ ex j þ c ¼ x  ln ð1 þ ex Þ þ c u

Choice (2) is the answer. 8.24. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

First, we need to find the intersection points of the curves as follows:   y2 ¼ y1 ⟹ x3 þ 2x2 þ x ¼ 0 ⟹ x x2 þ 2x þ 1 ¼ xðx þ 1Þ2 ¼ 0 ⟹ x ¼ 0,  1,  1 S¼

ð x2

ðy2  y1 Þdx ¼

x1

ð0 1



 4    x 2 3 x2  0 1 2 1  þ x þ 2x þ x dx ¼ þ x þ ¼0  4 3 2 4 3 2 1 3

2

¼



  38þ6 1 ¼ 12 12

The surface area must be a positive quantity. Therefore,   1  1 S ¼   ¼ 12 12 Choice (1) is the answer.

ð2Þ

132

8

Solutions of Problems: Definite and Indefinite Integrals

8.25. As we know, the average value of a function can be determined as follows: 1 ba

ðb

f ðxÞdx

a

In addition, from the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

The problem can be solved as follows. 1 52

    1 a 2 5 1 25a 4a  ðax þ bÞdx ¼ x þ bx  ¼ þ 5b   2b 2 3 2 3 2 2 2

ð5

7 The average value ¼ a þ b 2 Choice (4) is the answer. 8.26. The problem can be solved as follows. ð dx xþ1 y2 x2 0 0 y ¼ ) y  yy ¼ x þ 1¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )y  ¼ þ x þ c 1y 2 2 0

ðx, yÞ ¼ ð1, 1Þ 1 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 )1  ¼ þ 1 þ c ) c ¼ 1 2 2 )y

y2 x2 ¼ þ x  1 ) x2 þ y2 þ 2x  2y  2 ¼ 0 2 2

Choice (1) is the answer. 8.27. As we know, the average value of a function can be determined as follows: 1 ba

ðb

f ðxÞdx

a

Moreover, from the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

The problem can be solved as follows. 1 λ1

ðλ 1

2x dx ¼

1 λ1

ðλ



1

  λ λ  3x2  2x dx ⟹ x2  ¼ x3  x2  1 1

    ⟹ λ2  1 ¼ λ3  λ2  0 ⟹ λ3  2λ2 þ 1 ¼ 0 ⟹ ðλ  1Þ λ2  λ  1 ¼ 0 ⟹λ¼

pffiffiffi pffiffiffi 1 5 1þ 5 , ,1 2 2

8

Solutions of Problems: Definite and Indefinite Integrals pffiffi 5

However, just 1þ2

133

is acceptable because the others are not within the range. ⟹λ¼

pffiffiffi 1þ 5 2

Choice (1) is the answer. 8.28. The problem can be solved as follows. 3x 3 ⟹ 2yy0 ¼ 3x ⟹ y2 ¼ x2 þ c 2y 2

y0 ¼

ðx, yÞ ¼ ð1, 1Þ 3 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 )1 ¼ þ c ) c ¼ 2 2 3 1 ) y2 ¼ x2  ) 2y2  3x2 þ 1 ¼ 0 2 2 Choice (1) is the answer. 8.29. Based on the information given in the problem, we have: ð

pffiffiffiffiffiffiffiffiffiffiffiffiffi 3x pffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ A x2 þ 1 þ c x2 þ 1

ð1Þ

From the list of integral of trigonometric functions, we know that: ð

pffiffiffi 1 pffiffiffi du ¼ 2 u þ c u

The problem can be solved by changing the variable of the integral as follows. d

dx 1 x2 þ 1 ≜ u ) 2xdx ¼ du ) xdx ¼ du 2

ð

ð ð pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3  12 du 3 1 3x 3 pffiffiffi du ¼  2 u ¼ 3 x2 þ 1 þ c pffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ 2 2 u u x2 þ 1

Therefore, pffiffiffiffiffiffiffiffiffiffiffiffiffi ð1Þ, ð2Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) A x2 þ 1 þ c ¼ 3 x2 þ 1 þ c ⟹ A ¼ 3 Choice (4) is the answer. 8.30. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

The problem can be solved as follows. ð2 1

jxjdx ¼

ð0 1

jxjdx þ

Choice (2) is the answer.

ð2 0

jxjdx ¼

ð0 1

ðxÞdx þ

ð2 0

xdx ¼ 

    x2  0 x2  2 1 5 þ ð2  0Þ ¼  þ  ¼ 0 2 2 2 1 2 0

ð2Þ

134

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Solutions of Problems: Definite and Indefinite Integrals

8.31. From trigonometry, we know that: 1  cos ð2xÞ ¼ 2 sin 2 ðxÞ Furthermore, from the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

ð cos ðaxÞdx ¼

1 sin ðaxÞ þ c a

The problem can be solved as follows.  ðπ    π   2 1 cos ð2xÞ 1 1 1 π π  dx ¼  x  sin ð2xÞ  2 ¼   0  ð 0Þ ¼ sin ðxÞdx ¼ 2 2 2 4 2 2 4 0 0 0

ðπ

2

2

Choice (3) is the answer. 8.32. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

The problem can be solved by changing the variable of the integral as follows. d  dx  tan ðxÞ ≜ u ) 1 þ tan 2 ðxÞ dx ¼ du

ð 3π 4



 tan 5 ðxÞ þ tan 7 ðxÞ dx ¼

0

ð 3π 4

  tan 5 ðxÞ 1 þ tan 2 ðxÞ dx

0

 3π  1 6 u2 1 1 1 6 u du ¼ u  ¼ tan ðxÞ 4 ¼ ð1Þ6  0 ¼ ¼ 6 6 6 6 u 0 1 u1 ð u2

5

Choice (4) is the answer. 8.33. The problem can be solved as follows. y0 ¼ y2 cos ðxÞ )

y0 1 ¼ sin ðxÞ þ c ¼ cos ðxÞ ) y y2

ðx, yÞ ¼ ðπ, 1Þ 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 ¼ ¼ ¼ )  1 ¼ 0 þ c ) c ¼ 1 ) ¼ sin ðxÞ  1 y We need to check each choice as follows:   3π ðx, yÞ ¼ ,2   2 1 3π 1 Choice 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 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) ¼ sin 1) 6¼ 2 2 2 2   π , 1 ðx, yÞ ¼   2 1 π ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) ¼ sin  1 ) 1 6¼ 0 Choice 2 : ¼ 1 2

8

Solutions of Problems: Definite and Indefinite Integrals

135

  π ,1 ðx, yÞ ¼   2 1 π  1 )  1 6¼ 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 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) ¼ sin Choice 3 : ¼ 1 2 ðx, yÞ ¼ ð0, 1Þ 1 Choice 4 : ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) ¼ sin ð0Þ  1 ) 1 ¼ 1 1 Choice (4) is the answer. 8.34. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

Based on the information given in the problem, we have: I2 ¼

ð5

1 dx ¼ m x  2 3

ð1Þ

The problem can be solved as follows. ð5

3x dx ¼ I1 ¼ x 3 2

ð5

3x  6 þ 6 dx ¼ x2 3

ð5

3dx þ 6

3

ð5

1 dx x  2 3

ð2Þ

Solving (1) and (2):  5 3x dx ¼ 3x þ 6m ¼ 15  9 þ 6m ¼ 6 þ 6m I1 ¼ x  2 3 3 ð5

Choice (3) is the answer. 8.35. From the list of integral of trigonometric functions, we know that: ð

1 du ¼ ln juj þ c u

The problem can be solved as follows. ð3

x 1 dx ¼ 21 2 x 2

ð3

2x dx 21 x 2

ð1Þ

Now, we need to change the variable of the integral as follows. d dx

x2  1 ≜ u ) 2xdx ¼ du Solving (1) and (2): 1 2

ð u2 u1

rffiffiffi   u2 1  2 3 1 1 1 1 1 8 8 du ¼ ln juj ¼ ln x  1 ¼ ln 8  ln 3 ¼ ln ¼ ln u 2 2 2 3 3 u1 2 2 2

Choice (4) is the answer.

ð2Þ

136

8

Solutions of Problems: Definite and Indefinite Integrals

8.36. The volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and x-axis, is calculated as follows: V ¼π

ð x2

ð f ðxÞÞ2 dx

x1

Moreover, from trigonometry, we know that: 1  cos ð2xÞ ¼ 2 sin 2 ðxÞ In addition, from the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 þc x nþ1

ð cos ðaxÞdx ¼

1 sin ðaxÞ þ c a

Therefore, V ¼π

ð 2π

sin ðxÞdx ¼ π

ð 2π 

2

0

0

   1 cos ð2xÞ x 1 2π dx ¼ π  sin ð2xÞ  ¼ π ðπ  0Þ ¼ π 2  2 2 2 4 0

Choice (1) is the answer. 8.37. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

As we know: F ðxÞ ¼

ð uðxÞ vðxÞ

f ðxÞdx ⟹ F 0 ðxÞ ¼ u0 ðxÞF ðuðxÞÞ  v0 ðxÞF ðvðxÞÞ

Therefore, f ð xÞ ¼

ðx

pffi pffiffiffi t dt ⟹ f 0 ðxÞ ¼ x

a

ð4

0

f ðxÞdx ¼

0

ð4

1

x2 dx ¼

0



 2 32 4 16 x  ¼ 0 3 3

Choice (2) is the answer. 8.38. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 þc x nþ1

8

Solutions of Problems: Definite and Indefinite Integrals

137

The problem can be solved by changing the variable of the integral as follows. d  dx  tan ðxÞ ≜ u ) 1 þ tan 2 ðxÞ dx ¼ du

ð

  8 tan 6 ðxÞ þ tan 8 ðxÞ dx ¼ 8

ð

  tan 6 ðxÞ 1 þ tan 2 ðxÞ dx

ð 8 8 ¼ 8 u6 du ¼ u7 þ c ¼ tan 7 ðxÞ þ c 7 7 Choice (4) is the answer. 8.39. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

The problem can be solved as follows. ð1 1

  ðx þ 1Þ x2 þ 2x þ 3 dx ¼

ð1 1

ðx þ 1Þ

 

 x þ 1Þ2 þ 2 dx ¼

ð1  1

 ðx þ 1Þ3 þ 2ðx þ 1Þ dx

ð1Þ

Now, we should change the variable of the integral as follows. x þ 1 ≜ u ⟹ dx ¼ du Solving (1) and (2): ð1



1



u4 u þ 2u dx ¼ þ u2 ¼ 4 3



 ð x þ 1Þ 4 16 2  1 ¼ þ40¼8 þ ð x þ 1Þ  1 4 4

Choice (3) is the answer. 8.40. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

The problem can be solved as follows.   2  ð1 ð1h i 1  1 1 1 1 1 x 3 1 ¼  ¼  3 dx ¼ 1  x3 dx ¼  ð2Þ ¼ 2 2 2  2 1 x 1 x 2x2  12 2

2

Choice (4) is the answer. 8.41. As we know: F ðxÞ ¼

ð uðxÞ vðxÞ

f ðxÞdx ⟹ F 0 ðxÞ ¼ u0 ðxÞF ðuðxÞÞ  v0 ðxÞF ðvðxÞÞ

ð2Þ

138

8

Solutions of Problems: Definite and Indefinite Integrals

Therefore, y0x ¼ u0x þ v0x ¼

 2x

   sin ðx2 Þ sin ðx2 Þ  0 þ 0  2x ¼0 x2 x2

Choice (3) is the answer. 8.42. From the list of integral of trigonometric functions, we know that: ð xn du ¼

1 nþ1 x þc nþ1

First, we need to find the intersection points of the curves as follows: (

S¼

ð1 1





y1 ¼ x 2 þ 1 ⟹ x2 þ 1 ¼ 2 ⟹ x2 ¼ 1 ⟹ x ¼ 1 y2 ¼ 2 

2  x þ 1 dx ¼ 2 2

ð1 0



     x3  1 1 4 0¼ 1  x dx ¼ 2 x   ¼2 1 3 3 3 0 2

Choice (4) is the answer. 8.43. Based on the information given in the problem, we know that: ð hð x Þ ¼

f 0 ð3x þ 2Þdx

ð1Þ

hð 0Þ ¼ 1

ð2Þ

f ð 2Þ ¼ 3

ð3Þ

We should change the variable of the integral of h(x) as follows. f ð3x þ 2Þ ≜ u ⟹ 3 f 0 ð3x þ 2Þdx ¼ du

ð4Þ

Solving (1) and (4): ð hðxÞ ¼

1 1 1 du ¼ u þ c ¼ f ð3x þ 2Þ þ c 3 3 3

Solving (2) and (5): Using ð3Þ 1 1 1 1 ¼ f ð 2Þ þ c ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )1 ¼  3 þ c ⟹ c ¼ 0 ⟹ hðxÞ ¼ f ð3x þ 2Þ ⟹ f ð3x þ 2Þ ¼ 3hðxÞ 3 3 3 Choice (3) is the answer. 8.44. From the list of integral of trigonometric functions, we know that: ð

   1 x   dx ¼ a arc tan  x 2 a 1þ a

ð5Þ

8

Solutions of Problems: Definite and Indefinite Integrals

139

The problem can be solved as follows. ð pffiffi2

1 S ¼ pffiffi dx ¼ 2 2 þ x2  2

¼

ð p2ffiffi 0



1  2  dx ¼ 1 þ pxffiffi

ð p2ffiffi 0

1 dx ¼ 2 2 þ x2

ð p2ffiffi 0

1   2  dx 2 1 þ pxffiffi2

pffiffiffi    pffiffiffi  pffiffiffi  2 pffiffiffiπ x π 2  ¼ 2  0 ¼ 2 arc tan pffiffiffi  0 4 4 2

2

Choice (1) is the answer. 8.45. From the list of integral of trigonometric functions, we know that: ð xn du ¼

1 nþ1 x þc nþ1

Based on the information given in the problem, we know that: y00 ðxÞ ¼ 2x þ 1

ð1Þ

The curve is tangent to y ¼ x in the origin. Thus: y 0 ð 0Þ ¼ 1

ð2Þ

y ð 0Þ ¼ 0

ð3Þ

dx ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )y0 ðxÞ ¼ x2 þ x þ a

ð4Þ

dx x3 x2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) yðxÞ ¼ þ þ ax þ b 3 2

ð5Þ

a¼1

ð6Þ

0¼0þ0þ0þb⟹b¼0

ð7Þ

Applying integral operation on (1): ð

Applying integral operation on (4): ð

Solving (2) and (4):

Solving (3) and (5):

Solving (5)–(7): yð xÞ ¼

x3 x2 þ þx 3 2

140

8

Solutions of Problems: Definite and Indefinite Integrals

Now, we need to check the choices as follows: y ð 1Þ ¼ y ð 2Þ ¼

1 1 2 þ 3 þ 6 11 þ þ1¼ ¼ 3 2 6 6

2 3 22 16 þ 12 þ 12 20 ¼ þ þ2¼ 6 3 3 2

Choice (1) is the answer. 8.46. From trigonometry, we know that: 1 þ cos ð2xÞ ¼ 2 cos 2 ðxÞ In addition, from the list of integral of trigonometric functions, we know that: ð sin ðaxÞdx ¼ 

1 cos ðaxÞ þ c a

The problem can be solved as follows. ð

ð sin ð2xÞ cos ð4xÞdx ¼

  sin ð2xÞ 2 cos 2 ð2xÞ  1 dx

ð ¼

ð cos ð2xÞ  2 sin ð2xÞdx  2

sin ð2xÞdx

ð1Þ

Now, we should change the variable of the integral as follows. cos ð2xÞ ≜ u ⟹  2 sin ð2xÞdx ¼ du Solving (1) and (2): ð

ð

 u du  2

1 1 1 1 sin ð2xÞdx ¼  u3 þ cos ð2xÞ þ c ¼  cos 3 ð2xÞ þ cos ð2xÞ þ c 3 2 3 2

Choice (2) is the answer. 8.47. From the list of integral of trigonometric functions, we know that: ð xn du ¼

1 nþ1 x þc nþ1

The problem can be solved as follows. ð1 h i ð0 ð1  x 0 dx ¼ ð1Þdx þ 0dx ¼ ðxÞ þ 0 ¼ 0  ð1Þ ¼ 1 3 1 1 1 0 Choice (3) is the answer.

ð2Þ

8

Solutions of Problems: Definite and Indefinite Integrals

141

8.48. The problem can be heuristically solved as follows. xy0 þ y ¼ 1 ) ðxyÞ0 ¼ 1 ) xy ¼ x þ c ðx, yÞ ¼ ð1, 2Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )1  2 ¼ 1 þ c ) c ¼ 1 ) xy ¼ x þ 1 ) y ¼ 1 þ

1 x

Choice (1) is the answer. 8.49. The problem can be solved as follows. ð

pffiffiffi ð pffiffiffi f 0 ð 3 xÞ 1 p ffiffiffiffi ffi p ffiffiffiffiffi f 0 3 x dx dx ¼ 3 3 2 3 2 x 3 x

ð1Þ

Now, we should change the variable of the integral as follows. f

p pffiffiffi ffiffiffi 1 3 ffiffiffiffiffi f 0 3 x dx ¼ du x ≜u⟹ p 3 2 3 x

Solving (1) and (2): ð pffiffiffi 3 du ¼ 3u þ c ¼ 3f 3 x þ c Choice (4) is the answer. 8.50. From integration by parts (partial integration), we know that: ð

ð uðxÞdv ¼ uðxÞvðxÞ  vðxÞdu

In addition, from the list of integral of trigonometric functions, we know that: ð

1 du ¼ ln juj þ c u

The problem can be solved as follows. (

ð ln xdx ⟹ ð ⟹ Choice (1) is the answer.

uðxÞ ¼ ln x dv ¼ dx

8 < du ¼ dx x ⟹ : vð xÞ ¼ x

ð ln xdx ¼ x ln x  dx ¼ x ln x  x þ c

ð2Þ

142

8

Solutions of Problems: Definite and Indefinite Integrals

8.51. From trigonometry, we know that: 1 ¼ 1 þ tan 2 ðxÞ cos 2 ðxÞ sin ðxÞ ¼ tan ðxÞ cos ðxÞ Moreover, from the list of integral of trigonometric functions, we know that: ð

1 du ¼ ln juj þ c u

The problem can be solved as follows. ð

ð

1 dx ¼ sin ðxÞ cos ðxÞ

1 sin ðxÞ cos ðxÞ 

ð dx ¼ cosðxÞ cos ðxÞ

sinðxÞ cos ðxÞ

ð 1 þ tan 2 ðxÞ 1 dx ¼ dx tan ðxÞ cos 2 ðxÞ

ð1Þ

Now, we should change the variable of the integral as follows.   tan ðxÞ ≜ u ⟹ 1 þ tan 2 ðxÞ dx ¼ du

ð2Þ

Solving (1) and (2): ð

1 du ¼ ln juj þ c ¼ ln j tan ðxÞj þ c u

Choice (2) is the answer. 8.52. From trigonometry, we know that: 1 þ tan 2 ðxÞ ¼

1 cos 2 ðxÞ

The problem can be solved as follows. ðπ

4

0

1 dx ¼ cos 4 ðxÞ

ðπ

4



  1 þ tan 2 ðxÞ 1 þ tan 2 ðxÞ dx

ð1Þ

0

Now, we should change the variable of the integral as follows. d  dx  tan ðxÞ ≜ u) 1 þ tan 2 ðxÞ dx ¼ du

Solving (1) and (2): ð u2 u1



   π    1 3 u2 1 1 4  0¼ 1 þ u du ¼ u þ u  ¼ tan ðxÞ þ tan 3 ðxÞ  4 ¼ 1 þ 3 3 3 3 u1 0

Choice (4) is the answer.

2



ð2Þ

8

Solutions of Problems: Definite and Indefinite Integrals

143

8.53. From trigonometry, we know that: 1 þ tan 2 ðxÞ ¼ tan ðxÞ ¼

1 cos 2 ðxÞ

sin ðxÞ cos ðxÞ

The problem can be solved as follows. ðπ

1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ 3 2 0 sin ðxÞ cos 4 ðxÞ 4

ðπ

1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi dx ¼ cos 3 2 0 sin ðxÞ cos 4 ðxÞ  cos 2ðxðxÞÞ 4

ðπ ¼

4

π 6

ðπ

4

0

  2 tan 3 ðxÞ 1 þ tan 2 ðxÞ dx

1 cos 2 ðxÞ

qffiffiffiffiffiffiffiffiffiffiffi ffi dx 2 3

sin ðxÞ cos 2 ðxÞ

ð1Þ

Now, we should change the variable of the integral as follows. d  dx  tan ðxÞ ≜ u ) 1 þ tan 2 ðxÞ dx ¼ du

ð2Þ

Solving (1) and (2):    π 2 1 u 1  u3 du ¼ 3u3  2 ¼ 3 tan 3 ðxÞ  4 ¼ 3ð1  0Þ ¼ 3 u 0 1 u1

ð u2

Choice (3) is the answer. 8.54. From the list of integral of trigonometric functions, we know that: ð

1 du ¼ ln juj þ c u

Based on the information given in the problem, we have: f ð 1Þ ¼ 0

ð1Þ

d   2  6 ¼ f x dx x

ð2Þ

  d   2  ¼ 2x f 0 x2 f x dx

ð3Þ

ð2Þ, ð3Þ 6     3 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) ¼ 2x f 0 x2 ) f 0 x2 ¼ 2 x x

ð4Þ

The problem can be solved as follows.

By changing the variable of the integral, we have: x2 ≜ t

ð5Þ

144

8

Solutions of Problems: Definite and Indefinite Integrals

ð dt ð4Þ, ð5Þ 0 3 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) f ðt Þ ¼ 3 ln jt j þ c ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) f ðt Þ ¼ ¼ t

ð6Þ

ð1Þ, ð6Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 0 ¼ 3  0 þ c ) c ¼ 0 ) f ðt Þ ¼ 3 ln jt j ) f ðeÞ ¼ 3 ln ðeÞ ¼ 3  1 ¼ 3 Choice (3) is the answer. 8.55. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

From trigonometry, we know that: 1 þ cos ð2xÞ ¼ 2 cos 2 ðxÞ Moreover, based on the information given in the problem, we have: f ð 1Þ ¼ 1

ð1Þ

  f 0 cos 2 ðxÞ ¼ cos ð2xÞ

ð2Þ

  f 0 cos 2 ðxÞ ¼ cos ð2xÞ ¼ 2 cos 2 ðxÞ  1

ð3Þ

The problem can be solved as follows.

By changing the variable of the integral, we have: cos 2 ðxÞ ≜ t

ð4Þ

ð dt ð3Þ, ð4Þ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) f ðt Þ ¼ 2t  1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) f ðt Þ ¼ t 2  t þ c ð1Þ, ð5Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )1 ¼ 1  1 þ c ) c ¼ 1 ) f ðt Þ ¼ t 2  t þ 1 ) f ð1Þ ¼ ð1Þ2  ð1Þ þ 1 ¼ 3 Choice (3) is the answer. 8.56. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

From trigonometry, we know that: sin ð2xÞ ¼ 2 sin ðxÞ cos ðxÞ

ð5Þ

8

Solutions of Problems: Definite and Indefinite Integrals

145

The problem can be solved as follows. ð

ð ð ð cos ð2xÞ cos ð2xÞ cos ð2xÞ dx ¼  dx ¼ 4 cos ð2xÞð sin ð2xÞÞ2 dx 2 dx ¼ 1 2 1 sin ð 2x Þ sin 2 ðxÞ cos 2 ðxÞ sin ð 2x Þ 4 2

Now, we need to change the variable of the integral as follows. d dx

sin ð2xÞ ≜ u ) 2 cos ð2xÞdx ¼ du ð )

2 2 2u2 du ¼  þ c ¼ þc u sin ð2xÞ

Choice (1) is the answer. 8.57. From integration by parts (partial integration), we know that: ð ln ðxÞdx ¼ x ln jxj  x or, in general: ð

ð uðxÞdv ¼ uðxÞvðxÞ  vðxÞdu

In addition, from the list of integral of trigonometric functions, we know that: ð

1 du ¼ ln juj þ c u

ð xn dx ¼

1 nþ1 þc x nþ1

The problem can be solved as follows. ðe

ð2x þ ln ðxÞÞdx ¼

ðe

1

2x dx þ

1

ðe

   e e ln ðxÞdx ¼ x2  þ ðx ln jxj  xÞ 1 1 1

¼ e2  1 þ ð e  e Þ  ð 0  1Þ ¼ e 2 Choice (1) is the answer. 8.58. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

The problem can be solved as follows. ð

 50 sin ð2xÞ 2 þ cos 2 ðxÞ dx

ð1Þ

146

8

Solutions of Problems: Definite and Indefinite Integrals

We should change the variable of the integral as follows. 2 þ cos 2 ðxÞ ≜ u ⟹  2 cos ðxÞ sin ðxÞdx ¼ du ⟹  sin ð2xÞ ¼ du

ð2Þ

Solving (1) and (2): ð  u50 du ¼ 

51 u51 1  2 þ cos 2 ðxÞ þ c þc¼ 51 51

Choice (2) is the answer. 8.59. From the list of integral of trigonometric functions, we know that: ð un du ¼

1 unþ1 þ c nþ1

The problem can be solved as follows. ðe 1

ln ðxÞ dx ¼ x

ðe

ln ðxÞ

1

  1 dx x

ð1Þ

Now, we should change the variable of the integral as follows. ln ðxÞ ≜ u ⟹

1 dx ¼ du x

ð2Þ

Solving (1) and (2):   1 u 1 1 e 1 udu ¼ u2  2 ¼ ð ln ðxÞÞ2  ¼  0 ¼ 2 2 1 2 2 u 1 u1

ð u2

Choice (2) is the answer. 8.60. From the list of integral of trigonometric functions, we know that: ð un du ¼

1 unþ1 þ c nþ1

The problem can be solved as follows. ð y0 ¼ 

dx 2x þ 2 ) 4yy0 þ y0 ¼ 2x  2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 2y2 þ y ¼ x2  2x þ c 4y þ 1 ðx, yÞ ¼ ð0, 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 ¼ ¼ ¼ ¼ ¼ ) 2þ1¼0þc)c¼3

Solving (1) and (2): 2y2 þ y ¼ x2  2x þ 3 ) x2 þ 2y2 ¼ y  2x þ 3 Choice (4) is the answer.

ð1Þ

ð2Þ

8

Solutions of Problems: Definite and Indefinite Integrals

147

8.61. From the list of integral of trigonometric functions, we know that: ð un du ¼

1 unþ1 þ c nþ1

Moreover, from trigonometry, we know that: cot ðxÞ ¼

cos ðxÞ sin ðxÞ

1  cos ð2xÞ ¼ 2 sin 2 ðxÞ The problem can be solved as follows. ðπ

cot ðxÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ π 1  cos ð2xÞ 6 2

ðπ

cos ðxÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ π 6 2 sin 2 ðxÞ sin ðxÞ 2

ðπ

2

π 6

cos ðxÞ pffiffiffi dx 2j sin ðxÞj sin ðxÞ

pffiffiffi ð π 2 2 ð sin ðxÞÞ2 cos ðxÞdx ¼ 2 π

ð1Þ

6

Now, we should change the variable of the integral as follows. sin ðxÞ ≜ u ⟹ cos ðxÞdx ¼ du Solving (1) and (2):  pffiffiffi ð u2 pffiffiffi pffiffiffi pffiffiffi  pffiffiffi π  2 2 1 u2 2 2 2 1 2  2 ¼ u  ¼   ð 1  2Þ ¼ u du ¼  2 u1 2 2 2 2 u1 sin ðxÞ π6 Choice (2) is the answer. 8.62. From the list of integral of trigonometric functions, we know that: ð un du ¼

1 unþ1 þ c nþ1

The problem can be solved as follows. ð6

xþ2 pffiffiffiffiffiffiffiffiffiffiffi dx ¼ x2 3

¼

ð6

x2þ4 pffiffiffiffiffiffiffiffiffiffiffi dx ¼ x2 3

ð6  1 1 ðx  2Þ2 þ 4ðx  2Þ2 dx 3

      3 1 6 1 1 2 2 32 2 3 ðx  2Þ2 þ 4  2ðx  2Þ2  ¼ ð4Þ þ 4  2ð4Þ2  ð1Þ2 þ 4  2ð1Þ2 3 3 3 3  ¼

Choice (2) is the answer.

   16 2 14 38 þ 16  þ8 ¼ þ8¼ 3 3 3 3

ð2Þ

148

8

Solutions of Problems: Definite and Indefinite Integrals

8.63. In addition, from the list of integral of trigonometric functions, we know that: ð un du ¼

1 unþ1 þ c nþ1

The problem can be solved as follows. pffiffiffi ð 4 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð4  pffiffiffi1 1þ x 1 pffiffiffi dx ¼ 2 1 þ x 2  pffiffiffi dx x 2 x 1 1

ð1Þ

Now, we should change the variable of the integral as follows. 1þ

pffiffiffi 1 x ≜ u ⟹ pffiffiffi dx ¼ du 2 x

ð2Þ

Solving (1) and (2): pffiffiffi      pffiffiffi pffiffiffi 2 2 pffiffiffi3 4 4  pffiffiffi 1 2 3 u 4 1þ x 2  ¼ 3 32 2 ¼4 u2 du ¼ 2  u2  2 ¼ 3 1 3 3 3 3 u1 u1

ð u2 2

Choice (4) is the answer. 8.64. In addition, from the list of integral of trigonometric functions, we know that: ð un du ¼

1 unþ1 þ c nþ1

From trigonometry, we know that: cot ðxÞ ¼

cos ðxÞ sin ðxÞ

The problem can be solved as follows. ð

ð ð pffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 cos ðxÞ 2 cot ðxÞ sin ðxÞdx ¼ ð sin ðxÞÞ dx ¼ ð sin ðxÞÞ 2 cos ðxÞdx sin ðxÞ

ð1Þ

Now, we should change the variable of the integral as follows. sin ðxÞ ≜ u ⟹ cos ðxÞdx ¼ du Solving (1) and (2): ð

1

1

u 2 du ¼ 2u2 þ c ¼ 2

pffiffiffiffiffiffiffiffiffiffiffiffiffi sin ðxÞ þ c

Choice (2) is the answer. 8.65. From the list of integral of trigonometric functions, we know that: ð xn du ¼

1 nþ1 þc x nþ1

ð2Þ

8

Solutions of Problems: Definite and Indefinite Integrals

149

The volume resulted from the rotation of a surface area around y-axis, enclosed between the curve of f(x) and x-axis, is calculated as follows: V ¼π

ð y2

x2 dy

y1

Since x-axis is the boundary, y1 ¼ 0. Another boundary for y can be determined as follows: 1 x ¼ 0 ⟹ y2 ¼ 1   0 ¼ 1 4 Therefore, ð1 ð1  1 V ¼ π x2 dy ¼ π ð4  4yÞdy ¼ π 4y  2y2  ¼ π ð4  2Þ ¼ 2π 0 0 0 Choice (2) is the answer. 8.66. From trigonometry, we know that: sec ðxÞ ¼

1 cos ðxÞ

tan ðxÞ ¼

sin ðxÞ cos ðxÞ

The problem can be solved as follows. ðπ

3

ðπ sec ðxÞ tan ðxÞdx ¼

0

3

0

sin ðxÞ 1 dx ¼ cos ðxÞ cos ðxÞ

ðπ

3

0

sin ðxÞ dx cos 2 ðxÞ

ð1Þ

Now, we should change the variable of the integral as follows. cos ðxÞ ≜ u ⟹  sin ðxÞdx ¼ du Solving (1) and (2): 

ð u2 u1

  1 1 u2 1  π3 1 1 ¼ du ¼  ¼  ¼1 u  u1 cos ðxÞ 0 12 1 u2

Choice (1) is the answer. 8.67. From trigonometry, we know that: cscðxÞ ¼

1 sin ðxÞ

cot ðxÞ ¼

cos ðxÞ sin ðxÞ

ð2Þ

150

8

Solutions of Problems: Definite and Indefinite Integrals

The problem can be solved as follows. ðπ

4

π 6

ðπ

4

cscðxÞ cot ðxÞdx ¼

π 6

cos ðxÞ 1 dx ¼ sin ðxÞ sin ðxÞ

ðπ

4

π 3

cos ðxÞ dx sin 2 ðxÞ

ð1Þ

Now, we should change the variable of the integral as follows. sin ðxÞ ≜ u ⟹ cos ðxÞdx ¼ du

ð2Þ

Solving (1) and (2): ð u2 u1

!

  1 1 u2 1 π4 ¼ du ¼   ¼  u u1 sin ðxÞ π6 u2

1 1 pffiffi  1 2 2

pffiffiffi  pffiffiffi ¼ 22 ¼2 2

2

Choice (2) is the answer. 8.68. From trigonometry, we know that: 1 þ cot 2 ðxÞ ¼

1 sin 2 ðxÞ

1 þ tan 2 ðxÞ ¼

1 cos 2 ðxÞ

tan ðxÞ cot ðxÞ ¼ 1 Moreover, from the list of integral of trigonometric functions, we know that: ð ð





 1 þ tan 2 ðxÞ dx ¼ tan ðxÞ þ c

 1 þ cot 2 ðxÞ dx ¼  cot ðxÞ þ c

The problem can be solved as follows. ðπ

4

π 6

ðπ ¼

4

π 6



1 dx ¼ sin 2 ðxÞ cos 2 ðxÞ

ðπ

4

π 6

   1 þ cot 2 ðxÞ 1 þ tan 2 ðxÞ dx 

1 þ tan ðxÞ þ cot ðxÞ þ cot ðxÞ tan ðxÞ dx ¼ 2

ðπ

4

π 6



2

2

 1 þ tan 2 ðxÞ dx þ

ðπ

4

π 6



2

4

π 6



 1 þ tan 2 ðxÞ þ cot 2 ðxÞ þ 1 dx

π   1 þ cot 2 ðxÞ dx ¼ ð tan ðxÞ  cot ðxÞÞπ4

¼ ð 1  1Þ  Choice (2) is the answer.

ðπ

6

pffiffiffi pffiffiffi  3 pffiffiffi 2 3  3 ¼ 3 3

8

Solutions of Problems: Definite and Indefinite Integrals

151

8.69. From the list of integral of trigonometric functions, we know that: ð xn dx ¼

1 nþ1 x þc nþ1

The problem can be solved as follows. ð ð tan ðxÞ  cot ðxÞÞð tan ðxÞ þ cot ðxÞÞ5 dx ð ¼ ð tan ðxÞ  cot ðxÞÞð tan ðxÞ þ cot ðxÞÞð tan ðxÞ þ cot ðxÞÞ4 dx ð ¼ ð ¼





 tan 2 ðxÞ  cot 2 ðxÞ ð tan ðxÞ þ cot ðxÞÞ4 dx

  1 þ tan 2 ðxÞ  1 þ cot 2 ðxÞ ð tan ðxÞ þ cot ðxÞÞ4 dx

ð1Þ

Now, we need to change the variable of the integral as follows. d   dx  tan ðxÞ þ cot ðxÞ ≜ u) 1 þ tan 2 ðxÞ  1 þ cot 2 ðxÞ dx ¼ du

ð2Þ

Solving (1) and (2): ð u4 du ¼

u5 1 þ c ¼ ð tan ðxÞ þ cot ðxÞÞ5 þ c 5 5

Choice (2) is the answer. 8.70. The volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and x-axis, is calculated as follows: V ¼π

ð x2

ð f ðxÞÞ2 dx

x1

Moreover, from trigonometry, we know that: 1 þ cot 2 ðxÞ ¼

1 sin 2 ðxÞ

In addition, from the list of integral of trigonometric functions, we know that: ð un du ¼ ð

1 unþ1 þ c nþ1

  1 þ cot 2 ðxÞ dx ¼  cot ðxÞ

Therefore, ðπ V ¼π

2

π 4

ðπ 2   1 dx ¼ π 1 þ cot 2 ðxÞ 1 þ cot 2 ðxÞ dx 4 π ð Þ sin x 4

ð1Þ

152

8

Solutions of Problems: Definite and Indefinite Integrals

Now, we should change the variable of the integral as follows.   cot ðxÞ ≜ u ⟹  1 þ cot 2 ðxÞ dx ¼ du

ð2Þ

Solving (1) and (2): V ¼ π

ð u2 u1



 π   1 3 u2 1 3 1 þ u du ¼ π u þ u  ¼ π cot ðxÞ þ cot ðxÞ π2 3 3 u1 4 2



   1 4 V ¼ π ð0 þ 0Þ  1 þ ¼ π 3 3 Choice (4) is the answer. 8.71. From trigonometry, we know that: 1 þ cos ð2xÞ ¼ 2 cos 2 ðxÞ

ð1Þ

sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1

ð2Þ

The problem can be solved as follows. ð I¼

sin ðxÞ cos ðxÞdx

ð3Þ

Now, we should change the variable of the integral as follows. sin ðxÞ ≜ u ⟹ cos ðxÞdx ¼ du

ð4Þ

Solving (3) and (4): ð

1 1 I ¼ udu ¼ u2 þ c ¼ sin 2 ðxÞ þ c 2 2

ð5Þ

Solving (2) and (5): I¼

 1 1 1 1 1  cos 2 ðxÞ þ c ¼  cos 2 ðxÞ þ þ c ¼  cos 2 ðxÞ þ c0 2 2 2 2

ð6Þ

Solving (1) and (6): I¼

  1 1 1 1 1 1 þ cos ð2xÞ þ c0 ¼  cos ð2xÞ þ c0  ¼  cos ð2xÞ þ c00 2 2 2 4 4 4

From (5), (6), and (7), choice (2) is the answer.

Reference 1. Rahmani-Andebili, M. (2020). Precalculus: Practice problems, methods, and solutions, Springer Nature, 2020.

ð7Þ

Index

A Absolute extrema, 88, 90, 101, 104 Alternate notation, 10, 11, 35, 36 Applications of derivatives, 85, 88, 89, 95, 100, 102 Applications of integrals, 111, 113, 115, 117, 118, 121, 129, 133, 134, 136, 139, 141, 146 Application, Taylor series in limits, 64–66, 77–82 Area between curves, 111–113, 117, 123, 129–131, 138, 151 Average value, 109, 115, 122, 127, 135, 148 C Chain rule, 83–87, 91–94, 96, 99 Co-function formulas, 4, 16, 25, 47 Critical points, 88, 101 D Definite integral, 107, 108, 110, 112, 114, 116, 125–127, 130, 134, 137 Definition of derivative, 89, 103 Degrees to radians formulas, 1, 4, 9, 17, 19, 24, 33, 49 Derivatives of exponential, 83, 88, 91, 101 Derivatives of hyperbolic functions, 86, 90, 96, 104 Derivatives of inverse trigonometric functions, 85, 87, 94, 99 Derivatives of trigonometric functions, 85, 94–96 Differentiation formulas, 84, 87, 88, 93, 99, 101 Double angle formulas, 14, 41, 42 E Even and odd formulas, 5, 16, 26, 27, 48 H Half angle formulas, 4, 8, 24, 31 Higher-order derivatives, 83, 92 I Implicit differentiation, 84, 86, 88, 89, 92, 93, 97, 98, 100, 102 Indefinite integral, 107, 108, 111, 118, 123, 125, 130, 141, 151 Integrals involving quadratics, 111, 116, 128, 137 Integrals involving roots, 108–110, 114, 121, 126–128, 133, 147 Integrals involving trigonometric functions, 108, 110, 111, 115, 120, 126, 128, 135, 136, 144 Integration by parts, 109, 112, 114, 121–123, 126, 130, 133, 134, 146, 148, 150

Integration techniques, 112, 113, 118, 120, 131, 132, 140, 141, 145 Integration using partial fractions, 117, 119, 138, 143 Inverse properties, 5, 6, 9, 11, 12, 14, 27, 28, 32, 37, 38, 43 Inverse trigonometric functions, 4, 5, 7, 10, 14, 26, 30, 33, 34, 43 L Least common multiple (LCM), 21 Limits and continuity, 64, 79 Limits at infinity, 54, 55, 57, 58, 60, 61, 67, 68, 70, 71, 73, 74 Limits by direct substitution, 53, 56, 58, 67, 69, 71 Limits by factoring, 54, 55, 60–63, 68, 73–75, 77 Limits by L’Hopital’s rule, 58, 59, 71, 72 Limits by rationalization, 57, 59, 61, 63, 70, 72, 73, 76 Limits involving Euler’s number, 58, 60, 62–65, 71, 73–76, 78, 80 Limits of absolute value functions, 53–55, 58, 67, 68, 72 M Minimum and maximum values, 86, 89, 90, 97, 102, 103 P Period, 2, 3, 21, 23 Periodic formulas, 2, 4, 20, 21, 25 Product rule, 84, 93 Product to sum formulas, 3, 11, 17, 18, 24, 36, 50, 51 Pythagorean identities, 15, 17, 44, 46, 49, 50 Q Quotient rule, 87, 98 R Range, 2, 7, 12, 13, 16, 20, 30, 39, 40, 48 Rates of change, 89 Reciprocal identities, 14, 15, 17, 42, 44, 49 S Sine and cosine identities, 1, 16, 20, 47 Substitution rule for integrals, 113, 116, 123, 132, 137, 149 Sum and difference to product formulas, 7, 18, 30, 51 T Tangent and cotangent identities, 1, 19

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Calculus, https://doi.org/10.1007/978-3-030-64980-7

153

154 Trigonometric equations, 3, 8, 11, 12, 15, 16, 18, 22, 31, 32, 36, 38, 40, 45, 46, 48, 52 Trigonometric identities, 3, 8, 22, 31 Trigonometric limits, 56, 57, 59, 60, 62, 63, 65, 66, 69, 72, 73, 75–77, 79, 81 Trigonometric substitutions, 109, 116, 118–120, 122, 124, 127, 136, 140, 142–145, 149, 152

Index U Unit circle, 6, 7, 28, 29 V Volume of solid of revolution, 122, 148