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
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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
11
12
1
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
13
14
1
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
1
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
22
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
26
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
30
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
32
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
>
<x ¼
þ
3x ¼ 2kπ þ π x ⟹ 4x ¼ 2kπ þ π
2 4
⟹
>
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<1
Problems: Limits
3
Problems: Limits
1)
2)
3)
4)
59
+1
1
1
1
3.25. Determine the value of the limit below.
sin ð3xÞ þ sin ð7xÞ
3x þ tan ð2xÞ
lim
x!0
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 2
3) 1
4) 2
3.26. Calculate the value of the following limit.
pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi
xþ3 3
lim
x
x!0
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
pffiffi
1) 33
pffiffi
2) 63
pffiffi
3) 23
pffiffi
4) 93
3.27. Calculate the value of the following limit.
lim
x!0
1 cos ðxÞ
sin ðxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 1
3) 1
pffiffiffi
4) 2
3.28. Calculate the value of the limit below.
lim
x!0
5x sin ðxÞ
2x þ cos ðxÞ 1
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
60
3
1)
2)
3)
4)
1
2
1
2
3.29. Determine the value of the limit below.
lim
x!2
x3 8
þ 5x
jx 2j
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 2
2) 2
3) 1
4) 1
3.30. Calculate the value of the limit below.
lim þ
x!ðπ2Þ
sin ðxÞ þ cos ðxÞ
cos ðxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 1
3) 0
4) 1
3.31. Calculate the value of the following limit.
lim
x!0
3x4 þ 2x3
ðarcð sin ðxÞÞÞ3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 2
3) 0
4) 1
3.32. Determine the value of the limit below.
h
lim
x!1
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 2
3) 0
4) 1
i
2
x
xþ1
Problems: Limits
3
Problems: Limits
61
3.33. Calculate the value of the following limit.
x4
lim pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
x 4x þ 3
x!3þ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 1
3) 1
4) 1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
3.34. Calculate the value of lim x þ x2 þ 4x 10 .
x!1
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 2
2) 2
3) 1
4) 1
3.35. Determine the value of the limit below.
lim
x!2
4 x2
pffiffiffiffiffiffiffiffiffiffiffiffiffi
6 2 x2 þ 5
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 2
3) 3
4) 1
3.36. Calculate the value of the following limit.
sin ð2xÞ
lim pffiffiffiffiffiffiffiffiffiffiffi
xþ11
x!0
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 2
2) 4
3) 3
4) 1
3.37. Calculate the limit of
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x4 þ 2x2 þ x x2 if x ! 1.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) +1
3) 0
4) 1
62
3
3.38. Calculate the value of the limit below.
lim
2
x 9
x!3
xþ3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 6
2) 6
3) 3
4) Undefined
3.39. Calculate the value of the following limit.
lim
x!12
tan πx
2 1
cos ðπxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) 1
3) 2
4) 2
3.40. Calculate the limit of
pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi
x þ 5 x þ 1 if x ! 1.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 4
2) 2
3) 0
4) 1
3.41. Calculate the value of the following limit.
limπ
x!2
tan ð2xÞ cos ðxÞ
1 þ cos ð2xÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) 12
3) 1
4) 12
3.42. Determine the value of the following limit.
tan ð2xÞ
lim pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 cos ðxÞ
x!0
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
Problems: Limits
3
Problems: Limits
1)
2)
3)
4)
63
pffiffiffi
2 2
pffiffiffi
2
pffiffiffi
2
pffiffiffi
2 2
3.43. Calculate the value of lim
x!1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p
3
n þ 1000 3 n 20 .
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 2
2) 0
3) 10
4) 20
3.44. Calculate the value of the limit below.
limþ
x!π
sin ðπ sin ðxÞÞ sin
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ cos ðxÞ
x
2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
pffiffiffi
1) π 2
2) 2π
3) π 2
pffiffiffi
4) π 2 2
3.45. Calculate the value of the following limit.
lim
x!0
p
ffiffiffiffiffiffiffiffiffiffiffiffiffi p
ffiffiffiffiffiffiffiffiffiffiffiffiffi
3
1 þ x2 4 1 2x
2x2 þ 2x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 14
2) 14
3) 1
4) 1
3.46. Calculate the limit of the function below if x ! 0.
f ð xÞ ¼
sin 2 ðxÞ þ sin ðxÞ þ cos 2 ðxÞ cos ðxÞ
sin 2 ðxÞ sin ðxÞ þ cos 2 ðxÞ cos ðxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) 1
3) 2
4) 2
64
3
3.47. Determine the value of the following limit.
lim
x!0
cos ðmxÞ cos ðnxÞ
x2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) n2 + m2
2) n2 m2
2
2
3) n m
2
4)
n2 þm2
2
3.48. Calculate the value of the limit below.
lim
x!0
sin ðxÞ x
tan ðxÞ x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 12
2) 12
3) 14
4) 14
3.49. Calculate the value of the following limit.
lim
x!π
1 þ cos 3 ðxÞ
1 cos 2 ðxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 32
2) 32
3) 3
4) 3
3.50. Calculate the limit of the function below if n ! + 1.
3n2
f ðnÞ ¼ pffiffiffiffinffi
5
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 1
3) 1
4) p3ffiffi
5
Problems: Limits
3
Problems: Limits
65
3.51. Determine the value of the limit below.
lim
x!0
x3 sin ðxÞð1 cos ðxÞÞ
x3
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 0
2) 12
3) 1
4) 1
3.52. Calculate the value of the following limit.
lim
x!1
arcð cos xÞ
pffiffiffiffiffiffiffiffiffiffiffi
1x
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
pffiffiffi
1) 2
pffiffiffi
2) 2
pffiffi
3) 22
pffiffi
4) 22
3.53. Determine the value of n in the following equation.
1
lim x2 1 cot ðxn 1Þ ¼
2
x!1
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 8
2) 4
3) 18
4) 14
3.54. Calculate the limit of sin(4x)(cot(2x) cot (x)) if x ! 0.
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 4
2) 2
3) 2
4) 4
3.55. Calculate the value of the following limit.
lim 1
x!0
2
sin ðxÞ x
sin ð2xÞ x cos ðxÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
66
3
1)
2)
3)
4)
0
1
1
1
3.56. Calculate the limit of the following function if x ! π4.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 3 tan ðxÞ
f ð xÞ ¼
1 2 sin 2 ðxÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 13
2) 3
3) 12
4) 2
3.57. Calculate the value of the limit below.
lim
x!0
1 cos 3 ðxÞ
sin ðxÞ tan ð2xÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 4
2) 4
3) 34
4) 34
3.58. Determine the value of the following limit.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 cos ðxÞ
pffiffiffi
lim
x!0þ 1 cos ð xÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ○ Normal ● Large
1) 0
2) 1
3) 12
4) 2
Reference
1. Rahmani-Andebili, M. (2020). Precalculus: Practice problems, methods, and solutions, Springer Nature, 2020.
Problems: Limits
4
Solutions of Problems: Limits
Abstract
In this chapter, the problems of the third chapter are fully solved, in detail, step-by-step, and with different methods. 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.
4.1. The problem can be solved as follows [1].
lim
x!ð1Þþ
½x þ 1 ð1Þ þ 1
0
¼ ¼0
¼ 1 1
0
x2 1
Choice (2) is the answer.
4.2. The problem can be solved as follows.
lim
x!2þ
xþ4
2þ4
¼
¼ 1
½x 3 3 3
Choice (2) is the answer.
4.3. The problem can be solved as follows.
lim
x!0
xþ2 0þ2
¼
¼ 2
1
½ x
Choice (2) is the answer.
4.4. The problem can be solved as follows.
lim
½x þ 3x
4x
¼ lim ð2Þ ¼ 2
¼ lim
3x x!1 2x x!1
x!1 ½x
Choice (2) is the answer.
# Springer Nature Switzerland AG 2021
M. Rahmani-Andebili, Calculus, https://doi.org/10.1007/978-3-030-64980-7_4
67
68
4
Solutions of Problems: Limits
4.5. The problem can be solved as follows.
ffiffiffiffiffi
p
ffiffiffip
3 2
3
ffiffiffi
p
þ
1
x
x
3
xþ x
¼ lim
pffiffiffi ¼ lim pffiffiffipffiffiffiffiffi
lim
x!0 x 3 x
x!0 3
x!0
x 3 x2 1
pffiffiffiffiffi
3 2
x þ1
pffiffiffiffiffi
¼ 1
3 2
x 1
Choice (3) is the answer.
4.6. The problem can be solved as follows.
lim
x!0
½x 1
¼ ¼ þ1
x
0
Choice (1) is the answer.
4.7. The problem can be solved as follows.
limþ
x!0
pffiffiffi
ðx2 1Þ x
ð x 2 1Þ
1
pffiffiffi
pffiffiffi ¼
¼ 1
¼ limþ pffiffiffi
ðx x þ 1Þx x!0 ðx x þ 1Þ x 0þ
Choice (2) is the answer.
4.8. The problem can be solved as follows.
limþ
x!0
2
1 1
x 1
1
3 ¼ limþ
¼ þ ¼ 1
x x
x3
0
x!0
Choice (2) is the answer.
4.9. The problem can be solved as follows.
limþ f ðxÞ lim f ðxÞ ¼ limþ
x!1
x!1
x!1
2x
2x
2x
2x
lim
¼ lim
lim
½2x þ 2 x!1 ½2x þ 2 x!1þ 2 þ 2 x!1 1 þ 2
x
2
1 2 1
¼ limþ lim x ¼ ¼
2
3
2 3
6
x!1
x!1
Choice (2) is the answer.
4.10. The problem can be solved as follows.
lim ð½x 2Þ½x ¼ limþ ð2 2Þ 2 ¼ limþ 0 ¼ 0
x!2þ
x!2
x!2
Choice (3) is the answer.
4.11. The problem can be solved as follows.
lim
½ x 4
34
1
¼ ¼ þ1
¼ lim 16
0
16
x!4
16
x!4 x2
Choice (3) is the answer.
4
Solutions of Problems: Limits
69
4.12. From trigonometry and calculus, we know that:
arcð cos ð1 ÞÞ ¼ 0þ
d
1
ðarcð cos xÞÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
dx
1 x2
The problem can be solved as follows.
lim
x!1
d
pffiffiffiffiffiffiffiffiffiffiffiffiffi
ð 1 x3 Þ
1 x3
0þ H
3x2
¼ lim 1 ¼ lim 3x2 1 x2 ¼ 0
¼ þ ) lim d dx
x!1
x!1 pffiffiffiffiffiffiffiffi
x!1
arcð cos ðxÞÞ 0
2
dx ðarcð cos ðxÞÞÞ
1x
Choice (3) is the answer.
4.13. From the application of the Taylor series in limit, we know that:
lim tan ðxÞ x þ
x!0
x3
3
The problem can be solved as follows.
3
3
3
x þ x3 3x þ ð3x3Þ þ 2x þ ð2x3Þ
tan ðxÞ tan ð3xÞ þ tan ð2xÞ
6x3
lim
¼ lim
¼ lim ð6Þ ¼ 6
lim
3
3
x!0
x!0
x!0
x!0
x
x3
x
Choice (1) is the answer.
4.14. The problem can be solved as follows.
lim
x!3
9 x2
0 H
pffiffiffiffiffiffiffiffiffiffiffi ¼ ) lim
x!3
2 xþ1 0
d
2
pffiffiffiffiffiffiffiffiffiffiffi
2x
dx ð9 x Þ
p
ffiffiffiffiffiffiffiffiffiffi
ffi ¼ lim 1 ¼ lim 4x x þ 1 ¼ 24
d
x!3 pffiffiffiffiffiffi
x!3
xþ1
dx 2 2 xþ1
Choice (4) is the answer.
4.15. From trigonometry, we know that:
1 sin ðxÞ 1
The problem can be solved as follows.
lim
x!þ1
sin ðxÞ
1
¼ lim sin ðxÞ ¼ ðBounded quantityÞ 0 ¼ 0
x!þ1
x
x
Choice (2) is the answer.
4.16. From the application of the Taylor series in limit, we know that:
lim tan ðxÞ x
x!0
70
4
Solutions of Problems: Limits
The problem can be solved as follows.
lim
x!0
½ x2 x2
0 x2
x2
¼ lim ð1Þ ¼ 1
¼ lim
lim
x tan ðxÞ x!0 x tan ðxÞ x!0 x x x!0
Choice (2) is the answer.
4.17. From the application of the Taylor series in limit, we know that:
lim sin
x!þ1
1
1
x
x
The problem can be solved as follows.
lim x sin
x!þ1
1
1
¼ lim x ¼ lim 1 ¼ 1
x!þ1
x
x x!þ1
Choice (1) is the answer.
4.18. From calculus, we know that:
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
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
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lim
¼ lim
¼ lim ¼
lim
x!þ1 2x 1
x!þ1 2x
x!þ1 2
2
x2 þ x 1 x!þ1 x 1 þ x þ 12
2
Choice (4) is the answer.
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.
lim
x!2
x3 8
ðx3 8Þ 0þ
pffiffiffiffiffi ¼ lim
pffiffiffiffiffi ¼ 0
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
x2 9
x2 9
¼ lim þ
xþ3
xþ3
x!ð3Þ
ð1Þ
4
Solutions of Problems: Limits
75
lim
x2 9
x2 9
¼ lim ¼ lim ðx 3Þ ¼ 6
xþ3
x!ð3Þ x þ 3
x!ð3Þ
ð2Þ
x2 9
ð x 2 9Þ
¼ lim þ
¼ lim þ ðx 3Þ ¼ 6
xþ3
xþ3
x!ð3Þ
x!ð3Þ
ð3Þ
x!ð3Þ
lim
x!ð3Þþ
ð1Þ, ð2Þ, ð3Þ
x2 9
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.
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<x
4
π
π
<x<
4
2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) 12
3) 12
4) 1
5.20. The function below is differentiable everywhere on ℝ domain. Determine the value of b.
f ð xÞ ¼
ax þ b
x < 1
x þa
x 1
2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
5
Problems: Derivatives and Its Applications
1)
2)
3)
4)
87
2
1
2
3
5.21. Calculate the derivative of the function below:
f ð xÞ ¼
ð2x 1Þ2
2x2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 2x1
2x3
2) 2x1
x3
2xþ1
3) x3
4) 2xþ1
2x3
5.22. Calculate the derivative of the function below.
f ð xÞ ¼
sin ðxÞ
1 þ tan 2 ðxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 54
2) 54
3) 58
4) 58
5.23. For the function below, calculate the value of f 0(x ¼ 2).
1 f ðxÞ ¼ x2 5x þ 6 arc sin
x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) π3
2) 2π
3
3) π6
4) π4
5.24. For the following function, calculate the value of f 0(x ¼ 3).
f ðxÞ ¼ x2 þ 2x 3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
gð x þ 2Þ
ðx3 þ 1Þgð2x þ 5Þ
88
5
Problems: Derivatives and Its Applications
1) 13
2
2) 13
2
2
3) 13
2
4) 13
5.25. For what value of m the line of y ¼ 2x + 1 is tangent to a curve with the following function.
y¼
1 þ x2
mþx
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 34
pffiffi
2) 83
3) 19
pffiffi
4) 23
5.26. Determine the third derivate of f(x) ¼ x4 |x|.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 60x2
2) 60x2
3) 60x|x|
4) 60x|x|
5.27. Determine the value of the parameter of “a” if the derivative of
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 2
2) 1
3) 1
4) 2
pffiffiffiffiffiffiffiffiffiffiffi
x þ a for x ¼ 2 is 14.
pffiffiffiffiffiffiffiffiffiffi
5.28. Calculate the derivative of y ¼ ln e sin ðxÞ at x ¼ π6.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
pffiffi
1) 83
pffiffi
2) 86
pffiffi
3) 43
pffiffi
4) 46
5.29. Determine the maximum value of the function of y(x) ¼ x3 3x2 9x + 5 in the range of [2, 2].
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 9
2) 10
3) 12
4) 17
5
Problems: Derivatives and Its Applications
89
5.30. Which one of the choices is correct about the function below in its one period.
yð xÞ ¼
1 sin ðxÞ
cos ðxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) The function is always ascending.
2) The function is always descending.
3) The function has one minimum point.
4) The function has one maximum point.
5.31. Determine the value of f 0(x)g(x) f(x)g0(x) if we have the following functions.
f ð xÞ ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffi
5
1
1 þ x2 x , gðxÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
5
1 þ x2 þ x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) 0
3) 1
4) 2
5.32. On a curve with the function of y ¼ x3 6x + 12, two tangent lines, parallel to x-axis, have been drawn. Determine the
distance between these two lines.
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 14
2) 6
pffiffiffi
3) 4 2
pffiffiffi
4) 8 2
5.33. For the function below, calculate the value of f 0(x ¼ 1).
8
5
< ð x þ 1Þ
f ð xÞ ¼
j x þ 1j
:
0
x 6¼ 1
x ¼ 1
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 0
2) 1
3) 1
4) 5
pffiffiffiffiffiffiffiffiffiffiffi
5.34. The point M(x, y) is moving on the curve of y ¼ x þ 8. Determine the changing rate of the distance of the point from
the origin when x ¼ 7.
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
90
5
1)
2)
3)
4)
Problems: Derivatives and Its Applications
15
16
15
8
3
7
5
4
5.35. Determine the derivative of f
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
jxj þ 3 if we have the relation below.
lim
x!2
f ðxÞ f ð2Þ
1
¼
x2
3
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 16
1
2) 12
3) 16
1
4) 12
5.36. Determine the value of f 0(1) if we have:
f ð xÞ ¼
ðx þ 1ÞhðxÞ
, hð1Þ 6¼ 0
ð2x þ 1Þhð2x þ 1Þ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 2
2) 1
3) 1
4) 2
5.37. Determine the value of the parameter of “a” so that the function of f ðxÞ ¼ cos 2 ðxÞ þ
point with the width of y ¼ 34 in the range of 0 < x < π2.
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) 12
3) 12
4) 1
pffiffiffi
3 sin ðxÞ þ a has an extremum
Reference
1. Rahmani-Andebili, M. (2020). Precalculus: Practice problems, methods, and solutions, Springer Nature, 2020.
6
Solutions of Problems: Derivatives
and its Applications
Abstract
In this chapter, the problems of the fifth chapter are fully solved, in detail, step-by-step, and with different methods. 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.
6.1. From the list of derivative rules, we know that [1]:
f ð xÞ ¼ ex ⟹ f 0 ð xÞ ¼ ex
f ðxÞ ¼ uðxÞvðxÞ ⟹ f 0 ðxÞ ¼ u0 ðxÞvðxÞ þ uðxÞv0 ðxÞ
Based on the information given in the problem, we have:
f ðxÞ ¼ xex ex
The problem can be solved as follows.
f 0 ðxÞ ¼ ex þ xex ex ¼ xex
f 0 ð1Þ ¼ 1e1 ⟹ f 0 ð1Þ ¼ e
Choice (4) is the answer.
6.2. From the list of derivative rules, we know that:
hðxÞ ¼ gðuðxÞÞ ⟹ h0 ðxÞ ¼ u0 ðxÞg0 ðuðxÞÞ
Based on the information given in the problem, we have:
f 0 ð 1Þ ¼ 2
f ðxÞ þ g x3 ¼ 5x 1
# Springer Nature Switzerland AG 2021
M. Rahmani-Andebili, Calculus, https://doi.org/10.1007/978-3-030-64980-7_6
91
92
6 Solutions of Problems: Derivatives and its Applications
The problem can be solved as follows.
d
f 0 ð 1Þ ¼ 2
3
dx
x¼1 0
0
2 0 3
0
f ðxÞ þ g x ¼ 5x 1 ) f ðxÞ þ 3x g x ¼ 5 ¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) f ð1Þ þ 3g ð1Þ ¼ 5 ¼
¼
¼
¼
¼
¼
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 þ 3g0 ð1Þ ¼ 5
⟹ g0 ð 1Þ ¼ 1
Choice (1) is the answer.
6.3. From the list of derivative rules, we know that:
f ðxÞ ¼ axn ⟹ f 0 ðxÞ ¼ anxn1
A function is ascending in a given range if its derivative is positive. Therefore,
yðxÞ ¼ 1 4x2 ⟹ y0 ðxÞ ¼ 8x > 0 ⟹ x < 0
Choice (1) is the answer.
6.4. From the list of derivative rules, we know that:
f ð xÞ ¼
pffiffiffi
1
x ⟹ f 0 ðxÞ ¼ pffiffiffi
2 x
First, we should simplify the function as follows:
f ð xÞ ¼
pffiffiffi pffiffiffi pffiffiffi
pffiffiffi
x ð x 1Þ
x x
pffiffiffi ¼
pffiffiffi ¼ x
1 x
1 x
Therefore,
1
1
1
¼ qffiffi ¼ 1
⟹ f 0 ðxÞ ¼ pffiffiffi ⟹ f 0
4
2 x
2 1
4
Choice (1) is the answer.
6.5. From the list of derivative rules, we know that:
f ðxÞ ¼ un ðxÞ ⟹ f 0 ðxÞ ¼ nu0 ðxÞun1 ðxÞ
Therefore
10
f ðxÞ ¼ x100 þ x50 þ 50x2 þ 50x þ 1
9
⟹ f 0 ðxÞ ¼ 10 100x99 þ 50x49 þ 100x þ 50 x100 þ x50 þ 50x2 þ 50x þ 1
⟹ f 0 ð0Þ ¼ 10ð0 þ 0 þ 0 þ 50Þð0 þ 0 þ 0 þ 0 þ 1Þ9 ¼ 500
Choice (4) is the answer.
6
Solutions of Problems: Derivatives and its Applications
93
6.6. From the list of derivative rules, we know that:
f ðxÞ ¼ tan n ðuðxÞÞ ⟹ f 0 ðxÞ ¼ nu0 ðxÞ 1 þ tan 2 ðuðxÞÞ tan n1 ðuðxÞÞ
Therefore,
f ðxÞ ¼ tan 3 ð2xÞ ⟹ f 0 ðxÞ ¼ 3 2 1 þ tan 2 ð2xÞ tan 2 ð2xÞ
⟹ f0
π
π
π
1
1 8
¼ 3 2 1 þ tan 2
tan 2
¼6 1þ
¼
12
6
6
3
3 3
Choice (3) is the answer.
6.7. Based on the information given in the problem, we have:
f ðxÞ ¼ x3 3x þ a
The derivative of an absolute value equation does not exist at its roots. Therefore, we need to solve the equation below:
f ð 2Þ ¼ 0
⟹ 23 3 2 þ a ¼ 0 ⟹ 8 6 þ a ¼ 0 ⟹ 2 þ a ¼ 0 ⟹ a ¼ 2
Choice (2) is the answer.
6.8. From the list of derivative rules, we know that:
f ð x Þ ¼ j uð x Þ j ⟹ f 0 ð x Þ ¼
u0 ðxÞuðxÞ
j uð x Þ j
Based on the information given in the problem, we have:
f ð x Þ ¼ x 2 6
Therefore,
⟹ f 0 ð xÞ ¼
2x ðx2 6Þ
4 ð 4 6Þ
⟹ f 0 ð 2Þ ¼
¼ 4,
j4 6j
j x 2 6j
f 0 ð 4Þ ¼
⟹ f 0 ð2Þ þ f 0 ð4Þ ¼ 4 þ 8 ¼ 4
Choice (4) is the answer.
6.9. From the list of derivative rules, we know that:
f ðxÞ ¼ gðhðxÞÞ ⟹ f 0 ðxÞ ¼ h0 ðxÞg0 ðhðxÞÞ
Based on the information given in the problem, we have:
f 0 ð xÞ ¼
5
x
8 ð16 6Þ
¼8
j16 6j
94
6 Solutions of Problems: Derivatives and its Applications
The problem can be solved as follows.
5 0
0 25
5
¼ 5x4 f 0 x5 ¼ 5x4 5 ⟹ f x5 ¼
f x
x
x
Choice (3) is the answer.
6.10. From the list of derivative rules, we have:
ð f ðgðxÞÞÞ0 ¼ g0 ðxÞ f 0 ðgðxÞÞ
Based on the information given in the problem, we have:
ð f ð sin ðxÞÞÞ0 ¼ cos 3 ðxÞ
The problem can be solved as follows.
ð f ð sin ðxÞÞÞ0 ¼ cos 3 ðxÞ ⟹ cos ðxÞ f 0 ð sin ðxÞÞ ¼ cos 3 ðxÞ ⟹ f 0 ð sin ðxÞÞ ¼ cos 2 ðxÞ
⟹ f 0 ð sin ðxÞÞ ¼ 1 sin 2 ðxÞ ⟹ f 0 ðxÞ ¼ 1 x2
Choice (2) is the answer.
6.11. From the list of derivative rules, we know that:
f ðxÞ ¼ arcð tan ðuðxÞÞÞ ⟹ f 0 ðxÞ ¼
u0 ð x Þ
1 þ u2 ð xÞ
Therefore,
f ðxÞ ¼ arcð tan ð3xÞÞ ⟹ f 0 ðxÞ ¼
f0
3
1 þ 9x2
1
3
3
¼
12 ¼ 2
3
1þ9
3
Choice (1) is the answer.
6.12. From the list of derivative rules, we know that:
f ðxÞ ¼ axn ⟹ f 0 ðxÞ ¼ anxn1
f ðxÞ ¼ gðhðxÞÞ ⟹ f 0 ðxÞ ¼ h0 ðxÞg0 ðhðxÞÞ
Based on the information given in the problem, we have:
f
pffi 1
þ g t ¼ t2 þ 1
t
g0 ð1Þ ¼ 5
6
Solutions of Problems: Derivatives and its Applications
95
The problem can be solved as follows.
f
d
pffi pffi dx
1
1
1
1
þ g t ¼ t2 þ 1 ) 2 f 0
þ pffi g0 t ¼ 2t
t
t
t
2 t
g0 ð 1Þ ¼ 5
1
5
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) f 0 ð1Þ þ ¼ 2 ⟹ f 0 ð1Þ ¼
t ¼ 1 ⟹ f 0 ð1Þ þ g0 ð1Þ ¼ 2 ¼
2
2
2
Choice (3) is the answer.
6.13. From the derivative rules, we know that:
f ðx, yÞ ¼ 0 ⟹ y0x ¼ d
f 0x ðx, yÞ
dx f ðx, yÞ
¼
d
f 0y ðx, yÞ
dy f ðx, yÞ
Based on the information given in the problem, we have:
2 cos ðyÞ sin ðx þ yÞ þ 2 ¼ 0
The problem can be solved as follows.
ð2 cos ðyÞ sin ðx þ yÞ þ 2Þ
cos ðx þ yÞ
cos ðx þ yÞ
¼
¼
2
sin
ð
y
Þ
cos
ð
x
þ
y
Þ
2
sin
ð
y
Þ þ cos ðx þ yÞ
ð
2
cos
ð
y
Þ
sin
ð
x
þ
y
Þ
þ
2
Þ
dy
d
y0x ¼ dx
d
ðx, yÞ ¼ ð0, π Þ 0
1
) y0x 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) yx ¼ 01
Choice (3) is the answer.
6.14. From the derivative rules, we know that:
f ðx, yÞ ¼ 0 ⟹ y0x ¼ d
f ðx, yÞ
f 0x ðx, yÞ
¼ dx
0
d
f y ðx, yÞ
dy f ðx, yÞ
Based on the information given in the problem, we have:
x3 þ y3 ¼ 16
The problem can be solved as follows.
y0x ¼ y0 ¼ 3x2
x2
¼
3y2
y2
2
x2
y0 ¼ 2
2
2xy
2y
xy2 x2
y 00
2xðy3 þ x3 Þ
2xy 2yy x
2xy3 þ 2x4
00
⟹y ¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) y ¼ ¼
¼
4
4
5
y
y
y5
y
2
0 2
x3 þ y3 ¼ 16 00
2x 16
32x
) y00 ¼ 5
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) y ¼ y5
y
Choice (4) is the answer.
96
6 Solutions of Problems: Derivatives and its Applications
6.15. From the derivative rules, we know that:
(
y ¼ yðt Þ
⟹ y0x ¼
x ¼ xðt Þ
y0t dtd yðt Þ
¼
x0t dtd xðt Þ
Based on the information given in the problem, we have:
x ¼ 2 þ 3 sin ðt Þ
y ¼ 3 2 cos ðt Þ
Hence,
y0x ¼
y0t 2 sin ðt Þ 2
¼
¼ tan ðt Þ
x0t 3 cos ðt Þ 3
pffiffiffi
pffiffiffi
3
π
2
2 3
0
0
⟹ yx ¼
t ¼ ⟹ yx ¼ 3
6
3
9
Choice (1) is the answer.
6.16. From the derivative rules, we know that:
(
y ¼ yð t Þ
x ¼ xð t Þ
⟹ y0x ¼
y0t dtd yðt Þ 0 x0t dtd xðt Þ
,x ¼ ¼
¼
x0t dtd xðt Þ y y0t dtd yðt Þ
Based on the information given in the problem, we have:
x ¼ t2 þ t
y ¼ t 2 2t
Therefore,
8
y0 2t 2
>
>
> 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<0
x0
x3
x
x<0
3
⟹ f ð0 Þ ¼ f ð0þ Þ
ð1Þ
Now, we can determine its first and second derivatives as follows:
⟹ f 0 ð xÞ ¼
⟹ f 00 ðxÞ ¼
3x
x0
3x2
x<0
2
6x
x0
6x
x<0
⟹ f 0 ð0Þ ¼
⟹ f 00 ð0Þ ¼
0
x0
0
x<0
0
x0
0
x<0
⟹ f 0 ð0 Þ ¼ f 0 ð0þ Þ
ð2Þ
⟹ f 00 ð0 Þ ¼ f 00 ð0þ Þ
ð3Þ
From (1) and (2), we can conclude that the first derivative of the function exits. Likewise, from (1), (2), and (3), we can
say that the second derivative of the function does exist. Choice (4) is the answer.
6.19. Based on the information given in the problem, we have:
f ð xÞ ¼
8
< sin 2 ðxÞ cos ð2xÞ
: a tan ðxÞ þ b sin ð2xÞ
π
0<x
4
π
π
<x<
4
2
ð1Þ
The function is differentiable at x ¼ π4. Therefore, we can conclude that:
f
þ
π
π
¼f
4
4
ð2Þ
f0
þ
π
π
¼ f0
4
4
ð3Þ
Solving (1) and (2):
sin 2
π
π
π
π
1
cos 2 ¼ a tan
þ b sin 2 ⟹aþb¼
4
4
4
4
2
ð4Þ
98
6 Solutions of Problems: Derivatives and its Applications
Solving (1) and (3):
2 sin
π
π
π
π
π
cos
þ 2 sin 2 ¼ a 1 þ tan 2
þ 2b cos 2 4
4
4
4
4
⟹ 1 þ 2 ¼ 2a þ 0 ⟹ a ¼
3
2
ð5Þ
Solving (4) and (5):
b ¼ 1
Choice (1) is the answer.
6.20. Based on the information given in the problem, we have:
f ð xÞ ¼
ax þ b
x2 þ a
x < 1
x 1
ð1Þ
The function is differentiable everywhere on ℝ domain including x ¼ 1. Hence:
f ðð1Þ Þ ¼ f ð1Þþ
ð2Þ
f 0 ðð1Þ Þ ¼ f 0 ð1Þþ
ð3Þ
a þ b ¼ 1 þ a ⟹ 2a þ b ¼ 1
ð4Þ
a ¼ 2
ð5Þ
Solving (1) and (2):
Solving (1) and (3):
Solving (4) and (5):
b ¼ 3
Choice (4) is the answer.
6.21. From 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
Therefore,
f ð xÞ ¼
ð2x 1Þ2
4ð2x 1Þð2x2 Þ 4xð2x 1Þ2
0
⟹
f
ð
x
Þ
¼
2x2
4x4
⟹ f 0 ðxÞ ¼
Choice (2) is the answer.
16x3 8x2 16x3 þ 16x2 4x 8x2 4x 2x 1
¼
¼
4x4
4x4
x3
6
Solutions of Problems: Derivatives and its Applications
99
6.22. From the list of derivative rules, we know that:
f ðxÞ ¼ uðxÞvðxÞ ⟹ f 0 ðxÞ ¼ u0 ðxÞvðxÞ þ uðxÞv0 ðxÞ
From trigonometry, we have:
1 þ tan 2 ðxÞ ¼
1
cos 2 ðxÞ
Based on the information given in the problem, we have:
f ðxÞ ¼
sin ðxÞ
sin ðxÞ
⟹ f ðxÞ ¼ 1 ¼ sin ðxÞ cos 2 ðxÞ
1 þ tan 2 ðxÞ
cos 2 ðxÞ
⟹ f 0 ðxÞ ¼ cos ðxÞ cos 2 ðxÞ sin ðxÞ 2 sin ðxÞ cos ðxÞ ¼ cos 3 ðxÞ 2 sin 2 ðxÞ cos ðxÞ
⟹ f0
pffiffiffi
3
3
π
1
¼
2
2
3
2
2
1 1 3
5
¼ ¼
2 8 4
8
Choice (4) is the answer.
6.23. From the list of derivative rules, we know that:
f ðxÞ ¼ uðxÞvðxÞ ⟹ f 0 ðxÞ ¼ u0 ðxÞvðxÞ þ uðxÞv0 ðxÞ
f ðxÞ ¼ axn ⟹ f 0 ðxÞ ¼ anxn1
From trigonometry, we know that:
1
π
¼
arc sin
2
6
Based on the information given in the problem, we have:
1 f ðxÞ ¼ x2 5x þ 6 arc sin
x
Therefore,
10
1
f 0 ðxÞ ¼ ð2x 5Þarc sin
þ x2 5x þ 6 arc sin
x
x
0
1
1
π
π
þ 0 arc sin
f 0 ð2Þ ¼ ð2 2 5Þarc sin
¼ ð1Þ ⟹ f 0 ð2Þ ¼ 2
x
6
6
0
As can be seen, we did not need to calculate the value of arc sin 1x
. Choice (3) is the answer.
6.24. From the list of derivative rules, we know that:
f ðxÞ ¼ uðxÞvðxÞ ⟹ f 0 ðxÞ ¼ u0 ðxÞvðxÞ þ uðxÞv0 ðxÞ
100
6 Solutions of Problems: Derivatives and its Applications
f ðxÞ ¼ axn ⟹ f 0 ðxÞ ¼ anxn1
Based on the information given in the problem, we have:
f ðxÞ ¼ x2 þ 2x 3 gð x þ 2Þ
≜uðxÞ vðxÞ
ðx3 þ 1Þgð2x þ 5Þ
The problem can be solved as follows.
f 0 ðxÞ ¼ ð2x þ 2Þ
) f 0 ðx ¼ 3Þ ¼ ð4Þ ð x3
gð x þ 2Þ
þ x2 þ 2x 3 v0 ðxÞ
þ 1Þgð2x þ 5Þ
gð1Þ
1
þ ð0Þ v0 ðx ¼ 3Þ ¼ 4 26
26gð1Þ
) f 0 ðx ¼ 3Þ ¼
2
13
As can be seen, we did not need to calculate the value of v0(x ¼ 3). Choice (4) is the answer.
6.25. Since the line is tangent to the curve, equating their equations and solving them will result in a new equation that will
have repeated roots. In other words, the discriminant of the new equation must be zero (Δ ¼ 0).
1 þ x2
¼ 2x þ 1 ⟹ 1 þ x2 ¼ 2x2 þ x þ 2mx þ m ⟹ x2 þ ð2m þ 1Þx þ m þ 1 ¼ 0
mþx
pffiffiffi
3
Δ ¼ 0 ⟹ ð2m þ 1Þ 4ðm þ 1Þ ¼ 0 ⟹ 4m 3 ¼ 0 ⟹ m ¼ 2
2
2
Choice (4) is the answer.
6.26. From the list of derivative rules, we know that:
f ðxÞ ¼ axn ⟹ f 0 ðxÞ ¼ anxn1
Based on the information given in the problem, we have:
f ðxÞ ¼ x4 jxj
Therefore,
(
f ð xÞ ¼
8 4
8
,x > 0
,x > 0
5x
20x3
>
>
>
>
<
<
x ,x 0
, x ¼ 0 ⟹ f 00 ðxÞ ¼ 0
,x ¼ 0
⟹ f 0 ðxÞ ¼ 0
>
>
>
>
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):
0pffiffiffiffiffiffiffiffiffiffiffiffiffi
!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
8
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