Mehdi Rahmani-Andebili
Calculus I
Practice Problems, Methods, and Solutions
Second Edition
Calculus I
Mehdi Rahmani-Andebili
Calculus I
Practice Problems, Methods, and Solutions
Second Edition
Mehdi Rahmani-Andebili
Electrical Engineering Department
Arkansas Tech University
Russellville, AR, USA
ISBN 978-3-031-45027-3
ISBN 978-3-031-45028-0
https://doi.org/10.1007/978-3-031-45028-0
(eBook)
# The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021,
2023
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Preface
Calculus is one of the most important courses of many majors, including engineering and
science, and even some non-engineering majors like economics and business, which is taught in
three successive courses at universities and colleges worldwide. Moreover, in many universities
and colleges, a precalculus course is mandatory for under-prepared students as the prerequisite
course of Calculus 1.
Unfortunately, some students do not have a solid background and knowledge in math and
calculus when they start their education in universities or colleges. This issue prevents them
from learning calculus-based courses such as physics and engineering courses. Sometimes, the
problem escalates, so they give up and leave the university. Based on my real professorship
experience, students do not have a serious issue comprehending physics and engineering
courses. In fact, it is the lack of enough knowledge of calculus that hinders them from
understanding those courses.
Therefore, a series of calculus textbooks covering Precalculus, Calculus 1, Calculus 2, and
Calculus 3 have been prepared to help students succeed in their major. This book, Calculus 1:
Practice Problems, Methods, and Solutions, is the second edition of the book Calculus:
Practice Problems, Methods, and Solutions, which was published in 2021. In the new version
of the book, many new problems have been added to each chapter. The subjects of the calculus
series books are as follows.
Precalculus: Practice Problems, Methods, and Solution
.
.
.
.
.
.
.
Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities
Systems of Equations
Quadratic Equations
Functions, Algebra of Functions, and Inverse Functions
Factorization of Polynomials
Trigonometric and Inverse Trigonometric Functions
Arithmetic and Geometric Sequences
Calculus 1: Practice Problems, Methods, and Solution
.
.
.
.
.
Characteristics of Functions
Trigonometric Equations and Identities
Limits and Continuities
Derivatives and Their Applications
Definite and Indefinite Integrals
v
vi
Preface
Calculus 2: Practice Problems, Methods, and Solution
.
.
.
.
Applications of Integration
Sequences and Series and Their Applications
Polar Coordinate System
Complex Numbers
Calculus 3: Practice Problems, Methods, and Solution
.
.
.
.
.
.
Linear Algebra and Analytical Geometry
Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System
Multivariable Functions
Double Integrals and their Applications
Triple Integrals and their Applications
Line Integrals and Their Applications
The textbooks include basic and advanced calculus problems with very detailed problem
solutions. They can be used as practicing study guides by students and as supplementary
teaching sources by instructors. Since the problems have very detailed solutions, the textbooks
are helpful for under-prepared students. In addition, they are beneficial for knowledgeable
students because they include advanced problems.
In preparing the problems and solutions, care has been taken to use methods typically found
in the primary instructor-recommended textbooks. By considering this key point, the textbooks
are in the direction of instructors’ lectures, and the instructors will not see any untaught and
unusual problem solutions in their students’ answer sheets.
To help students study in the most efficient way, the problems have been categorized into
nine different levels. In this regard, for each problem, a difficulty level (easy, normal, or hard)
and a calculation amount (small, normal, or large) have been assigned. Moreover, problems
have been ordered in each chapter from the easiest problem with the smallest calculations to the
most difficult problems with the largest ones. Therefore, students are suggested to start studying
the textbooks from the easiest problems and continue practicing until they reach the normal and
then the hardest ones. This classification can also help instructors choose their desirable
problems to conduct a quiz or a test. Moreover, the classification of computation amount can
help students manage their time during future exams, and instructors assign appropriate
problems based on the exam duration.
Russellville, AR, USA
Mehdi Rahmani-Andebili
The Other Works Published by the Author
The author has already published the books and textbooks below with Springer Nature.
Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2023.
Calculus III – Practice Problems, Methods, and Solutions, Springer Nature, 2023.
Calculus II – Practice Problems, Methods, and Solutions, Springer Nature, 2023.
Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2023.
Planning and Operation of Electric Vehicles in Smart Grid, Springer Nature, 2023.
Applications of Artificial Intelligence in Planning and Operation of Smart Grid, Springer
Nature, 2022.
AC Electric Machines – Practice Problems, Methods, and Solutions, Springer Nature, 2022.
DC Electric Machines, Electromechanical Energy Conversion Principles, and Magnetic Circuit
Analysis- Practice Problems, Methods, and Solutions, Springer Nature, 2022.
Applications of Fuzzy Logic in Planning and Operation of Smart Grids, Springer Nature, 2021.
Differential Equations – Practice Problems, Methods, and Solutions, Springer Nature, 2022.
Feedback Control Systems Analysis and Design – Practice Problems, Methods, and Solutions,
Springer Nature, 2022.
Power System Analysis – Practice Problems, Methods, and Solutions, Springer Nature, 2022.
Advanced Electrical Circuit Analysis – Practice Problems, Methods, and Solutions, Springer
Nature, 2022.
Design, Control, and Operation of Microgrids in Smart Grids, Springer Nature, 2021.
Applications of Fuzzy Logic in Planning and Operation of Smart Grids, Springer Nature, 2021.
Operation of Smart Homes, Springer Nature, 2021.
AC Electrical Circuit Analysis – Practice Problems, Methods, and Solutions, Springer
Nature, 2021.
Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
DC Electrical Circuit Analysis – Practice Problems, Methods, and Solutions, Springer
Nature, 2020.
Planning and Operation of Plug-in Electric Vehicles: Technical, Geographical, and Social
Aspects, Springer Nature, 2019.
vii
Contents
1
Problems: Characteristics of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
2
Solutions of Problems: Characteristics of Functions . . . . . . . . . . . . . . . . . . . .
13
3
Problems: Trigonometric Equations and Identities . . . . . . . . . . . . . . . . . . . . .
39
4
Solutions of Problems: Trigonometric Equations and Identities . . . . . . . . . . .
61
5
Problems: Limits and Continuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
6
Solutions of Problems: Limits and Continuities . . . . . . . . . . . . . . . . . . . . . . . . 119
7
Problems: Derivatives and Their Applications . . . . . . . . . . . . . . . . . . . . . . . . 139
8
Solutions of Problems: Derivatives and Their Applications . . . . . . . . . . . . . . . 151
9
Problems: Definite and Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
10
Solutions of Problems: Definite and Indefinite Integrals . . . . . . . . . . . . . . . . . 195
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
1
Problems: Characteristics of Functions
Abstract
In this chapter, the basic and advanced problems of functions and inverse functions, algebra of functions, characteristics of
functions such as domain of functions, range of functions, axis of symmetry of functions, and types of functions in terms of
being odd or even are presented. To help students study the chapter in the most efficient way, the problems are categorized
in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and
large). Moreover, the problems are ordered from the easiest problem with the smallest computations to the most difficult
problems with the largest calculations.
1.1 Determine the reflection of the graph of the function below with respect to the origin [1, 2].
y ¼ log
x-1
xþ1
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
x-1
1) y ¼ log
xþ1
1-x
2) y ¼ log
1þx
1þx
3) y ¼ log
1-x
xþ1
4) y ¼ log
x-1
1.2 Determine the relation of f (g(h(x))) for the information below.
f ðxÞ ¼ ln x9
gð x Þ ¼
p
3
x2
h ð x Þ ¼ ex
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 6x
2) 9x
3) 10x
4) e9x
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_1
1
2
1
1.3 If f
1
x
¼
2x - 1
x2
and g(x) ¼ 2cos2(x), calculate the value of fog
π
3
:
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
1
2)
2p
3
3)
2
4) 2
p
1.4 Determine the value of fog(3) if f ðxÞ ¼ x - 2 and g(x) ¼ x + 1.
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 1
3) 2
4) 3
1.5 If f (x) ¼ 2x - 2 and g(x) ¼ x2 - 1, solve the equation of fog(x) ¼ 0.
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
p
1) ± 2
2) ±2
p
3) ± 3
p
4) 2
1.6 In the function below, calculate the value of f (-2) + f (2).
f ð xÞ ¼
2x2 þ 4
x≥2
½ x - 4
x<2
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 8
2) 6
3) 10
4) 5
p
1.7 Determine the domain of y ¼ 2 - x2 .
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
p
p
1) x ≤ - 2, x ≥ 2
p
p
2) - 2 ≤ x ≤ 2
3) x ¼ 0
p
p
4) - 2 < x < 2
Problems: Characteristics of Functions
1
Problems: Characteristics of Functions
3
1.8 Calculate fog(x) if f (x) ¼ 1 - x2 and g (x) ¼ sin (x).
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) cos2(x)
2) cos(x)
3) sin(1 - x2)
4) sin(cos(x))
1
1.9 What is the inverse function of f ðxÞ ¼ ?
x
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) x
1
2)
x
1
3) px
4) x
1.10 Which one of the following choices is correct about the graph of the relation below?
sinhðxyÞ - coshðxyÞ þ 1 ¼ 0
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) The graph is symmetric with respect to y-axis.
2) The graph is symmetric with respect to the origin and y-axis.
3) The graph is symmetric with respect to x-axis.
4) The graph is symmetric with respect to the origin.
1-x
.
1þx
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1þx 2
1)
1-x
2) 1
3) x
1-x 2
4)
1þx
1.11 Calculate fof (x) if f ðxÞ ¼
1.12 Calculate the inverse function of the following function if x ≥ 1.
f ðxÞ ¼ x4 - 2x2 þ 1
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
p
1) f - 1 ðxÞ ¼ 1 þ x
p
2) f - 1 ðxÞ ¼ 1 - x
p
3) f - 1 ðxÞ ¼ - 1 þ x
p
4) f - 1 ðxÞ ¼ - 1 - x
4
1
Problems: Characteristics of Functions
1.13 Calculate the inverse function of the function below.
f ð xÞ ¼
2ex þ 1
ex - 3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
3x - 1
1) f - 1 ðxÞ ¼
xþ2
x-2
-1
2) f ðxÞ ¼
3x - 1
x-2
3) f - 1 ðxÞ ¼
3x þ 1
3x þ 1
4) f - 1 ðxÞ ¼
x-2
1.14 Consider the function below.
f ðxÞ ¼
4x þ 5
2x - 3
Which one of the following choices includes a quantity that is not in the domain of f -1 (x)?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) -2
2) 2
5
3)
3
3
4)
2
1.15 Calculate the value of cos(π sinh (ln3)).
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) -1
1
2) p2
2
3)
2
p
3
4)
2
1.16 Determine the inverse function of f (x) ¼ x2 - 2x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
p
1) 1 þ x þ 1
p
2) 1 - x þ 1
p
3) 1 þ x - 1
p
4) 1 - x - 1
1.17 Determine the value of f (x) if f (x + 1) ¼ x2 - 2x + 1.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1
Problems: Characteristics of Functions
1)
2)
3)
4)
5
(x - 2)2
(x - 1)2
x2 - 2x
(x + 2)2
1.18 Which one of the terms below is not a function?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) y2 ¼ x
2) y3 ¼ x
p
3) y ¼ x2
4) y ¼
x2
x≥0
1
x<0
p
p
1.19 If f ð xÞ ¼ x þ x, calculate the value of f (2) + f (1).
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 6
2) 7
3) 8
4) 9
1.20 Calculate the value of f ( f (0)) if:
f ðxÞ ¼
x2 þ 1
2x þ 3
x≥1
x<1
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 3
2) 5
3) 10
4) 26
1.21 Determine the domain of the function below.
f ðxÞ ¼
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) ℝ
2) x ≤ 1
3) x ≥ 1
4) -1 ≤ x ≤ 1
1 - jxj
1 þ j xj
6
1
1.22 Determine the domain of the function below.
f ðxÞ ¼
x-1
þ
x-3
2-x
x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) (0,1]
2) [0,1]
3) (0,2]
4) (1,3)
1.23 Determine the domain of the function below.
p
x
f ð xÞ ¼
j xj - 1
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) ℝ
2) [0, 1) - {1}
3) ℝ - {1}
4) [0, 1)
1.24 Determine the domain of the function below.
p
xþ1
f ðxÞ ¼ p
x xþ1
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) ℝ - {0}
2) [1, 1)
3) [0, 1)
4) ℝ - {-1, 0}
1.25 Which number does not exist in the domain of the function below?
f ðxÞ ¼
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) -2
2) 1
3) 2
4) 0
1.26 Determine g (x) if f (x) ¼ 2x and f ( g(x)) ¼ 2x + 2.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1-x
4x þ x3
Problems: Characteristics of Functions
1
Problems: Characteristics of Functions
1)
2)
3)
4)
x-1
x+2
x+1
x-2
1.27 Determine g (x) if f (x) ¼ x - 1 and f (g (x)) ¼ x.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) x + 1
2) x2 - x
3) 2x - 1
4) x2 - 1
p
1.28 Calculate the inverse function of f ðxÞ ¼ 1 - x.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) f -1(x) ¼ 1 - x2, x ≥ 0
1
2) f - 1 ðxÞ ¼ p
1
p -x
3) f - 1 ðxÞ ¼ 1 þ x
4) f -1(x) ¼ 1 - x2
1.29 What is the inverse function of f (x) ¼ sin (x) - 2.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 2arc(sin(x))
2) -2arc(sin(x))
3) arc(sin(x - 2))
4) arc(sin(x + 2))
1.30 Calculate the inverse function of fog (x) if f (x) ¼ 3x - 2 and g (x) ¼ 2 + x.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1
4
1) x 3
3
2) 3x - 4
1
4
3) x þ
3
3
4) 3x + 4
1.31 Calculate the inverse function of f (x) ¼ x3 + 3x2 + 3x + 2.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
p
1) 1 - 3 x - 1
p
2) 1 - 3 x þ 1
p
3) - 1 þ 3 x - 1
p
4) - 1 - 3 x þ 1
1.32 Calculate the value of tanh (lnx) in which x > 0.
○ Easy ● Normal ○ Hard
Difficulty level
Calculation amount ○ Small ● Normal ○ Large
7
8
1
1) -1
2) 1
x2 - 1
3) 2
x þ1
x2
4) 2
x þ1
1.33 If f (x) + x f (-x) ¼ x2 + 1, then what is the value of f (2)?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) -1
2) -2
3) 3
4) 4
1.34 Calculate the value of f ( f ( f (2 cos (x)))) if f (x) ¼ x2 - 2.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 2 sin8(x)
2) 2 cos8(x)
3) 2 sin (8x)
4) 2 cos (8x)
1.35 Determine the domain of the following function.
f
p
x-1
¼ 2x - 1
x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) [-1, 0)
2) [-1, 1]
3) [-1, 1)
4) [1, 1)
p
1.36 Determine the domain of f ðxÞ ¼ 1 - x - 1.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) x ≥ 1
2) 1 ≤ x ≤ 2
3) x ≤ 2
5
7
4) ≤ x ≤
4
4
1.37 Determine the domain of the function of f ðxÞ ¼ jxj - 1 þ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) ℝ - [-1, 1]
2) ℝ
3) [-1, 1]
4) ℝ - (-1, 1)
jxj þ 1
Problems: Characteristics of Functions
1
Problems: Characteristics of Functions
9
1.38 Calculate the value of f (3) if:
f xþ
1
1
¼ x2 þ 2
x
x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
28
1)
3
1
2)
7
3) 7
3
4)
28
1.39 For what value of a, the function of f (x) ¼ jx + 2j + ajx - 2j is even?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) -1
2) 0
3) 1
4) 2
1.40 The function of f (x) ¼ x2 + (A - 1) x and g (x) ¼ (B + 2) x2 + sin (x) are even and odd functions, respectively. Calculate
the value of A + B.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ○ Normal ● Large
1) -2
2) -1
3) 1
4) 2
1.41 Which one of the following functions is odd?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ○ Normal ● Large
1) arc(cos(x))
p
p
2) 1 - x - 1 þ x
3) x4 + x
4) x sin (x)
1.42 Which one of the functions below is odd?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ○ Normal ● Large
1) j x - 1j + j x + 1j
2) sin(j x j)
3) x3 + x2
4) j x - 1j - j x + 1j
1.43 Which one of the functions below is even?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ○ Normal ● Large
1) j x - 1j + j x + 1j + j x j
2) (x + 1)4
p
3) f 2 ðxÞ þ 3 x - 1 ¼ 0
4) f (x) ¼ [x] + 1
10
1
Problems: Characteristics of Functions
1
1
- 4 and gðxÞ ¼ x - . Determine f (x).
x
x2
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1) x2 - 4
2) x2 - 2
3) x2
4) x2 + 2
1.44 We know that f ðgðxÞÞ ¼ x2 þ
x2 þ 1 x > 0
.
1
x≤0
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) x + 1
3) x2 + 1
4) (x2 + 1)2 + 1
1.45 Calculate the value of f (-f (x)) if f ðxÞ ¼
1.46 For what value of the parameter of “a,” the graph of the term below is symmetric with respect to the line of y ¼ x.
3x2 þ 4xy þ ð2a - 1Þy2 þ a2 - 4 x ¼ 7
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) -1
2) -2
3) -2, 2
4) 2
1.47 Calculate the value of gog(x) for x ¼
p
2 - 1 based on the following information:
f ðxÞ ¼ cos x
gof ðxÞ ¼ 1 þ tan 2 x
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
p
1) 3 - 2 2
p
2) 5 2 - 7
p
3) 13 - 9 2
p
4) 17 - 12 2
1.48 Determine the domain of the function below.
f ðxÞ ¼
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1 < x < 4
2) 0 < x < 5
log
5x - x2
4
1
Problems: Characteristics of Functions
11
3) 1 ≤ x ≤ 4
4) 0 ≤ x ≤ 5
1.49 Determine the relation of g
1
for the information below.
x
f ðxÞ ¼
2x
xþ2
g ð f ð xÞ Þ ¼ x
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
2x
1)
2-x
2
2)
2x - 1
x-2
3)
2x
2
4)
1 þ 2x
1.50 Determine the relation of f (x) if we have:
f
1 - cos 2x
¼ cot x
1 þ cos 2x
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
p
1) x
1
2) p
x
3) x
1
4)
x
1.51 Determine the domain of f ðxÞ ¼ log x ðx2 þ 9Þ.
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) (-1, 1)
2) (0, 1)
3) [-3, 3]
4) (0, 1) - {1}
1.52 Calculate the range of the function of f (x) ¼ 2x - 2[x] + 1.
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) [0, 2]
2) [1, 3)
3) [0, 2)
4) [0, 3]
12
1
1.53 Calculate the range of fog (x) if f (x) ¼ x2 + 1 and gðxÞ ¼
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) [0, 1)
2) [1, 1)
3) [-1, 1)
4) ℝ
p
Problems: Characteristics of Functions
x - 1.
p
1.54 Calculate the range of f ðxÞ ¼ x2 - 2x þ 3.
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
p
1)
2, 1
p
2)
3, 1
3) [0, 1)
4) [1, 1)
1.55 Which one of the following functions is equivalent to f (x) ¼ j x - 2j?
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ○ Normal ● Large
x2 - 3x þ 2
1) g1 ðxÞ ¼
x-1
2
x -4
2) g2 ðxÞ ¼
xþ2
ð x - 2Þ 2
j x - 2j
j6x - 12j
4) g4 ðxÞ ¼
6
3) g3 ðxÞ ¼
1.56 Which one of the choices is the axis of symmetry of the following function?
f ðxÞ ¼
p
p
3
x- 3 x þ 2
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ○ Normal ● Large
1) x ¼ - 2
2) x ¼ - 1
3) x ¼ 1
4) x ¼ 2.
References
1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
2
Solutions of Problems: Characteristics
of Functions
Abstract
In this chapter, the problems of the first chapter are fully solved, in detail, step-by-step, and with different methods.
2.1 Based on the information given in the problem, the function is as follows [1, 2]:
y ¼ log
x-1
xþ1
The reflection of the graph of a function in the form of f(x, y) with respect to the origin can be achieved by changing
x to -x and y to -y. In other words, f(-x, -y) is the reflection of f(x, y) with respect to the origin. Therefore:
-y ¼ log
) - y ¼ log
-x - 1
-x þ 1
xþ1
xþ1
x-1
) y ¼ - log
) y ¼ log
x-1
x-1
xþ1
In the calculations, the rule below was used.
- log a ¼ log
1
a
Choice (1) is the answer.
2.2 Based on the information given in the problem, we have:
f ðxÞ ¼ ln x9
gð x Þ ¼
p
3
x2
h ð x Þ ¼ ex
The problem can be solved as follows:
f ðgðhðxÞÞÞ ¼ f ðgðex ÞÞ ¼ f
3
ð ex Þ 2
2
2
¼ f e3x ¼ ln e3x
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_2
9
¼ ln e6x
13
14
2
Solutions of Problems: Characteristics of Functions
) f ðgðhðxÞÞÞ ¼ 6x
In the calculations, the rules below were used:
n
ðf ðxÞÞn ¼ ðf ðxÞÞm
m
ln e f ðxÞ ¼ f ðxÞ
Choice (1) is the answer.
2.3 Based on the information given in the problem, we have:
f
2x - 1
x2
1
¼
x
gðxÞ ¼ 2 cos 2 ðxÞ
The problem can be solved as follows:
fog
π
π
¼f g
3
3
¼ f 2cos 2
π
3
) fog
¼f 2
1
2
2
¼f
1
¼
2
p
3
π
¼
2
3
Choice (3) is the answer.
2.4 Based on the information given in the problem, we have:
f ðxÞ ¼
p
x-2
gðxÞ ¼ x þ 1
The problem can be solved as follows:
fogð3Þ ¼ f ðgð3ÞÞ ¼ f ð3 þ 1Þ ¼ f ð4Þ ¼
) fogð3Þ ¼ 0
Choice (1) is the answer.
2.5 Based on the information given in the problem, we have:
f ðxÞ ¼ 2x - 2
gðxÞ ¼ x2 - 1
p
4-2
2ð 2Þ - 1
22
2
Solutions of Problems: Characteristics of Functions
15
The problem can be solved as follows:
fogðxÞ ¼ f ðgðxÞÞ ¼ f x2 - 1 ¼ 2 x2 - 1 - 2 ¼ 2x2 - 4
fogðxÞ ¼ 0 ) 2x2 - 4 ¼ 0 ) x2 ¼ 2
p
)x¼ ± 2
Choice (1) is the answer.
2.6 Based on the information given in the problem, we have:
f ðxÞ ¼
2x2 þ 4 x ≥ 2
½ x] - 4
x<2
The problem can be solved as follows:
f ð2Þ ¼ 2ð2Þ2 þ 4 ¼ 12
f ð-2Þ ¼ ½-2] - 4 ¼ -6
) f ð2Þ þ f ð-2Þ ¼ 12 þ ð-6Þ
) f ð2Þ þ f ð-2Þ ¼ 6
Choice (2) is the answer.
2.7 Based on the information given in the problem, we have:
y¼
p
2 - x2
The domain of a function in the radical form with an even root is determined by considering the radicand equal and
greater than zero. Therefore:
2 - x2 ≥ 0 ) x 2 ≤ 2
p
p
) - 2≤x≤ 2
Choice (2) is the answer.
2.8 From trigonometry, we know that:
sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1
Based on the information given in the problem, we have:
f ð x Þ ¼ 1 - x2
gðxÞ ¼ sinðxÞ
16
2
Solutions of Problems: Characteristics of Functions
Therefore:
fogðxÞ ¼ f ðgðxÞÞ ¼ f ðsinðxÞÞ ¼ 1 - ðsinðxÞÞ2
) fogðxÞ ¼ cos 2 ðxÞ
Choice (1) is the answer.
2.9 Based on the information given in the problem, we have:
f ð xÞ ¼
1
x
To calculate the inverse function of a function, we need to determine x based on y. After that, we must replace x by y and
vice versa. Note that the domain of f -1(x) is the same as the range of f (x).
Therefore:
y¼
1
1
1
)x¼ )y¼
x
y
x
) f -1 ðxÞ ¼
1
x
Choice (2) is the answer.
2.10 Based on the information given in the problem, we have:
sinhðxyÞ - coshðxyÞ þ 1 ¼ 0
A function in the form of f (x, y) ¼ 0 is symmetric with respect to x-axis if it does not change by the conversion of
y → -y, that is, f (x, y) ¼ f (x, -y).
Moreover, a function in the form of f (x, y) ¼ 0 is symmetric with respect to y-axis if it does not change by the conversion
of x → -x, that is, f (x, y) ¼ f (-x, y).
In addition, a function in the form of f (x, y) ¼ 0 is symmetric with respect to the origin if it does not change by the
conversions of x → -x and y → -y, that is, f (x, y) ¼ f (-x, -y).
Therefore:
y ! -y ) sinhð-xyÞ - coshð-xyÞ þ 1 ¼ 0
) -sinhðxyÞ - coshðxyÞ þ 1 ¼ 0
) f ðx, yÞ ≠ f ðx, -yÞ
Therefore, the relation is not symmetric with respect to x-axis.
x ! -x ) sinhð-xyÞ - coshð-xyÞ þ 1 ¼ 0
) - sinhðxyÞ - coshðxyÞ þ 1 ¼ 0
) f ðx, yÞ ≠ f ð-x, yÞ
Thus, the relation is not symmetric with respect to y-axis.
2
Solutions of Problems: Characteristics of Functions
17
x ! -x
) sinhðð -xÞð -yÞÞ - coshðð -xÞð -yÞÞ þ 1 ¼ 0
y ! -y
) sinhðxyÞ - coshðxyÞ þ 1 ¼ 0
) f ðx, yÞ ¼ f ð-x, -yÞ
Hence, the relation is symmetric with respect to the origin.
In the calculations, the rules below were used:
sinhð -aÞ ¼ - sinhðaÞ
coshð -aÞ ¼ coshðaÞ
Choice (4) is the answer.
2.11 Based on the information given in the problem, we have:
f ðxÞ ¼
1-x
1þx
Therefore:
fof ðxÞ ¼ f ðf ðxÞÞ ¼ f
1þx-ð1-xÞ
2x
1 - 1-x
1-x
1þx
1þx
1þx
¼
¼
¼
2
1þxþ1-x
1þx
1 þ 1-x
1þx
1þx
1þx
) fof ðxÞ ¼ x
Choice (3) is the answer.
2.12 Based on the information given in the problem, we have:
y ¼ x4 - 2x2 þ 1
x≥1
To calculate the inverse function of a function, we need to determine x based on y. After that, we must replace x by y and
vice versa. Note that the domain of f -1(x) is the same as the range of f (x).
Therefore:
y ¼ x4 - 2x2 þ 1 ) y ¼ x2 - 1
2
Since x ≥ 1, the value of y is positive in the last equation. Hence, its square root can be determined.
p
y ¼ x2 - 1
)1þ
p
y ¼ x2
18
2
)x¼
1þ
)y¼
1þ
) f -1 ðxÞ ¼
p
p
Solutions of Problems: Characteristics of Functions
y
x
1þ
p
x
Choice (1) is the answer.
2.13 Based on the information given in the problem, we have:
y¼
2ex þ 1
ex - 3
To calculate the inverse function of a function, we need to determine x based on y. After that, we must replace x by y and
vice versa. Note that the domain of f -1(x) is the same as the range of f (x).
Therefore:
y¼
2ex þ 1
) yex - 3y ¼ 2ex þ 1
ex - 3
) ex ðy - 2Þ ¼ 1 þ 3y ) ex ¼
1 þ 3y
y-2
ln
1 þ 3y
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) x ¼ ln
y-2
) y ¼ ln
1 þ 3x
x-2
) f -1 ðxÞ ¼ ln
3x þ 1
x-2
In the calculations, the rule below was used:
ln e f ðxÞ ¼ f ðxÞ
Choice (4) is the answer.
2.14 Based on the information given in the problem, we have:
f ðxÞ ¼
4x þ 5
2x - 3
To calculate the inverse function of a function, we need to determine x based on y. After that, we must replace x by y and
vice versa. Note that the domain of f -1(x) is the same as the range of f (x).
Therefore:
y¼
4x þ 5
) 2xy - 3y ¼ 4x þ 5 ) xð2y - 4Þ ¼ 3y þ 5
2x - 3
2
Solutions of Problems: Characteristics of Functions
19
)x¼
3y þ 5
2y - 4
)y¼
3x þ 5
2x - 4
) f -1 ðxÞ ¼
3x þ 5
2x - 4
The domain of f -1(x) can be determined as follows:
2x - 4 ¼ 0 ) x ¼ 2
) Df -1 ðxÞ ¼ ℝ - f2g
Therefore, x ¼ 2 is not in the in the domain of f -1(x). Choice (2) is the answer.
2.15 From trigonometry, we know that:
sinh x ¼
ex - e-x
2
Therefore:
e ln 3 - e- ln 3 3 - 3 8 4
¼
¼ ¼
2
2
6 3
1
sinhðln 3Þ ¼
) cosðπ sinhðln 3ÞÞ ¼ cos
π
4
π
¼ - cos
π ¼ cos π þ
3
3
3
) cosðπ sinhðln 3ÞÞ ¼ -
1
2
In the calculations, the rules below were used:
e ln f ðxÞ ¼ f ðxÞ
cos
π
1
¼
3
2
Choice (2) is the answer.
2.16 Based on the information given in the problem, we have:
f ðxÞ ¼ x2 - 2x
To calculate the inverse function of a function, we need to determine x based on y. After that, we must replace x by y and
vice versa. Note that the domain of f -1(x) is the same as the range of f (x).
First, we need to define the function in a square form as follows:
y ¼ x2 - 2x þ 1 - 1 ) y ¼ ðx - 1Þ2 - 1
20
2
) y þ 1 ¼ ðx - 1Þ2 )
Solutions of Problems: Characteristics of Functions
yþ1¼x-1)x¼1þ
)y¼1þ
p
yþ1
xþ1
) f -1 ðxÞ ¼ 1 þ
p
xþ1
Choice (1) is the answer.
2.17 Based on the information given in the problem, we have:
f ðx þ 1Þ ¼ x2 - 2x þ 1
The problem can be solved as follows:
f ð x þ 1Þ ¼ ð x - 1Þ 2
x!x-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) f ððx - 1Þ þ 1Þ ¼ ððx - 1Þ - 1Þ2
) f ð x Þ ¼ ð x - 2Þ 2
Choice (1) is the answer.
2.18 A mathematical relation is a function if for any value of x, one value of y is achieved at most. Or, a function is a binary
relation between two sets that associates every element of the first set to exactly one element of the second set. Herein,
y2 ¼ x is not a function because for x ¼ 1, y ¼ - 1, 1 are achieved.
Choice (1) is the answer.
2.19 Based on the information given in the problem, we have:
f
p
x ¼xþ
p
x
The problem can be solved as follows:
f
p
x ¼xþ
p
x¼
p
x
2
þ
p
x
p
x!x
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) f ðxÞ ¼ x2 þ x
¼
¼
¼
¼
Now, the function is in a standard form.
f ð2Þ þ f ð1Þ ¼ 22 þ 2 þ 12 þ 1
) f ð 2Þ þ f ð 1Þ ¼ 8
Choice (3) is the answer.
2.20 Based on the information given in the problem, we have:
f ðxÞ ¼
x2 þ 1
2x þ 3
x≥1
x<1
2
Solutions of Problems: Characteristics of Functions
21
The problem can be solved as follows:
f ð 0Þ ¼ 2 x 0 þ 3 ¼ 3
) f ðf ð0ÞÞ ¼ f ð3Þ ¼ 32 þ 1
) f ðf ð0ÞÞ ¼ 10
Choice (3) is the answer.
2.21 Based on the information given in the problem, we have:
1 - j xj
1 þ j xj
f ðxÞ ¼
The domain of a function in radical form, including an even root, is determined by considering the radicand equal and
greater than zero. Therefore:
1 - j xj
≥0
1 þ j xj
1 þ jxj > 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) 1 - jxj ≥ 0 ) jxj ≤ 1 ) 1 ≤ x ≤ 1
Choice (4) is the answer.
2.22 Based on the information given in the problem, we have:
f ðxÞ ¼
x-1
þ
x-3
2-x
x
The domain of a function in radical form that includes an even root is determined by considering the radicand equal and
greater than zero. Therefore:
x-1
≥0
x-3
)
2-x
≥0
x
x-1
≥0
x-3
)
x-2
≤0
x
x ≤ 1, x > 3 \
¼
¼) 0 < x ≤ 1
0<x≤2
Choice (1) is the answer.
2.23 Based on the information given in the problem, we have:
f ðxÞ ¼
p
x
j xj - 1
The domain of a function in radical form, including even root, is determined by considering the radicand equal and
greater than zero. Moreover, the value of those x that make the denominator zero must be excluded from the
domain. Thus:
x≥0
)
jxj - 1 ≠ 0
Choice (2) is the answer.
x≥0
)
j xj ≠ 1
x≥0
\
¼
¼) Df ¼ ½0, 1Þ - f1g
x≠ ±1
22
2
Solutions of Problems: Characteristics of Functions
2.24 Based on the information given in the problem, we have:
p
xþ1
f ðxÞ ¼ p
x xþ1
The domain of a function in radical form, including even root, is determined by considering the radicand equal and
greater than zero. Moreover, the value of those x that make the denominator zero must be excluded from the
domain. Thus:
x≥0
x x þ 1≠0
p
p
Note that x x þ 1 ≠ 0 is true for any x. Hence:
x ≥ 0 ) Df ¼ ½0,1Þ
Choice (3) is the answer.
2.25 Based on the information given in the problem, we have:
f ðxÞ ¼
1-x
4x þ x3
The value of those x that make the denominator zero are not in the domain. Thus:
4x þ x3 ≠ 0 ) x 4 þ x2 ≠ 0
Note that 4 + x2 ≠ 0 for any x. Hence:
x¼0
Choice (4) is the answer.
2.26 Based on the information given in the problem, we have:
f ðxÞ ¼ 2x
ð1Þ
f ðgðxÞÞ ¼ 2x þ 2
ð2Þ
f ðgðxÞÞ ¼ 2gðxÞ
ð3Þ
Therefore:
Solving (2) and (3):
2gðxÞ ¼ 2x þ 2 ) gðxÞ ¼ x þ 1
Choice (3) is the answer.
2
Solutions of Problems: Characteristics of Functions
23
2.27 Based on the information given in the problem, we have:
f ðxÞ ¼ x - 1
ð1Þ
f ð gð x Þ Þ ¼ x
ð2Þ
f ð gð x Þ Þ ¼ g ð x Þ - 1
ð3Þ
Thus:
Solving (2) and (3):
gð x Þ - 1 ¼ x ) gð x Þ ¼ x þ 1
Choice (1) is the answer.
2.28 Based on the information given in the problem, we have:
f ðxÞ ¼
p
1-x
To calculate the inverse function of a function, we need to determine x based on y. After that, we must replace x by y and
vice versa. Note that the domain of f -1(x) is the same as the range of f (x).
Therefore:
y¼
p
1 - x ) y2 ¼ 1 - x ) x ¼ 1 - y2
) y ¼ 1 - x2
) f -1 ðxÞ ¼ 1 - x2
Since the domain of f -1(x) is the same as the range of f (x), which is [0, 1), we need to add x ≥ 0 to f -1(x) as its
domain. Thus:
f -1 ðxÞ ¼ 1 - x2 , x ≥ 0
Choice (1) is the answer.
2.29 Based on the information given in the problem, we have:
f ðxÞ ¼ sinðxÞ - 2
To calculate the inverse function of a function, we need to determine x based on y. After that, we must replace x by y and
vice versa. Note that the domain of f -1(x) is the same as the range of f (x).
Therefore:
y ¼ sinðxÞ - 2 ) y þ 2 ¼ sinðxÞ ) x ¼ arcðsinðy þ 2ÞÞ
) y ¼ arcðsinðx þ 2ÞÞ
) f -1 ðxÞ ¼ arcðsinðx þ 2ÞÞ
Choice (4) is the answer.
24
2
Solutions of Problems: Characteristics of Functions
2.30 Based on the information given in the problem, we have:
f ðxÞ ¼ 3x - 2
gðxÞ ¼ 2 þ x
First, we need to determine fog(x) as follows:
fogðxÞ ¼ f ðgðxÞÞ ¼ f ð2 þ xÞ ¼ 3ð2 þ xÞ - 2 ¼ 3x þ 4
To calculate the inverse function of a function, we need to determine x based on y. After that, we must replace x by y and
vice versa. Note that the domain of f -1(x) is the same as the range of f (x).
Therefore:
y ¼ 3x þ 4 ) 3x ¼ y - 4
1
4
)x¼ y3
3
1
4
)y¼ x3
3
1
4
) f -1 ðxÞ ¼ x 3
3
Choice (1) is the answer.
2.31 Based on the information given in the problem, we have:
f ðxÞ ¼ x3 þ 3x2 þ 3x þ 2
To calculate the inverse function of a function, we need to determine x based on y. After that, we must replace x by y and
vice versa. Note that the domain of f -1(x) is the same as the range of f (x).
First, we need to define the function in a cube form as follows:
) f ðxÞ ¼ x3 þ 3x2 þ 3x þ 1 þ 1 ¼ ðx þ 1Þ3 þ 1
1
) y ¼ ð x þ 1Þ 3 þ 1 ) y - 1 ¼ ð x þ 1Þ 3 ) ð y - 1Þ 3 ¼ x þ 1
1
) x ¼ ð y - 1Þ 3 - 1
1
) y ¼ ð x - 1Þ 3 - 1
) f -1 ðxÞ ¼ -1 þ
Choice (3) is the answer.
p
3
x-1
2
Solutions of Problems: Characteristics of Functions
25
2.32 Based on the information given in the problem, we have:
x>0
From trigonometry, we know that:
sinh t ¼
et - e-t
2
cosh t ¼
et þ e-t
2
tanh t ¼
sinh t
sinh t
Therefore:
sinhðln xÞ ¼
e ln x - e- ln x x - x
¼
2
2
coshðln xÞ ¼
e ln x þ e- ln x x þ x
¼
2
2
1
1
x-1
x
x - 1x
sinhðln xÞ
tanhðln xÞ ¼
¼ xþ2 1 ¼
sinhðln xÞ
x
x þ 1x
2
) tanhðln xÞ ¼
x2 - 1
x2 þ 1
In the calculations, the rule below was used:
e ln f ðxÞ ¼ f ðxÞ
Choice (3) is the answer.
2.33 Based on the information given in the problem, we have:
f ðxÞ þ x f ð-xÞ ¼ x2 þ 1
The problem can be solved as follows:
x¼2
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼)
x ¼ -2
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼)
f ð2Þ þ 2f ð-2Þ ¼ 22 þ 1
2
f ð-2Þ þ ð-2Þf ð2Þ ¼ ð-2Þ þ 1
)
xð-2Þ
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼)
)
f ð2Þ þ 2f ð-2Þ ¼ 5
f ð-2Þ - 2f ð2Þ ¼ 5
f ð2Þ þ 2f ð-2Þ ¼ 5
-2f ð-2Þ þ 4f ð2Þ ¼ -10
þ
¼
¼
¼) 5f ð2Þ ¼ -5 ) f ð2Þ ¼ -1
Choice (1) is the answer.
26
2
Solutions of Problems: Characteristics of Functions
2.34 From trigonometry, we know that:
1 þ cosð2xÞ ¼ 2 cos 2 ðxÞ
Based on the information given in the problem, we have:
f ð x Þ ¼ x2 - 2
The problem can be solved as follows:
) f ð2cosðxÞÞ ¼ ð2cosðxÞÞ2 - 2 ¼ 4 cos 2 ðxÞ - 2 ¼ 2ð1 þ cosð2xÞÞ - 2 ¼ 2 cosð2xÞ
) f ðf ð2cosðxÞÞÞ ¼ ð2cosð2xÞÞ2 - 2 ¼ 4 cos 2 ð2xÞ - 2 ¼ 2ð1 þ cosð4xÞÞ - 2 ¼ 2 cosð4xÞ
) f ðf ðf ð2cosðxÞÞÞÞ ¼ ð2cosð4xÞÞ2 - 2 ¼ 4 cos 2 ð4xÞ - 2 ¼ 2ð1 þ cosð8xÞÞ - 2
) f ðf ðf ð2cosðxÞÞÞÞ ¼ 2 cosð8xÞ
Choice (4) is the answer.
2.35 Based on the information given in the problem, we have:
f
p
x-1
¼ 2x - 1
x
First, we need to determine the f (x) as follows:
x-1
¼t
x
)x¼
) f ðt Þ ¼
2x
1
1-t
1
-1¼
1-t
t!x
¼
¼
¼
¼
¼
¼
¼
¼
¼) f ðxÞ ¼
¼
¼
¼
¼
¼
¼
1þt
1-t
1þx
1-x
The domain of a function in radical form, including even root, is determined by considering the radicand equal and
greater than zero. Therefore:
1þx
xþ1
≥0 )
≤ 0 ) -1 ≤ x < 1
1-x
x-1
Note, that x ¼ 1 must be excluded from the domain, since it makes the denominator zero. Choice (3) is the answer.
2.36 Based on the information given in the problem, we have:
f ðxÞ ¼
1-
p
x-1
The domain of a function in radical form that includes an even root is determined by considering the radicand equal and
greater than zero. Therefore:
2
Solutions of Problems: Characteristics of Functions
p
1 - x - 1≥0
)
x - 1≥0
27
p
x - 1≤1
)
x≥1
x - 1≤1
)
x≥1
x≤2 \
) 1≤x≤2
x≥1
Choice (2) is the answer.
2.37 Based on the information given in the problem, we have:
f ðxÞ ¼
j xj - 1 þ
jxj þ 1
The domain of a function in radical form, including even root, is determined by considering the radicand equal and
greater than zero.
j xj - 1 ≥ 0
)
j xj þ 1 ≥ 0
jxj ≥ 1
)
x2
x ≤ - 1, x ≥ 1 \
) x ≤ - 1, x ≥ 1 ) Df ¼ ℝ - ð-1, 1Þ
x2
Note that jxj + 1 ≥ 0 is true for any x. Choice (4) is the answer.
2.38 Based on the information given in the problem, we have:
f xþ
1
1
¼ x2 þ 2
x
x
The problem can be solved as follows:
)f xþ
1
1
1
¼ x2 þ 2 þ 2 - 2 ¼ x þ
x
x
x
2
-2
1
xþ !x
x
¼
¼
¼
¼
¼
¼
¼
¼
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 ðxÞ ¼ x2 - 2 ) f ð3Þ ¼ 32 - 2
) f ð 3Þ ¼ 7
Choice (3) is the answer.
2.39 Based on the information given in the problem, we have:
f ð x Þ ¼ j x þ 2j þ aj x - 2j
ð1Þ
Based on the definition, the function of f (x) is an even function if its domain is symmetric and:
f ð-xÞ ¼ f ðxÞ
ð2Þ
) f ð-xÞ ¼ j-x þ 2j þ aj-x - 2j ¼ j-ðx - 2Þj þ aj-ðx þ 2Þj ¼ jx - 2j þ ajx þ 2j
ð3Þ
Therefore:
Solving (1), (2), and (3):
jx - 2j þ ajx þ 2j ¼ jx þ 2j þ ajx - 2j ) a ¼ 1
Choice (3) is the answer.
28
2
Solutions of Problems: Characteristics of Functions
2.40 Based on the information given in the problem, we have:
f ðxÞ ¼ x2 þ ðA - 1Þx is an even function
ð1Þ
gðxÞ ¼ ðB þ 2Þx2 þ sinðxÞis an odd function
ð2Þ
Based on the definition, the function of f (x) is an even function if its domain is symmetric and:
f ð-xÞ ¼ f ðxÞ
ð3Þ
Additionally, the function of f (x) is an odd function if its domain is symmetric and:
f ð-xÞ ¼ -f ðxÞ
ð4Þ
Solving (1) and (3):
ð-xÞ2 þ ðA - 1Þð-xÞ ¼ x2 þ ðA - 1Þx ) 2ðA - 1Þx ¼ 0 ) A ¼ 1
Solving (2) and (4):
ðB þ 2Þð-xÞ2 þ sinð-xÞ ¼ - ðB þ 2Þx2 þ sinðxÞ
) ðB þ 2Þx2 - sinðxÞ ¼ -ðB þ 2Þx2 - sinðxÞ ) 2ðB þ 2Þx2 ¼ 0 ) B ¼ -2
Therefore:
A þ B ¼ 1 þ ð-2Þ ¼ -1
Choice (2) is the answer.
2.41 Based on the definition, the function of f (x) is an odd function if its domain is symmetric and:
f ð-xÞ ¼ -f ðxÞ
Choice (1):
f ðxÞ ¼ arcðcosðxÞÞ
) f ð-xÞ ¼ arcðcosð-xÞÞ ¼ π - arcðcosðxÞÞ ≠ f-f ðxÞ, f ðxÞg
) Not even nor odd
Choice (2):
f ðxÞ ¼
) f ð-xÞ ¼
1 - ð-xÞ -
1 þ ð-xÞ ¼
p
p
1-x-
1þx) Odd
p
p
1þx
1-x¼-
p
1-x-
p
1 þ x ¼ -f ðxÞ
2
Solutions of Problems: Characteristics of Functions
29
Choice (3):
f ð xÞ ¼ x4 þ x
) f ð-xÞ ¼ ð-xÞ4 þ ð-xÞ ¼ x4 - x ≠ f-f ðxÞ, f ðxÞg
) Not even nor odd
Choice (4):
f ðxÞ ¼ x sinðxÞ
) f ð-xÞ ¼ -x sinð-xÞ ¼ x sinðxÞ ¼ f ðxÞ
) Even
Choice (2) is the answer.
2.42 Based on the definition, the function of f (x) is an odd function if its domain is symmetric and:
f ð-xÞ ¼ -f ðxÞ
Choice (1):
f ð x Þ ¼ j x - 1j þ j x þ 1j
) f ð-xÞ ¼ j-x - 1j þ j-x þ 1j ¼ j-ðx þ 1Þj þ j-ðx - 1Þj ¼ jx þ 1j þ jx - 1j ¼ f ðxÞ
) Even
Choice (2):
f ðxÞ ¼ sinðjxjÞ
) f ð-xÞ ¼ sinðj-xjÞ ¼ sinðjxjÞ ¼ f ðxÞ
) Even
Choice (3):
f ðxÞ ¼ x3 þ x2 ) f ð-xÞ ¼ ð-xÞ3 þ ð-xÞ2 ¼ -x3 þ x2 ≠ f-f ðxÞ, f ðxÞg
) Not even nor odd
Choice (4):
f ð x Þ ¼ j x - 1j - j x þ 1j
) f ð-xÞ ¼ j-x - 1j - j-x þ 1j ¼ j-ðx þ 1Þj - j-ðx - 1Þj ¼ jx þ 1j - jx - 1j ¼ -f ðxÞ
) Odd
Choice (4) is the answer.
30
2
Solutions of Problems: Characteristics of Functions
2.43 Based on the definition, the function of f (x) is an even function if its domain is symmetric and:
f ð-xÞ ¼ f ðxÞ
Choice (1):
f ð x Þ ¼ j x - 1j þ j x þ 1j þ j x j
) f ð-xÞ ¼ j-x - 1j þ j-x þ 1j þ j-xj ¼ j-ðx þ 1Þj þ j-ðx - 1Þj þ j-xj ¼ jx þ 1j þ jx - 1j þ jxj ¼ f ðxÞ
) Even
Choice (2):
f ð x Þ ¼ ð x þ 1Þ 4
) f ð-xÞ ¼ ð-x þ 1Þ4 ¼ ðx - 1Þ4 ≠ f-f ðxÞ, f ðxÞg
) Not even nor odd
Choice (3):
f 2 ð xÞ þ
p
3
x - 1 ¼ 0 ) Not a function
Choice (4):
f ð xÞ ¼ ½ x] þ 1
) f ð-xÞ ¼ ½-x] þ 1 ≠ f-f ðxÞ, f ðxÞg
) Not even nor odd
Choice (1) is the answer.
2.44 Based on the information given in the problem, we have:
gðxÞ ¼ x f ð gð x Þ Þ ¼ x 2 þ
1
x
ð1Þ
1
-4
x2
ð2Þ
The problem can be solved as follows:
f ðgðxÞÞ ¼ x2 þ
1
1
-2-2¼ xx
x2
Solving (1) and (3):
f ð gð x Þ Þ ¼ ð g ð x Þ Þ 2 - 2
2
-2
ð3Þ
2
Solutions of Problems: Characteristics of Functions
31
) f ðxÞ ¼ x2 - 2
Choice (2) is the answer.
2.45 Based on the information given in the problem, we have:
f ðxÞ ¼
x2 þ 1
x>0
1
x≤0
As can be noticed from f (x), the value of function is always positive. Therefore, the value of -f (x) is always negative.
Hence:
f ð-f ðxÞÞ ¼ 1
Choice (1) is the answer.
2.46 A function in the form of f (x, y) ¼ 0 is symmetric with respect to the line of y ¼ x if f (x, y) ¼ f(y, x).
Based on the information given in the problem, we have:
3x2 þ 4xy þ ð2a - 1Þy2 þ a2 - 4 x ¼ 7
) f ðx, yÞ ¼ 3x2 þ 4xy þ ð2a - 1Þy2 þ a2 - 4 x - 7
Moreover, from f (x, y) ¼ f(y, x), we have:
) 3x2 þ 4xy þ ð2a - 1Þy2 þ a2 - 4 x - 7 ¼ 3y2 þ 4yx þ ð2a - 1Þx2 þ a2 - 4 y - 7
) ð4 - 2aÞx2 þ ð2a - 4Þy2 þ a2 - 4 x - a2 - 4 y ¼ 0
4 - 2a ¼ 0
)
2a - 4 ¼ 0
a2 - 4 ¼ 0
) a ¼ f2g \ f2g \ f-2, 2g \ f-2, 2g ) a ¼ 2
a2 - 4 ¼ 0
Choice (4) is the answer.
2.47 Based on the information given in the problem, we have:
p
2-1
ð1Þ
f ðxÞ ¼ cos x
ð2Þ
gof ðxÞ ¼ 1 þ tan 2 x
ð3Þ
gof ðxÞ ¼ gðf ðxÞÞ ¼ gðcos xÞ ¼ 1 þ tan 2 x
ð4Þ
x¼
Solving (2) and (3):
32
2
Solutions of Problems: Characteristics of Functions
From trigonometry, we know that:
1
cos 2 x
1 þ tan 2 x ¼
ð5Þ
Solving (4) and (5):
gðcos xÞ ¼
1
1
) gð x Þ ¼ 2
x
cos 2 x
) gogðxÞ ¼ gðgðxÞÞ ¼ g
ð6Þ
1
1
¼ x4
¼
1 2
x2
2
ð7Þ
x
Solving (1) and (7):
ðgogÞ
p
2-1 ¼
p
2-1
4
p
¼ 2-2 2þ1
) ðgogÞ
p
2
p
¼ 3-2 2
2
p
¼ 9 - 12 2 þ 8
p
2 - 1 ¼ 17 - 12 2
Choice (4) is the answer.
2.48 Based on the information given in the problem, we have:
f ðxÞ ¼
log
5x - x2
4
The domain of a function in radical form, including even root, is determined by considering the radicand equal and
greater than zero.
Moreover, the domain of a logarithmic function with the base of 10 can be determined by considering its argument
greater than zero. Therefore:
log
5x - x2
≥0
4
5x - x2
>0
4
)
)
log
5x - x2
≥ logð1Þ
4
)
2
x - 5x < 0
ð x - 4Þ ð x - 1 Þ ≤ 0
xð x - 5Þ < 0
)
5x - x2
≥1
)
4
xð x - 5Þ < 0
x2 - 5x þ 4 ≤ 0
x ð x - 5Þ < 0
1≤x≤4 \
¼) 1 ≤ x ≤ 4
0<x<5
Choice (3) is the answer.
2.49 Based on the information given in the problem, we have:
2x
xþ2
ð1Þ
gð f ð x Þ Þ ¼ x
ð2Þ
f ðxÞ ¼
2
Solutions of Problems: Characteristics of Functions
33
The problem can be solved as follows:
t≜
2x
xþ2
) tx þ 2t ¼ 2x ) 2x - tx ¼ 2t ) xð2 - t Þ ¼ 2t ) x ¼
ð3Þ
2t
2-t
ð4Þ
Solving (1) and (2):
gð f ð x Þ Þ ¼ g
2x
¼x
xþ2
ð5Þ
Solving (3) and (5):
gð t Þ ¼ x
ð6Þ
Solving (4) and (6):
2t
2-t
ð7Þ
2
2 1
1
x
¼ x 1 ¼ 2x-1
x
2-x
x
ð8Þ
gð t Þ ¼
)g
)g
1
2
¼
x
2x - 1
Choice (2) is the answer.
2.50 Based on the information given in the problem, we have:
f
1 - cos 2x
¼ cot x
1 þ cos 2x
ð1Þ
From trigonometry, we know that:
1 - cos 2x ¼ 2 sin 2 x
ð2Þ
1 þ cos 2x ¼ 2 cos 2 x
ð3Þ
tan x ¼
sin x
cos x
ð4Þ
cot x ¼
1
tan x
ð5Þ
Therefore:
1 - cos 2x 2 sin 2 x
¼ tan 2 x
¼
1 þ cos 2x 2 cos 2 x
ð6Þ
34
2
Solutions of Problems: Characteristics of Functions
Solving (5) and (1):
f tan 2 x ¼ cot x
ð7Þ
1
tan x
ð8Þ
Solving (7) and (5):
f tan 2 x ¼
Defining a new parameter:
t≜ tan 2 x
ð9Þ
Solving (8) and (9):
1
1
gðt Þ ¼ p ) gðxÞ ¼ p
x
t
Choice (2) is the answer.
2.51 Based on the information given in the problem, we have:
f ðxÞ ¼
log x ðx2 þ 9Þ
The domain of a function in radical form, including even root, is determined by considering the radicand equal and
greater than zero.
In addition, the domain of a logarithmic function can be determined by considering its argument greater than zero. In
addition, the base of the logarithm must be greater than zero but not equal to one. Therefore:
log x x2 þ 9 ≥ 0
x2 þ 9 > 0
x > 0, x ≠ 1
log x x2 þ 9 ≥ log x ð1Þ
)
x2 þ 9 > 0
x2 þ 9 ≥ 1
)
x > 0, x ≠ 1
x2 þ 9 > 0
x > 0, x ≠ 1
Note that x2 + 9 > 0 and x2 + 8 ≥ 0 are true for any x. Hence:
\
¼) x > 0, x ≠ 1 ) Df ¼ ð0, 1Þ - f1g
Choice (4) is the answer.
2.52 Based on the information given in the problem, we have:
f ðxÞ ¼ 2x - 2½x] þ 1
Based on the definition, we know that:
-½x]
þ1
x2
¼
¼
¼
¼
¼
¼
¼) 0 ≤ x - ½x] < 1 ¼
½x] ≤ x < ½x] þ 1 ¼
¼
¼
¼) 0 ≤ 2x - 2½x] < 2 ¼
¼
¼
¼) 1 ≤ 2x - 2½x] þ 1 < 3
) 1 ≤ f ðxÞ < 3 ) Rf ¼ ½1, 3Þ
Choice (2) is the answer.
2
Solutions of Problems: Characteristics of Functions
35
2.53 Based on the information given in the problem, we have:
f ð x Þ ¼ x2 þ 1
gð x Þ ¼
) fogðxÞ ¼ f ðgðxÞÞ ¼ f
p
p
x-1
x-1 ¼
p
x-1
2
þ1¼x-1þ1¼x
Next, we need to determine the domain of the function. As we know, the domain of a function in radical form, including
even root, is determined by considering the radicand equal and greater than zero.
x - 1 ≥ 0 ) x ≥ 1 ) Dfog ¼ ½1, 1Þ
Now, we can determine the range of the function based its domain as follows:
Dfog ¼ ½1, 1Þ
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) Rfog ¼ ½1, 1Þ
fogðxÞ ¼ x ¼
Choice (2) is the answer.
2.54 Based on the information given in the problem, we have:
f ðxÞ ¼
x2 - 2x þ 3
The problem can be solved as follows:
) f ðxÞ ¼
ð x - 1Þ 2 þ 2
As we know:
p
þ2
2
ðx - 1Þ ≥ 0 ¼
¼
¼) ðx - 1Þ þ 2 ≥ 2 ¼
¼
¼)
2
p
p
p
ðx - 1Þ2 þ 2 ≥ 2 ) f ðxÞ ≥ 2 ) Rf ðxÞ ¼
2, 1
Choice (1) is the answer.
2.55 Based on the information given in the problem, we have:
f ð x Þ ¼ j x - 2j ) D f ¼ ℝ
Based on the definition, two functions are equivalent if they are equal, and their domains are the same.
Choice (1):
g1 ð xÞ ¼
ð x - 1Þ ð x - 2Þ
x2 - 3x þ 2
¼
¼ j x - 2j
x-1
x-1
) D g 1 ¼ ℝ - f 1g
Therefore, the functions are not equivalent, since their domains are different. Note that x ¼ 1 makes the denominator
zero; thus, it is not in the domain.
36
2
Solutions of Problems: Characteristics of Functions
Choice (2):
g2 ð x Þ ¼
ðx - 2Þðx þ 2Þ
x2 - 4
¼
¼ j x - 2j
xþ2
xþ2
) Dg2 ¼ ℝ - f-2g
Therefore, the functions are not equivalent because their domains are not the same. Note that x ¼ - 2 makes the
denominator zero; thus, it must be excluded from the domain.
Choice (3):
g3 ð xÞ ¼
ð x - 2Þ 2 j x - 2j 2
¼
¼ j x - 2j
j x - 2j
jx - 2j
) Dg3 ¼ ℝ - f-2g
Therefore, the functions are not equivalent because their domains are not the same. Note that x ¼ - 2 makes the
denominator zero; thus, it must be excluded from the domain.
Choice (4):
g 4 ð xÞ ¼
j6x - 12j 6jx - 2j
¼
¼ jx - 2j
6
6
) D g4 ¼ ℝ
Therefore, the functions are equivalent because their functions and domains are the same. Choice (4) is the answer.
2.56 Based on the information given in the problem, we have:
f ðxÞ ¼
p
3
x-
p
3
xþ2
The line of x ¼ a is the axis of symmetry of the function of f (x, y) if f(a + x, y) ¼ f(a -x, y).
For Choice (1), we have x ¼ -2 ) a ¼ -2.
f ð-2 þ x, yÞ ¼
3
ð-2 þ xÞ -
3
ð-2 þ xÞ þ 2
f ð-2 - x, yÞ ¼
3
ð-2 - xÞ -
3
ð-2 - xÞ þ 2
)
p
3
x-2-
p
3
p
p
3
x ≠ -x - 2 þ 3 x
) f ð-2 þ x, yÞ ≠ f ð-2 - x, yÞ
For Choice (2), we have x ¼ -1 ) a ¼ -1.
f ð-1 þ x, yÞ ¼
3
ð-1 þ xÞ -
3
ð-1 þ xÞ þ 2
f ð-1 - x, yÞ ¼
3
ð-1 - xÞ -
3
ð-1 - xÞ þ 2
References
37
)
p
3
x-1-
p
p
p
3
xþ1¼-3 xþ1þ 3 x-1
) f ð-1 þ x, yÞ ¼ f ð-1 - x, yÞ
For Choice (3), we have x ¼ 1 ) a ¼ 1.
f ð1 þ x, yÞ ¼
3
ð 1 þ xÞ -
3
ð 1 þ xÞ þ 2
f ð1 - x, yÞ ¼
3
ð 1 - xÞ -
3
ð 1 - xÞ þ 2
)
p
3
xþ1-
p
3
x þ 3≠ -
p
3
x-1þ
p
3
x-3
) f ð1 þ x, yÞ ≠ f ð1 - x, yÞ
For Choice (4), we have x ¼ 2 ) a ¼ 2.
f ð2 þ x, yÞ ¼
3
ð 2 þ xÞ -
3
ð 2 þ xÞ þ 2
f ð2 - x, yÞ ¼
3
ð 2 - xÞ -
3
ð 2 - xÞ þ 2
)
p
3
xþ2-
p
3
x þ 4≠ -
p
3
x-2þ
p
3
x-4
) f ð2 þ x, yÞ ≠ f ð2 - x, yÞ
In the calculations, the rule below was used because n was an odd number:
n
-f ðxÞ ¼ - n f ðxÞ
Choice (2) is the answer.
References
1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
3
Problems: Trigonometric Equations and Identities
Abstract
In this chapter, the basic and advanced problems of trigonometric equations and trigonometric identities are presented. The
subjects include trigonometric equations, trigonometric identities, domain, range, period, half angle formulas, reciprocal
identities, Pythagorean identities, expressing sum of sine and cosine as a product, expressing product of sine and cosine as
a sum, even and odd functions, periodic functions, degrees to radians formula, cofunction formulas, unit circle, inverse
trigonometric functions, and domain and range of inverse trigonometric functions. To help students study the chapter in the
most efficient way, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard)
and calculation amounts (small, normal, and large). Moreover, the problems are ordered from the easiest problem with the
smallest computations to the most difficult problems with the largest calculations.
3.1. Calculate the value of sin x cos x(1 - 2sin2x) for x ¼ 7.5° [1, 2].
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
p
3
1)
8
1
2)
8
p
3
3)
4
1
4)
4
3.2. Calculate the value of tan3x + cot3x if tan x + cot x ¼ 3.
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 18
2) 9
3) 27
4) 3
3.3. Which one of the following choices is correct about the extension of hyperbolic functions?
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
coshða þ bÞ ¼ cosh a cosh b - sinh a sinh b
1)
sinhða þ bÞ ¼ sinh a cosh b þ cosh a sinh b
coshða þ bÞ ¼ cosh a cosh b - sinh a sinh b
2)
sinhða þ bÞ ¼ sinh a cosh b - cosh a sinh b
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_3
39
40
3
3)
4)
coshða þ bÞ ¼ cosh a cosh b þ sinh a sinh b
sinhða þ bÞ ¼ sinh a cosh b - cosh a sinh b
coshða þ bÞ ¼ cosh a cosh b þ sinh a sinh b
sinhða þ bÞ ¼ sinh a cosh b þ cosh a sinh b
3.4. Calculate the value of tan(2θ) if cot(θ) ¼ 5.
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
5
1)
12
5
2)
13
5
3) 12
5
4) 13
3.5. Determine the value of tan(-2100°).
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
p
1) 3
p
3
2)
3p
3) - 3
p
3
4) 3
3.6. Simplify and calculate the final value of the following term:
1 þ cos ð40 ° Þ
sin ð40 ° Þ
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) sin (20°)
2) cos (20°)
3) tan (20°)
4) cot (20°)
3m - 1
π
2π
and ≤ α ≤ .
4
6
3
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
p
2 3-1
1) 1,
3
p
2 3þ1
2) 1,
3
3) [1, 2]
5
4) 1,
3
3.7. Determine the range of m if sinðαÞ ¼
Problems: Trigonometric Equations and Identities
3
Problems: Trigonometric Equations and Identities
2m - 1
π
π
and - ≤ x ≤ .
6
3
3
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
7
1) 2,
2
3 7
2)
,
2 2
3.8. Determine the range of m if cosðxÞ ¼
3) 2,
4)
5
2
3 5
,
2 2
2x
?
3
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) π
2) 2π
3) 3π
4) 4π
3.9. What is the main period of cos 2 ðxÞ - 5 cos
3x
2x
?
þ cos 3
5
3
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 3π
2) 5π
3) 15π
4) 30π
3.10. What is the main period of sin 4
πx
þ cosðπxÞ þ 5.
3
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 2
3) 3
4) 6
3.11. Determine the main period of sin 4
3.12. Figure 3.1 illustrates part of the function of y ¼ sin (kx). Determine the value of k.
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
2
1)
3
3
2)
4
3
3)
2
4
4)
3
41
42
3
Problems: Trigonometric Equations and Identities
Figure 3.1 The graph of problem 3.12
3.13. Figure 3.2 illustrates part of the function of y ¼ cos
ax þ
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1
1)
2
3
2)
2
2
3)
3
7
4)
4
1
π . Determine the value of a.
2
Figure 3.2 The graph of problem 3.13
3.14. Which one of the following choices is correct if α + β ¼ 19π?
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) sin(α) ¼ sin (β)
2) cos(α) ¼ cos (β)
3) tan(α) ¼ tan (β)
4) cot(α) ¼ cot (β)
3.15. Calculate the final value of the following equation.
sinð5π þ xÞ þ sin x Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
π
7π
þ sin x þ
3
3
3
Problems: Trigonometric Equations and Identities
43
1) 0
π
3
π
3) 2 sin
3
π
4) - sin
3
2) sin
3.16. Calculate the value of the term below.
sin arccos
p
-1
- 3
þ arcsin
2
2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
p
3
1)
2
2) 1
1
3)
2
4) -1
1
.
2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
4
1) arctan
3
3
2) arctan
4
2
3) arctan
3
3
4) arctan
2
3.17. Calculate the value of 2 arctan
3.18. What is the value of cos(π sinh ln 3)?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) -1
1
2) p2
2
3)
2
p
3
4)
2
3.19. Calculate the value of cos(20°) if sin(50°) + sin (10°) ¼ m.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
m
1)
2
2) m
3) 2m
2m
4)
3
44
3
Problems: Trigonometric Equations and Identities
3.20. Simplify and calculate the final value of the following term:
ð1 þ tan 2 ð5 ° ÞÞ sinð10 ° Þ
ð1 - tan 2 ð5 ° ÞÞ tanð10 ° Þ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) tan(15°)
2) tan(25°)
3) tan(35°)
4) tan(45°)
3.21. Which one of the following relations is correct if cot(α) ¼ m and cos(α) ¼ n?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) m2(1 + n2) ¼ n2
2) m2(1 - n2) ¼ n2
3) m2(2 + n2) ¼ 1
4) m2(2 - n2) ¼ 1
3.22. Determine the main period of sin(3x) cos (5x) + 11.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) π
2) 2π
2π
3)
3
2π
4)
5
4π
.
3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
π
1)
6
5π
2)
6
π
3)
3
π
4) 6
17π
3.24. Calculate the value of arc sin sin
.
5
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
2π
1)
5
3π
2)
5
2π
3) 5
3π
4) 5
3.23. Calculate the value of arc cos sin
3
Problems: Trigonometric Equations and Identities
45
19π
.
5
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
π
1)
5
4π
2)
5
π
3) 5
4π
4) 5
3.25. Calculate the value of arc cos cos
1
.
2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
3
2)
4
4
3)
3
3
4)
5
3.26. Calculate the value of tan 2arc tan
3
3
þ arc tan
5
4
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
10
1)
13
9
2)
13
12
3)
35
24
4)
25
3.27. Calculate the final value of sin arc sin
4
3
- arc cot
3
4
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) π
2π
2)
3
π
3)
2
π
4)
3
3.28. Calculate the final value of arc cot -
3
.
2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
π
1)
4
π
2) 4
3.29. Calculate the final value of arcðtanð5ÞÞ þ arc tan
.
.
46
3
Problems: Trigonometric Equations and Identities
3π
4
5π
4)
4
3)
3
4
þ cos arc sin 5
5
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
7
1)
5
1
2) 5
1
3)
5
7
4) 5
3.30. Calculate the final value of sin arc cos
.
3.31. Figure 3.3 shows a unit circle. Which one of the choices shows the value of tan(θ) and cot(θ), respectively?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) OA, OB
2) HA, HB
3) OA, AB
4) OB, BH
Figure 3.3 The graph of problem 3.31
3.32. Figure 3.4 illustrates a unit circle. Which one of the choices shows the value of sec(θ)?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) HA
2) MB
3) OB
4) OM
3
Problems: Trigonometric Equations and Identities
47
Figure 3.4 The graph of problem 3.32
3.33. Figure 3.5 illustrates a unit circle. Which one of the choices shows the value of csc(θ)?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) MB
2) OB
3) HC
4) OM
Figure 3.5 The graph of problem 3.33
2
1
.
þ arc tan
3
5
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
π
1)
6
π
2)
4
π
3)
3
π
4)
2
3.34. Calculate the value of arc tan
3.35. Calculate the final value of the term below.
arcðtanðmÞÞ þ arc tan
1
m
þ arcðcotðmÞÞ þ arcðcotð- mÞÞ
48
3
Problems: Trigonometric Equations and Identities
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) π or 2π
π
3π
2) or
2
2
3π
3)
2
π
4)
2
3.36. Determine the range of x in the inequality below. Herein, x is an acute angle.
- 1 ≤ cosð4xÞ cosð2xÞ þ sinð4xÞ sinð2xÞ ≤ 0
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
π 3π
,
1)
6 8
π π
2)
,
8 4
π π
3)
,
6 3
π π
4)
,
4 2
3.37. Calculate the value of tan(2y) if tan(x + y) ¼ 5 and tan(x - y) ¼ 7.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1
1)
18
1
2) 18
1
3)
36
1
4) 36
3.38. Simplify and calculate the value of the following term:
5π
5π
þ cos
12
12
5π
5π
sin
- cos
12
12
sin
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
p
1) 3
p
3
2)
3 p
3) - 2 3
p
3
4) 3
3
Problems: Trigonometric Equations and Identities
49
3.39. Figure 3.6 illustrates part of the function of y ¼ a sin (bπx). Determine the value of a + b.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
4
1)
3
5
2)
3
7
3)
3
8
4)
3
Figure 3.6 The graph of problem 3.39
3.40. Figure 3.7 illustrates part of the function of y ¼ a sin (bπx). Determine the value of a b.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) -6
2) -3
9
3)
2
4) 6
Figure 3.7 The graph of problem 3.40
3.41. Figure 3.8 illustrates part of the function of y ¼ a sin
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 2
5
2)
2
bx þ
1
π . Determine the value of a b.
2
50
3
3) 3
7
4)
2
Figure 3.8 The graph of problem 3.41
3.42. Simplify and calculate the value of the following term:
cosð5 ° Þ cosð10 ° Þ cosð20 ° Þ
cosð50 ° Þ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1
1)
4 cosð85 ° Þ
1
2)
8 cosð85 ° Þ
1
3)
8 sinð85 ° Þ
1
4)
4 sinð85 ° Þ
x
7
if sinðxÞ þ cosðxÞ ¼ .
2
5
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 2 or 3
1
1
2) or
3
2
3
3) 2 or
5
2
4) 3 or
5
3.43. Calculate the value of tan
3.44. Simplify and calculate the value of the following term:
sin 4 ðαÞ - cos 4 ðαÞ
sinðαÞ cosðαÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
Problems: Trigonometric Equations and Identities
3
Problems: Trigonometric Equations and Identities
1)
2)
3)
4)
51
2 cot (2α)
-2 cot (2α)
2 tan (3α)
-2 tan (3α)
1
3.45. Calculate the value of cot2(2α) if sin 4 ðαÞ þ cos 4 ðαÞ ¼ .
2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) 2
3) 0
4) 3
3.46. Calculate the value of the following relation for x ¼
3π
:
8
sin 3 ðxÞ cosðxÞ - cos 3 ðxÞ sinðxÞ þ 3 sin 2 ðxÞ cos 2 ðxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
3
1)
8
5
2)
8
5
3) 8
3
4) 8
3.47. Calculate the value of the following relation for α ¼
π
:
15
sinð2αÞ þ sinð5αÞ þ sinð8αÞ
cosð2αÞ þ cosð5αÞ þ cosð8αÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
p
3
1) 3
p
2) - 3
p
3) 3
p
3
4)
3
3.48. Calculate the value of the following relation for x ¼
π
:
12
ðsinðxÞ - cosðxÞ þ 2ÞðsinðxÞ - cosðxÞ - 2Þ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
52
3
Problems: Trigonometric Equations and Identities
7
2
5
2)
2
1)
5
2
7
4) 2
3) -
3.49. Calculate the value of 4sin2(α)cos2(α)(tan(α) + cot (α))2.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) 2
3) 3
4) 4
3.50. Determine the number of roots of the equation below.
sinðπxÞ cos 2 ðπxÞ þ sin 2 ðπxÞ cosðπxÞ ¼ 0:
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 11
2) 12
3) 13
4) 14
1
16π
.
þ 2 cosðmxÞ. Determine the value of the function for x ¼
2
3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1
1) 2
1
2)
2
3) 1
4) 0
3.51. Figure 3.9 illustrates part of the function of y ¼
Figure 3.9 The graph of problem 3.51
3
Problems: Trigonometric Equations and Identities
3.52. Figure 3.10 shows part of the function of y ¼ 1 + sin (mx). Determine the value of the function for x ¼
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 0
1
2)
2
3) 1
4) 2
53
7π
.
6
Figure 3.10 The graph of problem 3.52
3.53. Figure 3.11 shows part of the function of y ¼ a - sin (bπx). Determine the value of the function for x ¼
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 2
5
2)
2
3) 3
7
4)
2
Figure 3.11 The graph of problem 3.53
25
.
3
54
3
Problems: Trigonometric Equations and Identities
π
x for 0 < x < 4. Determine the value of b.
2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) -2
2) -1
3) 1
4) 2
3.54. Figure 3.12 shows the function of y ¼ a þ b cos
Figure 3.12 The graph of problem 3.54
4
3.55. Figure 3.13 shows the function of y ¼ 1 + a sin (bπx) for 0 < x < . Determine the value of a + b.
3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 3
2) 4
3) 5
4) 6
Figure 3.13 The graph of problem 3.55
3.56. Figure 3.14 shows part the function of y ¼ a - 2 cos bx þ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1
1)
2
2) 1
3
3)
2
4) 2
π
. Determine the value of a + b.
2
3
Problems: Trigonometric Equations and Identities
55
Figure 3.14 The graph of problem 3.56
3
3.57. Calculate the value of cos(25 ° - α) if tanðα þ 20 ° Þ ¼ .
4
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 5
2) 6
3) 7
4) 8
π
3
þ α assuming that α is an acute angle and sinðαÞ ¼ .
4
5
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) -7
1
2) 7
1
3)
7
4) 7
3.58. Calculate the value of tan
2
π
π
- α if tan - α ¼ .
4
2
3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1
1) 3
1
2) 5
1
3)
5
1
4)
3
3.59. Calculate the value of tan
3.60. Calculate the value of tan(2a) while we know that tanða þ bÞ ¼
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1
1) 3
1
2) 2
3) 3
4) 1
2
3
and tanða - bÞ ¼ .
5
7
56
3
3.61. Calculate the value of tan(x) if we have:
π
4 ¼2
π
cos x 4
sin x -
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) -3
1
2)
3
2
3)
3
4) 3
π
3.62. Calculate the value of (1 + tan (α))(1 + tan (β)) if α þ β ¼ .
4
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) -2
2) 2
1
3)
3
1
4) 2
π
π
þ α - tan - α .
4
4
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 2 tan (2α)
2) 2 cos (2α)
3) 0
4) 2 sin (2α)
3.63. Calculate the value of tan
π
1
-α ¼ .
4
5
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1.5
2) 1.8
3) 2.4
4) 2.5
3.64. Calculate the value of tan(2α) if tan
1
3.65. Calculate the value of tan(2α - β) if tan(α) ¼ 2 and tanðβÞ ¼ .
3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) -3
2) -2
3) 0.5
4) 3
Problems: Trigonometric Equations and Identities
3
Problems: Trigonometric Equations and Identities
57
3.66. Determine the common solution of the equation of cos(3x) + cos (x) ¼ 0 assuming cos(x) ≠ 0.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
kπ π
1)
þ
2 4
kπ π
2)
þ
2 8
π
3) kπ 4
π
4) kπ þ
4
3.67. Calculate the sum of the positive acute roots of the equation of tan(4x) ¼ cot (x).
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
2π
1)
5
4π
2)
5
3π
3)
5
π
4)
5
3.68. Determine the common solution of the equation of 2sin2(x) + 3 cos (x) ¼ 0.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
2π
1) 2kπ ±
3
π
2) 2kπ ±
3
5π
3) 2kπ ±
6
π
4) kπ 3
3.69. Determine the common solution of the equation of 2sin2(x) ¼ 3 cos (x).
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
π
1) kπ ±
6
π
2) kπ ±
3
π
3) 2kπ ±
6
π
4) 2kπ ±
3
3.70. Two lines with the equations of x tan (α) + y cot (α) ¼ 2 and x tan (α) - y cot (α) ¼ 1 are intersecting each other at point
M. By changing the value of α, what is the position equation of the point?
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1
1) y ¼
x
3
2) y ¼
x
1
3) y ¼
4x
3
4) y ¼
4x
58
3
Problems: Trigonometric Equations and Identities
3.71. What is the position equation of the point of (2 - 3 sin (α), 1 + 4 cos (α)) if the value of α changes?
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1) Circle
2) Ellipse
3) Parabola
4) Hyperbola
3.72. What is the position equation of the point of (2 - 5 cos (α), 4) if the value of α changes?
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1) A horizontal line
2) A vertical line
3) A horizontal line segment
4) A vertical line segment
3.73. Calculate the value of y if 2 cos (x - y) + 3 sin (x + y) ¼ 5 and 0 < x, y < 2π.
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
π
2π
1) or
3
3
π
5π
2) or
4
4
π
5π
3) or
6
6
π
3π
4) or
2
2
π
3.74. Calculate the value of m if tan(α) ≠ tan (β), α þ β ¼ , and α and β are the two roots of the equation below.
4
tan 2 ðxÞ þ ðm þ 2Þ tanðxÞ þ 2m - 2 ¼ 0
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 3
3) 5
4) 7
3.75. Calculate the final value of the following relation:
sin 6 ðαÞ þ cos 6 ðαÞ þ 3 sin 2 ðαÞ cos 2 ðαÞ
sin 4 ðαÞ þ cos 4 ðαÞ þ 2 sin 2 ðαÞ cos 2 ðαÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1) sin2(α)
2) cos2(α)
3) sin2(α) - cos2(α)
4) 1
3
Problems: Trigonometric Equations and Identities
59
3.76. Calculate the final value of the relation below.
sinð135 ° Þ cosð210 ° Þ þ cosð135 ° Þ sinð420 ° Þ
tanð210 ° Þ cotð420 ° Þ þ cotð120 ° Þ tanð330 ° Þ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
p
6
1) 4
p
3 6
2) 4
p
6
3) 2
p
3 6
4) 2
π
3.77. Calculate the final value of (1 + cot (x))(1 + cot ( y)) if x þ y ¼ kπ þ .
4
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) tan(x) tan ( y)
2) 2 tan (x) tan ( y)
3) cot(x) cot ( y)
4) 2 cot (x) cot ( y)
3.78. Determine the common solution of the equation below.
ðsinðxÞ - tanðxÞÞ tan
3π
4π
- x ¼ cos
2
3
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
π
1) kπ 6
π
2) kπ þ
3
π
3) 2kπ ±
3
π
4) 2kπ ±
6
3.79. Calculate the sum of the roots of the equation below for x 2 [0, π].
sinð2xÞðsinðxÞ þ cosðxÞÞ ¼ cosð2xÞðcosðxÞ - sinðxÞÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
3π
1)
4
5π
2)
4
3π
3)
2
7π
4)
4
60
3
3.80. Determine the common solution of the equation of
p
2 sin
Problems: Trigonometric Equations and Identities
π
5π
þx .
- x ¼ 1 þ sin
4
2
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
π
1) kπ þ
2
π
2) 2kπ 4
π
3) 2kπ 2
π
4) 2kπ þ
2
3.81. Which one of the following choices shows one of the common solutions of the equation of cosð2xÞ þ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
π
1) kπ 6
π
2) kπ 3
π
3) kπ þ
6
π
4) kπ þ
3
3.82. Calculate the value of the following term:
cos 3α þ sin α sin 2α sin α
sin 3α - sin 2α cos α cos α
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) tanα
2) cotα
3) 1
4) -1
References
1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
p
3 sinð2xÞ ¼ 1?
4
Solutions of Problems: Trigonometric Equations
and Identities
Abstract
In this chapter, the problems of the third chapter are fully solved, in detail, step-by-step, and with different methods.
4.1. From trigonometry, we know that [1, 2]:
sin 2x = 2 sinðxÞ cosðxÞ
1 -cos 2x = 2 sin 2 x
sin 30 ° =
1
2
Therefore:
sin x cos x 1 - 2 sin 2 x =
1
1
sin 2x ðcos 2xÞ = sin 4x
4
2
For x = 7.5°, we have:
1
1
1
× sinð4 × 7:5 ° Þ = sin 30 ° =
4
4
8
Choice (2) is the answer.
4.2. Based on the information given in the problem, we have:
tan x þ cot x = 3
From algebra, we know that:
a3 þ b3 = ða þ bÞ3 - 3abða þ bÞ
In addition, from trigonometry, we know that:
tan x cot x = 1
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_4
61
62
4 Solutions of Problems: Trigonometric Equations and Identities
Therefore:
tan 3 x þ cot 3 x = ðtan x þ cot xÞ3 - 3 tan x cot xðtan x þ cot xÞ = ð3Þ3 - 3ð1Þð3Þ
) tan 3 x þ cot 3 x = 18
Choice (1) is the answer.
4.3. From trigonometry of hyperbolic functions, we know that:
coshða ± bÞ = cosh a cosh b ± sinh a sinh b
sinhða ± bÞ = sinh a cosh b ± cosh a sinh b
Choice (4) is the answer.
4.4. From trigonometry, we know that:
tanðθÞ =
tanð2θÞ =
1
cotðθÞ
2 tanðθÞ
1 - tan 2 ðθÞ
Based on the information given in the problem:
cotðθÞ = 5 ) tanðθÞ =
1
5
Therefore:
2 tanðθÞ
tanð2θÞ =
=
1 - tan 2 ðθÞ
) tanð2θÞ =
1
5
1
15
2×
2
2
= 5
24
25
5
12
Choice (1) is the answer.
4.5. From trigonometry, we know that:
tanðα þ nπ Þ = tanðαÞ, 8n 2 ℤ
tanð- αÞ = - tanðαÞ
p
tan 60 ° = 3
Therefore:
tan - 2100 ° = -tan 2100 ° = -tan 12 × 180 - 60 ° = -tan - 60 ° = tan 60 °
p
) tan - 2100 ° = 3
Choice (1) is the answer.
4
Solutions of Problems: Trigonometric Equations and Identities
63
4.6. From trigonometry, we know that:
1 þ cosðθÞ = 2 cos 2
sinðθÞ = 2 sin
cotðθÞ =
θ
2
θ
θ
cos
2
2
cosðθÞ
sinðθÞ
Therefore:
1 þ cos 40 °
2 cos 2 20 °
cos 20 °
=
=
sin 40 °
2 sin 20 ° cos 20 °
sin 20 °
)
1 þ cos 40 °
= cot 20 °
sin 40 °
Choice (4) is the answer.
4.7. For the given range of α, we can conclude that:
π
2π
1
≤α≤
) ≤ sinðαÞ ≤ 1
6
3
2
Therefore, based on the given information, i.e., sinðαÞ =
3m - 1
4 ,
we can write:
1 3m - 1
≤
≤ 1 ) 2 ≤ 3m - 1 ≤ 4 ) 3 ≤ 3m ≤ 5
2
4
) 1≤m≤
5
3
Choice (4) is the answer.
4.8. For the given range of x, we can conclude that:
-
π
π
1
≤ x ≤ ) ≤ cosðxÞ ≤ 1
3
3
2
Therefore, based on the given information, i.e., cosðxÞ =
2m - 1
6 ,
we can write:
1 2m - 1
≤
≤ 1 ) 3 ≤ 2m - 1 ≤ 6 ) 4 ≤ 2m ≤ 7
2
6
) 2≤m≤
Choice (1) is the answer.
7
2
64
4 Solutions of Problems: Trigonometric Equations and Identities
4.9. From trigonometry, we know that:
f 1 ðxÞ = cos 2n ðaxÞ, 8n 2 ℤ ) T 1 =
π
j aj
f 2 ðxÞ = cos 2nþ1 ðaxÞ, 8n 2 ℤ ) T 2 =
2π
jaj
Therefore:
f 1 ðxÞ = cos 2 ðxÞ ) T 1 =
f 2 ðxÞ = - 5 cos
π
=π
1
2x
2π
) T 2 = 2 = 3π
3
3
The main period of the given expression is the least common multiple (LCM) of the main periods of the terms, as can be
seen in the following:
T = LCMðπ, 3π Þ
) T = 3π
Choice (3) is the answer.
4.10. From trigonometry, we know that:
f 1 ðxÞ = sin 2n ðaxÞ, 8n 2 ℤ ) T 1 =
π
jaj
f 2 ðxÞ = cos 2nþ1 ðaxÞ, 8n 2 ℤ ) T 2 =
2π
jaj
Therefore:
f 1 ðxÞ = sin 4
π 5π
3x
) T1 = 3 =
3
5
5
f 2 ðxÞ = cos 3
2π
2x
) T 2 = 2 = 3π
3
3
The main period of the given expression is: the least common multiple (LCM) of the main periods of the terms as
follows:
T = LCM
5π
, 3π
3
) T = 15π
Choice (3) is the answer.
4
Solutions of Problems: Trigonometric Equations and Identities
65
4.11. From trigonometry, we know that:
f 1 ðxÞ = sin 2n ðaxÞ, 8n 2 ℤ ) T 1 =
π
jaj
f 2 ðxÞ = cos 2nþ1 ðaxÞ, 8n 2 ℤ ) T 2 =
2π
jaj
Therefore:
f 1 ðxÞ = sin 4
π
πx
) T1 = π = 3
3
3
f 2 ðxÞ = cosðπxÞ ) T 2 =
2π
=2
π
The main period of the given expression is: the least common multiple (LCM) of the main periods of the terms, as can
be seen in the following:
T = LCMð3, 2Þ
) T =6
Choice (4) is the answer.
4.12. From trigonometry, we know that:
y = sinðkxÞ ) T =
2π
jk j
Therefore:
)
3π 1 2π
=
4
2 jk j
) jk j =
4
4
) k= ±
3
3
Based on the graph and the function, the positive value of k is acceptable.
) k=
4
3
Choice (4) is the answer (Fig. 4.1).
Figure 4.1 The graph of solution of problem 4.12
66
4 Solutions of Problems: Trigonometric Equations and Identities
4.13. From trigonometry, we know that:
y = cos πax þ
π
= -sinðπaxÞ
2
y = sinðmxÞ ) T =
2π
jmj
Therefore:
) 1- -
1
2π
4
2
3
=
) =
) a= ±
3
3
2
jπaj
j aj
Based on the graph and y = - sin (πax), the positive value of a is acceptable.
) a=
3
2
Choice (2) is the answer (Fig. 4.2).
Figure 4.2 The graph of solution of problem 4.13
4.14. From trigonometry, we know that:
sinðα þ 2nπ Þ = sinðαÞ, 8n 2 ℤ
cosðα þ 2nπ Þ = cosðαÞ, 8n 2 ℤ
tanðα þ nπ Þ = tanðαÞ, 8n 2 ℤ
cotðα þ nπ Þ = cotðαÞ, 8n 2 ℤ
sinðπ - αÞ = sinðαÞ
cosðπ - αÞ = -cosðαÞ
tanð- αÞ = -tanðαÞ
cotð- αÞ = -cotðαÞ
4
Solutions of Problems: Trigonometric Equations and Identities
67
Based on the information given in the problem, we have:
α þ β = 19π ) α = 19π - β
Therefore:
sinðαÞ = sinð19π - βÞ = sinðπ - βÞ = sinðβÞ
cosðαÞ = cosð19π - βÞ = cosðπ - βÞ = -cosðβÞ
tanðαÞ = tanð19π - βÞ = tanð- βÞ = -tanðβÞ
cotðαÞ = cotð19π - βÞ = cotð- βÞ = -cotðβÞ
Choice (1) is the answer.
4.15. From trigonometry, we know that:
sinðα þ 2nπ Þ = sinðαÞ, 8n 2 ℤ
cosðα þ 2nπ Þ = cosðαÞ, 8n 2 ℤ
sinðα þ π Þ = -sinðαÞ
sinðαÞ þ sinðβÞ = 2 sin
αþβ
α-β
cos
2
2
Therefore:
sinð5π þ xÞ þ sin x -
π
7π
π
π
þ sin x þ
= sinðx þ π Þ þ sin x þ sin x þ
3
3
3
3
π
= -sinðxÞ þ 2 sinðxÞ cos
= -sinðxÞ þ sinðxÞ
3
) sinð5π þ xÞ þ sin x -
π
7π
þ sin x þ
=0
3
3
Choice (1) is the answer.
4.16. Let us assume:
cos
2π
1
1
≜ - ) arc cos 3
2
2
p
-π
1
- 3
sin
≜ - ) arc sin
3
2
2
=
2π
3
=
-π
3
Therefore:
1
sin arc cos 2
p
- 3
þ arc sin
2
= sin
2π
-π
þ
3
3
= sin
π
3
68
4 Solutions of Problems: Trigonometric Equations and Identities
p
p
- 3
2
) sin arc cos -
1
2
þ arc sin
arc tan
1
2
≜ α ) tanðαÞ =
=
3
2
Choice (1) is the answer.
4.17. Let us assume:
We need to find the value of 2 arctan
1
2
1
which is equal to 2α.
2
From trigonometry, we know that:
tanð2αÞ =
2 tan α
1 - tan 2 α
Hence:
tanð2αÞ =
2×
1-
1
2
1 2
2
) 2α = arc tan
=
4
3
4
3
Choice (1) is the answer.
4.18. As we know:
e ln a = a
Moreover, from trigonometry of hyperbolic functions, we know that:
cos
π
1
=
3
2
sinh x =
ex - e - x
2
Thus, for x = ln 3, we can write:
sinh ln 3 =
e ln 3 - e - ln 3 3 =
2
2
1
3
=
4
3
Therefore,
cosðπ sinh ln 3Þ = cos π ×
4
π
π
= cos π þ
= -cos
3
3
3
) cosðπ sinh ln 3Þ = Choice (2) is the answer.
1
2
4
Solutions of Problems: Trigonometric Equations and Identities
69
4.19. From trigonometry, we know that:
sin 30 ° =
1
2
αþβ
α-β
cos
2
2
sinðαÞ þ sinðβÞ = 2 sin
Therefore:
sin 50 ° þ sin 10 ° = m
) 2 sin 30 ° cos 20 ° = m ) 2 ×
1
cos 20 ° = m ) cos 20 ° = m
2
Choice (2) is the answer.
4.20. From trigonometry, we know that:
cosðθÞ =
1 - tan 2
1 þ tan 2
tanðθÞ =
θ
2
θ
2
sinðθÞ
cosðθÞ
tan 45 ° = 1
Therefore:
1 þ tan 2 5 °
sin 10 °
°
tan 10 °
1 - tan 2 5
)
=
sin 10 °
1
×
=1
°
sinð10 ° Þ
cos 10
°
cosð10 Þ
1 þ tan 2 5 °
sin 10 °
1 - tan 2 5 °
tan 10 °
= tan 45 °
Choice (4) is the answer.
4.21. From trigonometry, we know that:
1
= 1 þ tan 2 ðαÞ
cos 2 ðαÞ
tanðαÞ =
1
cotðαÞ
Based on the information given in the problem, we have:
cotðαÞ = m
cosðαÞ = n
70
4 Solutions of Problems: Trigonometric Equations and Identities
Therefore:
1
1
=1 þ
cos 2 ðαÞ
cot 2 ðαÞ
1
1 × m 2 n2 2
=1 þ 2 ¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) m = m2 n2 þ n2
2
n
m
)
) m2 1 - n2 = n2
Choice (2) is the answer.
4.22. From trigonometry, we can know that:
sin 2nþ1 ðaxÞ, 8n 2 ℤ ) T =
sinðαÞ cosðβÞ =
2π
jaj
1
ðsinðα þ βÞ þ sinðα - βÞÞ
2
We need to change the product expression to the summation one, as follows:
y = sinð3xÞ cosð5xÞ þ 11 ) y =
1
1
sinð8xÞ - sinð2xÞ þ 11
2
2
Then:
2π π
1
=
sinð8xÞ ) T 1 =
8
4
2
-
1
2π
sinð2xÞ ) T 2 =
=π
2
2
The main period of the given expression is: the least common multiple (LCM) of the main periods of the terms, as is
presented in the following:
) T = LCM
π
,π
4
) T =π
Choice (1) is the answer.
4.23. From trigonometry, we know that:
sin
p
4π
π
π
3
= sin π þ
= -sin
=3
3
2
3
arcðcosð- αÞÞ = π - arcðcosðαÞÞ
p
arc cos
3
2
=
π
6
4
Solutions of Problems: Trigonometric Equations and Identities
71
Therefore:
arc cos sin
p
4π
3
= arc cos -
p
3
2
) arc cos sin
= π - arc cos
4π
3
=
3
2
=π-
π
6
5π
6
Choice (2) is the answer.
4.24. From trigonometry, we know that:
17π
3π
3π
= sin 4π = sin 5
5
5
sin
sin -
3π
3π
≜ α ) arcðsinðαÞÞ = 5
5
Therefore:
arc sin sin
17π
5
= arc sin sin -
) arc sin sin
17π
5
3π
5
=-
= arcðsinðαÞÞ
3π
5
Choice (4) is the answer.
4.25. From trigonometry, we know that:
cos
19π
π
π
= cos 4π = cos 5
5
5
cos
π
π
≜ α ) arcðcosðαÞÞ =
5
5
Therefore:
arc cos cos
19π
5
= arc cos cos -
π
5
) arc cos cos
= arc cos cos
19π
5
=
π
5
Choice (1) is the answer.
4.26. From trigonometry, we know that:
tanð2αÞ =
arc tan
1
2
2 tanðαÞ
1 - tan 2 ðαÞ
≜ α ) tanðαÞ =
1
2
π
5
= arcðcosðαÞÞ
72
4 Solutions of Problems: Trigonometric Equations and Identities
Therefore:
1
2
tan 2arc tan
= tanð2αÞ =
2 × 12
2 tanðαÞ
1
= 3
=
2
1 - tan ðαÞ 1 - 1 2
4
2
1
2
) tan 2arc tan
=
4
3
Choice (3) is the answer.
4.27. From trigonometry, we know that:
sin 2 ðαÞ þ cos 2 ðαÞ = 1
1 þ tan 2 ðαÞ =
1
cos 2 ðαÞ
Therefore:
arc sin
arc tan
) sin arc sin
3
5
3
4
þ arc tan
3
5
≜ α ) sinðαÞ =
≜ β ) tanðβÞ =
3
4
3
4
) cosðαÞ =
5
5
3
4
3
) cosðβÞ = ) sinðβÞ =
4
5
5
= sinðα þ βÞ = sinðαÞ cosðβÞ þ cosðαÞ sinðβÞ =
3
5
) sin arc sin
þ arc tan
3
4
=
3 4 4 3
× þ ×
5 5 5 5
24
25
Choice (4) is the answer.
4.28. From trigonometry, we know that:
arcðcotð- αÞÞ = π - arcðcotðαÞÞ
arcðcotðαÞÞ þ arc cot
1
α
=
π
2
Therefore:
arc cot -
4
3
- arc cot
3
4
= π - arc cot
4
3
- arc cot
) arc cot Choice (3) is the answer.
4
3
3
4
- arc cot
= π - arc cot
3
4
=
π
2
4
3
þ arc cot
3
4
=π-
π
2
4
Solutions of Problems: Trigonometric Equations and Identities
73
4.29. From trigonometry, we know that:
tanðα þ βÞ =
tanðαÞ þ tanðβÞ
1 -tanðαÞ tanðβÞ
arcðtanð5ÞÞ≜ α ) tanðαÞ = 5
arc tan
3
2
≜ β ) tanðβÞ =
arcðtanð- 1ÞÞ = tan - 1 ð- 1Þ =
3
2
3π
4
Therefore:
tan arcðtanð5ÞÞ þ arc tan
3
2
= tanðα þ βÞ =
) arcðtanð5ÞÞ þ arc tan
13
5 þ 32
tanðαÞ þ tanðβÞ
2
=
=
= -1
1 -tanðαÞ tanðβÞ 1 - 15
- 13
2
2
3
2
) arcðtanð5ÞÞ þ arc tan
= tan - 1 ð- 1Þ
3
2
=
3π
4
Choice (3) is the answer.
4.30. From trigonometry, we know that:
sin 2 ðαÞ þ cos 2 ðαÞ = 1
3
5
arc cos
3
5
≜ α ) cosðαÞ =
arc sin -
4
5
≜ β ) sinðβÞ = -
4
5
Therefore:
sin arc cos
3
5
þ cos arc sin -
4
5
= sinðαÞ þ cosðβÞ =
) sin arc cos
3
5
þ cos arc sin -
1-
3
5
2
4
5
=
7
5
Choice (1) is the answer.
4.31. From trigonometry, we know that:
cotðθÞ = tan
π
-θ
2
Based on the definition of tan(θ) and cot(θ), we can write:
tanðθÞ =
Opposite for θ
HA
HA
=
=
= HA
Adjacent for θ OH
1
þ
1- -
4
5
2
=
4 3
þ
5 5
74
4 Solutions of Problems: Trigonometric Equations and Identities
cotðθÞ = tan HOB =
Opposite for HOB
HB
HB
=
=
= HB
OH
1
Adjacent for HOB
Choice (2) is the answer (Fig. 4.3).
Figure 4.3 The graph of solution of problem 4.31
4.32. Based on the definition of sec(θ) and cos(θ), we can write:
secðθÞ =
1
=
cosðθÞ
1
Adjacent for θ
Hypotenuse for θ
=
1
OA
OB
=
1
1
OB
= OB
Choice (3) is the answer (Fig. 4.4).
Figure 4.4 The graph of solution of problem 4.32
4.33. From trigonometry, we know that:
sinðθÞ = cos
π
-θ
2
Based on the definition of csc(θ) and sin(θ), we can write:
cscðθÞ =
1
1
=
=
sinðθÞ
cos COB
1
Adjacent for COB
Hypotenuse for COB
Choice (2) is the answer (Fig. 4.5).
=
1
OC
OB
=
1
1
OB
= OB
4
Solutions of Problems: Trigonometric Equations and Identities
75
Figure 4.5 The graph of solution of problem 4.33
4.34. From trigonometry, we know that:
tanðα þ βÞ =
tanðαÞ þ tanðβÞ
1 -tanðαÞ tanðβÞ
arc tan
2
3
≜ α ) tanðαÞ =
2
3
arc tan
1
5
≜ β ) tanðβÞ =
1
5
π
4
arcðtanð1ÞÞ =
Therefore:
tanðα þ βÞ =
2
þ1
tanðαÞ þ tanðβÞ
= 3 52 =
1 -tanðαÞ tanðβÞ 1 - 15
13
15
13
15
=1
) α þ β = arcðtanð1ÞÞ
) αþβ=
π
4
Choice (2) is the answer.
4.35. From trigonometry, we know that:
1
arcðtanðαÞÞ þ arc tan
α
=
π
2
π
2
if α > 0
if α < 0
arcðcotðαÞÞ þ arcðcotð- αÞÞ = π
Therefore:
1
arcðtanðmÞÞ þ arc tan
m
þ arcðcotðmÞÞ þ arcðcotð- mÞÞ = π þ
π
2
π
2
if m > 0
if m < 0
76
4 Solutions of Problems: Trigonometric Equations and Identities
1
) arcðtanðmÞÞ þ arc tan
m
þ arcðcotðmÞÞ þ arcðcotð- mÞÞ =
3π
2
π
2
if m > 0
if m < 0
Choice (2) is the answer.
4.36. From trigonometry, we know that:
cosðα - βÞ = cosðαÞ cosðβÞ þ sinðαÞ sinðβÞ
Therefore:
- 1 ≤ cosð4xÞ cosð2xÞ þ sinð4xÞ sinð2xÞ ≤ 0
) - 1 ≤ cosð4x - 2xÞ ≤ 0 ) - 1 ≤ cosð2xÞ ≤ 0
Since x is an acute angle:
)
π
π
π
≤ 2x ≤ π ) ≤ x ≤
2
4
2
Choice (4) is the answer.
4.37. From trigonometry, we know that:
tanðα - βÞ =
tanðαÞ -tanðβÞ
1 þ tanðαÞ tanðβÞ
Therefore:
tanð2yÞ = tanððx þ yÞ - ðx - yÞÞ =
tanðx þ yÞ -tanðx - yÞ
5-7
-2
=
=
1 þ tanðx þ yÞ tanðx - yÞ 1 þ 5 × 7 1 þ 35
) tanð2yÞ =
-1
18
Choice (2) is the answer.
4.38. From trigonometry, we know that:
p
π
sinðαÞ þ cosðαÞ = 2 sin α þ
4
p
π
sinðαÞ -cosðαÞ = 2 sin α 4
p
2π
3
sin
=
3
2
sin
π
1
=
6
2
4
Solutions of Problems: Trigonometric Equations and Identities
77
Therefore:
p
5π
2 sin
sin 5π
12 þ cos 12
=p
5π
5π
sin 12 -cos 12
2 sin
)
5π
12
5π
12
þ π4
-
π
4
sin 2π
3
=
=
sin π6
p
3
2
1
2
5π
p
sin 5π
12 þ cos 12
= 3
5π
5π
sin 12 -cos 12
Choice (1) is the answer.
4.39. From trigonometry, we know that:
y = a sinðmxÞ ) T =
2π
jmj
Therefore:
) 6=
2π
1
1
) jbj = ) b = ±
3
3
jbπ j
Based on the graph and the function, the positive value of b is acceptable.
) b=
1
3
Moreover, based on y = a sin (bπx) and the given graph, it is concluded that a = 2. Therefore:
) a þ b=2 þ
) a þ b=
1
3
7
3
Choice (3) is the answer (Fig. 4.6).
Figure 4.6 The graph of solution of problem 4.39
78
4 Solutions of Problems: Trigonometric Equations and Identities
4.40. From trigonometry, we know that:
y = a sinðmxÞ ) T =
2π
jmj
Therefore:
) 3=3×
2π
= 1 ) jbj = 2 ) b = ± 2
jbπ j
Based on the graph and the given function, the negative value of b is accepted.
) b= -2
In addition, based on y = a sin (bπx) and the given graph, it is clear that a = 3. Therefore:
) a × b = 3 × ð- 2Þ
) a×b= -6
Choice (1) is the answer (Fig. 4.7).
Figure 4.7 The graph of solution of problem 4.40
4.41. From trigonometry, we know that:
y = a sin
π
þ bπx = a cosðbπxÞ
2
y = a cosðmxÞ ) T =
2π
jmj
Therefore:
) 3:5 - ð- 2:5Þ = 3 ×
2π
6
) 6=
) j bj = 1 ) b = ± 1
jbπ j
j bj
Based on the graph and y = a cos (bπx), the positive value of b is accepted.
) b=1
4
Solutions of Problems: Trigonometric Equations and Identities
79
In addition, based on y = a cos (bπx) and the given graph, it is clear that a = 2. Therefore:
) a×b=2×1
) a×b=2
Choice (1) is the answer (Fig. 4.8).
Figure 4.8 The graph of solution of problem 4.41
4.42. From trigonometry, we know that:
cosðαÞ = sin
π
-α
2
sinðαÞ = cos
π
-α
2
sinð2αÞ = 2 sinðαÞ cosðαÞ
Therefore:
cos 5 ° cos 10 ° cos 20 °
cos 50 °
=
=
cos 5 ° cos 10 ° cos 20 °
cos 5 ° cos 10 ° cos 20 °
=
sin 40 °
2 sin 20 ° cos 20 °
cos 5 ° cos 10 °
cos 5 °
1
=
=
2 × 2 sin 10 ° cos 10 °
4 × 2 sin 5 ° cos 5 °
8 sin 5 °
)
cos 5 ° cos 10 ° cos 20 °
cos 50
°
=
Choice (2) is the answer.
4.43. From trigonometry, we know that:
sinðxÞ =
2 tan 2x
1 þ tan 2
x
2
cosðxÞ =
1 - tan 2
1 þ tan 2
x
2
x
2
1
8 cos 85 °
80
4 Solutions of Problems: Trigonometric Equations and Identities
Therefore:
sinðxÞ þ cosðxÞ =
) 12 tan
2
2 tan 2x
7
)
5
1 þ tan 2
x
2
þ
1 - tan 2
1 þ tan 2
x
2
x
2
2 tan 2x þ 1 - tan 2
7
)
5
1 þ tan 2 2x
=
x
2
=
p
x
x
x
10 ± 102 - 4 × 12 × 2 10 ± 2
- 10 tan
þ 2 = 0 ) tan
=
=
24
2
2
2
24
) tan
x
1
1
= or
2
2
3
Choice (2) is the answer.
4.44. From trigonometry, we know that:
sinð2αÞ = 2 sinðαÞ cosðαÞ
sin 2 ðαÞ þ cos 2 ðαÞ = 1
cos 2 ðαÞ - sin 2 ðαÞ = cosð2αÞ
cotð2αÞ =
cosð2αÞ
sinð2αÞ
In addition, from factoring rule, we know that:
a4 - b 4 = a 2 - b 2 a 2 þ b 2
Therefore:
sin 2 ðαÞ - cos 2 ðαÞ sin 2 ðαÞ þ cos 2 ðαÞ
sin 4 ðαÞ - cos 4 ðαÞ
-cosð2αÞ × 1
= 1
=
sinðαÞ cosðαÞ
sinðαÞ cosðαÞ
2 sinð2αÞ
)
sin 4 ðαÞ - cos 4 ðαÞ
= - 2 cotð2αÞ
sinðαÞ cosðαÞ
Choice (2) is the answer.
4.45. Based on the information given in the problem, we have:
sin 4 ðαÞ þ cos 4 ðαÞ =
1
2
From trigonometry, we know that:
sinð2αÞ = 2 sinðαÞ cosðαÞ
sin 2 ðαÞ þ sin 2 ðαÞ = 1 ) sin 2 ðαÞ þ sin 2 ðαÞ
2
=1
) sin 4 ðαÞ þ cos 4 ðαÞ þ 2 sin 2 ðαÞ cos 2 ðαÞ = 1
7
5
4
Solutions of Problems: Trigonometric Equations and Identities
81
Therefore:
1
þ 2 sin 2 ðαÞ cos 2 ðαÞ = 1 ) 4 sin 2 ðαÞ cos 2 ðαÞ = 1 ) ð2sinðαÞ cosðαÞÞ2 = 1
2
) sin 2 ð2αÞ = ± 1 ) cos 2 ð2αÞ = 0 )
cos 2 ð2αÞ
=0
sin 2 ð2αÞ
) cot 2 ð2αÞ = 0
Choice (3) is the answer.
4.46. From trigonometry, we know that:
sinð2xÞ = 2 sinðxÞ cosðxÞ
cosð2xÞ = cos 2 ðxÞ - sin 2 ðxÞ
Therefore:
sin 3 ðxÞ cosðxÞ - cos 3 ðxÞ sinðxÞ þ 3 sin 2 ðxÞ cos 2 ðxÞ
= sinðxÞ cosðxÞ sin 2 ðxÞ - cos 2 ðxÞ þ
=
1
3
sinð2xÞð -cosð2xÞÞ þ sin 2 ð2xÞ
4
2
=-
For x =
3
× 4 sin 2 ðxÞ cos 2 ðxÞ
4
1
3
sinð4xÞ þ sin 2 ð2xÞ
4
4
3π
, we have:
8
3π
1
3
1
3π
3
= - ð- 1Þ þ
þ sin 2 2 ×
- sin 4 ×
8
4
4
8
4
4
p
2
2
2
=
1 3 5
þ =
4 8 8
Choice (2) is the answer.
4.47. From trigonometry, we know that:
sinðαÞ þ sinðβÞ = 2 sin
αþβ
α-β
cos
2
2
cosðαÞ þ cosðβÞ = 2 cos
α-β
αþβ
cos
2
2
Therefore:
sinð2αÞ þ sinð5αÞ þ sinð8αÞ
sinð8αÞ þ sinð2αÞ þ sinð5αÞ
=
cosð2αÞ þ cosð5αÞ þ cosð8αÞ
cosð8αÞ þ cosð2αÞ þ cosð5αÞ
=
sinð5αÞð2cosð3αÞ þ 1Þ
2 sinð5αÞ cosð3αÞ þ sinð5αÞ
=
= tanð5αÞ
2 cosð5αÞ cosð3αÞ þ cosð5αÞ
cosð5αÞð2cosð3αÞ þ 1Þ
82
4 Solutions of Problems: Trigonometric Equations and Identities
π
α=
p
15
π
π
= tan
= 3
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) tan 5 ×
¼
15
3
Choice (3) is the answer.
4.48. From trigonometry, we know that:
sin 2 ðxÞ þ cos 2 ðxÞ = 1
2 sinðxÞ cosðxÞ = sinð2xÞ
In addition, from factoring rule, we know that:
ð a þ bÞ ð a - b Þ = a 2 - b 2
Therefore:
ðsinðxÞ -cosðxÞ þ 2ÞðsinðxÞ -cosðxÞ - 2Þ
= ðsinðxÞ -cosðxÞÞ2 - 4 = sin 2 ðxÞ þ cos 2 ðxÞ
- 2 sinðxÞ cosðxÞ - 4 = 1 -sinð2xÞ - 4
= - 3 -sinð2xÞ
π
x=
12
π
1
7
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) - 3 -sin
= -3- = 6
2
2
Choice (4) is the answer.
4.49. From trigonometry, we know that:
tanðαÞ cotðαÞ = 1
1 þ tan 2 ðαÞ =
1
cos 2 ðαÞ
1 þ cot 2 ðαÞ =
1
sin 2 ðαÞ
sin 2 ðαÞ þ cos 2 ðαÞ = 1
Therefore:
4 sin 2 ðαÞ cos 2 ðαÞðtanðαÞ þ cotðαÞÞ2
= 4 sin 2 ðαÞ cos 2 ðαÞ tan 2 ðαÞ þ cot 2 ðαÞ þ 2 tanðαÞ cotðαÞ
= 4 sin 2 ðαÞ cos 2 ðαÞ 1 þ tan 2 ðαÞ þ 1 þ cot 2 ðαÞ
4
Solutions of Problems: Trigonometric Equations and Identities
83
= 4 sin 2 ðαÞ cos 2 ðαÞ
1
1
þ
2
cos ðαÞ sin 2 ðαÞ
= 4 sin 2 ðαÞ cos 2 ðαÞ
sin 2 ðαÞ þ cos 2 ðαÞ
sin 2 ðαÞ cos 2 ðαÞ
= 4 sin 2 ðαÞ cos 2 ðαÞ
1
=4
sin ðαÞ cos 2 ðαÞ
2
Choice (4) is the answer.
4.50. From trigonometry, we know the common solution of the equations below.
sinðαÞ = 0 ) α = kπ, 8k 2 ℤ
π
cosðαÞ = 0 ) α = kπ þ , 8k 2 ℤ
2
tanðαÞ = - 1 ) α = kπ -
π
, 8k 2 ℤ
4
Hence:
sinðπxÞ cos 2 ðπxÞ þ sin 2 ðπxÞ cosðπxÞ = 0 ) sinðπxÞ cosðπxÞðcosðπxÞ þ sinðπxÞÞ = 0
)
-2≤x≤2
sinðπxÞ = 0 ) πx = kπ ) x = k ¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) x = - 2, - 1, 0, 1, 2
3
1 1 3
π
1 -2≤x≤2
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) x = - , - , ,
cosðπxÞ = 0 ) πx = kπ þ ) x = k þ ¼
¼
¼
¼
¼
¼
¼
2
2 2 2
2
2
5
1 3 7
π
1 -2≤x≤2
sinðπxÞ þ cosðπxÞ = 0 ) tanðπxÞ = - 1 ) πx = kπ - ) x = k - ¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) x = - , - , ,
¼
¼
¼
¼
¼
¼
¼
¼
4
4 4 4
4
4
Therefore, the number of roots of the equation is: 5 + 4 + 4 = 13.
Choice (3) is the answer.
4.51. From trigonometry, we know that:
y = a þ b cosðmxÞ ) T =
2π
jmj
Therefore:
) 4π =
2π
1
1
) jm j = ) m = ±
2
2
jmj
Based on the graph and the function, the positive value of m is accepted.
) m=
The value of function for x =
y
16π
3
1
1
1
) y = þ 2 cos x
2
2
2
is:
1 16π
1
8π
1
2π
1
2π
16π
1
= þ 2 cos
= þ 2 cos 2π þ
= þ 2 cos
= þ 2 cos
×
2
3
2
3
2
3
2
3
3
2
84
4 Solutions of Problems: Trigonometric Equations and Identities
=
1
π
1
π
1
1
1
= - 2 cos
= -2
=þ 2 cos π 2
3
2
3
2
2
2
Choice (1) is the answer (Fig. 4.9).
Figure 4.9 The graph of solution of problem 4.51
4.52. From trigonometry, we know that:
y = a þ b sinðmxÞ ) T =
2π
jmj
Therefore:
)
2π
2π
=
) jmj = 3 ) m = ± 3
3
jmj
Based on the graph and the given function, the positive value of m is accepted.
) m = 3 ) y = 1 -sinð3xÞ
The value of function for x =
y
7π
6
is:
7π
7π
7π
3π
3π
= 1 -sin 3 ×
= 1 -sin
= 1 -sin 2π þ
= 1 -sin
= 1 - ð- 1Þ = 2
6
6
2
2
2
Choice (4) is the answer (Fig. 4.10).
Figure 4.10 The graph of solution of problem 4.52
4
Solutions of Problems: Trigonometric Equations and Identities
85
4.53. From trigonometry, we know that:
y = a þ b sinðmxÞ ) T =
2π
jmj
Therefore:
) 5-1=
2π
1
) b= ±
2
jbπ j
Based on the graph and the given function, the positive value of m is accepted.
) b= -
1
π
) y = a þ sin - x
2
2
By testing the point of (0, 3) in the function, we have:
3 = a þ sin The value of function for x =
y
25
3
π
π
× 0 ) a = 3 ) y = 3 þ sin - x
2
2
is:
25
π 25
π
= 3 þ sin - ×
= 3 þ sin - 4π 3
2
3
6
= 3 þ sin -
π
1
= 3 - = 2:5
6
2
Choice (2) is the answer (Fig. 4.11).
Figure 4.11 The graph of solution of problem 4.53
86
4 Solutions of Problems: Trigonometric Equations and Identities
4.54. By testing the point of (0, 0) in the function, we have:
0 = a þ b cos
π
×0 ) a þ b=0
2
ð1Þ
Based on the function and the graph given in the problem, we can write:
ymax = a þ jbj ) a þ jbj = 4
ð2Þ
The assumption of b < 0 is not acceptable because it results in the equations with an impossible solution, as can be seen
in the following:
Using ð1Þ, ð2Þ a þ b = 0
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼)
) Impossible
a þ b=4
However, for the assumption of b > 0, we have:
Using ð1Þ, ð2Þ
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼)
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
a þ b=0
a-b=4
) 2b = - 4 ) b = - 2
Choice (1) is the answer (Fig. 4.12).
Figure 4.12 The graph of solution of problem 4.54
4.55. From trigonometry, we know that:
y = 1 þ a sinðmxÞ ) T =
2π
jmj
Therefore:
)
4
2π
=2×
) jbj = 3 ) b = ± 3
3
jbπ j
Based on the function and the graph given in the problem, we can write:
ymin = 1 - jaj ) - 1 = 1 - jaj ) jaj = 2 ) a = ± 2
Based on the graph and the given function, both of a and b must be either positive or negative. Hence:
4
Solutions of Problems: Trigonometric Equations and Identities
87
a=2
) a þ b=5
b=3
a= -2
) a þ b= -5
b= -3
Only a + b = 5 exists in the choices. Choice (3) is the answer (Fig. 4.13).
Figure 4.13 The graph of solution of problem 4.55
4.56. From trigonometry, we know that:
cos α þ
π
= -sinðαÞ
2
Therefore:
y = a - 2 cos bx þ
π
= a þ 2 sinðbxÞ
2
In addition, from trigonometry, we know that:
y = a þ 2 sinðbxÞ ) T =
2π
13π
π
2π
)
=
) jbj = 3 ) b = ± 3
18
18 jbj
jbj
Based on the graph and the simplified function, i.e., y = a + 2 sin (bx), the positive value of b is acceptable.
) b=3
Based on the simplified function and the given graph, we can write:
ymax = a þ 2 ) 1 = a þ 2 ) a = - 1
Hence:
a þ b= -1 þ 3=2
Choice (4) is the answer (Fig. 4.14).
88
4 Solutions of Problems: Trigonometric Equations and Identities
Figure 4.14 The graph of solution of problem 4.56
4.57. From trigonometry, we know that:
tan 45 ° = 1
tanðα - βÞ =
tanðαÞ -tanðβÞ
1 þ tanðαÞ tanðβÞ
cotðαÞ =
1
tanðαÞ
In addition, based on the information given in the problem, we have:
tan α þ 20 ° =
3
4
Therefore:
cot 25 ° - α =
=
1
1
=
tan 45 ° - α þ 20 °
tan 25 ° - α
1 þ tan 45 ° tan α þ 20 °
tan 45
°
-tan α þ 20 °
=
1 þ tan α þ 20 °
1 þ 34
=
°
1 - 34
1 -tan α þ 20
) cot 25 ° - α = 7
Choice (3) is the answer.
4.58. From trigonometry, we know that:
cosðαÞ = -
1 - sin 2 ðαÞ for an obtuse angle
tanðαÞ =
sinðαÞ
cosðαÞ
4
Solutions of Problems: Trigonometric Equations and Identities
89
tan
tanðα þ βÞ =
π
=1
4
tanðαÞ þ tanðβÞ
1 -tanðαÞ tanðβÞ
In addition, based on the information given in the problem, we have:
sinðαÞ =
3
5
Therefore:
cosðαÞ = -
1 - sin 2 ðαÞ = -
tanðαÞ =
tan
1-
3
5
2
=-
4
5
3
sinðαÞ
3
= 54=4
cosðαÞ
-5
tan π4 þ tanðaÞ
1 þ tanðαÞ 1 þ - 34
π
=
=
=
þα =
π
4
1 -tanðαÞ
1 -tan 4 tanðaÞ
1 - - 34
) tan
π
1
þα =
4
7
Choice (3) is the answer.
4.59. From trigonometry, we know that:
tan
π
- α = cotðαÞ
2
cotðαÞ =
tan
tanðα - βÞ =
1
tanðαÞ
π
=1
4
tanðαÞ -tanðβÞ
1 þ tanðαÞ tanðβÞ
In addition, based on the information given in the problem, we have:
tan
π
2
-α =
2
3
Therefore:
2
π
1
3
) tanðαÞ =
= tan - α = cotðαÞ =
3
2
2
tanðαÞ
1
4
7
4
90
4 Solutions of Problems: Trigonometric Equations and Identities
tan
tan π4 -tanðaÞ
-1
1 - 32
1 -tanðαÞ
π
= 52
=
=
-α =
3
π
4
1 þ tan 4 tanðaÞ 1 þ tanðαÞ 1 þ 2
2
) tan
π
1
-α = 4
5
Choice (2) is the answer.
4.60. From trigonometry, we know that:
tanðα þ βÞ =
tanðαÞ þ tanðβÞ
1 -tanðαÞ tanðβÞ
In addition, based on the information given in the problem, we have:
tanða þ bÞ =
2
5
tanða - bÞ =
3
7
Therefore:
tanð2aÞ = tanðða þ bÞ þ ða - bÞÞ =
2
þ3
tanða þ bÞ þ tanða - bÞ
= 5 27
1 -tanða þ bÞ tanða - bÞ 1 - 5 ×
) tanð2aÞ = 1
Choice (4) is the answer.
4.61. From trigonometry, we know that:
sinðαÞ = cos
π
-α
2
cosðαÞ = cosð- αÞ
tan
tanðα - βÞ =
π
=1
4
tanðαÞ -tanðβÞ
1 þ tanðαÞ tanðβÞ
Therefore:
sin
π
π
π
-x
þ x = cos 2
4
4
) 2=
= cos
π
π
- x = cos x 4
4
sin x - π4
sin x - π4
=
sin x þ π4
cos x - π4
3
7
=
29
35
29
35
4
Solutions of Problems: Trigonometric Equations and Identities
) tan x -
)
91
π
tanðxÞ -tan 4
π
=2 )
=2
4
1 þ tanðxÞ tan π4
tanðxÞ - 1
= 2 ) tanðxÞ - 1 = 2 þ 2 tanðxÞ
1 þ tanðxÞ
) tanðxÞ = - 3
Choice (1) is the answer.
4.62. From trigonometry, we know that:
π
=1
4
tan
tanðαÞ þ tanðβÞ
1 -tanðαÞ tanðβÞ
tanðα þ βÞ =
Moreover, based on the information given in the problem, we have:
αþβ=
π
4
If we calculate the tangent value of each side of the abovementioned relation, we will have:
tanðα þ βÞ = tan
tanðαÞ þ tanðβÞ
π
)
= 1 ) tanðαÞ þ tanðβÞ = 1 -tanðαÞ tanðβÞ
4
1 -tanðαÞ tanðβÞ
Therefore:
ð1 þ tanðαÞÞð1 þ tanðβÞÞ = 1 þ tanðαÞ þ tanðβÞ þ tanðαÞ tanðβÞ = 1 þ ð1 -tanðαÞ tanðβÞÞ þ tanðαÞ tanðβÞ
) ð1 þ tanðαÞÞð1 þ tanðβÞÞ = 2
Choice (2) is the answer.
4.63. From trigonometry, we know that:
π
=1
4
tan
tanðα þ βÞ =
tanðαÞ þ tanðβÞ
1 -tanðαÞ tanðβÞ
tanðα - βÞ =
tanðαÞ -tanðβÞ
1 þ tanðαÞ tanðβÞ
tanð2αÞ =
2 tanðαÞ
1 - tan 2 ðαÞ
92
4 Solutions of Problems: Trigonometric Equations and Identities
Therefore:
tan
tan π4 -tanðαÞ
tan π4 þ tanðαÞ
π
π
þ α -tan - α =
π
4
4
1 -tan 4 tanðαÞ 1 þ tan π4 tanðαÞ
=
=
1 þ tanðαÞ
1 -tanðαÞ
1 -tanðαÞ
1 þ tanðαÞ
ð1 þ tanðαÞÞ2 - ð1 -tanðαÞÞ2
1 - tan 2 ðαÞ
=
) tan
4 tanðαÞ
1 - tan 2 ðαÞ
π
π
þ α -tan - α = 2 tanð2αÞ
4
4
Choice (1) is the answer.
4.64. From trigonometry, we know that:
π
=1
4
tan
tanðα - βÞ =
tanðαÞ -tanðβÞ
1 þ tanðαÞ tanðβÞ
tanð2αÞ =
2 tanðαÞ
1 - tan 2 ðαÞ
Moreover, based on the information given in the problem, we have:
tan
tan π4 -tanðαÞ
1 -tanðαÞ
π
1
1
=
=
-α = )
4
5
1 þ tan π4 tanðαÞ 1 þ tanðαÞ 5
) 5 - 5 tanðαÞ = 1 þ tanðαÞ ) tanðαÞ =
) tanð2αÞ =
2
3
2 × 23
2 tanðαÞ
12
=
=
5
1 - tan 2 ðαÞ 1 - 2 2
3
) tanð2αÞ = 2:4
Choice (3) is the answer.
4.65. From trigonometry, we know that:
tanð2αÞ =
tanðα - βÞ =
2 tanðαÞ
1 - tan 2 ðαÞ
tanðαÞ -tanðβÞ
1 þ tanðαÞ tanðβÞ
4
Solutions of Problems: Trigonometric Equations and Identities
93
Moreover, based on the information given in the problem, we have:
tanðαÞ = 2
tanðβÞ =
1
3
Therefore:
tanð2αÞ =
2 tanðαÞ
2×2
4
=) tanð2αÞ =
3
1 - tan 2 ðαÞ
1 - 22
tanð2α - βÞ =
tanð2αÞ -tanðβÞ
1 þ tanð2αÞ tanðβÞ
) tanð2α - βÞ =
- 43 1 þ - 43
1
3
1
3
=
-
5
3
5
9
) tanð2α - βÞ = - 3
Choice (1) is the answer.
4.66. From trigonometry, we know that:
cosðπ - xÞ = -cosðxÞ
cosðαÞ = cosðα0 Þ ) α = 2kπ ± α0 , 8k 2 ℤ
Moreover, based on the information given in the problem, we have:
cosðxÞ ≠ 0
Therefore:
cosð3xÞ þ cosðxÞ = 0 ) cosð3xÞ = -cosðxÞ ) cosð3xÞ = cosðπ - xÞ
) 3x = 2kπ ± ðπ - xÞ )
3x = 2kπ þ π - x ) 4x = 2kπ þ π
)
3x = 2kπ - π þ x ) 2x = 2kπ - π
However:
cosðxÞ ≠ 0 ) x =
Choice (1) is the answer.
kπ π
þ
2 4
kπ π
þ
2 4
π
x = kπ 2
x=
94
4 Solutions of Problems: Trigonometric Equations and Identities
4.67. From trigonometry, we know that:
cotðαÞ = tan
π
-α
2
tanðαÞ = tanðα0 Þ ) α = kπ þ α0 , 8k 2 ℤ
Therefore:
tanð4xÞ = cotðxÞ ) tanð4xÞ = tan
) 4x = kπ þ
)
π
-x
2
π
π
kπ π
- x ) 5x = kπ þ ) x =
þ
2
2
5 10
π
is not a positive angle
k = - 1 ) x1 = 10
π
k = 0 ) x2 =
is an acute angle
2π
10
) x2 þ x 3 =
3π
5
k = 1 ) x3 =
is an acute angle
10
π
k = 2 ) x4 = is not an acute angle
2
Choice (1) is the answer.
4.68. From trigonometry, we know that:
sin 2 ðxÞ þ cos 2 ðxÞ = 1
cosðxÞ = cosðx0 Þ ) x = 2kπ ± x0
Therefore:
2 sin 2 ðxÞ þ 3 cosðxÞ = 0 ) 2 1 - cos 2 ðxÞ þ 3 cosðxÞ = 0 ) 2 cos 2 ðxÞ - 3 cosðxÞ - 2 = 0
) cos 2 ðxÞ -
3
1
cosðxÞ - 1 = 0 ) cosðxÞ þ ðcosðxÞ - 2Þ = 0
2
2
)
1
2π
) x = 2kπ ±
2
3
cosðxÞ = 2 ) not acceptable
cosðxÞ = -
Choice (1) is the answer.
4.69. From trigonometry, we know that:
sin 2 ðxÞ þ cos 2 ðxÞ = 1
cosðxÞ = cosðx0 Þ ) x = 2kπ ± x0
Therefore:
2 sin 2 ðxÞ = 3 cosðxÞ ) 2 1 - cos 2 ðxÞ - 3 cosðxÞ = 0 ) 2 cos 2 ðxÞ þ 3 cosðxÞ - 2 = 0
4
Solutions of Problems: Trigonometric Equations and Identities
) cos 2 ðxÞ þ
95
3
1
cosðxÞ - 1 = 0 ) cosðxÞ - ðcosðxÞ þ 2Þ = 0
2
2
)
1
π
) x = 2kπ ±
2
3
cosðxÞ = - 2 ) not acceptable
cosðxÞ =
Choice (4) is the answer.
4.70. From trigonometry, we know that:
tanðαÞ: cotðαÞ = 1
Now, let us find the intersection point of the lines, as follows:
x tanðαÞ - y cotðαÞ = 1
)
x tanðαÞ þ y cotðαÞ = 2
) xy =
3
2 tanðαÞ
1
2y cotðαÞ = 1 ) y =
2 cotðαÞ
2x tanðαÞ = 3 ) x =
3
1
3
3
×
= ) y=
4x
2 tanðαÞ 2 cotðαÞ 4
Choice (4) is the answer.
4.71. From trigonometry, we know that:
sin 2 ðαÞ þ cos 2 ðαÞ = 1
Based on the information given in the problem, we have:
x-2
-3
y-1
y = 1 þ 4 cosðαÞ ) cosðαÞ =
4
x = 2 - 3 sinðαÞ ) sinðαÞ =
Therefore:
)
ðx - 2Þ2 ðy - 1Þ2
þ
=1
9
16
which is the equation of an ellipse. Choice (2) is the answer.
4.72. Based on the information given in the problem, we have:
x = 2 - 5 cosðαÞ
y=4
From trigonometry, we know that:
- 1 ≤ cosðαÞ ≤ 1 ) - 1 ≤
2-x
≤1
5
96
4 Solutions of Problems: Trigonometric Equations and Identities
) -5≤2-x≤5 ) -7≤ -x≤3 ) -3≤x≤7
Therefore:
)
-3≤x≤7
y=4
which is the equation of a horizontal line segment. Choice (3) is the answer.
4.73. From trigonometry, we know that the maximum value of cos(.) and sin(.) is one. Therefore, the only solution of the
given equation is:
cosðx - yÞ = 1
sinðx þ yÞ = 1
The common solution of the equations can be calculated as follows:
)
x - y = 2kπ
0 < x, y < 2π
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼)
π¼
x þ y = 2kπ þ
2
x-y=0
π
5π
π
5π ) y = 4 or 4
x þ y = or
2
2
Choice (2) is the answer.
4.74. From trigonometry, we know that:
tanðα þ βÞ =
tanðαÞ þ tanðβÞ
1 -tanðαÞ tanðβÞ
In addition, we know that the sum and the product of the roots of a quadratic equation in the form of ax2 + bx + c = 0 are
- ba and ac, respectively.
Based on the information given in the problem, we have:
tan 2 ðxÞ þ ðm þ 2Þ tanðxÞ þ 2m - 2 = 0
αþβ=
π
4
Therefore:
tanð:Þ
tanðαÞ þ tanðβÞ
=1
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) tanðα þ βÞ = 1 )
¼
¼
1 -tanðαÞ tanðβÞ
- ðmþ2Þ
)
sum of the roots of the quadratic equation
-m-2
1
=
=
=1
1 - product of the roots of the quadratic equation 1 - 2m1- 2
3 - 2m
) - m - 2 = 3 - 2m ) m = 5
Choice (3) is the answer.
4
Solutions of Problems: Trigonometric Equations and Identities
97
4.75. From trigonometry, we know that:
sin 6 ðαÞ þ cos 6 ðαÞ = 1 - 3 sin 2 ðαÞ cos 2 ðαÞ
sin 4 ðαÞ þ cos 4 ðαÞ = 1 - 2 sin 2 ðαÞ cos 2 ðαÞ
Therefore:
sin 6 ðαÞ þ cos 6 ðαÞ þ 3 sin 2 ðαÞ cos 2 ðαÞ 1 - 3 sin 2 ðαÞ cos 2 ðαÞ þ 3 sin 2 ðαÞ cos 2 ðαÞ
=
sin 4 ðαÞ þ cos 4 ðαÞ þ 2 sin 2 ðαÞ cos 2 ðαÞ 1 - 2 sin 2 ðαÞ cos 2 ðαÞ þ 2 sin 2 ðαÞ cos 2 ðαÞ
)
sin 6 ðαÞ þ cos 6 ðαÞ þ 3 sin 2 ðαÞ cos 2 ðαÞ
=1
sin 4 ðαÞ þ cos 4 ðαÞ þ 2 sin 2 ðαÞ cos 2 ðαÞ
Choice (4) is the answer.
4.76. From trigonometry, we know that:
sin 135 ° = sin 180 ° - 45 ° = sin 45 °
cos 210 ° = cos 180 ° þ 30 ° = -cos 30 °
cos 135 ° = cos 180 ° - 45 ° = -cos 45 °
sin 420 ° = sin 360 ° þ 60 ° = sin 60 °
tan 210 ° = tan 180 ° þ 30 ° = tan 30 °
cot 420 ° = cot 360 ° þ 60 ° = cot 60 °
cot 120 ° = cot 180 ° - 60 ° = -cot 60 °
tan 330 ° = tan 360 ° - 30 ° = -tan 30 °
Therefore:
sin 45 ° -cos 30 ° þ -cos 45 ° sin 60 °
=
-tan 30 °
tan 30 ° cot 60 ° þ -cot 60 °
p
p
- 3
2
2 ×
2
p
p
3
3
×
3 þ
3
þ -
p
3
3
p
sin 45 ° -cos 30 ° þ -cos 45 ° sin 60 °
-3 6
=
)
4
-tan 30 °
tan 30 ° cot 60 ° þ -cot 60 °
Choice (2) is the answer.
p
2
2
-
p
p
3
2
3
3
98
4 Solutions of Problems: Trigonometric Equations and Identities
4.77. From trigonometry, we know that:
cotðx þ yÞ =
cotðxÞ cotðyÞ - 1
cotðxÞ þ cotðyÞ
Based on the information given in the problem, we have:
π k=0
π cotð:Þ cotðxÞ cotðyÞ - 1
x þ y = kπ þ ¼
=1
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) x þ y = ¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼)
4
4
cotðxÞ þ cotðyÞ
) 1 þ cotðxÞ þ cotðyÞ = cotðxÞ cotðyÞ
ð1Þ
ð1 þ cotðxÞÞð1 þ cotðyÞÞ = ð1 þ cotðxÞ þ cotðyÞÞ þ cotðxÞ cotðyÞ
ð2Þ
On the other hand, we can write:
Solving (1) and (2):
ð1 þ cotðxÞÞð1 þ cotðyÞÞ = cotðxÞ cotðyÞ þ cotðxÞ cotðyÞ
) ð1 þ cotðxÞÞð1 þ cotðyÞÞ = 2 cotðxÞ cotðyÞ
Choice (4) is the answer.
4.78. From trigonometry, we know that:
tan
cos
3π
- x = cotðxÞ
2
4π
π
1
= -cos
=3
3
2
tanðxÞ cotðxÞ = 1
cotðxÞ =
cosðxÞ
sinðxÞ
cosðxÞ = cosðx0 Þ ) x = 2kπ ± x0
Therefore:
ðsinðxÞ -tanðxÞÞ tan
3π
4π
1
) ðsinðxÞ -tanðxÞÞ cotðxÞ = - x = cos
2
3
2
) sinðxÞ cotðxÞ -tanðxÞ cotðxÞ = -
1
1
1
) cosðxÞ - 1 = - ) cosðxÞ =
2
2
2
) x = 2kπ ±
Choice (3) is the answer.
π
3
4
Solutions of Problems: Trigonometric Equations and Identities
99
4.79. From trigonometry, we know that:
sinðα þ βÞ = sinðαÞ cosðβÞ þ sinðαÞ cosðβÞ
cosðα þ βÞ = cosðαÞ cosðβÞ -sinðαÞ sinðβÞ
tanðxÞ =
sinðxÞ
cosðxÞ
tanðxÞ = tanðx0 Þ ) x = kπ þ x0
Therefore:
sinð2xÞðsinðxÞ þ cosðxÞÞ = cosð2xÞðcosðxÞ -sinðxÞÞ
) sinð2xÞ sinðxÞ þ sinð2xÞ cosðxÞ = cosð2xÞ cosðxÞ -cosð2xÞ sinðxÞ
) sinð2xÞ cosðxÞ þ cosð2xÞ sinðxÞ = cosð2xÞ cosðxÞ -sinð2xÞ sinðxÞ
1
×
& cosð3xÞ ≠ 0
cosð3xÞ
¼
¼
¼
¼
¼
) sinð2x þ xÞ = cosð2x þ xÞ ) sinð3xÞ = cosð3xÞ¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) tanð3xÞ = 1
) 3x = kπ þ
π
kπ π k = 0, 1, 2&x 2 ½0, π ]
π
5π
9π
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) x1 =
þ ¼
¼
¼
¼
¼
¼
¼
¼
) x=
,x =
,x =
4
3 12
12 2 12 3 12
) x 1 þ x2 þ x3 =
5π
4
Choice (2) is the answer.
4.80. From trigonometry, we know that:
sin
5π
π
þ x = sin þ x = cosðxÞ
2
2
sinðα þ βÞ = sinðαÞ cosðβÞ þ sinðαÞ cosðβÞ
sin
p
π
π
2
= cos
=
4
4
2
sinðxÞ = sinðx0 Þ )
x = 2kπ þ x0
x = 2kπ þ π - x0
Therefore:
)
p
2 sin
p
π
π
π
cosðxÞ -cos
sinðxÞ = 1 þ cosðxÞ
- x = 1 þ cosðxÞ ) 2 sin
4
4
4
100
4 Solutions of Problems: Trigonometric Equations and Identities
) cosðxÞ -sinðxÞ = 1 þ cosðxÞ ) sinðxÞ = - 1 )
x = 2kπ þ -
π
2
x = 2kπ þ π - -
π
2
π
2
3π
x = 2kπ þ
2
x = 2kπ -
)
) x = 2kπ -
π
2
Choice (3) is the answer.
4.81. From trigonometry, we know that:
tan
p
π
= 3
3
tanðxÞ =
cos
sinðxÞ
cosðxÞ
π
1
=
3
2
cosðα - βÞ = cosðαÞ cosðβÞ þ sinðαÞ sinðβÞ
cosðxÞ = cosðx0 Þ ) x = 2kπ ± x0
Therefore:
cosð2xÞ þ
p
3 sinð2xÞ = 1 ) cosð2xÞ þ tan
) cosð2xÞ cos
sin
π
sinð2xÞ = 1 ) cosð2xÞ þ
3
cos
π
3
π
3
sinð2xÞ = 1
π
π
π
π
π
þ sin
sinð2xÞ = cos
) cos 2x = cos
3
3
3
3
3
π
π
) 2x - = 2kπ ± )
3
3
π
π
π
= 2kπ þ ) x = kπ þ
3
3
3
π
π
2x - = 2kπ - ) x = kπ
3
3
2x -
Choice (4) is the answer.
4.82. From trigonometry, we know that:
sin a sin b =
1
½cosða - bÞ -cosða þ bÞ]
2
sin a cos b =
1
½sinða þ bÞ þ sinða - bÞ]
2
cos a þ cos b = 2 cos
aþb
a-b
cos
2
2
References
101
sin a -sin b = 2 cos
aþb
a-b
sin
2
2
Therefore:
cos 3α þ 12 ½cos α -cos 3α] sin α
cos 3α þ sin α sin 2α sin α
×
×
=
sin 3α -sin 2α cos α
cos α
sin 3α - 12 ½sin 3α þ sin α] cos α
=
1
2 cos 3α þ
1
2 sin 3α -
1
2
1
2
cos α sin α
cos 3α þ cos α sin α
2 cos 2α cos α sin α
×
=
×
=
×
cos α
sin 3α -sin α
cos α
2 cos 2α sin α cos α
sin α
)
cos 3α þ sin α sin 2α sin α
×
=1
sin 3α -sin 2α cos α
cos α
Choice (3) is the answer.
References
1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
5
Problems: Limits and Continuities
Abstract
In this chapter, the basic and advanced problems of limits and continuities are presented. The subjects include limits by
direct substitution, limits by factoring, limits by rationalization, limits at infinity, trigonometric limits, limits of absolute
value functions, limits involving Euler’s number, limits by L’Hopital’s rule, application of Taylor series in limits, and
limits and continuity. To help students study the chapter in the most efficient way, the problems are categorized in different
levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large).
Moreover, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems
with the largest calculations.
5.1. Determine the continuity status of the following function [1, 2]:
f ð xÞ ¼
10jxj x ≠ 0
0
x¼0
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) It is continuous everywhere except from the right-hand side of x ¼ 0.
2) It is continuous everywhere except from the left-hand side of x ¼ 0.
3) It is continuous everywhere.
4) It is continuous everywhere except at x ¼ 0.
5.2. What is the continuity status of the function below?
f ðxÞ ¼
j xj x ≠ 0
1 x¼0
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) It is continuous everywhere except from the right-hand side of x ¼ 0.
2) It is continuous everywhere except from the left-hand side of x ¼ 0.
3) It is continuous everywhere.
4) It is continuous everywhere except at x ¼ 0.
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_5
103
104
5
5.3. Calculate the value of k if the function below is continuous at x ¼ 2.
f ðxÞ ¼
ð x þ 2Þ ½ - x x < 2
xþk
x≥2
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 10
2) -10
3) 6
4) -8
5.4. For which value of the parameter of “a” the function below is continuous at ¼ -2 ?
f ðxÞ ¼
jxj½x þ a
j xj þ ½ x
x < -2
x ≥ -2
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 2
3) 3
4) 6
5.5. Calculate the value of the following limit:
lim
þ
x → ð- 1Þ
½ x þ 1
x2 - 1
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1
1) 2
2) 0
1
3)
2
4) 1
5.6. Calculate the limit of the following function if x → 2+.
f ðxÞ ¼
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) -1
3) 2
4) -2
xþ4
½ -x -3
Problems: Limits and Continuities
5
Problems: Limits and Continuities
105
5.7. Determine the value of the following limit:
lim
x → 0-
xþ2
½ x
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 2
2) -2
3) 1
4) -1
5.8. Determine the value of the limit below.
lim
x→ -1
½x þ 3x
½x - 3x
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 2
2) -2
3) 4
4) -4
5.9. Calculate the limit of the function below if x → 0.
f ðxÞ ¼
p
xþ 3 x
p
x- 3 x
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 1
3) -1
4) 1
5.10. Calculate the value of the following limit:
lim-
x→0
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) -1
3) 0
4) 1
½ x
x
106
5
5.11. Determine the limit of the function below if x → 0+.
f ðxÞ ¼
p
ðx2 - 1Þ x
p
ðx x þ 1Þx
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) -1
3) 0
4) -1
5.12. Determine the value of the following limit:
lim
x → 0þ
1
1
x x3
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) -1
3) 0
4) 1
5.13. For the function below, calculate the value of limþ f ðxÞ - lim- f ðxÞ.
x→1
x→1
f ðxÞ ¼
2x
½2x þ 2
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) -1
1
2) 6
2
3)
3
4) 1
5.14. Calculate the value of limþ ð½x - 2Þ½x.
x→2
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) -2
2) -1
3) 0
4) 1
5.15. Calculate the limit of the following function if x → 4-:
f ðxÞ ¼
½ x - 4
x2 - 16
Problems: Limits and Continuities
5
Problems: Limits and Continuities
107
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
1
2)
8
3) 1
4) -1
5.16. Determine the value of the limit below.
lim-
x→1
1 - x3
arcðcosðxÞÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) -1
3) 0
4) -3
5.17. Calculate the value of the limit below.
lim
x→0
tanðxÞ - tanð3xÞ þ tanð2xÞ
x3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) -6
2) 6
3) 10
4) -10
5.18. Calculate the value of the following limit:
9 - x2
p
x→3 2 - x þ 1
lim
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 6
2) 12
3) 18
4) 24
5.19. Determine the limit of the function below if x → + 1.
f ð xÞ ¼
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
sinðxÞ
x
108
5
1)
2)
3)
4)
Undefined
0
1
1
5.20. Determine the value of the limit below.
½ x2 - x2
x tanðxÞ
lim
x→0
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) -1
3) 2
4) -2
5.21. Determine the value of the following limit:
lim x sin
x → þ1
1
x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) -1
3) 0
4) Undefined
5.22. Calculate the value of the following limit:
lim
x→ -1
x2 þ x - 1
p
- 3x þ 4 - x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1
1)
3
1
2) 3
3) 1
4) -1
5.23. Calculate the value of the limit below.
limþ
x→0
p
ðx þ 1Þ x
x2 - x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
Problems: Limits and Continuities
5
Problems: Limits and Continuities
109
2) -1
3) 1
4) -1
5.24. Determine the limit of the following function if x → + 1:
f ðxÞ ¼
x-1 þ
x
p
x2 þ x - 1
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 0
1
3) 2
1
4)
2
5.25. Calculate the value of lim x cotðxÞ.
x→0
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 1
3) 1
4) 2
5.26. For what value of “a” the following function has a definite limit at x ¼ 1?
f ðxÞ ¼
x2 þ ax
x-3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 3
3) -3
4) -2
5.27. Determine the value of the limit below.
lim-
x→2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) -24
2) -16
3) 16
4) 24
x3 - 8
p
x - 2x
x>1
x<1
110
5
5.28. Calculate the value of the limit below.
½ x þ x
½ - x þ x
lim
x → 0-
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) +1
2) -1
3) 1
4) -1
5.29. Determine the value of the limit below.
lim
x→0
sinð3xÞ þ sinð7xÞ
3x þ tanð2xÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 2
3) -1
4) -2
5.30. Calculate the value of the following limit:
p
lim
x→0
p
x þ 3- 3
x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
p
3
1)
3
p
3
2)
6
p
3
3)
2
p
3
4)
9
5.31. Calculate the value of the following limit:
lim
x→0
1 - cosðxÞ
sinðxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 1
3) -1
Problems: Limits and Continuities
5
Problems: Limits and Continuities
4)
p
111
2
5.32. Calculate the value of the limit below.
lim
x→0
5x - sinðxÞ
2x þ cosðxÞ - 1
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 2
3) -1
4) -2
5.33. 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
5.34. Calculate the value of the limit below.
lim þ
x → ðπ2Þ
sinðxÞ þ cosðxÞ
cosðxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) -1
3) 0
4) 1
5.35. Calculate the value of the following limit:
lim
x→0
3x4 þ 2x3
ðarcðsinðxÞÞÞ3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 2
3) 0
4) 1
5.36. Determine the value of the limit below.
112
5
lim
x→ -1
2
x
xþ1
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 2
3) 0
4) -1
5.37. Calculate the value of the following limit:
limþ p
x→3
x2
x-4
- 4x þ 3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) -1
3) 1
4) -1
5.38. Calculate the value of lim
x→ -1
xþ
p
x2 þ 4x - 10 .
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 2
2) -2
3) 1
4) -1
5.39. Determine the value of the limit below.
4 - x2
p
x → 2 6 - 2 x2 þ 5
lim
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 2
3) 3
4) 1
5.40. Calculate the value of the following limit:
lim p
x→0
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 2
2) 4
sinð2xÞ
x þ 1-1
Problems: Limits and Continuities
5
Problems: Limits and Continuities
113
3) 3
4) 1
p
5.41. Calculate the limit of 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
5.42. Calculate the value of the limit below.
lim
x→ -3
x2 - 9
xþ3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 6
2) -6
3) 3
4) Undefined
5.43. Calculate the value of the following limit:
lim1
x→2
tan πx
2 -1
cosðπxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) -1
3) 2
4) -2
p
p
5.44. Calculate the limit of 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
5.45. Calculate the value of the following limit:
limπ
x→2
tanð2xÞ cosðxÞ
1 þ cosð2xÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
1
2)
2
3) -1
114
5
4) -
1
2
5.46. Determine the value of the following limit:
lim
x → 0-
tanð2xÞ
1 - cosðxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
p
1) - 2 2
p
2) - 2
p
3) 2
p
4) 2 2
5.47. Calculate the value of x →lim
-1
p
3
p
n þ 1000 - 3 n - 20 .
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 2
2) 0
3) 10
4) 20
5.48. Calculate the value of the limit below.
limþ
sinðπ sinðxÞÞ sin
x
2
1 þ cosðxÞ
x→π
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
p
1) - π 2
2) -2π
3) π 2
p
4) π 2 2
5.49. Calculate the value of the following limit.
p
3
lim
x→0
p
1 þ x2 - 4 1 - 2x
2x2 þ 2x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1
1)
4
1
2) 4
3) 1
4) -1
5.50. Calculate the limit of the function below if x → 0.
Problems: Limits and Continuities
5
Problems: Limits and Continuities
115
f ðxÞ ¼
sin 2 ðxÞ þ sinðxÞ þ cos 2 ðxÞ - cosðxÞ
sin 2 ðxÞ - sinðxÞ þ cos 2 ðxÞ - cosðxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) -1
3) 2
4) -2
5.51. Determine the value of the following limit:
lim
x→0
cosðmxÞ - cosðnxÞ
x2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) n2 + m2
2) n2 - m2
n2 - m2
3)
2
n2 þ m 2
4)
2
5.52. Calculate the value of the limit below.
lim
x→0
sinðxÞ - x
tanðxÞ - x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1
1) 2
1
2)
2
1
3) 4
1
4)
4
5.53. 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
3
1)
2
3
2) 2
116
5
Problems: Limits and Continuities
3) 3
4) -3
5.54. For the following function, we have f (0) ¼ 0. Which one of the choices below is correct about the continuity of the
function at x ¼ 0?
f ðxÞ ¼ xð- 1Þ½x ,
1
x 2 ℝ - f 0g
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1) The function has only right-hand side continuity at x ¼ 0.
2) The function is only left continuous at x ¼ 0.
3) The function is continuous at x ¼ 0.
4) The function is not continuous at x ¼ 0.
5.55. Calculate the limit of the function below if n → + 1.
3n2
f ð nÞ ¼ p n
5
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 1
3) 1
3
4) p
5
5.56. Determine the value of the limit below.
lim
x→0
x3 - sinðxÞð1 - cosðxÞÞ
x3
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 0
1
2)
2
3) 1
4) -1
5.57. Calculate the value of the following limit:
lim
x → 1-
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
p
1) 2
p
2) - 2
p
2
3) 2
arcðcos xÞ
p
1-x
5
Problems: Limits and Continuities
117
p
4)
2
2
5.58. Determine the value of n in the following equation:
lim x2 - 1 cotðxn - 1Þ ¼
x→1
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 8
2) 4
1
3)
8
1
4)
4
5.59. 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
5.60. Calculate the value of the following limit:
lim
x → 0- 1
2
sinðxÞ - x
sinð2xÞ - x cosðxÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 0
2) 1
3) 1
4) -1
π
5.61. Calculate the limit of the following function if x → :
4
f ðxÞ ¼
1 - 3 tanðxÞ
1 - 2 sin 2 ðxÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1
1)
3
2) 3
1
2
118
5
Problems: Limits and Continuities
1
2
4) 2
3)
5.62. Calculate the value of the limit below.
lim
x→0
1 - cos 3 ðxÞ
sinðxÞ tanð2xÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 4
2) -4
3
3)
4
3
4) 4
5.63. Determine the value of the following limit:
lim
x → 0þ
1 - cosðxÞ
p
1 - cosð xÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ○ Normal ● Large
1) 0
2) 1
1
3)
2
4) 2
References
1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
6
Solutions of Problems: Limits and Continuities
Abstract
In this chapter, the problems of the fifth chapter are fully solved, in detail, step-by-step, and with different methods.
6.1. Based on the information given in the problem, we have [1, 2]:
f ðxÞ ¼
10jxj
x≠0
0
x¼0
The function of 10jxj is continuous everywhere on real numbers (ℝ); however, the continuity of the f (x) must be checked
at x ¼ 0.
A function is continuous at the given point of x0 if:
lim f ðxÞ ¼ lim - f ðxÞ ¼ f ðx0 Þ
x → x0
x → x0 þ
As can be seen, for the f (x), we have:
lim 10jxj ¼ 0
x → 0þ
lim 10jxj ¼ 0
x → 0-
f ð 0Þ ¼ 0
) limþ 10jxj ¼ lim- 10jxj ¼ f ð0Þ
x→0
x→0
Therefore, the function is continuous everywhere on ℝ.
Choice (3) is the answer.
6.2. Based on the information given in the problem, we have:
f ðxÞ ¼
j xj x ≠ 0
1 x¼0
The function of jxj is continuous everywhere on ℝ; however, the continuity of the f (x) must be checked at x ¼ 0.
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_6
119
120
6
Solutions of Problems: Limits and Continuities
A function is continuous at the given point of x0 if:
lim f ðxÞ ¼ lim - f ðxÞ ¼ f ðx0 Þ
x → x0
x → x0 þ
As can be seen, for the abovementioned function, we have:
lim jxj ¼ 0
x → 0þ
lim jxj ¼ 0
x → 0-
f ð 0Þ ¼ 1
) limþ jxj ¼ lim- jxj ≠ f ð0Þ
x→0
x→0
Therefore, the function is continuous everywhere on ℝ except at x ¼ 0.
Choice (4) is the answer.
6.3. Based on the information given in the problem, the following function is continuous at x ¼ 2.
f ðxÞ ¼
ð x þ 2Þ ½ - x ] x < 2
xþk
x≥2
Therefore, we must have:
lim f ðxÞ ¼ lim- f ðxÞ ¼ f ð2Þ
x→2
x → 2þ
For f (x), we have:
lim f ðxÞ ¼ limþ ðx þ kÞ ¼ 2 þ k
x → 2þ
)
x→2
lim f ðxÞ ¼ lim- ðx þ 2Þ½ - x] ¼ ð2 - þ 2Þ½ - 2 - ] ¼ - 8
x → 2-
x→2
f ð 2Þ ¼ 2 þ k
) 2 þ k ¼ -8
) k ¼ - 10
Choice (2) is the answer.
6.4. Based on the information given in the problem, the function is continuous at x ¼ -2.
f ðxÞ ¼
jxj½x] þ a x < - 2
j xj þ ½ x] x ≥ - 2
Therefore, we must have:
lim f ðxÞ ¼
x → - 2þ
lim f ðxÞ ¼ f ð- 2Þ
x → - 2-
6
Solutions of Problems: Limits and Continuities
121
For f (x), we can write:
lim f ðxÞ ¼
lim ðjxj þ ½x]Þ ¼ 2 þ ð - 2Þ ¼ 0
x → - 2þ
)
x → - 2þ
lim f ðxÞ ¼
lim jxj½x] þ a ¼ ð2Þð - 3Þ þ a ¼ - 6 þ a
x → - 2-
x → - 2-
f ð - 2Þ ¼ j - 2j þ ½ - 2] ¼ 0
) -6 þ a ¼ 0
)a¼6
Choice (4) is the answer.
6.5. The problem can be solved as follows:
lim
x → ð- 1Þ
þ
½x] þ 1 ð- 1Þ þ 1
0
¼ ¼ - ¼0
1 -1
0
x2 - 1
Choice (2) is the answer.
6.6. The problem can be solved as follows:
lim
x → 2þ
xþ4
2þ4
¼
¼ -1
½ - x] - 3 - 3 - 3
Choice (2) is the answer.
6.7. The problem can be solved as follows:
lim
x → 0-
xþ2 0þ2
¼
¼ -2
-1
½ x]
Choice (2) is the answer.
6.8. The problem can be solved as follows:
lim
x→ -1
½x] þ 3x
4x
¼ lim
¼ lim ð- 2Þ ¼ - 2
x
→
1
2x x → - 1
½x] - 3x
Choice (2) is the answer.
6.9. The problem can be solved as follows:
p
p
p
3 2
3
p
x þ1
x 3 x2 þ 1
xþ 3 x
p
¼
lim
¼ -1
¼
lim
lim
p
p
p
3 2
x→0 x - 3 x
x→0 3
x→0
x 3 x2 - 1
x -1
Choice (3) is the answer.
122
6
Solutions of Problems: Limits and Continuities
6.10. The problem can be solved as follows:
lim
x → 0-
½ x] - 1
¼ - ¼ þ1
x
0
Choice (1) is the answer.
6.11. The problem can be solved as follows:
p
ð x2 - 1Þ x
ð x 2 - 1Þ
-1
p ¼ þ ¼ -1
¼ limþ p
limþ p
0
x → 0 ðx x þ 1Þx
x → 0 ðx x þ 1 Þ x
Choice (2) is the answer.
6.12. The problem can be solved as follows:
1
1
x2 - 1
- 3 ¼ limþ
x x
x3
x→0
limþ
x→0
¼
-1
¼ -1
0þ
Choice (2) is the answer.
6.13. The problem can be solved as follows:
lim f ðxÞ - lim- f ðxÞ ¼ limþ
x → 1þ
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
¼ limþ
x→1
x
2
1 2 -1
- lim x ¼ - ¼
2 x → 1- 3
2 3
6
Choice (2) is the answer.
6.14. The problem can be solved as follows:
lim ð½x] - 2Þ½x] ¼ limþ ð2 - 2Þ x 2 ¼ limþ 0 ¼ 0
x → 2þ
x→2
x→2
Choice (3) is the answer.
6.15. The problem can be solved as follows:
lim
x → 4-
½ x] - 4
-1
3-4
¼ þ1
¼ lim
¼
x2 - 16 x → 4 - 16 - - 16 0 -
Choice (3) is the answer.
6.16. From trigonometry and calculus, we know that:
arcðcosð1 - ÞÞ ¼ 0þ
d
-1
ðarcðcos xÞÞ ¼ p
dx
1 - x2
6
Solutions of Problems: Limits and Continuities
123
The problem can be solved as follows:
lim-
x→1
H
¼
¼
¼
¼
¼
¼) limx→1
d
3
dx ð1 - x Þ
d
dx ðarcðcosðxÞÞÞ
0þ
1 - x3
¼ þ
arcðcosðxÞÞ 0
¼ limx→1
p
¼ lim- 3x2 1 - x2 ¼ 0
- 3x2
p-1
1 - x2
x→1
Choice (3) is the answer.
6.17. From application of Taylor series in limit, we know that:
lim tanðxÞ - x þ
x→0
x3
3
The problem can be solved as follows:
x þ x3 - 3x þ ð3x3Þ
tanðxÞ - tanð3xÞ þ tanð2xÞ
lim
- lim
x→0
x→0
x3
x3
3
¼ lim
x→0
3
- 6x3
¼ lim ð- 6Þ ¼ -6
x→0
x3
Choice (1) is the answer.
6.18. The problem can be solved as follows:
9 - x2
0
p
¼
0
x→3 2 - x þ 1
lim
H
¼
¼
¼
¼) lim
x→3 d
dx
2
d
dx ð9 - x Þ
p
2- x þ 1
¼ lim
- 2x
x → 3 p- 1
2 xþ1
p
¼ lim 4x x þ 1 ¼ 24
x→3
Choice (4) is the answer.
6.19. From trigonometry, we know that:
- 1 ≤ sinðxÞ ≤ 1
The problem can be solved as follows:
lim
x → þ1
Choice (2) is the answer.
3
þ 2x þ ð2x3Þ
sinðxÞ
1
¼ lim sinðxÞ x ¼ ðBounded quantityÞ x 0 ¼ 0
x → þ1
x
x
124
6
Solutions of Problems: Limits and Continuities
6.20. From application of Taylor series in limit, we know that:
lim tanðxÞ - x
x→0
The problem can be solved as follows:
lim
x→0
½ x2 ] - x2
0 - x2
- x2
¼ lim
- lim
¼ lim ð- 1Þ ¼ -1
x
x tanðxÞ x → 0 x tanðxÞ x → 0 x x x → 0
Choice (2) is the answer.
6.21. From application of 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 x ¼ lim 1 ¼ 1
x → þ1
x
x x → þ1
Choice (1) is the answer.
6.22. From calculus, we know that:
lim
x→ ±1
am x m þ am - 1 x m - 1 þ . . . þ a2 x 2 þ a1 x þ a0 - am x m
or:
lim ðam xm þ an xn Þ - am xm
x→ ±1
if m > n
Therefore:
lim
x→ -1
x2 þ x - 1
p
- 3x þ 4 - x
- lim
x→ -1
x
x2
¼ þ1
¼ lim 3
- 3x x → - 1
Choice (3) is the answer.
6.23. From calculus, we know that:
lim am xm þ am - 1 xm - 1 þ . . . þ am - n xm - n þ am - n - 1 xm - n - 1 - am - n - 1 xm - n - 1
x→0
or:
lim ðam xm þ an xn Þ - an xn
x→0
if m > n
The problem can be solved as follows:
p
p
p
p
ð x þ 1Þ x
x
x xþ x
-1
¼ limþ
- limþ
limþ
¼ limþ p ¼ - 1
2-x
2-x
x
x
x
x
x→0
x→0
x→0
x→0
Choice (4) is the answer.
6
Solutions of Problems: Limits and Continuities
125
6.24. From calculus, we know that:
lim
x → þ1
x2 þ ax þ b - x þ
a
2
The problem can be solved as follows:
lim
x → þ1 x - 1
þ
x
x
x
x
1 1
p
¼ lim
- lim
- lim
¼ lim
¼
x → þ1 2x - 1
x → þ1 2x
x → þ1 2
2
x2 þ x - 1 x → þ1 x - 1 þ x þ 12
2
Choice (4) is the answer.
6.25. From trigonometry, we know that:
cotðxÞ ¼
1
tanðxÞ
From application of Taylor series in limit, we know that:
lim tanðxÞ - x
x→0
The problem can be solved as follows:
lim x cotðxÞ ¼ lim
x→0
x→0
x
x
- lim ¼ 1
tanðxÞ x → 0 x
Choice (3) is the answer.
6.26. As we know, the limit of a function at the point of x0 exits if:
lim f ðxÞ ¼ lim þ f ðxÞ ) lim- f ðxÞ ¼ limþ f ðxÞ
ð1Þ
lim f ðxÞ ¼ lim- ðx - 3Þ ¼ 1 - 3 ¼ - 2
ð2Þ
lim f ðxÞ ¼ limþ x2 þ ax ¼ 1 þ a
ð3Þ
x → x0 -
x→1
x → x0
x→1
Therefore:
x → 1-
x→1
x → 1þ
x→1
Using ð1Þ, ð2Þ, ð3Þ
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) - 2 ¼ 1 þ a ) a ¼ -3
¼
¼
¼
¼
¼
Choice (3) is the answer.
6.27. The problem can be solved as follows:
lim-
x→2
x3 - 8
- ðx3 - 8Þ 0þ
p
p ¼ lim¼ 0
x - 2x x → 2 x - 2x
d
ð- ðx3 - 8ÞÞ
H
- 3x2
- 3 x 22
- 12
p
¼
¼
¼
¼
¼
¼) lim- dxd
¼
¼ 1 ¼ -24
¼ lim1
1
p
p
x→2
x
→
2
1
1
2
dx x - 2x
2x
2x2
Choice (1) is the answer.
126
6
Solutions of Problems: Limits and Continuities
6.28. The problem can be solved as follows:
lim-
x→0
½x] þ x
- 1 þ x - 1 þ 0-1
¼ lim¼ - ¼ þ1
¼
0 þ 00þx
0
½ - x] þ x x → 0
Choice (1) is the answer.
6.29. The problem can be solved as follows:
sinð3xÞ þ sinð7xÞ 0
¼
0
3x þ tanð2xÞ
lim
x→0
d
dx ðsinð3xÞ
d
x→0
dx ð3x þ
H
¼
¼
¼
¼
¼
¼) lim
þ sinð7xÞÞ
3 cosð3xÞ þ 7 cosð7xÞ 3 þ 7
¼
¼2
¼ lim
3þ2
x → 0 3 þ 2ð1 þ tan 2 ð2xÞÞ
tanð2xÞÞ
Choice (2) is the answer.
6.30. The problem can be solved as follows:
p
lim
x→0
p
p
p
p
p
x þ 3-3
x þ 3- 3
xþ3- 3
xþ3þ 3
p
p ¼ lim p
¼ lim
xp
x
x
x→0
x
→
0
x xþ3þ 3
xþ3þ 3
¼ lim p
x→0
p
1
1
1
3
p ¼p
p ¼ p ¼
6
xþ3þ 3
3þ 3 2 3
Choice (2) is the answer.
6.31. The problem can be solved as follows:
lim
x→0
1 - cosðxÞ 0
¼
0
sinðxÞ
d
dx ð1 - cosðxÞÞ
d
x→0
dx sinðxÞ
H
¼
¼
¼
¼
¼
¼) lim
¼ lim
x→0
sinðxÞ 0
¼ ¼0
cosðxÞ 1
Choice (1) is the answer.
6.32. The problem can be solved as follows:
lim
x→0
5x - sinðxÞ
0
¼
2x þ cosðxÞ - 1 0
d
dx ð5x - sinðxÞÞ
d
x→0
dx ð2x þ cosðxÞ - 1Þ
H
¼
¼
¼
¼
¼
¼) lim
Choice (2) is the answer.
¼ lim
x→0
5 - cosðxÞ 5 - 1
¼
¼2
2 - sinðxÞ 2 - 0
6
Solutions of Problems: Limits and Continuities
127
6.33. The problem can be solved as follows:
lim-
x→2
ðx - 2Þðx2 þ 2x þ 4Þ
x3 - 8
þ 5x
þ 5x ¼ limx→2
- ð x - 2Þ
j x - 2j
¼ lim- - x2 - 2x - 4 þ 5x
x→2
lim - x2 þ 3x - 4 ¼ - 4 þ 6 - 4 ¼ - 2
x → 2-
Choice (2) is the answer.
6.34. The problem can be solved as follows:
sinðxÞ þ cosðxÞ 1 - þ 0 - 1 ¼
¼ - ¼ -1
00
cosðxÞ
lim þ
x → ðπ2Þ
Choice (2) is the answer.
6.35. The problem can be solved as follows:
lim arcðsinðxÞÞ - x
x→0
lim 3x4 þ 2x3 - 2x3
x→0
lim
x→0
3x4 þ 2x3
2x3
- lim 3 ¼ lim 2 ¼ 2
3
x→0 x
x→0
ðarcðsinðxÞÞÞ
Choice (2) is the answer.
6.36. The problem can be solved as follows:
lim
x→ -1
2
x ¼ lim ½0 - ]x ¼ ð- 1Þð- 1Þ ¼ þ1
x→ -1
xþ1
Choice (1) is the answer.
6.37. The problem can be solved as follows:
lim p
x → 3þ
x-4
¼ lim
x2 - 4x þ 3 x → 3þ
x-4
-1
¼ þ ¼ -1
0
ð x - 3Þ ð x - 1Þ
Choice (2) is the answer.
6.38. From calculus, we know that:
lim
x→ ±1
x2 þ ax þ b - x þ
a
2
The problem can be solved as follows:
lim
x→ -1
xþ
x2 þ 4x - 10 - lim ðx þ jx þ 2jÞ ¼ lim ðx - x - 2Þ ¼ lim ð- 2Þ ¼ - 2
Choice (2) is the answer.
x→ -1
x→ -1
x→ -1
128
6
Solutions of Problems: Limits and Continuities
6.39. The problem can be solved as follows:
0
4 - x2
p
¼
2
0
x→2 6 - 2 x þ 5
lim
H
¼
¼
¼
¼
¼
¼) lim
x→2 d
dx
2
d
dx ð4 - x Þ
p
6 - 2 x2 þ
5
¼ lim
x→2
- 2x
¼ lim
x→2
- 2 x 2p2x
x2 þ5
x2 þ 5 ¼ 3
Choice (3) is the answer.
6.40. The problem can be solved as follows:
lim p
x→0
d
dx ðsinð2xÞÞ
H
¼
¼
¼
¼
¼
¼) lim
p
x→0 d
dx
x þ 1-1
sinð2xÞ
0
¼
x þ 1-1 0
¼ lim
2 cosð2xÞ
x→0
p1
2 xþ1
¼
2x1
1
2
¼4
Choice (2) is the answer.
6.41. From calculus, we know that:
lim
x→ -1
lim
x→ ±1
x4 þ 2x2 þ x - x2
am x m þ am - 1 x m - 1 þ . . . þ a2 x 2 þ a1 x þ a0 - am x m
or:
lim ðam xm þ an xn Þ - am xm
x→ ±1
if m > n
The problem can be solved as follows:
p
lim
x→ -1
x4 þ 2x2 þ x - x ¼ x →lim
-1
2
x4 þ 2x2 þ x - x
2
x p
x4 þ 2x2 þ x þ x2
x4 þ 2x2 þ x þ x2
x4 þ 2x2 þ x - x4
2x2 þ x
2x2
p
lim
¼ x →lim
1¼1
¼ x →lim
¼
lim
2
2
-1
x → - 1 ðx þ x Þ
x → - 1 2x2
-1
x4 þ 2x2 þ x þ x2
Choice (1) is the answer.
6.42. From calculus, we know that the limit of a function at a specific point (x0) exits if:
lim f ðxÞ ¼ lim þ f ðxÞ
x → x0 -
x → x0
Therefore, we must have:
lim
x → ð- 3Þ
-
x2 - 9
x2 - 9
¼ lim þ
xþ3
xþ3
x → ð- 3Þ
ð1Þ
6
Solutions of Problems: Limits and Continuities
lim
x → ð- 3Þ
lim
x → ð- 3Þþ
129
x2 - 9
x2 - 9
¼ lim ¼ lim - ðx - 3Þ ¼ - 6
xþ3
x → ð- 3Þ x þ 3
x → ð- 3Þ
ð2Þ
x2 - 9
- ð x2 - 9Þ
¼ lim þ
¼ lim þ - ðx - 3Þ ¼ 6
xþ3
xþ3
x → ð- 3Þ
x → ð- 3Þ
ð3Þ
-
ð1Þ, ð2Þ, ð3Þ
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) - 6 ≠ 6 ) lim
x→ -3
x2 - 9
¼ Undefined
xþ3
Choice (4) is the answer.
6.43. The problem can be solved as follows:
lim
x → 12
H
¼
¼
¼
¼
¼
¼) lim
d
dx
x → 12
tan πx
0
2 -1
¼
0
cosðπxÞ
tan πx
2 -1
¼ lim
d
x → 12
dx ðcosðπxÞÞ
π
2
1 þ tan 2 πx
2
- π sinðπxÞ
π
ð 1 þ 1Þ
¼2
¼ -1
-π x 1
Choice (2) is the answer.
6.44. The problem can be solved as follows:
lim
p
x → þ1
p
x þ 5 - x þ 1 ¼ lim
x → þ1
¼ lim
x → þ1
p
p
p
x þ 5- x þ 1 x p
xþ5þ
xþ5þ
p
p
xþ1
xþ1
x þ 5 - ð x þ 1Þ
4
p
p
p
¼0
¼ lim p
x
→
þ1
xþ5þ xþ1
xþ5þ xþ1
Choice (3) is the answer.
6.45. From trigonometry, we know that:
1 þ cosð2xÞ ¼ 2 cos 2 ðxÞ
tanðxÞ ¼
sinðxÞ
cosðxÞ
sinð2xÞ ¼ 2 sinðxÞ cosðxÞ
The problem can be solved as follows:
limπ
x→2
tanð2xÞ cosðxÞ
sinð2xÞ cosðxÞ
2 sinðxÞ cos 2 ðxÞ
sinðxÞ
1
¼ limπ
¼
¼ limπ
¼ limπ
¼ -1
2
-1
x → 2 cosð2xÞ x 2 cos ðxÞ
x → 2 cosð2xÞ x 2 cos 2 ðxÞ
x → 2 cosð2xÞ
1 þ cosð2xÞ
Choice (3) is the answer.
130
6
Solutions of Problems: Limits and Continuities
6.46. From calculus and trigonometry, we know that:
1 - cos 2 ðxÞ ¼ sin 2 ðxÞ
Moreover, from application of Taylor series in limit, we know that:
lim sinðxÞ - x
x → 0-
lim tanðxÞ - x
x → 0-
The problem can be solved as follows:
tanð2xÞ
¼ lim
1 - cosðxÞ x → 0 -
lim
x → 0-
¼ lim-
tanð2xÞ x
x→0
p
2
sin 2 ðxÞ
tanð2xÞ
x
1 - cosðxÞ
1 þ cosðxÞ
1 þ cosðxÞ
¼ lim-
tanð2xÞ
1 þ cosðxÞ
1 - cos 2 ðxÞ
x→0
p
p
p
tanð2xÞ x 2
tanð2xÞ x 2
2x x 2
¼ lim¼ lim¼ lim-x
x→0
x→0
x→0
- sinðxÞ
j sinðxÞj
p
p
¼ lim- - 2 2 ¼ - 2 2
x→0
Choice (1) is the answer.
6.47. The problem can be solved as follows:
lim
p
3
p
n þ 1000 - 3 n - 20
3
ðn þ 1000Þ2 þ
3
ðn þ 1000Þðn - 20Þ þ
3
ðn - 20Þ2
3
ðn þ 1000Þ2 þ
3
ðn þ 1000Þðn - 20Þ þ
3
ðn - 20Þ2
x→ -1
¼ lim
x→ -1
p
3
p
3
n þ 1000 - n - 20 x
¼ lim
x→ -1
n þ 1000 - ðn - 20Þ
3
ðn þ 1000Þ2 þ
3
ðn þ 1000Þðn - 20Þ þ
Choice (2) is the answer.
6.48. From application of Taylor series in limit, we know that:
lim sinðxÞ - x
x→0
In addition, from trigonometry, we know that:
1 þ cosðxÞ ¼ 2 cos 2
sinðxÞ ¼ 2 sin
x
2
x
x
cos
2
2
3
ðn - 20Þ2
¼
1020
¼0
þ1
6
Solutions of Problems: Limits and Continuities
131
The problem can be solved as follows:
limþ
sinðπ sinðxÞÞ sin
x
2
1 þ cosðxÞ
x→π
¼ limþ
x→π
- limþ
π sinðxÞ sin
x→π
π x 2 sin 2 2x cos
p
2 - cos 2x
x
2
2 cos 2
x
2
¼ limþ
x→π
x
2
π x 2 sin 2x cos 2x sin
p
2 cos 2x
x
2
p
p
¼ -π 2 x 1 ¼ -π 2
p
x
¼ limþ - π 2 sin 2
2
x→π
Choice (1) is the answer.
6.49. From application of Taylor series in limit, we know that:
lim
α→0
p
n
1 þ α - lim 1 þ
α→0
α
n
lim am xm þ am - 1 xm - 1 þ . . . þ am - n xm - n þ am - n - 1 xm - n - 1 - am - n - 1 xm - n - 1
x→0
or:
lim ðam xm þ an xn Þ - an xn
x→0
if m > n
The problem can be solved as follows:
p
3
lim
x→0
p
2
1 þ x3 - 1 1 þ x2 - 4 1 - 2x
- lim
x→0
2x2 þ 2x
2x2 þ 2x
2x
4
¼
x2
þx
lim 32 2
x → 0 2x þ 2x
- lim
x
2
x → 0 2x
¼
1
4
Choice (1) is the answer.
6.50. From trigonometry, we know that:
sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1
The problem can be solved as follows:
lim
x→0
sin 2 ðxÞ þ sinðxÞ þ cos 2 ðxÞ - cosðxÞ
1 þ sinðxÞ - cosðxÞ 0
¼ lim
¼
sin 2 ðxÞ - sinðxÞ þ cos 2 ðxÞ - cosðxÞ x → 0 1 - sinðxÞ - cosðxÞ 0
d
dx ð1 þ
x → 0 d ð1 dx
H
¼
¼
¼
¼
¼
¼) lim
Choice (2) is the answer.
sinðxÞ - cosðxÞÞ
cosðxÞ þ sinðxÞ
1þ0
¼
¼ lim
¼ -1
sinðxÞ - cosðxÞÞ x → 0 - cosðxÞ þ sinðxÞ - 1 þ 0
132
6
Solutions of Problems: Limits and Continuities
6.51. From application of Taylor series in limit, we know that:
lim sinðuðxÞÞ - uðxÞ
uðxÞ → 0
The problem can be solved as follows:
lim
x→0
d
dx ðcosðmxÞ - cosðnxÞÞ
d
2
x→0
dx ðx Þ
H
¼
¼
¼
¼
¼
¼) lim
cosðmxÞ - cosðnxÞ 0
¼
0
x2
- m sinðmxÞ þ n sinðnxÞ
- mðmxÞ þ nðnxÞ
- lim
2x
2x
x→0
¼ lim
x→0
- m2 þ n2 n2 - m2
¼
2
2
¼ lim
x→0
Choice (3) is the answer.
6.52. From application of Taylor series in limit, we know that:
lim sinðxÞ - x
x→0
lim tanðxÞ - x
x→0
The problem can be solved as follows:
lim
x→0
d
dx ðsinðxÞ - xÞ
d
x → 0 ðtanðxÞ - xÞ
dx
H
¼
¼
¼
¼
¼
¼) lim
d
dx ðcosðxÞ - 1Þ
x → 0 d ðtan 2 ðxÞÞ
dx
H
¼
¼
¼
¼
¼
¼) lim
¼ lim
x→0
sinðxÞ - x 0
¼
tanðxÞ - x 0
¼ lim
x→0 1
cosðxÞ - 1
cosðxÞ - 1 0
¼ lim
¼
0
þ tan 2 ðxÞ - 1 x → 0 tan 2 ðxÞ
- sinðxÞ
-x
-1
¼ lim
- lim
2 tanðxÞð1 þ tan 2 ðxÞÞ x → 0 2xð1 þ x2 Þ x → 0 2ð1 þ x2 Þ
¼
-1
-1
¼
2
2ð 1 þ 0 Þ
Choice (1) is the answer.
6.53. From trigonometry, we know that:
1 þ cos 3 ðxÞ ¼ ð1 þ cosðxÞÞ 1 - cosðxÞ þ cos 2 ðxÞ
1 - cos 2 ðxÞ ¼ ð1 þ cosðxÞÞð1 - cosðxÞÞ
The problem can be solved as follows:
1 þ cos 3 ðxÞ 0
¼
x → π 1 - cos 2 ðxÞ
0
lim
6
Solutions of Problems: Limits and Continuities
) lim
x→π
133
1 þ cos 3 ðxÞ
ð1 þ cosðxÞÞð1 - cosðxÞ þ cos 2 ðxÞÞ
1 - cosðxÞ þ cos 2 ðxÞ
¼ lim
¼ lim
2
x
→
π
x
→
π
ð1 þ cosðxÞÞð1 - cosðxÞÞ
1 - cosðxÞ
1 - cos ðxÞ
¼
1 - ð- 1Þ þ ð- 1Þ2 3
¼
2
1 - ð- 1Þ
Choice (1) is the answer.
6.54. Based on the information given in the problem, we have:
f ð 0Þ ¼ 0
1
f ðxÞ ¼ xð- 1Þ½x] ,
x 2 ℝ - f 0g
A function is right continuous at this given point of x0 if:
lim f ðxÞ ¼ f ðx0 Þ
x → x0 þ
Moreover, a function is left continuous at this given point of x0 if:
lim f ðxÞ ¼ f ðx0 Þ
x → x0 -
In addition, a function is continuous at this given point of x0 if:
lim f ðxÞ ¼ lim - f ðxÞ ¼ f ðx0 Þ
x → x0 þ
x → x0
For the given function, we have:
1
lim- f ðxÞ ¼ lim- xð- 1Þ½x] ¼ 0 x ðfinite quantityÞ ¼ 0
x→0
x→0
1
limþ f ðxÞ ¼ limþ xð- 1Þ½x] ¼ 0 x ðfinite quantityÞ ¼ 0
x→0
x→0
f ð 0Þ ¼ 0
) limþ f ðxÞ ¼ lim- f ðxÞ ¼ f ð0Þ
x→0
x→0
Thus, the function is continuous at x ¼ 0.
Choice (3) is the answer.
6.55. As we know from calculus:
nk
¼0
n → þ1 an
If a > 1, k 2 N ) lim
Hence:
3n2
n2
lim p n ¼ lim 3 p
n → þ1 5
n → þ1
5
Choice (1) is the answer.
n
¼3x0¼0
134
6
Solutions of Problems: Limits and Continuities
6.56. From trigonometry, we know that:
x
2
1 - cosðxÞ ¼ 2 sin 2
Moreover, from application of Taylor series in limit, we know that:
lim sin n ðxÞ - xn
x→0
Thus:
lim
x→0
x3 - sinðxÞ 2sin 2
x3 - sinðxÞð1 - cosðxÞÞ
¼
lim
x→0
x3
x3
¼ lim
x→0
x3 x3
x3
2
x3
2
x → 0 x3
¼ lim
x
2
- lim
x→0
x3 - x x 2
x3
x 2
2
1 1
¼
2 2
¼ lim
x→0
Choice (2) is the answer.
6.57. From trigonometry and calculus, we know that:
arcðcosð1 - ÞÞ ¼ 0þ
d
-1
ðarcðcos xÞÞ ¼ p
dx
1 - x2
-1
d p
1-x ¼ p
dx
2 1-x
The problem can be solved as follows:
lim-
x→1
H
¼
¼
¼
¼
¼
¼) limx→1
d
dx ðarcðcos xÞÞ
p
d
1-x
dx
¼ limx→1
arcðcos xÞ 0þ
p
¼ þ
0
1-x
p-1
1 - x2
p- 1
2 1-x
p
¼ limx→1
1
ð1 - xÞð1þxÞ
p1
2 1-x
¼ lim- p
x→1
p
2
¼ 2
1þx
Choice (1) is the answer.
6.58. Based on the information given in the problem, we have:
lim x2 - 1 cotðxn - 1Þ ¼
x→1
From calculus, we know that:
cotðxÞ ¼
1
tanðxÞ
From application of Taylor series in limit, we know that:
lim
uðxÞ → 0
tanðuðxÞÞ - uðxÞ
1
2
ð1Þ
6
Solutions of Problems: Limits and Continuities
135
The problem can be solved as follows:
x2 - 1
x2 - 1 0
- lim n
¼
n
tanðx - 1Þ x → 1 x - 1 0
lim x2 - 1 cotðxn - 1Þ ¼ lim
x→1
x→1
H
¼
¼
¼
¼
¼
¼) lim
x→1
2
2x
¼
n
1
n
nx
ð2Þ
Solving (1) and (2):
2 1
¼ )n¼4
n 2
Choice (2) is the answer.
6.59. From trigonometry, we know that:
sinð4xÞ ¼ 2 sinð2xÞ cosð2xÞ
cosðxÞ
sinðxÞ
cotðxÞ ¼
sinðx - yÞ ¼ sinðxÞ cosðyÞ - cosðxÞ sinðyÞ
The problem can be solved as follows:
lim sinð4xÞðcotð2xÞ - cotðxÞÞ ¼ lim 2 sinð2xÞ cosð2xÞ
x→0
x→0
¼ lim 2 sinð2xÞ cosð2xÞ
x→0
sinðxÞ cosð2xÞ - cosðxÞ sinð2xÞ
sinð2xÞ sinðxÞ
sinðx - 2xÞ
sinð2xÞ sinðxÞ
¼ lim 2 sinð2xÞ cosð2xÞ
x→0
cosð2xÞ
cosðxÞ
sinð2xÞ
sinðxÞ
¼ lim ð- 2 cosð2xÞÞ ¼ - 2
x→0
Choice (3) is the answer.
6.60. From application of Taylor series in limit, we know that:
lim sinðxÞ - x -
x3
6
lim cosðxÞ - 1 -
x2
2
x→0
x→0
The problem can be solved as follows:
lim
x → 0- 1
2
sinðxÞ - x
- lim
sinð2xÞ - x cosðxÞ x → 0 -
Choice (3) is the answer.
1
2
2x -
ð2xÞ3
6
x3
6
-x 1-
x2
2
¼ limx→0
-
x3
6
x3
6
¼ lim- 1 ¼ 1
x→0
136
6
Solutions of Problems: Limits and Continuities
6.61. The problem can be solved as follows:
limπ
x→4
1 - 3 tanðxÞ 0
¼
1 - 2 sin 2 ðxÞ 0
tan ðxÞ
p
- 1þ
3
2
2
H
¼
¼
¼
¼
¼
¼) limπ
x→4
d
dx
d
dx
1-
3
tanðxÞ
1 - 2 sin ðxÞ
2
¼ limπ
x→4
3
tan ðxÞ
- 4 sinðxÞ cosðxÞ
¼
4x
1þ1
3x1 p
p
2
2
2 x 2
¼
1
3
Choice (1) is the answer.
6.62. From calculus and trigonometry, we know that:
sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1
1 - cos 3 ðxÞ ¼ ð1 - cosðxÞÞ 1 þ cosðxÞ þ cos 2 ðxÞ
From application of Taylor series in limit, we know that:
lim sinðxÞ - x
x→0
lim tanðuðxÞÞ - uðxÞ
u ð xÞ → 0
The problem can be solved as follows:
lim
x→0
) lim
x→0
¼ lim
1 - cos 3 ðxÞ
0
¼
sinðxÞ tanð2xÞ 0
1 - cos 3 ðxÞ
ð1 - cosðxÞÞð1 þ cosðxÞ þ cos 2 ðxÞÞ ð1 - cosðxÞÞ
¼ lim
x
sinðxÞ tanð2xÞ x → 0
sinðxÞ tanð2xÞ
ð1 þ cosðxÞÞ
x→0
ð1 - cos 2 ðxÞÞð1 þ cosðxÞ þ cos 2 ðxÞÞ
sin 2 ðxÞ x ð1 þ 1 þ 1Þ
¼ lim
x → 0 sinðxÞ tanð2xÞ x ð1 þ 1Þ
sinðxÞ tanð2xÞð1 þ cosðxÞÞ
3 sin 2 ðxÞ
3x2
3 3
- lim
¼ lim ¼
4
x → 0 2 sinðxÞ tanð2xÞ
x → 0 2x x 2x
x→0 4
lim
Choice (3) is the answer.
6.63. From trigonometry, we know that:
1 - cosðxÞ ¼ 2 sin 2
x
2
sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1
From application of Taylor series in limit, we know that:
lim sin n ðuðxÞÞ -
uðxÞ → 0þ
lim ðuðxÞÞn
uðxÞ → 0þ
References
137
The problem can be solved as follows:
limþ
x→0
1 - cosðxÞ
1 - cosðxÞ 1 þ
p ¼ lim
p x
1 - cosð xÞ x → 0þ 1 - cosð xÞ 1 þ
cosðxÞ
cosðxÞ
x
p
1 þ cosð xÞ
p
1 þ cosð xÞ
¼ limþ
p
ð1 - cosðxÞÞð1 þ cosð xÞÞ
ð1 - cosðxÞÞð1 þ 1Þ
p
¼ limþ
p
2
ð
1
- cos 2 ð xÞÞð1 þ 1Þ
x
→
0
ð1 - cos ð xÞÞ 1 þ cosðxÞ
¼ limþ
2 sin 2 2x
2 x
1 - cosðxÞ
x
p ¼ limþ
p - limþ p2 2 ¼ limþ ¼ 0
2
2
1 - cos ð xÞ x → 0 sin ð xÞ x → 0 ð xÞ
x→0 2
x→0
2
x→0
Choice (1) is the answer.
References
1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
7
Problems: Derivatives and Their Applications
Abstract
In this chapter, the basic and advanced problems of derivatives and their applications are presented. The subjects include
the definition of derivative, differentiation formulas, product rule, quotient rule, chain rule, derivatives of trigonometric
functions, derivatives of exponential functions, derivatives of logarithm functions, derivatives of inverse trigonometric
functions, derivatives of hyperbolic functions, implicit differentiation, higher-order derivatives, logarithmic differentiation,
applications of derivatives, rates of change, critical points, minimum and maximum values, and absolute extrema. To help
students study the chapter in the most efficient way, the problems are categorized in different levels based on their difficulty
levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Moreover, the problems are ordered
from the easiest problem with the smallest computations to the most difficult problems with the largest calculations.
7.1. The current population of a specific animal in a jungle is about 820. How long will it take for the population to be 3280
if the growth constant is about 0.2 [1, 2]?
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 4 ln 2
2) 10 ln 2
3) 2 ln 2
4) 2 ln 10
7.2. Which one of the following choices presents the nondifferentiable point(s) of the function below?
f ð x Þ ¼ x ð x þ 2Þ 2 ð x - 3Þ 3
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) x ¼ -2
2) x ¼ 3
3) x ¼ 0
4) x ¼ -2, 0, 3
7.3. Calculate the value of f ′(x ¼ 1) if f (x) ¼ xex - ex.
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 0
3) -e
4) e
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_7
139
140
7
Problems: Derivatives and Their Applications
7.4. If f(x) + g(x3) ¼ 5x - 1 and f ′(1) ¼ 2, calculate the value of g′(1).
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) -1
3) 2
4) -2
7.5. Determine the range of x where the function of y(x) ¼ 1 - 4x2 is ascending.
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) x < 0
2) x > 0
3) -2 < x < 2
4) -4 < x < 4
1
7.6. Determine the derivative of the function below at x ¼ .
4
p
x- x
p
f ð xÞ ¼
1- x
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) -1
1
2) 2
1
3)
2
4) 1
7.7. Determine the first derivative of the function of (x100 + x50 + 50x2 + 50x + 1)10 at x ¼ 0.
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 100
2) 200
3) 400
4) 500
7.8. Calculate the derivative of the function of f (x) ¼ tan3(2x) at
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
4
1)
3
4
2)
9
8
3)
3
8
4)
9
π
.
12
7
Problems: Derivatives and Their Applications
141
7.9. What is the first derivative of the inverse function of f (x) ¼ x3 + x - 2 at a point with the length of zero on the inverse
function?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) -1
2) 0
3) 1
1
4)
4
7.10. If f(x) ¼ x5 + 3x3 + x + 1, then calculate the first derivative of the inverse function at a point with the length of six on the
inverse function.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1
1)
5
1
2)
15
1
3)
6
1
4)
16
7.11. For the following function, calculate the value of ( f-1(x))′ for x ¼ 2.
f ðxÞ ¼
4x3
þ1
x2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
112
1)
25
25
2)
112
1
3)
4
4) 4
7.12. Calculate the value of the limit below if f(x) ¼ x tan x.
lim
f ðxÞ - f
x⟶π4
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
π
2) 1 4
π
3) 1 þ
4
π
4) 1 þ
2
x-
π
4
π
4
142
7
Problems: Derivatives and Their Applications
7.13. If the function of f (x) ¼ jx3 - 3x + aj does not have a derivate at x ¼ 2, calculate the value of a.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 2
2) -2
3) 1
4) -1
7.14. Calculate the value of f ′(2) + f ′(4) if f (x) ¼ jx2 - 6j.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) -8
2) 8
3) -4
4) 4
7.15. If f 0 ðxÞ ¼ 5x, calculate the first derivative of f (x5).
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
5
1) x
25
2) x
25
3)
x
5
4) 5
x
7.16. If the first derivative of f (sin (x)) is equal to cos3(x), determine the value of f ′(x).
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1 + x2
2) 1 - x2
3) x3
4) -x3
1
7.17. Calculate the derivative of the function of f (x) ¼ arc(tan(3x)) at x ¼ .
3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
3
1)
2
4
2)
3
2
3)
3
3
4)
4
p
1
þ g t ¼ t 2 þ 1 and g′(1) ¼ 5, calculate the value of f ′(1).
t
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
7.18. If f
7
Problems: Derivatives and Their Applications
1) 1
2) 2
1
3)
2
4) -
1
2
7.19. If 2 cos ( y) - sin (x + y) + 2 ¼ 0, calculate the value of y0x at (0, π).
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1
1)
2
1
2) 2
3) -1
4) 1
7.20. The equation of a curve is given by x3 + y3 ¼ 16. Calculate the second derivate of y with respect to x.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
16y
1) - 5
x
16x
2) 5
y
32y
3) - 5
x
32x
4) - 5
y
π
7.21. If x ¼ 2 + 3 sin (t) and y ¼ 3 - 2 cos (t), calculate the value of y0x for t ¼ .
6
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
p
2 3
1)
9
p
2 3
2)
3
p
2 2
3)
3
p
4 2
4)
3
7.22. If x ¼ t2 + t and y ¼ t2 - 2t, calculate the value of x0y þ y0x for t ¼ -1.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
11
1)
4
13
2)
4
15
3)
4
17
4)
4
143
144
7
Problems: Derivatives and Their Applications
7.23. Calculate the value of f ′(4) if we know that:
lim
h→0
p
f ð x þ hÞ - f ð x - hÞ
¼2 x
h
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
2
1)
3
4
2)
3
3) 4
4) 2
7.24. Which one of the choices is true about the function of f (x) ¼ x2jxj at x ¼ 0?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) The first derivative exists, but the second derivative does not.
2) The second derivative exists, but the first derivative does not.
3) The first and second derivatives do not exist.
4) The first and second derivatives exist.
π
7.25. The function below is differentiable at x ¼ . Determine the value of b.
4
f ðxÞ ¼
sin 2 ðxÞ - cosð2xÞ
a tanðxÞ þ b sinð2xÞ
π
0<x≤
4
π
π
<x<
4
2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) -1
1
2) 2
1
3)
2
4) 1
7.26. The function below is differentiable everywhere on ℝ domain. Determine the value of b.
f ðxÞ ¼
ax þ b
x2 þ a
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 2
2) 1
3) -2
4) -3
x< -1
x≥ -1
7
Problems: Derivatives and Their Applications
145
7.27. Calculate the derivative of the function below.
f ðxÞ ¼
ð2x - 1Þ2
2x2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
2x - 1
1)
2x3
2x - 1
2)
x3
2x þ 1
3)
x3
2x þ 1
4)
2x3
7.28. Calculate the derivative of the function below.
f ðxÞ ¼
sinðxÞ
1 þ tan 2 ðxÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
5
1)
4
5
2) 4
5
3)
8
5
4) 8
7.29. For the function below, calculate the value of f ′(x ¼ 2).
f ðxÞ ¼ x2 - 5x þ 6 arc sin
1
x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
π
1) 3
2π
2)
3
π
3) 6
π
4)
4
7.30. For the following function, calculate the value of f ′(x ¼ -3).
f ðxÞ ¼ x2 þ 2x - 3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
ðx3
g ð x þ 2Þ
þ 1Þgð2x þ 5Þ
146
7
Problems: Derivatives and Their Applications
13
2
13
2)
2
2
3) 13
2
4)
13
1) -
7.31. 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
3
1)
4p
3
2) ±
8
1
3)
9p
3
4) ±
2
7.32. Determine the third derivate of f (x) ¼ x4 jxj.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) -60x2
2) 60x2
3) -60xjxj
4) 60xjxj
7.33. 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
p
π
at x ¼ .
6
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
p
3
1)
8
p
6
2)
8
p
3
3)
4
p
6
4)
4
7.34. Calculate the derivative of y ¼ ln e
sinðxÞ
p
1
x þ a for x ¼ 2 is .
4
7
Problems: Derivatives and Their Applications
147
7.35. 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
7.36. Which one of the choices is correct about the function below in its one period?
yð xÞ ¼
1 - sinðxÞ
cosðxÞ
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.
7.37. Determine the value of f ′(x)g(x) - f(x)g′(x) if we have the following functions:
f ðxÞ ¼
5
1
1 þ x2 - x , gðxÞ ¼ p
1 þ x2 þ x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) -1
2) 0
3) 1
4) 2
4
7.38. Calculate the first derivative of the following function for x ¼ :
3
yð xÞ ¼ p
1þ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ○ Normal ● Large
- 27
1)
125
-9
2)
25
18
3)
25
54
4)
125
x2
1
p
x þ 1 þ x2
5
148
7
Problems: Derivatives and Their Applications
7.39. 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
p
3) 4 2
p
4) 8 2
7.40. For the function below, calculate the value of f ′(x ¼ -1).
f ðxÞ ¼
ðx þ 1Þ5
j x þ 1j
0
x≠ -1
x ¼ -1
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 0
2) 1
3) -1
4) 5
p
7.41. 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
15
1)
16
15
2)
8
3
3)
7
5
4)
4
7.42. Determine the derivative of f
j - xj þ 3 if we have the relation below.
lim
x→2
f ðxÞ - f ð2Þ
1
¼ x-2
3
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1
1)
6
1
2)
12
1
3) 6
1
4) 12
7
Problems: Derivatives and Their Applications
149
7.43. Determine the value of f ′(-1) if we have:
f ðxÞ ¼
ðx þ 1ÞhðxÞ
, hð- 1Þ ≠ 0
ð2x þ 1Þhð2x þ 1Þ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) -2
2) -1
3) 1
4) 2
7.44. Determine the value of the parameter of “a” so that the function of f ðxÞ ¼ cos 2 ðxÞ þ
3
π
point with the width of y ¼ in the range of 0 < x < .
4
2
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
1
2)
2
1
3) 2
4) -1
7.45. Calculate the first derivative of the function of y(x) ¼ xx for x ¼ e.
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) e
2) ee - 1
3) 2ee
4) ee
7.46. Calculate the first derivative of the function of y(x) ¼ xlnx.
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) xlnx ln x
2) 2xlnx ln x
ln x
3) 2x x ln x
4) (2xlnx ln x)-x
7.47. Calculate the n-th derivative of the function below.
yð xÞ ¼
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 0
n!
2) nþ1
x
n!
3) ð- 1Þn n
x
n!
4) ð- 1Þn nþ1
x
1
x
p
3 sinðxÞ þ a has an extremum
150
7
Problems: Derivatives and Their Applications
7.48. Determine the equation of the line which is perpendicular on the curve of y(x) ¼ x2x at (1, 1).
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) x + 2y - 3 ¼ 0
2) 2x + y - 3 ¼ 0
3) x + y - 2 ¼ 0
4) x - y ¼ 0
7.49. Determine the angle between the right and left tangent lines of the function below at x ¼ 1.
f ðxÞ ¼
x3
p
x>1
x
x≤1
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
π
1)
4
2π
2)
3
π
3)
3
3π
4)
4
References
1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
8
Solutions of Problems: Derivatives
and Their Applications
Abstract
In this chapter, the problems of the seventh chapter are fully solved, in detail, step-by-step, and with different methods.
8.1. Based on the information given in the problem, we have [1, 2]:
yð0Þ ¼ 820
yðt Þ ¼ 3280
k ¼ 0:2
The population of the animal in the jungle at time t can be calculated as follows:
yðt Þ ¼ yð0Þekt
By putting the quantities in the formula, we have:
3280 ¼ 820e0:2t
) 4 ¼ e0:2t ) ln 4 ¼ 0:2t ) t ¼
ln 4
¼ 5 ln 4
0:2
) t ¼ 10 ln 2
Choice (2) is the answer.
In this problem, the rules below were used.
ln eb ¼ b
ln ab ¼ b ln a
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_8
151
152
8 Solutions of Problems: Derivatives and Their Applications
8.2. An absolute function is nondifferentiable at the simple roots of the equation inside the absolute notation.
As can be noticed from the equation inside the absolute notation, x ¼ 0 is the only simple root. Thus, the absolute
function is not differentiable at this point.
f ðxÞ ¼ xðx þ 2Þ2 ðx - 3Þ3
Choice (3) is the answer.
8.3. From list of derivative rules, we know that:
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.
8.4. From 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
The problem can be solved as follows:
f ð xÞ þ g x
3
d
dx 0
¼ 5x - 1¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) f ðxÞ þ 3x2 g0 x3 ¼ 5
x¼1 0
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) f ð1Þ þ 3g0 ð1Þ ¼ 5
f 0 ð 1Þ ¼ 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
¼
¼
¼
¼
¼) 2 þ 3g0 ð1Þ ¼ 5
) g0 ð 1Þ ¼ 1
Choice (1) is the answer.
8
Solutions of Problems: Derivatives and Their Applications
153
8.5. From list of derivative rules, we know that:
f ðxÞ ¼ axn ) f 0 ðxÞ ¼ anxn - 1
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.
8.6. From list of derivative rules, we know that:
f ðxÞ ¼
p
1
x ) f 0 ð xÞ ¼ p
2 x
First, we should simplify the function as follows:
f ðxÞ ¼
p p
p
p
x ð x - 1Þ
x- x
p ¼
p
¼ - x
1- x
1- x
Therefore:
1
1
1
¼ ) f 0 ð xÞ ¼ - p ) f 0
¼ -1
4
2 x
2 14
Choice (1) is the answer.
8.7. From list of derivative rules, we know that:
f ðxÞ ¼ un ðxÞ ) f 0 ðxÞ ¼ nu0 ðxÞun - 1 ðxÞ
Therefore:
f ðxÞ ¼ x100 þ x50 þ 50x2 þ 50x þ 1
10
) f 0 ðxÞ ¼ 10 100x99 þ 50x49 þ 100x þ 50 x100 þ x50 þ 50x2 þ 50x þ 1
) f 0 ð0Þ ¼ 10ð0 þ 0 þ 0 þ 50Þð0 þ 0 þ 0 þ 0 þ 1Þ9 ¼ 500
Choice (4) is the answer.
8.8. From list of derivative rules, we know that:
f ðxÞ ¼ tan n ðuðxÞÞ ) f 0 ðxÞ ¼ nu0 ðxÞ 1 þ tan 2 ðuðxÞÞ tan n - 1 ðuðxÞÞ
Therefore:
f ðxÞ ¼ tan 3 ð2xÞ ) f 0 ðxÞ ¼ 3 2 1 þ tan 2 ð2xÞ tan 2 ð2xÞ
) f0
Choice (3) is the answer.
π
π
¼ 3 2 1 þ tan 2
12
6
tan 2
π
1
1 8
¼6 1þ
¼
6
3
3 3
9
154
8 Solutions of Problems: Derivatives and Their Applications
8.9. The derivative of the inverse function of f(x) at a point with the length of “b” on the inverse function can be calculated as
follows:
0
f - 1 ðbÞ ¼
1
f 0 ð aÞ
On the other hand, we know that:
f ð a Þ ¼ b , f - 1 ð bÞ ¼ a
Therefore, for the function of f(x) ¼ x3 + x - 2 and the point with the length of zero (b ¼ 0) on the inverse function,
we can calculate “a” as follows:
0 ¼ a3 þ a - 2 ) a ¼ 1
Also, we have:
f ðxÞ ¼ x3 þ x - 2 ) f 0 ðxÞ ¼ 3x2 þ 1
Hence:
0
f - 1 ð 0Þ ¼
1
1
1
¼
¼
f 0 ð 1Þ 3 1 2 þ 1 4
Choice (4) is the answer.
8.10. As we know, the derivative of the inverse function of f(x) at a point with the length of “b” on the inverse function can be
calculated as follows:
0
f - 1 ðbÞ ¼
1
f 0 ð aÞ
Moreover:
f ð a Þ ¼ b , f - 1 ð bÞ ¼ a
Thus, for the function of f(x) ¼ x5 + 3x3 + x + 1 and the point with the length of six (b ¼ 6) on the inverse function,
we can calculate “a” as follows:
6 ¼ a5 þ 3a3 þ a þ 1 ) a ¼ 1
On the other hand, we have:
f ðxÞ ¼ x5 þ 3x3 þ x þ 1 ) f 0 ðxÞ ¼ 5x4 þ 9x2 þ 1
Hence:
0
f - 1 ð 6Þ ¼
Choice (2) is the answer.
1
1
1
¼
¼
f 0 ð1Þ 5 14 þ 9 12 þ 1 15
8
Solutions of Problems: Derivatives and Their Applications
155
8.11. As we know, the derivative of the inverse function of f(x) at a point with the length of “b” on the inverse function can be
calculated as follows:
0
f - 1 ðbÞ ¼
1
f 0 ð aÞ
In addition, we know that:
f ð a Þ ¼ b , f - 1 ð bÞ ¼ a
3
Hence, for the function of f ðxÞ ¼ x4x
2 þ1 and the point with the length of two (b ¼ 2) on the inverse function, we can
calculate “a” as follows:
2¼
4a3
)a¼1
a2 þ 1
On the other hand, we can write:
f ðxÞ ¼
12x2 ðx2 þ 1Þ - ð2xÞð4x3 Þ
4x3
) f 0 ð xÞ ¼
þ1
ðx2 þ 1Þ2
x2
Thus:
0
f - 1 ð 2Þ ¼
2
12 þ 1
1
4
1
¼
¼
¼
0
2
2
24 - 8 4
f ð1Þ 12 1 1 þ 1 - ð2 1Þ 4 13
Choice (3) is the answer.
8.12. We know that:
d
ðuðxÞvðxÞÞ ¼ u0 ðxÞvðxÞ þ uðxÞv0 ðxÞ
dx
ð1Þ
d
ðtan xÞ ¼ 1 þ tan 2 x
dx
tan
π
¼1
4
The first derivative of a function is defined as follows:
lim
x⟶x0
f ð xÞ - f ð x0 Þ
x - x0
Therefore, the value of the following limit for the function of f(x) ¼ x tan x is equal to the first derivative of the function
at x0 ¼ π4.
f ðxÞ - f
x - π4
x⟶4
limπ
π
4
¼ f 0 ð x0 Þ
156
8 Solutions of Problems: Derivatives and Their Applications
The first derivative of the function can be calculated as follows:
f 0 ðxÞ ¼ tan x þ 1 þ tan 2 x x
For x0 ¼ π4, we have:
f0
π
π
π π
π
¼ tan þ 1 þ tan 2
¼1þ
4
4
4 4
2
Choice (4) is the answer.
8.13. Based on the information given in the problem, we have:
f ðxÞ ¼ x3 - 3x þ a
A derivative of an absolute function does not exist at its simple roots. Therefore, we need to solve the equation below:
f ð 2Þ ¼ 0
) 23 - 3 2 þ a ¼ 0 ) 8 - 6 þ a ¼ 0 ) a ¼ - 2
Choice (2) is the answer.
8.14. From list of derivative rules, we know that:
f ð x Þ ¼ j uð x Þ j ) f 0 ð xÞ ¼
u0 ðxÞuðxÞ
j uð x Þ j
Based on the information given in the problem, we have:
f ð x Þ ¼ x2 - 6
Therefore:
) f 0 ð xÞ ¼
) f 0 ð 2Þ ¼
2x ðx2 - 6Þ
j x 2 - 6j
4 ð 4 - 6Þ
8 ð16 - 6Þ
¼ - 4 and f 0 ð4Þ ¼
¼8
j 4 - 6j
j16 - 6j
) f 0 ð 2Þ þ f 0 ð 4Þ ¼ - 4 þ 8 ¼ 4
Choice (4) is the answer.
8.15. From list of derivative rules, we know that:
f ðxÞ ¼ gðhðxÞÞ ) f 0 ðxÞ ¼ h0 ðxÞg0 ðhðxÞÞ
Based on the information given in the problem, we have:
f 0 ðxÞ ¼
5
x
8
Solutions of Problems: Derivatives and Their Applications
157
The problem can be solved as follows:
f x5
0
¼ 5x4 f 0 x5 ¼ 5x4 5
) f x5
x5
0
¼
25
x
Choice (3) is the answer.
8.16. From list of derivative rules, we have:
ðf ðgðxÞÞÞ0 ¼ g0 ðxÞf 0 ðgðxÞÞ
Based on the information given in the problem, we have:
ðf ðsinðxÞÞÞ0 ¼ cos 3 ðxÞ
The problem can be solved as follows:
ðf ðsinðxÞÞÞ0 ¼ cos 3 ðxÞ ) cosðxÞ f 0 ðsinðxÞÞ ¼ cos 3 ðxÞ ) f 0 ðsinðxÞÞ ¼ cos 2 ðxÞ
) f 0 ðsinðxÞÞ ¼ 1 - sin 2 ðxÞ ) f 0 ðxÞ ¼ 1 - x2
Choice (2) is the answer.
8.17. From list of derivative rules, we know that:
f ðxÞ ¼ arcðtanðuðxÞÞÞ ) f 0 ðxÞ ¼
u0 ð x Þ
1 þ u2 ð xÞ
Therefore:
f ðxÞ ¼ arcðtanð3xÞÞ ) f 0 ðxÞ ¼
f0
1
3
¼
3
1þ9
1 2
3
¼
3
1 þ 9x2
3
2
Choice (1) is the answer.
8.18. From list of derivative rules, we know that:
f ðxÞ ¼ axn ) f 0 ðxÞ ¼ anxn - 1
f ðxÞ ¼ gðhðxÞÞ ) f 0 ðxÞ ¼ h0 ðxÞg0 ðhðxÞÞ
Based on the information given in the problem, we have:
f
p
1
þ g t ¼ t2 þ 1
t
g0 ð1Þ ¼ 5
158
8 Solutions of Problems: Derivatives and Their Applications
The problem can be solved as follows:
d
p
p
dx
1
1
1
1
þ p g0 t ¼ 2t
f
þ g t ¼ t 2 þ 1¼
¼
¼
¼
¼
¼
¼
¼) - 2 f 0
t
t
t
2 t
g0 ð 1Þ ¼ 5
1
5
t ¼ 1 ) - f 0 ð1Þ þ g0 ð1Þ ¼ 2¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
) - f 0 ð 1Þ þ ¼ 2
2
2
) f 0 ð 1Þ ¼
1
2
Choice (3) is the answer.
8.19. From derivative rules, we know that:
f ðx, yÞ ¼ 0 ) y0x ¼ -
f 0x ðx, yÞ
¼ f 0y ðx, yÞ
d
dx f ðx, yÞ
d
dy f ðx, yÞ
Based on the information given in the problem, we have:
2 cosðyÞ - sinðx þ yÞ þ 2 ¼ 0
The problem can be solved as follows:
y0x ¼ -
d
dx ð2cosðyÞ d
dy ð2cosðyÞ -
sinðx þ yÞ þ 2Þ
- cosðx þ yÞ
cosðx þ yÞ
¼ ¼ 2
sin
ð
y
Þ
cos
ð
x
þ
y
Þ
2
sin
ð
y
Þ þ cosðx þ yÞ
sinðx þ yÞ þ 2Þ
ðx, yÞ ¼ ð0, π Þ 0
-1
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) yx ¼ 0-1
) y0x ¼ - 1
Choice (3) is the answer.
8.20. From derivative rules, we know that:
f ðx, yÞ ¼ 0 ) y0x ¼ -
f 0x ðx, yÞ
¼ f 0y ðx, yÞ
Based on the information given in the problem, we have:
x3 þ y3 ¼ 16
The problem can be solved as follows:
y0x ¼ y0 ¼ ) y00 ¼ -
3x2
x2
¼ - 2
2
3y
y
2xy2 - 2yy0 x2
y4
d
dx f ðx, yÞ
d
dy f ðx, yÞ
8
Solutions of Problems: Derivatives and Their Applications
159
x2
y0 ¼ - 2
2xy2 - 2y y 00
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) y ¼ y4
x2
y2
x2
¼ -
2xðy3 þ x3 Þ
2xy3 þ 2x4
¼ 5
y5
y
x3 þ y3 ¼ 16 00
2x 16
32x
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) y ¼ ) y00 ¼ - 5
y5
y
Choice (4) is the answer.
8.21. From derivative rules, we know that:
y ¼ yð t Þ
y0 d yð t Þ
) y0x ¼ t0 ¼ dtd
xt dt xðt Þ
x ¼ xð t Þ
Based on the information given in the problem, we have:
x ¼ 2 þ 3 sinðt Þ
y ¼ 3 - 2 cosðt Þ
Hence:
y0x ¼
y0t 2 sinðt Þ 2
¼
¼ tanðt Þ
x0t 3 cosðt Þ 3
p
p
2
3
2 3
π
0
0
) yx ¼
t ¼ ) yx ¼ 3
3
9
6
Choice (1) is the answer.
8.22. From derivative rules, we know that:
y ¼ yð t Þ
y0 d yð t Þ
x0 d xð t Þ
) y0x ¼ t0 ¼ dtd
, x0y ¼ t0 ¼ dtd
xt dt xðt Þ
yt dt yðt Þ
x ¼ xð t Þ
Based on the information given in the problem, we have:
x ¼ t2 þ t
y ¼ t 2 - 2t
Therefore:
)
y0t 2t - 2
¼
x0t 2t þ 1
t ¼ -1
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼)
0
x
2t
þ
1
t
0
xy ¼ 0 ¼
yt 2t - 2
y0x ¼
Choice (4) is the answer.
-2-2
¼4
1
17
-2 þ 1
) y0x þ x0x ¼ 4 þ ) y0x þ x0x ¼
4
4
2
þ
1
1
¼
x0y ¼
-2-2 4
y0x ¼
160
8 Solutions of Problems: Derivatives and Their Applications
8.23. Based on the information given in the problem, we have:
lim
h→0
p
f ð x þ hÞ - f ð x - hÞ
¼2 x
h
ð1Þ
From definition of derivative, we know:
f 0 ðxÞ ¼ lim
h→0
f ð x þ hÞ - f ð x Þ
f ð x Þ - f ð x - hÞ
¼ lim
h
h
h→0
The problem can be solved as follows:
lim
h→0
f ð x þ hÞ - f ð x - hÞ
f ð x þ h Þ - f ð x Þ þ f ð x Þ - f ð x - hÞ
¼ lim
h
h
h→0
¼ lim
h→0
f ðx þ h Þ - f ðxÞ
f ð x Þ - f ð x - hÞ
þ lim
¼ 2f 0 ðxÞ
h
h
h→0
ð2Þ
Solving (1) and (2):
p
p
p
2f 0 ðxÞ ¼ 2 x ) f 0 ðxÞ ¼ x ) f 0 ð4Þ ¼ 4 ¼ 2
Choice (4) is the answer.
8.24. The function can be simplified as follows:
f ðxÞ ¼ x2 jxj ¼
) f ð 0Þ ¼
0
0
x≥0
x3
-x
3
x<0
x≥0
) f ð 0 - Þ ¼ f ð 0þ Þ
x<0
ð1Þ
Now, we can determine its first and second derivatives as follows:
) f 0 ð xÞ ¼
) f 00 ðxÞ ¼
x≥0
3x2
- 3x
2
x<0
) f 0 ð 0Þ ¼
6x
x≥0
) f 00 ð0Þ ¼
- 6x x < 0
0 x≥0
) f 0 ð 0 - Þ ¼ f 0 ð 0þ Þ
ð2Þ
x≥0
) f 00 ð0 - Þ ¼ f 00 ð0þ Þ
x<0
ð3Þ
0 x<0
0
0
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.
8.25. Based on the information given in the problem, we have:
f ðxÞ ¼
sin 2 ðxÞ - cosð2xÞ
a tanðxÞ þ b sinð2xÞ
π
0<x≤
4
π
π
<x<
4
2
ð1Þ
8
Solutions of Problems: Derivatives and Their Applications
161
The function is differentiable at x ¼ π4. Therefore, we can conclude that:
f
π
4
-
f0
π
4
-
¼f
π
4
¼ f0
π
4
þ
ð2Þ
þ
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
Solving (1) and (3):
2 sin
π
π
π
π
¼ a 1 þ tan 2
cos
þ 2 sin 2 4
4
4
4
) 1 þ 2 ¼ 2a þ 0 ) a ¼
þ 2b cos 2 3
2
π
4
ð5Þ
Solving (4) and (5):
b ¼ -1
Choice (1) is the answer.
8.26. Based on the information given in the problem, we have:
f ðxÞ ¼
ax þ b
x< -1
x þa
x≥ -1
2
ð1Þ
The function is differentiable everywhere on ℝ domain including x ¼ - 1. Hence:
f ðð- 1Þ - Þ ¼ f ð- 1Þþ
ð2Þ
f 0 ðð- 1Þ - Þ ¼ f 0 ð- 1Þþ
ð3Þ
Solving (1) and (2):
- a þ b ¼ 1 þ a ) - 2a þ b ¼ 1
ð4Þ
a ¼ -2
ð5Þ
Solving (1) and (3):
Solving (4) and (5):
b ¼ -3
Choice (4) is the answer.
162
8 Solutions of Problems: Derivatives and Their Applications
8.27. From list of derivative rules, we know that:
f ðxÞ ¼
gð x Þ
g0 ðxÞhðxÞ - h0 ðxÞgðxÞ
) f 0 ð xÞ ¼
hð x Þ
h 2 ð xÞ
Therefore:
f ðxÞ ¼
ð2x - 1Þ2
4ð2x - 1Þð2x2 Þ - 4xð2x - 1Þ2
) f 0 ð xÞ ¼
2
2x
4x4
) f 0 ð xÞ ¼
16x3 - 8x2 - 16x3 þ 16x2 - 4x 8x2 - 4x 2x - 1
¼
¼
4x4
4x4
x3
Choice (2) is the answer.
8.28. From 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
sinðxÞ
sinðxÞ
) f ðxÞ ¼ 1 ¼ sinðxÞ cos 2 ðxÞ
1 þ tan 2 ðxÞ
cos 2 ðxÞ
) f 0 ðxÞ ¼ cosðxÞ cos 2 ðxÞ - sinðxÞ 2 sinðxÞ cosðxÞ ¼ cos 3 ðxÞ - 2 sin 2 ðxÞ cosðxÞ
π
1
¼
)f
3
2
0
3
p
-2
3
2
2
1 1 3
5
¼ - ¼ 2 8 4
8
Choice (4) is the answer.
8.29. From 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 anxn - 1
From trigonometry, we know that:
arc sin
1
2
¼
π
6
Based on the information given in the problem, we have:
f ðxÞ ¼ x2 - 5x þ 6 arc sin
1
x
8
Solutions of Problems: Derivatives and Their Applications
163
Therefore:
f 0 ðxÞ ¼ ð2x - 5Þarc sin
f 0 ð2Þ ¼ ð2 2 - 5Þarc sin
1
2
1
x
þ x2 - 5x þ 6
þ 0 arc sin
0
1
x
As can be seen, we did not need to calculate the value of arc sin
arc sin
¼ ð- 1Þ 1
x
0
1
x
0
π
π
) f 0 ð2Þ ¼ 6
6
. Choice (3) is the answer.
8.30. From 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 anxn - 1
Based on the information given in the problem, we have:
f ðxÞ ¼ x2 þ 2x - 3 ð x3
g ð x þ 2Þ
≜uðxÞ vðxÞ
þ 1Þgð2x þ 5Þ
The problem can be solved as follows:
f 0 ðxÞ ¼ ð2x þ 2Þ
) f 0 ðx ¼ - 3Þ ¼ ð- 4Þ g ð x þ 2Þ
þ x2 þ 2x - 3 v0 ðxÞ
ðx3 þ 1Þgð2x þ 5Þ
gð- 1Þ
1
þ ð 0 Þ v 0 ð x ¼ - 3Þ ¼ - 4 26
- 26gð- 1Þ
) f 0 ðx ¼ - 3Þ ¼
2
13
As can be seen, we did not need to calculate the value of v′(x ¼ - 3). Choice (4) is the answer.
8.31. 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
p
Δ ¼ 0 ) ð2m þ 1Þ2 - 4ðm þ 1Þ ¼ 0 ) 4m2 - 3 ¼ 0 ) m ¼ ±
Choice (4) is the answer.
8.32. From list of derivative rules, we know that:
f ðxÞ ¼ axn ) f 0 ðxÞ ¼ anxn - 1
Based on the information given in the problem, we have:
f ðxÞ ¼ x4 jxj
3
2
164
8 Solutions of Problems: Derivatives and Their Applications
Therefore:
f ðxÞ ¼
x5 ,
- x5 ,
x≥0
) f 0 ð xÞ ¼
x<0
000
) f ð xÞ ¼
5x4 ,
0,
x>0
x ¼ 0 ) f 00 ðxÞ ¼
20x3 ,
0,
x>0
x¼0
- 5x4 ,
x<0
- 20x3 ,
x<0
60x2 ,
x>0
0,
- 60x2 ,
x ¼ 0 ) f 000 ðxÞ ¼ 60xjxj
x<0
Choice (4) is the answer.
8.33. Based on the information given in the problem, we have:
f ðxÞ ¼
p
1
4
ð1Þ
u0 ð x Þ
2 uð x Þ
ð2Þ
x þ a ) f 0 ð 2Þ ¼
From list of derivative rules, we know that:
f ðxÞ ¼
uð x Þ ) f 0 ð xÞ ¼
Solving (1) and (2):
1
2 xþa
p
¼
x¼2
p
1
1
1
¼ ) 2þa¼2)a¼2
) p
4
2 2þa 4
Choice (4) is the answer.
8.34. First, we should simplify the function as follows:
yðxÞ ¼ ln e
p
sinðxÞ
¼
sinðxÞ
ð1Þ
u0 ð x Þ
2 uð x Þ
ð2Þ
From list of derivative rules, we know that:
f ðxÞ ¼
uð x Þ ) f 0 ð xÞ ¼
Solving (1) and (2):
cos π6
cosðxÞ
π
¼
) y ð xÞ ¼
) y0
6
2 sinðxÞ
2 sin
0
Choice (4) is the answer.
π
6
¼
p
3
2
2
p
6
¼
4
1
2
8
Solutions of Problems: Derivatives and Their Applications
165
8.35. From list of derivative rules, we know that:
f ðxÞ ¼ axn ) f 0 ðxÞ ¼ anxn - 1
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:
yð- 2Þ ¼ ð- 2Þ3 - 3ð- 2Þ2 - 9ð- 2Þ þ 5 ¼ 3
yð- 1Þ ¼ ð- 1Þ3 - 3ð- 1Þ2 - 9ð- 1Þ 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.
8.36. From list of derivative rules, we know that:
f ðxÞ ¼
g0 ðxÞhðxÞ - h0 ðxÞgðxÞ
gð x Þ
) f 0 ð xÞ ¼
hð x Þ
h 2 ð xÞ
ð1Þ
First, we need to take its derivative as follows:
yð xÞ ¼
1 - sinðxÞ
- cosðxÞ cosðxÞ - ð- sinðxÞÞð1 - sinðxÞÞ
) y0 ðxÞ ¼
cosðxÞ
cos 2 ðxÞ
y0 ð x Þ ¼
- cos 2 ðxÞ þ sinðxÞ - sin 2 ðxÞ - 1 þ sinðxÞ
¼
cos 2 ðxÞ
cos 2 ðxÞ
The y′(x) is always nonpositive because sin(x) ≤ 1. Hence, the function is a descending function. Choice (2) is the
answer.
8.37. Based on the information given in the problem, we have:
f ðxÞ ¼
5
1
1 þ x2 - x , gðxÞ ¼ p
1 þ x2 þ x
ð1Þ
5
From list of derivative rules, we know that:
f ðxÞ
gð x Þ
0
¼
f 0 ðxÞgðxÞ - g0 ðxÞf ðxÞ
) f 0 ðxÞgðxÞ - g0 ðxÞf ðxÞ ¼
ðgðxÞÞ2
0
f ð xÞ
ðgðxÞÞ2
gðxÞ
ð2Þ
166
8 Solutions of Problems: Derivatives and Their Applications
Solving (1) and (2):
p
0
0
f ðxÞgðxÞ - g ðxÞf ðxÞ ¼
ð
¼
1 þ x2 - x 2
5 0
p
1
1þ
x2
þx
10
5
1 þ x2 - x
0
p
1 þ x2 þ x
1
p
5
1þx2 þxÞ
¼ ð 1Þ 0 p
2
1
1
1þ
x2
þx
10
5
¼0 p
1
1 þ x2 þ x
10
¼0
Choice (2) is the answer.
8.38. From list of derivative rules, we know that:
f 0 ð xÞ
d
ðln f ðxÞÞ ¼
dx
f ðxÞ
d
dx
f ðxÞ ¼
ð1Þ
f 0 ðxÞ
2 f ðxÞ
ð1Þ
In addition, we know that:
ln 1 ¼ 0
ln
m
f ðxÞ ¼
1
ln f ðxÞ
m
lnðf ðxÞgðxÞÞ ¼ lnðf ðxÞÞ þ lnðgðxÞÞ
The problem can be solved as follows:
yð xÞ ¼ p
ln
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) ln yðxÞ ¼ ln 1 - ln
1þ
1 þ x2 x þ
) ln yðxÞ ¼ -
x2
1
p
x þ 1 þ x2
1 þ x2
¼ 0 - ln
1
ln 1 þ x2 - ln x þ
2
1 þ x2 - ln x þ
1 þ x2
d
1 þ 2p2x
2
dx y0 ðxÞ
1 2x
p 1þx
¼ ¼
¼
¼
¼
¼
¼
¼
¼
¼
¼)
2 1 þ x2 x þ 1 þ x 2
yðxÞ
1 þ 2p1þx2
1 2x
p
2 1 þ x 2 x þ 1 þ x2
2x
) y 0 ð xÞ ¼ yð xÞ -
) y 0 ð xÞ ¼
p
1
p
2
1 þ x x þ 1 þ x2
1 þ 2p1þx2
1 2x
p
2 1 þ x2 x þ 1 þ x2
2x
-
1 þ x2
8
Solutions of Problems: Derivatives and Their Applications
167
Now, for x ¼ 43, we have:
y0
4
27
¼ 3
125
Choice (1) is the answer.
8.39. 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
p
2 ) y1 ¼
p
3
p
p
p
p
- 6 2 þ 12 ¼ 2 2 - 6 2 þ 12 ¼ - 4 2 þ 12
3
p
p
p
p
- 6 - 2 þ 12 ¼ - 2 2 þ 6 2 þ 12 ¼ 4 2 þ 12
2
p
p
x2 ¼ - 2 ) y 2 ¼ - 2
p
p
2, - 2
p
p
p
y2 - y1 ¼ 4 2 þ 12 - - 4 2 þ 12 ) y2 - y1 ¼ 8 2
Choice (4) is the answer.
8.40. Based on the information given in the problem, we have:
ðx þ 1Þ5
j x þ 1j
0
f ðxÞ ¼
x≠ -1
x ¼ -1
This problem can be solved by using the definition of derivative of a function as follows:
f 0 ðx0 Þ ¼ lim
x → x0
0
) f ð- 1Þ ¼ lim
x → ð- 1Þ
ðxþ1Þ5
jxþ1j
-0
x - ð- 1Þ
¼ lim
f ð xÞ - f ð x0 Þ
x - x0
ðxþ1Þ5
jxþ1j
x → ð- 1Þ x
þ1
ðx þ 1Þ4
¼ lim jx þ 1j3 ¼ 0
x → ð- 1Þ jx þ 1j
x → ð- 1Þ
¼ lim
Choice (1) is the answer.
8.41. From list of derivative rules, we know that:
f ðxÞ ¼
uð x Þ ) f 0 ð xÞ ¼
u0 ð x Þ
2 uð x Þ
ð1Þ
The distance of the point from the origin can be calculated as follows:
D ðxÞ ¼
ð x - 0Þ 2 þ
p
x þ 8-0
2
¼
x2 þ x þ 8
In addition, the changing rate of the distance can be determined as follows:
D0 ðxÞ ¼
d
d
DðxÞ ¼
dx
dx
x2 þ x þ 8
ð2Þ
168
8 Solutions of Problems: Derivatives and Their Applications
Solving (1) and (2):
15
2x þ 1
27þ1
¼
D0 ðxÞ ¼ p
) D0 ð7Þ ¼
2
2
2 x þxþ8
2 7 þ 7 þ 8 16
Choice (1) is the answer.
8.42. From list of derivative rules, we know that:
f ðxÞ ¼ gðhðxÞÞ ) f 0 ðxÞ ¼ h0 ðxÞg0 ðhðxÞÞ
f ðxÞ ¼
uð x Þ ) f 0 ð xÞ ¼
u0 ð x Þ
2 uð x Þ
The problem should be solved using the definition of derivative of a function as follows:
f 0 ðx0 Þ ¼ lim
x → x0
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 x-2
3
ð2Þ
Solving (1) and (3):
f 0 ð 2Þ ¼ -
1
3
ð3Þ
Therefore:
d
f
dx
j - xj þ 3
x¼ - 1
¼
p
d
-x þ 3
f
dx
x¼ - 1
p
-1
) p
f0 -x þ 3
2 -x þ 3
x¼ - 1
¼ -
1 0
f ð 2Þ
4
ð4Þ
Solving (3) and (4):
d
f
dx
j - xj þ 3
x¼ - 1
¼ -
1
1
1
¼
4
3
12
Choice (2) is the answer.
8.43. Based on the information given in the problem, we have:
f ðxÞ ¼
ðx þ 1ÞhðxÞ
, hð- 1Þ ≠ 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 ð xÞ - f ð x0 Þ
f ðxÞ - f ð- 1Þ
) f 0 ð- 1Þ ¼ lim
x - x0
x → - 1 x - ð- 1Þ
ð2Þ
8
Solutions of Problems: Derivatives and Their Applications
169
Solving (1) and (2):
f 0 ð- 1Þ ¼ lim
ðxþ1ÞhðxÞ
ð2xþ1Þhð2xþ1Þ
-
ð- 1þ1Þhð- 1Þ
ð- 2þ1Þhð- 2þ1Þ
x - ð- 1Þ
x→ -1
f 0 ð- 1Þ ¼ lim
x → - 1 ð2x
¼ lim
ðxþ1ÞhðxÞ
ð2xþ1Þhð2xþ1Þ
x→ -1
-0
xþ1
hð x Þ
hð- 1Þ
¼
¼ -1
þ 1Þhð2x þ 1Þ ð- 1Þhð- 1Þ
Choice (2) is the answer.
8.44. Based on the information given in the problem, the width of the extremum point is as follows:
yð xM 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:
f 0 ðxÞ ¼ 0
f ðxÞ ¼ cos 2 ðxÞ þ
) f 0 ðxÞ ¼ - 2 sinðxÞ cosðxÞ þ
p
p
ð2Þ
3 sinðxÞ þ a
3 cosðxÞ ¼ cosðxÞ - 2 sinðxÞ þ
ð3Þ
p
3
ð4Þ
Solving (2) and (4):
cosðxÞ - 2 sinðxÞ þ
p
3 ¼0)
cosðxÞ ¼ 0
p
3
sinðxÞ ¼
2
ð 5Þ
ð 6Þ
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, using xM ¼ π3 and (1) in (3), we have:
p
3
π
π
3 1 3
þ 3 sin
þ a ) ¼ þ þ a ) a ¼ -1
¼ cos 2
4
3
3
4 4 2
Choice (4) is the answer.
8.45. From list of derivative rules, we know that:
f 0 ð xÞ
d
ðln f ðxÞÞ ¼
dx
f ðxÞ
ð1Þ
d
ðuðxÞvðxÞÞ ¼ u0 ðxÞvðxÞ þ uðxÞv0 ðxÞ
dx
ð2Þ
Moreover, we know that:
ln ab ¼ a ln b
ln e ¼ 1
170
8 Solutions of Problems: Derivatives and Their Applications
The problem can be solved as follows:
yðxÞ ¼ xx
ln
¼
¼
¼
¼
¼
¼) ln yðxÞ ¼ ln xx ) ln yðxÞ ¼ x ln x
d
dx y0 ðxÞ
1
¼ ln x þ 1
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼)
¼ ln x þ x
x
yð xÞ
) y0 ðxÞ ¼ yðxÞðln x þ 1Þ ) y0 ðxÞ ¼ xx ðln x þ 1Þ
) y0 ðeÞ ¼ ee ðln e þ 1Þ ¼ 2ee
Choice (3) is the answer.
8.46. From list of derivative rules, we know that:
f 0 ð xÞ
d
ðln f ðxÞÞ ¼
dx
f ð xÞ
ð1Þ
d
ððf ðxÞÞn Þ ¼ nf 0 ðxÞðf ðxÞÞn - 1
dx
ð2Þ
Also, we know that:
ln f ðxÞgðxÞ ¼ gðxÞ ln f ðxÞ
The problem can be solved as follows:
yðxÞ ¼ x ln x
ln
¼
¼
¼
¼
¼
¼) ln yðxÞ ¼ ln x ln x ¼ ln x ln x ) ln yðxÞ ¼ ðln xÞ2
d
dx y0 ðxÞ
1
¼2
ln x
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼)
x
yð xÞ
) y0 ð x Þ ¼
2x ln x ln x
x
Choice (3) is the answer.
8.47. From list of derivative rules, we know that:
d f ðxÞ
dx gðxÞ
¼
f 0 ðxÞgðxÞ - f ðxÞg0 ðxÞ
ð gð x Þ Þ 2
We can find a formula for the n-th derivative of the function below as follows:
yð xÞ ¼
1
x
ð1Þ
8
Solutions of Problems: Derivatives and Their Applications
171
d
dx 0
1
0-1 1
1
¼ - 2 ¼ ð- 1Þ1 1þ1
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) y ðxÞ ¼
x
x2
x
d2
dx2 00
21
5 - 2x 1 2
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) y ðxÞ ¼ ¼ 3 ¼ ð- 1Þ2 2þ1
2
2
x
x
ðx Þ
d3
dx3 000
321
5 - 3x2 2 - 6
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) y ðxÞ ¼
¼ 4 ¼ ð- 1Þ3
3
3
x3þ1
x
ðx Þ
⋮
dn
dxn ðnÞ
n!
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) y ðxÞ ¼ ð- 1Þn nþ1
x
Choice (4) is the answer.
8.48. The equation of a line which is tangent on a curve at the point of (x0, y0), located on the curve, can be calculated as
follows:
y - y0 ¼ mðx - x0 Þ
Moreover, the equation of a line which is perpendicular on a curve at the point of (x0, y0), located on the curve, can be
calculated as follows:
y - y0 ¼ m 0 ð x - x 0 Þ
where m and m′ are the slope of the tangent and perpendicular lines. In addition, we have:
m ¼ y0 ð x 0 Þ
m0 ¼ -
1
m
Therefore, first, we need to determine the first derivative of y(x).
yðxÞ ¼ x2x
ln
¼
¼
¼
¼
¼
¼) ln yðxÞ ¼ ln x2x ¼ 2x ln x
d
dx y0 ðxÞ
¼ 2 ln x þ 2
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼)
yð xÞ
) y0 ðxÞ ¼ yðxÞð2ln x þ 2Þ ¼ x2x ð2ln x þ 2Þ
The slope of the tangent line:
m ¼ y0 ðx0 ¼ 1Þ ¼ 12 ð2ln 1 þ 2Þ ¼ 2
172
8 Solutions of Problems: Derivatives and Their Applications
The slope of the perpendicular line:
m0 ¼ -
1
1
¼ m
2
The equation of the perpendicular line:
y-1 ¼ )y¼ -
1
ð x - 1Þ
2
1
3
x þ ) x þ 2y - 3 ¼ 0
2
2
Choice (1) is the answer.
In this problem, the rules below were used.
ln ab ¼ b ln a
f 0 ð xÞ
d
ðln f ðxÞÞ ¼
dx
f ð xÞ
8.49. The angle between the right and left tangent lines of a function can be determined as follows:
θ ¼ π - tan - 1
m - m0
1 þ mm0
where m and m′ are the slope of the right and left tangent lines.
For the following function at the given point of x0 ¼ 1, we have:
f ðxÞ ¼
x3
p
x>1
x
x≤1
m ¼ f 0 ð1þ Þ ¼ 3x2 x ¼ 1þ ¼ 3
0
1
1
m0 ¼ f 0 ð1 - Þ ¼ p
¼
2
2 x x0 ¼ 1 Therefore:
θ ¼ π - tan - 1
3 - 12
π
¼ π - tan - 1 ð1Þ ¼ π 4
1 þ 3 12
θ¼
Choice (4) is the answer.
3π
4
References
173
In this problem, the rules below were used.
m ¼ f 0 ð x0 Þ
tan - 1 ð1Þ ¼
π
4
References
1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
9
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,
and integrals involving quadratics. To help students study the chapter in the most efficient way, the problems are
categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small,
normal, and large). Moreover, the problems are ordered from the easiest problem with the smallest computations to the
most difficult problems with the largest calculations.
9.1. Calculate the value of the indefinite integral below [1, 2].
I¼
ð3x þ 5Þ17 dx
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) (3x + 5)18 + c
ð3x þ 5Þ18
2)
þc
54
17
ð3x þ 5Þ
þc
3)
3
18
ð3x þ 5Þ
þc
4)
18
9.2. Calculate the value of the following indefinite integral.
I¼
cosð1 þ πxÞdx
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) sin(1 + πx) + c
1
sinð1 þ πxÞ þ c
2)
1þπ
3) cos(1 + πx) + c
1
4) sinð1 þ πxÞ þ c
π
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_9
175
176
9 Problems: Definite and Indefinite Integrals
9.3. If F(x) ¼ f (x)dx, calculate the value of f (ax + b)dx.
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) aF(ax + b)
1
2) F ðxÞ
a
3) aF(x)
1
4) F ðax þ bÞ
a
9.4. Calculate the value of the definite integral below.
2
1
xþ4
dx
x3
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
2) 2
3) 3
4) 4
9.5. Solve the following indefinite integral:
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
9.6. Calculate the value of f ′′(1) if we know that:
f ðxÞ ¼
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 2
2) 4
3) 6
4) 8
π
if f (x) ¼ cos3(x)dx.
2
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 1
9.7. Calculate the value of f 00
x3 þ 5x dx
9
Problems: Definite and Indefinite Integrals
177
3) -1
p
3 2
4)
2
9.8. Calculate the value of the definite integral below.
-1 3
-2
x þ x2 - 1
dx
x2
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) -1
1
3)
2
1
4) 2
9.9. Solve the indefinite integral below.
x-2
p dx
x
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
2p
xð x þ 6Þ þ c
1)
3
2p
2)
xð x - 6Þ þ c
3
1p
xð x þ 6Þ þ c
3)
3
2p
4)
xð x - 6Þ þ c
3
9.10. Calculate the value of the following definite integral:
I¼
1
0 ð x3
Difficulty level
● Easy ○ Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
5
1)
36
7
2)
36
5
3)
72
7
4)
72
x2
dx
þ 1Þ 4
178
9 Problems: Definite and Indefinite Integrals
9.11. Calculate the value of the following indefinite integral.
I¼
cos 2 x
dx
p
1 þ tan x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1
1)
þ tan x þ c
cos x
1
2)
þ cot gx þ c
cos x
p
3) 2 1 þ tan x þ c
4) 2(1 + tan x) + c
9.12. Calculate the value of the indefinite integral below.
dx
p
x ln x
I¼
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
2
1) p
þc
ln x
p
ln x
2)
þc
x
p
3) 2 ln x þ c
p
4) lnðln xÞ
9.13. Calculate the integral of the function below for the range of 1 < x < + 1.
f ðxÞ ¼
1
x2 þ 4
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
π
1)
2
2) π
3π
3)
2
4) 2π
9.14. Calculate the value of the definite integral below.
4
1
p
x- x
p dx
x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1
1)
3
2
2)
3
9
Problems: Definite and Indefinite Integrals
179
4
3
5
4)
3
3)
x3
1
with respect to x.
, determine the first derivate of f
6
x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1
1) - 3
x
x
2) 6
1
3) 2
1
4) 3
x
9.15. If the primary function of f (x) is equal to
9.16. Calculate the value of F′(λ ¼ 0) if:
λ
F ðλÞ ¼
0
1
dx
x4 þ 2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 1
1
2)
2
1
3)
3
4) 0
9.17. Calculate the value of the definite integral below.
1
x2
arcðtanðxÞÞdx
2
-1 1 þ x
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 1
π
3)
4
π
4)
2
9.18. Calculate the integral of the function below for the range of f ðxÞ ¼ p
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1
1 - x2
1
1
<x< .
2
2
180
9 Problems: Definite and Indefinite Integrals
π
6
π
2)
3
π
3) 6
π
4) 3
1)
9.19. Calculate the value of the definite integral below.
1
p
0
1
dx
2x - x2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
π
2)
4
π
3)
2
4) π
9.20. Calculate the value of the definite integral below.
1
-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
9.21. Solve the following indefinite integral:
sinðxÞ
dx
1 þ cosðcosðxÞÞ
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1
1) - tan cosðxÞ þ c
2
2) tan(cos(x)) + c
3) - tan (cos(x)) + c
1
4) tan cosðxÞ þ c
2
x
9.22. Determine the function of a curve that passes from the point of (3, 4) and its derivative is - .
y
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 2x2 + y2 ¼ 34
2) x2 + y2 ¼ 16
9
Problems: Definite and Indefinite Integrals
181
3) y2 ¼ 4x + 4
4) x2 + y2 ¼ 25
9.23. Solve the following indefinite integral:
p
x
dx
x-1
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
3
1
2
1) ðx - 1Þ2 - 2ðx - 1Þ2 þ c
3
3
1
2
2) ðx - 1Þ2 þ 2ðx - 1Þ2 þ c
3
3
1
1
3) ðx - 1Þ2 - 2ðx - 1Þ2 þ c
3
3
1
1
4) - ðx - 1Þ2 - 2ðx - 1Þ2 þ c
3
9.24. Calculate the value of the definite integral below.
5
-2
jx - 3jdx
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
25
1)
2
27
2)
2
29
3)
2
31
4)
2
9.25. Calculate the value of the definite integral below.
2
1
x ð x þ 1Þ 2 þ 2
dx
ð x þ 1Þ 2
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
12
1)
5
9
2)
5
11
3)
6
7
4)
4
182
9 Problems: Definite and Indefinite Integrals
9.26. Solve the following indefinite integral:
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 lnð1 þ ex Þ þ c
4) 12 x2 - lnð1 þ ex Þ þ c
9.27. Determine the function of a curve that passes from the point of (1, 1) and its derivative is as follows:
y0 ¼
xþ1
1-y
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
9.28. What is the function of a curve that passes from the point of (1, 1) and the relation below holds?
y0 ¼
3x
2y
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
9.29. In the equation below, determine the value of A.
p
3x
dx ¼ A
x2 þ 1
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1
1)
2
2) 1
3
3)
2
4) 3
x2 þ 1 þ c
9
Problems: Definite and Indefinite Integrals
183
9.30. Calculate the value of the definite integral below.
2
-1
jxjdx
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
3
1)
2
5
2)
2
7
3)
2
9
4)
2
9.31. Calculate the value of the following definite integral.
π
2
sin 2 ðxÞdx
0
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 0
π
2)
2
π
3)
4
π
4)
8
9.32. Calculate the value of the definite integral below.
3π
4
tan 5 ðxÞ þ tan 7 ðxÞ dx
0
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 0
1
2)
2
1
3)
3
1
4)
6
9.33. Which one of the points below is on a curve that passes from the point of (π, 1) and y′ ¼ y2 cos (x) holds for that?
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
3π
1)
,2
2
π
2)
, -1
2
184
9 Problems: Definite and Indefinite Integrals
π
,1
2
4) (0, 1)
3)
9.34. Calculate the value of the definite integral of I1 if I2 ¼ m.
I1 ¼
5
3x
dx
x-2
5
1
dx
x-2
3
I2 ¼
3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) m + 2
2) 4m - 6
3) 6m + 6
4) 6m - 4
9.35. Calculate the value of the definite integral below.
3
2
x
dx
x2 - 1
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
8
1) ln
3
p
2 3
2) arc sin
5
3
3) arc tan
2
4) ln
8
3
9.36. Calculate the value of the definite integral of 0 f 0 ðxÞdx if we have f ðxÞ ¼
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
8
1)
3
16
2)
3
8
3) 3
16
4) 3
4
9.37. Solve the following indefinite integral:
8 tan 6 ðxÞ þ tan 8 ðxÞ dx
xp
a t dt.
9
Problems: Definite and Indefinite Integrals
185
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) tan7(x) + c
1
2) tan 8 ðxÞ þ c
7
8
3) tan 5 ðxÞ þ c
5
8
4) tan 7 ðxÞ þ c
7
9.38. Calculate the value of the definite integral below.
1
-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
9.39. Calculate the value of the definite integral below.
1
1
2
1 1
dx
x x3
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) 2
1
3)
2
3
4)
2
9.40. Calculate the value of y0x if we have:
x2
y ¼ u þ v, u ¼
1
sinðt Þ
dt, v ¼
t
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ● Normal ○ Large
4 sinðx2 Þ
1)
x
4 sinðx2 Þ
2) x
3) 0
4) 1
1
x2
sinðuÞ
du
u
186
9 Problems: Definite and Indefinite Integrals
9.41. Calculate the value of f(3x + 2) if we have:
hðxÞ 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
9.42. 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
11
1) 1,
6
13
2) 1,
6
11
3) 2,
6
13
4) 2,
6
9.43. Solve the indefinite integral of sin (2x) cos (4x)dx.
Difficulty level
○ Easy ● Normal ○ Hard
Calculation amount ○ Small ○ Normal ● Large
1
1
1) cos 3 ð2xÞ - cosð2xÞ þ c
2
3
1
1
3
2) - cos ð2xÞ þ cosð2xÞ þ c
3
2
1
1
3
3) cos ð2xÞ þ cosð2xÞ þ c
3
2
1
1
3
4) - cos ð2xÞ - cosð2xÞ þ c
3
2
9.44. Calculate the value of the following indefinite integral.
earc tan x
dx
1 þ x2
I¼
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1) earc tanx + c
2) 2earc tanx + c
1
3) earc tan x þ c
2
4) arc tan(ex + 1) + c
9.45. Calculate the value of the definite integral below.
1
-1
x
dx
3
9
Problems: Definite and Indefinite Integrals
187
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1) 0
2) 1
3) -1
4) -3
9.46. What is the function of a curve that passes from the point of (1, 2) and the relation of xy′ + y ¼ 1 holds.
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1
1) y ¼ 1 þ
x
1
2) y ¼ 2 x
3
3) y ¼ - 1
x
3
4) y ¼ þ 1
x
9.47. Solve the indefinite integral below.
p
f 0 ð 3 xÞ
p
dx
3 2
x
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ● Small ○ Normal ○ Large
1 p
1) f 3 x þ c
3
2 p
2) f 3 x þ c
3
p
3) f ð 3 xÞ þ c
p
4) 3f ð 3 xÞ þ c
9.48. Determine the value of the following definite integral:
1
I¼
0
dx
1 þ eax
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1
1)
a
2) a ln 2
1
3) ln 2
a
4) 1
9.49. Solve the indefinite integral below.
ln xdx
188
9 Problems: Definite and Indefinite Integrals
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) x ln x - x + c
2) x ln x + x + c
3) -x ln x + x + c
4) -x ln x - x + c
9.50. Solve the following indefinite integral.
1
dx
sinðxÞ cosðxÞ
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
9.51. Calculate the value of the definite integral below.
π
4
0
1
dx
cos 4 ðxÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1
1)
3
2
2)
3
3) 1
4
4)
3
9.52. Calculate the value of the definite integral below.
π
4
0
1
3
dx
sin ðxÞ cos 4 ðxÞ
2
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) 2
3) 3
4) 4
9.53. 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
9
Problems: Definite and Indefinite Integrals
189
3) 3
4) 6
9.54. Calculate the value of f(x ¼ - 1) if f ′(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
9.55. Solve the following indefinite integral.
cosð2xÞ
dx
sin 2 ðxÞ cos 2 ðxÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
2
þc
1) sinð2xÞ
1
2) þc
sinð2xÞ
2
3)
þc
sinð2xÞ
1
4)
þc
sinð2xÞ
9.56. Calculate the value of the definite integral below.
e
ð2x þ lnðxÞÞdx
1
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) e2
2) 1
3) 1 + e
4) e - 1
9.57. Solve the indefinite integral below.
sinð2xÞ 2 þ cos 2 ðxÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
51
1
2 þ cos 2 ðxÞ þ c
1)
51
51
1
2) 2 þ cos 2 ðxÞ þ c
51
50
1
3)
2 þ cos 2 ðxÞ þ c
51
50
1
4) 2 þ cos 2 ðxÞ þ c
51
50
dx
190
9 Problems: Definite and Indefinite Integrals
9.58. Calculate the value of the definite integral below.
e
1
lnðxÞ
dx
x
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
1
2)
2
3) 2
1
4)
e
9.59. 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
9.60. Calculate the value of the definite integral below.
π
2
cotðxÞ
dx
1 - cosð2xÞ
π
6
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
p
1) 2
p
2
2)
2
p
3) 3
p
3
4)
2
9.61. Calculate the value of the definite integral below.
6
3
xþ2
p
dx
x-2
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
25
1)
3
38
2)
3
9
Problems: Definite and Indefinite Integrals
191
23
3
34
4)
3
3)
9.62. Calculate the value of the definite integral below.
4
1
p
1þ x
p
dx
x
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
p
1
1) 2 3 þ
3
p
1
2) 2 3 3
p
p
2 2
3) 4
3þ
3
p
p
2 2
4) 4
33
9.63. Solve the following indefinite integral if we know that 0 < x < π.
cotðxÞ
sinðxÞdx
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
cotðxÞ þ c
1)
2) 2 sinðxÞ þ c
3) sinðxÞ sinðxÞ þ c
1
4)
sinðxÞ þ c
2
9.64. Calculate the value of the definite integral below.
π
3
secðxÞ tanðxÞdx
0
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
1) 1
2) 2
1
3)
2
3
4)
2
192
9 Problems: Definite and Indefinite Integrals
9.65. Calculate the value of the definite integral below.
π
4
π
6
cscðxÞ cotðxÞdx
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ● Normal ○ Large
p
1) 2 þ 2
p
2) 2 - 2
p
p
3) 3 - 2
p
p
4) 3 þ 2
9.66. Calculate the value of the definite integral below.
π
4
π
6
1
dx
sin 2 ðxÞ cos 2 ðxÞ
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ○ Normal ● Large
p
3
1)
3p
2 3
2)
p3
3) 3
p
4) 2 3
9.67. Solve the following indefinite integral.
ðtanðxÞ - cotðxÞÞðtanðxÞ þ cotðxÞÞ5 dx
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ○ Normal ● Large
1
1) ðtanðxÞ þ cotðxÞÞ4 þ c
4
1
2) ðtanðxÞ þ cotðxÞÞ5 þ c
5
1
3) ðtanðxÞ þ cotðxÞÞ3 þ c
3
1
4) ðtanðxÞ - cotðxÞÞ5 þ c
5
9.68. Which one of the choices is not an acceptable solution for the indefinite integral of
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ○ Normal ● Large
1
1) - cosð2xÞ þ c
4
1
2) - sinð2xÞ þ c
4
1
3) - cos 2 ðxÞ þ c
2
1
4) sin 2 ðxÞ þ c
2
sin (x) cos (x)dx?
References
193
9.69. Calculate the value of the definite integral below.
I¼
ln 3
ln 2
1 - e - 2x
dx
1 þ e - 2x
Difficulty level
○ Easy ○ Normal ● Hard
Calculation amount ○ Small ○ Normal ● Large
1) ln3 - ln 2
1
2) 3 ln 2 - ln 3
2
3) 2 ln 2 - ln 3
4) 3 ln 2 - 2 ln 3
References
1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
Solutions of Problems: Definite and Indefinite
Integrals
10
Abstract
In this chapter, the problems of the ninth chapter are fully solved, in detail, step-by-step, and with different methods.
10.1. From list of integral of different functions, we know that [1, 2]:
un ðxÞdu ¼
1
unþ1 ðxÞ þ c
nþ1
The problem can be solved by defining a new variable as follows:
uðxÞ ¼ 3x þ 5
) du ¼ 3dx ) dx ¼
du
3
Therefore:
I¼
ð3x þ 5Þ17 dx ¼
u17 ðxÞ
du
3
)I¼
1 u18 ðxÞ
þc
x
18
3
)I¼
ð3x þ 5Þ18
þc
54
¼
1
3
u17 ðxÞdu
Choice (2) is the answer.
10.2. From list of integral of different functions, we know that:
cos uðxÞdu ¼ sin uðxÞ þ c
The problem can be solved by defining a new variable as follows:
uðxÞ ¼ 1 þ πx
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0_10
195
196
10
) du ¼ πdx ) dx ¼
Solutions of Problems: Definite and Indefinite Integrals
du
π
Thus:
I¼
cosð1 þ πxÞdx ¼
ðcos uðxÞÞ
)I¼
)I¼
du 1
¼
π
π
cos uðxÞdu
1
sin uðxÞ þ c
π
1
sinð1 þ πxÞ þ c
π
Choice (4) is the answer.
10.3. From the problem, we have:
F ð xÞ ¼
f ðxÞdx
The problem can be solved by defining a new variable as follows:
ax þ b ¼ uðxÞ
) adx ¼ du ) dx ¼
du
a
Therefore:
I¼
f ðax þ bÞdx ¼
)I¼
1
a
f ðuÞdu
F ð uÞ
a
1
) I ¼ F ðax þ bÞ
a
Choice (4) is the answer.
10.4. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
þc
x
nþ1
The problem can be solved as follows:
2
1
xþ4
dx ¼
x3
Choice (2) is the answer.
2
1
1
1 2
4
1
2 2
¼ - - - ð- 1 - 2Þ ¼ 2
þ
dx ¼ - - 2
2 4
x x 1
x 2 x3
10
Solutions of Problems: Definite and Indefinite Integrals
197
10.5. From list of integral of functions, we know that:
eu du ¼ eu
The problem can be solved as follows:
2
2
ex þ 2xex dx ¼ ex þ ex þ c
Choice (4) is the answer.
10.6. Based on the information given in the problem, we have:
x3 þ 5x dx
f ðxÞ ¼
d
d
dx
dx 0
¼
¼
¼
¼
¼) f 00 ðxÞ ¼ 3x2 þ 5
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) f ðxÞ ¼ x3 þ 5x ¼
x ¼ 1 00
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) f ð1Þ ¼ 3 þ 5 ¼ 8
Choice (4) is the answer.
10.7. The problem can be solved as follows:
cos 3 ðxÞdx ) f 0 ðxÞ ¼ cos 3 ðxÞ ) f 00 ðxÞ ¼ - 3 cos 2 ðxÞ sinðxÞ
f ðxÞ ¼
f 00
π
¼ -3 x 0 x 1 ¼ 0
2
Choice (1) is the answer.
10.8. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
x
þc
nþ1
The problem can be solved as follows:
-1 3
-2
x þ x2 - 1
dx ¼
x2
¼
Choice (2) is the answer.
-1
-2
x þ 1-
1
dx ¼
x2
x2
1
þxþ
2
x
1
1
-1-1 - 2-2¼ -1
2
2
-1
-2
198
10
Solutions of Problems: Definite and Indefinite Integrals
10.9. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
þc
x
nþ1
The problem can be solved as follows:
1
1
1
2p
2 3
x ð x - 6Þ þ c
x2 - 2x - 2 dx ¼ x2 - 4x2 þ c ¼
3
3
x-2
p dx ¼
x
Choice (2) is the answer.
10.10. The problem can be solved by changing the variable of the integral as follows:
d
dx
du
x3 þ 1 ≜ u ¼
¼
¼
¼
¼) 3x2 dx ¼ du ) x2 dx ¼
3
1
I¼
0
x2
dx ¼
3
ðx þ 1Þ4
)I¼ -
)I¼ -
u2
u-4
u1
1 3
x þ1
9
du 1 u - 3 u2
¼
3
3 - 3 u1
-3
1
0
1
1 1 -1 þ 8
1
þ ¼
ð 1 þ 1Þ - 3 þ ð 0 þ 1Þ - 3 ¼ 9
72 9
72
9
)I¼
7
72
Choice (4) is the answer.
10.11. From list of integral of different functions, we know that:
du
¼2
uð x Þ
uð x Þ
The problem can be solved by defining a new variable as follows:
1 þ tan x ¼ uðxÞ
) 1 þ tan 2 x dx ¼ du )
dx
¼ du
cos 2 x
Therefore:
I¼
)I¼
dx
p
cos 2 x 1 þ tan x
1
du
p ¼ 2u2 þ c
u
p
) I ¼ 2 1 þ tan x þ c
Choice (3) is the answer.
10
Solutions of Problems: Definite and Indefinite Integrals
199
10.12. From list of integral of different functions, we know that:
du
¼2
uð x Þ
uð x Þ
The problem can be solved by defining a new variable as follows:
uðxÞ ¼ ln x
) du ¼
dx
x
Therefore:
dx
p
x ln x
I¼
du
¼2
uð x Þ
)I¼
uð x Þ þ c
p
) I ¼ 2 ln x þ c
Choice (3) is the answer.
10.13. From list of integral of functions, we know that:
1
x
1
dx ¼ arc tan
þc
a
a
x2 þ a2
Therefore:
þ1
-1
1
1
x
dx ¼ arc tan
2
2
x2 þ 4
þ1
1 π
π
¼
- -1 2 2
2
¼
π
2
Choice (1) is the answer.
10.14. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
þc
x
nþ1
The problem can be solved as follows:
4
1
p
x- x
p dx ¼
x
Choice (4) is the answer.
4
1
1
x2 - 1 dx ¼
2 32
4
¼
x -x
3
1
16
2
5
-1 ¼
-4 3
3
3
200
10
Solutions of Problems: Definite and Indefinite Integrals
10.15. Based on the information given in the problem, we have:
d
x3 dx
x2
f ðxÞdx ¼ ¼
¼
¼
¼
¼
¼) f ðxÞ ¼
6
2
Therefore:
1 2
)f
)
d
1
f
dx
x
1
1
¼ x ¼ 2
x
2
2x
¼
d 1
- 4x - 1
¼ 3
¼
dx 2x2
x
4x4
Choice (1) is the answer.
10.16. As we know:
F ð xÞ ¼
uðxÞ
vðxÞ
f ðxÞdx ) F 0 ðxÞ ¼ u0 ðxÞF ðuðxÞÞ - v0 ðxÞF ðvðxÞÞ
The problem can be solved as follows:
λ
F ðλÞ ¼
0
1
1
1
1
dx ) F 0 ðλÞ ¼ 1 x 4
-0 ¼ 4
) F 0 ð0Þ ¼
2
x4 þ 2
λ þ2
λ þ2
Choice (2) is the answer.
10.17. Since the function is an odd function and the range of the integral is symmetric, the final answer is zero.
1
x2
arcðtanðxÞÞdx ¼ 0
2
-1 1 þ x
Choice (1) is the answer.
10.18. From list of integral of functions, we know that:
1
p
dx ¼ arcðsinðxÞÞ
1 - x2
Therefore:
1
2
- 12
Choice (2) is the answer.
p
1
1
π
π
π
¼
dx ¼ arcðsinðxÞÞ 2 1 ¼ - -2 6
3
6
1 - x2
10
Solutions of Problems: Definite and Indefinite Integrals
201
10.19. From list of integral of functions, we know that:
1
p
dx ¼ arcðsinðuÞÞ
1 - u2
The problem can be solved by changing the variable of the integral as follows:
d
dx
x-1 ≜ u¼
¼
¼
¼
¼
¼) dx ¼ du
1
0
p
1
1
dx ¼
2x - x2
0
1
1 - ð x - 1Þ
2
u2
dx ¼
¼ ðarcðsinðx - 1ÞÞÞ
p
u1
1
u
du ¼ ðarcðsinðuÞÞÞ 2
2
u
1
1-u
1
π
π
¼ 0- ¼
0
2
2
Choice (3) is the answer.
10.20. The final answer is zero, since the function is an odd function and the range of the integral is symmetric.
1
-1
x2 þ 1 x3 þ 3x dx ¼ 0
Choice (2) is the answer.
10.21. From trigonometry, we know that:
1 þ cosðuÞ ¼ 2 cos 2
u
2
1
cos 2
u
2
¼ 1 þ tan 2
u
2
In addition, from list of integral of functions, we know that:
1 þ tan 2
u
a
du ¼ a tan
u
þc
a
The problem can be solved by changing the variable of the integral as follows:
cosðxÞ ≜ u )
sinðxÞ
dx ¼ 1 þ cosðcosðxÞÞ
)
¼ -
d
d
cosðxÞ ¼ u ) - sinðxÞdx ¼ du
dx
dx
1
2
Choice (1) is the answer.
1 þ tan 2
u
2
1
du ¼ 1 þ cosðuÞ
du ¼ - tan
1
2 cos 2
u
2
du
cosðxÞ
u
þ c ¼ - tan
þc
2
2
202
10
Solutions of Problems: Definite and Indefinite Integrals
10.22. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
þc
x
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:
dx
y 2 x2
x
0
¼
¼
¼
¼
¼
¼
¼
¼
¼)
y ¼ - ) yy þ x ¼ 0¼
þ ¼ c 0 ) y 2 þ x2 ¼ c
2
2
y
ð1Þ
y ð 3Þ ¼ 4 2
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) 4 þ 32 ¼ c ) c ¼ 25
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
ð2Þ
0
ð1Þ, ð2Þ 2
¼
¼
¼
¼
¼
¼
¼) y þ x2 ¼ 25
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
Choice (4) is the answer.
10.23. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
þc
x
nþ1
The problem can be solved as follows:
x
p
dx ¼
x-1
x-1 þ 1
p
dx ¼
x-1
1
1
3
1
2
ðx - 1Þ2 þ ðx - 1Þ - 2 dx ¼ ðx - 1Þ2 þ 2ðx - 1Þ2 þ c
3
Choice (2) is the answer.
10.24. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
x
þc
nþ1
The problem can be solved as follows:
5
-2
¼ 9-
jx - 3jdx ¼
3
-2
5
ð3 - xÞdx þ
3
ðx - 3Þdx ¼
3x -
x2
2
3
x2
þ
- 3x
-2
2
5
3
9
25
9
9
5 9 9 þ 16 - 5 þ 9 29
- ð- 6 - 2Þ þ
- 15 -9 ¼ þ 8- þ ¼
¼
2
2
2
2
2 2
2
2
Choice (3) is the answer.
10
Solutions of Problems: Definite and Indefinite Integrals
203
10.25. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
þc
x
nþ1
The problem can be solved as follows:
2
1
x ð x þ 1Þ 2 þ 2
dx ¼
ð x þ 1Þ 2
2
2
x2
2
xþ1
x þ 2ðx þ 1Þ - 2 dx ¼
1
2
2
1
4 1 11
¼ 2-1 ¼ þ ¼
1
3
2
3 2
6
Choice (3) is the answer.
10.26. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
þc
x
nþ1
1
du ¼ lnjuj þ c
u
The problem can be solved as follows:
1
dx ¼
1 þ ex
1 þ ex - e x
dx ¼
1 þ ex
¼ x þ c0 -
1-
ex
dx ¼
1 þ ex
1dx -
ex
dx
1 þ ex
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 -
1
du ¼ x þ c0 - lnjuj þ c00 ¼ x - lnj1 þ ex j þ c ¼ x - lnð1 þ ex Þ þ c
u
Choice (2) is the answer.
10.27. The problem can be solved as follows:
y0 ¼
dx
xþ1
y 2 x2
¼
¼
¼
¼
¼
¼
¼
¼
¼) y - ¼ þ x þ c
) y0 - yy0 ¼ x þ 1¼
1-y
2
2
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) 1 - ¼ þ 1 þ c ) c ¼ - 1
¼
2 2
) yChoice (1) is the answer.
y2 x 2
¼ þ x - 1 ) x2 þ y2 þ 2x - 2y - 2 ¼ 0
2
2
ð2Þ
204
10
Solutions of Problems: Definite and Indefinite Integrals
10.28. The problem can be solved as follows:
y0 ¼
3x
3
) 2yy0 ¼ 3x ) y2 ¼ x2 þ c
2y
2
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
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.
10.29. Based on the information given in the problem, we have:
p
3x
dx ¼ A
x2 þ 1
x2 þ 1 þ c
ð1Þ
From list of integral of functions, we know that:
p
1
p du ¼ 2 u þ c
u
The problem can be solved by changing the variable of the integral as follows:
d
dx
1
¼
¼
¼
¼) 2xdx ¼ du ) xdx ¼ du
x2 þ 1 ≜ u ¼
2
p
3x
dx ¼
x2 þ 1
p
1
3
p du ¼ x 2 u ¼ 3
2
u
3 x 12 du 3
p
¼
2
u
x2 þ 1 þ c
ð2Þ
Therefore:
ð1Þ, ð2Þ
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) A
¼
¼
¼
¼
¼
¼
¼
¼
x2 þ 1 þ c ¼ 3
x2 þ 1 þ c ) A ¼ 3
Choice (4) is the answer.
10.30. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
x
þc
nþ1
The problem can be solved as follows:
2
-1
jxjdx ¼
0
-1
2
jxjdx þ
jxjdx ¼
0
¼ - 0-
0
-1
2
ð- xÞdx þ
0
xdx ¼ -
x2 0
x2 2
þ
2 -1 2 0
1
5
þ ð 2 - 0Þ ¼
2
2
Choice (2) is the answer.
Note that for this problem, a graphical method can be used which is faster than the abovementioned method. Think
about it!
10
Solutions of Problems: Definite and Indefinite Integrals
205
10.31. From trigonometry, we know that:
1 - cosð2xÞ ¼ 2 sin 2 ðxÞ
Furthermore, from list of integral of functions, we know that:
xn dx ¼
1 nþ1
þc
x
nþ1
cosðaxÞdx ¼
1
sinðaxÞ þ c
a
The problem can be solved as follows:
π
2
π
2
sin ðxÞdx ¼
2
0
π
cosð2xÞ
1
1
1
1 π
π
dx ¼
x - sinð2xÞ 2 ¼
x - 0 - ð 0Þ ¼
2
0
2
2
4
2 2
4
0
Choice (3) is the answer.
10.32. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
x
þc
nþ1
The problem can be solved by changing the variable of the integral as follows:
d
dx
tanðxÞ ≜ u¼
¼
¼
¼
¼
¼) 1 þ tan 2 ðxÞ dx ¼ du
3π
4
0
¼
3π
4
tan 5 ðxÞ þ tan 7 ðxÞ dx ¼
tan 5 ðxÞ 1 þ tan 2 ðxÞ dx
0
u2
u1
3π
1
1
1 u
1
u5 du ¼ u6 2 ¼ tan 6 ðxÞ 4 ¼ ð- 1Þ6 - 0 ¼
0 6
6 u1 6
6
Choice (4) is the answer.
10.33. The problem can be solved as follows:
y0 ¼ y2 cosðxÞ )
y0
-1
¼ cosðxÞ )
¼ sinðxÞ þ c
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) - 1 ¼ 0 þ c ) c ¼ - 1 )
¼ sinðxÞ - 1
y
We need to check each choice as follows:
3π
,2
ðx, yÞ ¼
2
-1
3π
-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)
-1 )
Choice 1 : ¼
¼ sin
≠ -2
2
2
2
206
10
Solutions of Problems: Definite and Indefinite Integrals
π
ðx, yÞ ¼
, -1
2
-1
π
Choice 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
¼
¼
¼
¼
¼
¼
¼
¼
¼
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 ) 1≠0
¼ sin
-1
2
π
ðx, yÞ ¼
,1
2
-1
π
Choice 3 : ¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
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 ) -1≠0
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼ sin
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 sinð0Þ - 1 ) - 1 ¼ - 1
1
Choice (4) is the answer.
10.34. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
þc
x
nþ1
Based on the information given in the problem, we have:
5
I2 ¼
3
1
dx ¼ m
x-2
ð1Þ
The problem can be solved as follows:
5
I1 ¼
3
3x
dx ¼
x-2
5
3
3x - 6 þ 6
dx ¼
x-2
5
5
3dx þ 6
3
3
1
dx
x-2
ð2Þ
Solving (1) and (2):
5
I1 ¼
3
3x
5
dx ¼ 3x þ 6m ¼ 15 - 9 þ 6m ¼ 6 þ 6m
x-2
3
Choice (3) is the answer.
10.35. From list of integral of functions, we know that:
1
du ¼ lnjuj þ c
u
The problem can be solved as follows:
3
2
1
x
dx ¼
2
x2 - 1
3
2
2x
dx
x2 - 1
ð1Þ
Now, we need to change the variable of the integral as follows:
d
dx
x - 1 ≜ u¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) 2xdx ¼ du
2
ð2Þ
10
Solutions of Problems: Definite and Indefinite Integrals
207
Solving (1) and (2):
1
2
u2
u1
1
1 8
1
1
1
3 1
u
du ¼ lnjuj 2 ¼ ln x2 - 1 ¼ ln 8 - ln 3 ¼ ln ¼ ln
2
2 3
u1 2
u
2
2 2
Choice (4) is the answer.
10.36. From list of integral of functions, we know that:
1 nþ1
þc
x
nþ1
xn dx ¼
As we know:
F ð xÞ ¼
uðxÞ
vðxÞ
f ðxÞdx ) F 0 ðxÞ ¼ u0 ðxÞF ðuðxÞÞ - v0 ðxÞF ðvðxÞÞ
Therefore:
xp
t dt ) f 0 ðxÞ ¼
f ðxÞ ¼
p
x
a
4
4
f 0 ðxÞdx ¼
0
1
x2 dx ¼
0
2 32 4 16
¼
x
0
3
3
Choice (2) is the answer.
10.37. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
þc
x
nþ1
The problem can be solved by changing the variable of the integral as follows:
d
dx
tanðxÞ ≜ u¼
¼
¼
¼
¼) 1 þ tan 2 ðxÞ dx ¼ du
¼
¼
¼
¼
¼
8 tan 6 ðxÞ þ tan 8 ðxÞ dx ¼ 8
¼8
tan 6 ðxÞ 1 þ tan 2 ðxÞ dx
8
8
u6 du ¼ u7 þ c ¼ tan 7 ðxÞ þ c
7
7
Choice (4) is the answer.
10.38. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
þc
x
nþ1
8
3
208
10
Solutions of Problems: Definite and Indefinite Integrals
The problem can be solved as follows:
1
-1
ðx þ 1Þ x2 þ 2x þ 3 dx ¼
1
-1
ð x þ 1Þ
x þ 1Þ2 þ 2 dx ¼
1
-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
ð2Þ
Solving (1) and (2):
1
-1
¼
u3 þ 2u dx ¼
ð x þ 1Þ 4
þ ð x þ 1Þ 2
4
u4
þ u2
4
16
1
¼
þ 4-0 ¼ 8
4
-1
Choice (3) is the answer.
10.39. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
x
þc
nþ1
The problem can be solved as follows:
1
1
2
1
1
x 3 dx ¼
x
x
1
1
2
1 x x - 3 dx ¼
x-2
-2
1
1
2
¼
-1 1 -1
3
¼
- ð- 2Þ ¼
2
2
2x2 12
Choice (4) is the answer.
10.40. As we know:
F ð xÞ ¼
uðxÞ
vðxÞ
f ðxÞdx ) F 0 ðxÞ ¼ u0 ðxÞF ðuðxÞÞ - v0 ðxÞF ðvðxÞÞ
Therefore:
y0x ¼ u0x þ v0x ¼
2x
sinðx2 Þ
sinðx2 Þ
0
þ
0
2x
x2
x2
¼0
Choice (3) is the answer.
10.41. Based on the information given in the problem, we know that:
hð x Þ ¼
f 0 ð3x þ 2Þdx
h ð 0Þ ¼ 1
ð1Þ
ð2Þ
10
Solutions of Problems: Definite and Indefinite Integrals
209
f ð 2Þ ¼ 3
ð3Þ
We should change the variable of the integral of h(x) as follows:
f ð3x þ 2Þ ≜ u ) 3f 0 ð3x þ 2Þdx ¼ du
ð4Þ
Solving (1) and (4):
hð x Þ ¼
1
1
1
du ¼ u þ c ¼ f ð3x þ 2Þ þ c
3
3
3
ð5Þ
Solving (2) and (5):
Using ð3Þ
1
1
1
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) 1 ¼ x 3 þ c ) c ¼ 0 ) hðxÞ ¼ f ð3x þ 2Þ
¼
¼
¼
¼
¼
¼
¼
¼
¼
1 ¼ f ð 2 Þ þ c¼
3
3
3
) f ð3x þ 2Þ ¼ 3hðxÞ
Choice (3) is the answer.
10.42. From list of integral of functions, we know that:
xn du ¼
1 nþ1
þc
x
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:
y0 ð0Þ ¼ 1
ð2Þ
y ð 0Þ ¼ 0
ð3Þ
dx
¼
¼
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
¼
¼) yðxÞ ¼ þ þ ax þ b
3
2
ð5Þ
a¼1
ð6Þ
Applying integral operation on (1):
Applying integral operation on (4):
Solving (2) and (4):
210
10
Solutions of Problems: Definite and Indefinite Integrals
Solving (3) and (5):
0¼0þ0þ0þb)b¼0
ð7Þ
Solving (5)–(7):
yð xÞ ¼
x3 x2
þ þx
3
2
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
16 þ 12 þ 12 20
2 3 22
þ þ2¼
¼
3
2
6
3
Choice (1) is the answer.
10.43. From trigonometry, we know that:
1 þ cosð2xÞ ¼ 2 cos 2 ðxÞ
In addition, from list of integral of functions, we know that:
sinðaxÞdx ¼ -
1
cosðaxÞ þ c
a
The problem can be solved as follows:
sinð2xÞ cosð4xÞdx ¼
¼
sinð2xÞ 2cos 2 ð2xÞ - 1 dx
cos 2 ð2xÞ x 2 sinð2xÞdx -
sinð2xÞdx
ð1Þ
Now, we should change the variable of the integral as follows:
cosð2xÞ ≜ u ) - 2 sinð2xÞdx ¼ du
Solving (1) and (2):
-
u2 du -
sinð2xÞdx ¼ -
1 3 1
1
1
u þ cosð2xÞ þ c ¼ - cos 3 ð2xÞ þ cosð2xÞ þ c
3
2
3
2
Choice (2) is the answer.
10.44. From list of integral of different functions, we know that:
euðxÞ du ¼ euðxÞ þ c
The problem can be solved by defining a new variable as follows:
arctan x ¼ uðxÞ
ð2Þ
10
Solutions of Problems: Definite and Indefinite Integrals
211
)
dx
¼ du
1 þ x2
Therefore:
e arctan x
dx
1 þ x2
)I¼
eu du ¼ eu þ c
)I¼
) I ¼ e arctan x þ c
Choice (1) is the answer.
10.45. From list of integral of functions, we know that:
xn du ¼
1 nþ1
x
þc
nþ1
The problem can be solved as follows:
1
-1
x
dx ¼
3
0
-1
1
ð- 1Þdx þ
0dx ¼ ð- xÞ
0
0
þ 0 ¼ 0 - ð 1Þ ¼ - 1
-1
Choice (3) is the answer.
10.46. 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) 1 x 2 ¼ 1 þ c ) c ¼ 1
) xy ¼ x þ 1 ) y ¼ 1 þ
1
x
Choice (1) is the answer.
10.47. The problem can be solved as follows:
p
f 0 ð 3 xÞ
p
dx ¼ 3
3 2
x
1 0 p
p
f 3 x dx
3 2
3 x
ð1Þ
Now, we should change the variable of the integral as follows:
f
p
1 0 p
3
x ≜u) p
f 3 x dx ¼ du
3
3 x2
Solving (1) and (2):
3
Choice (4) is the answer.
du ¼ 3u þ c ¼ 3f
p
3
x þc
ð2Þ
212
10
Solutions of Problems: Definite and Indefinite Integrals
10.48. The problem can be solved as follows:
1
I¼
0
e - ax
x - ax
e
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) I ¼
dx
1 þ eax
1
0
e - ax
dx
e - ax þ 1
From list of integral of different functions, we know that:
du
¼ ln uðxÞ
uð x Þ
The problem can be solved by defining a new variable as follows:
e - ax þ 1 ¼ uðxÞ
) - ae - ax dx ¼ du ) e - ax dx ¼
du
-a
Therefore:
1
I¼
0
)I¼ -
1
a
u2
u1
e - ax
dx
e - ax þ 1
du
1
¼ - ½ln uðxÞ]uu21
a
uð x Þ
) I ¼ ½lnðe - ax þ 1Þ]1
0
)I¼ -
1
½ln 1 - ln 2]
a
)I¼
1
ln 2
a
Choice (3) is the answer.
10.49. From integration by parts (partial integration), we know that:
uðxÞdv ¼ uðxÞvðxÞ In addition, from list of integral of functions, we know that:
1
du ¼ lnjuj þ c
u
The problem can be solved as follows:
vðxÞdu
10
Solutions of Problems: Definite and Indefinite Integrals
213
)
dx
x
vð xÞ ¼ x
uðxÞ ¼ ln x
)
dv ¼ dx
ln xdx )
ln xdx ¼ x ln x -
du ¼
dx ¼ x ln x - x þ c
Choice (1) is the answer.
10.50. From trigonometry, we know that:
1
¼ 1 þ tan 2 ðxÞ
cos 2 ðxÞ
sinðxÞ
¼ tanðxÞ
cosðxÞ
Moreover, from list of integral of functions, we know that:
1
du ¼ lnjuj þ c
u
The problem can be solved as follows:
1
dx ¼
sinðxÞ cosðxÞ
1
sinðxÞ cosðxÞ x
cosðxÞ
cosðxÞ
dx ¼
sinðxÞ
cosðxÞ
1
dx ¼
cos 2 ðxÞ
1 þ tan 2 ðxÞ
dx
tanðxÞ
ð1Þ
Now, we should change the variable of the integral as follows:
tanðxÞ ≜ u ) 1 þ tan 2 ðxÞ dx ¼ du
ð2Þ
Solving (1) and (2):
1
du ¼ lnjuj þ c ¼ lnj tanðxÞj þ c
u
Choice (2) is the answer.
10.51. From trigonometry, we know that:
1 þ tan 2 ðxÞ ¼
1
cos 2 ðxÞ
The problem can be solved as follows:
π
4
0
1
dx ¼
cos 4 ðxÞ
π
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
tanðxÞ ≜ u¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) 1 þ tan 2 ðxÞ dx ¼ du
ð2Þ
214
10
Solutions of Problems: Definite and Indefinite Integrals
Solving (1) and (2):
π
u
1
1
1
4
1 þ u2 du ¼ u þ u3 2 ¼ tanðxÞ þ tan 3 ðxÞ 4 ¼ 1 þ - 0 ¼
u1
0
3
3
3
3
u2
u1
Choice (4) is the answer.
10.52. From trigonometry, we know that:
1 þ tan 2 ðxÞ ¼
tanðxÞ ¼
1
cos 2 ðxÞ
sinðxÞ
cosðxÞ
The problem can be solved as follows:
π
4
0
1
3
sin 2 ðxÞ cos 4 ðxÞ
π
4
dx ¼
0
¼
1
3
π
4
tan
π
6
sin 2 ðxÞ cos 4 ðxÞ x
-2
3
cos 2 ðxÞ
cos 2 ðxÞ
π
4
dx ¼
0
ðxÞ 1 þ tan 2 ðxÞ dx
1
cos 2 ðxÞ
3
sin 2 ðxÞ
cos 2 ðxÞ
dx
ð1Þ
Now, we should change the variable of the integral as follows:
d
dx
tanðxÞ ≜ u¼
¼
¼
¼
¼
¼) 1 þ tan 2 ðxÞ dx ¼ du
ð2Þ
Solving (1) and (2):
u2
u - 3 du ¼ 3u3
2
u1
1
π
1
u2
¼ 3 tan 3 ðxÞ 4 ¼ 3ð1 - 0Þ ¼ 3
u1
0
Choice (3) is the answer.
10.53. From list of integral of functions, we know that:
1
du ¼ lnjuj þ c
u
Based on the information given in the problem, we have:
f ð 1Þ ¼ 0
d
f x2
dx
The problem can be solved as follows:
¼
ð1Þ
6
x
ð2Þ
10
Solutions of Problems: Definite and Indefinite Integrals
215
d
f x2
dx
¼ 2xf 0 x2
ð2Þ, ð3Þ 6
3
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) ¼ 2xf 0 x2 ) f 0 x2 ¼ 2
x
x
ð3Þ
ð4Þ
By changing the variable of the integral, we have:
x2 ≜ t
ð5Þ
dt
ð4Þ, ð5Þ 0
3
¼
¼
¼
¼
¼
¼
¼
¼
¼) f ðt Þ ¼ 3 lnjt j þ c
¼
¼) f ðt Þ ¼ ¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
t
ð6Þ
ð1Þ, ð6Þ
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) 0 ¼ 3 x 0 þ c ) c ¼ 0 ) f ðt Þ ¼ 3 lnjt j
) f ðeÞ ¼ 3 lnðeÞ ¼ 3 x 1 ¼ 3
Choice (3) is the answer.
10.54. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
x
þc
nþ1
From trigonometry, we know that:
1 þ cosð2xÞ 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 cosð2xÞ
ð2Þ
f 0 cos 2 ðxÞ ¼ cosð2xÞ 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
dt
ð3Þ, ð4Þ 0
¼
¼
¼
¼
¼
¼
¼
¼
¼) f ðt Þ ¼ t 2 - t þ c
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) f ðt Þ ¼ 2t - 1¼
ð5Þ
ð1Þ, ð5Þ
¼
¼
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 ð- 1Þ ¼ ð- 1Þ2 - ð- 1Þ þ 1 ¼ 3
Choice (3) is the answer.
216
10
Solutions of Problems: Definite and Indefinite Integrals
10.55. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
þc
x
nþ1
From trigonometry, we know that:
sinð2xÞ ¼ 2 sinðxÞ cosðxÞ
The problem can be solved as follows:
cosð2xÞ
cosð2xÞ
dx ¼
sin ðxÞ cos 2 ðxÞ
2
2
1
2 sinð2xÞ
dx ¼
1
4
cosð2xÞ
dx ¼
sin 2 ð2xÞ
4 cosð2xÞðsinð2xÞÞ - 2 dx
Now, we need to change the variable of the integral as follows:
d
dx
sinð2xÞ ≜ u¼
¼
¼
¼
¼
¼) 2 cosð2xÞdx ¼ du
2u - 2 du ¼ -
)
2
-2
þc
þc¼
u
sinð2xÞ
Choice (1) is the answer.
10.56. From integration by parts (partial integration), we know that:
lnðxÞdx ¼ x lnjxj - x
or, in general:
uðxÞdv ¼ uðxÞvðxÞ -
vðxÞdu
In addition, from list of integral of functions, we know that:
1
du ¼ lnjuj þ c
u
xn dx ¼
1 nþ1
þc
x
nþ1
The problem can be solved as follows:
e
e
ð2x þ lnðxÞÞdx ¼
1
1
e
2x dx þ
lnðxÞdx ¼ x2
1
¼ e 2 - 1 þ ð e - e Þ - ð 0 - 1Þ ¼ e 2
Choice (1) is the answer.
e
e
þ ðxlnjxj - xÞ
1
1
10
Solutions of Problems: Definite and Indefinite Integrals
217
10.57. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
þc
x
nþ1
The problem can be solved as follows:
sinð2xÞ 2 þ cos 2 ðxÞ
50
ð1Þ
dx
We should change the variable of the integral as follows:
2 þ cos 2 ðxÞ ≜ u ) - 2 cosðxÞ sinðxÞdx ¼ du ) - sinð2xÞ ¼ du
ð2Þ
Solving (1) and (2):
-
u50 du ¼ -
1
u51
þc¼ 2 þ cos 2 ðxÞ
51
51
51
þc
Choice (2) is the answer.
10.58. From list of integral of functions, we know that:
un du ¼
1
unþ1 þ c
nþ1
The problem can be solved as follows:
e
1
lnðxÞ
dx ¼
x
e
lnðxÞ
1
1
dx
x
ð1Þ
Now, we should change the variable of the integral as follows:
lnðxÞ ≜ u )
1
dx ¼ du
x
Solving (1) and (2):
u2
u1
1 u
e 1
1
1
udu ¼ u2 2 ¼ ðlnðxÞÞ2 ¼ - 0 ¼
2 u1 2
1 2
2
Choice (2) is the answer.
10.59. From list of integral of functions, we know that:
un du ¼
The problem can be solved as follows:
1
unþ1 þ c
nþ1
ð2Þ
218
10
y0 ¼ -
Solutions of Problems: Definite and Indefinite Integrals
dx
2x þ 2
¼
¼
¼
¼
¼
¼
¼
¼
¼) 2y2 þ y ¼ - x2 - 2x þ c
) 4yy0 þ y0 ¼ - 2x - 2 ¼
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) 2 þ 1 ¼ 0 þ c ) c ¼ 3
ð1Þ
ð2Þ
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.
10.60. From list of integral of functions, we know that:
un du ¼
1
unþ1 þ c
nþ1
Moreover, from trigonometry, we know that:
cotðxÞ ¼
cosðxÞ
sinðxÞ
1 - cosð2xÞ ¼ 2 sin 2 ðxÞ
The problem can be solved as follows:
π
2
cotðxÞ
dx ¼
1 - cosð2xÞ
π
6
π
2
cosðxÞ
π
6
2 sin 2 ðxÞ sinðxÞ
p
2
¼
2
π
2
π
6
dx ¼
π
2
π
6
p
cosðxÞ
dx
2j sinðxÞj sinðxÞ
ðsinðxÞÞ - 2 cosðxÞdx
ð1Þ
Now, we should change the variable of the integral as follows:
sinðxÞ ≜ u ) cosðxÞdx ¼ du
ð2Þ
Solving (1) and (2):
p
2
2
u2
u
-2
u1
p
p
2
1
2 - 1 u2
¼ du ¼ x
u
u1
2
2
sinðxÞ
Choice (2) is the answer.
10.61. From list of integral of functions, we know that:
un du ¼
The problem can be solved as follows:
1
unþ1 þ c
nþ1
π
2
π
6
p
p
2
2
¼ ð 1 - 2Þ ¼
2
2
10
Solutions of Problems: Definite and Indefinite Integrals
6
3
¼
6
xþ2
p
dx ¼
x-2
3
219
x-2 þ 4
p
dx ¼
x-2
6
ðx - 2Þ2 þ 4ðx - 2Þ - 2 dx
1
1
3
3
1
1
1
2
6
2 32
2 3
¼
ðx - 2Þ2 þ 4 x 2ðx - 2Þ2
ð4Þ þ 4 x 2ð4Þ2 - ð1Þ2 þ 4 x 2ð1Þ2
3
3
3
3
¼
16
2
14
38
þ 16 - þ 8 ¼
þ8¼
3
3
3
3
Choice (2) is the answer.
10.62. In addition, from list of integral of functions, we know that:
un du ¼
1
unþ1 þ c
nþ1
The problem can be solved as follows:
4
1
p
1þ x
p
dx ¼ 2
x
4
1þ
1
p 12
1
x x p dx
2 x
ð1Þ
Now, we should change the variable of the integral as follows:
1þ
p
1
x ≜ u ) p dx ¼ du
2 x
ð2Þ
Solving (1) and (2):
u2
2
u1
p
4
2 3 u
u du ¼ 2 x u2 2 ¼
1þ x
u1
3
3
1
2
3
2
p
p
p
4 4 p
2 2
¼
33 3-2 2 ¼ 4
1 3
3
Choice (4) is the answer.
10.63. In addition, from list of integral of functions, we know that:
un du ¼
1
unþ1 þ c
nþ1
From trigonometry, we know that:
cotðxÞ ¼
cosðxÞ
sinðxÞ
The problem can be solved as follows:
cotðxÞ
sinðxÞdx ¼
1
cosðxÞ
ðsinðxÞÞ2 dx ¼
sinðxÞ
-1
ðsinðxÞÞ 2 cosðxÞdx
ð1Þ
Now, we should change the variable of the integral as follows:
sinðxÞ ≜ u ) cosðxÞdx ¼ du
Solving (1) and (2):
ð2Þ
220
10
-1
1
u 2 du ¼ 2u2 þ c ¼ 2
Solutions of Problems: Definite and Indefinite Integrals
sinðxÞ þ c
Choice (2) is the answer.
10.64. From trigonometry, we know that:
secðxÞ ¼
1
cosðxÞ
tanðxÞ ¼
sinðxÞ
cosðxÞ
The problem can be solved as follows:
π
3
π
3
secðxÞ tanðxÞdx ¼
0
0
sinðxÞ
1
dx ¼
cosðxÞ cosðxÞ
π
3
0
sinðxÞ
dx
cos 2 ðxÞ
ð1Þ
Now, we should change the variable of the integral as follows:
cosðxÞ ≜ u ) - sinðxÞdx ¼ du
ð2Þ
Solving (1) and (2):
-
u2
u1
π
1
1
1 u2
3 ¼ 1 - 1 ¼ 1
¼
du
¼
u u1
1
cosðxÞ 0 12
u2
Choice (1) is the answer.
10.65. From trigonometry, we know that:
cscðxÞ ¼
1
sinðxÞ
cotðxÞ ¼
cosðxÞ
sinðxÞ
The problem can be solved as follows:
π
4
π
6
cscðxÞ cotðxÞdx ¼
π
4
π
6
cosðxÞ
1
dx ¼
sinðxÞ sinðxÞ
π
4
π
3
cosðxÞ
dx
sin 2 ðxÞ
ð1Þ
Now, we should change the variable of the integral as follows:
sinðxÞ ≜ u ) cosðxÞdx ¼ du
ð2Þ
Solving (1) and (2):
u2
u1
1
1
1 u2
¼ du ¼ u u1
sinðxÞ
u2
Choice (2) is the answer.
π
4
π
6
¼
1
p
2
2
-
1
1
2
¼ -
p
p
2-2 ¼ 2- 2
10
Solutions of Problems: Definite and Indefinite Integrals
221
10.66. 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
tanðxÞ cotðxÞ ¼ 1
Moreover, from list of integral of functions, we know that:
1 þ tan 2 ðxÞ dx ¼ tanðxÞ þ c
1 þ cot 2 ðxÞ dx ¼ - cotðxÞ þ c
The problem can be solved as follows:
π
4
π
6
¼
π
4
π
6
1
dx ¼
2
sin ðxÞ cos 2 ðxÞ
π
4
π
6
1 þ cot 2 ðxÞ 1 þ tan 2 ðxÞ dx
1 þ tan 2 ðxÞ þ cot 2 ðxÞ þ cot 2 ðxÞ tan 2 ðxÞ dx ¼
π
4
π
6
1 þ tan 2 ðxÞ dx þ
π
4
π
6
π
4
π
6
1 þ tan 2 ðxÞ þ cot 2 ðxÞ þ 1 dx
1 þ cot 2 ðxÞ dx ¼ ðtanðxÞ - cotðxÞÞ
p
3 p
¼ ð 1 - 1Þ - 3
3
p
2 3
¼
3
Choice (2) is the answer.
10.67. From list of integral of functions, we know that:
xn dx ¼
1 nþ1
x
þc
nþ1
The problem can be solved as follows:
ðtanðxÞ - cotðxÞÞðtanðxÞ þ cotðxÞÞ5 dx
¼
ðtanðxÞ - cotðxÞÞðtanðxÞ þ cotðxÞÞðtanðxÞ þ cotðxÞÞ4 dx
¼
tan 2 ðxÞ - cot 2 ðxÞ ðtanðxÞ þ cotðxÞÞ4 dx
π
4
π
6
222
10
¼
Solutions of Problems: Definite and Indefinite Integrals
1 þ tan 2 ðxÞ - 1 þ cot 2 ðxÞ ðtanðxÞ þ cotðxÞÞ4 dx
ð1Þ
Now, we need to change the variable of the integral as follows:
d
dx
tanðxÞ þ cotðxÞ ≜ u¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) 1 þ tan 2 ðxÞ - 1 þ cot 2 ðxÞ dx ¼ du
ð2Þ
Solving (1) and (2):
u4 du ¼
1
u5
þ c ¼ ðtanðxÞ þ cotðxÞÞ5 þ c
5
5
Choice (2) is the answer.
10.68. From trigonometry, we know that:
1 þ cosð2xÞ ¼ 2 cos 2 ðxÞ
ð1Þ
sin 2 ðxÞ þ cos 2 ðxÞ ¼ 1
ð2Þ
I¼
ð3Þ
The problem can be solved as follows:
sinðxÞ cosðxÞdx
Now, we should change the variable of the integral as follows:
sinðxÞ ≜ u ) cosðxÞdx ¼ du
ð4Þ
Solving (3) and (4):
I¼
1
1
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
1
1
1
1 1 1
þ cosð2xÞ þ c0 ¼ - cosð2xÞ þ c0 - ¼ - cosð2xÞ þ c00
4
4
4
2 2 2
ð7Þ
Solving (1) and (6):
I¼ -
From (5), (6), and (7), choice (2) is the answer.
10
Solutions of Problems: Definite and Indefinite Integrals
223
10.69. From list of integral of different functions, we know that:
1
du ¼ lnðuðxÞÞ
uð x Þ
The problem can be solved as follows:
ln 3
I¼
ln 2
ex
x x
e
¼
¼
¼
¼
¼
¼
¼
¼
¼
¼) I ¼
1 - e - 2x
dx
1 þ e - 2x
ln 3
ln 2
ex - e - x
dx
ex þ e - x
By defining the new variable, we have:
u ð x Þ ¼ ex þ e - x
) du ¼ ðex - e - x Þdx
Therefore:
I¼
u2
u1
) I ¼ ½lnðex þ e - x Þ]
u2
1
du ¼ ½lnðuðxÞÞ]
uð x Þ
u1
ln 3
¼ ln eln 3 þ e - ln 3 - ln eln 2 þ e - ln 2
ln 2
) I ¼ ln eln 3 þ e ln 3
-1
- ln eln 2 þ e ln 2
-1
I ¼ ln 3 þ 3 - 1 - ln 2 þ 2 - 1
) I ¼ ln 3 þ
10
5
20
4
1
1
- ln 2 þ
¼ ln
- ln ¼ ln
¼ ln ¼ ln 4 - ln 3
3
2
3
2
15
3
) I ¼ 2 ln 2 - ln 3
Choice (3) is the answer.
In this problem, the rules below were applied.
e ln a ¼ a
e - ln a ¼ e ln a
-1
¼ a-1
ln a - ln b ¼ ln
ln an ¼ n ln a
a
b
224
10
Solutions of Problems: Definite and Indefinite Integrals
References
1. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
2. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
Index
A
Algebra of functions, 61
Application of Taylor series in limit, 123–125, 130–132, 134–136
Applications of derivatives, 139–173
Ascending, 140, 147, 153
Axis of symmetry, 12, 36
L
Left continuous, 116, 133
Left-hand side continuity, 103
Limits and continuity, 103–119
N
Non-differentiable point, 139, 152
C
Continuity status, 103
Critical points, 165
D
Definite integrals, 175–193, 195
Definition of derivative, 160, 167, 168
Descending, 147, 165
Domain, 2, 5, 6, 8, 10, 11, 15–19, 21–24, 26–30, 32, 34–36, 144, 161
Domain of function, 2, 4–6, 8, 10, 15, 21, 22, 26, 27, 32, 34, 35
E
Even function, 27, 28, 30
F
Function, 1, 13, 39, 62, 103, 119, 139, 152, 178, 195
G
Growth constant, 139
H
Hyperbolic functions, 39, 62, 68
I
Indefinite integrals, 175–193, 195
Integration by parts, 212, 216
Inverse function, 3, 4, 7, 16–19, 23, 24, 141, 154, 155
Inverse trigonometric functions, v
O
Odd functions, 9, 28, 29, 200, 201
P
Period, 41, 44, 64, 65, 70, 147
R
Range, 11, 12, 16–19, 23, 24, 35, 40, 48, 63, 140, 147, 149, 165, 169,
178, 179, 200, 201
Reflection of the graph, 1, 13
Right continuous, 133
Right-hand side continuity, 103, 116
S
Symmetric with respect to the line, 10, 31
Symmetric with respect to the origin, 3, 16, 17
Symmetric with respect to x-axis, 3, 16
Symmetric with respect to y-axis, 3, 16
T
Trigonometric equations, 39–101
Trigonometric identities, 39–101
U
Unit circle, 46, 47
# The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023
M. Rahmani-Andebili, Calculus I, https://doi.org/10.1007/978-3-031-45028-0
225