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Mehdi Rahmani-Andebili
Practice Problems, Methods, and Solutions
Second Edition
Calculus I
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 (eBook) 
https://doi.org/10.1007/978-3-031-45028-0 
# The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021, 
2023 
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https://doi.org/10.1007/978-3-031-45028-0
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 ElectricalCircuit 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
þ
þ
þ
p
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 
x 1 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) y ¼ log x- 1 
x 1 
2) y ¼ log 1- x 
1 x 
3) y ¼ log 1 þ x 
1- x 
4) y ¼ log xþ 1 
x- 1 
1.2 Determine the relation of f (g(h(x))) for the information below. 
f xð Þ ¼ ln x9 
g xð Þ ¼ x23 
h xð 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
http://crossmark.crossref.org/dialog/?doi=10.1007/978-3-031-45028-0_1&domain=pdf
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p
ð Þ ¼ p ¼
¼ ¼ ¼
p
p
p
1.7
p
p p
p p
¼p p
2 1 Problems: Characteristics of Functions
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 
2) 
1 
2 
3) 
3 
2 
4) 2 
1.4 Determine the value of fog(3) if f x x- 2 and g (x) 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) 2x - 2 and g(x) x2 - 1, solve the equation of fog(x) 0. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) ± 2 
2) ±2 
3) ± 3 
4) 2 
1.6 In the function below, calculate the value of f (-2) + f (2). 
f xð Þ ¼ 2x
2 þ 4 x≥ 2 
x½ �- 4 x< 2 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 8 
2) 6 
3) 10 
4) 5 
Determine the domain of y 2- x2.¼ 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) x≤ - 2, x≥ 2 
2) - 2≤ x≤ 2 
3) x 0 
4) - 2< x< 2
¼ ¼
p
ð Þ ð Þ þ ¼
þ
þ
ð Þ ¼ þ
ð Þ ¼ þ
ð Þ ¼
ð Þ ¼ þ
ð Þ ¼
1 Problems: Characteristics of Functions 3
1.8 Calculate fog(x) if f (x) 1 - x2 and g (x) 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)) 
1.9 What is the inverse function of f xð Þ ¼ 1 
x 
? 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) x 
2) 
1 
x 
3) -
1 
x 
4) x 
1.10 Which one of the following choices is correct about the graph of the relation below? 
sinh xy - cosh xy 1 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. 
1.11 Calculate fof (x) if f xð Þ ¼ 1- x 
1 x 
. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
1þ x 
1- x 
2 
2) 1 
3) x 
4) 
1- x 
1 x 
2 
1.12 Calculate the inverse function of the following function if x ≥ 1. 
f x x4 - 2x2 1 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) f - 1 x 1 x
p 
2) f - 1 x 1- x
p 
3) f - 1 x - 1 x
p 
4) f - 1 x - 1- x
p
þ
þ
p
p
¼
þ þp
þp
þpp
¼
4 1 Problems: Characteristics of Functions
1.13 Calculate the inverse function of the function below. 
f xð Þ ¼ 2e
x þ 1 
ex - 3 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) f - 1 xð Þ ¼ 3x- 1 
x 2 
2) f - 1 xð Þ ¼ x- 2 
3x- 1 
3) f - 1 xð Þ ¼ x- 2 
3x 1 
4) f - 1 xð Þ ¼ 3x þ 1 
x- 2 
1.14 Consider the function below. 
f xð Þ ¼ 4xþ 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 
3) 
5 
3 
4) 
3 
2 
1.15 Calculate the value of cos(π sinh (ln3)). 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) -1 
2) -
1 
2 
3) 
2 
2 
4) 
3 
2 
1.16 Determine the inverse function of f (x) x2 - 2x 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 1 x 1 
2) 1- x 1 
3) 1 x- 1 
4) 1- x- 1 
1.17 Determine the value of f (x) if f (x + 1) x2 - 2x + 1. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large
¼
¼
¼
p
pð Þ ¼ þp
1 Problems: Characteristics of Functions 5
1) (x - 2)2 
2) (x - 1)2 
3) x2 - 2x 
4) (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 x 
2) y3 x 
3) y x2 
4) y ¼ x
2 x≥ 0 
1 x< 0 
1.19 If f x x 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 xð Þ ¼ x
2 þ 1 x≥ 1 
2xþ 3 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 xð Þ ¼ 1- xj j 
1þ xj j 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) ℝ 
2) x ≤ 1 
3) x ≥ 1 
4) -1 ≤ x ≤ 1
j j
1
1
þ
1
1
þ
¼ ¼
6 1 Problems: Characteristics of Functions
1.22 Determine the domain of the function below. 
f xð Þ ¼ x- 1 
x- 3
þ 2- x 
x 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculationamount ● Small ○ Normal ○ Large 
1) (0,1] 
2) [0,1] 
3) (0,2] 
4) (1,3) 
1.23 Determine the domain of the function below. 
f xð Þ ¼ x
p 
x - 1 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) ℝ 
2) [0, ) - {1} 
3) ℝ - {1} 
4) [0, ) 
1.24 Determine the domain of the function below. 
f xð Þ ¼ x
p þ 1 
x x
p
1 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) ℝ - {0} 
2) [1, ) 
3) [0, ) 
4) ℝ - {-1,0} 
1.25 Which number does not exist in the domain of the function below? 
f xð Þ ¼ 1- x 
4x x3 
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) 2x and f (g (x)) 2x + 2. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large
¼ ¼
ð Þ ¼ p
¼
ð Þ ¼ þp
¼
¼
¼ ¼
¼
p
þp
þp
þp
1 Problems: Characteristics of Functions 7
1) x - 1 
2) x + 2 
3) x + 1 
4) x - 2 
1.27 Determine g (x) if f (x) x - 1 and f (g (x)) x. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) x + 1 
2) x2 - x 
3) 2x - 1 
4) x2 - 1 
1.28 Calculate the inverse function of f x 1- x. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) f-1 (x) 1 - x2 , x ≥ 0 
2) f - 1 xð Þ ¼ 1 
1- x
p 
3) f - 1 x 1 x 
4) f-1 (x) 1 - x2 
1.29 What is the inverse function of f (x) 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) 3x - 2 and g (x) 2 + x. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1 
3 
x-
4 
3 
2) 3x - 4 
3) 
1 
3 
xþ 4 
3 
4) 3x + 4 
1.31 Calculate the inverse function of f (x) x3 + 3x2 + 3x + 2. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 1- x- 13 
2) 1- x 13 
3) - 1 x- 13 
4) - 1- x 13 
1.32 Calculate the value of tanh (lnx) in which x > 0. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large
þ
þ
¼
¼
1
ð Þ ¼
ð Þ ¼ j j þ j j þ
8 1 Problems: Characteristics of Functions
1) -1 
2) 1 
3) 
x2 - 1 
x2 1 
4) 
x2 
x2 1 
1.33 If f (x) + x f (-x) 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) 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 
x- 1 
x 
¼ 2x- 1 
p 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) [-1, 0) 
2) [-1, 1] 
3) [-1, 1) 
4) [1, ) 
1.36 Determine the domain of f x 1- x- 1
p
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) x ≥ 1 
2) 1 ≤ x ≤ 2 
3) x ≤ 2 
4) 
5 
4 
≤ x≤ 7 
4 
1.37 Determine the domain of the function of f x x - 1 x 1 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) ℝ - [-1, 1] 
2) ℝ 
3) [-1, 1] 
4) ℝ - (-1, 1)
þ ¼ þ
¼ j j j j
p þp
j j j
j j
j j j
j j j j j
ð Þ þ p ¼
¼
1 Problems: Characteristics of Functions 9
1.38 Calculate the value of f (3) if: 
f x 
1 
x 
x2 
1 
x2 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
28 
3 
2) 
1 
7 
3) 7 
4) 
3 
28 
1.39 For what value of a, the function of f (x) x + 2 + a x - 2 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)) 
2) 1- x- 1 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 - 1 + x + 1 
2) sin( x ) 
3) x3 + x2 
4) j x - 1 - x + 1 
1.43 Which one of the functions below is even? 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) j x - 1 + x + 1 + x 
2) (x + 1)4 
3) f 2 x x- 13 0 
4) f (x) [x] + 1
1.44 ð Þð Þ ¼ þ
x2
ð Þ ¼
x
Calculate the value of f (-f (x)) if f xð Þ ¼ þ
1 x≤ 0
.
¼
¼ p
ð Þ ¼ þ
p
p
p
p
10 1 Problems: Characteristics of Functions
We know that f g x x2 
1
- 4 and g x x-
1 
. Determine f (x). 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) x2 - 4 
2) x2 - 2 
3) x2 
4) x2 + 2 
1.45 
x2 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 
1.46 For what value of the parameter of “a,” the graph of the term below is symmetric with respect to the line of y x. 
3x2 þ 4xy þ 2a- 1ð Þy2 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 2- 1 based on the following information: 
f xð Þ ¼ cos x 
gof x 1 tan 2 x 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 3- 2 2 
2) 5 2- 7 
3) 13- 9 2 
4) 17- 12 2 
1.48 Determine the domain of the function below. 
f xð Þ ¼ log 5x- x
2 
4 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 1 < x < 4 
2) 0 < x < 5
þ
þ
p
ð Þ ¼ þð Þ
1 1
1
1
¼
1 Problems: Characteristics of Functions 11
3) 1 ≤ x ≤ 4 
4) 0 ≤ x ≤ 5 
1.49 Determine the relation of g 
1 
x 
for the information below. 
f xð Þ ¼ 2x 
xþ 2 
g f xð Þð Þ ¼ x 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
2x 
2- x 
2) 
2 
2x- 1 
3) 
x- 2 
2x 
4) 
2 
1 2x 
1.50 Determine the relation of f (x) if we have: 
f 
1- cos 2x 
1 cos 2x 
¼ cot x 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) x 
2) 
1 
x
p 
3) x 
4) 
1 
x 
1.51 Determine the domain of f x log x x2 9 . 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) (- , ) 
2) (0, ) 
3) [-3, 3] 
4) (0, ) - {1} 
1.52 Calculate the range of the function of f (x) 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]
¼ ð Þ ¼ p
1
1
1
ð Þ ¼ þp
p 1p 1
1
1
¼ j j
þ
j j
ð Þ ¼ þ
¼
¼
¼
¼
12 1 Problems: Characteristics of Functions
1.53 Calculate the range of fog (x) if f (x) x2 + 1 and g x x- 1. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) [0, ) 
2) [1, ) 
3) [-1, ) 
4) ℝ 
1.54 Calculate the range of f x x2 - 2x 3. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2, 
2) 3, 
3) [0, ) 
4) [1, ) 
1.55 Which one of the following functions is equivalent to f (x) x - 2 ? 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) g1 xð Þ ¼ x
2 - 3xþ 2 
x- 1 
2) g2 xð Þ ¼ x
2 - 4 
x 2 
3) g3 xð Þ ¼ 
x- 2ð Þ2 
x- 2 
4) g4 xð Þ ¼ 
6x- 12j j 
6 
1.56 Which one of the choices is the axis of symmetry of the following function? 
f x x3
p
- x 23
p 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) x - 2 
2) x - 1 
3) x 1 
4) x 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.
Solutions of Problems: Characteristics 
of Functions 2 
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 
x þ 1 
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 -x- 1-xþ 1 
) - y ¼ log xþ 1 
x- 1 ) y ¼ - log 
xþ 1 
x- 1 ) y ¼ log 
x- 1 
xþ 1 
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 xð Þ ¼ ln x9 
g xð Þ ¼ x23
p 
h xð Þ ¼ ex 
The problem can be solved as follows: 
f g h xð Þð Þð Þ ¼ f g exð Þð Þ ¼ f exð Þ23 ¼ f e 2 3x ¼ ln e 2 3x 
9 
¼ ln e6x 
# 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
13
http://crossmark.crossref.org/dialog/?doi=10.1007/978-3-031-45028-0_2&domain=pdf
https://doi.org/10.1007/978-3-031-45028-0_2#DOI
14 2 Solutions of Problems: Characteristics of Functions
) f g h xð Þð Þð Þ ¼ 6x 
In the calculations, the rules below were used: 
f xð Þð Þnm ¼ f xð Þð Þn m 
ln e f xð Þ ¼ f xð Þ 
Choice (1) is the answer. 
2.3 Based on the information given in the problem, we have: 
f 
1 
x 
¼ 2x- 1 
x2 
g xð Þ ¼ 2 cos 2 xð Þ 
The problem can be solved as follows: 
fog 
π 
3 
¼ f g π 
3 
¼ f 2cos 2 π 
3 
¼ f 2 1 
2 
2 
¼ f 1 
2 
¼ 2 2ð Þ - 1 
22 
) fog π 
3 
¼ 3
p 
2 
Choice (3) is the answer. 
2.4 Based on the information given in the problem, we have: 
f xð Þ ¼ xp - 2 
g xð Þ ¼ xþ 1 
The problem can be solved as follows: 
fog 3ð Þ ¼ f g 3ð Þð Þ ¼ f 3 þ 1ð Þ ¼ f 4ð Þ ¼ 4 
p
- 2 
) fog 3ð Þ ¼ 0 
Choice (1) is the answer. 
2.5 Based on the information given in the problem, we have: 
f xð Þ ¼ 2x- 2 
g xð Þ ¼ x2 - 1
Based on the information given in the problem, we have:
2 Solutions of Problems: Characteristics of Functions 15
The problem can be solved as follows: 
fog xð Þ ¼ f g xð Þð Þ ¼ f x2 - 1 ¼ 2 x2 - 1 - 2 ¼ 2x2 - 4 
fog xð Þ ¼ 0 ) 2x2 - 4 ¼ 0 ) x2 ¼ 2 
) x ¼ ± 2 
p 
Choice (1) is the answer. 
2.6 Based on the information given in the problem, we have: 
f xð Þ ¼ 2x
2 þ 4 x≥ 2 
x½ ] - 4 x< 2 
The problem can be solved as follows: 
f 2ð Þ ¼ 2 2ð Þ2 þ 4 ¼ 12 
f -2ð Þ ¼ -2½ ] - 4 ¼ -6 
) f 2ð Þ þ f -2ð Þ ¼ 12þ -6ð Þ 
) f 2ð Þ þ f -2ð Þ ¼ 6 
Choice (2) is the answer. 
2.7 Based on the information given in the problem, we have: 
y ¼ 2- x2 
p 
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 ) x2 ≤ 2 
) - 2 
p 
≤ x≤ 2 
p 
Choice (2) is the answer. 
2.8 From trigonometry, we know that: 
sin 2 xð Þ þ cos 2 xð Þ ¼ 1 
f xð Þ ¼ 1- x2 
g xð Þ ¼ sin xð Þ
16 2 Solutions of Problems: Characteristics of Functions
Therefore: 
fog xð Þ ¼ f g xð Þð Þ ¼ f sin xð Þð Þ ¼ 1- sin xð Þð Þ2 
) fog xð Þ ¼ cos 2 xð Þ 
Choice (1) is the answer. 
2.9 Based on the information given in the problem, we have: 
f xð Þ 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: 
y ¼ 1 
x 
) x ¼ 1 
y 
) y ¼ 1 
x 
) f-1 xð Þ ¼ 1 
x 
Choice (2) is the answer. 
2.10 Based on the information given in the problem, we have: 
sinh xyð Þ - cosh xyð Þ 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 ) sinh-xyð Þ - cosh-xyð Þ þ 1 ¼ 0 
) -sinh xyð Þ - cosh xyð Þ þ 1 ¼ 0 
) f x, yð Þ≠ f x,-yð Þ 
Therefore, the relation is not symmetric with respect to x-axis. 
x ! -x ) sinh-xyð Þ - cosh-xyð Þ þ 1 ¼ 0 
) - sinh xyð Þ - cosh xyð Þ þ 1 ¼ 0 
) f x, yð Þ≠ f -x, yð Þ 
Thus, the relation is not symmetric with respect to y-axis.
2 Solutions of Problems: Characteristics of Functions 17
x ! -x 
y ! -y ) sinh -xð Þ -yð Þð Þ - cosh -xð Þ -yð Þð Þ þ 1 ¼ 0 
) sinh xyð Þ - cosh xyð Þ þ 1 ¼ 0 
) f x, yð Þ ¼ f -x,-yð Þ 
Hence, the relation is symmetric with respect to the origin. 
In the calculations, the rules below were used: 
sinh -að Þ ¼ - sinh að Þ 
cosh -að Þ ¼ cosh að Þ 
Choice (4) is the answer. 
2.11 Based on the information given in the problem, we have: 
f xð Þ ¼ 1- x 
1þ x 
Therefore: 
fof xð Þ ¼ f f xð Þð Þ ¼ f 1- x 
1þ x ¼ 
1- 1-x 1þx 
1þ 1-x 1þx 
¼ 
1þx- 1-xð Þ 
1þx 
1þxþ1-x 
1þx 
¼ 
2x 
1þx 
2 
1þx 
) fof xð Þ 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. 
y
p ¼ x2 - 1 
) 1 þ yp ¼ x2
18 2 Solutions of Problems: Characteristics of Functions
) x ¼ 1 þ yp 
) y ¼ 1 þ xp 
) f-1 xð Þ ¼ 1þ xp 
Choice (1) is the answer. 
2.13 Based on the information given in the problem, we have: 
y ¼ 2e
x 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: 
y ¼ 2e
x þ 1 
ex - 3 ) ye
x - 3y ¼ 2ex þ 1 
) ex y- 2ð Þ ¼ 1 þ 3y ) ex ¼ 1þ 3y 
y- 2 
ln¼¼¼¼¼¼¼¼¼¼) x ¼ ln 1 þ 3y 
y- 2 
) y ¼ ln 1þ 3x 
x- 2 
) f-1 xð Þ ¼ ln 3xþ 1 
x- 2 
In the calculations, the rule below was used: 
ln e f xð Þ ¼ f xð Þ 
Choice (4) is the answer. 
2.14 Based on the information given in the problem, we have: 
f xð Þ ¼ 4xþ 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: 
y ¼ 4xþ 5 
2x- 3 ) 2xy- 3y ¼ 4xþ 5 ) x 2y- 4ð Þ ¼ 3yþ 5
2 Solutions of Problems: Characteristics of Functions 19
) x ¼ 3y þ 5 
2y- 4 
) y ¼ 3xþ 5 
2x- 4 
) f-1 xð Þ ¼ 3xþ 5 
2x- 4 
The domain of f-1 (x) can be determined as follows: 
2x- 4 ¼ 0 ) x ¼ 2 
) Df-1 xð Þ ¼ ℝ- 2f g 
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 e
x - e-x 
2 
Therefore: 
sinh ln 3ð Þ ¼ e
ln 3 - e- ln 3 
2 
¼ 3-
1 
3 
2 
¼ 8 
6 
¼ 4 
3 
) cos π sinh ln 3ð Þð Þ ¼ cos 4 
3 
π ¼ cos π þ π 
3 
¼ - cos π 
3 
) cos π sinh ln 3ð Þð Þ ¼ - 1 
2 
In the calculations, the rules below were used: 
e ln f xð Þ ¼ f xð Þ 
cos 
π 
3 
¼ 1 
2 
Choice (2) is the answer. 
2.16 Based on the information given in the problem, we have: 
f xð Þ 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 ina square form as follows: 
y ¼ x2 - 2xþ 1- 1 ) y ¼ x- 1ð Þ2 - 1
20 2 Solutions of Problems: Characteristics of Functions
) y þ 1 ¼ x- 1ð Þ2 ) yþ 1 ¼ x- 1 ) x ¼ 1þ yþ 1 
) y ¼ 1 þ xþ 1p 
) f-1 xð Þ ¼ 1þ xþ 1p 
Choice (1) is the answer. 
2.17 Based on the information given in the problem, we have: 
f x þ 1ð Þ ¼ x2 - 2x þ 1 
The problem can be solved as follows: 
f xþ 1ð Þ ¼ x- 1ð Þ2 
x ! x- 1¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) f x- 1ð Þ þ 1ð Þ ¼ x- 1ð Þ - 1ð Þ2 
) f xð Þ ¼ x- 2ð Þ2 
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 x
p ¼ xþ xp 
The problem can be solved as follows: 
f x
p ¼ xþ xp ¼ xp 2 þ xp 
x
p ! x¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) f xð Þ ¼ x2 þ x 
Now, the function is in a standard form. 
f 2ð Þ þ f 1ð Þ ¼ 22 þ 2þ 12 þ 1 
) f 2ð Þ þ f 1ð Þ ¼ 8 
Choice (3) is the answer. 
2.20 Based on the information given in the problem, we have: 
f xð Þ ¼ x
2 þ 1 x≥ 1 
2xþ 3 x< 1
2 Solutions of Problems: Characteristics of Functions 21
The problem can be solved as follows: 
f 0ð Þ ¼ 2x 0þ 3 ¼ 3 
) f f 0ð Þð Þ ¼ f 3ð Þ ¼ 32 þ 1 
) f f 0ð Þð Þ ¼ 10 
Choice (3) is the answer. 
2.21 Based on the information given in the problem, we have: 
f xð Þ ¼ 1- xj j 
1þ xj j 
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- xj j 
1 þ xj j ≥ 0 
1þ xj j> 0¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 1- xj j≥ 0 ) xj j≤ 1 ) 1 ≤ x≤ 1 
Choice (4) is the answer. 
2.22 Based on the information given in the problem, we have: 
f xð Þ ¼ x- 1 
x- 3þ 
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 
x- 3 ≥ 0 
2- x 
x 
≥ 0 
) 
x- 1 
x- 3 ≥ 0 
x- 2 
x 
≤ 0 
) x≤ 1, x> 3 
0< x≤ 2 
\¼¼) 0< x≤ 1 
Choice (1) is the answer. 
2.23 Based on the information given in the problem, we have: 
f xð Þ ¼ x
p 
xj j - 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 
xj j - 1≠ 0 ) 
x≥ 0 
xj j≠ 1 ) 
x≥ 0 
x≠ ± 1 
\¼¼) Df ¼ 0,1½ Þ - 1f g 
Choice (2) is the answer.
Choice (3) is the answer.
22 2 Solutions of Problems: Characteristics of Functions
2.24 Based on the information given in the problem, we have: 
f xð Þ ¼ x
p þ 1 
x x
p þ 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 
x x
p þ 1≠ 0 
Note that x x
p þ 1≠ 0 is true for any x. Hence: 
x≥ 0 ) Df ¼ 0,1½ Þ 
2.25 Based on the information given in the problem, we have: 
f xð Þ ¼ 1- x 
4xþ x3 
The value of those x that make the denominator zero are not in the domain. Thus: 
4xþ x3 ≠ 0 ) x 4þ x2 ≠ 0 
Note that 4 + x2 ≠ 0 for any x. Hence: 
x ¼ 0 
Choice (4) is the answer. 
2.26 Based on the information given in the problem, we have: 
f xð Þ ¼ 2x ð1Þ 
f g xð Þð Þ ¼ 2xþ 2 ð2Þ 
Therefore: 
f g xð Þð Þ ¼ 2g xð Þ ð3Þ 
Solving (2) and (3): 
2g xð Þ ¼ 2x þ 2 ) g xð Þ ¼ xþ 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 xð Þ ¼ x- 1 ð1Þ 
f g xð Þð Þ ¼ x ð2Þ 
Thus: 
f g xð Þð Þ ¼ g xð Þ - 1 ð3Þ 
Solving (2) and (3): 
g xð Þ - 1 ¼ x ) g xð Þ ¼ xþ 1 
Choice (1) is the answer. 
2.28 Based on the information given in the problem, we have: 
f xð Þ ¼ 1- x 
p 
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 1- x 
p 
) y2 ¼ 1- x ) x ¼ 1- y2 
) y ¼ 1- x2 
) f-1 xð Þ 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 xð Þ ¼ 1- x2 , x≥ 0 
Choice (1) is the answer. 
2.29 Based on the information given in the problem, we have: 
f xð Þ ¼ sin xð 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). 
Therefore: 
y ¼ sin xð Þ - 2 ) y þ 2 ¼ sin xð Þ ) x ¼ arc sin yþ 2ð Þð Þ 
) y ¼ arc sin xþ 2ð Þð Þ 
) f-1 xð Þ ¼ arc sin x þ 2ð Þð Þ 
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 xð Þ ¼ 3x- 2 
g xð Þ ¼ 2þ x 
First, we need to determine fog(x) as follows: 
fog xð Þ ¼ f g xð Þð Þ ¼ f 2 þ xð Þ ¼ 3 2 þ xð Þ - 2 ¼ 3xþ 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 3xþ 4 ) 3x ¼ y- 4 
) x ¼ 1 
3 
y- 4 
3 
) y ¼ 1 
3 
x- 4 
3 
) f -1 xð Þ ¼ 1 
3 
x- 4 
3 
Choice (1) is the answer. 
2.31 Based on the information given in the problem, we have: 
f xð Þ ¼ x3 þ 3x2 þ 3xþ 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 xð Þ ¼ x3 þ 3x2 þ 3xþ 1 þ 1 ¼ xþ 1ð Þ3 þ 1 
) y ¼ xþ 1ð Þ3 þ 1 ) y- 1 ¼ x þ 1ð Þ3 ) y- 1ð Þ1 3 ¼ xþ 1 
) x ¼ y- 1ð Þ1 3 - 1 
) y ¼ x- 1ð Þ1 3 - 1 
) f-1 xð Þ ¼ -1 þ x- 13
p 
Choice (3) is the answer.
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 e
t - e-t 
2 
cosh t ¼ e
t þ e-t 
2 
tanh t ¼ sinh t 
sinh t 
Therefore: 
sinh ln xð Þ ¼ e
ln x - e- ln x 
2 
¼ x-
1 
x 
2 
cosh ln xð Þ ¼ e
ln x þ e- ln x 
2 
¼ xþ
1 
x 
2 
tanh ln xð Þ ¼ sinh ln xð Þ 
sinh ln xð Þ ¼ 
x-1 x 
2 
xþ1 x 
2 
¼ x-
1 
x 
xþ 1 x 
) tanh ln xð Þ ¼ x
2 - 1 
x2 þ 1 
In the calculations, the rule below was used: 
e ln f xð Þ ¼ f xð Þ 
Choice (3) is the answer. 
2.33 Based on the information given in the problem, we have: 
f xð Þ þ x f -xð Þ ¼ x2 þ 1 
The problem can be solved as follows: 
x ¼ 2¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 
x ¼ -2¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 
f 2ð Þ þ 2f -2ð Þ ¼ 22 þ 1 
f -2ð Þ þ -2ð Þf 2ð Þ ¼ -2ð Þ2 þ 1 
) 
f 2ð Þ þ 2f -2ð Þ ¼ 5 
f -2ð Þ - 2f 2ð Þ ¼ 5 
)
x-2ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 
f 2ð Þ þ 2f -2ð Þ ¼ 5
-2f -2ð Þ þ 4f 2ð Þ ¼ -10 
þ¼¼¼) 5f 2ð Þ ¼ -5 ) f 2ð Þ ¼ -1 
Choice (1) is the answer.
26 2 Solutions of Problems: Characteristics of Functions
2.34 From trigonometry, we know that: 
1þ cos 2xð Þ ¼ 2 cos 2 xð Þ 
Based on the information given in the problem, we have: 
f xð Þ ¼ x2 - 2 
The problem can be solved as follows: 
) f 2cos xð Þð Þ ¼ 2cos xð Þð Þ2 - 2 ¼ 4 cos 2 xð Þ - 2 ¼ 2 1þ cos 2xð Þð Þ - 2 ¼ 2 cos 2xð Þ 
) f f 2cos xð Þð Þð Þ ¼ 2cos 2xð Þð Þ2 - 2 ¼ 4 cos 2 2xð Þ - 2 ¼ 2 1þ cos 4xð Þð Þ - 2 ¼ 2 cos 4xð Þ 
) f f f 2cos xð Þð Þð Þð Þ ¼ 2cos 4xð Þð Þ2 - 2 ¼ 4 cos 2 4xð Þ - 2 ¼ 2 1þ cos 8xð Þð Þ - 2 
) f f f 2cos xð Þð Þð Þð Þ ¼ 2 cos 8xð Þ 
Choice (4) is the answer. 
2.35 Based on the information given in the problem, we have: 
f 
x- 1 
x 
¼ 2x- 1 
p 
First, we needto determine the f (x) as follows: 
x- 1 
x 
¼ t 
) x ¼ 1 
1- t 
) f tð Þ ¼ 2x 1 
1- t - 1 ¼ 
1þ t 
1- t 
t ! x¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) f xð Þ ¼ 1þ x 
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: 
1 þ x 
1- x ≥ 0 ) 
x þ 1 
x- 1 ≤ 0 ) -1≤ 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 xð Þ ¼ 1- x- 1 
p 
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 27
1- x- 1 
p 
≥ 0 
x- 1≥ 0 )
x- 1 
p 
≤ 1 
x≥ 1 
) x- 1≤ 1 
x≥ 1 
) x≤ 2 
x≥ 1 
)\ 1≤ x≤ 2 
Choice (2) is the answer. 
2.37 Based on the information given in the problem, we have: 
f xð Þ ¼ xj j - 1þ xj j þ 1 
The domain of a function in radical form, including even root, is determined by considering the radicand equal and 
greater than zero. 
xj j - 1≥ 0 
xj j þ 1≥ 0 ) 
xj j≥ 1 
x 2  ) 
x≤ - 1, x≥ 1 
x 2  )
\ 
x≤ - 1, x≥ 1 ) Df ¼ ℝ- -1, 1ð Þ 
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 
x 
¼ x2 þ 1 
x2 
The problem can be solved as follows: 
) f xþ 1 
x 
¼ x2 þ 1 
x2 
þ 2- 2 ¼ xþ 1 
x 
2
- 2 
x þ 1 
x 
! x 
¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) f xð Þ ¼ x2 - 2 ) f 3ð Þ ¼ 32 - 2 
) f 3ð Þ ¼ 7 
Choice (3) is the answer. 
2.39 Based on the information given in the problem, we have: 
f xð Þ ¼ xþ 2j j þ a x- 2j j ð1Þ 
Based on the definition, the function of f (x) is an even function if its domain is symmetric and: 
f -xð Þ ¼ f xð Þ ð2Þ 
Therefore: 
) f -xð Þ ¼ -xþ 2j j þ a -x- 2j j ¼ - x- 2ð Þj j þ a - xþ 2ð Þj j ¼ x- 2j j þ a xþ 2j j ð3Þ 
Solving (1), (2), and (3): 
x- 2j j þ a xþ 2j j ¼ xþ 2j j þ a x- 2j j ) a ¼ 1 
Choice (3) is the answer.
Choice (2):
28 2 Solutions of Problems: Characteristics of Functions
2.40 Based on the information given in the problem, we have: 
f xð Þ ¼ x2 þ A- 1ð Þx is an even function ð1Þ 
g xð Þ ¼ Bþ 2ð Þx2 þ sin xð Þis an odd function ð2Þ 
Based on the definition, the function of f (x) is an even function if its domain is symmetric and: 
f -xð Þ ¼ f xð Þ ð3Þ 
Additionally, the function of f (x) is an odd function if its domain is symmetric and: 
f -xð Þ ¼ -f xð Þ ð4Þ 
Solving (1) and (3):
-xð Þ2 þ A- 1ð Þ-xð Þ ¼ x2 þ A- 1ð Þx ) 2 A- 1ð Þx ¼ 0 ) A ¼ 1 
Solving (2) and (4): 
B þ 2ð Þ-xð Þ2 þ sin-xð Þ ¼ - Bþ 2ð Þx2 þ sin xð Þ 
) Bþ 2ð Þx2 - sin xð Þ ¼ - Bþ 2ð Þx2 - sin xð Þ ) 2 Bþ 2ð Þx2 ¼ 0 ) B ¼ -2 
Therefore: 
A þ B ¼ 1þ -2ð Þ 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 -xð Þ ¼ -f xð Þ 
Choice (1): 
f xð Þ ¼ arc cos xð Þð Þ 
) f -xð Þ ¼ arc cos-xð Þð Þ ¼ π - arc cos xð Þð Þ≠ -f xð Þ, f xð Þf g 
) Not even nor odd 
f xð Þ ¼ 1- x 
p
- 1þ xp 
) f -xð Þ ¼ 1- -xð Þ - 1þ -xð Þ ¼ 1 þ xp - 1- x 
p 
¼ - 1- x 
p
- 1 þ xp ¼ -f xð Þ 
) Odd
2 Solutions of Problems: Characteristics of Functions 29
Choice (3): 
f xð Þ ¼ x4 þ x 
) f -xð Þ ¼ -xð Þ4 þ -xð Þ ¼ x4 - x≠ -f xð Þ, f xð Þf g 
) Not even nor odd 
Choice (4): 
f xð Þ ¼ x sin xð Þ 
) f -xð Þ ¼ -x sin-xð Þ ¼ x sin xð Þ ¼ f xð Þ 
) 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 -xð Þ ¼ -f xð Þ 
Choice (1): 
f xð Þ ¼ x- 1j j þ xþ 1j j 
) f -xð Þ ¼ -x- 1j j þ -xþ 1j j ¼ - xþ 1ð Þj j þ - x- 1ð Þj j ¼ xþ 1j j þ x- 1j j ¼ f xð Þ 
) Even 
Choice (2): 
f xð Þ ¼ sin xj jð Þ 
) f -xð Þ ¼ sin -xj jð Þ ¼ sin xj jð Þ ¼ f xð Þ 
) Even 
Choice (3): 
f xð Þ ¼ x3 þ x2 ) f -xð Þ ¼ -xð Þ3 þ -xð Þ2 ¼ -x3 þ x2 ≠ -f xð Þ, f xð Þf g 
) Not even nor odd 
Choice (4): 
f xð Þ ¼ x- 1j j - x þ 1j j 
) f -xð Þ ¼ -x- 1j j - -x þ 1j j ¼ - xþ 1ð Þj j - - x- 1ð Þj j ¼ xþ 1j j - x- 1j j ¼ -f xð Þ 
) 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 -xð Þ ¼ f xð Þ 
Choice (1): 
f xð Þ ¼ x- 1j j þ x þ 1j j þ xj j 
) f -xð Þ ¼ -x- 1j j þ -xþ 1j j þ -xj j ¼ - x þ 1ð Þj j þ - x- 1ð Þj j þ -xj j ¼ x þ 1j j þ x- 1j j þ xj j ¼ f xð Þ 
) Even 
Choice (2): 
f xð Þ ¼ xþ 1ð Þ4 
) f -xð Þ ¼ -xþ 1ð Þ4 ¼ x- 1ð Þ4 ≠ -f xð Þ, f xð Þf g 
) Not even nor odd 
Choice (3): 
f 2 xð Þ þ x- 13
p 
¼ 0 ) Not a function 
Choice (4): 
f xð Þ ¼ x½ ] þ 1 
) f -xð Þ ¼ -x½ ] þ 1≠ -f xð Þ, f xð Þf g 
) Not even nor odd 
Choice (1) is the answer. 
2.44 Based on the information given in the problem, we have: 
g xð Þ ¼ x- 1 
x
ð1Þ 
f g xð Þð Þ ¼ x2 þ 1 
x2
- 4 ð2Þ 
The problem can be solved as follows: 
f g xð Þð Þ ¼ x2 þ 1 
x2
- 2- 2 ¼ x- 1 
x 
2
- 2 ð3Þ 
Solving (1) and (3): 
f g xð Þð Þ ¼ g xð Þð Þ2 - 2
Solving and : (3)(2) 
2 Solutions of Problems: Characteristics of Functions 31
) f xð Þ ¼ x2 - 2 
Choice (2) is the answer. 
2.45 Based on the information given in the problem, we have: 
f xð Þ ¼ x
2 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 -f xð Þð Þ 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 4xyþ 2a- 1ð Þy2 þ a2 - 4 x ¼ 7 
) f x, yð Þ ¼ 3x2 þ 4xyþ 2a- 1ð Þy2 þ a2 - 4 x- 7 
Moreover, from f (x, y) ¼ f(y, x), we have: 
) 3x2 þ 4xyþ 2a- 1ð Þy2 þ a2 - 4 x- 7 ¼ 3y2 þ 4yxþ 2a- 1ð Þx2 þ a2 - 4 y- 7 
) 4- 2að Þx2 þ 2a- 4ð Þy2 þ a2 - 4 x- a2 - 4 y ¼ 0 
) 
4- 2a ¼ 0 
2a- 4 ¼ 0 
a2 - 4 ¼ 0 
a2 - 4 ¼ 0 
) a ¼ 2f g \ 2f g \ -2, 2f g \ -2, 2f g ) a ¼ 2 
Choice (4) is the answer. 
2.47 Based on the information given in the problem, we have: 
x ¼ 2 
p
- 1 ð1Þ 
f xð Þ ¼ cos x ð2Þ 
gof xð Þ ¼ 1 þ tan 2 x ð3Þ 
gof xð Þ ¼ g f xð Þð Þ ¼ g cos xð Þ ¼ 1þ tan 2 x ð4Þ
2.48
32 2 Solutions of Problems: Characteristics of Functions
From trigonometry, we know that: 
1þ tan 2 x ¼ 1 
cos 2x
ð5Þ 
Solving (4) and (5): 
g cos xð Þ ¼ 1 
cos 2x 
) g xð Þ ¼ 1 
x2
ð6Þ 
) gog xð Þ ¼ g g xð Þð Þ ¼ g 1 
x2 
¼ 1 
1 
x2 
2 ¼ x4 ð7Þ 
Solving (1) and (7): 
gogð Þ 2 
p
- 1 ¼ 2 
p
- 1 
4 
¼ 2- 2 2 
p 
þ 1 
2 
¼ 3- 2 2 
p 2 
¼ 9- 12 2 
p 
þ 8 
) gogð Þ 2 
p
- 1 ¼ 17- 12 2 
p 
Choice (4) is the answer. 
2.49 
Based on the information given in the problem, we have: 
f xð Þ ¼ log 5x- x
2 
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 
4
≥ 0 
5x- x2 
4 
> 0 
) log 
5x- x2 
4
≥ log 1ð Þ 
x2 - 5x< 0 
) 
5x- x2 
4 
≥ 1 
x x- 5ð Þ< 0 
) x
2 - 5xþ 4≤ 0 
x x- 5ð Þ< 0 
) x- 4ð Þ x- 1ð Þ≤ 0 
x x- 5ð Þ< 0 ) 
1≤ x≤ 4 
0< x< 5 
\¼) 1≤ x≤ 4 
Choice (3) is the answer. 
Based on the information given in the problem, we have: 
f xð Þ ¼ 2x 
xþ 2 ð1Þ 
g f xð Þð Þ ¼ x ð2Þ
Therefore:
2 Solutions of Problems: Characteristics of Functions 33
The problem can be solved as follows: 
t≜ 2x 
xþ 2 ð3Þ 
) txþ 2t ¼ 2x ) 2x- tx ¼ 2t ) x 2- tð Þ ¼ 2t ) x ¼ 2t 
2- t ð4Þ 
Solving (1) and (2): 
g f xð Þð Þ ¼ g 2x 
xþ 2 ¼ x ð5Þ 
Solving (3) and (5): 
g tð Þ ¼ x ð6Þ 
Solving (4) and (6): 
g tð Þ ¼ 2t 
2- t ð7Þ 
) g 1 
x 
¼ 2 
1 
x 
2- 1 x 
¼ 
2 
x 
2x-1 
x 
ð8Þ 
) g 1 
x 
¼ 2 
2x- 1 
Choice (2) is the answer. 
2.50 Based on the information given in the problem, we have: 
f 
1-cos 2x 
1þ cos 2x ¼ cot x ð1Þ 
From trigonometry, we know that: 
1- cos 2x ¼ 2 sin 2 x ð2Þ 
1þ cos 2x ¼ 2 cos 2 x ð3Þ 
tan x ¼ sin x 
cos x
ð4Þ 
cot x ¼ 1 
tan x
ð5Þ 
1- cos 2x 
1þ cos 2x ¼ 
2 sin 2 x 
2 cos 2x 
¼ tan 2 x ð6Þ
34 2 Solutions of Problems: Characteristics of Functions
Solving (5) and (1): 
f tan 2 x ¼ cot x ð7Þ 
Solving (7) and (5): 
f tan 2 x ¼ 1 
tan x
ð8Þ 
Defining a new parameter: 
t≜ tan 2 x ð9Þ 
Solving (8) and (9): 
g tð Þ ¼ 1 
t
p ) g xð Þ ¼ 1 
x
p 
Choice (2) is the answer. 
2.51 Based on the information given in the problem, we have: 
f xð Þ ¼ log x x2 þ 9ð Þ 
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 x
2 þ 9 ≥ 0 
x2 þ 9> 0 
x> 0, x≠ 1 
) 
log x x
2 þ 9 ≥ log x 1ð Þ 
x2 þ 9> 0 
x> 0, x≠ 1 
) 
x2 þ 9≥ 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 0,1ð Þ - 1f g 
Choice (4) is the answer. 
2.52 Based on the information given in the problem, we have: 
f xð Þ ¼ 2x- 2 x½ ] þ 1 
Based on the definition, we know that: 
x½ ]≤ x< x½ ] þ 1 - x½ ]¼¼¼¼¼¼¼¼) 0≤ x- x½ ]< 1 x2¼¼¼¼) 0≤ 2x- 2 x½ ]< 2 þ1¼¼¼¼) 1≤ 2x- 2 x½ ] þ 1< 3 
) 1≤ f xð Þ< 3 ) Rf ¼ 1, 3½ Þ 
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 xð Þ ¼ x2 þ 1 
g xð Þ ¼ x- 1 
p 
) fog xð Þ ¼ f g xð Þð Þ ¼ f x- 1 
p 
¼ x- 1 
p 2 
þ 1 ¼ x- 1þ 1 ¼ x 
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 1,1½ Þ 
Now, we can determine the range of the function based its domain as follows: 
fog xð Þ ¼ x 
Dfog ¼ 1,1½ Þ ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) Rfog ¼ 1,1½ Þ 
Choice (2) is the answer. 
2.54 Based on the information given in the problem, we have: 
f xð Þ ¼ x2 - 2xþ 3 
The problem can be solved as follows: 
) f xð Þ ¼ x- 1ð Þ2 þ 2 
As we know: 
x- 1ð Þ2 ≥ 0 þ2¼¼¼) x- 1ð Þ2 þ 2≥ 2 
p 
¼¼¼) x- 1ð Þ2 þ 2≥ 2 
p 
) f xð Þ≥ 2 
p 
) Rf xð Þ ¼ 2 
p 
,1 
Choice (1) is the answer. 
2.55 Based on the information given in the problem, we have: 
f xð Þ ¼ x- 2j j ) Df ¼ ℝ 
Based on the definition, two functions are equivalent if they are equal, and their domains are the same. 
Choice (1): 
g1 xð Þ ¼ x
2 - 3xþ 2 
x- 1 ¼ 
x- 1ð Þ x- 2ð Þ 
x- 1 ¼ x- 2j j 
) Dg1 ¼ ℝ- 1f g 
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 xð Þ ¼ x
2 - 4 
xþ 2 ¼ 
x- 2ð Þ xþ 2ð Þ 
xþ 2 ¼ x- 2j j 
) Dg2 ¼ ℝ- -2f g 
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 xð Þ ¼ 
x- 2ð Þ2 
x- 2j j ¼ 
x- 2j j2 
x- 2j j ¼ x- 2j j 
) Dg3 ¼ ℝ- -2f g 
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): 
g4 xð Þ ¼ 
6x- 12j j 
6
¼ 6 x- 2j j 
6 
¼ x- 2j j 
) Dg4 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 xð Þ ¼ x3p - xþ 23p 
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 -2þ x, yð Þ ¼ -2þ xð Þ3 - -2þ xð Þ þ 23 
f -2- x, yð Þ ¼ -2- xð Þ3 - -2- xð Þ þ 23 
) x- 23
p
- x3p ≠ -x- 23
p 
þ x3p 
) f -2þ x, yð Þ≠ f -2- x, yð Þ 
For Choice (2), we have x ¼ -1 ) a ¼ -1. 
f -1þ x, yð Þ ¼ -1þ xð Þ3 - -1þ xð Þ þ 23 
f -1- x, yð Þ ¼ -1- xð Þ3 - -1- xð Þ þ 23
References 37
) x- 13
p
- xþ 13p ¼ - xþ 13p þ x- 13
p 
) f -1þ x, yð Þ ¼ f -1- x, yð Þ 
For Choice (3), we have x ¼ 1 ) a ¼ 1. 
f 1þ x, yð Þ ¼ 1þ xð Þ3 - 1þ xð Þ þ 23 
f 1- x, yð Þ ¼ 1- xð Þ3 - 1- xð Þ þ 23 
) xþ 13p - xþ 33p ≠ - x- 13
p 
þ x- 33
p 
) f 1þ x, yð Þ≠ f 1- x, yð Þ 
For Choice (4), we have x ¼ 2 ) a ¼ 2. 
f 2þ x, yð Þ ¼ 2þ xð Þ3 - 2þ xð Þ þ 23 
f 2- x, yð Þ ¼ 2- xð Þ3 - 2- xð Þ þ 23 
) xþ 23p - xþ 43p ≠ - x- 23
p 
þ x- 43
p 
) f 2þ x, yð Þ≠ f 2- x, yð Þ 
In the calculations, the rule below was used because n was an odd number:
-f xð Þn ¼ - f xð Þn 
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.
¼
p
p
¼
þð Þ ¼ þ
þð Þ ¼
Problems: Trigonometric Equations and Identities 3 
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 - 2sin2 x) for x 7.5° [1, 2]. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
3 
8 
2) 
1 
8 
3) 
3 
4 
4) 
1 
4 
3.2. Calculate the value of tan3 x + cot3 x if tan x + cot x 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 
1) 
cosh aþ bð Þ ¼ cosh a cosh b- sinh a sinh b 
sinh a b sinh a cosh b cosh a sinh b 
2) 
cosh aþ bð Þ ¼ cosh a cosh b- sinh a sinh b 
sinh a b 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
http://crossmark.crossref.org/dialog/?doi=10.1007/978-3-031-45028-0_3&domain=pdf
https://doi.org/10.1007/978-3-031-45028-0_3#DOI
þð Þ ¼
þð Þ ¼ þ
¼
p
p
p
p
ð Þ
ð Þ ¼
p
p
40 3 Problems: Trigonometric Equations and Identities
3) 
cosh aþ bð Þ ¼ cosh a cosh bþ sinh a sinh b 
sinh a b sinh a cosh b- cosh a sinh b 
4) 
cosh aþ bð Þ ¼ cosh a cosh bþ sinh a sinh b 
sinh a b sinh a cosh b cosh a sinh b 
3.4. Calculate the value of tan(2θ) if cot(θ) 5. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
5 
12 
2) 
5 
13 
3) -
5 
12 
4) -
5 
13 
3.5. Determine the value of tan(-2100°). 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 3 
2) 
3 
3 
3) - 3 
4) -
3 
3 
3.6. Simplify and calculate the final value of the following term: 
1þ cos 40 °ð Þ 
sin 40 ° 
Difficulty level ● Easy ○ Normal ○Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) sin (20°) 
2) cos (20°) 
3) tan (20°) 
4) cot (20°) 
3.7. Determine the range of m if sin α 
3m- 1 
4 
and 
π 
6 
≤ α≤ 2π 
3 
. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 1, 
2 3- 1 
3 
2) 1, 
2 3þ 1 
3 
3) [1, 2] 
4) 1, 
5 
3
ð Þ
þ
¼
3 Problems: Trigonometric Equations and Identities 41
3.8. Determine the range of m if cos xð Þ ¼ 2m- 1 
6 
and -
π 
3 
≤ x≤ π 
3 
. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 2, 
7 
2 
2) 
3 
2 
, 
7 
2 
3) 2, 
5 
2 
4) 
3 
2 
, 
5 
2 
3.9. What is the main period of cos 2 x - 5 cos 
2x 
3 
? 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) π 
2) 2π 
3) 3π 
4) 4π 
3.10. What is the main period of sin 4 
3x 
5 
cos 3 
2x 
3 
? 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 3π 
2) 5π 
3) 15π 
4) 30π 
3.11. Determine the main period of sin 4 
πx 
3 
þ cos πxð Þ þ 5. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 1 
2) 2 
3) 3 
4) 6 
3.12. Figure 3.1 illustrates part of the function of y sin (kx). Determine the value of k. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
2 
3 
2) 
3 
4 
3) 
3 
2 
4) 
4 
3
3.13. ¼ þ
¼
¼
¼
¼
¼
42 3 Problems: Trigonometric Equations and Identities
Figure 3.1 The graph of problem 3.12 
Figure 3.2 illustrates part of the function of y cos ax 
1 
2 
π . Determine the value of a. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1 
2 
2) 
3 
2 
3) 
2 
3 
4) 
7 
4 
Figure 3.2 The graph of problem 3.13 
3.14. Which one of the following choices is correct if α + β 19π? 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) sin(α) sin (β) 
2) cos(α) cos (β) 
3) tan(α) tan (β) 
4) cot(α) cot (β) 
3.15. Calculate the final value of the following equation. 
sin 5π þ xð Þ þ sin x- π 
3 
þ sin xþ 7π 
3 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large
p
p
p
¼
3 Problems: Trigonometric Equations and Identities 43
1) 0 
2) sin 
π 
3 
3) 2 sin 
π 
3 
4) - sin 
π 
3 
3.16. Calculate the value of the term below. 
sin arccos
- 1 
2 
þ arcsin - 3
p 
2 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
3 
2 
2) 1 
3) 
1 
2 
4) -1 
3.17. Calculate the value of 2 arctan 
1 
2 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) arctan 
4 
3 
2) arctan 
3 
4 
3) arctan 
2 
3 
4) arctan 
3 
2 
3.18. What is the value of cos(π sinh ln 3)? 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) -1 
2) -
1 
2 
3) 
2 
2 
4) 
3 
2 
3.19. Calculate the value of cos(20°) if sin(50°) + sin (10°) m. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
m 
2 
2) m 
3) 2m 
4) 
2m 
3
ð Þð Þ ð Þ
¼ ¼
¼
¼
¼
¼
44 3 Problems: Trigonometric Equations and Identities
3.20. Simplify and calculate the final value of the following term: 
1þ tan 2 5 °ð Þð Þ sin 10 °ð Þ 
1- tan 2 5 ° tan 10 ° 
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) m and cos(α) n? 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) m2 (1 + n2 ) n2 
2) m2 (1 - n2 ) n2 
3) m2 (2 + n2 ) 1 
4) m2 (2 - n2 ) 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 
3) 
2π 
3 
4) 
2π 
5 
3.23. Calculate the value of arc cos sin 
4π 
3 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
π 
6 
2) 
5π 
6 
3) 
π 
3 
4) -
π 
6 
3.24. Calculate the value of arc sin sin 
17π 
5 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
2π 
5 
2) 
3π 
5 
3) -
2π 
5 
4) -
3π 
5
þ
ð Þð Þ þ
3 Problems: Trigonometric Equations and Identities 45
3.25. Calculate the value of arc cos cos 
19π 
5 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
π 
5 
2) 
4π 
5 
3) -
π 
5 
4) -
4π 
5 
3.26. Calculate the value of tan 2arc tan 
1 
2 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 1 
2) 
3 
4 
3) 
4 
3 
4) 
3 
5 
3.27. Calculate the final value of sin arc sin 
3 
5 
arc tan 
3 
4 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
10 
13 
2) 
9 
13 
3) 
12 
35 
4) 
24 
25 
3.28. Calculate the final value of arc cot -
4 
3
- arc cot 
3 
4 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) π 
2) 
2π 
3 
3) 
π 
2 
4) 
π 
3 
3.29. Calculate the final value of arc tan 5 arc tan 
3 
2 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
π 
4 
2) -
π 
4
þ
3.32.
46 3 Problems: Trigonometric Equations and Identities
3) 
3π 
4 
4) 
5π 
4 
3.30. Calculate the final value of sin arc cos 
3 
5 
cos arc sin -
4 
5 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
7 
5 
2) -
1 
5 
3) 
1 
5 
4) -
7 
5 
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 
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
þ
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 
3.34. Calculate the value of arc tan 
2 
3 
arc tan 
1 
5 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
π 
6 
2) 
π 
4 
3) 
π 
3 
4) 
π 
2 
3.35. Calculate the final value of the term below. 
arc tan mð Þð Þ þ arc tan 1 
m 
þ arc cot mð Þð Þ þ arc cot -mð Þð Þ
ð Þ ð Þ þ ð Þ ð Þ
p
p
p
p
48 3 Problems: Trigonometric Equations and Identities
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) π or 2π 
2) 
π 
2 
or 
3π 
2 
3) 
3π 
2 
4) 
π 
2 
3.36. Determine the range of x in the inequality below. Herein, x is an acute angle.
- 1≤ cos 4x cos 2x sin 4x sin 2x ≤ 0 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
π 
6 
, 
3π 
8 
2) 
π 
8 
, 
π 
4 
3) 
π 
6 
, 
π 
3 
4) 
π 
4 
, 
π 
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 
2) -
1 
18 
3) 
1 
36 
4) -
1 
36 
3.38. Simplify and calculate the value of the following term: 
sin 
5π 
12 
þ cos 5π 
12 
sin 
5π 
12
- cos 
5π 
12 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 3 
2) 
3 
3 
3) - 2 3 
4) -
3 
3
¼
¼ �
¼ þ �
3 Problems: Trigonometric Equations and Identities 49
3.39. Figure 3.6 illustrates part of the function of y a sin (bπx). Determine the value of a + b. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large1) 
4 
3 
2) 
5 
3 
3) 
7 
3 
4) 
8 
3 
Figure 3.6 The graph of problem 3.39 
3.40. Figure 3.7 illustrates part of the function of y a sin (bπx). Determine the value of a b. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) -6 
2) -3 
3) 
9 
2 
4) 6 
Figure 3.7 The graph of problem 3.40 
3.41. Figure 3.8 illustrates part of the function of y a sin bx 
1 
2 
π . Determine the value of a b. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 2 
2) 
5 
2
ð Þ
ð Þ
ð Þ
ð Þ
ð Þ
ð Þ ð Þ
50 3 Problems: Trigonometric Equations and Identities
3) 3 
4) 
7 
2 
Figure 3.8 The graph of problem 3.41 
3.42. Simplify and calculate the value of the following term: 
cos 5 °ð Þ cos 10 °ð Þ cos 20 °ð Þ 
cos 50 ° 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
1 
4 cos 85 ° 
2) 
1 
8 cos 85 ° 
3) 
1 
8 sin 85 ° 
4) 
1 
4 sin 85 ° 
3.43. Calculate the value of tan 
x 
2 
if sin xð Þ þ cos xð Þ ¼ 7 
5 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2 or 3 
2) 
1 
2 
or 
1 
3 
3) 2 or 
3 
5 
4) 3 or 
2 
5 
3.44. Simplify and calculate the value of the following term: 
sin 4 αð Þ- cos 4 αð Þ 
sin α cos α 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large
ð Þ ð Þ ð Þ ð Þ þ ð Þ ð Þ
ð Þ þ ð Þ þ ð Þ
p
p
p
p
ð Þ ð Þ þð Þ ð Þ ð Þð Þ
3 Problems: Trigonometric Equations and Identities 51
1) 2 cot (2α) 
2) -2 cot (2α) 
3) 2 tan (3α) 
4) -2 tan (3α) 
3.45. Calculate the value of cot2 (2α) if sin 4 αð Þ þ cos 4 αð Þ ¼ 1 
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 x cos x - cos 3 x sin x 3 sin 2 x cos 2 x 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
3 
8 
2) 
5 
8 
3) -
5 
8 
4) -
3 
8 
3.47. Calculate the value of the following relation for α ¼ π 
15 
: 
sin 2αð Þ þ sin 5αð Þ þ sin 8αð Þ 
cos 2α cos 5α cos 8α 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -
3 
3 
2) - 3 
3) 3 
4) 
3 
3 
3.48. Calculate the value of the following relation for x ¼ π 
12 
: 
sin x - cos x 2 sin x - cos x - 2 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large
1)
ð Þ ð Þ þ ð Þ ð Þ ¼
¼ þ ð Þ ¼
52 3 Problems: Trigonometric Equations and Identities
7 
2 
2) 
5 
2 
3) -
5 
2 
4) -
7 
2 
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. 
sin πx cos 2 πx sin 2 πx cos πx 0: 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 11 
2) 12 
3) 13 
4) 14 
3.51. Figure 3.9 illustrates part of the function of y 
1 
2 
2 cos mx . Determine the value of the function for x 
16π 
3 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -
1 
2 
2) 
1 
2 
3) 1 
4) 0 
Figure 3.9 The graph of problem 3.51
¼ ¼
¼ ¼
3 Problems: Trigonometric Equations and Identities 53
3.52. Figure 3.10 shows part of the function of y 1 + sin (mx). Determine the value of the function for x 
7π 
6 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 0 
2) 
1 
2 
3) 1 
4) 2 
Figure 3.10 The graph of problem 3.52 
3.53. Figure 3.11 shows part of the function of y a - sin (bπx). Determine the value of the function for x 
25 
3 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2 
2) 
5 
2 
3) 3 
4) 
7 
2 
Figure 3.11 The graph of problem 3.53
54 3 Problems: Trigonometric Equations and Identities
3.54. Figure 3.12 shows the function of y ¼ aþ b cos π 
2 
x for 0 < x < 4. Determine the value of b. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -2 
2) -1 
3) 1 
4) 2 
Figure 3.12 The graph of problem 3.54 
3.55. Figure 3.13 shows the function of y ¼ 1 + a sin (bπx) for 0< x< 4 
3 
. Determine the value of a + b. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 3 
2) 4 
3) 5 
4) 6 
Figure 3.13 The graph of problem 3.55 
3.56. Figure 3.14 shows part the function of y ¼ a- 2 cos bxþ π 
2 
. Determine the value of a + b. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
1 
2 
2) 1 
3) 
3 
2 
4) 2
3 Problems: Trigonometric Equations and Identities 55
Figure 3.14 The graph of problem 3.56 
3.57. Calculate the value of cos(25 ° - α) if tan α þ 20 °ð Þ ¼ 3 
4 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 5 
2) 6 
3) 7 
4) 8 
3.58. Calculate the value of tan 
π 
4
þ α assuming that α is an acute angle and sin αð Þ ¼ 3 
5 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -7 
2) -
1 
7 
3) 
1 
7 
4) 7 
3.59. Calculate the value of tan 
π 
4
- α if tan 
π 
2
- α ¼ 2 
3 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -
1 
3 
2) -
1 
5 
3) 
1 
5 
4) 
1 
3 
3.60. Calculate the value of tan(2a) while we know that tan aþ bð Þ ¼ 2 
5 
and tan a- bð Þ ¼ 3 
7 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -
1 
3 
2) -
1 
2 
3) 3 
4) 1
56 3 Problems: Trigonometric Equations and Identities
3.61. Calculate the value of tan(x) if we have: 
sin x-
π 
4 
cos x-
π 
4 
¼ 2 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -3 
2) 
1 
3 
3) 
2 
3 
4) 3 
3.62. Calculate the value of (1 + tan (α))(1 + tan (β)) if αþ β ¼ π 
4 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -2 
2) 2 
3) 
1 
3 
4) -
1 
2 
3.63. Calculate the value of tan 
π 
4
þ α - tan π 
4
- α . 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2 tan (2α) 
2) 2 cos (2α) 
3) 0 
4) 2 sin (2α) 
3.64. Calculate the value of tan(2α) if tan 
π 
4
- α ¼ 1 
5 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 1.5 
2) 1.8 
3) 2.4 
4) 2.5 
3.65. Calculate the value of tan(2α - β) if tan(α) ¼ 2 and tan βð Þ ¼ 1 
3 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -3 
2) -2 
3) 0.5 
4) 3
¼
¼
¼
¼
3 Problems: Trigonometric Equations and Identities 57
3.66. Determine the common solution of the equation of cos(3x) + cos (x) 0 assuming cos(x) ≠ 0. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
kπ 
2 
þ π 
4 
2) 
kπ 
2 
þ π 
8 
3) kπ-
π 
4 
4) kπ þ π 
4 
3.67. Calculate the sum of the positive acute roots of the equation of tan(4x) cot (x). 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
2π 
5 
2) 
4π 
5 
3) 
3π 
5 
4) 
π 
5 
3.68. Determine the common solution of the equation of 2sin2 (x) + 3 cos (x) 0. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2kπ ± 
2π 
3 
2) 2kπ ± 
π 
3 
3) 2kπ ± 
5π 
6 
4) kπ-
π 
3 
3.69. Determine the common solution of the equation of 2sin2 (x) 3 cos (x). 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) kπ ± 
π 
6 
2) kπ ± 
π 
3 
3) 2kπ ± 
π 
6 
4) 2kπ ± 
π 
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) y ¼ 1 
x 
2) y ¼ 3 
x 
3) y ¼ 1 
4x 
4) y ¼ 3 
4x
¼
ð Þ þ þð Þ ð Þ þ ¼
ð Þ þ ð Þ þ ð Þ ð Þ
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) 5 and 0 < x, y < 2π. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
π 
3 
or 
2π 
3 
2) 
π 
4 
or 
5π 
4 
3) 
π 
6 
or 
5π 
6 
4) 
π 
2 
or 
3π 
2 
3.74. Calculate the value of m if tan(α) ≠ tan (β), αþ β ¼ π 
4 
, and α and β are the two roots of the equation below. 
tan 2 x m 2 tan x 2m- 2 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 αð Þ þ cos 6 αð Þ þ 3 sin 2 αð Þ cos 2 αð Þ 
sin 4 α cos 4 α 2 sin 2 α cos 2 α 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) sin2 (α) 
2) cos2 (α) 
3) sin2 (α) - cos2 (α) 
4) 1
ð Þ ð Þ þ ð Þ ð Þ
p
p
p
p
ð Þ ð Þ þ ð Þð Þ ¼ ð Þ ð Þ ð Þð Þ
3 Problems: Trigonometric Equations and Identities 59
3.76. Calculate the final value of the relation below. 
sin 135 °ð Þ cos 210 °ð Þ þ cos 135 °ð Þ sin 420 °ð Þ 
tan 210 ° cot 420 ° cot 120 ° tan 330 ° 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -
6 
4 
2) -
3 6 
4 
3) -
6 
2 
4) -
3 6 
2 
3.77. Calculate the final value of (1 + cot (x))(1 + cot (y)) if xþ y ¼ kπ þ π 
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. 
sin xð Þ- tan xð Þð Þ tan 3π 
2
- x ¼ cos 4π 
3 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) kπ-
π 
6 
2) kπ þ π 
3 
3) 2kπ ± 
π 
3 
4) 2kπ ± 
π 
6 
3.79. Calculate the sum of the roots of the equation below for x 2 [0, π]. 
sin 2x sin x cos x cos 2x cos x - sin x 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
3π 
4 
2) 
5π 
4 
3) 
3π 
2 
4) 
7π 
4
¼ þ þ
ð Þ þ p ð Þ ¼
60 3 Problems: Trigonometric Equations and Identities
3.80. Determine the common solution of the equation of 2 
p 
sin 
π 
4
- x 1 sin 
5π 
2 
x . 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) kπ þ π 
2 
2) 2kπ-
π 
4 
3) 2kπ-
π 
2 
4) 2kπ þ π 
2 
3.81. Which one of the following choices shows one of the common solutions of the equation of cos 2x 3 sin 2x 1? 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) kπ-
π 
6 
2) kπ-
π 
3 
3) kπ þ π 
6 
4) kπ þ π 
3 
3.82. Calculate the value of the following term: 
cos 3αþ sin α sin 2α 
sin 3α- sin 2α cos α
� sin 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.
Solutions of Problems: Trigonometric Equations
and Identities 4
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 sin xð Þ cos xð Þ
1-cos 2x= 2 sin 2x
sin 30 ° =
1
2
Therefore:
sin x cos x 1- 2 sin 2x =
1
2
sin 2x cos 2xð Þ= 1
4
sin 4x
For x = 7.5°, we have:
1
4
× sin 4 × 7:5 °ð Þ= 1
4
sin 30 ° =
1
8
Choice (2) is the answer.
4.2. Based on the information given in the problem, we have:
tan xþ cot x= 3
From algebra, we know that:
a3 þ b3 = aþ bð Þ3 - 3ab aþ bð Þ
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
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https://doi.org/10.1007/978-3-031-45028-0_4#DOI
ð Þ
p
p
62 4 Solutions of Problems: Trigonometric Equations and Identities
Therefore:
tan 3xþ cot 3x= tan xþ cot xð Þ3 - 3 tan x cot x tan xþ cot xð Þ= 3ð Þ3 - 3 1ð Þ 3ð Þ
) tan 3xþ cot 3x= 18
Choice (1) is the answer.
4.3. From trigonometry of hyperbolic functions, we know that:
cosh a± b = cosh a cosh b± sinh a sinh b
sinh a± bð Þ= sinh a cosh b± cosh a sinh b
Choice (4) is the answer.
4.4. From trigonometry, we know that:
tan θð Þ= 1
cot θð Þ
tan 2θð Þ= 2 tan θð Þ
1- tan 2 θð Þ
Based on the information given in the problem:
cot θð Þ= 5 ) tan θð Þ= 1
5
Therefore:
tan 2θð Þ= 2 tan θð Þ
1- tan 2 θð Þ =
2×
1
5
1-
1
5
2 =
2
5
24
25
) tan 2θð Þ= 5
12
Choice (1) is the answer.
4.5. From trigonometry, we know that:
tan αþ nπð Þ= tan αð Þ,8n 2 ℤ
tan - αð Þ= - tan αð Þ
tan 60 ° = 3
Therefore:
tan - 2100 ° = - tan 2100 ° = - tan 12× 180- 60 ° = - tan - 60 ° = tan 60 °
) tan - 2100 ° = 3
Choice (1) is the answer.
ð Þ
ð Þ
4 Solutions of Problems: Trigonometric Equations and Identities 63
4.6. From trigonometry, we know that:
1þ cos θð Þ= 2 cos 2 θ
2
sin θð Þ= 2 sin θ
2
cos
θ
2
cot θð Þ= cos θð Þ
sin θð Þ
Therefore:
1þ cos 40 °
sin 40 °
=
2 cos 2 20 °
2 sin 20 ° cos 20 °
=
cos 20 °
sin 20 °
) 1þ cos 40
°
sin 40 °
= cot 20 °
Choice (4) is the answer.
4.7. For the given range of α, we can conclude that:
π
6
≤ α≤ 2π
3
) 1
2
≤ sin αð Þ≤ 1
Therefore, based on the given information, i.e., sin α = 3m- 14 , we can write:
1
2
≤ 3m- 1
4
≤ 1 ) 2≤ 3m- 1≤ 4 ) 3≤ 3m≤ 5
) 1≤m≤ 5
3
Choice (4) is the answer.
4.8. For the given range of x, we can conclude that:
-
π
3
≤ x≤ π
3
) 1
2
≤ cos xð Þ≤ 1
Therefore, based on the given information, i.e., cos x = 2m- 16 , we can write:
1
2
≤ 2m- 1
6
≤ 1 ) 3≤ 2m- 1≤ 6 ) 4≤ 2m≤ 7
) 2≤m≤ 7
2
Choice (1) is the answer.
64 4 Solutions of Problems: Trigonometric Equations and Identities
4.9. From trigonometry, we know that:
f 1 xð Þ= cos 2n axð Þ,8n 2 ℤ ) T1 = πaj j
f 2 xð Þ= cos 2nþ1 axð Þ, 8n 2 ℤ ) T2 = 2πaj j
Therefore:
f 1 xð Þ= cos 2 xð Þ ) T1 = π1 = π
f 2 xð Þ= - 5 cos 2x3 ) T2 =
2π
2
3
= 3π
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 =LCM π, 3πð Þ
) T = 3π
Choice (3) is the answer.
4.10. From trigonometry, we know that:
f 1 xð Þ= sin 2n axð Þ, 8n 2 ℤ ) T1 = πaj j
f 2 xð Þ= cos 2nþ1 axð Þ, 8n 2 ℤ ) T2 = 2πaj j
Therefore:
f 1 xð Þ= sin 4 3x5 ) T1 =
π
3
5
=
5π
3
f 2 xð Þ= cos 3 2x3 ) T2 =
2π
2
3
= 3π
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
3
, 3π
) 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 xð Þ= sin 2n axð Þ, 8n 2 ℤ ) T1 = πaj j
f 2 xð Þ= cos 2nþ1 axð Þ, 8n 2 ℤ ) T2 = 2πaj j
Therefore:
f 1 xð Þ= sin 4 πx3 ) T1 =
π
π
3
= 3
f 2 xð Þ= cos πxð Þ ) T2 = 2ππ = 2
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 =LCM 3, 2ð Þ
) T = 6
Choice (4) is the answer.
4.12. From trigonometry, we know that:
y= sin kxð Þ ) T = 2π
kj j
Therefore:
) 3π
4
=1
2
2π
kj j ) kj j=
4
3
) k= ± 4
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 πaxþ π
2
= -sin πaxð Þ
y= sin mxð Þ ) T = 2π
mj j
Therefore:
) 1- - 1
3
=
2π
πaj j )
4
3
=
2
aj j ) a= ±
3
2
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:
sin αþ 2nπð Þ= sin αð Þ, 8n 2 ℤ
cos αþ 2nπð Þ= cos αð Þ,8n 2 ℤ
tan αþ nπð Þ= tan αð Þ,8n 2 ℤ
cot αþ nπð Þ= cot αð Þ,8n 2 ℤ
sin π- αð Þ= sin αð Þ
cos π- αð Þ= -cos αð Þ
tan - αð Þ= - tan αð Þ
cot - αð Þ= -cot αð Þ
p
4 Solutions of Problems: Trigonometric Equations and Identities 67
Based on the information given in the problem, we have:
αþ β= 19π ) α= 19π- β
Therefore:
sin αð Þ= sin 19π- βð Þ= sin π- βð Þ= sin βð Þ
cos αð Þ= cos 19π- βð Þ= cos π- βð Þ= -cos βð Þ
tan αð Þ= tan 19π- βð Þ= tan - βð Þ= - tan βð Þ
cot αð Þ= cot 19π- βð Þ= cot - βð Þ= -cot βð Þ
Choice (1) is the answer.
4.15. From trigonometry, we know that:
sin αþ 2nπð Þ= sin αð Þ, 8n 2 ℤ
cos αþ 2nπð Þ= cos αð Þ,8n 2 ℤ
sin αþ πð Þ= -sin αð Þ
sin αð Þ þ sin βð Þ= 2 sin αþ β
2
cos
α- β
2
Therefore:
sin 5π þ xð Þ þ sin x- π
3
þ sin xþ 7π
3
= sin xþ πð Þ þ sin x- π
3
þ sin xþ π
3
= -sin xð Þ þ 2 sin xð Þ cos π
3
= -sin xð Þ þ sin xð Þ
) sin 5π þ xð Þ þ sin x- π
3
þ sin xþ 7π
3
= 0
Choice (1) is the answer.
4.16. Let us assume:
cos
2π
3
≜- 1
2
) arc cos - 1
2
=
2π
3
sin
- π
3
≜- 1
2
) arc sin - 3
2
=
- π
3
Therefore:
sin arc cos -
1
2
þ arc sin - 3
p
2
= sin
2π
3
þ - π
3
= sin
π
3
68 4 Solutions of Problems: Trigonometric Equations and Identities
) sin arc cos - 1
2
þ arc sin - 3
p
2
=
3
p
2
Choice (1) is the answer.
4.17. Let us assume:
arc tan
1
2
≜ α ) tan αð Þ= 1
2
We need to find the value of 2 arctan
1
2
which is equal to 2α.
From trigonometry, we know that:
tan 2αð Þ= 2 tan α
1- tan 2α
Hence:
tan 2αð Þ= 2×
1
2
1- 12
2 =
4
3
) 2α= arc tan 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
3
=
1
2
sinh x=
ex - e- x
2
Thus, for x = ln 3, we can write:
sinh ln 3=
e ln 3 - e- ln 3
2
=
3- 13
2
=
4
3
Therefore,
cos π sinh ln 3ð Þ= cos π × 4
3
= cos π þ π
3
= -cos
π
3
) cos π sinh ln 3ð Þ= - 1
2
Choice (2) is the answer.
4 Solutions of Problems: Trigonometric Equations and Identities 69
4.19. From trigonometry, we know that:
sin 30 ° =
1
2
sin αð Þ þ sin βð Þ= 2 sin αþ β
2
cos
α- β
2
Therefore:
sin 50 ° þ sin 10 ° =m
) 2 sin 30 ° cos 20 ° =m ) 2× 1
2
cos 20 ° =m ) cos 20 ° =m
Choice (2) is the answer.
4.20. From trigonometry, we know that:
cos θð Þ= 1- tan
2 θ
2
1þ tan 2 θ2
tan θð Þ= sin θð Þ
cos θð Þ
tan 45 ° = 1
Therefore:
1þ tan 2 5 ° sin 10 °
1- tan 2 5 ° tan 10 °
=
1
cos 10 °
×
sin 10 °
sin 10 °ð Þ
cos 10 °ð Þ
= 1
) 1þ 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
cos 2 αð Þ = 1þ tan
2 αð Þ
tan αð Þ= 1
cot αð Þ
Based on the information given in the problem, we have:
cot αð Þ=m
cos αð Þ= n
p
70 4 Solutions of Problems: Trigonometric Equations and Identities
Therefore:
1
cos 2 αð Þ = 1þ
1
cot 2 αð Þ
) 1
n2
= 1þ 1
m2
×m2n2¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) m2 =m2n2 þ n2
) m2 1- n2 = n2
Choice (2) is the answer.
4.22. From trigonometry, we can know that:
sin 2nþ1 axð Þ,8n 2 ℤ ) T = 2π
aj j
sin αð Þ cos βð Þ= 1
2
sin αþ βð Þ þ sin α- βð Þð Þ
We need to change the product expression to the summation one, as follows:
y= sin 3xð Þ cos 5xð Þ þ 11 ) y= 1
2
sin 8xð Þ- 1
2
sin 2xð Þ þ 11
Then:
1
2
sin 8xð Þ ) T1 = 2π8 =
π
4
-
1
2
sin 2xð Þ ) T2 = 2π2 = π
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
4
, π
) T = π
Choice (1) is the answer.
4.23. From trigonometry, we know that:
sin
4π
3
= sin π þ π
3
= -sin
π
3
= -
3
p
2
arc cos - αð Þð Þ= π- arc cos αð Þð Þ
arc cos
3
2
=
π
6
4 Solutions of Problems: Trigonometric Equations and Identities 71
Therefore:
arc cos sin
4π
3
= arc cos -
3
p
2
= π- arc cos
3
p
2
= π-
π
6
) arc cos sin 4π
3
=
5π
6
Choice (2) is the answer.
4.24. From trigonometry, we know that:
sin
17π
5
= sin 4π-
3π
5
= sin -
3π
5
sin -
3π
5
≜ α ) arc sin αð Þð Þ= - 3π
5
Therefore:
arc sin sin
17π
5
= arc sin sin -
3π
5
= arc sin αð Þð Þ
) arc sin sin 17π
5
= -
3π
5
Choice (4) is the answer.
4.25. From trigonometry, we know that:
cos
19π
5
= cos 4π-
π
5
= cos -
π
5
cos
π
5
≜ α ) arc cos αð Þð Þ= π
5
Therefore:
arc cos cos
19π
5
= arc cos cos -
π
5
= arc cos cos
π
5
= arc cos αð Þð Þ
) arc cos cos 19π
5
=
π
5
Choice (1) is the answer.
4.26. From trigonometry, we know that:
tan 2αð Þ= 2 tan αð Þ
1- tan 2 αð Þ
arc tan
1
2
≜ α ) tan αð Þ= 1
2
72 4 Solutions of Problems: Trigonometric Equations and Identities
Therefore:
tan 2arc tan
1
2
= tan 2αð Þ= 2 tan αð Þ
1- tan 2 αð Þ =
2 × 12
1- 12
2 =
1
3
4
) tan 2arc tan 1
2
=
4
3
Choice (3) is the answer.
4.27. From trigonometry, we know that:
sin 2 αð Þ þ cos 2 αð Þ= 1
1þ tan 2 αð Þ= 1
cos 2 αð Þ
Therefore:
arc sin
3
5
≜ α ) sin αð Þ= 3
5
) cos αð Þ= 4
5
arc tan
3
4
≜ β ) tan βð Þ= 3
4
) cos βð Þ= 4
5
) sin βð Þ= 3
5
) sin arc sin 3
5
þ arc tan 3
4
= sin αþ βð Þ= sin αð Þ cos βð Þ þ cos αð Þ sin βð Þ= 3
5
×
4
5
þ 4
5
×
3
5
) sin arc sin 3
5
þ arc tan 3
4
=
24
25
Choice (4) is the answer.
4.28. From trigonometry, we know that:
arc cot - αð Þð Þ= π- arc cot αð Þð Þ
arc cot αð Þð Þ þ arc cot 1
α
=
π
2
Therefore:
arc cot -
4
3
- arc cot
3
4
= π- arc cot
4
3
- arc cot
3
4
= π- arc cot
4
3
þ arc cot 3
4
= π-
π
2
) arc cot - 4
3
- arc cot
3
4
=
π
2
Choice (3) is the answer.
4 Solutions of Problems: Trigonometric Equations and Identities 73
4.29. From trigonometry, we know that:
tan αþ βð Þ= tan αð Þ þ tan βð Þ
1- tan αð Þ tan βð Þ
arc tan 5ð Þð Þ≜ α ) tan αð Þ= 5
arc tan
3
2
≜ β ) tan βð Þ= 3
2
arc tan - 1ð Þð Þ= tan - 1 - 1ð Þ= 3π
4
Therefore:
tan arc tan 5ð Þð Þ þ arc tan 3
2
= tan αþ βð Þ= tan αð Þ þ tan βð Þ
1- tan αð Þ tan βð Þ =
5þ 32
1- 152
=
13
2
- 132
= - 1
) arc tan 5ð Þð Þ þ arc tan 3
2
= tan - 1 - 1ð Þ
) arc tan 5ð Þð Þ þ arc tan 3
2
=
3π
4
Choice (3) is the answer.
4.30. From trigonometry, we know that:
sin 2 αð Þ þ cos 2 αð Þ= 1
arc cos
3
5
≜ α ) cos αð Þ= 3
5
arc sin -
4
5
≜ β ) sin βð Þ= - 4
5
Therefore:
sin arc cos
3
5
þ cos arc sin - 4
5
= sin αð Þ þ cos βð Þ= 1- 3
5
2
þ 1- - 4
5
2
=
4
5
þ 3
5
) sin arc cos 3
5
þ cos arc sin - 4
5
=
7
5
Choice (1) is the answer.
4.31. From trigonometry, we know that:
cot θð Þ= tan π
2
- θ
Based on the definition of tan(θ) and cot(θ), we can write:
tan θð Þ= Opposite for θ
Adjacent for θ
=
HA
OH
=
HA
1
=HA
74 4 Solutions of Problems: Trigonometric Equations and Identities
cot θð Þ= tan HOB = Opposite for HOB
Adjacent for HOB
=
HB
OH
=
HB
1
=HB
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:
sec θð Þ= 1
cos θð Þ =
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:
sin θð Þ= cos π
2
- θ
Based on the definition of csc(θ) and sin(θ), we can write:
csc θð Þ= 1
sin θð Þ =
1
cos COB
=
1
Adjacent for COB
Hypotenuse for COB
=
1
OC
OB
=
1
1
OB
=OB
Choice (2) is the answer (Fig. 4.5).
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:
tan αþ βð Þ= tan αð Þ þ tan βð Þ
1- tan αð Þ tan βð Þ
arc tan
2
3
≜ α ) tan αð Þ= 2
3
arc tan
1
5
≜ β ) tan βð Þ= 1
5
arc tan 1ð Þð Þ= π
4
Therefore:
tan αþ βð Þ= tan αð Þ þ tan βð Þ
1- tan αð Þ tan βð Þ =
2
3 þ 15
1- 215
=
13
15
13
15
= 1
) αþ β= arc tan 1ð Þð Þ
) αþ β= π
4
Choice (2) is the answer.
4.35. From trigonometry, we know that:
arc tan αð Þð Þ þ arc tan 1
α
=
π
2
if α> 0
-
π
2
if α< 0
arc cot αð Þð Þ þ arc cot - αð Þð Þ= π
Therefore:
arc tan mð Þð Þ þ arc tan 1
m
þ arc cot mð Þð Þ þ arc cot -mð Þð Þ= π þ
π
2
if m> 0
-
π
2
if m< 0
p
76 4 Solutions of Problems: Trigonometric Equations and Identities
) arc tan mð Þð Þ þ arc tan 1
m
þ arc cot mð Þð Þ þ arc cot -mð Þð Þ=
3π
2
if m> 0
π
2
if m< 0
Choice (2) is the answer.
4.36. From trigonometry, we know that:
cos α- βð Þ= cos αð Þ cos βð Þ þ sin αð Þ sin βð Þ
Therefore:
- 1≤ cos 4xð Þ cos 2xð Þ þ sin 4xð Þ sin 2xð Þ≤ 0
) - 1≤ cos 4x- 2xð Þ≤ 0 ) - 1≤ cos 2xð Þ≤ 0
Since x is an acute angle:
) π
2
≤ 2x≤ π ) π
4
≤ x≤ π
2
Choice (4) is the answer.
4.37. From trigonometry, we know that:
tan α- βð Þ= tan αð Þ- tan βð Þ
1þ tan αð Þ tan βð Þ
Therefore:
tan 2yð Þ= tan xþ yð Þ- x- yð Þð Þ= tan xþ yð Þ- tan x- yð Þ
1þ tan xþ yð Þ tan x- yð Þ =
5- 7
1þ 5× 7 =
- 2
1þ 35
) tan 2yð Þ= - 1
18
Choice (2) is the answer.
4.38. From trigonometry, we know that:
sin αð Þ þ cos αð Þ= 2
p
sin αþ π
4
sin αð Þ-cos αð Þ= 2
p
sin α-
π
4
sin
2π
3
=
3
2
sin
π
6
=
1
2
4 Solutions of Problems: Trigonometric Equations and Identities 77
Therefore:
sin 5π12 þ cos 5π12
sin 5π12 -cos
5π
12
=
2
p
sin 5π12 þ π4
2
p
sin 5π12 -
π
4
=
sin 2π3
sin π6
=
3
p
2
1
2
) sin
5π
12 þ cos 5π12
sin 5π12 -cos
5π
12
= 3
p
Choice (1) is the answer.
4.39. From trigonometry, we know that:
y= a sin mxð Þ ) T = 2π
mj j
Therefore:
) 6= 2π
bπj j ) bj j=
1
3
) b= ± 1
3
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:
) aþ b= 2þ 1
3
) aþ b= 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 sin mxð Þ ) T = 2π
mj j
Therefore:
) 3= 3× 2π
bπj j = 1 ) bj j= 2 ) b= ± 2
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× - 2ð Þ
) 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
2
þ bπx = a cos bπxð Þ
y= a cos mxð Þ ) T = 2π
mj j
Therefore:
) 3:5- - 2:5ð Þ= 3× 2π
bπj j ) 6=
6
bj j ) bj j= 1 ) b= ± 1
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:
cos αð Þ= sin π
2
- α
sin αð Þ= cos π
2
- α
sin 2αð Þ= 2 sin αð Þ cos αð Þ
Therefore:
cos 5 ° cos 10 ° cos 20 °
cos 50 °
=
cos 5 ° cos 10 ° cos 20 °
sin 40 °
=
cos 5 ° cos 10 ° cos 20 °
2 sin 20 ° cos 20 °
=
cos 5 ° cos 10 °
2× 2 sin 10 ° cos 10 °
=
cos 5 °
4 × 2 sin 5 ° cos 5 °
=
1
8 sin 5 °
) cos 5
° cos 10 ° cos 20 °
cos 50 °
=
1
8 cos 85 °
Choice (2) is the answer.
4.43. From trigonometry, we know that:
sin xð Þ= 2 tan
x
2
1þ tan 2 x2
cos xð Þ= 1- tan
2 x
2
1þ tan 2 x2
p
80 4 Solutions of Problems: Trigonometric Equations and Identities
Therefore:
sin xð Þ þ cos xð Þ= 7
5
) 2 tan
x
2
1þ tan 2 x2
þ 1- tan
2 x
2
1þ tan 2 x2
=
7
5
) 2 tan
x
2 þ 1- tan 2 x2
1þ tan 2 x2
=
7
5
) 12 tan 2 x
2
- 10 tan
x
2
þ 2= 0 ) tan x
2
=
10± 102 - 4× 12× 2
24
=
10± 2
24
) tan x
2
=
1
2
or
1
3
Choice (2) is the answer.
4.44. From trigonometry, we know that:
sin 2αð Þ= 2 sin αð Þ cos αð Þ
sin 2 αð Þ þ cos 2 αð Þ= 1
cos 2 αð Þ- sin 2 αð Þ= cos 2αð Þ
cot 2αð Þ= cos 2αð Þ
sin 2αð Þ
In addition, from factoring rule, we know that:
a4 - b4 = a2 - b2 a2 þ b2
Therefore:
sin 4 αð Þ- cos 4 αð Þ
sin αð Þ cos αð Þ =
sin 2 αð Þ- cos 2 αð Þ sin 2 αð Þ þ cos 2 αð Þ
sin αð Þ cos αð Þ =
-cos 2αð Þ× 1
1
2 sin 2αð Þ
) sin
4 αð Þ- cos 4 αð Þ
sin αð Þ cos αð Þ = - 2 cot 2αð Þ
Choice (2) is the answer.
4.45. Based on the information given in the problem, we have:
sin 4 αð Þ þ cos 4 αð Þ= 1
2
From trigonometry, we know that:
sin 2αð Þ= 2 sin αð Þ cos αð Þ
sin 2 αð Þ þ sin 2 αð Þ= 1 ) sin 2 αð Þ þ sin 2 αð Þ 2 = 1
) sin 4 αð Þ þ cos 4 αð Þ þ 2 sin 2 αð Þ cos 2 αð Þ= 1
8
4 Solutions of Problems: Trigonometric Equations and Identities 81
Therefore:
1
2
þ 2 sin 2 αð Þ cos 2 αð Þ= 1 ) 4 sin 2 αð Þ cos 2 αð Þ= 1 ) 2sin αð Þ cos αð Þð Þ2 = 1
) sin 2 2αð Þ= ± 1 ) cos 2 2αð Þ= 0 ) cos
2 2αð Þ
sin 2 2αð Þ = 0
) cot 2 2αð Þ= 0
Choice (3) is the answer.
4.46. From trigonometry, we know that:
sin 2xð Þ= 2 sin xð Þ cos xð Þ
cos 2xð Þ= cos 2 xð Þ- sin 2 xð Þ
Therefore:
sin 3 xð Þ cos xð Þ- cos 3 xð Þ sin xð Þ þ 3 sin 2 xð Þ cos 2 xð Þ
= sin xð Þ cos xð Þ sin 2 xð Þ- cos 2 xð Þ þ 3
4
× 4 sin 2 xð Þ cos 2 xð Þ
=
1
2
sin 2xð Þ -cos 2xð Þð Þ þ 3
4
sin 2 2xð Þ
= -
1
4
sin 4xð Þ þ 3
4
sin 2 2xð Þ
For x=
3π
, we have:
-
1
4
sin 4×
3π
8
þ 3
4
sin 2 2 ×
3π
8
= -
1
4
- 1ð Þ þ 3
4
2
p
2
2
=
1
4
þ 3
8
=
5
8
Choice (2) is the answer.
4.47. From trigonometry, we know that:
sin αð Þ þ sin βð Þ= 2 sin αþ β
2
cos
α- β
2
cos αð Þ þ cos βð Þ= 2 cos αþ β
2
cos
α- β
2
Therefore:
sin 2αð Þ þ sin 5αð Þ þ sin 8αð Þ
cos 2αð Þ þ cos 5αð Þ þ cos 8αð Þ =
sin 8αð Þ þ sin 2αð Þ þ sin 5αð Þ
cos 8αð Þ þ cos 2αð Þ þ cos 5αð Þ
=
2 sin 5αð Þ cos 3αð Þ þ sin 5αð Þ
2 cos 5αð Þ cos 3αð Þ þ cos 5αð Þ =
sin 5αð Þ 2cos 3αð Þ þ 1ð Þ
cos 5αð Þ 2cos 3αð Þ þ 1ð Þ = tan 5αð Þ
82 4 Solutions of Problems: Trigonometric Equations and Identities
α=
π
15¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) tan 5× π
15
= tan
π
3
= 3
p
Choice (3) is the answer.
4.48. From trigonometry, we know that:
sin 2 xð Þ þ cos 2 xð Þ= 1
2 sin xð Þ cos xð Þ= sin 2xð Þ
In addition, from factoring rule, we know that:
aþ bð Þ a- bð Þ= a2 - b2
Therefore:
sin xð Þ-cos xð Þ þ 2ð Þ sin xð Þ-cos xð Þ- 2ð Þ
= sin xð Þ-cos xð Þð Þ2 - 4= sin 2 xð Þ þ cos 2 xð Þ
- 2 sin xð Þ cos xð Þ- 4= 1-sin 2xð Þ- 4
= - 3-sin 2xð Þ
x=
π
12¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) - 3-sin π
6
= - 3-
1
2
= -
7
2
Choice (4) is the answer.
4.49. From trigonometry, we know that:
tan αð Þ cot αð Þ= 1
1þ tan 2 αð Þ= 1
cos 2 αð Þ
1þ cot 2 αð Þ= 1
sin 2 αð Þ
sin 2 αð Þ þ cos 2 αð Þ= 1
Therefore:
4 sin 2 αð Þ cos 2 αð Þ tan αð Þ þ cot αð Þð Þ2
= 4 sin 2 αð Þ cos 2 αð Þ tan 2 αð Þ þ cot 2 αð Þ þ 2 tan αð Þ cot αð Þ
= 4 sin 2 αð Þ cos 2 αð Þ 1þ tan 2 αð Þ þ 1þ cot 2 αð Þ
4 Solutions of Problems: Trigonometric Equations and Identities 83
= 4 sin 2 αð Þ cos 2 αð Þ 1
cos 2 αð Þ þ
1
sin 2 αð Þ
= 4 sin 2 αð Þ cos 2 αð Þ sin
2 αð Þ þ cos 2 αð Þ
sin 2 αð Þ cos 2 αð Þ
= 4 sin 2 αð Þ cos 2 αð Þ 1
sin 2 αð Þ cos 2 αð Þ = 4
Choice (4) is the answer.
4.50. From trigonometry, we know the common solution of the equations below.
sin αð Þ= 0 ) α= kπ, 8k 2 ℤ
cos αð Þ= 0 ) α= kπ þ π
2
,8k 2 ℤ
tan αð Þ= - 1 ) α= kπ- π
4
,8k 2 ℤ
Hence:
sin πxð Þ cos 2 πxð Þ þ sin 2 πxð Þ cos πxð Þ= 0 ) sin πxð Þ cos πxð Þ cos πxð Þ þ sin πxð Þð Þ= 0
)
sin πxð Þ= 0 ) πx= kπ ) x= k - 2≤ x≤ 2¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) x= - 2, - 1, 0, 1, 2
cos πxð Þ= 0 ) πx= kπ þ π
2
) x= k þ 1
2
- 2≤ x≤ 2¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) x= - 3
2
, -
1
2
,
1
2
,
3
2
sin πxð Þ þ cos πxð Þ= 0 ) tan πxð Þ= - 1 ) πx= kπ- π
4
) x= k- 1
4
- 2≤ x≤ 2¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) x= - 5
4
, -
1
4
,
3
4
,
7
4
Therefore, the number of roots of the equation is: 5 + 4 + 4 = 13.
Choice (3) is theanswer.
4.51. From trigonometry, we know that:
y= aþ b cos mxð Þ ) T = 2π
mj j
Therefore:
) 4π= 2π
mj j ) mj j=
1
2
) m= ± 1
2
Based on the graph and the function, the positive value of m is accepted.
) m= 1
2
) y= 1
2
þ 2 cos 1
2
x
The value of function for x= 16π3 is:
y
16π
3
=
1
2
þ 2 cos 1
2
×
16π
3
=
1
2
þ 2 cos 8π
3
=
1
2
þ 2 cos 2π þ 2π
3
=
1
2
þ 2 cos 2π
3
The value of function for x= 7π6 is:
84 4 Solutions of Problems: Trigonometric Equations and Identities
=
1
2
þ 2 cos π- π
3
=
1
2
- 2 cos
π
3
=
1
2
- 2
1
2
= -
1
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= aþ b sin mxð Þ ) T = 2π
mj j
Therefore:
) 2π
3
=
2π
mj j ) mj j= 3 ) m= ± 3
Based on the graph and the given function, the positive value of m is accepted.
) m= 3 ) y= 1-sin 3xð Þ
Choice (4) is the answer (Fig. 4.10).
y
7π
6
= 1-sin 3×
7π
6
= 1-sin
7π
2
= 1-sin 2π þ 3π
2
= 1-sin
3π
2
= 1- - 1ð Þ= 2
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= aþ b sin mxð Þ ) T = 2π
mj j
Therefore:
) 5- 1= 2π
bπj j ) b= ±
1
2
Based on the graph and the given function, the positive value of m is accepted.
) b= - 1
2
) y= aþ sin - π
2
x
By testing the point of (0, 3) in the function, we have:
3= aþ sin - π
2
× 0 ) a= 3 ) y= 3þ sin - π
2
x
The value of function for x= 253 is:
y
25
3
= 3þ sin - π
2
×
25
3
= 3þ sin - 4π- π
6
= 3þ sin - π
6
= 3-
1
2
= 2:5
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= aþ b cos π
2
× 0 ) aþ b= 0 ð1Þ
Based on the function and the graph given in the problem, we can write:
ymax = aþ bj j ) aþ bj j= 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 1ð Þ, 2ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) aþ b= 0
aþ b= 4 ) Impossible
However, for the assumption of b > 0, we have:
Using 1ð Þ, 2ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) aþ 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= 1þ a sin mxð Þ ) T = 2π
mj j
Therefore:
) 4
3
= 2 ×
2π
bπj j ) bj j= 3 ) b= ± 3
Based on the function and the graph given in the problem, we can write:
ymin = 1- aj j ) - 1= 1- aj j ) aj j= 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
b= 3
) aþ b= 5
a= - 2
b= - 3
) aþ b= - 5
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 αþ π
2
= -sin αð Þ
Therefore:
y= a- 2 cos bxþ π
2
= aþ 2 sin bxð Þ
In addition, from trigonometry, we know that:
y= aþ 2 sin bxð Þ ) T = 2π
bj j )
13π
18
-
π
18
=
2π
bj j ) bj j= 3 ) b= ± 3
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 = aþ 2 ) 1= aþ 2 ) a= - 1
Hence:
aþ b= - 1þ 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
tan α- βð Þ= tan αð Þ- tan βð Þ
1þ tan αð Þ tan βð Þ
cot αð Þ= 1
tan αð Þ
In addition, based on the information given in the problem, we have:
tan αþ 20 ° = 3
4
Therefore:
cot 25 ° - α =
1
tan 25 ° - α
=
1
tan 45 ° - αþ 20 °
=
1þ tan 45 ° tan αþ 20 °
tan 45 ° - tan αþ 20 ° =
1þ tan αþ 20 °
1- tan αþ 20 ° =
1þ 34
1- 34
) cot 25 ° - α = 7
Choice (3) is the answer.
4.58. From trigonometry, we know that:
cos αð Þ= - 1- sin 2 αð Þ for an obtuse angle
tan αð Þ= sin αð Þ
cos αð Þ
4 Solutions of Problems: Trigonometric Equations and Identities 89
tan
π
4
= 1
tan αþ βð Þ= tan αð Þ þ tan βð Þ
1- tan αð Þ tan βð Þ
In addition, based on the information given in the problem, we have:
sin αð Þ= 3
5
Therefore:
cos αð Þ= - 1- sin 2 αð Þ= - 1- 3
5
2
= -
4
5
tan αð Þ= sin αð Þ
cos αð Þ =
3
5
- 45
= -
3
4
tan
π
4
þ α = tan
π
4 þ tan að Þ
1- tan π4 tan að Þ
=
1þ tan αð Þ
1- tan αð Þ =
1þ - 34
1- - 34
=
1
4
7
4
) tan π
4
þ α = 1
7
Choice (3) is the answer.
4.59. From trigonometry, we know that:
tan
π
2
- α = cot αð Þ
cot αð Þ= 1
tan αð Þ
tan
π
4
= 1
tan α- βð Þ= tan αð Þ- tan βð Þ
1þ tan αð Þ tan βð Þ
In addition, based on the information given in the problem, we have:
tan
π
2
- α =
2
3
Therefore:
2
3
= tan
π
2
- α = cot αð Þ= 1
tan αð Þ ) tan αð Þ=
3
2
90 4 Solutions of Problems: Trigonometric Equations and Identities
tan
π
4
- α =
tan π4 - tan að Þ
1þ tan π4 tan að Þ
=
1- tan αð Þ
1þ tan αð Þ =
1- 32
1þ 32
=
- 12
5
2
) tan π
4
- α = -
1
5
Choice (2) is the answer.
4.60. From trigonometry, we know that:
tan αþ βð Þ= tan αð Þ þ tan βð Þ
1- tan αð Þ tan βð Þ
In addition, based on the information given in the problem, we have:
tan aþ bð Þ= 2
5
tan a- bð Þ= 3
7
Therefore:
tan 2að Þ= tan aþ bð Þ þ a- bð Þð Þ= tan aþ bð Þ þ tan a- bð Þ
1- tan aþ bð Þ tan a- bð Þ =
2
5 þ 37
1- 25 ×
3
7
=
29
35
29
35
) tan 2að Þ= 1
Choice (4) is the answer.
4.61. From trigonometry, we know that:
sin αð Þ= cos π
2
- α
cos αð Þ= cos - αð Þ
tan
π
4
= 1
tan α- βð Þ= tan αð Þ- tan βð Þ
1þ tan αð Þ tan βð Þ
Therefore:
sin
π
4
þ x = cos π
2
-
π
4
- x = cos
π
4
- x = cos x-
π
4
) 2= sin x-
π
4
sin xþ π4
=
sin x- π4
cos x- π4
4 Solutions of Problems: Trigonometric Equations and Identities 91
) tan x- π
4
= 2 ) tan xð Þ- tan
π
4
1þ tan xð Þ tan π4
= 2
) tan xð Þ- 1
1þ tan xð Þ = 2 ) tan xð Þ- 1= 2þ 2 tan xð Þ
) tan xð Þ= - 3
Choice (1) is the answer.
4.62. From trigonometry, we know that:
tan
π
4
= 1
tan αþ βð Þ= tan αð Þ þ tan βð Þ
1- tan αð Þ tan βð Þ
Moreover, based on the information given in the problem, we have:
αþ β= π
4
If we calculate the tangent value of each side of the abovementioned relation, we will have:
tan αþ βð Þ= tan π
4
) tan αð Þ þ tan βð Þ
1- tan αð Þ tan βð Þ = 1 ) tan αð Þ þ tan βð Þ= 1- tan αð Þ tan βð Þ
Therefore:
1þ tan αð Þð Þ 1þ tan βð Þð Þ= 1þ tan αð Þ þ tan βð Þ þ tan αð Þ tan βð Þ= 1þ 1- tan αð Þ tan βð Þð Þ þ tan αð Þ tan βð Þ
) 1þ tan αð Þð Þ 1þ tan βð Þð Þ= 2
Choice (2) is the answer.
4.63. From trigonometry, we know that:
tan
π
4
= 1
tan αþ βð Þ= tan αð Þ þ tan βð Þ
1- tan αð Þ tan βð Þ
tan α- βð Þ= tan αð Þ- tan βð Þ
1þ tan αð Þ tan βð Þ
tan 2αð Þ= 2 tan αð Þ
1- tan 2 αð Þ
92 4 Solutions of Problems: Trigonometric Equations and Identities
Therefore:
tan
π
4
þ α - tan π
4
- α =
tan π4 þ tan αð Þ
1- tan π4 tan αð Þ
-
tan π4 - tan αð Þ
1þ tan π4 tan αð Þ
=
1þ tan αð Þ
1- tan αð Þ -
1- tan αð Þ
1þ tan αð Þ
=
1þ tan αð Þð Þ2 - 1- tan αð Þð Þ2
1- tan 2 αð Þ
=
4 tan αð Þ
1- tan 2 αð Þ
) tan π
4
þ α - tan π
4
- α = 2 tan 2αð Þ
Choice (1) is the answer.
4.64. From trigonometry, we know that:
tan
π
4
= 1
tan α- βð Þ= tan αð Þ- tan βð Þ
1þ tan αð Þ tan βð Þ
tan 2αð Þ= 2 tan αð Þ
1- tan 2 αð Þ
Moreover, based on the information given in the problem, we have:
tan
π
4
- α =
1
5
) tan
π
4 - tan αð Þ
1þ tan π4 tan αð Þ
=
1- tan αð Þ
1þ tan αð Þ =
1
5
) 5- 5 tan αð Þ= 1þ tan αð Þ ) tan αð Þ= 2
3
) tan 2αð Þ= 2 tan αð Þ
1- tan 2 αð Þ =
2 × 23
1- 23
2 =
12
5
) tan 2αð Þ= 2:4
Choice (3) is the answer.
4.65. From trigonometry, we know that:
tan 2αð Þ= 2 tan αð Þ
1- tan 2 αð Þ
tan α- βð Þ= tan αð Þ- tan βð Þ
1þ tan αð Þ tan βð Þ
4 Solutions of Problems: Trigonometric Equations and Identities 93
Moreover, based on the informationgiven in the problem, we have:
tan αð Þ= 2
tan βð Þ= 1
3
Therefore:
tan 2αð Þ= 2 tan αð Þ
1- tan 2 αð Þ ) tan 2αð Þ=
2 × 2
1- 22
= -
4
3
tan 2α- βð Þ= tan 2αð Þ- tan βð Þ
1þ tan 2αð Þ tan βð Þ
) tan 2α- βð Þ= -
4
3 -
1
3
1þ - 43 13
=
- 53
5
9
) tan 2α- βð Þ= - 3
Choice (1) is the answer.
4.66. From trigonometry, we know that:
cos π- xð Þ= -cos xð Þ
cos αð Þ= cos α0ð Þ ) α= 2kπ ± α0,8k 2 ℤ
Moreover, based on the information given in the problem, we have:
cos xð Þ≠ 0
Therefore:
cos 3xð Þ þ cos xð Þ= 0 ) cos 3xð Þ= -cos xð Þ ) cos 3xð Þ= cos π- xð Þ
) 3x= 2kπ ± π- xð Þ ) 3x= 2kπ þ π- x ) 4x= 2kπ þ π
3x= 2kπ- π þ x ) 2x= 2kπ- π )
x=
kπ
2
þ π
4
x= kπ-
π
2
However:
cos xð Þ≠ 0 ) x= kπ
2
þ π
4
Choice (1) is the answer.
94 4 Solutions of Problems: Trigonometric Equations and Identities
4.67. From trigonometry, we know that:
cot αð Þ= tan π
2
- α
tan αð Þ= tan α0ð Þ ) α= kπ þ α0, 8k 2 ℤ
Therefore:
tan 4xð Þ= cot xð Þ ) tan 4xð Þ= tan π
2
- x
) 4x= kπ þ π
2
- x ) 5x= kπ þ π
2
) x= kπ
5
þ π
10
)
k= - 1 ) x1 = - π10 is not a positive angle
k= 0 ) x2 = π10 is an acute angle
k= 1 ) x3 = 3π10 is an acute angle
k= 2 ) x4 = π2 is not an acute angle
) x2 þ x3 = 2π5
Choice (1) is the answer.
4.68. From trigonometry, we know that:
sin 2 xð Þ þ cos 2 xð Þ= 1
cos xð Þ= cos x0ð Þ ) x= 2kπ ± x0
Therefore:
2 sin 2 xð Þ þ 3 cos xð Þ= 0 ) 2 1- cos 2 xð Þ þ 3 cos xð Þ= 0 ) 2 cos 2 xð Þ- 3 cos xð Þ- 2= 0
) cos 2 xð Þ- 3
2
cos xð Þ- 1= 0 ) cos xð Þ þ 1
2
cos xð Þ- 2ð Þ= 0
) cos xð Þ= -
1
2
) x= 2kπ ± 2π
3
cos xð Þ= 2 ) not acceptable
Choice (1) is the answer.
4.69. From trigonometry, we know that:
sin 2 xð Þ þ cos 2 xð Þ= 1
cos xð Þ= cos x0ð Þ ) x= 2kπ ± x0
Therefore:
2 sin 2 xð Þ= 3 cos xð Þ ) 2 1- cos 2 xð Þ - 3 cos xð Þ= 0 ) 2 cos 2 xð Þ þ 3 cos xð Þ- 2= 0
4 Solutions of Problems: Trigonometric Equations and Identities 95
) cos 2 xð Þ þ 3
2
cos xð Þ- 1= 0 ) cos xð Þ- 1
2
cos xð Þ þ 2ð Þ= 0
) cos xð Þ=
1
2
) x= 2kπ ± π
3
cos xð Þ= - 2 ) not acceptable
Choice (4) is the answer.
4.70. From trigonometry, we know that:
tan αð Þ: cot αð Þ= 1
Now, let us find the intersection point of the lines, as follows:
x tan αð Þ- y cot αð Þ= 1
x tan αð Þ þ y cot αð Þ= 2 )
2x tan αð Þ= 3 ) x= 3
2 tan αð Þ
2y cot αð Þ= 1 ) y= 1
2 cot αð Þ
) xy= 3
2 tan αð Þ ×
1
2 cot αð Þ =
3
4
) y= 3
4x
Choice (4) is the answer.
4.71. From trigonometry, we know that:
sin 2 αð Þ þ cos 2 αð Þ= 1
Based on the information given in the problem, we have:
x= 2- 3 sin αð Þ ) sin αð Þ= x- 2
- 3
y= 1þ 4 cos αð Þ ) cos αð Þ= y- 1
4
Therefore:
) x- 2ð Þ
2
9
þ y- 1ð Þ
2
16
= 1
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 cos αð Þ
y= 4
From trigonometry, we know that:
- 1≤ cos αð Þ≤ 1 ) - 1≤ 2- x
5
≤ 1
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:
cos x- yð Þ= 1
sin xþ yð Þ= 1
The common solution of the equations can be calculated as follows:
)
x- y= 2kπ
xþ y= 2kπ þ π
2
0< x, y< 2π¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼)
x- y= 0
xþ y= π
2
or
5π
2
) y= π
4
or
5π
4
Choice (2) is the answer.
4.74. From trigonometry, we know that:
tan αþ βð Þ= tan αð Þ þ tan βð Þ
1- tan αð Þ tan βð Þ
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
c
a, respectively.
Based on the information given in the problem, we have:
tan 2 xð Þ þ mþ 2ð Þ tan xð Þ þ 2m- 2= 0
αþ β= π
4
Therefore:
tan :ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) tan αþ βð Þ= 1 ) tan αð Þ þ tan βð Þ
1- tan αð Þ tan βð Þ = 1
) sum of the roots of the quadratic equation
1- product of the roots of the quadratic equation
=
- mþ2ð Þ
1
1- 2m- 21
=
-m- 2
3- 2m
= 1
) -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 αð Þ þ cos 6 αð Þ= 1- 3 sin 2 αð Þ cos 2 αð Þ
sin 4 αð Þ þ cos 4 αð Þ= 1- 2 sin 2 αð Þ cos 2 αð Þ
Therefore:
sin 6 αð Þ þ cos 6 αð Þ þ 3 sin 2 αð Þ cos 2 αð Þ
sin 4 αð Þ þ cos 4 αð Þ þ 2 sin 2 αð Þ cos 2 αð Þ =
1- 3 sin 2 αð Þ cos 2 αð Þ þ 3 sin 2 αð Þ cos 2 αð Þ
1- 2 sin 2 αð Þ cos 2 αð Þ þ 2 sin 2 αð Þ cos 2 αð Þ
) sin
6 αð Þ þ cos 6 αð Þ þ 3 sin 2 αð Þ cos 2 αð Þ
sin 4 αð Þ þ cos 4 αð Þ þ 2 sin 2 αð Þ cos 2 αð Þ = 1
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 ° cot 60 ° þ -cot 60 ° - tan 30 ° =
2
p
2 ×
- 3
p
2 þ - 2
p
2
3
p
2
3
p
3 ×
3
p
3 þ - 3
p
3 -
3
p
3
) sin 45
° -cos 30 ° þ -cos 45 ° sin 60 °
tan 30 ° cot 60 ° þ -cot 60 ° - tan 30 ° =
- 3 6
p
4
Choice (2) is the answer.
98 4 Solutions of Problems: Trigonometric Equations and Identities
4.77. From trigonometry, we know that:
cot xþ yð Þ= cot xð Þ cot yð Þ- 1
cot xð Þ þ cot yð Þ
Based on the information given in the problem, we have:
xþ y= kπ þ π
4
k= 0¼¼¼¼¼¼¼¼¼¼¼¼) xþ y= π
4
cot :ð Þ¼¼¼¼¼¼¼¼¼¼¼¼) cot xð Þ cot yð Þ- 1
cot xð Þ þ cot yð Þ = 1
) 1þ cot xð Þ þ cot yð Þ= cot xð Þ cot yð Þ ð1Þ
On the other hand, we can write:
1þ cot xð Þð Þ 1þ cot yð Þð Þ= 1þ cot xð Þ þ cot yð Þð Þ þ cot xð Þ cot yð Þ ð2Þ
Solving (1) and (2):
1þ cot xð Þð Þ 1þ cot yð Þð Þ= cot xð Þ cot yð Þ þ cot xð Þ cot yð Þ
) 1þ cot xð Þð Þ 1þ cot yð Þð Þ= 2 cot xð Þ cot yð Þ
Choice (4) is the answer.
4.78. From trigonometry, we know that:
tan
3π
2
- x = cot xð Þ
cos
4π
3
= -cos
π
3
= -
1
2
tan xð Þ cot xð Þ= 1
cot xð Þ= cos xð Þ
sin xð Þ
cos xð Þ= cos x0ð Þ ) x= 2kπ ± x0
Therefore:
sin xð Þ- tan xð Þð Þ tan 3π
2
- x = cos
4π
3
) sin xð Þ- tan xð Þð Þ cot xð Þ= - 1
2
) sin xð Þ cot xð Þ- tan xð Þ cot xð Þ= - 1
2
) cos xð Þ- 1= - 1
2
) cos xð Þ= 1
2
) x= 2kπ ± π
3
Choice (3) is the answer.
p
4 Solutions of Problems: Trigonometric Equations and Identities 99
4.79. From trigonometry, we know that:
sin αþ βð Þ= sin αð Þ cos βð Þ þ sin αð Þ cos βð Þ
cos αþ βð Þ= cos αð Þ cos βð Þ-sin αð Þ sin βð Þ
tan xð Þ= sin xð Þ
cos xð Þ
tan xð Þ= tan x0ð Þ ) x= kπ þ x0
Therefore:
sin 2xð Þ sin xð Þ þ cos xð Þð Þ= cos 2xð Þ cos xð Þ-sin xð Þð Þ
) sin 2xð Þ sin xð Þ þ sin 2xð Þ cos xð Þ= cos 2xð Þ cos xð Þ-cos 2xð Þ sin xð Þ
) sin 2xð Þ cos xð Þ þ cos 2xð Þ sin xð Þ= cos 2xð Þ cos xð Þ-sin 2xð Þ sin xð Þ
) sin 2xþ xð Þ= cos 2xþ xð Þ ) sin 3xð Þ= cos 3xð Þ
×
1
cos 3xð Þ& cos 3xð Þ≠ 0¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) tan 3xð Þ= 1
) 3x= kπ þ π
4
) x= kπ
3
þ π
12
k= 0, 1, 2&x 2 0, π½ ]¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) x1 = π12 , x2 =
5π
12
, x3 =
9π
12
) x1 þ x2 þ x3 = 5π4
Choice (2) is the answer.
4.80. From trigonometry, we know that:
sin
5π
2
þ x = sin π
2
þ x = cos xð Þ
sin αþ βð Þ= sin αð Þ cos βð Þ þ sin αð Þ cos βð Þ
sin
π
4
= cos
π
4
=
2
2
sin xð Þ= sin x0ð Þ )
x= 2kπ þ x0
x= 2kπ þ π- x0
Therefore:
) 2
p
sin
π
4
- x = 1þ cos xð Þ ) 2
p
sin
π
4
cos xð Þ-cos π
4
sin xð Þ = 1þ cos xð Þ
100 4 Solutions of Problems: Trigonometric Equations and Identities
) cos xð Þ-sin xð Þ= 1þ cos xð Þ ) sin xð Þ= - 1 )
x= 2kπ þ - π
2
x= 2kπ þ π- - π
2
)
x= 2kπ-
π
2
x= 2kπ þ 3π
2
) x= 2kπ- π
2
Choice (3) is the answer.
4.81. From trigonometry, we know that:
tan
π
3
= 3
p
tan xð Þ= sin xð Þ
cos xð Þ
cos
π
3
=
1
2
cos α- βð Þ= cos αð Þ cos βð Þ þ sinαð Þ sin βð Þ
cos xð Þ= cos x0ð Þ ) x= 2kπ ± x0
Therefore:
cos 2xð Þ þ 3
p
sin 2xð Þ= 1 ) cos 2xð Þ þ tan π
3
sin 2xð Þ= 1 ) cos 2xð Þ þ sin
π
3
cos π3
sin 2xð Þ= 1
) cos 2xð Þ cos π
3
þ sin π
3
sin 2xð Þ= cos π
3
) cos 2x- π
3
= cos
π
3
) 2x- π
3
= 2kπ ±
π
3
)
2x-
π
3
= 2kπ þ π
3
) x= kπ þ π
3
2x-
π
3
= 2kπ-
π
3
) x= kπ
Choice (4) is the answer.
4.82. From trigonometry, we know that:
sin a sin b=
1
2
cos a- bð Þ-cos aþ bð Þ½ ]
sin a cos b=
1
2
sin aþ bð Þ þ sin a- bð Þ½ ]
cos aþ cos b= 2 cos aþ b
2
cos
a- b
2
References 101
sin a-sin b= 2 cos
aþ b
2
sin
a- b
2
Therefore:
cos 3αþ sin α sin 2α
sin 3α-sin 2α cos α
×
sin α
cos α
=
cos 3αþ 12 cos α-cos 3α½ ]
sin 3α- 12 sin 3αþ sin α½ ]
×
sin α
cos α
=
1
2 cos 3αþ 12 cos α
1
2 sin 3α-
1
2 sin α
×
sin α
cos α
=
cos 3αþ cos α
sin 3α-sin α
×
sin α
cos α
=
2 cos 2α cos α
2 cos 2α sin α
×
sin α
cos α
) cos 3αþ sin α sin 2α
sin 3α-sin 2α cos α
×
sin α
cos α
= 1
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.
5Problems: 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 xð Þ ¼ 10 xj j x≠ 0 
0 x ¼ 0 
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 xð Þ ¼ xj j x≠ 0 
1 x ¼ 0 
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
http://crossmark.crossref.org/dialog/?doi=10.1007/978-3-031-45028-0_5&domain=pdf
https://doi.org/10.1007/978-3-031-45028-0_5#DOI
104 5 Problems: Limits and Continuities
5.3. Calculate the value of k if the function below is continuous at x ¼ 2. 
f xð Þ ¼ xþ 2ð Þ - x½ � x< 2 
xþ k 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 xð Þ ¼ xj j x½ � þ a x< -2 
xj j þ x½ � 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 
x→ - 1ð Þþ 
x½ � þ 1 
x2 - 1 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) -
1 
2 
2) 0 
3) 
1 
2 
4) 1 
5.6. Calculate the limit of the following function if x → 2+ . 
f xð Þ ¼ x þ 4
-x½ �-3 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 1 
2) -1 
3) 2 
4) -2
5 Problems: Limits and Continuities 105
5.7. Determine the value of the following limit: 
lim 
x→ 0-
x þ 2 
x½ �
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 
x½ � þ 3x 
x½ �- 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 xð Þ ¼ xþ x
3
p 
x- x3
p 
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-
x½ �
x 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 1 
2) -1 
3) 0 
4) 1
106 5 Problems: Limits and Continuities
5.11. Determine the limit of the function below if x → 0+ . 
f xð Þ ¼ x
2 - 1ð Þ xp 
x x
p þ 1ð Þx 
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 
x
-
1 
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 lim 
x→ 1þ 
f xð Þ- lim 
x→ 1-
f xð Þ. 
f xð Þ ¼ 2x 
2x½ � þ 2 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) -1 
2) -
1 
6 
3) 
2 
3 
4) 1 
5.14. Calculate the value of lim 
x→ 2þ 
x½ �- 2ð Þ x½ �. 
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 xð Þ ¼ x½ �- 4 
x2 - 16
2)
5 Problems: Limits and Continuities 107
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 0 
1 
8 
3) 1 
4) -1 
5.16. Determine the value of the limit below. 
lim 
x→ 1-
1- x3 
arc cos xð Þð Þ 
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 
tan xð Þ- tan 3xð Þ þ tan 2xð Þ 
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: 
lim 
x→ 3 
9- x2 
2- xþ 1p 
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 xð Þ ¼ sin xð Þ 
x 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large
108 5 Problems: Limits and Continuities
1) Undefined 
2) 0 
3) 1 
4) 1 
5.20. Determine the value of the limit below. 
lim 
x→ 0 
x2½ �- x2 
x tan xð Þ 
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→þ1x sin 
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
- 3xþ 4 - xp 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1 
3 
2) -
1 
3 
3) 1 
4) -1 
5.23. Calculate the value of the limit below. 
lim 
x→ 0þ 
xþ 1ð Þ xp 
x2 - x 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 0
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 xð Þ ¼ x 
x- 1þ x2 þ x- 1p 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 1 
2) 0 
3) -
1 
2 
4) 
1 
2 
5.25. Calculate the value of lim 
x→0 
x cot xð Þ. 
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 xð Þ ¼ x
2 þ ax x> 1 
x- 3 x< 1 
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-
x3 - 8 
x- 2x
p 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) -24 
2) -16 
3) 16 
4) 24
110 5 Problems: Limits and Continuities
5.28. Calculate the value of the limit below. 
lim 
x→ 0-
x½ � þ x
- x½ � þ x 
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 
sin 3xð Þ þ sin 7xð Þ 
3xþ tan 2xð Þ 
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: 
lim 
x→ 0 
xþ 3p - 3p 
x 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
3
p 
3 
2) 
3
p 
6 
3) 
3
p 
2 
4) 
3
p 
9 
5.31. Calculate the value of the following limit: 
lim 
x→ 0 
1- cos xð Þ 
sin xð Þ 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 0 
2) 1 
3) -1
5 Problems: Limits and Continuities 111
4) 2
p 
5.32. Calculate the value of the limit below. 
lim 
x→ 0 
5x- sin xð Þ 
2x þ cos xð Þ- 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 
x- 2j j þ 5x 
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 
x→ π 2ð Þþ 
sin xð Þ þ cos xð Þ 
cos xð Þ 
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 
arc sin xð Þð Þð Þ3 
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.
2
112 5 Problems: Limits and Continuities
lim
x→ -1 xþ 1 x 
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: 
lim 
x→ 3þ 
x- 4 
x2 - 4xþ 3p 
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 x2 þ 4x- 10
p
. 
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. 
lim 
x→ 2 
4- x2 
6- 2 x2 þ 5p 
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 
x→ 0 
sin 2xð Þ 
x þ 1p - 1 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 2 
2) 4
5 Problems: Limits and Continuities 113
3) 3 
4) 1 
5.41. Calculate the limit of x4 þ 2x2 þ xp - 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 
xþ 3 
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: 
lim 
x→ 1 2 
tan πx 2 - 1 
cos πxð Þ 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 1 
2) -1 
3) 2 
4) -2 
5.44. Calculate the limit of xþ 5p - xþ 1p 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: 
lim 
x→ π 2 
tan 2xð Þ cos xð Þ 
1þ cos 2xð Þ 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 1 
2) 
1 
2 
3) -1
1)
2)
114 5 Problems: Limits and Continuities
4) -
1 
2 
5.46. Determine the value of the following limit: 
lim 
x→ 0-
tan 2xð Þ 
1- cos xð Þ 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) - 2 2
p 
2) - 2
p 
3) 2
p 
4) 2 2
p 
5.47. Calculate the value of lim
x→ -1 nþ 1000
3
p
- n- 203
p
. 
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. 
lim 
x→ πþ 
sin π sin xð Þð Þ sin x 2 
1þ cos xð Þ 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) - π 2
p 
2) -2π 
3) π2 
4) π2 2
p 
5.49. Calculate the value of the following limit. 
lim 
x→ 0 
1 þ x23p - 1- 2x4p 
2x2 þ 2x 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1 
4
-
1 
4 
3) 1 
4) -1 
5.50. Calculate the limit of the function below if x → 0.
3)
4)
5 Problems: Limits and Continuities 115
f xð Þ ¼ sin
2 xð Þ þ sin xð Þ þ cos 2 xð Þ- cos xð Þ 
sin 2 xð Þ- sin xð Þ þ cos 2 xð Þ- cos xð Þ 
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 
cos mxð Þ- cos nxð Þ 
x2 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) n2 + m2 
2) n2 - m2 
n2 -m2 
2 
n2 þ m2 
2 
5.52. Calculate the value of the limit below. 
lim 
x→ 0 
sin xð Þ- x 
tan xð Þ- x 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -
1 
2 
2) 
1 
2 
3) -
1 
4 
4) 
1 
4 
5.53. Calculate the value of the following limit: 
lim
x→ π 
1þ cos 3 xð Þ 
1- cos 2 xð Þ 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
3 
2 
2) -
3 
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 xð Þ ¼ x - 1ð Þ 1 x½ �, x 2 ℝ- 0f g 
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. 
f nð Þ ¼ 3n
2 
5n
p 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 0 
2) 1 
3) 1 
4) 
3 
5
p 
5.56. Determine the value of the limit below. 
lim 
x→ 0 
x3 - sin xð Þ 1- cos xð Þð Þ 
x3 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 0 
2) 
1 
2 
3) 1 
4) -1 
5.57. Calculate the value of the following limit: 
lim 
x→ 1-
arc cos xð Þ 
1- x
p 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2
p 
2) - 2
p 
3) -
2
p 
2
5 Problems: Limits and Continuities 117
4) 
2
p 
2 
5.58. Determine the value of n in the following equation: 
lim 
x→ 1 
x2 - 1 cot xn - 1ð Þ ¼ 1 
2 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 8 
2) 4 
3) 
1 
8 
4) 
1 
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-
sin xð Þ- x 
1 
2 sin 2xð Þ- x cos xð Þ 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 02) 1 
3) 1 
4) -1 
5.61. Calculate the limit of the following function if x→ 
π 
4 
: 
f xð Þ ¼ 1- tan xð Þ
3 
1- 2 sin 2 xð Þ 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
1 
3 
2) 3
118 5 Problems: Limits and Continuities
3) 
1 
2 
4) 2 
5.62. Calculate the value of the limit below. 
lim 
x→ 0 
1- cos 3 xð Þ 
sin xð Þ tan 2xð Þ 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 4 
2) -4 
3) 
3 
4 
4) -
3 
4 
5.63. Determine the value of the following limit: 
lim 
x→ 0þ 
1- cos xð Þ 
1- cos x
pð Þ 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) 0 
2) 1 
3) 
1 
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.
6Solutions 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 xð Þ ¼ 10 xj j x≠ 0 
0 x ¼ 0 
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 
x→ x0þ 
f xð Þ ¼ lim
x→ x0 -
f xð Þ ¼ f x0ð Þ 
As can be seen, for the f (x), we have: 
lim 
x→ 0þ 
10 xj j ¼ 0 
lim 
x→ 0-
10 xj j ¼ 0 
f 0ð Þ ¼ 0 
) lim 
x→ 0þ 
10 xj j ¼ lim 
x→ 0-
10 xj j ¼ f 0ð Þ 
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 xð Þ ¼ xj j x≠ 0 
1 x ¼ 0 
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
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120 6 Solutions of Problems: Limits and Continuities
A function is continuous at the given point of x0 if: 
lim 
x→ x0þ 
f xð Þ ¼ lim
x→ x0 -
f xð Þ ¼ f x0ð Þ 
As can be seen, for the abovementioned function, we have: 
lim 
x→ 0þ 
xj j ¼ 0 
lim 
x→ 0-
xj j ¼ 0 
f 0ð Þ ¼ 1 
) lim 
x→ 0þ 
xj j ¼ lim 
x→ 0-
xj j≠ f 0ð Þ 
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 xð Þ ¼ xþ 2ð Þ - x½ ] x< 2 
xþ k x≥ 2 
Therefore, we must have: 
lim 
x→ 2þ 
f xð Þ ¼ lim 
x→ 2-
f xð Þ ¼ f 2ð Þ 
For f (x), we have: 
) 
lim 
x→ 2þ 
f xð Þ ¼ lim 
x→ 2þ 
x þ kð Þ ¼ 2þ k 
lim 
x→ 2-
f xð Þ ¼ lim 
x→ 2-
x þ 2ð Þ - x½ ] ¼ 2- þ 2ð Þ - 2-½ ] ¼ - 8 
f 2ð Þ ¼ 2þ 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 xð Þ ¼ xj j x½ ] þ a x< - 2 
xj j þ x½ ] x≥ - 2 
Therefore, we must have: 
lim 
x→ - 2þ 
f xð Þ ¼ lim 
x→ - 2-
f xð Þ ¼ f - 2ð Þ
6 Solutions of Problems: Limits and Continuities 121
For f (x), we can write: 
) 
lim 
x→ - 2þ 
f xð Þ ¼ lim 
x→ - 2þ 
xj j þ x½ ]ð Þ ¼ 2þ - 2ð Þ ¼ 0 
lim 
x→ - 2-
f xð Þ ¼ lim 
x→ - 2-
xj j x½ ] þ a ¼ 2ð Þ - 3ð Þ þ a ¼ - 6 þ a 
f - 2ð Þ ¼ - 2j j þ - 2½ ] ¼ 0 
) - 6þ a ¼ 0 
) a ¼ 6 
Choice (4) is the answer. 
6.5. The problem can be solved as follows: 
lim 
x→ - 1ð Þþ 
x½ ] þ 1 
x2 - 1 
¼ - 1ð Þ þ 1 
1- - 1 
¼ 0 
0-
¼ 0 
Choice (2) is the answer. 
6.6. The problem can be solved as follows: 
lim 
x→ 2þ 
xþ 4
- x½ ]- 3 ¼ 
2þ 4
- 3- 3 
¼ -1 
Choice (2) is the answer. 
6.7. The problem can be solved as follows: 
lim 
x→ 0-
xþ 2 
x½ ] ¼ 
0 þ 2
- 1 
¼ -2 
Choice (2) is the answer. 
6.8. The problem can be solved as follows: 
lim
x→ -1 
x½ ] þ 3x 
x½ ]- 3x ¼ limx→ -1 
4x
- 2x 
¼ lim
x→ -1 - 2ð Þ ¼ - 2 
Choice (2) is the answer. 
6.9. The problem can be solved as follows: 
lim 
x→ 0 
xþ x3p 
x- x3
p ¼ lim 
x→ 0 
x3
p 
x23
p 
þ 1 
x3
p
x23
p
- 1 
¼ lim 
x→ 0 
x23
p 
þ 1 
x23
p
- 1 
¼ -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-
x½ ]
x 
¼ - 1 
0-
¼ þ1 
Choice (1) is the answer. 
6.11. The problem can be solved as follows: 
lim 
x→ 0þ 
x2 - 1ð Þ xp 
x x
p þ 1ð Þx ¼ lim x→ 0þ 
x2 - 1ð Þ 
x x
p þ 1ð Þ xp ¼
- 1 
0þ 
¼ -1 
Choice (2) is the answer. 
6.12. The problem can be solved as follows: 
lim 
x→ 0þ 
1 
x
-
1 
x3 
¼ lim 
x→ 0þ 
x2 - 1 
x3 
¼ - 1 
0þ 
¼ -1 
Choice (2) is the answer. 
6.13. The problem can be solved as follows: 
lim 
x→ 1þ 
f xð Þ- lim 
x→ 1-
f xð Þ ¼ lim 
x→ 1þ 
2x 
2x½ ] þ 2 - lim x→ 1-
2x 
2x½ ] þ 2 ¼ lim x→ 1þ 
2x 
2þ 2 - lim x→ 1-
2x 
1þ 2 
¼ lim 
x→ 1þ 
x 
2
- lim 
x→ 1-
2 
3 
x ¼ 1 
2
-
2 
3
¼ -1 
6 
Choice (2) is the answer. 
6.14. The problem can be solved as follows: 
lim 
x→ 2þ 
x½ ]- 2ð Þ x½ ] ¼ lim 
x→ 2þ 
2- 2ð Þ x 2 ¼ lim 
x→ 2þ 
0 ¼ 0 
Choice (3) is the answer. 
6.15. The problem can be solved as follows: 
lim 
x→ 4-
x½ ]- 4 
x2 - 16 
¼ lim 
x→ 4-
3- 4 
16- - 16 
¼ - 1 
0-
¼ þ1 
Choice (3) is the answer. 
6.16. From trigonometry and calculus, we know that: 
arc cos 1-ð Þð Þ ¼ 0þ 
d 
dx 
arc cos xð Þð Þ ¼ - 1 
1- x2
p
6 Solutions of Problems: Limits and Continuities 123
The problem can be solved as follows: 
lim 
x→ 1-
1- x3 
arc cos xð Þð Þ ¼ 
0þ 
0þ 
H¼¼¼¼¼¼) lim 
x→ 1-
d 
dx 1- x
3ð Þ 
d 
dx arc cos xð Þð Þð Þ
¼ lim 
x→ 1-
- 3x2
- 1 
1- x2
p ¼ lim x→ 1- 3x
2 1- x2 
p 
¼ 0 
Choice (3) is the answer. 
6.17. From application of Taylor series in limit, we know that: 
lim 
x→ 0 
tan xð Þ - xþ x
3 
3 
The problem can be solved as follows: 
lim 
x→ 0 
tan xð Þ- tan 3xð Þ þ tan 2xð Þ 
x3
- lim 
x→ 0 
xþ x3 3 - 3xþ 3xð Þ
3 
3 þ 2xþ 2xð Þ
3 
3 
x3 
¼ lim 
x→ 0
- 6x3 
x3 
¼ lim 
x→ 0
- 6ð Þ ¼ -6 
Choice (1) is the answer. 
6.18. The problem can be solved as follows: 
lim 
x→ 3 
9- x2 
2- xþ 1p ¼ 
0 
0 
H¼¼¼¼) lim 
x→ 3 
d 
dx 9- x
2ð Þ 
d 
dx 2- x þ 1
p ¼ lim 
x→ 3
- 2x
- 1 
2 xþ1p 
¼ lim 
x→ 3 
4x xþ 1p ¼ 24 
Choice (4) is the answer. 
6.19. From trigonometry, we know that:
- 1≤ sin xð Þ≤ 1 
The problem can be solved as follows: 
lim 
x→þ1 
sin xð Þ 
x 
¼ lim 
x→þ1 sin xð Þ x
1 
x 
¼ Bounded quantityð Þ x 0 ¼ 0 
Choice (2) is the answer.
124 6 Solutions of Problems: Limits and Continuities
6.20. From application of Taylor series in limit, we know that: 
lim 
x→ 0 
tan xð Þ - x 
The problem can be solved as follows: 
lim 
x→ 0 
x2½ ]- x2 
x tan xð Þ ¼ lim x→ 0 
0- x2 
x tan xð Þ - lim x→ 0
- x2 
xx x ¼ lim x→ 0 - 1ð Þ ¼ -1 
Choice (2) is the answer. 
6.21. From application of Taylor series in limit, we know that: 
lim 
x→þ1 sin 
1 
x
- 1 
x 
The problem can be solved as follows: 
lim 
x→þ1x sin 
1 
x 
¼ lim 
x→þ1xx
1 
x 
¼ lim 
x→þ11 ¼ 1 
Choice (1) is the answer. 
6.22. From calculus, we know that: 
lim
x→ ±1 amx
m þ am- 1xm- 1 þ . . .þ a2x2 þ a1xþ a0 - amxm 
or: 
lim
x→ ±1 amx
m þ anxnð Þ - amxm if m> n 
Therefore: 
lim
x→ -1 
x2 þ x- 1
- 3x þ 4 - xp - limx→ -1 
x2
- 3x 
¼ lim
x→ -1 -
x 
3 
¼ þ1 
Choice (3) is the answer. 
6.23. From calculus, we know that: 
lim 
x→ 0 
amx
m þ am- 1xm- 1 þ . . .þ am- nxm- n þ am- n- 1xm- n- 1 - am- n- 1xm- n- 1 
or: 
lim 
x→ 0 
amx
m þ anxnð Þ - anxn if m> n 
The problem can be solved as follows: 
lim 
x→ 0þ 
x þ 1ð Þ xp 
x2 - x 
¼ lim 
x→ 0þ 
x x
p þ xp 
x2 - x
- lim 
x→ 0þ 
x
p
- x 
¼ lim 
x→ 0þ
- 1 
x
p ¼ -1 
Choice (4) is the answer.
6 Solutions of Problems: Limits and Continuities125
6.24. From calculus, we know that: 
lim 
x→þ1 x
2 þ ax þ b - xþ a 
2 
The problem can be solved as follows: 
lim 
x→þ1 
x 
x- 1þ x2 þ x- 1p - lim x→þ1 
x 
x- 1þ xþ 1 2 
¼ lim 
x→þ1 
x 
2x- 1 2
- lim 
x→þ1 
x 
2x 
¼ lim 
x→þ1 
1 
2 
¼ 1 
2 
Choice (4) is the answer. 
6.25. From trigonometry, we know that: 
cot xð Þ ¼ 1 
tan xð Þ 
From application of Taylor series in limit, we know that: 
lim 
x→ 0 
tan xð Þ - x 
The problem can be solved as follows: 
lim 
x→ 0 
x cot xð Þ ¼ lim 
x→ 0 
x 
tan xð Þ - lim x→ 0 
x 
x 
¼ 1 
Choice (3) is the answer. 
6.26. As we know, the limit of a function at the point of x0 exits if: 
lim
x→ x0 -
f xð Þ ¼ lim 
x→ x0þ 
f xð Þ ) lim 
x→ 1-
f xð Þ ¼ lim 
x→ 1þ 
f xð Þ ð1Þ 
Therefore: 
lim 
x→ 1-
f xð Þ ¼ lim 
x→ 1-
x- 3ð Þ ¼ 1- 3 ¼ - 2 ð2Þ 
lim 
x→ 1þ 
f xð Þ ¼ lim 
x→ 1þ 
x2 þ ax ¼ 1þ a ð3Þ 
Using 1ð Þ, 2ð Þ, 3ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) - 2 ¼ 1 þ a ) a ¼ -3 
Choice (3) is the answer. 
6.27. The problem can be solved as follows: 
lim 
x→ 2-
x3 - 8 
x- 2x
p ¼ lim 
x→ 2-
- x3 - 8ð Þ 
x- 2x
p ¼ 0
þ 
0-
H¼¼¼¼¼¼) lim 
x→ 2-
d 
dx - x
3 - 8ð Þð Þ 
d 
dx x- 2x
p ¼ lim 
x→ 2-
- 3x2 
1- 1 
2x
p ¼
- 3x 22 
1- 1 
2x2p 
¼ - 12 1 
2 
¼ -24 
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-
x½ ] þ x
- x½ ] þ x ¼ lim x→ 0-
- 1þ x 
0þ x ¼
- 1þ 0-
0þ 0- ¼
- 1 
0-
¼ þ1 
Choice (1) is the answer. 
6.29. The problem can be solved as follows: 
lim 
x→ 0 
sin 3xð Þ þ sin 7xð Þ 
3xþ tan 2xð Þ ¼ 
0 
0 
H¼¼¼¼¼¼) lim 
x→ 0 
d 
dx sin 3xð Þ þ sin 7xð Þð Þ 
d 
dx 3x þ tan 2xð Þð Þ 
¼ lim 
x→ 0 
3 cos 3xð Þ þ 7 cos 7xð Þ 
3þ 2 1 þ tan 2 2xð Þð Þ ¼ 
3þ 7 
3 þ 2 ¼ 2 
Choice (2) is the answer. 
6.30. The problem can be solved as follows: 
lim 
x→ 0 
x þ 3p - 3p 
x
¼ lim 
x→ 0 
xþ 3p - 3p 
x
x x þ 3
p þ 3p 
xþ 3p þ 3p ¼ lim x→ 0 
xþ 3- 3 
x xþ 3p þ 3p 
¼ lim 
x→ 0 
1 
xþ 3p þ 3p ¼
1 
3
p þ 3p ¼ 
1 
2 3
p ¼ 3
p 
6 
Choice (2) is the answer. 
6.31. The problem can be solved as follows: 
lim 
x→ 0 
1- cos xð Þ 
sin xð Þ ¼ 
0 
0 
H¼¼¼¼¼¼) lim 
x→ 0 
d 
dx 1- cos xð Þð Þ 
d 
dx sin xð Þ 
¼ lim 
x→ 0 
sin xð Þ 
cos xð Þ ¼ 
0 
1 
¼ 0 
Choice (1) is the answer. 
6.32. The problem can be solved as follows: 
lim 
x→ 0 
5x- sin xð Þ 
2x þ cos xð Þ- 1 ¼ 
0 
0 
H¼¼¼¼¼¼) lim 
x→ 0 
d 
dx 5x- sin xð Þð Þ 
d 
dx 2xþ cos xð Þ- 1ð Þ
¼ lim 
x→ 0 
5- cos xð Þ 
2- sin xð Þ ¼ 
5- 1 
2- 0 
¼ 2 
Choice (2) is the answer.
6 Solutions of Problems: Limits and Continuities 127
6.33. The problem can be solved as follows: 
lim 
x→ 2-
x3 - 8 
x- 2j j þ 5x ¼ lim x→ 2-
x- 2ð Þ x2 þ 2xþ 4ð Þ
- x- 2ð Þ þ 5x ¼ lim x→ 2- - x
2 - 2x- 4þ 5x 
lim 
x→ 2-
- x2 þ 3x- 4 ¼ - 4þ 6- 4 ¼ - 2 
Choice (2) is the answer. 
6.34. The problem can be solved as follows: 
lim 
x→ π2ð Þþ 
sin xð Þ þ cos xð Þ 
cos xð Þ ¼
1- þ 0-
0-
¼ 1
-
0-
¼ -1 
Choice (2) is the answer. 
6.35. The problem can be solved as follows: 
lim 
x→ 0 
arc sin xð Þð Þ - x 
lim 
x→ 0 
3x4 þ 2x3 - 2x3 
lim 
x→ 0 
3x4 þ 2x3 
arc sin xð Þð Þð Þ3 - lim x→ 0 
2x3 
x3 
¼ lim 
x→ 0 
2 ¼ 2 
Choice (2) is the answer. 
6.36. The problem can be solved as follows: 
lim
x→ -1 
2 
xþ 1 x ¼ limx→ -1 0
-½ ]x ¼ - 1ð Þ -1ð Þ ¼ þ1 
Choice (1) is the answer. 
6.37. The problem can be solved as follows: 
lim 
x→ 3þ 
x- 4 
x2 - 4xþ 3p ¼ lim x→ 3þ 
x- 4 
x- 3ð Þ x- 1ð Þ ¼
- 1 
0þ 
¼ -1 
Choice (2) is the answer. 
6.38. From calculus, we know that: 
lim
x→ ±1 x
2 þ axþ b - xþ a 
2 
The problem can be solved as follows: 
lim
x→ -1 x þ x2 þ 4x- 10 - limx→ -1 x þ xþ 2j jð Þ ¼ limx→ -1 x- x- 2ð Þ ¼ limx→ -1 - 2ð Þ ¼ - 2 
Choice (2) is the answer.
128 6 Solutions of Problems: Limits and Continuities
6.39. The problem can be solved as follows: 
lim 
x→ 2 
4- x2 
6- 2 x2 þ 5p ¼ 
0 
0 
H¼¼¼¼¼¼) lim 
x→ 2 
d 
dx 4- x
2ð Þ 
d 
dx 6- 2 x
2 þ 5p ¼ lim x→ 2
- 2x
- 2x 2x 
2 x2þ5p 
¼ lim 
x→ 2 
x2 þ 5 ¼ 3 
Choice (3) is the answer. 
6.40. The problem can be solved as follows: 
lim 
x→ 0 
sin 2xð Þ 
x þ 1p - 1 ¼ 
0 
0 
H¼¼¼¼¼¼) lim 
x→ 0 
d 
dx sin 2xð Þð Þ 
d 
dx xþ 1
p
- 1 
¼ lim 
x→ 0 
2 cos 2xð Þ 
1 
2 xþ1p 
¼ 2x 1 1 
2 
¼ 4 
Choice (2) is the answer. 
6.41. From calculus, we know that: 
lim
x→ -1 x
4 þ 2x2 þ x - x2 
lim
x→ ±1 amx
m þ am- 1xm- 1 þ . . .þ a2x2 þ a1xþ a0 - amxm 
or: 
lim
x→ ±1 amx
m þ anxnð Þ - amxm if m> n 
The problem can be solved as follows: 
lim
x→ -1 x
4 þ 2x2 þ x- x2 ¼ lim
x→ -1 x
4 þ 2x2 þ x- x2 x x
4 þ 2x2 þ xp þ x2 
x4 þ 2x2 þ xp þ x2 
¼ lim
x→ -1 
x4 þ 2x2 þ x- x4 
x4 þ 2x2 þ xp þ x2 - limx→ -1 
2x2 þ x 
x2 þ x2ð Þ ¼ limx→ -1 
2x2 
2x2 
¼ lim
x→ -1 1 ¼ 1 
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
x→ x0 -
f xð Þ ¼ lim 
x→ x0þ 
f xð Þ 
Therefore, we must have: 
lim 
x→ - 3ð Þ-
x2 - 9 
xþ 3 ¼ lim x→ - 3ð Þþ 
x2 - 9 
xþ 3 ð1Þ
6 Solutions of Problems: Limits and Continuities 129
lim 
x→ - 3ð Þ-
x2 - 9 
xþ 3 ¼ lim x→ - 3ð Þ-
x2 - 9 
xþ 3 ¼ lim x→ - 3ð Þ- x- 3ð Þ ¼ - 6 ð2Þ 
lim 
x→ - 3ð Þþ 
x2 - 9 
x þ 3 ¼ lim x→ - 3ð Þþ
- x2 - 9ð Þ 
xþ 3 ¼ lim x→ - 3ð Þþ - x- 3ð Þ ¼ 6 ð3Þ 
1ð Þ, 2ð Þ, 3ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) - 6≠ 6 ) lim 
x→ - 3 
x2 - 9 
xþ 3 ¼ Undefined 
Choice (4) is the answer. 
6.43. The problem can be solved as follows: 
lim 
x→ 1 2 
tan πx 2 - 1 
cos πxð Þ ¼
0 
0 
H¼¼¼¼¼¼) lim 
x→ 1 2 
d 
dx tan 
πx 
2 - 1 
d 
dx cos πxð Þð Þ 
¼ lim 
x→ 1 2 
π
2 1 þ tan 2 πx 2
- π sin πxð Þ ¼ 
π
2 1þ 1ð Þ
- π x 1 ¼ -1 
Choice (2) is the answer. 
6.44. The problem can be solved as follows: 
lim 
x→þ1 xþ 5
p
- x þ 1p ¼ lim 
x→þ1 xþ 5
p
- xþ 1p x xþ 5
p þ x þ 1p 
xþ 5p þ xþ 1p 
¼ lim 
x→þ1 
xþ 5- xþ 1ð Þ 
xþ 5p þ xþ 1p ¼ lim x→þ1 
4 
xþ 5p þ xþ 1p ¼ 0 
Choice (3) is the answer. 
6.45. From trigonometry, we know that: 
1þ cos 2xð Þ ¼ 2 cos 2 xð Þ 
tan xð Þ ¼ sin xð Þ 
cos xð Þ 
sin 2xð Þ ¼ 2 sin xð Þ cos xð Þ 
The problem can be solved as follows: 
lim 
x→ π2 
tan 2xð Þ cos xð Þ 
1 þ cos 2xð Þ ¼ lim x→ π2 
sin 2xð Þ cos xð Þ 
cos 2xð Þ x 2 cos 2 xð Þ ¼ lim x→ π2 
2 sin xð Þ cos 2 xð Þ 
cos 2xð Þ x 2 cos 2 xð Þ ¼ lim x→ π2 
sin xð Þ 
cos 2xð Þ ¼ 
1
- 1 
¼ -1 
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 xð Þ ¼ sin 2 xð Þ 
Moreover, from application of Taylor series in limit, we know that: 
lim 
x→ 0-
sin xð Þ - x 
lim 
x→ 0-
tan xð Þ - x 
The problem can be solved as follows: 
lim 
x→ 0-
tan 2xð Þ 
1- cos xð Þ ¼ lim x→ 0-
tan 2xð Þ 
1- cos xð Þ x
1 þ cos xð Þ 
1 þ cos xð Þ ¼ lim x→ 0-
tan 2xð Þ 1þ cos xð Þ 
1- cos 2 xð Þ 
¼ lim 
x→ 0-
tan 2xð Þ x 2p 
sin 2 xð Þ 
¼ lim 
x→ 0-
tan 2xð Þ x 2p 
sin xð Þj j ¼ lim x→ 0-
tan 2xð Þ x 2p
- sin xð Þ ¼ lim x→ 0-
2xx 2p
- x 
¼ lim 
x→ 0-
- 2 2 
p 
¼ - 2 2 
p 
Choice (1) is the answer. 
6.47. The problem can be solved as follows: 
lim
x→ -1 nþ 1000
3
p
- n- 203
p 
¼ lim
x→ -1 nþ 1000
3
p
- n- 203
p
x
n þ 1000ð Þ23 þ nþ 1000ð Þ n- 20ð Þ3 þ n- 20ð Þ23 
nþ 1000ð Þ23 þ nþ 1000ð Þ n- 20ð Þ3 þ n- 20ð Þ23 
¼ lim
x→ -1 
n þ 1000- n- 20ð Þ 
nþ 1000ð Þ23 þ nþ 1000ð Þ n- 20ð Þ3 þ n- 20ð Þ23 
¼ 1020 þ1 ¼ 0 
Choice (2) is the answer. 
6.48. From application of Taylor series in limit, we know that: 
lim 
x→ 0 
sin xð Þ - x 
In addition, from trigonometry, we know that: 
1 þ cos xð Þ ¼ 2 cos 2 x 
2 
sin xð Þ ¼ 2 sin x 
2 
cos 
x 
2
6 Solutions of Problems: Limits and Continuities 131
The problem can be solved as follows: 
lim 
x→ πþ 
sin π sin xð Þð Þ sin x 2 
1 þ cos xð Þ - lim x→ πþ 
π sin xð Þ sin x 2 
2 cos 2 x 2 
¼ lim 
x→ πþ 
π x 2 sin x 2 cos x 2 sin x 2 
2
p 
cos x 2 
¼ lim 
x→ πþ 
π x 2 sin 2 x 2 cos x 2 
2
p
- cos x 2 
¼ lim 
x→ πþ
- π 2 
p 
sin 2 
x 
2 
¼ -π 2 
p
x 1 ¼ - π 2 
p 
Choice (1) is the answer. 
6.49. From application of Taylor series in limit, we know that: 
lim 
α→ 0 
1þ αnp - lim 
α→ 0 
1þ α
n 
lim 
x→ 0 
amx
m þ am- 1xm- 1 þ . . .þ am- nxm- n þ am- n- 1xm- n- 1 - am- n- 1xm- n- 1 
or: 
lim 
x→ 0 
amx
m þ anxnð Þ - anxn if m> n 
The problem can be solved as follows: 
lim 
x→ 0 
1þ x23p - 1- 2x4p 
2x2 þ 2x - lim x→ 0 
1þ x2 3 - 1- 2x 4 
2x2 þ 2x ¼ lim x→ 0 
x2 
3 þ x 2 
2x2 þ 2x - lim x→ 0 
x 
2 
2x 
¼ 1 
4 
Choice (1) is the answer. 
6.50. From trigonometry, we know that: 
sin 2 xð Þ þ cos 2 xð Þ ¼ 1 
The problem can be solved as follows: 
lim 
x→ 0 
sin 2 xð Þ þ sin xð Þ þ cos 2 xð Þ- cos xð Þ 
sin 2 xð Þ- sin xð Þ þ cos 2 xð Þ- cos xð Þ ¼ lim x→ 0 
1þ sin xð Þ- cos xð Þ 
1- sin xð Þ- cos xð Þ ¼ 
0 
0 
H¼¼¼¼¼¼) lim 
x→ 0 
d 
dx 1þ sin xð Þ- cos xð Þð Þ 
d 
dx 1- sin xð Þ- cos xð Þð Þ 
¼ lim 
x→ 0 
cos xð Þ þ sin xð Þ
- cos xð Þ þ sin xð Þ ¼ 
1þ 0
- 1þ 0 ¼ - 1 
Choice (2) is the answer.
132 6 Solutions of Problems: Limits and Continuities
6.51. From application of Taylor series in limit, we know that: 
lim 
u xð Þ→ 0 
sin u xð Þð Þ - u xð Þ 
The problem can be solved as follows: 
lim 
x→ 0 
cos mxð Þ- cos nxð Þ 
x2
¼ 0 
0 
H¼¼¼¼¼¼) lim 
x→ 0 
d 
dx cos mxð Þ- cos nxð Þð Þ 
d 
dx x
2ð Þ ¼ lim x→ 0
-m sin mxð Þ þ n sin nxð Þ 
2x
- lim 
x→ 0
-m mxð Þ þ n nxð Þ 
2x 
¼ lim 
x→ 0
-m2 þ n2 
2 
¼ n
2 -m2 
2 
Choice (3) is the answer. 
6.52. From application of Taylor series in limit, we know that: 
lim 
x→ 0 
sin xð Þ - x 
lim 
x→ 0 
tan xð Þ - x 
The problem can be solved as follows: 
lim 
x→ 0 
sin xð Þ- x 
tan xð Þ- x ¼ 
0 
0 
H¼¼¼¼¼¼) lim 
x→ 0 
d 
dx sin xð Þ- xð Þ 
d 
dx tan xð Þ- xð Þ
¼ lim 
x→ 0 
cos xð Þ- 1 
1þ tan 2 xð Þ- 1 ¼ lim x→ 0 
cos xð Þ- 1 
tan 2 xð Þ ¼ 
0 
0 
H¼¼¼¼¼¼) lim 
x→ 0 
d 
dx cos xð Þ- 1ð Þ 
d 
dx tan
2 xð Þð Þ ¼ lim x→ 0
- sin xð Þ 
2 tan xð Þ 1þ tan 2 xð Þð Þ - lim x→ 0
- x 
2x 1þ x2ð Þ ¼ lim x→ 0
- 1 
2 1þ x2ð Þ 
¼ - 1 
2 1þ 0ð Þ ¼
- 1 
2 
Choice (1) is the answer. 
6.53. From trigonometry, we know that: 
1þ cos 3 xð Þ ¼ 1þ cos xð Þð Þ 1- cos xð Þ þ cos 2 xð Þ 
1- cos 2 xð Þ ¼ 1þ cos xð Þð Þ 1- cos xð Þð Þ 
The problem can be solved as follows: 
lim
x→ π
1þ cos 3 xð Þ 
1- cos 2 xð Þ ¼ 
0 
0
6 Solutions of Problems: Limits and Continuities 133
) lim
x→ π
1þ cos 3 xð Þ 
1- cos 2 xð Þ ¼ limx→ π
1 þ cos xð Þð Þ 1- cos xð Þ þ cos 2 xð Þð Þ 
1þ cos xð Þð Þ 1- cos xð Þð Þ ¼ limx→ π
1- cos xð Þ þ cos 2 xð Þ 
1- cos xð Þ 
¼ 1- - 1ð Þ þ - 1ð Þ
2 
1- - 1ð Þ ¼ 
3 
2 
Choice (1) is the answer. 
6.54. Based on the information given in the problem, we have: 
f 0ð Þ ¼ 0 
f xð Þ ¼ x - 1ð Þ 1 x½ ], x 2 ℝ- 0f g 
A function is right continuous at this given point of x0 if: 
lim 
x→ x0þ 
f xð Þ ¼ f x0ð Þ 
Moreover, a function is left continuous at this given point of x0 if: 
lim
x→ x0 -
f xð Þ ¼ f x0ð Þ 
In addition, a function is continuous at this given point of x0 if: 
lim 
x→ x0þ 
f xð Þ ¼ lim
x→ x0 -
f xð Þ ¼ f x0ð Þ 
For the given function, we have: 
lim 
x→ 0-
f xð Þ ¼ lim 
x→ 0-
x - 1ð Þ 1 x½ ] ¼ 0x finite quantityð Þ ¼ 0 
lim 
x→ 0þ 
f xð Þ ¼ lim 
x→ 0þ 
x - 1ð Þ 1 x½ ] ¼ 0x finite quantityð Þ ¼ 0 
f 0ð Þ ¼ 0 
) lim 
x→ 0þ 
f xð Þ ¼ lim 
x→ 0-
f xð Þ ¼ f 0ð Þ 
Thus, the function is continuous at x ¼ 0. 
Choice (3) is the answer. 
6.55. As we know from calculus: 
If a> 1, k 2 N ) lim 
n→þ1 
nk 
an 
¼ 0 
Hence: 
lim 
n→þ1 
3n2 
5n
p ¼ lim 
n→þ13 
n2 
5
p n ¼ 3x 0 ¼ 0 
Choice (1) is the answer.
134 6 Solutions of Problems: Limits and Continuities
6.56. From trigonometry, we know that: 
1- cos xð Þ ¼ 2 sin 2 x 
2 
Moreover, from application of Taylor series in limit, we know that: 
lim 
x→ 0 
sin n xð Þ - xn 
Thus: 
lim 
x→ 0 
x3 - sin xð Þ 1- cos xð Þð Þ 
x3
¼ lim 
x→ 0 
x3 - sin xð Þ 2sin 2 x 2 
x3
- lim 
x→ 0 
x3 - xx 2 x 2 
2 
x3 
¼ lim 
x→ 0 
x3 - x
3 
2 
x3 
¼ lim 
x→ 0 
x3 
2 
x3 
¼ lim 
x→ 0 
1 
2 
¼ 1 
2 
Choice (2) is the answer. 
6.57. From trigonometry and calculus, we know that: 
arc cos 1-ð Þð Þ ¼ 0þ 
d 
dx 
arc cos xð Þð Þ ¼ - 1 
1- x2
p 
d 
dx 
1- x 
p 
¼ - 1 
2 1- x
p 
The problem can be solved as follows: 
lim 
x→ 1-
arc cos xð Þ 
1- x
p ¼ 0
þ 
0þ 
H¼¼¼¼¼¼) lim 
x→ 1-
d 
dx arc cos xð Þð Þ 
d 
dx 1- x
p ¼ lim 
x→ 1-
- 1 
1- x2
p
- 1 
2 1- x
p ¼ lim x→ 1-
1 
1- xð Þ 1þxð Þ 
p 
1 
2 1- x
p ¼ lim x→ 1-
2 
1þ xp ¼ 2 
p 
Choice (1) is the answer. 
6.58. Based on the information given in the problem, we have: 
lim 
x→ 1 
x2 - 1 cot xn - 1ð Þ ¼ 1 
2
ð1Þ 
From calculus, we know that: 
cot xð Þ ¼ 1 
tan xð Þ 
From application of Taylor series in limit, we know that: 
lim 
u xð Þ→ 0 
tan u xð Þð Þ - u xð Þ
6 Solutions of Problems: Limits and Continuities 135
The problem can be solved as follows: 
lim 
x→ 1 
x2 - 1 cot xn - 1ð Þ ¼ lim 
x→ 1 
x2 - 1 
tan xn - 1ð Þ - lim x→ 1 
x2 - 1 
xn - 1
¼ 0 
0 
H¼¼¼¼¼¼) lim 
x→ 1 
2x 
nxn- 1 
¼ 2 
n
ð2Þ 
Solving (1) and (2): 
2 
n 
¼ 1 
2 
) n ¼ 4 
Choice (2) is the answer. 
6.59. From trigonometry, we know that: 
sin 4xð Þ ¼ 2 sin 2xð Þ cos 2xð Þ 
cot xð Þ ¼ cos xð Þ 
sin xð Þ 
sin x- yð Þ ¼ sin xð Þ cos yð Þ- cos xð Þ sin yð Þ 
The problem can be solved as follows: 
lim 
x→ 0 
sin 4xð Þ cot 2xð Þ- cot xð Þð Þ ¼ lim 
x→ 0 
2 sin 2xð Þ cos 2xð Þ cos 2xð Þ 
sin 2xð Þ -
cos xð Þ 
sin xð Þ 
¼ lim 
x→ 0 
2 sin 2xð Þ cos 2xð Þ sin xð Þ cos 2xð Þ- cos xð Þ sin 2xð Þ 
sin 2xð Þ sin xð Þ 
¼ lim 
x→ 0 
2 sin 2xð Þ cos 2xð Þ sin x- 2xð Þ 
sin 2xð Þ sin xð Þ ¼ lim x→ 0 - 2 cos 2xð Þð Þ ¼ - 2 
Choice (3) is the answer. 
6.60. From application of Taylor series in limit, we know that: 
lim 
x→ 0 
sin xð Þ - x- x
3 
6 
lim 
x→ 0 
cos xð Þ - 1- x
2 
2 
The problem can be solved as follows: 
lim 
x→ 0-
sin xð Þ- x 
1 
2 sin 2xð Þ- x cos xð Þ
- lim 
x→ 0-
- x
3 
6 
1 
2 2x-
2xð Þ3 
6 - x 1-
x2 
2 
¼ lim 
x→ 0-
- x
3 
6
- x3 6 
¼ lim 
x→ 0-
1 ¼ 1 
Choice (3) is the answer.
136 6 Solutions of Problems: Limits and Continuities
6.61. The problem can be solved as follows: 
lim 
x→ π4 
1- tan xð Þ3 
1- 2 sin 2 xð Þ ¼ 
0 
0 
H¼¼¼¼¼¼) lim 
x→ π4 
d 
dx 1- tan xð Þ3 
d 
dx 1- 2 sin
2 xð Þ ¼ lim x→ π4
- 1þ tan
2 xð Þ 
3 tan 2 xð Þ3
p
- 4 sin xð Þ cos xð Þ ¼ 
1þ1 
3x1 
4x 2
p 
2 x 2
p 
2 
¼ 1 
3 
Choice (1) is the answer. 
6.62. From calculus and trigonometry, we know that: 
sin 2 xð Þ þ cos 2 xð Þ ¼ 1 
1- cos 3 xð Þ ¼ 1- cos xð Þð Þ 1þ cos xð Þ þ cos 2 xð Þ 
From application of Taylor series in limit, we know that: 
lim 
x→ 0 
sin xð Þ - x 
lim 
u xð Þ→ 0 
tan u xð Þð Þ - u xð Þ 
The problem can be solved as follows: 
lim 
x→ 0 
1- cos 3 xð Þ 
sin xð Þ tan 2xð Þ ¼ 
0 
0 
) lim 
x→ 0 
1- cos 3 xð Þ 
sin xð Þ tan 2xð Þ ¼ lim x→ 0 
1- cos xð Þð Þ 1þ cos xð Þ þ cos 2 xð Þð Þ 
sin xð Þ tan 2xð Þ x
1- cos xð Þð Þ 
1þ cos xð Þð Þ 
¼ lim 
x→ 0 
1- cos 2 xð Þð Þ 1þ cos xð Þ þ cos 2 xð Þð Þ 
sin xð Þ tan 2xð Þ 1þ cos xð Þð Þ ¼ lim x→ 0 
sin 2 xð Þ x 1þ 1þ 1ð Þ 
sin xð Þ tan 2xð Þ x 1þ 1ð Þ 
lim 
x→ 0 
3 sin 2 xð Þ 
2 sin xð Þ tan 2xð Þ - lim x→ 0 
3x2 
2xx 2x ¼ lim x→ 0 
3 
4 
¼ 3 
4 
Choice (3) is the answer. 
6.63. From trigonometry, we know that: 
1- cos xð Þ ¼ 2 sin 2 x 
2 
sin 2 xð Þ þ cos 2 xð Þ ¼ 1 
From application of Taylor series in limit, we know that: 
lim 
u xð Þ→ 0þ 
sin n u xð Þð Þ - lim 
u xð Þ→ 0þ 
u xð Þð Þn
References 137
The problem can be solved as follows: 
lim 
x→ 0þ 
1- cos xð Þ 
1- cos x
pð Þ ¼ lim x→ 0þ 
1- cos xð Þ 
1- cos x
pð Þ x
1þ cos xð Þ 
1þ cos xð Þ x
1þ cos xpð Þ 
1þ cos xpð Þ 
¼ lim 
x→ 0þ 
1- cos xð Þð Þ 1þ cos xpð Þð Þ 
1- cos 2 x
pð Þð Þ 1þ cos xð Þ ¼ lim x→ 0þ 
1- cos xð Þð Þ 1þ 1ð Þ 
1- cos 2 x
pð Þð Þ 1þ 1ð Þ 
¼ lim 
x→ 0þ 
1- cos xð Þ 
1- cos 2 x
pð Þ ¼ lim x→ 0þ 
2 sin 2 x 2 
sin 2 x
pð Þ - lim x→ 0þ 
2 x 2 
2 
x
pðÞ2 
¼ lim 
x→ 0þ 
x 
2 
¼ 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
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¼
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¼
¼ ¼
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 xð Þ ¼ x xþ 2ð Þ2 x- 3ð Þ3 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) x -2 
2) x 3 
3) x 0 
4) x -2, 0, 3 
7.3. Calculate the value of f ′(x 1) if f (x) 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
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¼ ¼
¼
¼
140 7 Problems: Derivatives and Their Applications
7.4. If f(x) + g(x3 ) 5x - 1 and f ′(1) 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) 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 
7.6. Determine the derivative of the function below at x ¼ 1 
4 
. 
f xð Þ ¼ x- x
p 
1- x
p 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) -1 
2) -
1 
2 
3) 
1 
2 
4) 1 
7.7. Determine the first derivative of the function of (x100 + x50 + 50x2 + 50x + 1)10 at x 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 π 
12 
. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
4 
3 
2) 
4 
9 
3) 
8 
3 
4) 
8 
9
þ
¼
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 
4) 
1 
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 
2) 
1 
15 
3) 
1 
6 
4) 
1 
16 
7.11. For the following function, calculate the value of ( f-1 (x))′ for x ¼ 2. 
f xð Þ ¼ 4x
3 
x2 1 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
112 
25 
2) 
25 
112 
3) 
1 
4 
4) 4 
7.12. Calculate the value of the limit below if f(x) x tan x. 
lim 
x⟶π 4 
f xð Þ- f π 
4 
x-
π 
4 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 1 
2) 1-
π 
4 
3) 1þ π 
4 
4) 1þ π 
2
¼ j j ¼
¼ j j
ð Þ ¼
t
þ ¼ þ ¼
142 7 Problems: Derivatives and Their Applications
7.13. If the function of f (x) x3 - 3x + a does not have a derivate at x 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) x2 - 6 . 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) -8 
2) 8 
3) -4 
4) 4 
7.15. If f 0 x 5 x, calculate the first derivative of f (x
5 ). 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) -
5 
x 
2) -
25 
x 
3) 
25 
x 
4) 
5 
x5 
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 
7.17. Calculate the derivative of the function of f (x) ¼ arc(tan(3x)) at x ¼ 1 
3 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
3 
2 
2) 
4 
3 
3) 
2 
3 
4) 
3 
4 
7.18. If f 
1 
g t
p
t2 1 and g′(1) 5, calculate the value of f ′(1). 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large
¼
p
p
p
p
¼ ¼ þ ¼
7 Problems: Derivatives and Their Applications 143
1) 1 
2) 2 
3) 
1 
2 
4) -
1 
2 
7.19. If 2 cos ( y) - sin (x + y) + 2 0, calculate the value of y0 x at (0, π). 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
1 
2 
2) -
1 
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 
1) -
16y 
x5 
2) 
16x 
y5 
3) -
32y 
x5 
4) -
32x 
y5 
7.21. If x ¼ 2 + 3 sin (t) and y ¼ 3 - 2 cos (t), calculate the value of y0 x for t ¼ 
π 
6 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
2 3 
9 
2) 
2 3 
3 
3) 
2 2 
3 
4) 
4 2 
3 
7.22. If x t2 + t and y t2 - 2t, calculate the value of x0 y y
0 
x for t -1. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
11 
4 
2) 
13 
4 
3) 
15 
4 
4) 
17 
4
¼ j j ¼
þ
144 7 Problems: Derivatives and Their Applications
7.23. Calculate the value of f ′(4) if we know that: 
lim 
h→ 0 
f xþ hð Þ- f x- hð Þ 
h
¼ 2 xp 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
2 
3 
2) 
4 
3 
3) 4 
4) 2 
7.24. Which one of the choices is true about the function of f (x) x2 x at x 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. 
7.25. The function below is differentiable at x ¼ π 
4 
. Determine the value of b. 
f xð Þ ¼ 
sin 2 xð Þ- cos 2xð Þ 0< x≤ π 
4 
a tan xð Þ þ b sin 2xð Þ π 
4 
< x< 
π 
2 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -1 
2) -
1 
2 
3) 
1 
2 
4) 1 
7.26. The function below is differentiable everywhere on ℝ domain. Determine the value of b. 
f xð Þ ¼ axþ b x<- 1 
x2 a x≥ - 1 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2 
2) 1 
3) -2 
4) -3
þ ð Þ
¼
¼
þð Þ þð Þ
7 Problems: Derivatives and Their Applications 145
7.27. Calculate the derivative of the function below. 
f xð Þ ¼ 2x- 1ð Þ
2 
2x2 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
2x- 1 
2x3 
2) 
2x- 1 
x3 
3) 
2xþ 1 
x3 
4) 
2xþ 1 
2x3 
7.28. Calculate the derivative of the function below. 
f xð Þ ¼ sin xð Þ 
1 tan 2 x 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
5 
4 
2) -
5 
4 
3) 
5 
8 
4) -
5 
8 
7.29. For the function below, calculate the value of f ′(x 2). 
f xð Þ ¼ x2 - 5xþ 6 arc sin 1 
x 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -
π 
3 
2) 
2π 
3 
3) -
π 
6 
4) 
π 
4 
7.30. For the following function, calculate the value of f ′(x -3). 
f xð Þ ¼ x2 þ 2x- 3 g xþ 2ð Þ 
x3 1 g 2x 5 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large
1)
¼
þ
p
p
¼ j j
j j
j j
p
p
p
p
p
146 7 Problems: Derivatives and Their Applications
-
13 
2 
2) 
13 
2 
3) -
2 
13 
4) 
2 
13 
7.31. For what value of m, the line of y 2x + 1 is tangent to a curve with the following function: 
y ¼ - 1þ x
2 
m x 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
3 
4 
2) ± 
3 
8 
3) 
1 
9 
4) ± 
3 
2 
7.32. Determine the third derivate of f (x) x4 x . 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -60x2 
2) 60x2 
3) -60x x 
4) 60x x 
7.33. Determine the value of the parameter of “a” if the derivative of xþ ap for x ¼ 2 is 1 
4 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -2 
2) -1 
3) 1 
4) 2 
7.34. Calculate the derivative of y ¼ ln e sin xð Þ at x ¼ π 
6 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
3 
8 
2) 
6 
8 
3) 
3 
4 
4) 
6 
4
¼
ð Þ
7 Problems: Derivatives and Their Applications 147
7.35. Determine the maximum value of the function of y(x) 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? 
y xð Þ ¼ 1- sin xð Þ 
cos x 
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 xð Þ ¼ 1þ x2 - x 
5 
, g xð Þ ¼ 1 
1þ x2p þ x 5 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -1 
2) 0 
3) 1 
4) 2 
7.38. Calculate the first derivative of the following function for x ¼ 4 
3 
: 
y xð Þ ¼ 1 
1þ x2p xþ 1þ x2p 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ○ Normal ● Large 
1)
- 27 
125 
2)
- 9 
25 
3) 
18 
25 
4) 
54 
125
p
8 2
p
¼
¼
p
¼
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 
3) 4 2 
4) 
7.40. For the function below, calculate the value of f ′(x -1). 
f xð Þ ¼ 
xþ 1ð Þ5 
xþ 1j j x≠ - 1 
0 x - 1 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 0 
2) 1 
3) -1 
4) 5 
7.41. The point M(x, y) is moving on the curve of y ¼ xþ 8. Determine the changing rate of the distance of the point from 
the origin when x 7. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
15 
16 
2) 
15 
8 
3) 
3 
7 
4) 
5 
4 
7.42. Determine the derivative of f - xj j þ 3 if we have the relation below. 
lim 
x→ 2 
f xð Þ- f 2ð Þ 
x- 2 
¼ - 1 
3 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
1 
6 
2) 
1 
12 
3) -
1 
6 
4) -
1 
12
þð Þ þð Þ
p
¼
4 2
¼ ¼
¼
7 Problems: Derivatives and Their Applications 149
7.43. Determine the value of f ′(-1) if we have: 
f xð Þ ¼ xþ 1ð Þh xð Þ 
2x 1 h 2x 1 
, h - 1ð Þ≠ 0 
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 xð Þ ¼ cos 2 xð Þ þ 3 sin xð Þ þ a has an extremum 
point with the width of y 
3 
in the range of 0< x< 
π 
. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 1 
2) 
1 
2 
3) -
1 
2 
4) -1 
7.45. Calculate the first derivative of the function of y(x) xx for x 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) xlnx . 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) xlnx ln x 
2) 2xlnx ln x 
3) 2x
ln x ln x 
x 
4) (2xlnx ln x)-x 
7.47. Calculate the n-th derivative of the function below. 
y xð Þ ¼ 1 
x 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 0 
2) 
n! 
xnþ1 
3) - 1ð Þn n! 
xn 
4) - 1ð Þn n! 
xnþ1
7.48. ¼
¼
¼
¼
¼
¼
150 7 Problems: Derivatives and Their Applications
Determine the equation of the line which is perpendicular on the curve of y(x) x2x at (1, 1). 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) x + 2y - 3 0 
2) 2x + y - 3 0 
3) x + y - 2 0 
4) x - y 0 
7.49. Determine the angle between the right and left tangent lines of the function below at x 1. 
f xð Þ ¼ x
3 x> 1 
x
p
x≤ 1 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
π 
4 
2) 
2π 
3 
3) 
π 
3 
4) 
3π 
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.
Solutions of Problems: Derivatives 
and Their Applications 8 
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]: 
y 0ð Þ ¼ 820 
y tð Þ ¼ 3280 
k ¼ 0:2 
The population of the animal in the jungle at time t can be calculated as follows: 
y tð Þ ¼ y 0ð Þekt 
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 
0:2 
¼ 5 ln 4 
) 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
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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 xð Þ ¼ x xþ 2ð Þ2 x- 3ð Þ3 
Choice (3) is the answer. 
8.3. From list of derivative rules, we know that: 
f xð Þ ¼ ex ) f 0 xð Þ ¼ ex 
f xð Þ ¼ u xð Þv xð Þ ) f 0 xð Þ ¼ u0 xð Þv xð Þ þ u xð Þv0 xð Þ 
Based on the information given in the problem,we have: 
f xð Þ ¼ xex - ex 
The problem can be solved as follows: 
f 0 xð Þ ¼ ex þ xex - ex ¼ xex 
f 0 1ð Þ ¼ 1e1 ) f 0 1ð Þ ¼ e 
Choice (4) is the answer. 
8.4. From list of derivative rules, we know that: 
h xð Þ ¼ g u xð Þð Þ ) h0 xð Þ ¼ u0 xð Þg0 u xð Þð Þ 
Based on the information given in the problem, we have: 
f 0 1ð Þ ¼ 2 
f xð Þ þ g x3 ¼ 5x- 1 
The problem can be solved as follows: 
f xð Þ þ g x3 ¼ 5x- 1 
d 
dx¼¼¼¼¼¼¼¼¼¼) f 0 xð Þ þ 3x2 g0 x3 ¼ 5 
x ¼ 1¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) f 0 1ð Þ þ 3g0 1ð Þ ¼ 5 
f 0 1ð Þ ¼ 2¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 2þ 3g0 1ð Þ ¼ 5 
) g0 1ð Þ 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 xð Þ ¼ axn ) f 0 xð Þ ¼ anxn- 1 
A function is ascending in a given range if its derivative is positive. Therefore: 
y xð Þ ¼ 1- 4x2 ) y0 xð Þ ¼ - 8x> 0 ) x< 0 
Choice (1) is the answer. 
8.6. From list of derivative rules, we know that: 
f xð Þ ¼ xp ) f 0 xð Þ ¼ 1 
2 x
p 
First, we should simplify the function as follows: 
f xð Þ ¼ x- x
p 
1- x
p ¼ x
p
x
p
- 1ð Þ 
1- x
p ¼ - xp 
Therefore: 
) f 0 xð Þ ¼ - 1 
2 x
p ) f 0 1 
4 
¼ - 1 
2 1 4 
¼ - 1 
Choice (1) is the answer. 
8.7. From list of derivative rules, we know that: 
f xð Þ ¼ un xð Þ ) f 0 xð Þ ¼ nu0 xð Þun- 1 xð Þ 
Therefore: 
f xð Þ ¼ x100 þ x50 þ 50x2 þ 50xþ 1 10 
) f 0 xð Þ ¼ 10 100x99 þ 50x49 þ 100x þ 50 x100 þ x50 þ 50x2 þ 50xþ 1 9 
) f 0 0ð Þ ¼ 10 0þ 0 þ 0 þ 50ð Þ 0þ 0þ 0þ 0þ 1ð Þ9 ¼ 500 
Choice (4) is the answer. 
8.8. From list of derivative rules, we know that: 
f xð Þ ¼ tan n u xð Þð Þ ) f 0 xð Þ ¼ nu0 xð Þ 1þ tan 2 u xð Þð Þ tan n- 1 u xð Þð Þ 
Therefore: 
f xð Þ ¼ tan 3 2xð Þ ) f 0 xð Þ ¼ 3� 2 1þ tan 2 2xð Þ tan 2 2xð Þ 
) f 0 π 
12 
¼ 3� 2 1þ tan 2 π 
6 
tan 2 
π 
6 
¼ 6� 1þ 1 
3
� 1 
3 
¼ 8 
3 
Choice (3) is the answer.
8.9.
154 8 Solutions of Problems: Derivatives and Their Applications
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: 
f - 1 
0 
bð Þ ¼ 1 
f 0 að Þ 
On the other hand, we know that: 
f að Þ ¼ b , f - 1 bð Þ 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 xð Þ ¼ x3 þ x- 2 ) f 0 xð Þ ¼ 3x2 þ 1 
Hence: 
f - 1 
0 
0ð Þ ¼ 1 
f 0 1ð Þ ¼
1 
3� 12 þ 1 ¼ 
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: 
f - 1 
0 
bð Þ ¼ 1 
f 0 að Þ 
Moreover: 
f að Þ ¼ b , f - 1 bð Þ 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 aþ 1 ) a ¼ 1 
On the other hand, we have: 
f xð Þ ¼ x5 þ 3x3 þ xþ 1 ) f 0 xð Þ ¼ 5x4 þ 9x2 þ 1 
Hence: 
f - 1 
0 
6ð Þ ¼ 1 
f 0 1ð Þ ¼
1 
5� 14 þ 9� 12 þ 1 ¼ 
1 
15 
Choice (2) is the answer.
Choice (3) is the answer.
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: 
f - 1 
0 
bð Þ ¼ 1 
f 0 að Þ 
In addition, we know that: 
f að Þ ¼ b , f - 1 bð Þ ¼ a 
Hence, for the function of f xð Þ ¼ 4x3 x2þ1 and the point with the length of two (b ¼ 2) on the inverse function, we can 
calculate “a” as follows: 
2 ¼ 4a
3 
a2 þ 1 ) a ¼ 1 
On the other hand, we can write: 
f xð Þ ¼ 4x
3 
x2 þ 1 ) f
0 xð Þ ¼ 12x
2 x2 þ 1ð Þ- 2xð Þ 4x3ð Þ 
x2 þ 1ð Þ2 
Thus: 
f - 1 
0 
2ð Þ ¼ 1 
f 0 1ð Þ ¼
12 þ 1 2 
12� 12 12 þ 1 - 2� 1ð Þ 4� 13 ¼ 
4 
24- 8
¼ 1 
4 
8.12. We know that: 
d 
dx 
u xð Þv xð Þð Þ ¼ u0 xð Þv xð Þ þ u xð Þv0 xð Þ ð1Þ 
d 
dx 
tan xð Þ ¼ 1þ tan 2 x 
tan 
π 
4 
¼ 1 
The first derivative of a function is defined as follows: 
lim
x⟶x0 
f xð Þ- f x0ð 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 π 4. 
lim 
x⟶π 4 
f xð Þ- f π 4 
x- π 4 
¼ f 0 x0ð Þ
¼
156 8 Solutions of Problems: Derivatives and Their Applications
The first derivative of the function can be calculated as follows: 
f 0 xð Þ ¼ tan xþ 1þ tan 2 x x 
For x0 π 4, we have: 
f 0 π 
4 
¼ tan π 
4
þ 1 þ tan 2 π 
4 
π 
4 
¼ 1þ π 
2 
Choice (4) is the answer. 
8.13. Based on the information given in the problem, we have: 
f xð Þ ¼ x3 - 3xþ a 
A derivative of an absolute function does not exist at its simple roots. Therefore, we need to solve the equation below: 
f 2ð Þ ¼ 0 
) 23 - 3� 2þ a ¼ 0 ) 8- 6þ a ¼ 0 ) a ¼ - 2 
Choice (2) is the answer. 
8.14. From list of derivative rules, we know that: 
f xð Þ ¼ u xð Þj j ) f 0 xð Þ ¼ u
0 xð Þu xð Þ 
u xð Þj j 
Based on the information given in the problem, we have: 
f xð Þ ¼ x2 - 6 
Therefore: 
) f 0 xð Þ ¼ 2x� x
2 - 6ð Þ 
x2 - 6j j 
) f 0 2ð Þ ¼ 4� 4- 6ð Þ 
4- 6j j ¼ - 4 and f
0 4ð Þ ¼ 8� 16- 6ð Þ 
16- 6j j ¼ 8 
) f 0 2ð Þ þ f 0 4ð Þ ¼ - 4þ 8 ¼ 4 
Choice (4) is the answer. 
8.15. From list of derivative rules, we know that: 
f xð Þ ¼ g h xð Þð Þ ) f 0 xð Þ ¼ h0 xð Þg0 h xð Þð Þ 
Based on the information given in the problem, we have: 
f 0 xð Þ 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 
x5 
) f x5 0 ¼ 25 
x 
Choice (3) is the answer. 
8.16. From list of derivative rules, we have: 
f g xð Þð Þð Þ0 ¼ g0 xð Þf 0 g xð Þð Þ 
Based on the information given in the problem, we have: 
f sin xð Þð Þð Þ0 ¼ cos 3 xð Þ 
The problem can be solved as follows: 
f sin xð Þð Þð Þ0 ¼ cos 3 xð Þ ) cos xð Þ � f 0 sin xð Þð Þ ¼ cos 3 xð Þ ) f 0 sin xð Þð Þ ¼ cos 2 xð Þ 
) f 0 sin xð Þð Þ ¼ 1- sin 2 xð Þ ) f 0 xð Þ ¼ 1- x2 
Choice (2) is the answer. 
8.17. From list of derivative rules, we know that: 
f xð Þ ¼ arc tan u xð Þð Þð Þ ) f 0 xð Þ ¼ u
0 xð Þ 
1þ u2 xð Þ 
Therefore: 
f xð Þ ¼ arc tan 3xð Þð Þ ) f 0 xð Þ ¼ 3 
1þ 9x2 
f 0 1 
3 
¼ 3 
1þ 9 1 3 
2 ¼ 
3 
2 
Choice (1) is the answer. 
8.18. From list of derivative rules, we know that: 
f xð Þ ¼ axn ) f 0 xð Þ ¼ anxn- 1 
f xð Þ ¼ g h xð Þð Þ ) f 0 xð Þ ¼ h0 xð Þg0 h xð Þð Þ 
Based on the information given in the problem, we have: 
f 
1 
t 
þ g tp ¼ t2 þ 1 
g0 1ð Þ ¼ 5
158 8 Solutions of Problems: Derivatives and Their Applications
The problem can be solved as follows: 
f 
1 
t 
þ g tp ¼ t2 þ 1 
d 
dx¼¼¼¼¼¼¼¼) - 1 
t2
� f 0 1 
t 
þ 1 
2 t
p � g0 tp ¼ 2t 
t ¼ 1 ) - f 0 1ð Þ þ 1 
2 
g0 1ð Þ ¼ 2 g
0 1ð Þ ¼ 5¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼)- f 0 1ð Þ þ 5 
2 
¼ 2 
) f 0 1ð Þ ¼ 1 
2 
Choice (3) is the answer. 
8.19. From derivative rules, we know that: 
f x, yð Þ ¼ 0 ) y0 x ¼ -
f 0 x x, yð Þ 
f 0 y x, yð Þ
¼ -
d 
dx f x, yð Þ 
d 
dy f x, yð Þ 
Based on the information given in the problem, we have: 
2 cos yð Þ- sin xþ yð Þ þ 2 ¼ 0 
The problem can be solved as follows: 
y0 x ¼ -
d 
dx 2cos yð Þ- sin x þ yð Þ þ 2ð Þ 
d 
dy 2cos yð Þ- sin x þ yð Þ þ 2ð Þ
¼ - - cos xþ yð Þ
- 2 sin yð Þ- cos xþ yð Þ ¼ -
cos xþ yð Þ 
2 sin yð Þ þ cos xþ yð Þ 
x, yð Þ ¼ 0, πð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) y0 x ¼ -
- 1 
0- 1 
) y0 x ¼ - 1 
Choice (3) is the answer. 
8.20. From derivative rules, we know that: 
f x, yð Þ ¼ 0 ) y0 x ¼ -
f 0 x x, yð Þ 
f 0 y x, yð Þ
¼ -
d 
dx f x, yð Þ 
d 
dy f x, yð Þ 
Based on the information given in the problem, we have: 
x3 þ y3 ¼ 16 
The problem can be solved as follows: 
y0 x ¼ y0 ¼ -
3x2 
3y2 
¼ - x
2 
y2 
) y00 ¼ - 2xy
2 - 2yy0x2 
y4
p p
8 Solutions of Problems: Derivatives and Their Applications 159
y0 ¼ - x
2 
y2 ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) y00 ¼-
2xy2 - 2y - x
2 
y2 x
2 
y4
¼ - 2xy
3 þ 2x4 
y5 
¼ - 2x y
3 þ x3ð Þ 
y5 
x3 þ y3 ¼ 16¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) y00 ¼ - 2x� 16 
y5 
) y00 ¼ - 32x 
y5 
Choice (4) is the answer. 
8.21. From derivative rules, we know that: 
y ¼ y tð Þ 
x ¼ x tð Þ ) y
0 
x ¼ 
y0 t 
x0t 
¼ 
d 
dt y tð Þ 
d 
dt x tð Þ 
Based on the information given in the problem, we have: 
x ¼ 2þ 3 sin tð Þ 
y ¼ 3- 2 cos tð Þ 
Hence: 
y0 x ¼
y0 t 
x0t 
¼ 2 sin tð Þ 
3 cos tð Þ ¼
2 
3 
tan tð Þ 
t ¼ π 
6 
) y0 x ¼
2 
3
� 3 
3 
) y0 x ¼ 
2 3 
9 
Choice (1) is the answer. 
8.22. From derivative rules, we know that: 
y ¼ y tð Þ 
x ¼ x tð Þ ) y
0 
x ¼
y0 t 
x0t 
¼ 
d 
dt y tð Þ 
d 
dt x tð Þ
, x0 y ¼
x0 t 
y0t 
¼ 
d 
dt x tð Þ 
d 
dt y tð Þ 
Based on the information given in the problem, we have: 
x ¼ t2 þ t 
y ¼ t2 - 2t 
Therefore: 
) 
y0 x ¼ 
y0 t 
x0t 
¼ 2t- 2 
2t þ 1 
x0 y ¼ 
x0 t 
y0t 
¼ 2t þ 1 
2t- 2 
t ¼ - 1¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 
y0 x ¼
- 2- 2
- 2 þ 1 ¼ 4 
x0 y ¼
- 2 þ 1
- 2- 2 
¼ 1 
4 
) y0 x þ x0 x ¼ 4þ 
1 
4 
) y0 x þ x0 x ¼ 
17 
4 
Choice (4) is the answer.
160 8 Solutions of Problems: Derivatives and Their Applications
8.23. Based on the information given in the problem, we have: 
lim 
h→ 0 
f xþ hð Þ- f x- hð Þ 
h
¼ 2 xp ð1Þ 
From definition of derivative, we know: 
f 0 xð Þ ¼ lim 
h→ 0 
f xþ hð Þ- f xð Þ 
h
¼ lim 
h→ 0 
f xð Þ- f x- hð Þ 
h 
The problem can be solved as follows: 
lim 
h→ 0 
f xþ hð Þ- f x- hð Þ 
h
¼ lim 
h→ 0 
f xþ hð Þ- f xð Þ þ f xð Þ- f x- hð Þ 
h 
¼ lim 
h→ 0 
f xþ hð Þ- f xð Þ 
h
þ lim 
h→ 0 
f xð Þ- f x- hð Þ 
h
¼ 2f 0 xð Þ ð2Þ 
Solving (1) and (2): 
2f 0 xð Þ ¼ 2 xp ) f 0 xð Þ ¼ xp ) f 0 4ð Þ ¼ 4 
p 
¼ 2 
Choice (4) is the answer. 
8.24. The function can be simplified as follows: 
f xð Þ ¼ x2 xj j ¼ x
3 x≥ 0
- x3 x< 0 
) f 0ð Þ ¼ 0 x≥ 0 
0 x< 0 
) f 0-ð Þ ¼ f 0þð Þ ð1Þ 
Now, we can determine its first and second derivatives as follows: 
) f 0 xð Þ ¼ 3x
2 x≥ 0
- 3x2 x< 0 
) f 0 0ð Þ ¼ 0 x≥ 0 
0 x< 0 
) f 0 0-ð Þ ¼ f 0 0þð Þ ð2Þ 
) f 00 xð Þ ¼ 6x x≥ 0
- 6x x< 0 
) f 00 0ð Þ ¼ 0 x≥ 0 
0 x< 0 
) f 00 0-ð Þ ¼ f 00 0þð Þ ð3Þ 
From (1) and (2), we can conclude that the first derivative of the function exits. Likewise, from (1), (2), and (3), we can 
say that the second derivative of the function does exist. Choice (4) is the answer. 
8.25. Based on the information given in the problem, we have: 
f xð Þ ¼ 
sin 2 xð Þ- cos 2xð Þ 0< x≤ π 
4 
a tan xð Þ þ b sin 2xð Þ π 
4 
< x< 
π 
2 
ð1Þ
¼
¼
8 Solutions of Problems: Derivatives and Their Applications 161
The function is differentiable at x π 4. Therefore, we can conclude that: 
f 
π 
4
-
¼ f π 
4 
þ 
ð2Þ 
f 0 π 
4
-
¼ f 0 π 
4 
þ 
ð3Þ 
Solving (1) and (2): 
sin 2 
π 
4
- cos 2� π 
4 
¼ a tan π 
4 
þ b sin 2� π 
4 
) aþ b ¼ 1 
2
ð4Þ 
Solving (1) and (3): 
2 sin 
π 
4 
cos 
π 
4 
þ 2 sin 2� π 
4 
¼ a 1þ tan 2 π 
4 
þ 2b cos 2� π 
4 
) 1þ 2 ¼ 2aþ 0 ) a ¼ 3 
2
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 xð Þ ¼ axþ b x< - 1 
x2 þ a x≥ - 1 ð1Þ 
The function is differentiable everywhere on ℝ domain including x - 1. Hence: 
f - 1ð Þ-ð Þ ¼ f - 1ð Þþ ð2Þ 
f 0 - 1ð Þ-ð Þ ¼ f 0 - 1ð Þþ ð3Þ 
Solving (1) and (2):
- a þ b ¼ 1þ a ) - 2aþ b ¼ 1 ð4Þ 
Solving (1) and (3): 
a ¼ - 2 ð5Þ 
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 xð Þ ¼ g xð Þ 
h xð Þ ) f
0 xð Þ ¼ g
0 xð Þh xð Þ- h0 xð Þg xð Þ 
h2 xð Þ 
Therefore: 
f xð Þ ¼ 2x- 1ð Þ
2 
2x2 
) f 0 xð Þ ¼ 4 2x- 1ð Þ 2x
2ð Þ- 4x 2x- 1ð Þ2 
4x4 
) f 0 xð Þ ¼ 16x
3 - 8x2 - 16x3 þ 16x2 - 4x 
4x4
¼ 8x
2 - 4x 
4x4 
¼ 2x- 1 
x3 
Choice (2) is the answer. 
8.28. From list of derivative rules, we know that: 
f xð Þ ¼ u xð Þv xð Þ ) f 0 xð Þ ¼ u0 xð Þv xð Þ þ u xð Þv0 xð Þ 
From trigonometry, we have: 
1þ tan 2 xð Þ ¼ 1 
cos 2 xð Þ 
Based on the information given in the problem, we have: 
f xð Þ ¼ sin xð Þ 
1 þ tan 2 xð Þ ) f xð Þ ¼ 
sin xð Þ 
1 
cos 2 xð Þ 
¼ sin xð Þ cos 2 xð Þ 
) f 0 xð Þ ¼ cos xð Þ cos 2 xð Þ- sin xð Þ � 2 sin xð Þ cos xð Þ ¼ cos 3 xð Þ- 2 sin 2 xð Þ cos xð Þ 
) f 0 π 
3 
¼ 1 
2 
3
- 2 
3
p 
2 
2 
1 
2 
¼ 1 
8
-
3 
4 
¼ - 5 
8 
Choice (4) is the answer. 
8.29. From list of derivative rules, we know that: 
f xð Þ ¼ u xð Þv xð Þ ) f 0 xð Þ ¼ u0 xð Þv xð Þ þ u xð Þv0 xð Þ 
f xð Þ ¼ axn ) f 0 xð Þ 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 xð Þ ¼ x2 - 5xþ 6 arc sin 1 
x
¼
¼
p
8 Solutions of Problems: Derivatives and Their Applications 163
Therefore: 
f 0 xð Þ ¼ 2x- 5ð Þarc sin 1 
x 
þ x2 - 5xþ 6 arc sin 1 
x 
0 
f 0 2ð Þ ¼ 2� 2- 5ð Þarc sin 1 
2 
þ 0� arc sin 1 
x 
0 
¼ - 1ð Þ � π 
6 
) f 0 2ð Þ ¼ - π 
6 
As can be seen, we did not need to calculate the value of arc sin 1 x 
0
. Choice (3) is the answer. 
8.30. From list of derivative rules, we know that: 
f xð Þ ¼ u xð Þv xð Þ ) f 0 xð Þ ¼ u0 xð Þv xð Þ þ u xð Þv0 xð Þ 
f xð Þ ¼ axn ) f 0 xð Þ ¼ anxn- 1 
Based on the information given in the problem, we have: 
f xð Þ ¼ x2 þ 2x- 3 � g xþ 2ð Þ 
x3 þ 1ð Þg 2x þ 5ð Þ ≜u xð Þ � v xð Þ 
The problem can be solved as follows: 
f 0 xð Þ ¼ 2xþ 2ð Þ g xþ 2ð Þ 
x3 þ 1ð Þg 2xþ 5ð Þ þ x
2 þ 2x- 3 v0 xð Þ 
) f 0 x ¼ - 3ð Þ ¼ - 4ð Þ � g - 1ð Þ
- 26g - 1ð Þ þ 0ð Þ � v
0 x ¼ - 3ð Þ ¼ - 4� - 1 
26 
) f 0 x ¼ - 3ð Þ ¼ 2 
13 
As can be seen, we did not need to calculate the value of v′(x - 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 0).
- 1þ x2 
mþ x ¼ 2xþ 1 ) - 1þ x
2 ¼ 2x2 þ xþ 2mxþ m ) x2 þ 2mþ 1ð Þxþ m þ 1 ¼ 0 
Δ ¼ 0 ) 2m þ 1ð Þ2 - 4 m þ 1ð Þ ¼ 0 ) 4m2 - 3 ¼ 0 ) m ¼ ± 3 
2 
Choice (4) is the answer. 
8.32. From list of derivative rules, we know that: 
f xð Þ ¼ axn ) f 0 xð Þ ¼ anxn- 1 
Based on the information given in the problem, we have: 
f xð Þ ¼ x4 xj j
164 8 Solutions of Problems: Derivatives and Their Applications
Therefore: 
f xð Þ ¼ x
5, x≥ 0
- x5, x< 0 
) f 0 xð Þ ¼ 
5x4, x> 0 
0, x ¼ 0
- 5x4, x< 0 
) f 00 xð Þ ¼ 
20x3, x> 0 
0, x ¼ 0
- 20x3, x< 0 
) f 000 xð Þ ¼ 
60x2, x> 0 
0, x ¼ 0
- 60x2, x< 0 
) f 000 xð Þ ¼ 60x xj j 
Choice (4) is the answer. 
8.33. Based on the information given in the problem, we have: 
f xð Þ ¼ xþ ap ) f 0 2ð Þ ¼ 1 
4
ð1Þ 
From list of derivative rules, we know that: 
f xð Þ ¼ u xð Þ ) f 0 xð Þ ¼ u
0 xð Þ 
2 u xð Þ ð2Þ 
Solving (1) and (2): 
1 
2 xþ ap x¼2 
¼ 1 
4 
) 1 
2 2 þ ap ¼ 
1 
4 
) 2þ ap ¼ 2 ) a ¼ 2 
Choice (4) is the answer. 
8.34. First, we should simplify the function as follows: 
y xð Þ ¼ ln e sin xð Þ 
p 
¼ sin xð Þ ð1Þ 
From list of derivative rules, we know that: 
f xð Þ ¼ u xð Þ ) f 0 xð Þ ¼ u
0 xð Þ 
2 u xð Þ ð2Þ 
Solving (1) and (2): 
) y0 xð Þ ¼ cos xð Þ 
2 sin xð Þ ) y
0 π 
6 
¼ cos 
π 
6 
2 sin π 6 
¼ 
3
p 
2 
2 1 2 
¼ 6
p 
4 
Choice (4) is the answer.
8 Solutions of Problems: Derivatives and Their Applications 165
8.35. From list of derivative rules, we know that: 
f xð Þ ¼ axn ) f 0 xð Þ 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. 
y xð Þ ¼ x3 - 3x2 - 9xþ 5 ) y0 xð Þ ¼ 3x2 - 6x- 9 
To find the extremum points of the function: 
) y0 xð Þ 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 asfollows: 
y - 2ð Þ ¼ - 2ð Þ3 - 3 - 2ð Þ2 - 9 - 2ð Þ þ 5 ¼ 3 
y - 1ð Þ ¼ - 1ð Þ3 - 3 - 1ð Þ2 - 9 - 1ð Þ þ 5 ¼ 10 
y 2ð Þ ¼ 2ð Þ3 - 3 2ð Þ2 - 9 2ð Þ 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 xð Þ ¼ g xð Þ 
h xð Þ ) f
0 xð Þ ¼ g
0 xð Þh xð Þ- h0 xð Þg xð Þ 
h2 xð Þ ð1Þ 
First, we need to take its derivative as follows: 
y xð Þ ¼ 1- sin xð Þ 
cos xð Þ ) y
0 xð Þ ¼ - cos xð Þ cos xð Þ- - sin xð Þð Þ 1- sin xð Þð Þ 
cos 2 xð Þ 
y0 xð Þ ¼ - cos
2 xð Þ þ sin xð Þ- sin 2 xð Þ 
cos 2 xð Þ ¼
- 1þ sin xð Þ 
cos 2 xð Þ 
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 xð Þ ¼ 1þ x2 - x 
5 
, g xð Þ ¼ 1 
1þ x2p þ x 5
ð1Þ 
From list of derivative rules, we know that: 
f xð Þ 
g xð Þ 
0 
¼ f
0 xð Þg xð Þ- g0 xð Þf xð Þ 
g xð Þð Þ2 ) f
0 xð Þg xð Þ- g0 xð Þf xð Þ ¼ f xð Þ 
g xð Þ 
0 
g xð Þð Þ2 ð2Þ
166 8 Solutions of Problems: Derivatives and Their Applications
Solving (1) and (2): 
f 0 xð Þg xð Þ- g0 xð Þf xð Þ ¼ 1þ x
2
p
- x 
5 
1 
1þx2p þxð Þ5 
0 
1 
1þ x2p þ x 5 
2 
¼ 1þ x2 - x2 5 
0 1 
1þ x2p þ x 10 
¼ 1ð Þ0 � 1 
1þ x2p þ x 10 
¼ 0� 1 
1þ x2p þ x 10 
¼ 0 
Choice (2) is the answer. 
8.38. From list of derivative rules, we know that: 
d 
dx 
ln f xð Þð Þ ¼ f
0 xð Þ 
f xð Þ ð1Þ 
d 
dx 
f xð Þ ¼ f
0 xð Þ 
2 f xð Þ ð1Þ 
In addition, we know that: 
ln 1 ¼ 0 
ln f xð Þm ¼ 1 
m 
ln f xð Þ 
ln f xð Þg xð Þð Þ ¼ ln f xð Þð Þ þ ln g xð Þð Þ 
The problem can be solved as follows: 
y xð Þ ¼ 1 
1þ x2p xþ 1þ x2p 
ln¼¼¼¼¼¼¼¼¼¼) ln y xð Þ ¼ ln 1- ln 1þ x2 xþ 1 þ x2 ¼ 0- ln 1þ x2 - ln xþ 1þ x2 
) ln y xð Þ ¼ - 1 
2 
ln 1 þ x2 - ln x þ 1þ x2 
d 
dx¼¼¼¼¼¼¼¼¼¼) y
0 xð Þ 
y xð Þ ¼ -
1 
2 
2x 
1þ x2 -
1þ 2x 
2 1þx2p 
xþ 1þ x2p 
) y0 xð Þ ¼ y xð Þ - 1 
2 
2x 
1þ x2 -
1þ 2x 
2 1þx2p 
xþ 1þ x2p 
) y0 xð Þ ¼ 1 
1 þ x2p xþ 1þ x2p -
1 
2 
2x 
1 þ x2 -
1 þ 2x 
2 1þx2p 
xþ 1þ x2p
¼
Choice (1) is the answer.
8 Solutions of Problems: Derivatives and Their Applications 167
Now, for x 4 3, we have: 
y0 4 
3 
¼ - 27 
125 
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 2 
p 
, - 2 
p 
x1 ¼ 2 
p 
) y1 ¼ 2 
p 3
- 6 2 
p 
þ 12 ¼ 2 2 
p
- 6 2 
p 
þ 12 ¼ - 4 2 
p 
þ 12 
x2 ¼ - 2 
p 
) y2 ¼ - 2 
p 3
- 6 - 2 
p 
þ 12 ¼ - 2 2 
p 
þ 6 2 
p 
þ 12 ¼ 4 2 
p 
þ 12 
y2 - y1 ¼ 4 2 
p 
þ 12 - - 4 2 
p 
þ 12 ) y2 - y1 ¼ 8 2 
p 
Choice (4) is the answer. 
8.40. Based on the information given in the problem, we have: 
f xð Þ ¼ 
xþ 1ð Þ5 
xþ 1j j x≠ - 1 
0 x ¼ - 1 
This problem can be solved by using the definition of derivative of a function as follows: 
f 0 x0ð Þ ¼ limx→ x0 
f xð Þ- f x0ð Þ 
x- x0 
) f 0 - 1ð Þ ¼ lim 
x→ - 1ð Þ 
xþ1ð Þ5 
xþ1j j - 0 
x- - 1ð Þ ¼ lim x→ - 1ð Þ 
xþ1ð Þ5 
xþ1j j 
xþ 1 ¼ lim x→ - 1ð Þ 
xþ 1ð Þ4 
xþ 1j j ¼ lim x→ - 1ð Þ xþ 1j j
3 ¼ 0 
Choice (1) is the answer. 
8.41. From list of derivative rules, we know that: 
f xð Þ ¼ u xð Þ ) f 0 xð Þ ¼ u
0 xð Þ 
2 u xð Þ ð1Þ 
The distance of the point from the origin can be calculated as follows: 
D xð Þ ¼ x- 0ð Þ2 þ x þ 8p - 0 2 ¼ x2 þ xþ 8 
In addition, the changing rate of the distance can be determined as follows: 
D0 xð Þ ¼ d 
dx 
D xð Þ ¼ d 
dx 
x2 þ xþ 8 ð2Þ
168 8 Solutions of Problems: Derivatives and Their Applications
Solving (1) and (2): 
D0 xð Þ ¼ 2xþ 1 
2 x2 þ xþ 8p ) D
0 7ð Þ ¼ 2� 7þ 1 
2 72 þ 7þ 8 
¼ 15 
16 
Choice (1) is the answer. 
8.42. From list of derivative rules, we know that: 
f xð Þ ¼ g h xð Þð Þ ) f 0 xð Þ ¼ h0 xð Þg0 h xð Þð Þ 
f xð Þ ¼ u xð Þ ) f 0 xð Þ ¼ u
0 xð Þ 
2 u xð Þ 
The problem should be solved using the definition of derivative of a function as follows: 
f 0 x0ð Þ ¼ limx→ x0 
f xð Þ- f x0ð Þ 
x- x0 
ð1Þ 
Based on the information given in the problem, we have: 
lim 
x→ 2 
f xð Þ- f 2ð Þ 
x- 2 
¼ - 1 
3
ð2Þ 
Solving (1) and (3): 
f 0 2ð Þ ¼ - 1 
3
ð3Þ 
Therefore: 
d 
dx 
f - xj j þ 3 
x¼- 1 
¼ d 
dx 
f - x þ 3p 
x¼- 1 )
- 1 
2 - xþ 3p f
0 - xþ 3p 
x¼- 1 
¼ - 1 
4 
f 0 2ð Þ ð4Þ 
Solving (3) and (4): 
d 
dx 
f - xj j þ 3 
x¼- 1 
¼ - 1 
4
� - 1 
3 
¼ 1 
12 
Choice (2) is the answer. 
8.43. Based on the information given in the problem, we have: 
f xð Þ ¼ x þ 1ð Þh xð Þ 
2xþ 1ð Þh 2xþ 1ð Þ , h - 1ð Þ≠ 0 ð1Þ 
The derivative of this function should be solved by using the definition of derivative of a function as follows: 
f 0 x0ð Þ ¼ limx→ x0 
f xð Þ- f x0ð Þ 
x- x0 
) f 0 - 1ð Þ ¼ lim 
x→ - 1 
f xð Þ- f - 1ð Þ 
x- - 1ð Þ ð2Þ
p
¼
¼
8 Solutions of Problems: Derivatives and Their Applications 169
Solving (1) and (2): 
f 0 - 1ð Þ ¼ lim 
x→ - 1 
xþ1ð Þh xð Þ 
2xþ1ð Þh 2xþ1ð Þ -
- 1þ1ð Þh- 1ð Þ
- 2þ1ð Þh- 2þ1ð Þ 
x- - 1ð Þ ¼ lim x→ - 1 
xþ1ð Þh xð Þ 
2xþ1ð Þh 2xþ1ð Þ - 0 
xþ 1 
f 0 - 1ð Þ ¼ lim 
x→ - 1 
h xð Þ 
2x þ 1ð Þh 2x þ 1ð Þ ¼ 
h - 1ð Þ
- 1ð Þh - 1ð Þ ¼ - 1 
Choice (2) is the answer. 
8.44. Based on the information given in the problem, the width of the extremum point is as follows: 
y xMð 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 xð Þ ¼ 0 ð2Þ 
f xð Þ ¼ cos 2 xð Þ þ 3 sin xð Þ þ a ð3Þ 
) f 0 xð Þ ¼ - 2 sin xð Þ cos xð Þ þ 3 
p 
cos xð Þ ¼ cos xð Þ - 2 sin xð Þ þ 3 
p 
ð4Þ 
Solving (2) and (4): 
cos xð Þ - 2 sin xð Þ þ 3 
p 
¼ 0 ) 
cos xð Þ ¼ 0 5ð Þ 
sin xð Þ ¼ 3
p 
2 
6ð Þ 
There is no answer for equation (5) in the range of 0< x< π 2. However, x 
π 
3 is only answer for equation (6). 
Therefore, using xM π 3 and (1) in (3), we have: 
3 
4
¼ cos 2 π 
3 
þ 3 
p 
sin 
π 
3 
þ a ) 3 
4 
¼ 1 
4
þ 3 
2
þ a ) a ¼ - 1 
Choice (4) is the answer. 
8.45. From list of derivative rules, we know that: 
d 
dx 
ln f xð Þð Þ ¼ f
0 xð Þ 
f xð Þ ð1Þ 
d 
dx 
u xð Þv xð Þð Þ ¼ u0 xð Þv xð Þ þ u xð Þv0 xð Þ 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: 
y xð Þ ¼ xx 
ln¼¼¼¼¼¼) ln y xð Þ ¼ ln xx ) ln y xð Þ ¼ x ln x 
d 
dx¼¼¼¼¼¼¼¼¼¼) y
0 xð Þ 
y xð Þ ¼ ln xþ x 
1 
x 
¼ ln xþ 1 
) y0 xð Þ ¼ y xð Þ ln xþ 1ð Þ ) y0 xð Þ ¼ xx ln xþ 1ð Þ 
) y0 eð Þ ¼ ee ln e þ 1ð Þ ¼ 2ee 
Choice (3) is the answer. 
8.46. From list of derivative rules, we know that: 
d 
dx 
ln f xð Þð Þ ¼ f
0 xð Þ 
f xð Þ ð1Þ 
d 
dx 
f xð Þð Þnð Þ ¼ nf 0 xð Þ f xð Þð Þn- 1 ð2Þ 
Also, we know that: 
ln f xð Þg xð Þ ¼ g xð Þ ln f xð Þ 
The problem can be solved as follows: 
y xð Þ ¼ x ln x 
ln¼¼¼¼¼¼) ln y xð Þ ¼ ln x ln x ¼ ln x� ln x ) ln y xð Þ ¼ ln xð Þ2 
d 
dx¼¼¼¼¼¼¼¼¼¼) y
0 xð Þ 
y xð Þ ¼ 2 
1 
x 
ln x 
) y0 xð Þ ¼ 2x
ln x ln x 
x 
Choice (3) is the answer. 
8.47. From list of derivative rules, we know that: 
d 
dx 
f xð Þ 
g xð Þ ¼ 
f 0 xð Þg xð Þ- f xð Þg0 xð Þ 
g xð Þð Þ2 ð1Þ 
We can find a formula for the n-th derivative of the function below as follows: 
y xð Þ ¼ 1 
x
8 Solutions of Problems: Derivatives and Their Applications 171
d 
dx¼¼¼¼¼¼¼¼¼¼) y0 xð Þ ¼ 0- 1� 1 
x2 
¼ - 1 
x2 
¼ - 1ð Þ1 1 
x1þ1 
d2 
dx2 ¼¼¼¼¼¼¼¼¼¼) y00 xð Þ ¼ - 5- 2x� 1 
x2ð Þ2 ¼ 
2 
x3 
¼ - 1ð Þ2 2� 1 
x2þ1 
d3 
dx3 ¼¼¼¼¼¼¼¼¼¼) y000 xð Þ ¼ 5- 3x
2 � 2 
x3ð Þ3 ¼
- 6 
x4 
¼ - 1ð Þ3 3� 2� 1 
x3þ1 
⋮ 
dn 
dxn ¼¼¼¼¼¼¼¼¼¼) y nð Þ xð Þ ¼ - 1ð Þn n! 
xnþ1 
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 m x- x0ð 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 m0 x- x0ð Þ 
where mand m′ are the slope of the tangent and perpendicular lines. In addition, we have: 
m ¼ y0 x0ð Þ 
m0 ¼ - 1 
m 
Therefore, first, we need to determine the first derivative of y(x). 
y xð Þ ¼ x2x 
ln¼¼¼¼¼¼) ln y xð Þ ¼ ln x2x ¼ 2x ln x 
d 
dx¼¼¼¼¼¼¼¼¼¼) y
0 xð Þ 
y xð Þ ¼ 2 ln x þ 2 
) y0 xð Þ ¼ y xð Þ 2ln x þ 2ð Þ ¼ x2x 2ln x þ 2ð Þ 
The slope of the tangent line: 
m ¼ y0 x0 ¼ 1ð Þ ¼ 12 2ln 1 þ 2ð Þ ¼ 2
¼
172 8 Solutions of Problems: Derivatives and Their Applications
The slope of the perpendicular line: 
m0 ¼ - 1 
m 
¼ - 1 
2 
The equation of the perpendicular line: 
y- 1 ¼ - 1 
2 
x- 1ð Þ 
) y ¼ - 1 
2 
xþ 3 
2 
) xþ 2y- 3 ¼ 0 
Choice (1) is the answer. 
In this problem, the rules below were used. 
ln ab ¼ b ln a 
d 
dx 
ln f xð Þð Þ ¼ f
0 xð Þ 
f xð Þ 
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-m
0 
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 1, we have: 
f xð Þ ¼ x
3 x> 1 
x
p
x≤ 1 
m ¼ f 0 1þð Þ ¼ 3x2 x0 ¼ 1þ ¼ 3 
m0 ¼ f 0 1-ð Þ ¼ 1 
2 x
p 
x0 ¼ 1-
¼ 1 
2 
Therefore: 
θ ¼ π- tan - 1 3-
1 
2 
1 þ 3� 1 2 
¼ π- tan - 1 1ð Þ ¼ π- π 
4 
θ ¼ 3π 
4 
Choice (4) is the answer.
References 173
In this problem, the rules below were used. 
m ¼ f 0 x0ð Þ 
tan - 1 1ð Þ 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
þ
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]. 
I ¼ 3x þ 5ð Þ17 dx 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) (3x + 5)18 + c 
2) 
3xþ 5ð Þ18 
54 
þ c 
3) 
3xþ 5ð Þ17 
3 
þ c 
4) 
3xþ 5ð Þ18 
18 
þ c 
9.2. Calculate the value of the following indefinite integral. 
I ¼ cos 1þ πxð Þdx 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large
1) sin(1 + πx) + c 
2) 
1 
1 π 
sin 1þ πxð Þ þ c 
3) cos(1 + πx) + c 
4) 
1 
π 
sin 1þ πxð Þ þ c
# 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
http://crossmark.crossref.org/dialog/?doi=10.1007/978-3-031-45028-0_9&domain=pdf
https://doi.org/10.1007/978-3-031-45028-0_9#DOI
¼
þ
þ þ
þ
þ þ
176 9 Problems: Definite and Indefinite Integrals
9.3. If F(x) 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) 
2) 
1 
a 
F xð Þ 
3) aF(x) 
4) 
1 
a 
F axþ bð Þ 
9.4. Calculate the value of the definite integral below. 
2 
1 
xþ 4 
x3 
dx 
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: 
ex þ 2xex2 dx 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) - ex - ex
2 
c 
2) - ex ex
2 
c 
3) ex - ex
2 
c 
4) ex ex
2 
c 
9.6. Calculate the value of f ′′ (1) if we know that: 
f xð Þ ¼ x3 þ 5x dx 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 2 
2) 4 
3) 6 
4) 8 
9.7. Calculate the value of f 00 π 
2 
if f (x) ¼ cos3 (x)dx. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 0 
2) 1
p
þð Þ
9 Problems: Definite and Indefinite Integrals 177
3) -1 
4) 
3 2 
2 
9.8. Calculate the value of the definite integral below.
- 1
- 2 
x3 þ x2 - 1 
x2
dx 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 1 
2) -1 
3) 
1 
2 
4) -
1 
2 
9.9. Solve the indefinite integral below. 
x- 2 
x
p dx 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
2 
3 
x
p 
xþ 6ð Þ þ c 
2) 
2 
3 
x
p 
x- 6ð Þ þ c 
3) 
1 
3 
x
p 
xþ 6ð Þ þ c 
4) 
2 
3 
x
p 
x- 6ð Þ þ c 
9.10. Calculate the value of the following definite integral: 
I ¼ 
1 
0 
x2 
x3 1 4 
dx 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
5 
36 
2) 
7 
36 
3) 
5 
72 
4) 
7 
72
þ
þp þ
p
p þpð Þ
1 1
þ
178 9 Problems: Definite and Indefinite Integrals
9.11. Calculate the value of the following indefinite integral. 
I ¼ dx 
cos 2x 1 tan x
p 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1 
cos x
þ tan xþ c 
2) 
1 
cos x
þ cot gxþ c 
3) 2 1 tan x c 
4) 2(1 + tan x) + c 
9.12. Calculate the value of the indefinite integral below. 
I ¼ dx 
x ln x
p 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
2 
ln x
p þ c 
2) 
ln x 
x 
þ c 
3) 2 ln x c 
4) ln ln x 
9.13. Calculate the integral of the function below for the range of < x < + . 
f xð Þ ¼ 1 
x2 4 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
π 
2 
2) π 
3) 
3π 
2 
4) 2π 
9.14. Calculate the value of the definite integral below. 
4 
1 
x- x
p 
x
p dx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1 
3 
2) 
2 
3
3)
¼
þ
þ
9 Problems: Definite and Indefinite Integrals 179
4 
3 
4) 
5 
3 
9.15. If the primary function of f (x) is equal to 
x3 
6 
, determine the first derivate of f 
1 
x 
with respect to x. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) -
1 
x3 
2) -
x 
6 
3) -
1 
2 
4) 
1 
x3 
9.16. Calculate the value of F′ (λ 0) if: 
F λð Þ ¼ 
λ 
0 
1 
x4 2 
dx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 1 
2) 
1 
2 
3) 
1 
3 
4) 0 
9.17. Calculate the value of the definite integral below. 
1
- 1 
x2 
1 x2 
arc tan xð Þð Þdx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 0 
2) 1 
3) 
π 
4 
4) 
π 
2 
9.18. Calculate the integral of the function below for the range of -
1 
2 
< x< 
1 
2 
. 
f xð Þ ¼ 1 
1- x2
p 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large
þ ð Þð Þ
¼
¼
180 9 Problems: Definite and Indefinite Integrals
1) 
π 
6 
2) 
π 
3 
3) -
π 
6 
4) -
π 
3 
9.19. Calculate the value of the definite integral below. 
1 
0 
1 
2x- x2
p dx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 0 
2) 
π 
4 
3) 
π 
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: 
sin xð Þ 
1 cos cos x 
dx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) - tan 
1 
2 
cos xð Þ þ c 
2) tan(cos(x)) + c 
3) - tan (cos(x)) + c 
4) tan 
1 
2 
cos xð Þ þ c 
9.22. Determine the function ofa curve that passes from the point of (3, 4) and its derivative is -
x 
y 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2x2 + y2 34 
2) x2 + y2 16
¼
¼
þð Þ
9 Problems: Definite and Indefinite Integrals 181
3) y2 4x + 4 
4) x2 + y2 25 
9.23. Solve the following indefinite integral: 
x 
x- 1
p dx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
2 
3 
x- 1ð Þ3 2 - 2 x- 1ð Þ1 2 þ c 
2) 
2 
3 
x- 1ð Þ3 2 þ 2 x- 1ð Þ1 2 þ c 
3) 
1 
3 
x- 1ð Þ3 2 - 2 x- 1ð Þ1 2 þ c 
4) -
1 
3 
x- 1ð Þ3 2 - 2 x- 1ð Þ1 2 þ c 
9.24. Calculate the value of the definite integral below. 
5
- 2 
x- 3j jdx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
25 
2 
2) 
27 
2 
3) 
29 
2 
4) 
31 
2 
9.25. Calculate the value of the definite integral below. 
2 
1 
x xþ 1ð Þ2 þ 2 
x 1 2 
dx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
12 
5 
2) 
9 
5 
3) 
11 
6 
4) 
7 
4
þ
þ þð Þ þ
þð Þ þ
¼
¼
¼
¼
¼
¼
¼
¼
þ
182 9 Problems: Definite and Indefinite Integrals
9.26. Solve the following indefinite integral: 
1 
1 ex 
dx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) x + ln (1 + ex ) + c 
2) 
1 
x - ln (1 + ex ) + c 
3) 2 x
2 ln 1 ex c 
4) 1 2 x
2 - ln 1 ex 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 x þ 1 
1- y 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) x2 + y2 + 2x - 2y - 2 0 
2) x2 - y2 + 4x - 4y + 1 0 
3) x2 + y2 - 2x + 2y - 2 0 
4) x2 - y2 + 3x - 2y - 1 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 0 
2) y2 - 2x2 + 1 0 
3) 2y2 + x2 - 3 0 
4) 2y2 - x2 - 1 0 
9.29. In the equation below, determine the value of A. 
3x 
x2 1
p dx ¼ A x2 þ 1þ c 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
1 
2 
2) 1 
3) 
3 
2 
4) 3
¼
9 Problems: Definite and Indefinite Integrals 183
9.30. Calculate the value of the definite integral below. 
2
- 1 
xj jdx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
3 
2 
2) 
5 
2 
3) 
7 
2 
4) 
9 
2 
9.31. Calculate the value of the following definite integral. 
π 
2 
0 
sin 2 xð Þdx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 0 
2) 
π 
2 
3) 
π 
4 
4) 
π 
8 
9.32. Calculate the value of the definite integral below. 
3π 
4 
0 
tan 5 xð Þ þ tan 7 xð Þ dx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 0 
2) 
1 
2 
3) 
1 
3 
4) 
1 
6 
9.33. Which one of the points below is on a curve that passes from the point of (π, 1) and y′ y2 cos (x) holds for that? 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
3π 
2 
, 2 
2) 
π 
2 
, - 1
¼
p
ð Þ ð Þ ¼
184 9 Problems: Definite and Indefinite Integrals
3) 
π 
2 
, 1 
4) (0, 1) 
9.34. Calculate the value of the definite integral of I1 if I2 m. 
I1 ¼ 
5 
3 
3x 
x- 2 
dx 
I2 ¼ 
5 
3 
1 
x- 2 
dx 
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 
x2 - 1 
dx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) ln 
8 
3 
2) arc sin 
2 3 
5 
3) arc tan 
3 
2 
4) ln 
8 
3 
9.36. Calculate the value of the definite integral of 4 0 f
0 x dx if we have f x x a t
p 
dt. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
8 
3 
2) 
16 
3 
3) -
8 
3 
4) -
16 
3 
9.37. Solve the following indefinite integral: 
8 tan 6 xð Þ þ tan 8 xð Þ dx
9 Problems: Definite and Indefinite Integrals 185
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) tan7 (x) + c 
2) 
1 
7 
tan 8 xð Þ þ c 
3) 
8 
5 
tan 5 xð Þ þ c 
4) 
8 
7 
tan 7 xð Þ þ c 
9.38. Calculate the value of the definite integral below. 
1
- 1 
xþ 1ð Þ x2 þ 2xþ 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 
x 
1 
x3 
dx 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 1 
2) 2 
3) 
1 
2 
4) 
3 
2 
9.40. Calculate the value of y0 x if we have: 
y ¼ uþ v, u ¼ 
x2 
1 
sin tð Þ 
t 
dt, v ¼ 
1 
x2 
sin uð Þ 
u 
du 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
4 sin x2ð Þ 
x 
2) -
4 sin x2ð Þ 
x 
3) 0 
4) 1
þ
186 9 Problems: Definite and Indefinite Integrals
9.41. Calculate the value of f(3x + 2) if we have: 
h xð Þ ¼ f 0 3xþ 2ð Þdx, h 0ð Þ ¼ 1, f 2ð Þ 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 
1) 1, 
11 
6 
2) 1, 
13 
6 
3) 2, 
11 
6 
4) 2, 
13 
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 
3 
cos 3 2xð Þ- 1 
2 
cos 2xð Þ þ c 
2) -
1 
3 
cos 3 2xð Þ þ 1 
2 
cos 2xð Þ þ c 
3) 
1 
3 
cos 3 2xð Þ þ 1 
2 
cos 2xð Þ þ c 
4) -
1 
3 
cos 3 2xð Þ- 1 
2 
cos 2xð Þ þ c 
9.44. Calculate the value of the following indefinite integral. 
I ¼ e
arc tan x 
1 x2 
dx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) earc tanx + c 
2) 2earc tanx + c 
3) 
1 
2 
earc tan x þ c 
4) arc tan(ex + 1) + c 
9.45. Calculate the value of the definite integral below. 
1
- 1 
x 
3 
dx
¼
pð Þ þpð Þ þ
þ
1
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 1 holds. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) y ¼ 1þ 1 
x 
2) y ¼ 2- 1 
x 
3) y ¼ 3 
x
- 1 
4) y ¼ 3 
x
þ 1 
9.47. Solve the indefinite integral below. 
f 0 x3
pð Þ 
x23
p dx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1 
3 
f x3
p þ c 
2) 
2 
3 
f x3
p þ c 
3) f x3 c 
4) 3f x3 c 
9.48. Determine the value of the following definite integral: 
I ¼ 
1 
0 
dx 
1 eax 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
1 
a 
2) a ln 2 
3) 
1 
a 
ln 2 
4) 
9.49. Solve the indefinite integral below. 
ln xdx
ð Þ ð Þ
ð Þ
¼ ¼ ¼
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 
sin x cos x 
dx 
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 
cos 4 x 
dx 
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 
sin 2 xð Þ cos 4 xð Þ3 
dx 
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 e) if the derivative of f(x2 ) with respect to x is 6 x and f(x 1) 0. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 0 
2) 1
¼ ¼ ¼ ¼
ð Þ ð Þ
ð Þ
ð Þ
ð Þ
ð Þ
9 Problems: Definite and Indefinite Integrals 189
3) 3 
4) 6 
9.54. Calculate the value of f(x - 1) if f ′ (cos2 (x)) cos (2x) and f(x 1) 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. 
cos 2xð Þ 
sin 2 x cos 2 x 
dx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -
2 
sin 2x 
þ c 
2) -
1 
sin 2x 
þ c 
3) 
2 
sin 2x 
þ c 
4) 
1 
sin 2x 
þ c 
9.56. Calculate the value of the definite integral below. 
e 
1 
2xþ ln xð Þð Þdx 
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. 
sin 2xð Þ 2þ cos 2 xð Þ 50 dx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
1 
51 
2þ cos 2 xð Þ 51 þ c 
2) -
1 
51 
2þ cos 2 xð Þ 51 þ c 
3) 
1 
51 
2þ cos 2 xð Þ 50 þ c 
4) -
1 
51 
2þ cos 2 xð Þ 50 þ c
þ
¼
¼
¼
¼
p
p
p
p
190 9 Problems: Definite and Indefinite Integrals
9.58. Calculate the value of the definite integral below. 
e 
1 
ln xð Þ 
x 
dx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 1 
2) 
1 
2 
3) 2 
4) 
1 
e 
9.59. Determine the function of a curve that passes from the point of (0, 1) and the relation below holds. 
y0 ¼ - 2xþ 2 
4y 1 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2x2 + y2 34 + 3x 
2) x2 - y2 - 7y + 5 
3) x2 + y2 4x + 4y - 1 
4) x2 + 2y2 - y - 2x + 3 
9.60. Calculate the value of the definite integral below. 
π 
2 
π 
6 
cot xð Þ 
1- cos 2xð Þ dx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2 
2) 
2 
2 
3) 3 
4) 
3 
2 
9.61. Calculate the value of the definite integral below. 
6 
3 
xþ 2 
x- 2
p dx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
25 
3 
2) 
38 
3
3)
p
p
ð Þ þ
ð Þ þ
ð Þ ð Þ þ
9 Problems: Definite and Indefinite Integrals 191
23 
3 
4) 
34 
3 
9.62. Calculate the value of the definite integral below. 
4 
1 
1 þ xp 
x
p dx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2 3 
p 
þ 1 
3 
2) 2 3 
p
-
1 
3 
3) 4 3 
p 
þ 2 2 
3 
4) 4 3 
p
-
2 2 
3 
9.63. Solve the following indefinite integral if we know that 0 < x < π. 
cot xð Þ sin xð Þdx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) cot x c 
2) 2 sin x c 
3) sin x sin x c 
4) 
1 
2 
sin xð Þ þ c 
9.64. Calculate the value of the definite integral below. 
π 
3 
0 
sec xð Þ tan xð Þdx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 1 
2) 2 
3) 
1 
2 
4) 
3 
2
þpp
p p
p þp
p
p
p
p
192 9 Problems: Definite and Indefinite Integrals
9.65. Calculate the value of the definite integral below. 
π 
4 
π 
6 
csc xð Þ cot xð Þdx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2 2 
2) 2- 2 
3) 3- 2 
4) 3 2 
9.66. Calculate the value of the definite integral below. 
π 
4 
π 
6 
1 
sin 2 xð Þ cos 2 xð Þ dx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) 
3 
3 
2) 
2 3 
3 
3) 3 
4) 2 3 
9.67. Solve the following indefinite integral. 
tan xð Þ- cot xð Þð Þ tan xð Þ þ cot xð Þð Þ5 dx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) 
1 
4 
tan xð Þ þ cot xð Þð Þ4 þ c 
2) 
1 
5 
tan xð Þ þ cot xð Þð Þ5 þ c 
3) 
1 
3 
tan xð Þ þ cot xð Þð Þ3 þ c 
4) 
1 
5 
tan xð Þ- cot xð Þð Þ5 þ c 
9.68. Which one of the choices is not an acceptable solution for the indefinite integral of sin (x) cos (x)dx? 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) -
1 
4 
cos 2xð Þ þ c 
2) -
1 
4 
sin 2xð Þ þ c 
3) -
1 
2 
cos 2 xð Þ þ c 
4) 
1 
2 
sin 2 xð Þ þ c
þ
References 193
9.69. Calculate the value of the definite integral below. 
I ¼ 
ln 3 
ln 2 
1- e- 2x 
1 e- 2x 
dx 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) ln3 - ln 2 
2) 3 ln 2-
1 
2 
ln 3 
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 xð Þdu ¼ 1 
nþ 1 u
nþ1 xð Þ þ c 
The problem can be solved by defining a new variable as follows: 
u xð Þ ¼ 3xþ 5 
) du ¼ 3dx ) dx ¼ du 
3 
Therefore: 
I ¼ 3xþ 5ð Þ17 dx ¼ u17 xð Þ du 
3 
¼ 1 
3 
u17 xð Þdu 
) I ¼ 1 
3
x u
18 xð Þ 
18 
þ c 
) I ¼ 3xþ 5ð Þ
18 
54
þ c 
Choice (2) is the answer. 
10.2. From list of integral of different functions, we know that: 
cos u xð Þdu ¼ sin u xð Þ þ c 
The problem can be solved by defining a new variable as follows: 
u xð Þ 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
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https://doi.org/10.1007/978-3-031-45028-0_10#DOI
196 10 Solutions of Problems: Definite and Indefinite Integrals
) du ¼ πdx ) dx ¼ du 
π 
Thus: 
I ¼ cos 1þ πxð Þdx ¼ cos u xð Þð Þ du 
π 
¼ 1 
π 
cos u xð Þdu 
) I ¼ 1 
π 
sin u xð Þ þ c 
) I ¼ 1 
π 
sin 1þ πxð Þ þ c 
Choice (4) is the answer. 
10.3. From the problem, we have: 
F xð Þ ¼ f xð Þdx 
The problem can be solved by defining a new variable as follows: 
axþ b ¼ u xð Þ 
) adx ¼ du ) dx ¼ du 
a 
Therefore: 
I ¼ f ax þ bð Þdx ¼ 1 
a 
f uð Þdu 
) I ¼ F uð Þ 
a 
) I ¼ 1 
a 
F axþ bð Þ 
Choice (4) is the answer. 
10.4. From list of integral of functions, we know that: 
xn dx ¼ 1 
nþ 1 x
nþ1 þ c 
The problem can be solved as follows: 
2 
1 
xþ 4 
x3 
dx ¼ 
2 
1 
1 
x2 
þ 4 
x3 
dx ¼ - 1 
x
-
2 
x2 
2 
1 
¼ - 1 
2
-
2 
4
- - 1- 2ð Þ ¼ 2
Choice (2) is the answer.
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: 
ex þ 2xex2 dx ¼ ex þ ex2 þ c 
Choice (4) is the answer. 
10.6. Based on the information given in the problem, we have: 
f xð Þ ¼ x3 þ 5x dx 
d 
dx¼¼¼¼¼¼¼¼¼¼) f 0 xð Þ ¼ x3 þ 5x 
d 
dx¼¼¼¼¼¼) f 00 xð Þ ¼ 3x2 þ 5 
x ¼ 1¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) f 00 1ð Þ ¼ 3þ 5 ¼ 8 
Choice (4) is the answer. 
10.7. The problem can be solved as follows: 
f xð Þ ¼ cos 3 xð Þdx ) f 0 xð Þ ¼ cos 3 xð Þ ) f 00 xð Þ ¼ - 3 cos 2 xð Þ sin xð Þ 
f 00 π 
2 
¼ - 3x 0x 1 ¼ 0 
Choice (1) is the answer. 
10.8. From list of integral of functions, we know that: 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
The problem can be solved as follows:
- 1
- 2 
x3 þ x2 - 1 
x2
dx ¼
- 1
- 2 
xþ 1- 1 
x2 
dx ¼ x
2 
2 
þ x þ 1 
x
- 1
- 2 
¼ 1 
2
- 1- 1 - 2- 2-
1 
2 
¼ - 1 
Choice (2) is the answer.
10.9.
p
198 10 Solutions of Problems: Definite and Indefinite Integrals
From list of integral of functions, we know that: 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
The problem can be solvedas follows: 
x- 2 
x
p dx ¼ x 1 2 - 2x- 1 2 dx ¼ 2 
3 
x 
3 
2 - 4x 
1 
2 þ c ¼ 2 
3 
x
p 
x- 6ð Þ þ c 
Choice (2) is the answer. 
10.10. The problem can be solved by changing the variable of the integral as follows: 
x3 þ 1≜u 
d 
dx¼¼¼¼¼) 3x2 dx ¼ du ) x2 dx ¼ du 
3 
I ¼ 
1 
0 
x2 
x3 þ 1ð Þ4 dx ¼ 
u2 
u1 
u- 4 
du 
3 
¼ 1 
3 
u- 3
- 3 
u2 
u1 
) I ¼ - 1 
9 
x3 þ 1 - 3 1 
0 
) I ¼ - 1 
9 
1 þ 1ð Þ- 3 þ 1 
9 
0 þ 1ð Þ- 3 ¼ - 1 
72
þ 1 
9 
¼ - 1þ 8 
72 
) I ¼ 7 
72 
Choice (4) is the answer. 
10.11. From list of integral of different functions, we know that: 
du 
u xð Þ ¼ 2 u xð Þ 
The problem can be solved by defining a new variable as follows: 
1 þ tan x ¼ u xð Þ 
) 1þ tan 2 x dx ¼ du ) dx 
cos 2x 
¼ du 
Therefore: 
I ¼ dx 
cos 2x 1þ tan xp 
) I ¼ du 
u
p ¼ 2u 1 2 þ c 
) I ¼ 2 1þ tan xþ c 
Choice (3) is the answer.
p
10 Solutions of Problems: Definite and Indefinite Integrals 199
10.12. From list of integral of different functions, we know that: 
du 
u xð Þ ¼ 2 u xð Þ 
The problem can be solved by defining a new variable as follows: 
u xð Þ ¼ ln x 
) du ¼ dx 
x 
Therefore: 
I ¼ dx 
x ln x
p 
) I ¼ du 
u xð Þ ¼ 2 u xð Þ þ c 
) I ¼ 2 ln xþ c 
Choice (3) is the answer. 
10.13. From list of integral of functions, we know that: 
1 
x2 þ a2 dx ¼ 
1 
a 
arc tan 
x 
a 
þ c 
Therefore: 
þ1
-1 
1 
x2 þ 4 dx ¼ 
1 
2 
arc tan 
x 
2 
þ1
-1 ¼ 
1 
2 
π 
2
- -
π 
2 
¼ π 
2 
Choice (1) is the answer. 
10.14. From list of integral of functions, we know that: 
xn dx ¼ 1 
nþ 1 x
nþ1 þ c 
The problem can be solved as follows: 
4 
1 
x- x
p 
x
p dx ¼ 
4 
1 
x 
1 
2 - 1 dx ¼ 2 
3 
x 
3 
2 - x 
4 
1
¼ 16 
3
- 4 -
2 
3
- 1 ¼ 5 
3 
Choice (4) is the answer.
200 10 Solutions of Problems: Definite and Indefinite Integrals
10.15. Based on the information given in the problem, we have: 
f xð Þdx ¼ x
3 
6 
d 
dx¼¼¼¼¼¼) f xð Þ ¼ x
2 
2 
Therefore: 
) f 1 
x 
¼ 
1 
x 
2 
2 
¼ 1 
2x2 
) d 
dx 
f 
1 
x 
¼ d 
dx 
1 
2x2 
¼ - 4x 
4x4 
¼ - 1 
x3 
Choice (1) is the answer. 
10.16. As we know: 
F xð Þ ¼ 
u xð Þ 
v xð Þ 
f xð Þdx ) F0 xð Þ ¼ u0 xð ÞF u xð Þð Þ- v0 xð ÞF v xð Þð Þ 
The problem can be solved as follows: 
F λð Þ ¼ 
λ 
0 
1 
x4 þ 2 dx ) F
0 λð Þ ¼ 1x 1 
λ4 þ 2 - 0 ¼ 
1 
λ4 þ 2 ) F
0 0ð Þ ¼ 1 
2 
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
- 1 
x2 
1þ x2 arc tan xð Þð Þdx ¼ 0 
Choice (1) is the answer. 
10.18. From list of integral of functions, we know that: 
1 
1- x2
p dx ¼ arc sin xð Þð Þ 
Therefore: 
1 
2
- 1 2 
1 
1- x2
p dx ¼ arc sin xð Þð Þ 
1 
2
- 1 2 
¼ π 
6
- -
π 
6 
¼ π 
3 
Choice (2) is the answer.
10 Solutions of Problems: Definite and Indefinite Integrals 201
10.19. From list of integral of functions, we know that: 
1 
1- u2
p dx ¼ arc sin uð Þð Þ 
The problem can be solved by changing the variable of the integral as follows: 
x- 1 ≜ u 
d 
dx¼¼¼¼¼¼) dx ¼ du 
1 
0 
1 
2x- x2
p dx ¼ 
1 
0 
1 
1- x- 1ð Þ2 
dx ¼ 
u2 
u1 
1 
1- u2
p du ¼ arc sin uð Þð Þð Þ u2 
u1 
¼ arc sin x- 1ð Þð Þð Þ 1 
0 
¼ 0- - π 
2 
¼ π 
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 cos uð Þ ¼ 2 cos 2 u 
2 
1 
cos 2 u 2 
¼ 1þ tan 2 u 
2 
In addition, from list of integral of functions, we know that: 
1þ tan 2 u 
a 
du ¼ a tan u 
a 
þ c 
The problem can be solved by changing the variable of the integral as follows: 
cos xð Þ ≜ u ) d 
dx 
cos xð Þ ¼ d 
dx 
u ) - sin xð Þdx ¼ du 
) sin xð Þ 
1þ cos cos xð Þð Þ dx ¼ -
1 
1þ cos uð Þ du ¼ -
1 
2 cos 2 u 2 
du 
¼ - 1 
2 
1 þ tan 2 u 
2 
du ¼ - tan u 
2 
þ c ¼ - tan cos xð Þ 
2 
þ c 
Choice (1) is the answer.
202 10 Solutions of Problems: Definite and Indefinite Integrals
10.22. From list of integral of functions, we know that: 
xn dx ¼ 1 
nþ 1 x
nþ1 þ c 
Based on the information given in the problem, we have: 
y 3ð Þ ¼ 4 
y0 ¼ - x 
y 
The problem can be solved as follows: 
y0 ¼ - x 
y 
) yy0 þ x ¼ 0 
dx 
¼¼¼¼¼¼¼¼¼¼) y
2 
2 
þ x
2 
2 
¼ c0 ) y2 þ x2 ¼ c ð1Þ 
y 3ð Þ ¼ 4¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 42 þ 32 ¼ c ) c ¼ 25 ð2Þ 
1ð Þ, 2ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) y2 þ x2 ¼ 25 
Choice (4) is the answer. 
10.23. From list of integral of functions, we know that: 
xn dx ¼ 1 
nþ 1 x
nþ1 þ c 
The problem can be solved as follows: 
x 
x- 1
p dx ¼ x- 1 þ 1 
x- 1
p dx ¼ x- 1ð Þ1 2 þ x- 1ð Þ- 1 2 dx ¼ 2 
3 
x- 1ð Þ3 2 þ 2 x- 1ð Þ1 2 þ c 
Choice (2) is the answer. 
10.24. From list of integral of functions, we know that: 
xn dx ¼ 1 
nþ 1 x
nþ1 þ c 
The problem can be solved as follows: 
5
- 2 
x- 3j jdx ¼ 
3
- 2 
3- xð Þdxþ 
5 
3 
x- 3ð Þdx ¼ 3x- x
2 
2 
3
- 2
þ x
2 
2
- 3x 
5 
3 
¼ 9- 9 
2
- - 6- 2ð Þ þ 25 
2
- 15 -
9 
2
- 9 ¼ 9 
2
þ 8- 5 
2
þ 9 
2
¼ 9þ 16- 5 þ 9 
2
¼ 29 
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 
nþ 1 x
nþ1 þ c 
The problem can be solved as follows: 
2 
1 
x xþ 1ð Þ2 þ 2 
xþ 1ð Þ2 dx ¼ 
2 
1 
x þ 2 xþ 1ð Þ- 2 dx ¼ x
2 
2
-
2 
xþ 1 
2 
1 
¼ 2- 2 
3
-
1 
2
- 1 ¼ 4 
3
þ 1 
2 
¼ 11 
6 
Choice (3) is the answer. 
10.26. From list of integral of functions, we know that: 
xn dx ¼ 1 
nþ 1 x
nþ1 þ c 
1 
u 
du ¼ ln uj j þ c 
The problem can be solved as follows: 
1 
1þ ex dx ¼ 
1þ ex - ex 
1 þ ex dx ¼ 1-
ex 
1þ ex dx ¼ 1dx-
ex 
1þ ex dx 
¼ xþ c0 - e
x 
1þ ex dx ð1Þ 
Now, we should change the variable of the integral as follows: 
1 þ ex ≜ u ) ex dx ¼ du ð2Þ 
Solving (1) and (2): 
xþ c0 - 1 
u 
du ¼ xþ c0 - ln uj j þ c00 ¼ x- ln 1þ exj j þ c ¼ x- ln 1 þ exð Þ þ c 
Choice (2) is the answer. 
10.27. The problem can be solved as follows: 
y0 ¼ x þ 1 
1- y 
) y0 - yy0 ¼ xþ 1 
dx 
¼¼¼¼¼¼¼¼¼¼) y- y
2 
2 
¼ x
2 
2 
þ xþ c 
x, yð Þ ¼ 1, 1ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 1- 1 
2 
¼ 1 
2
þ 1þ c ) c ¼ - 1 
) y- y
2 
2 
¼ x
2 
2 
þ x- 1 ) x2 þ y2 þ 2x- 2y- 2 ¼ 0 
Choice (1) is the answer.
204 10 Solutions of Problems: Definite and Indefinite Integrals
10.28. The problem can be solved as follows: 
y0 ¼ 3x 
2y 
) 2yy0 ¼ 3x ) y2 ¼ 3 
2 
x2 þ c 
x, yð Þ ¼ 1, 1ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 1 ¼ 3 
2
þ c ) c ¼ - 1 
2 
) y2 ¼ 3 
2 
x2 -
1 
2 
) 2y2 - 3x2 þ 1 ¼ 0 
Choice (1) is the answer. 
10.29. Based on the information given in the problem, we have: 
3x 
x2 þ 1p dx ¼ A x
2 þ 1þ c ð1Þ 
From list of integral of functions, we know that: 
1 
u
p du ¼ 2 up þ c 
The problem can be solved by changing the variable of the integral as follows: 
x2 þ 1 ≜ u 
d 
dx¼¼¼¼¼) 2xdx ¼ du ) xdx ¼ 1 
2 
du 
3x 
x2 þ 1p dx ¼ 
3x 1 2 du 
u
p ¼ 3 
2 
1 
u
p du ¼ 3 
2
x 2 up ¼ 3 x2 þ 1þ c ð2Þ 
Therefore: 
1ð Þ, 2ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) A x2 þ 1 þ c ¼ 3 x2 þ 1þ c ) A ¼ 3 
Choice (4) is the answer. 
10.30. From list of integral of functions, we know that: 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
The problem can be solved as follows: 
2
- 1 
xj jdx ¼ 
0
- 1 
xj jdxþ 
2 
0 
xj jdx ¼ 
0
- 1
- xð Þdxþ 
2 
0 
xdx ¼ - x
2 
2 
0
- 1
þ x
2 
2 
2 
0 
¼ - 0- 1 
2 
þ 2- 0ð Þ ¼ 5 
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- cos 2xð Þ ¼ 2 sin 2 xð Þ 
Furthermore, from list of integral of functions, we know that: 
xn dx ¼ 1 
nþ 1 x
nþ1 þ c 
cos axð Þdx ¼ 1 
a 
sin axð Þ þ c 
The problem can be solved as follows: 
π 
2 
0 
sin 2 xð Þdx ¼ 
π 
2 
0 
1 
2
-
cos 2xð Þ 
2 
dx ¼ 1 
2 
x-
1 
4 
sin 2xð Þ 
π 
2 
0
¼ 1 
2
x π2
- 0 - 0ð Þ ¼ π 
4 
Choice (3) is the answer. 
10.32. From list of integral of functions, we know that: 
xn dx ¼ 1 
nþ 1 x
nþ1 þ c 
The problem can be solved by changing the variable of the integral as follows: 
tan xð Þ ≜ u 
d 
dx¼¼¼¼¼¼) 1þ tan 2 xð Þ dx ¼ du 
3π 
4 
0 
tan 5 xð Þ þ tan 7 xð Þ dx ¼ 
3π 
4 
0 
tan 5 xð Þ 1þ tan 2 xð Þ dx 
¼ 
u2 
u1 
u5 du ¼ 1 
6 
u6 
u2 
u1 
¼ 1 
6 
tan 6 xð Þ 
3π 
4 
0 
¼ 1 
6
- 1ð Þ6 - 0 ¼ 1 
6 
Choice (4) is the answer. 
10.33. The problem can be solved as follows: 
y0 ¼ y2 cos xð Þ ) y
0 
y2 
¼ cos xð Þ ) - 1 
y 
¼ sin xð Þ þ c 
x, yð Þ ¼ π, 1ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) - 1 ¼ 0þ c ) c ¼ - 1 ) - 1 
y 
¼ sin xð Þ- 1 
We need to check each choice as follows: 
Choice 1 : 
x, yð Þ ¼ 3π 
2 
, 2 
¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) - 1 
2 
¼ sin 3π 
2
- 1 ) - 1 
2 
≠ - 2
206 10 Solutions of Problems: Definite and Indefinite Integrals
Choice 2 : 
x, yð Þ ¼ π 
2 
, - 1 
¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) - 1
- 1 
¼ sin π 
2
- 1 ) 1≠ 0 
Choice 3 : 
x, yð Þ ¼ π 
2 
, 1 
¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) - 1 
1 
¼ sin π 
2
- 1 ) - 1≠ 0 
Choice 4 : 
x, yð Þ ¼ 0, 1ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) - 1 
1 
¼ sin 0ð Þ- 1 ) - 1 ¼ - 1 
Choice (4) is the answer. 
10.34. From list of integral of functions, we know that: 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
Based on the information given in the problem, we have: 
I2 ¼ 
5 
3 
1 
x- 2 
dx ¼ m ð1Þ 
The problem can be solved as follows: 
I1 ¼ 
5 
3 
3x 
x- 2 
dx ¼ 
5 
3 
3x- 6 þ 6 
x- 2 
dx ¼ 
5 
3 
3dxþ 6 
5 
3 
1 
x- 2 
dx ð2Þ 
Solving (1) and (2): 
I1 ¼ 
5 
3 
3x 
x- 2 
dx ¼ 3x 5 
3
þ 6m ¼ 15- 9þ 6m ¼ 6þ 6m 
Choice (3) is the answer. 
10.35. From list of integral of functions, we know that: 
1 
u 
du ¼ ln uj j þ c 
The problem can be solved as follows: 
3 
2 
x 
x2 - 1 
dx ¼ 1 
2 
3 
2 
2x 
x2 - 1 
dx ð1Þ 
Now, we need to change the variable of the integral as follows: 
x2 - 1 ≜ u 
d 
dx¼¼¼¼¼¼¼¼¼¼) 2xdx ¼ du ð2Þ
10 Solutions of Problems: Definite and Indefinite Integrals 207
Solving (1) and (2): 
1 
2 
u2 
u1 
1 
u 
du ¼ 1 
2 
ln uj j u2 
u1 
¼ 1 
2 
ln x2 - 1 
3 
2 
¼ 1 
2 
ln 8-
1 
2 
ln 3 ¼ 1 
2 
ln 
8 
3 
¼ ln 8 
3 
Choice (4) is the answer. 
10.36. From list of integral of functions, we know that: 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
As we know: 
F xð Þ ¼ 
u xð Þ 
v xð Þ 
f xð Þdx ) F0 xð Þ ¼ u0 xð ÞF u xð Þð Þ- v0 xð ÞF v xð Þð Þ 
Therefore: 
f xð Þ ¼ 
x 
a 
t
p 
dt ) f 0 xð Þ ¼ xp 
4 
0 
f 0 xð Þdx ¼ 
4 
0 
x 
1 
2dx ¼ 2 
3 
x 
3 
2 
4 
0 
¼ 16 
3 
Choice (2) is the answer. 
10.37. From list of integral of functions, we know that: 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
The problem can be solved by changing the variable of the integral as follows: 
tan xð Þ ≜ u 
d 
dx¼¼¼¼¼¼¼¼¼¼) 1þ tan 2 xð Þ dx ¼ du 
8 tan 6 xð Þ þ tan 8 xð Þ dx ¼ 8 tan 6 xð Þ 1þ tan 2 xð Þ dx 
¼ 8 u6 du ¼ 8 
7 
u7 þ c ¼ 8 
7 
tan 7 xð Þ þ c 
Choice (4) is the answer. 
10.38. From list of integral of functions, we know that: 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c
208 10 Solutions of Problems: Definite and Indefinite Integrals
The problem can be solved as follows: 
1
- 1 
x þ 1ð Þ x2 þ 2xþ 3 dx ¼ 
1
- 1 
xþ 1ð Þ x þ 1Þ2 þ 2 dx ¼ 
1
- 1 
xþ 1ð Þ3 þ 2 x þ 1ð Þ dx ð1Þ 
Now, we should change the variable of the integral as follows: 
xþ 1 ≜ u ) dx ¼ du ð2Þ 
Solving (1) and (2): 
1
- 1 
u3 þ 2u dx ¼ u
4 
4 
þ u2 
¼ xþ 1ð Þ
4 
4 
þ xþ 1ð Þ2 1
- 1
¼ 16 
4 
þ 4- 0 ¼ 8 
Choice (3) is the answer. 
10.39. From list of integral of functions, we know that: 
xn dx ¼ 1 
nþ 1 x
nþ1 þ c 
The problem can be solved as follows: 
1 
1 
2 
1 
x
x 1 
x3 
dx ¼ 
1 
1 
2 
1x x- 3 dx ¼ x
- 2
- 2 
1 
1 
2 
¼ - 1 
2x2 
1 
1 
2 
¼ - 1 
2
- - 2ð Þ ¼ 3 
2 
Choice (4) is the answer. 
10.40. As we know: 
F xð Þ ¼ 
u xð Þ 
v xð Þ 
f xð Þdx ) F0 xð Þ ¼ u0 xð ÞF u xð Þð Þ- v0 xð ÞF v xð Þð Þ 
Therefore: 
y0 x ¼ u0 x þ v0 x ¼ 2x 
sin x2ð Þ 
x2
- 0 þ 0- 2x sin x
2ð Þ 
x2 
¼ 0 
Choice (3) is the answer. 
10.41. Based on the information given in the problem, we know that: 
h xð Þ ¼ f 0 3x þ 2ð Þdx ð1Þ 
h 0ð Þ ¼ 1 ð2Þ
¼
10 Solutions of Problems: Definite and Indefinite Integrals 209
f 2ð Þ ¼ 3 ð3Þ 
We should change the variable of the integral of h(x) as follows: 
f 3xþ 2ð Þ ≜ u ) 3f 0 3xþ 2ð Þdx ¼ du ð4Þ 
Solving (1) and (4): 
h xð Þ ¼ 1 
3 
du ¼ 1 
3 
u þ c ¼ 1 
3 
f 3xþ 2ð Þ þ c ð5Þ 
Solving (2) and (5): 
1 ¼ 1 
3 
f 2ð Þ þ c Using 3ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 1 ¼ 1 
3
x 3þ c ) c ¼ 0 ) h xð Þ ¼ 1 
3 
f 3xþ 2ð Þ 
) f 3xþ 2ð Þ ¼ 3h xð Þ 
Choice (3) is the answer. 
10.42. From list of integral of functions, we know that: 
xn du ¼ 1 
n þ 1 x
nþ1 þ c 
Based on the information given in the problem, we know that: 
y00 xð Þ ¼ 2x þ 1 ð1Þ 
The curve is tangent to y x in the origin. Thus: 
y0 0ð Þ ¼ 1 ð2Þ 
y 0ð Þ ¼ 0 ð3Þ 
Applying integral operation on (1): 
dx 
¼¼¼¼¼¼¼¼¼¼) y0 xð Þ ¼ x2 þ xþ a ð4Þ 
Applying integral operation on (4): 
dx 
¼¼¼¼¼¼¼¼¼¼) y xð Þ ¼ x
3 
3 
þ x
2 
2 
þ axþ b ð5Þ 
Solving (2) and (4): 
a ¼ 1 ð6Þ
210 10 Solutions of Problems: Definite and Indefinite Integrals
Solving (3) and (5): 
0 ¼ 0þ 0 þ 0þ b ) b ¼ 0 ð7Þ 
Solving (5)–(7): 
y xð Þ ¼ x
3 
3 
þ x
2 
2 
þ x 
Now, we need to check the choices as follows: 
y 1ð Þ ¼ 1 
3
þ 1 
2 
þ 1 ¼ 2þ 3þ 6 
6 
¼ 11 
6 
y 2ð Þ ¼ 2
3 
3 
þ 2
2 
2 
þ 2 ¼ 16þ 12þ 12 
6
¼ 20 
3 
Choice (1) is the answer. 
10.43. From trigonometry, we know that: 
1þ cos 2xð Þ ¼ 2 cos 2 xð Þ 
In addition, from list of integral of functions, we know that: 
sin axð Þdx ¼ - 1 
a 
cos axð Þ þ c 
The problem can be solved as follows: 
sin 2xð Þ cos 4xð Þdx ¼ sin 2xð Þ 2cos 2 2xð Þ- 1 dx 
¼ cos 2 2xð Þ x 2 sin 2xð Þdx- sin 2xð Þdx ð1Þ 
Now, we should change the variable of the integral as follows: 
cos 2xð Þ ≜ u ) - 2 sin 2xð Þdx ¼ du ð2Þ 
Solving (1) and (2):
- u2 du- sin 2xð Þdx ¼ - 1 
3 
u3 þ 1 
2 
cos 2xð Þ þ c ¼ - 1 
3 
cos 3 2xð Þ þ 1 
2 
cos 2xð Þ þ c 
Choice (2) is the answer. 
10.44. From list of integral of different functions, we know that: 
eu xð Þdu ¼ eu xð Þ þ c 
The problem can be solved by defining a new variable as follows: 
arctan x ¼ u xð Þ
10 Solutions of Problems: Definite and Indefinite Integrals 211
) dx 
1 þ x2 ¼ du 
Therefore: 
) I ¼ e
arctan x 
1þ x2 dx 
) I ¼ eu du ¼ eu þ c 
) 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 
nþ 1 x
nþ1 þ c 
The problem can be solved as follows: 
1
- 1 
x 
3 
dx ¼ 
0
- 1
- 1ð Þdx þ 
1 
0 
0dx ¼ - xð Þ 0
- 1
þ 0 ¼ 0- 1ð Þ ¼ - 1 
Choice (3) is the answer. 
10.46. The problem can be heuristically solved as follows: 
xy0 þ y ¼ 1 ) xyð Þ0 ¼ 1 ) xy ¼ xþ c 
x, yð Þ ¼ 1, 2ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 1x 2 ¼ 1þ c ) c ¼ 1 
) xy ¼ xþ 1 ) y ¼ 1þ 1 
x 
Choice (1) is the answer. 
10.47. The problem can be solved as follows: 
f 0 x3
pð Þ 
x23
p dx ¼ 3 1 
3 x23
p f 0 x3p dx ð1Þ 
Now, we should change the variable of the integral as follows: 
f x3
p 
≜ u ) 1 
3 x23
p f 0 x3p dx ¼ du ð2Þ 
Solving (1) and (2): 
3 du ¼ 3uþ c ¼ 3f x3p þ c 
Choice (4) is the answer.
212 10 Solutions of Problems: Definite and Indefinite Integrals
10.48. The problem can be solved as follows: 
I ¼ 
1 
0 
dx 
1þ eax
x e
- ax 
e- ax ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) I ¼ 
1 
0 
e- ax 
e- ax þ 1 dx 
From list of integral of different functions, we know that: 
du 
u xð Þ ¼ ln u xð Þ 
The problem can be solved by defining a new variable as follows: 
e- ax þ 1 ¼ u xð Þ 
) - ae- ax dx ¼ du ) e- ax dx ¼ du
- a 
Therefore: 
I ¼ 
1 
0 
e- ax 
e- ax þ 1 dx 
) I ¼ - 1 
a 
u2 
u1 
du 
u xð Þ ¼ -
1 
a 
ln u xð Þ½ ]u2 u1 
) I ¼ ln e- ax þ 1ð Þ½ ]1 0 
) I ¼ - 1 
a 
ln 1- ln 2½ ]
) I ¼ 1 
a 
ln 2 
Choice (3) is the answer. 
10.49. Fromintegration by parts (partial integration), we know that: 
u xð Þdv ¼ u xð Þv xð Þ- v xð Þdu 
In addition, from list of integral of functions, we know that: 
1 
u 
du ¼ ln uj j þ c 
The problem can be solved as follows:
10 Solutions of Problems: Definite and Indefinite Integrals 213
ln xdx ) u xð Þ ¼ ln x 
dv ¼ dx ) 
du ¼ dx 
x 
v xð Þ ¼ x 
) ln xdx ¼ x ln x- dx ¼ x ln x- xþ c 
Choice (1) is the answer. 
10.50. From trigonometry, we know that: 
1 
cos 2 xð Þ ¼ 1 þ tan
2 xð Þ 
sin xð Þ 
cos xð Þ ¼ tan xð Þ 
Moreover, from list of integral of functions, we know that: 
1 
u 
du ¼ ln uj j þ c 
The problem can be solved as follows: 
1 
sin xð Þ cos xð Þ dx ¼
1 
sin xð Þ cos xð Þ x cos xð Þ cos xð Þ 
dx ¼ 1 
sin xð Þ 
cos xð Þ cos
2 xð Þ 
dx ¼ 1þ tan
2 xð Þ 
tan xð Þ dx ð1Þ 
Now, we should change the variable of the integral as follows: 
tan xð Þ ≜ u ) 1þ tan 2 xð Þ dx ¼ du ð2Þ 
Solving (1) and (2): 
1 
u 
du ¼ ln uj j þ c ¼ ln tan xð Þj j þ c 
Choice (2) is the answer. 
10.51. From trigonometry, we know that: 
1þ tan 2 xð Þ ¼ 1 
cos 2 xð Þ 
The problem can be solved as follows: 
π 
4 
0 
1 
cos 4 xð Þ dx ¼ 
π 
4 
0 
1þ tan 2 xð Þ 1þ tan 2 xð Þ dx ð1Þ 
Now, we should change the variable of the integral as follows: 
tan xð Þ ≜ u 
d 
dx¼¼¼¼¼¼¼¼¼¼) 1þ tan 2 xð Þ dx ¼ du ð2Þ
214 10 Solutions of Problems: Definite and Indefinite Integrals
Solving (1) and (2): 
u2 
u1 
1þ u2 du ¼ u þ 1 
3 
u3 
u2 
u1 
¼ tan xð Þ þ 1 
3 
tan 3 xð Þ 
π 
4 
0 
¼ 1þ 1 
3
- 0 ¼ 4 
3 
Choice (4) is the answer. 
10.52. From trigonometry, we know that: 
1þ tan 2 xð Þ ¼ 1 
cos 2 xð Þ 
tan xð Þ ¼ sin xð Þ 
cos xð Þ 
The problem can be solved as follows: 
π 
4 
0 
1 
sin 2 xð Þ cos 4 xð Þ3 
dx ¼ 
π 
4 
0 
1 
sin 2 xð Þ cos 4 xð Þ x cos 2 xð Þ cos 2 xð Þ 3 
dx ¼ 
π 
4 
0 
1 
cos 2 xð Þ 
sin 2 xð Þ 
cos 2 xð Þ 
3 
dx 
¼ 
π 
4 
π 
6 
tan
- 2 
3 xð Þ 1 þ tan 2 xð Þ dx ð1Þ 
Now, we should change the variable of the integral as follows: 
tan xð Þ ≜ u 
d 
dx¼¼¼¼¼¼) 1þ tan 2 xð Þ dx ¼ du ð2Þ 
Solving (1) and (2): 
u2 
u1 
u-
2 
3du ¼ 3u 1 3 u2 
u1 
¼ 3 tan 1 3 xð Þ 
π 
4 
0 
¼ 3 1- 0ð Þ ¼ 3 
Choice (3) is the answer. 
10.53. From list of integral of functions, we know that: 
1 
u 
du ¼ ln uj j þ c 
Based on the information given in the problem, we have: 
f 1ð Þ ¼ 0 ð1Þ 
d 
dx 
f x2 ¼ 6 
x
ð2Þ 
The problem can be solved as follows:
10 Solutions of Problems: Definite and Indefinite Integrals 215
d 
dx 
f x2 ¼ 2xf 0 x2 ð3Þ 
2ð Þ, 3ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 6 
x 
¼ 2xf 0 x2 ) f 0 x2 ¼ 3 
x2
ð4Þ 
By changing the variable of the integral, we have: 
x2 ≜ t ð5Þ 
4ð Þ, 5ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) f 0 tð Þ ¼ 3 
t 
dt 
¼¼¼¼¼¼¼¼¼¼) f tð Þ ¼ 3 ln tj j þ c ð6Þ 
1ð Þ, 6ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 0 ¼ 3x 0þ c ) c ¼ 0 ) f tð Þ ¼ 3 ln tj j 
) f eð Þ ¼ 3 ln eð Þ ¼ 3x 1 ¼ 3 
Choice (3) is the answer. 
10.54. From list of integral of functions, we know that: 
xn dx ¼ 1 
nþ 1 x
nþ1 þ c 
From trigonometry, we know that: 
1þ cos 2xð Þ ¼ 2 cos 2 xð Þ 
Moreover, based on the information given in the problem, we have: 
f 1ð Þ ¼ 1 ð1Þ 
f 0 cos 2 xð Þ ¼ cos 2xð Þ ð2Þ 
The problem can be solved as follows: 
f 0 cos 2 xð Þ ¼ cos 2xð Þ ¼ 2 cos 2 xð Þ- 1 ð3Þ 
By changing the variable of the integral, we have: 
cos 2 xð Þ ≜ t ð4Þ 
3ð Þ, 4ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) f 0 tð Þ ¼ 2t- 1 
dt 
¼¼¼¼¼¼¼¼¼¼) f tð Þ ¼ t2 - t þ c ð5Þ 
1ð Þ, 5ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 1 ¼ 1- 1 þ c ) c ¼ 1 ) f tð Þ ¼ t2 - t þ 1 
) f - 1ð Þ ¼ - 1ð Þ2 - - 1ð Þ 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 
n þ 1 x
nþ1 þ c 
From trigonometry, we know that: 
sin 2xð Þ ¼ 2 sin xð Þ cos xð Þ 
The problem can be solved as follows: 
cos 2xð Þ 
sin 2 xð Þ cos 2 xð Þ dx ¼
cos 2xð Þ 
1 
2 sin 2xð Þ 
2 dx ¼
cos 2xð Þ 
1 
4 sin
2 2xð Þ dx ¼ 4 cos 2xð Þ sin 2xð Þð Þ
- 2 dx 
Now, we need to change the variable of the integral as follows: 
sin 2xð Þ ≜ u 
d 
dx¼¼¼¼¼¼) 2 cos 2xð Þdx ¼ du 
) 2u- 2 du ¼ - 2 
u
þ c ¼ - 2 
sin 2xð Þ þ c 
Choice (1) is the answer. 
10.56. From integration by parts (partial integration), we know that: 
ln xð Þdx ¼ x ln xj j- x 
or, in general: 
u xð Þdv ¼ u xð Þv xð Þ- v xð Þdu 
In addition, from list of integral of functions, we know that: 
1 
u 
du ¼ ln uj j þ c 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
The problem can be solved as follows: 
e 
1 
2xþ ln xð Þð Þdx ¼ 
e 
1 
2x dxþ 
e 
1 
ln xð Þdx ¼ x2 e 
1
þ xln xj j- xð Þ e 
1 
¼ e2 - 1 þ e- eð Þ- 0- 1ð Þ ¼ e2 
Choice (1) is the answer.
10 Solutions of Problems: Definite and Indefinite Integrals 217
10.57. From list of integral of functions, we know that: 
xn dx ¼ 1 
nþ 1 x
nþ1 þ c 
The problem can be solved as follows: 
sin 2xð Þ 2 þ cos 2 xð Þ 50 dx ð1Þ 
We should change the variable of the integral as follows: 
2þ cos 2 xð Þ ≜ u ) - 2 cos xð Þ sin xð Þdx ¼ du ) - sin 2xð Þ ¼ du ð2Þ 
Solving (1) and (2):
- u50 du ¼ - u
51 
51 
þ c ¼ - 1 
51 
2þ cos 2 xð Þ 51 þ c 
Choice (2) is the answer. 
10.58. From list of integral of functions, we know that: 
un du ¼ 1 
n þ 1 u
nþ1 þ c 
The problem can be solved as follows: 
e 
1 
ln xð Þ 
x 
dx ¼ 
e 
1 
ln xð Þ 1 
x 
dx ð1Þ 
Now, we should change the variable of the integral as follows: 
ln xð Þ ≜ u ) 1 
x 
dx ¼ du ð2Þ 
Solving (1) and (2): 
u2 
u1 
udu ¼ 1 
2 
u2 
u2 
u1 
¼ 1 
2 
ln xð Þð Þ2 e 
1 
¼ 1 
2
- 0 ¼ 1 
2 
Choice (2) is the answer. 
10.59. From list of integral of functions, we know that: 
un du ¼ 1 
n þ 1 u
nþ1 þ c 
The problem can be solved as follows:
218 10 Solutions of Problems: Definite and Indefinite Integrals
y0 ¼ - 2xþ 2 
4yþ 1 ) 4yy
0 þ y0 ¼ - 2x- 2 
dx 
¼¼¼¼¼¼¼¼¼¼) 2y2 þ y ¼ - x2 - 2xþ c ð1Þ 
x, yð Þ ¼ 0, 1ð Þ¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼¼) 2þ 1 ¼ 0þ c ) c ¼ 3 ð2Þ 
Solving (1) and (2): 
2y2 þ y ¼ - x2 - 2xþ 3 ) x2 þ 2y2 ¼ - y- 2xþ 3 
Choice (4) is the answer. 
10.60. From list of integral of functions, we know that: 
un du ¼ 1 
nþ 1 u
nþ1 þ c 
Moreover, from trigonometry, we know that: 
cot xð Þ ¼ cos xð Þ 
sin xð Þ 
1- cos 2xð Þ ¼ 2 sin 2 xð Þ 
The problem can be solved as follows: 
π 
2 
π 
6 
cot xð Þ 
1- cos 2xð Þ dx ¼ 
π 
2 
π 
6 
cos xð Þ 
2 sin 2 xð Þ sin xð Þ 
dx ¼ 
π 
2 
π 
6 
cos xð Þ 
2
p 
sin xð Þj j sin xð Þ dx 
¼ 2
p 
2 
π 
2 
π 
6 
sin xð Þð Þ- 2 cos xð Þdx ð1Þ 
Now, we should change the variable of the integral as follows: 
sin xð Þ ≜ u ) cos xð Þdx ¼ du ð2Þ 
Solving (1) and (2): 
2
p 
2 
u2 
u1 
u- 2 du ¼ - 2
p 
2 
u- 1 
u2 
u1 
¼ - 2
p 
2
x 1 
sin xð Þ 
π 
2 
π 
6 
¼ - 2
p 
2 
1- 2ð Þ ¼ 2
p 
2 
Choice (2) is the answer. 
10.61. From list of integral of functions, we know that: 
un du ¼ 1 
nþ 1 u
nþ1 þ c 
The problem can be solved as follows:
10 Solutions of Problems: Definite and Indefinite Integrals 219
6 
3 
xþ 2 
x- 2
p dx ¼ 
6 
3 
x- 2 þ 4 
x- 2
p dx ¼ 
6 
3 
x- 2ð Þ1 2 þ 4 x- 2ð Þ- 1 2 dx 
¼ 2 
3 
x- 2ð Þ3 2 þ 4x 2 x- 2ð Þ1 2 6 
3
¼ 2 
3 
4ð Þ3 2 þ 4x 2 4ð Þ1 2 - 2 
3 
1ð Þ3 2 þ 4x 2 1ð Þ1 2 
¼ 16 
3 
þ 16 - 2 
3
þ 8 ¼ 14 
3 
þ 8 ¼ 38 
3 
Choice (2) is the answer. 
10.62. In addition, from list of integral of functions, we know that: 
un du ¼ 1 
nþ 1 u
nþ1 þ c 
The problem can be solved as follows: 
4 
1 
1þ xp 
x
p dx ¼ 2 
4 
1 
1 þ xp 1 2 x 1 
2 x
p dx ð1Þ 
Now, we should change the variable of the integral as follows: 
1þ xp ≜ u ) 1 
2 x
p dx ¼ du ð2Þ 
Solving (1) and (2): 
2 
u2 
u1 
u 
1 
2du ¼ 2x 2 
3 
u 
3 
2 
u2 
u1 
¼ 4 
3 
1þ xp 3 2 4 
1 
¼ 4 
3 
3 3 
p
- 2 2 
p 
¼ 4 3 
p
-
2 2
p 
3 
Choice (4) is the answer. 
10.63. In addition, from list of integral of functions, we know that: 
un du ¼ 1 
nþ 1 u
nþ1 þ c 
From trigonometry, we know that: 
cot xð Þ ¼ cos xð Þ 
sin xð Þ 
The problem can be solved as follows: 
cot xð Þ sin xð Þdx ¼ cos xð Þ 
sin xðÞ sin xð Þð Þ
1 
2 dx ¼ sin xð Þð Þ- 1 2 cos xð Þdx ð1Þ 
Now, we should change the variable of the integral as follows: 
sin xð Þ ≜ u ) cos xð Þdx ¼ du ð2Þ 
Solving (1) and (2):
220 10 Solutions of Problems: Definite and Indefinite Integrals
u
- 1 
2 du ¼ 2u 1 2 þ c ¼ 2 sin xð Þ þ c 
Choice (2) is the answer. 
10.64. From trigonometry, we know that: 
sec xð Þ ¼ 1 
cos xð Þ 
tan xð Þ ¼ sin xð Þ 
cos xð Þ 
The problem can be solved as follows: 
π 
3 
0 
sec xð Þ tan xð Þdx ¼ 
π 
3 
0 
1 
cos xð Þ 
sin xð Þ 
cos xð Þ dx ¼ 
π 
3 
0 
sin xð Þ 
cos 2 xð Þ dx ð1Þ 
Now, we should change the variable of the integral as follows: 
cos xð Þ ≜ u ) - sin xð Þdx ¼ du ð2Þ 
Solving (1) and (2):
-
u2 
u1 
1 
u2 
du ¼ 1 
u 
u2 
u1 
¼ 1 
cos xð Þ 
π 
3 
0 
¼ 1 1 
2
-
1 
1 
¼ 1 
Choice (1) is the answer. 
10.65. From trigonometry, we know that: 
csc xð Þ ¼ 1 
sin xð Þ 
cot xð Þ ¼ cos xð Þ 
sin xð Þ 
The problem can be solved as follows: 
π 
4 
π 
6 
csc xð Þ cot xð Þdx ¼ 
π 
4 
π 
6 
1 
sin xð Þ 
cos xð Þ 
sin xð Þ dx ¼ 
π 
4 
π 
3 
cos xð Þ 
sin 2 xð Þ dx ð1Þ 
Now, we should change the variable of the integral as follows: 
sin xð Þ ≜ u ) cos xð Þdx ¼ du ð2Þ 
Solving (1) and (2): 
u2 
u1 
1 
u2 
du ¼ - 1 
u 
u2 
u1 
¼ - 1 
sin xð Þ 
π 
4 
π 
6 
¼ 1 
2
p 
2
-
1 
1 
2 
¼ - 2 
p
- 2 ¼ 2- 2 
p 
Choice (2) is the answer.
p p
10 Solutions of Problems: Definite and Indefinite Integrals 221
10.66. From trigonometry, we know that: 
1þ cot 2 xð Þ ¼ 1 
sin 2 xð Þ 
1þ tan 2 xð Þ ¼ 1 
cos 2 xð Þ 
tan xð Þ cot xð Þ ¼ 1 
Moreover, from list of integral of functions, we know that: 
1þ tan 2 xð Þ dx ¼ tan xð Þ þ c 
1þ cot 2 xð Þ dx ¼ - cot xð Þ þ c 
The problem can be solved as follows: 
π 
4 
π 
6 
1 
sin 2 xð Þ cos 2 xð Þ dx ¼ 
π 
4 
π 
6 
1þ cot 2 xð Þ 1þ tan 2 xð Þ dx 
¼ 
π 
4 
π 
6 
1þ tan 2 xð Þ þ cot 2 xð Þ þ cot 2 xð Þ tan 2 xð Þ dx ¼ 
π 
4 
π 
6 
1þ tan 2 xð Þ þ cot 2 xð Þ þ 1 dx 
π 
4 
π 
6 
1þ tan 2 xð Þ dx þ 
π 
4 
π 
6 
1þ cot 2 xð Þ dx ¼ tan xð Þ- cot xð Þð Þ 
π 
4 
π 
6 
¼ 1- 1ð Þ- 3 
3
- 3 
p 
¼ 2 3 
3 
Choice (2) is the answer. 
10.67. From list of integral of functions, we know that: 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
The problem can be solved as follows: 
tan xð Þ- cot xð Þð Þ tan xð Þ þ cot xð Þð Þ5 dx 
¼ tan xð Þ- cot xð Þð Þ tan xð Þ þ cot xð Þð Þ tan xð Þ þ cot xð Þð Þ4 dx 
¼ tan 2 xð Þ- cot 2 xð Þ tan xð Þ þ cot xð Þð Þ4 dx
222 10 Solutions of Problems: Definite and Indefinite Integrals
¼ 1þ tan 2 xð Þ- 1þ cot 2 xð Þ tan xð Þ þ cot xð Þð Þ4 dx ð1Þ 
Now, we need to change the variable of the integral as follows: 
tan xð Þ þ cot xð Þ ≜ u 
d 
dx¼¼¼¼¼¼¼¼¼¼) 1þ tan 2 xð Þ- 1 þ cot 2 xð Þ dx ¼ du ð2Þ 
Solving (1) and (2): 
u4 du ¼ u
5 
5 
þ c ¼ 1 
5 
tan xð Þ þ cot xð Þð Þ5 þ c 
Choice (2) is the answer. 
10.68. From trigonometry, we know that: 
1þ cos 2xð Þ ¼ 2 cos 2 xð Þ ð1Þ 
sin 2 xð Þ þ cos 2 xð Þ ¼ 1 ð2Þ 
The problem can be solved as follows: 
I ¼ sin xð Þ cos xð Þdx ð3Þ 
Now, we should change the variable of the integral as follows: 
sin xð Þ ≜ u ) cos xð Þdx ¼ du ð4Þ 
Solving (3) and (4): 
I ¼ udu ¼ 1 
2 
u2 þ c ¼ 1 
2 
sin 2 xð Þ þ c ð5Þ 
Solving (2) and (5): 
I ¼ 1 
2 
1- cos 2 xð Þ þ c ¼ - 1 
2 
cos 2 xð Þ þ 1 
2
þ c ¼ - 1 
2 
cos 2 xð Þ þ c0 ð6Þ 
Solving (1) and (6): 
I ¼ - 1 
2 
1 
2 
þ 1 
2 
cos 2xð Þ þ c0 ¼ - 1 
4 
cos 2xð Þ þ c0 - 1 
4 
¼ - 1 
4 
cos 2xð Þ þ c00 ð7Þ 
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 
u xð Þ du ¼ ln u xð Þð Þ 
The problem can be solved as follows: 
I ¼ 
ln 3 
ln 2 
1- e- 2x 
1þ e- 2x dx
x e
x 
ex ¼¼¼¼¼¼¼¼¼¼) I ¼ 
ln 3 
ln 2 
ex - e- x 
ex þ e- x dx 
By defining the new variable, we have: 
u xð Þ ¼ ex þ e- x 
) du ¼ ex - e- xð Þdx 
Therefore: 
I ¼ 
u2 
u1 
1 
u xð Þ du ¼ ln u xð Þð Þ½ ]
u2 
u1 
) I ¼ ln ex þ e- xð Þ½ ] ln 3 
ln 2 
¼ ln eln 3 þ e- ln 3 - ln eln 2 þ e- ln 2 
) I ¼ ln eln 3 þ e ln 3- 1 - ln eln 2 þ e ln 2- 1 
I ¼ ln 3þ 3- 1 - ln 2þ 2- 1 
) I ¼ ln 3þ 1 
3
- ln 2þ 1 
2 
¼ ln 10 
3
- ln 
5 
2 
¼ ln 20 
15 
¼ ln 4 
3 
¼ ln 4- ln 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 a 
b 
ln an ¼ n ln a
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 
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 
# 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
L 
Left continuous, 116, 133 
Left-hand side continuity, 103 
Limits and continuity, 103–119 
N 
Non-differentiable point, 139, 152 
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
https://doi.org/10.1007/978-3-031-45028-0#DOI
	Preface
	Precalculus: Practice Problems, Methods, and Solution
	Calculus 1: Practice Problems, Methods, and Solution
	Calculus 2: Practice Problems, Methods, and Solution
	Calculus 3: Practice Problems, Methods, and Solution
	The Other Works Published by the Author
	Contents
	1: Problems: Characteristics of Functions
	References
	2: Solutions of Problems: Characteristics of Functions
	References
	3: Problems: Trigonometric Equations and Identities
	References
	4: Solutions of Problems: Trigonometric Equations and Identities
	References
	5: Problems: Limits and Continuities
	References
	6: Solutions of Problems: Limits and Continuities
	References
	7: Problems: Derivatives and Their Applications
	References
	8: Solutions of Problems: Derivatives and Their Applications
	References
	9: Problems: Definite and Indefinite Integrals
	References
	10: Solutions of Problems: Definite and Indefinite Integrals
	References
	Index