Precalculus: Practice Problems, Methods, and Solutions 3030650553, 9783030650551

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Mehdi Rahmani-Andebili

Precalculus

Practice Problems, Methods, and Solutions

Precalculus

Mehdi Rahmani-Andebili

Precalculus Practice Problems, Methods, and Solutions

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

ISBN 978-3-030-65055-1 ISBN 978-3-030-65056-8 https://doi.org/10.1007/978-3-030-65056-8

(eBook)

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

Preface

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

Mehdi Rahmani-Andebili

v

Contents

1

Problems: Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Real Number Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Exponents and Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Absolute Values and Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

1 1 2 6

Solutions of Problems: Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities . . . . . . . . . . . . . . . . . . . 2.1 Real Number Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Exponents and Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Absolute Values and Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

9 9 10 17

3

Problems: Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21

4

Solutions of Problems: Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . .

27

5

Problems: Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33

6

Solutions of Problems: Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . .

39

7

Problems: Functions, Algebra of Functions, and Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49

Solutions of Problems: Functions, Algebra of Functions, and Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

57

9

Problems: Factorization of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

71

10

Solutions of Problems: Factorization of Polynomials . . . . . . . . . . . . . . . . . . . .

77

11

Problems: Trigonometric and Inverse Trigonometric Functions . . . . . . . . . . .

83

12

Solutions of Problems: Trigonometric and Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

89

2

8

13

Problems: Arithmetic and Geometric Sequences and Series . . . . . . . . . . . . . . 101

14

Solutions of Problems: Arithmetic and Geometric Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

vii

1

Problems: Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities

Abstract

In this chapter, the basic and advanced problems of real number systems, exponents, radicals, absolute values, and inequalities 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

Real Number Systems

1.1. Which one of the numbers below exists? Difficulty level Calculation amount 1) 2) 3) 4)

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large

The minimum integer number smaller than
1 The minimum irrational number larger than
1 The maximum integer number smaller than
1 The maximum rational number smaller than
1

1.2. As we know, ℝ is set of real numbers, ℤ is set of integer numbers, and ℕ is set of natural numbers. Which one of the choices is correct? Difficulty level Calculation amount 1) ℕ ⊂ ℤ ⊂ ℝ 2) ℝ ⊂ ℤ ⊂ ℕ 3) ℝ ⊂ ℕ ⊂ ℤ 4) ℤ ⊂ ℝ ⊂ ℕ

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large

1.3. If ab ¼ dc, then which one of the choices is correct? Difficulty level Calculation amount 1) 2) 3) 4)

b c b a a c b a

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large

¼ ab 6¼ dc 6¼ db ¼ dc

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Precalculus, https://doi.org/10.1007/978-3-030-65056-8_1

1

2

1

Problems: Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities

1.4. Which one of the choices has the greatest absolute value? Difficulty level Calculation pffiffiffiamount pffiffiffi 1)
1 þ 2
3 pffiffiffi pffiffiffi 2)
1
2 þ 3 pffiffiffi pffiffiffi 3)
1
2
3 pffiffiffi pffiffiffi 4) 1 þ 2
3

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

1.5. Which one of the choices below might be a rational number if respectively? Difficulty level Calculation amount 1) α + β4 2) α + β 3) 1þβpffiffiα

○ Easy ● Small

○ Normal ○ Normal

pffiffiffi α and β2 are rational and irrational numbers,

● Hard ○ Large

4) α2 + β

1.2

Exponents and Radicals

1.6. Calculate the value of x2 if x ¼ Difficulty level Calculation amount pffiffiffi 1) 2 pffiffiffi 2) 3 2 pffiffiffi 3) 3 4 4) 2

● Easy ● Small

 1.7. Calculate the value of Difficulty level Calculation pffiffiffiamount 1)
1 þ 2 pffiffiffi 2)
2 þ 2 pffiffiffi  3)
1 þ 2 pffiffiffi  4)
2 þ 2

p1ffiffi 2


1

p ffiffiffiffiffiffiffiffiffi pffiffiffi 3 2 2. ○ Normal ○ Normal


1

● Easy ● Small

.

○ Normal ○ Normal

1.8. Simplify and calculate the final answer of Difficulty level Calculation amount 1)
2x
2 2)
2 3) 2x + 2 4) 2

○ Easy ● Small

○ Hard ○ Large

○ Hard ○ Large

qffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffi 3 ð
xÞ3 þ x2 þ ð
2Þ2 if x > 0.

● Normal ○ Normal

○ Hard ○ Large

 pffiffiffiffiffi53 10 1.9. Calculate the final answer of
36 . Difficulty level

○ Easy

● Normal

○ Hard

(continued)

1.2 Exponents and Radicals

Calculation amount 1)
9 2) 9 3)
3 4) 3

3

● Small

○ Normal

○ Large

pffiffiffiffiffi pffiffiffiffiffi 1.10. Calculate the value of 2 3 x3 þ 4 x4 if x < 0. Difficulty level Calculation amount 1) 3x 2) x 3)
x 4)
3x

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

1.11. In the equation below, determine the value of x: 3x
1 ¼ Difficulty level Calculation amount 1) 24 2) 27 3) 31 4) 33

○ Easy ● Small

● Normal ○ Normal



1 81


8

○ Hard ○ Large

k

1.12. Calculate the value of 22 þ 1 for k ¼ 3. Difficulty level Calculation amount 1) 33 2) 65 3) 129 4) 257

○ Easy ● Small

● Normal ○ Normal

1.13. Solve the equation below:

Difficulty level Calculation amount 1)
1 2) 1 3) 37 4) 73

○ Easy ● Small

 3x
7  7x
3 3 7 ¼ 7 3 ● Normal ○ Normal

1.14. Calculate the value of the following term:

Difficulty level Calculation amount 1) 25 2) 2 3) 52

○ Easy ○ Small

○ Hard ○ Large



● Normal ● Normal

○ Hard ○ Large

8 25


3

 ð0:8Þ4  0:2

○ Hard ○ Large

4

1

Problems: Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities

4) 5

1.15. Calculate the value of the term below:

Difficulty level Calculation amount 1) 52 2) 10 3 3) 5 4) 15 2

○ Easy ○ Small

● Normal ● Normal

1.16. Calculate the value of the term below:

Difficulty level Calculation amount 1) 18 2) 2 3) 8 4) 4

 5 25 3   ð0:75Þ
3 90 2

○ Easy ○ Small

○ Hard ○ Large

pffiffiffi6 4 2 16   ð0:5Þ
6  8
3 2

● Normal ● Normal

○ Hard ○ Large

1.17. Calculate the value of the term below: 3  ð45Þ6 ð15Þ6  37 Difficulty level Calculation amount 1) 1 2) 3 3) 5 4) 15

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

1.18. Calculate the value of x if we have the following term: !1  1 12 2 p ffiffiffi 1 2 x ð16Þ3 2¼ Difficulty level Calculation amount 1) 5 2) 6 3) 9 4) 12

○ Easy ○ Small

● Normal ● Normal

1

1.19. Calculate the value of ðx þ x
1 Þ3 if x ¼ 1
Difficulty level Calculation amount pffiffiffi 1)
2 2)
1

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

pffiffiffi 2. ○ Hard ○ Large

1.2 Exponents and Radicals

5

pffiffiffi 3) 3 2 4) 1

1.20. Simplify and calculate the final answer of the term below: 1 1 1 pffiffiffi pffiffiffiffiffi þ pffiffiffiffiffi pffiffiffiffiffi þ pffiffiffiffiffi pffiffiffiffiffi 4 þ 11 11 þ 18 18 þ 25 Difficulty level Calculation amount 1) 27 2) 37 3) 13 4) 23

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

1.21. Determine the relation between A and B if: tþ1

1

A ¼ x t , B ¼ xtþ1 , t 6¼ 0,
1 Difficulty level Calculation amount t tþ1 1) Atþ1 ¼ B t t 2) Atþ1 ¼ Btþ1 1 3) Atþ1 ¼ Btþ1 1 4) Atþ1 ¼ Btþ1

○ Easy ○ Small

● Normal ○ Normal

○ Hard ● Large

1.22. Assume that a and b have different signs and a < b. Then, which one of the following statements is true? Difficulty level Calculation amount 1) a2 < b2 2) a3 < b3 3) b2 < a3 4) b3 < a3

○ Easy ● Small

1.23. Determine the final answer of Difficulty level Calculation amount pffiffiffi 1) 2 2) 2 pffiffiffi 3) 2 2 4) 4 1.24. Calculate the value of Difficulty level Calculation amount 1) 12 2) 1 3) 32 4) 2

○ Normal ○ Normal

● Hard ○ Large

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi p pffiffiffi 4 4
2 2  6 þ 4 2.

○ Easy ○ Small

○ Normal ● Normal

● Hard ○ Large

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 4 7
4 3 2 þ 3. ○ Easy ○ Small

○ Normal ● Normal

● Hard ○ Large

6

1

Problems: Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities

1.25. Which one of the numbers is greater? Difficulty level Calculation pffiffiffi amount 1) 1 þ 2 2 pffiffiffi 2) 1 þ 6 6 pffiffiffi 3) 1 þ 4 4 pffiffiffi 4) 1 þ 3 3

○ Easy ○ Small

○ Normal ● Normal

● Hard ○ Large

1.26. Determine the final answer of the term below: 40:75 pffiffiffi pffiffiffi þ 90:25 1þ 2þ 3 Difficulty level Calculation amount pffiffiffi 1) 2
1 2) 1 pffiffiffi 3) 2 pffiffiffi 4) 1 þ 2

1.3

○ Easy ○ Small

○ Normal ○ Normal

● Hard ● Large

Absolute Values and Inequalities

1.27. If |
x + 1| < 2, determine the range of x. Difficulty level Calculation amount 1)
3 < x <
1 2)
3 < x < 1 3)
1 < x < 3 4) 1 < x < 3

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large

1.28. What is the solution of
1  3x
2  1? Difficulty level Calculation amount 1) 13  x  1 2)
1  x  1 3)
1  x  13 4)
2  x  1

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large

1.29. Which one of the choices shows the range of the solution of |2x
3|  5? Difficulty level Calculation amount 1) [2, 6] 2) [
1, 4] 3) [
2, 2] 4) [
3, 1]

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large

1.3 Absolute Values and Inequalities

7

1.30. If the inequalities of |x
1| < 0.1 and A < 2x
3 < B are equivalent, calculate the value of A + B. Difficulty level Calculation amount 1)
2.1 2)
2 3)
1.1 4)
1

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

1.31. Which one of the choices is equivalent to |2x
3| < x? Difficulty level Calculation amount 1) |x
2| < 1 2) |x
1| < 2 3) |x
3| < 1 4) |x
2| < 2

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

1.32. Solve the inequality below: 1 1 > x
1 x
3 Difficulty level Calculation amount 1) x < 3 2) 1 < x < 3 3) 2 < x < 3 4)
2 < x < 3

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

1.33. Solve the inequality of |x
1|  |x
3|. Difficulty level Calculation amount 1) |x|  2 2) |x
2|  1 3) x  2 4) x  2

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

1.34. Solve the inequality of |x + 1|  |x
1|. Difficulty level Calculation amount 1) x  0 2) x  0 3) x  1 4) x 
1

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

1.35. Which one of the choices is equivalent to
5 < x
11 < 3? Difficulty level Calculation amount 1) |x + 5| < 2 2) |x| < 7 3) |x
9| < 5

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

8

1

Problems: Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities

4) |x
10| < 4 1.36. Calculate the value of |2x
1| + |2
x| if
1 < x < 0. Difficulty level Calculation amount 1)
3
3x 2) 3
3x 3)
3 + 3x 4) 1 + x

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

1.37. Determine the neighborhood radius for the deleted symmetric neighborhood of (3a
7, a + 5)
{3}. Difficulty level Calculation amount 1) 1 2) 2 3) 3 4) 4

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

1.38. If b < 0 < a and ja j > j bj, calculate the final answer of |a + b| + |a| + j bj. Difficulty level Calculation amount 1)
2b 2)
2a 3) 2a 4) 2b

○ Easy ● Small

○ Normal ○ Normal

● Hard ○ Large

1.39. Determine the number of answers of x in the equation of |x + 1| + |x
3| ¼ 2. Difficulty level Calculation amount 1) 0 2) 1 3) 2 4) 3

○ Easy ● Small

○ Normal ○ Normal

● Hard ○ Large

2

Solutions of Problems: Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities

Abstract

In this chapter, the problems of the first chapter are fully solved, in detail, step-by-step, and with different methods.

2.1

Real Number Systems

2.1. Choice (1) is not correct because set of integer numbers is not left-bounded. Choice (2) is not correct because between any pair of real numbers, infinite irrational numbers exist. Choice (3) is correct because
2 is the maximum integer number smaller than
1. Choice (4) is not correct because between any pair of real numbers, infinite rational numbers exist. Choice (3) is the answer. 2.2. Figure 2.1 shows the diagram of set of real numbers, irrational numbers, rational numbers, integer numbers, whole numbers, and natural numbers. Choice (1) is the answer.

Figure 2.1 The diagram of problem 2.2

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Precalculus, https://doi.org/10.1007/978-3-030-65056-8_2

9

10

2 Solutions of Problems: Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities

2.3. We know that: α¼β) Therefore:

1 1 ¼ α β

a c b d ¼ ) ¼ b d a c

Choice (4) is the answer. pffiffiffi pffiffiffi 2.4. As we know, 3 ¼ 1:73 and 2 ¼ 1:41, therefore: Choice (1):
1 þ

  pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi  2
3 < 0 ) 
1 þ 2
3 ¼

1 þ 2
3  1:31


1


  pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi  2 þ 3 < 0 ) 
1
2 þ 3 ¼

1
2 þ 3  0:68


1


  pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi  2
3 < 0 ) 
1
2
3 ¼

1
2
3  4:14

Choice (2):

Choice (3):

Choice (4): 1þ

 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi  2
3 > 0 ) 1 þ 2
3 ¼ 1 þ 2
3  0:68

Choice (3) is the answer. 2.5. To solve the problem, we need to recall the following points: Point 1: A rational number will remain a rational number even if it is squared. pffiffiffi Point 2: An irrational number might be changed to a rational number if it is squared (e.g., β ¼ 4 2 ) β4 ¼ 2). Point 3: An irrational number will remain an irrational number if its square root is taken. Point 4: The sum of two rational numbers is always a rational number. Point 5: The sum of a rational number and an irrational number is always an irrational number. Point 6: The product of a nonzero rational number and an irrational number is always an irrational number. Point 7: A rational number will remain a rational number if it is inversed. Choice (1) might be a rational number based on points 1, 2, and 4. Choice (2) is an irrational number based on points 1, 3, and 5. Choice (3) is an irrational number based on points 3, 4, 7, and 6. Choice (4) is an irrational number based on points 1, 3, and 5. Choice (1) is the answer.

2.2

Exponents and Radicals

2.6. We know that:

ffiffiffiffiffi p n m an ¼ am ðam Þn ¼ amn

Hence: Choice (4) is the answer.

qffiffiffiffiffiffiffiffiffi  1  1 2 pffiffiffi 3 3 3 1 x ¼ 2 2 ¼ 22 ¼ 22 ) x 2 ¼ 22 ¼ 2

2.2 Exponents and Radicals

11

2.7. We know that: ð a þ bÞ ð a
bÞ ¼ a2
b2 Thus:

pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi  
1 pffiffiffi  2 1þ 2 2 2 1 1 1þ 2 pffiffiffi
1 pffiffiffi ¼ pffiffiffi  pffiffiffi ¼ ¼ 1 ¼
2þ2 ¼ pffiffi
1 1
2 2 1
2 1
2 1þ 2 2

Choice (4) is the answer. 2.8. From exponent and radical rules, we know that: ffiffiffiffiffi p n an ¼ a, if n is an odd number ffiffiffiffiffi p n an ¼ jaj, if n is an even number Therefore:

qffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffi 3 ð
xÞ3 þ x2 þ ð
2Þ2 ¼ ð
xÞ þ jxj þ j
2j x>0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )
xþxþ2¼2

Choice (4) is the answer. 2.9. From exponent and radical rules, we know that: ðabÞn ¼ an bn ðam Þn ¼ amn Hence:

 p 53  6 53  13  6 5  ffiffiffiffiffi53  5 1 6 10 3103 ¼ ð
1Þ3 31 ¼
1  3 ¼
3
36 ¼
1  310 ¼ ð
1Þ3 310 ¼ ð
1Þ5

Choice (3) is the answer. 2.10. From exponent and radical rules, we know that: ffiffiffiffiffi p n an ¼ a, if n is an odd number ffiffiffiffiffi p n an ¼ jaj, if n is an even number Therefore:

ffiffiffiffiffi p p ffiffiffiffiffi 3 4 2 x3 þ x4 ¼ 2x þ jxj x>0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 2x
x ¼ x

Choice (2) is the answer. 2.11. We know that:  
n 1 ¼ an a Therefore:

12

2 Solutions of Problems: Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities



1 81


8

 8 ¼ 818 ¼ 34 ¼ 332

3x
1 ¼ 332 ) x
1 ¼ 32 ) x ¼ 33 Choice (4) is the answer. 2.12. From exponent rules, we know that: zfflfflfflfflfflfflffl}|fflfflfflfflfflfflffl{ ¼ an  . . .  n m times

a

nm

Therefore for k ¼ 3, we can write: k

3

22 þ 1 ¼ 22 þ 1 ¼ 28 þ 1 ¼ 256 þ 1 ¼ 257 Choice (4) is the answer. 2.13. From exponent rules, we know that: an ¼

 
n 1 a

an ¼ am ) n ¼ m Therefore:

 3x
7  7x
3  3x
7  
7xþ3 3 7 3 3 ¼ ) ¼ 7 3 7 7 ) 3x
7 ¼
7x þ 3 ) 10x ¼ 10 ) x ¼ 1

Choice (2) is the answer. 2.14. From exponent rules, we know that: an ¼

 
n 1 a

 n a an ¼ n b b ðam Þn ¼ amn an ¼ an
m am The problem can be solved as follows: 

8 25


3



8 ð0:8Þ ¼ 10 4

 ¼

4

¼

25 8

 2 3 5 56 ¼  3 ¼ 9 2 23

 3 4 2 ð 2  5Þ

0:2 ¼ Therefore:

3

4

¼

2 1 ¼ 10 5

212 28 ¼ 4 4 2 5 5 4

2.2 Exponents and Radicals

13



8 25


3

 ð0:8Þ4  0:2 ¼

56 28 1 56 28 1 5   ¼  ¼5 ¼ 2 2 29 54 5 55 29

Choice (3) is the answer.

2.15. From exponent rules, we know that: an ¼

 
n 1 a

 n a an ¼ n b b ðam Þn ¼ amn an ¼ an
m am Therefore:  2 3  5  5  
3 2 25 3 5 3 3 5 35 43 5 35    ð0:75Þ
3 ¼  ¼   ¼   90 2 18 2 4 33 2  32 25 33 2  32 25 ¼5

35 26 35 26  ¼ 5   ¼511¼5 32  33 2  25 35 26

Choice (3) is the answer. 2.16. From exponent rules, we know that: an ¼

 
n 1 a

 n a an ¼ n b b ðam Þn ¼ amn an ¼ an
m am Hence:

pffiffiffi6  6  
6 
4 2 1 1
6
43 4 16   ð0:5Þ  8 ¼ 2  1   23 3 2 2 22 ¼ 24 

1 1 24  26  26  4 ¼ 4 ¼ 24þ6
4
3 ¼ 23 ¼ 8 3 2  23 2 2

Choice (3) is the answer. 2.17. From exponent rules, we know that: ðam Þn ¼ amn ðabÞn ¼ an bn

14

2 Solutions of Problems: Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities

Thus:

 6 3  5  32 3  ð45Þ6 3  56  312 31þ12 ¼ ¼ ¼ 6þ7 ¼ 1 3 ð15Þ6  37 ð5  3Þ6  37 56  36  37

Choice (1) is the answer.

2.18. From exponent rules, we know that: ffiffiffi p 1 m 2 ¼ 2m ðam Þn ¼ amn an ¼ am ) n ¼ m Hence: ffiffiffi p 1 x 2 ¼ 2x 

1 3

ð16Þ

12 12

Therefore: p ffiffiffi x 2¼

!12 ¼

 1 !12  4 13 12 2 1 1 1 1 1 2 ¼ 243222 ¼ 26

!1  1 12 2 1 2 1 1 ð16Þ3 ) 2 x ¼ 26 ) x ¼ 6

Choice (2) is the answer. 2.19. We know that: ð a þ bÞ ð a
bÞ ¼ a2
b2 ðam Þn ¼ amn Based on the information given in the problem, we have: x¼1
Therefore: 

x þ x
1

13

¼

 pffiffiffi 1
2þ

pffiffiffi 2

pffiffiffi1  pffiffiffi1 13  pffiffiffi pffiffiffi 1 þ 2 3 1 1 1þ 2 3 pffiffiffi ¼ 1
2 þ pffiffiffi  pffiffiffi ¼ 1
2 þ 1
2 1
2 1
2 1þ 2



pffiffiffi pffiffiffi13  pffiffiffi13 ¼ 1
2
1
2 ¼
2 2 ¼ Choice (1) is the answer. 2.20. Based on the rule of difference of two squares, we know that:

 1 pffiffiffi3 3 pffiffiffi
2 ¼
2

2.2 Exponents and Radicals

15

pffiffiffi pffiffiffipffiffiffi pffiffiffi aþ b a
b ¼a
b To solve this problem, we need to change the denominator of each fraction to a rational number as follows: 1 1 1 pffiffiffi pffiffiffiffiffi þ pffiffiffiffiffi pffiffiffiffiffi þ pffiffiffiffiffi pffiffiffiffiffi 4 þ 11 11 þ 18 18 þ 25 pffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi 18
25 4
11 11
18 1 1 1 ¼ pffiffiffi pffiffiffiffiffi  pffiffiffi pffiffiffiffiffi þ pffiffiffiffiffi pffiffiffiffiffi  pffiffiffiffiffi pffiffiffiffiffi þ pffiffiffiffiffi pffiffiffiffiffi  pffiffiffiffiffi pffiffiffiffiffi 4 þ 11 11 þ 18 4
11 11
18 18
25 18 þ 25 pffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi 18
25 4
11 11
18 4
11 þ 11
18 þ 18
25 ¼ ¼ þ þ 18
25 4
11 11
18
7 ¼

pffiffiffi pffiffiffiffiffi 4
25 2
5 3 ¼ ¼
7
7 7

Choice (2) is the answer. 2.21. From the problem, we have: tþ1

1

A ¼ x t , B ¼ xtþ1 , t 6¼ 0,
1 From exponent rules, we know that: ðam Þn ¼ amn In choice (1):

8  tþ1 tþ1t t > < Atþ1 ¼ x t ¼x tþ1 t ) Atþ1 6¼ B t tþ1   > tþ1 1 1 t : B t ¼ xtþ1 ¼ xt

In choice (2):

8  tþ1 tþ1t t > < Atþ1 ¼ x t ¼x t ) Atþ1 ¼ Btþ1   tþ1 > tþ1 1 : B ¼ xtþ1 ¼x

In choice (3):

8 1  tþ1 tþ1 1 1 > < Atþ1 ¼ x t ¼ xt 1 ) Atþ1 6¼ Btþ1  1 tþ1 > : Btþ1 ¼ xtþ1 ¼x

In choice (4):

8  tþ1 tþ1 ðtþ1Þ2 > ¼x t < Atþ1 ¼ x t 1 ) Atþ1 6¼ Btþ1 1  1 tþ1 1 > 1 : Btþ1 ¼ xtþ1 ¼ xðtþ1Þ2

Choice (2) is the answer. 2.22. Based on the information given in the problem, we know that a and b have different signs and a < b. Therefore:

16

2 Solutions of Problems: Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities

a < 0, b > 0 ð Þ3

) a3 < 0, b3 > 0 ) a3 < b3

Choice (2) is the answer.

2.23. We know that: ð a þ bÞ ð a
bÞ ¼ a2
b2 p ffiffiffiffiffi n m an ¼ am The problem can be solved as follows: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   pffiffiffi pffiffiffi2 pffiffiffi pffiffiffi pffiffiffi 4 2 pffiffiffi2 pffiffiffi 4 4 2þ 2 4
2 2 6þ4 2¼ 4
2 2 2 þ 2 þ4 2¼ 2 2
2 

¼

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 2 2
2  2 þ 2 ¼ 2 2
2 2 þ 2 ¼ 2ð 4
2Þ ¼ 4 ¼ 2

Choice (2) is the answer. 2.24. We know that: ð a þ bÞ ð a
bÞ ¼ a2
b2 p ffiffiffiffiffi n m an ¼ am Therefore: ffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi    pffiffiffi2 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi2 pffiffiffi pffiffiffi 4  4 4 4 4 2þ 3 ¼ 7
4 3 2þ 3 ¼ 7
4 3 4þ4 3þ3 7
4 3 2þ 3¼ 7
4 3 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  pffiffiffi2ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi  pffiffiffi pffiffiffi 4 4 4 2 ¼ 7
4 3 7 þ 4 3 ¼ ð7Þ
4 3 ¼ 4 49
48 ¼ 1 ¼ 1 Choice (2) is the answer. 2.25. From radical rules, we know that:

pffiffiffiffiffi ffiffiffi p m a ¼ mn an

Now, we should define the radicals based on a common root (greatest common divisor (GCL)), as follows: GCLf2, 6, 4, 3g ¼ 12 Therefore: 1þ

pffiffiffiffiffi ffiffiffi p p ffiffiffiffiffi 26 2 12 2¼1þ 26 ¼ 1 þ 64

2.3 Absolute Values and Inequalities

As can be seen, 1 þ

17

1þ

pffiffiffiffiffi p p ffiffiffiffiffi ffiffiffi 62 6 12 62 ¼ 1 þ 36 6¼1þ

1þ

pffiffiffiffiffi p p ffiffiffiffiffi ffiffiffi 43 4 12 43 ¼ 1 þ 64 4¼1þ

1þ

pffiffiffiffiffi p pffiffiffiffiffi ffiffiffi 34 3 34 ¼ 1 þ 12 81 3¼1þ

p ffiffiffi 3 3 is the greatest number. Choice (4) is the answer.

2.26. We know that: ð a þ bÞ ð a
bÞ ¼ a2
b2 ðam Þn ¼ amn The problem can be solved as follows:  2 34 pffiffiffi 3 pffiffiffi  1 2 1 40:75 22 2 2 0:25 pffiffiffi pffiffiffi þ 9 pffiffiffi pffiffiffi þ 32 4 ¼ pffiffiffi pffiffiffi þ 32 ¼ pffiffiffi pffiffiffi þ 3 ¼ 1þ 2þ 3 1þ 2þ 3 1þ 2þ 3 1þ 2þ 3 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 2 2 1 þ 2
3 pffiffiffi 2 2 1 þ 2
3 pffiffiffi 2 2 1 þ 2
3 pffiffiffi pffiffiffi  pffiffiffi pffiffiffi þ 3 ¼  pffiffiffi þ 3 ¼ pffiffiffi 2 pffiffiffi 2 þ 3 ¼ 1þ 2þ 3 1þ 2
3 1þ2 2þ2
3 3 1þ 2
pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 2 2 1þ 2
3 pffiffiffi ¼ þ 3¼1þ 2
3þ 3¼1þ 2 2 2 Choice (4) is the answer.

2.3

Absolute Values and Inequalities

2.27. We know from the features of absolute value and inequality that: jx
aj < b ,
b < x
a < b The problem can be solved as follows: ð
1Þ
1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )
10) >0 >0) x
1 x
3 x
1 x
3 ð x
1Þ ð x
3Þ ðx
1Þðx
3Þ ) ðx
1Þðx
3Þ < 0 ) 1 < x < 3 Choice (2) is the answer. 2.33. The problem can be solved as follows: ð Þ2

jx
1j  jx
3j ) ðx
1Þ2  ðx
3Þ2 ) x2
2x þ 1  ) x2
6x þ 9 ) 4x  8 ) x  2 Choice (4) is the answer. 2.34. The problem can be solved as follows: ð Þ2

jx þ 1j  jx
1j ) ðx þ 1Þ2  ðx
1Þ2 ) x2 þ 2x þ 1  ) x2
2x þ 1 ) 4x  0 ) x  0 Choice (2) is the answer.

ð2Þ

2.3 Absolute Values and Inequalities

19

2.35. We know from the features of absolute value and inequality that: jx
aj < b ,
b < x
a < b The problem can be solved as follows: þ11
10
5 < x
11 < 3 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 6 < x < 14 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )
4 < x
10 < 4 ) jx
10j < 4 Choice (4) is the answer.

2.36. The problem can be solved as follows: 2


1


1 < x < 0 )
2 < 2x < 0 )
3 < 2x
1 <
1 ) j2x
1j < 1
2x ð
1Þ þ2
1 < x < 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 0 <
x < 1) 2 < 2
x < 3 ) j2
xj < 2
x ) j2x
1j þ j2
xj ¼ 1
2x þ 2
x ¼ 3
3x Choice (2) is the answer. 2.37. The neighborhood center (C) and neighborhood radius (R) for the symmetric neighborhood of (α1, α3)
{α2} can be determined as follows: α þ α3 C ¼ α2 ¼ 1 ð1Þ 2 α3
α1 2

ð2Þ

ð3a
7Þ þ ða þ 5Þ 4a
2 )3¼ )a¼2 2 2

ð3Þ

ða þ 5Þ
ð3a
7Þ
2a þ 12 ¼ 2 2

ð4Þ

R¼ Therefore, for (3a
7, a + 5)
{3}, we can write: 3¼

R¼ Solving (3) and (4):

R¼


2  2 þ 12 ¼4 2

Choice (4) is the answer. 2.38. The problem can be solved as follows:

Solving (1), (2), and (3):

a > 0 ) j aj ¼ a

ð1Þ

b < 0 ) jbj ¼
b

ð2Þ

ð1Þ, ð2Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) a >
b ) a þ b > 0 ) ja þ bj ¼ a þ b jaj > jbj ¼

ð3Þ

20

2 Solutions of Problems: Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities

ja þ bj þ jaj þ jbj ¼ a þ b þ a þ ð
bÞ ¼ 2a Choice (3) is the answer. 2.39. From the problem, we have: j x þ 1j þ j x
3j ¼ 2

ð1Þ

We know that: jAj þ jBj  jA
Bj Therefore: j x þ 1j þ j x
3j  j x þ 1
ð x
3Þ j ) j x þ 1j þ j x
3j  4 Solving (1) and (2): 2  4 ) Wrong Therefore, there is no answer for the problem. Choice (1) is the answer.

ð2Þ

3

Problems: Systems of Equations

Abstract

In this chapter, the basic and advanced problems of systems of equations 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. 3.1. For what value of m, the system of equations below has a unique solution? ( 2x
3y ¼ 5 x þ my ¼ 2 Difficulty level Calculation amount 1) m ¼ 32 2) m 6¼
32 3) m ¼
32 4) For any m

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large

3.2. If the value of x from the system of equations below is 2.5, determine the value of 2a + b: ( ax
y ¼ 1 bx þ 2y ¼ 3 Difficulty level Calculation amount 1)
2 2)
1 3) 1 4) 2

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

3.3. Calculate the value of y2 + y from the solution of the system of equations below: ( ðx þ 1Þ2 þ 3y ¼ 8 xðx þ 2Þ
5y ¼ 31 Difficulty level Calculation amount

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Precalculus, https://doi.org/10.1007/978-3-030-65056-8_3

21

22

3

1) 2) 3) 4)

12 6 10 8

3.4. For what value of a, the matrix below has a unique solution?    aþ1 2 x
1 Difficulty level ○ Easy Calculation amount ● Small 1) {
1, 1} 2) ℝ
{0, 1} 3) For no value of a 4) ℝ

● Normal ○ Normal

a
1

y

¼

  a 1

○ Hard ○ Large

3.5. Calculate the value of y from the solution of the system of equations below: 8 < x
y¼4 2 3 3 : 2x þ 3y ¼ 14 Difficulty level Calculation amount 1)
4 2) 2 3) 3 4) 4

○ Easy ● Small

● Normal ○ Normal

3.6. Solve the system of equations below:

Difficulty level ○ Easy Calculation amount ● Small 1) x ¼
3, y ¼
4 2) x ¼ 3, y ¼ 4 3) x ¼ 3, y ¼
4 4) x ¼
3, y ¼ 4

● Normal ○ Normal

○ Hard ○ Large

8 < x
1¼y
1 2 3 : x þ 2y ¼ 11 ○ Hard ○ Large

3.7. Calculate the value of x + y from the solution of the system of equations below: ( x þ 2y ¼
1 3x
y ¼ 4 Difficulty level Calculation amount 1)
1 2) 0 3) 1 4) 2

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

Problems: Systems of Equations

3

Problems: Systems of Equations

23

3.8. For what value of a, the system of equations below has an infinite number of solutions? ( 2x þ 3y ¼ 4 y ¼ aðx
2Þ Difficulty level Calculation amount 1)
32 2)
23 3) 23 4) 32

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

3.9. Calculate the value of y
x from the solution of the system of equations below: ( x þ 3y ¼
2 3x þ y ¼
14 Difficulty level Calculation amount 1)
4 2)
3 3) 3 4) 6

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

3.10. Calculate the value of y from the solution of the system of equations below: 8

> < 2x
1
y
2 ¼ 3 > > : Difficulty level Calculation amount 1) 1 2) 2 3) 3 4) 4

○ Easy ○ Small

● Normal ○ Normal

1 5 þ ¼2 2x
1 2
y ○ Hard ● Large

3.20. How many solution(s) does the system of equations below have? 5x þ y 3x
y 7x þ y ¼ ¼ 1 3 2 Difficulty level Calculation amount

○ Easy ○ Small

○ Normal ● Normal

● Hard ○ Large

4

Solutions of Problems: Systems of Equations

Abstract

In this chapter, the problems of the third chapter are fully solved, in detail, step-by-step, and with different methods. 4.1. The system of equations with the form presented in (1) has a unique solution (the lines intersect each other) if: a1 b1 6¼ a2 b2 (

a1 x þ b1 y ¼ c 1 a2 x þ b2 y ¼ c 2

Therefore:

(

2x  3y ¼ 5 x þ my ¼ 2

)

ð1Þ

2 3 3 6¼ ) m 6¼  1 m 2

Choice (2) is the answer. 4.2. The problem can be solved as follows: 2 ( ( (  2 5a  2y ¼ 2 þ ax  y ¼ 1 x ¼ 2:5 2:5a  y ¼ 1 ¼ 5 ¼ ¼ ¼ ) ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) ) 5a þ 2:5b ¼ 5 ¼ ¼ ¼ ¼ ) 2a þ b ¼ 2 bx þ 2y ¼ 3 2:5b þ 2y ¼ 3 ) 2:5b þ 2y ¼ 3 Choice (4) is the answer. 4.3. The problem can be solved as follows: ( ( x2 þ 2x þ 3y ¼ 7  ðx þ 1Þ2 þ 3y ¼ 8 ) 8y ¼ 24 ) y ¼ 3 ) y2 ¼ 9 ) xðx þ 2Þ  5y ¼ 31 x2 þ 2x  5y ¼ 31 ) y2 þ y ¼ 9 þ ð3Þ ¼ 6 Choice (2) is the answer. 4.4. The system of equations with the matrix form presented in (1) has a solution (the lines intersect each other) if its determinant is nonzero as follows: jAj 6¼ 0 ) a1 a4  a2 a3 6¼ 0

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Precalculus, https://doi.org/10.1007/978-3-030-65056-8_4

27

28

4 Solutions of Problems: Systems of Equations

 AX ¼ B )

a1 a3

a2 a4



x1 x2



 ¼

b1 b2

 ð1Þ

Therefore: 

aþ1

2

  x

1

a1

y

¼

  a 1

) jAj ¼ ða þ 1Þða  1Þ  ð2Þð1Þ ¼ a2  1 þ 2 6¼ 0 ) a2 þ 1 6¼ 0 ) a 2 ℝ

Choice (4) is the answer. 4.5. The problem can be solved as follows: 8 12 ( 6x  4y ¼ 16 < xy¼4 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) þ 2 3 3  ð 3 Þ 6x  9y ¼ 42 )  13y ¼ 26 ) y ¼ 2 : 2x þ 3y ¼ 14 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) Choice (2) is the answer. 4.6. The problem can be solved as follows: 8 8 ( < 3x  2y ¼ 1 þ < x  1 ¼ y  1 6 3x  3 ¼ 2y  2 ¼ ¼ ¼ ) 2 3 ¼ )4x ¼ 12 ) x ¼ 3 ) : x þ 2y ¼ 11 : x þ 2y ¼ 11 ) x þ 2y ¼ 11 x¼3 x þ 2y ¼ 11 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 3 þ 2y ¼ 11 ) y ¼ 4 Choice (2) is the answer. 4.7. The problem can be solved as follows: ( ( x þ 2y ¼ 1 ) x þ 2y ¼ 1 þ 2 ) 7x ¼ 7 ) x ¼ 1 3x  y ¼ 4 ) 6x  2y ¼ 8 x ¼1 3x  y ¼ 4 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 3  1  y ¼ 4 ) y ¼ 1 ) x þ y ¼ 1 þ ð1Þ ¼ 0 Choice (2) is the answer. 4.8. The system of equations with the form presented in (1) has an infinite number of solutions (the lines match each other) if: a 1 b1 c 1 ¼ ¼ a 2 b2 c 2 (

a1 x þ b1 y ¼ c 1 a2 x þ b2 y ¼ c 2

Thus:

(

2x þ 3y ¼ 4 y ¼ að x  2Þ

Choice (2) is the answer.

( )

2x þ 3y ¼ 4 ax  y ¼ 2a

)

2 3 4 2 ¼ ¼ )a¼  a 1 2a 3

ð1Þ

4

Solutions of Problems: Systems of Equations

29

4.9. The problem can be solved as follows: (

8 ð3Þ < 3x  9y ¼ 6 þ x þ 3y ¼ 2  ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) )  8y ¼ 8 ) y ¼ 1 3x þ y ¼ 14 ) : 3x þ y ¼ 14 y¼1 3x þ y ¼ 14 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 3x þ 1 ¼ 14 ) x ¼ 5 ) y  x ¼ 1  ð5Þ ¼ 6

Choice (4) is the answer. 4.10. The problem can be solved as follows: 8 8 (


4x þ ðm þ 1Þy ¼ 2 :m  3 ¼ m 4 2

ðm  3Þx þ 3y ¼ m

3 m ¼ ) m2 þ m ¼ 6 ) m2 þ m  6 ¼ 0 ) ðm þ 3Þðm  2Þ ¼ 0 ) m ¼ 3, 2 mþ1 2

However, m ¼ 2 does not satisfy (3), as is shown in the following: ð3Þ )

m3 m 23 2 1 ¼ ) ¼ )  6¼ 1 4 2 4 2 4

Therefore, only m ¼  3 is acceptable. Choice (2) is the answer. 4.18. The system of equations with the form presented in (1) does not have a solution (the lines are parallel) if:

ð 2Þ ð 3Þ

32

4 Solutions of Problems: Systems of Equations

a 1 b1 c 1 ¼ 6¼ a 2 b2 c 2 (

a1 x þ b1 y ¼ c 1

ð1Þ

a2 x þ b2 y ¼ c 2 Therefore:



my þ 2x ¼ 5 ) y þ ðm  1Þx ¼ 2m  3

 8 >
: m 6¼ 5 1 2m  3 ð2Þ )

ð 2Þ ð 3Þ

2 m ¼ ) m2  m ¼ 2 ) m2  m  2 ¼ 0 ) ðm  2Þðm þ 1Þ ¼ 0 ) m ¼ 1, 2 m1 1

However, m ¼  1 does not satisfy (3), as is proven in the following: m 5 1 5 6¼ ) 6¼ )  1 6¼ 1 ) Wrong 1 2m  3 1 2  ð1Þ  3 Therefore, only m ¼ 2 is acceptable. Choice (4) is the answer. 4.19. The problem can be solved as follows: 8 8 3 1 4 > > 15 þ 5 ¼  20 >  ¼ < 2x  1 y  2 3 ð5Þ > < 3 þ 15 1 20 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 2x  1 y  2 ) þ ¼ þ2 2x  1 2x  1 3 > > 1 5 > : 1 þ 5 ¼2 ) > : þ ¼2 2x  1 2  y 2x  1 2  y )

14 14 ¼ ) 2x  1 ¼ 3 ) x ¼ 2 2x  1 3

x¼2 1 5 1 5 5 5 þ ¼2¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) þ ¼2) ¼ ) 2  y ¼ 3 ) y ¼ 1 2x  1 2  y 221 2y 2y 3 ) x þ y ¼ 2 þ ð1Þ ¼ 1 Choice (1) is the answer. 4.20. The system of equations shows the equations of two lines. Therefore, we need to check each pair of equations: 8 ( 5x þ y 3x  y >  < ¼ 15x þ 3y ¼ 3x  y 12x þ 4y ¼ 0 5x þ y 3x  y 7x þ y 1 3 ) ) ¼ ¼ ) 1 3 2 3x  y 7x þ y > 6x  2y ¼ 21x þ 3y 15x þ 5y ¼ 0 : ¼ 3 2  )

3x þ y ¼ 0 3x þ y ¼ 0

Therefore, the equations of the lines are the same and the lines match each other. Hence, the system of equations has an infinite number of solutions. Choice (4) is the answer.

5

Problems: Quadratic Equations

Abstract

In this chapter, the basic and advanced problems of quadratic equations 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. 5.1. Two times of a positive quantity is 9 units fewer than the one-third of its square value. Determine this quantity. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 9 2) 12 3) 15 4) 18 5.2. For what value of a, the value of the term of ax2 + 2x + 4a is always positive? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) a
0.5 2) a   0.5 3) 0  a  0.5 4) 0.5  a  0.5 5.3. For what value of k, the roots of the quadratic equation of 2x2 + 3x  k ¼ 0 are two units fewer than the quadratic equation of 2x2  5x + 1 ¼ 0? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 2 3) 3 4) 4 5.4. For what value of m, the quadratic equation of 2x2  5x + m ¼ 0 has two real roots that the roots are inverse value of each other? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1 2) 2 3) 3 4) 4 # Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Precalculus, https://doi.org/10.1007/978-3-030-65056-8_5

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34

5

Problems: Quadratic Equations

5.5. Determine the value of m so that the quadratic equation of (m + 2)x2 + 4x + m  1 ¼ 0 has two real roots. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2 < m < 1 2) 1 < m < 2 3) 2 < m < 2 4) 3 < m < 2 5.6. For what value of m, the inequality of m3 x2 þ mx þ m1 < 0 is always true? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) |m| < 3 2) m < 0 3) m > 0 4) m 2 ℝ 5.7. Which one of the quadratic equations below has the roots that their values are nine times of the roots of the quadratic equation of x2 + x  3 ¼ 0? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) x2 + 9x  243 ¼ 0 2) x2 + 9x  27 ¼ 0 3) x2 + 18x  243 ¼ 0 4) x2 + 18x  27 ¼ 0 5.8. The sum of the roots of the quadratic equation of ax2 + bx + c ¼ 0 is equal to the product of the reciprocal of the roots. What relation exists between the parameters? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) a2 + bc ¼ 0 2) a2  bc ¼ 0 3) b2  ac ¼ 0 4) b2 + ac ¼ 0 5.9. For what value of a, the quadratic equation of 2x2 þ ax þ a  32 ¼ 0 has two distinct roots? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) a < 2 or a > 6 2) a < 3 or a > 4 3) 2 < a < 6 4) 3 < a < 4 5.10. Determine the range of m so that 4x2  2mx + 4m2
0. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) ℝ 2) 2) { } 3) |m|  2 4) |m|
2

5

Problems: Quadratic Equations

35

5.11. In the quadratic equation of 2x2 + (2k  1)x  k ¼ 0, for what value of k, the sum of the inverse value of the roots is equal to 73? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 4 2) 3 3) 3 4) 4 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 5.12. Determine the quadratic equation that its roots are 2 þ 4  a and 2  4  a. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) x2  4x + a ¼ 0 2) x2 + ax  4 ¼ 0 3) x2 + 4x  a ¼ 0 4) x2  ax + 4 ¼ 0 5.13. For what value of m, the sum of the square of the roots of 2x2  5x + m ¼ 0 is 37 4? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 2) 3 3) 2 4) 3 5.14. If x1 and x2 are the roots of the quadratic equation of mnx2 + n2x + m2 ¼ 0, determine the value of x21 x2 þ x1 x22 . Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 1 3) mþn mn 4) mn 5.15. For what value of m, the quadratic equation of (m + 1)x2 + m(m2  9)x  2 ¼ 0 has two real and symmetric (equal in magnitude, but different in sign) roots. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 3 3) 3 4) 9 5.16. What is the minimum value of x2  x + 2? Difficulty level ○ Easy ● Normal Calculation amount ○ Small ● Normal 1) 14 2) 34 3) 54 4) 74

○ Hard ○ Large

36

5

Problems: Quadratic Equations

5.17. Determine the value of k if the relation of x21 þ x22 ¼ 12 exists between the roots of the quadratic equation of x2  2kx  2 ¼ 0. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large pffiffiffi 1)  2 pffiffiffi 2)  3 3) 2 4) 3 5.18. The difference of the roots of the quadratic equation of 2x2  3x + m ¼ 0 is 52. Determine the value of m. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 2) 1 3) 1 4) 2 5.19. If the roots of the quadratic equation of x2 + ax  a2 ¼ 0 are x1 and x2, determine the quadratic equation that its roots are x1 + 1 and x2 + 1. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) x2 + (a  2)x + a2 + a  1 ¼ 0 2) x2 + (a + 2)x  a2  a  1 ¼ 0 3) x2 + (a  2)x  a2  a + 1 ¼ 0 4) x2 + (a + 2)x + a2 + a  1 ¼ 0 5.20. The sum of the square of two successive numbers is 925. Determine the sum of these two numbers. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) 41 2) 43 3) 45 4) 47 pffiffiffiffi pffiffiffiffiffi 5.21. If x0 and x00 are the roots of the quadratic equation of x2  4x + 1 ¼ 0, determine the value of  x0  x00 . Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large pffiffiffi 1) 2 pffiffiffi 2) 3 3) 2 4) 3 5.22. If the quadratic equations of x2  x  2a ¼ 0 and x2 + 2x + a ¼ 0 have a common root, determine this common root for a > 0. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 2) 1 3) 1

5

Problems: Quadratic Equations

37

4) 2

5.23. What is the solution of the following inequality? x1 > 2x xþ1 Difficulty level Calculation amount 1) x <  1 2) x >  1 3) 1 < x < 1 4) 2 < x <  1

○ Easy ○ Small

○ Normal ● Normal

● Hard ○ Large

5.24. How many distinctive real roots do the equation of (x2  2x)2  (x2  2x) ¼ 2 have? Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 2 3) 3 4) 4 5.25. For what value of m, the quadratic equation of mx2 + 5x + m2  6 ¼ 0 has two real roots that the roots are inverse value of each other? Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) 2, 3 2) 2 3) 2 4) 3 5.26. For what value of m, the sum of the square of the roots of 2x2  mx + m  1 ¼ 0 is 4? Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) 2, 6 2) 2 3) 2 4) 6

6

Solutions of Problems: Quadratic Equations

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, two times of a positive quantity is 9 units fewer than the one-third of its square value. In other words: x>0

ð1Þ

1 2x ¼ x2  9 3

ð2Þ

) x2  6x  27 ¼ 0 ) ðx  9Þðx þ 3Þ ¼ 0 ) x ¼ 3, 9

ð3Þ

Solving (1) and (3): x¼9 Choice (1) is the answer. 6.2. The value of the quadratic equation of ax2 + bx + c is always positive if a > 0 and Δ  0. Therefore, for the given quadratic equation of ax2 + 2x + 4a, we can write: a>0 Δ ¼ 22  4ðaÞð4aÞ  0 ) 4  16a2  0 ) a2 

ð1Þ 1 1 1 )a & a 4 2 2

ð2Þ

Solving (1) and (2): a

1 2

Choice (1) is the answer. 6.3. Based on the information given in the problem, we have: 2x2 þ 3x  k ¼ 0

ð1Þ

2X 2  5X þ 1 ¼ 0

ð2Þ

x¼X2)X ¼xþ2

ð3Þ

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Precalculus, https://doi.org/10.1007/978-3-030-65056-8_6

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6 Solutions of Problems: Quadratic Equations

Solving (2) and (3): 2ðx þ 2Þ2  5ðx þ 2Þ þ 1 ¼ 0 ) 2x2 þ 3x  1 ¼ 0

ð4Þ

Comparing (1) and (4): k¼1 Choice (1) is the answer. 6.4. The product of the roots of a quadratic equation can be determined as follows: ax2 þ bx þ c ¼ 0 ) Product of roots ¼

c a

Since the roots are inverse value of each other, the product of the roots is equal to one. Hence: 2x2  5x þ m ¼ 0 ) Product of roots ¼

m ¼1)m¼2 2

Choice (2) is the answer. 6.5. A quadratic equation has two real roots if its discriminant is equal or greater than zero. In other words: ax2 þ bx þ c ¼ 0 ) Δ ¼ b2  4ac  0 Therefore, for the quadratic equation of (m + 2)x2 + 4x + m  1 ¼ 0, we can write:
 Δ ¼ 42  4ðm þ 2Þðm  1Þ  0 ) 16  4 m2 þ m  2  0 ) m2 þ m  6  0 ) ð m þ 3Þ ð m  2Þ  0 )  3  m  2 Choice (4) is the answer. 6.6. The value of the quadratic equation of ax2 + bx + c is always negative if a < 0 and Δ < 0. Therefore, for the given quadratic equation of m3 x2 þ mx þ m1 , we can write: m3 < 0 ) m < 0

ð1Þ


 1  < 0 ) m2  4m2 < 0 )  3m2 < 0 ) m2 > 0 ) m 2 ℝ Δ ¼ m2  4 m3 m

ð2Þ

Solving (1) and (2): m 0 Therefore: 2x2 þ ax þ a 

  3 3 ¼ 0 ) Δ ¼ a2  4  2  a  > 0 ) a2  8a þ 12 > 0 2 2 ) ða  6Þða  2Þ > 0 ) a < 2 & a > 6

Choice (1) is the answer. 6.10. The value of the quadratic equation of ax2 + bx + c is always positive if a > 0 and Δ  0. Therefore, for the given quadratic equation of 4x2  2mx + 4m2  0, we can write: 4>0

ð1Þ


 Δ ¼ ð2mÞ2  4ð4Þ 4m2  0 ) 4m2  64m2  0 )  60m2  0 ) m2  0

ð2Þ

As can be seen in (1) and (2), both criteria are satisfied disregarding the value of m. Therefore: m2ℝ Choice (1) is the answer. 6.11. Based on the information given in the problem, we have: 1 1 7 x þ x2 7 þ ¼ ) 1 ¼ x1 x2 3 3 x1 x2 As we know, the sum and the product of the roots of a quadratic equation can be determined as follows: b c ax2 þ bx þ c ¼ 0 ) S ¼ sum of roots ¼  , P ¼ product of roots ¼ a a

ð1Þ

42

6 Solutions of Problems: Quadratic Equations

Hence, for the quadratic equation of 2x2 + (2k  1)x  k ¼ 0, we can write: x1 þ x2 ¼  x1 x2 ¼

ð2k  1Þ 2

ð2Þ

k 2

ð3Þ

Solving (1), (2), and (3): Þ  ð2k1 2 k 2

¼

7 2k  1 7 ) ¼ ) 6k  3 ¼ 7k ) k ¼  3 3 k 3

Choice (2) is the answer. 6.12. A quadratic equation has the following form, where S and P are the sum and the product of its roots:

x2  Sx þ P ¼ 0, S ¼ sum of roots, P ¼ product of roots Therefore, for the given roots of 2 þ

ð1Þ

pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 4  a and 2  4  a, we can write:

 pffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffiffiffi S¼ 2þ 4a þ 2 4a ¼4

ð2Þ

 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi P ¼ 2 þ 4  a 2  4  a ¼ 4  ð 4  aÞ ¼ a

ð3Þ

Solving (1), (2), and (3): x2  4x þ a ¼ 0 Choice (1) is the answer. 6.13. Based on the information given in the problem, we know that: x21 þ x22 ¼

37 37 ) ðx1 þ x2 Þ2  2x1 x2 ¼ 4 4

ð1Þ

On the other hand, we know that a quadratic equation has the form below, where S and P are the sum and the product of its roots: b c x2  Sx þ P ¼ 0, S ¼ sum of roots ¼  , P ¼ product of roots ¼ ð2Þ a a Solving (1) and (2): 37 S  2P ¼ ) 4 2



b  a

2 2

  c 37 ¼ a 4

Solving (3) for the given quadratic equation of 2x2  5x + m ¼ 0:  2 5 m 37 25 37 ) m¼ )m¼ 3 2 ¼  2 2 4 4 4 Choice (2) is the answer.

ð3Þ

6

Solutions of Problems: Quadratic Equations

43

6.14. The sum and the product of the roots of a quadratic equation can be determined as follows: b c ax2 þ bx þ c ¼ 0 ) S ¼ sum of roots ¼  , P ¼ product of roots ¼ a a

ð1Þ

The problem can be solved as follows: x21 x2 þ x1 x22 ¼ x1 x2 ðx1 þ x2 Þ Solving (1) and (2): x21 x2 þ x1 x22 ¼ P  S ¼

   b c  a a

ð2Þ ð3Þ

For the quadratic equation of mnx2 + n2x + m2 ¼ 0, we can write: x21 x2 þ x1 x22 ¼

  2  n2 m  ¼1 mn mn

Choice (1) is the answer. 6.15. Based on the information given in the problem, we have: x1 ¼ x2 ) x1 þ x2 ¼ 0

ð1Þ

As we know, the sum and the product of the roots of a quadratic equation are determined as follows: b c ax2 þ bx þ c ¼ 0 ) S ¼ sum of roots ¼  , P ¼ product of roots ¼ a a

ð2Þ

Solving (1) and (2): b S¼ ¼0)b¼0 a

ð3Þ

Hence, for the given quadratic equation of (m + 1)x2 + m(m2  9)x  2 ¼ 0, we can write:
 m m2  9 ¼ 0 ) m ¼ 0,  3, 3

ð4Þ

On the other hand, to have real roots for the given quadratic equation, the constraint below must be satisfied:
2 Δ  0 ) m2 m2  9 þ 8ðm þ 1Þ  0 ) 8ðm þ 1Þ  0 ) m  1 Solving (4) and (5):

ð5Þ

) m ¼ 0, 3

However, m ¼ 0 does not exist in the choices. Therefore, m ¼ 3 is the only acceptable choice. Choice (3) is the answer. 6.16. The problem can be solved as follows:    2   1 7 1 1 7 1 2 7 x 2  x þ 2 ¼ x2  x þ þ ¼ x2  2 þ ¼ x þ xþ 4 4 2 2 4 2 4
2 The term of x  12 is equal or greater than zero. Hence: 

x

Choice (4) is the answer.

1 2

2



   1 2 7 7 1 2 7 7 x ¼ 0) x þ  ) min þ 2 4 4 2 4 4

44

6 Solutions of Problems: Quadratic Equations

6.17. Based on the information given in the problem, we know that: x21 þ x22 ¼ 12 ) ðx1 þ x2 Þ2  2x1 x2 ¼ 12

ð1Þ

On the other hand, we know that a quadratic equation has the form below, where S and P are the sum and the product of its roots: b c x2  Sx þ P ¼ 0, S ¼ sum of roots ¼  , P ¼ product of roots ¼ ð2Þ a a Solving (1) and (2):

 2   b c S  2P ¼ 12 )  2 ¼ 12 a a 2

ð3Þ

Solving (3) for the given quadratic equation of x2  2kx  2 ¼ 0:  

2k 1

2 2



 pffiffiffi 2 ¼ 12 ) 4k 2 þ 4 ¼ 12 ) k2 ¼ 2 ) k ¼  2 1

Choice (1) is the answer. 6.18. We know that a quadratic equation has the form below, where S and P are the sum and the product of its roots: b c x2  Sx þ P ¼ 0, S ¼ sum of roots ¼  , P ¼ product of roots ¼ a a Therefore, for the given quadratic equation of 2x2  3x + m ¼ 0, we can write: x1 þ x 2 ¼ S ¼ 

3 3 ¼ 2 2

Based on the information given in the problem, we have: x1  x2 ¼ Hence:

5 2

8 3 > < x1 þ x2 ¼ þ 2 ) 2x ¼ 4 ) x ¼ 2 ) 1 1 > : x1  x2 ¼ 5 2 x1 þ x2 ¼ x 1 x2 ¼ P ¼

3 x1 ¼ 2 1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) x2 ¼  2 2

  c 1 m )2  ¼ )m¼ 2 a 2 2

Choice (1) is the answer. 6.19. Based on the information given in the problem, x1 and x2 are the roots of the quadratic equation of x2 + ax  a2 ¼ 0. To determine the quadratic equation that its roots are y1 ¼ x1 + 1 and y2 ¼ x2 + 1, we need to replace x by y  1 in the same quadratic equation, as follows: ðy  1 Þ2 þ aðy  1 Þ  a2 ¼ 0 ) y2  2y þ 1 þ ay  a  a2 ¼ 0

6

Solutions of Problems: Quadratic Equations

45

y!x 2 ) y2 þ ða  2Þy þ 1  a  a2 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) x þ ða  2Þx þ 1  a  a2 ¼ 0 Choice (3) is the answer.

6.20. Based on the information given in the problem, we know that the sum of the square of the two successive numbers (e.g., n and n + 1) is 925. In other words: n2 þ ðn þ 1Þ2 ¼ 925 ) n2 þ n2 þ 2n þ 1 ¼ 925 ) 2n2 þ 2n ¼ 924 ) nðn þ 1Þ ¼ 462

ð1Þ

We can consider 462 as follows: 462 ¼ 21  22

ð2Þ

Solving (1) and (2): nðn þ 1Þ ¼ 21  22 ) n ¼ 21 ) The sum of the two numbers ¼ n þ ðn þ 1Þ ¼ 21 þ 22 ¼ 43 Choice (2) is the answer. 6.21. The sum and the product of the roots of a quadratic equation can be determined as follows: b c ax2 þ bx þ c ¼ 0 ) S ¼ sum of roots ¼  , P ¼ product of roots ¼ a a The problem can be solved as follows: pffiffiffiffi pffiffiffiffiffi2 pffiffiffiffiffiffiffiffi x0  x00 ¼ x0 þ x00  2 x0 x00 b 4 c 1 x2  4x þ 1 ¼ 0 ) S ¼  ¼  ¼ 4, P ¼ ¼ ¼ 1 a 1 a 1

ð1Þ

ð2Þ ð3Þ

Solving (1), (2), and (3): pffiffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffiffi pffiffiffiffiffi2 pffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi   x0  x00 ¼ x0 þ x00  2 x0 x00 ¼ S  2 P ¼ 4  2 1 ¼ 2 )  x0  x00  ¼ 2 Choice (1) is the answer. 6.22. Based on the information given in the problem, we have: a>0

ð1Þ

If it is assumed that x1 is the common root, then we can write: (

x21 þ 2x1 þ a ¼ 0 x21

 x1  2a ¼ 0



) 3x1 þ 3a ¼ 0 ) x1 ¼ a

ð2Þ

By replacing x1 ¼  a in one of the equations, we have: x1 ¼ a ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) ðaÞ2 þ 2ðaÞ þ a ¼ 0 ) a2  a ¼ 0 ) aða  1Þ ¼ 0 ) a ¼ 0, 1 x21 þ 2x1 þ a ¼ 0 ¼ Solving (1) and (3):

ð3Þ

46

6 Solutions of Problems: Quadratic Equations ð2Þ

a ¼ 1 ) x1 ¼  1 Choice (2) is the answer.

6.23. The problem can be solved as follows: x1 x1 x  1  2x2  2x 2x2 þ x þ 1 > 2x )  2x > 0 ) >0) 0 and Δ < 0, as can be seen in the following: a¼2>0 Δ ¼ 1  4  2  1 ¼ 7 < 0 Therefore: xþ1 0 Therefore, we will have two distinctive real roots, since the discriminant is greater than zero. Moreover, regarding the equation of x2  2x + 1 ¼ 0: Δ ¼ b2  4ac ¼ ð2Þ2  4  1  1 ¼ 0 Thus, we will have just one distinctive real root, as the discriminant is equal to zero. In overall, we will have three distinctive real roots. Choice (3) is the answer. 6.25. A quadratic equation has two real distinct roots if its discriminant is greater than zero. In other words: ax2 þ bx þ c ¼ 0 ) Δ ¼ b2  4ac > 0 Therefore:


 mx2 þ 5x þ m2  6 ¼ 0 ) Δ ¼ 52  4m m2  6 > 0

ð1Þ

Moreover, the product of the roots of a quadratic equation can be determined as follows: ax2 þ bx þ c ¼ 0 ) Product of roots ¼

c a

Since the roots are inverse value of each other, the product of the roots is equal to one. Hence: mx2 þ 5x þ m2  6 ¼ 0 ) Product of roots ¼

m2  6 ¼ 1 ) m2  m  6 ¼ 0 m

) ðm þ 2Þðm  3Þ ¼ 0 ) m ¼ 2, 3 Solving (1) and (2) for m ¼ 3:
 m¼3 2
 Δ ¼ 52  4m m2  6 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 5  4  3 32  6 ¼ 25  12  3 ¼ 11 < 0 Therefore, m ¼ 3 is not acceptable. Solving (1) and (2) for m ¼  2:
 m ¼ 2 2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 5  4  ð2Þðð2Þ2  6Þ ¼ 25  16 ¼ 9 > 0 Δ ¼ 52  4m m2  6 ¼ Thus, m ¼  2 is not acceptable. Choice (2) is the answer. 6.26. Based on the information given in the problem, we know that:

ð2Þ

48

6 Solutions of Problems: Quadratic Equations

x21 þ x22 ¼ 4 ) ðx1 þ x2 Þ2  2x1 x2 ¼ 4

ð1Þ

On the other hand, we know that a quadratic equation has the form below, where S and P are the sum and the product of its roots: b c x2  Sx þ P ¼ 0, S ¼ sum of roots ¼  , P ¼ product of roots ¼ ð2Þ a a Solving (1) and (2):

 2   b c S  2P ¼ 4 )  ¼4 2 a a 2

Solving (3) for the given quadratic equation of 2x2  mx + m  1 ¼ 0: 



 m 2 m1 m2 m2 ¼4) 2 mþ1¼4)  m  3 ¼ 0 ) m2  4m  12 ¼ 0 2 2 4 4 ) ðm  6Þðm þ 2Þ ¼ 0 ) m ¼ 6,  2

m ¼ 6 is not acceptable because the equation will not have real roots, as can be seen in the following: m¼6 2 2x2  mx þ m  1 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 2x  6x þ 5 ¼ 0 ) Δ ¼ ð6Þ2  4ð2Þð5Þ ¼ 36  40 ¼ 4 < 0 m ¼  2 is acceptable because the equation will have two real roots, as is presented below: m ¼ 2 2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) 2x þ 2x  3 ¼ 0 ) Δ ¼ ð2Þ2  4ð2Þð3Þ ¼ 4 þ 24 ¼ 28 > 0 2x2  mx þ m  1 ¼ 0 ¼ Choice (2) is the answer.

ð3Þ

7

Problems: Functions, Algebra of Functions, and Inverse Functions

Abstract

In this chapter, the basic and advanced problems of functions, algebra of functions, and inverse functions 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. 7.1. If f


1 x

¼

qffiffiffiffiffiffiffiffi 2x1 x2

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

Difficulty level Calculation amount 1) 0 2) 12 pffiffi 3) 23 4) 2

● Easy ● Small

○ Normal ○ Normal


π  3

:

○ Hard ○ Large

pffiffiffi 7.2. Determine the value of fog(3) if f ðxÞ ¼ x  2 and g(x) ¼ x + 1. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 2) 1 3) 2 4) 3 7.3. If f(x) ¼ 2x  2 and g(x) ¼ x2  1, solve the equation of fog(x) ¼ 0. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large pffiffiffi 1)  2 2) 2 pffiffiffi 3)  3 4) 3

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Precalculus, https://doi.org/10.1007/978-3-030-65056-8_7

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7 Problems: Functions, Algebra of Functions, and Inverse Functions

7.4. In the function below, calculate the value of f(2) + f(2): ( 2 2x þ 4 f ð xÞ ¼ ½x  4 Difficulty level Calculation amount 1) 8 2) 6 3) 10 4) 5

● Easy ● Small

○ Normal ○ Normal

pffiffiffiffiffiffiffiffiffiffiffiffiffi 7.5. Determine the domain of y ¼ 2  x2 . Difficulty level ● Easy ○ Normal Calculation amount ● Small ○ Normal pffiffiffi pffiffiffi 1) x   2, x  2 pffiffiffi pffiffiffi 2)  2  x  2 3) x ¼ 0 pffiffiffi pffiffiffi 4)  2 < x < 2

○ Hard ○ Large

○ Hard ○ Large

7.6. Calculate fog(x) if f(x) ¼ 1  x2 and g(x) ¼ sin (x). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) cos2(x) 2) cos(x) 3) sin(1  x2) 4) sin(cos(x)) 7.7. What is the inverse function of f ðxÞ ¼ 1x? Difficulty level ● Easy ○ Normal Calculation amount ● Small ○ Normal 1) x 2) 1x 3)  1x pffiffiffi 4) x 7.8. Calculate fof(x) if f ðxÞ ¼ 1x 1þx. Difficulty level ● Easy Calculation amount ○ Small
2 1) 1þx 1x 2) 1 3) x  2 4) 1x 1þx

○ Normal ● Normal

○ Hard ○ Large

○ Hard ○ Large

7.9. Determine the inverse function of f(x) ¼ x2  2x Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large pffiffiffiffiffiffiffiffiffiffiffi 1) 1 þ x þ 1 pffiffiffiffiffiffiffiffiffiffiffi 2) 1  x þ 1 pffiffiffiffiffiffiffiffiffiffiffi 3) 1 þ x  1 pffiffiffiffiffiffiffiffiffiffiffi 4) 1  x  1

x2 x ( > > 0 < 3 \ x3 )0 > 0 > :2  x  0 :x  2  0 x x Note, that x ¼ 0 must be excluded from the domain, since it makes the denominator zero. Choice (1) is the answer.

8

Solutions of Problems: Functions, Algebra of Functions, and Inverse Functions

61

8.16. Based on the information given in the problem, we have: pffiffiffi x f ð xÞ ¼ j xj  1 The domain of a function in radical form, including even root, is determined by considering the radicand equal and greater than zero. Moreover, the value of those x that make the denominator zero must be excluded from the domain. Thus: x0 x0 x0 \ ) ) ) D f ¼ ½0, 1Þ  f1g x 6¼ 1 jxj  1 6¼ 0 jxj 6¼ 1 Choice (2) is the answer. 8.17. Based on the information given in the problem, we have: pffiffiffi xþ1 f ðxÞ ¼ pffiffiffi x xþ1 The domain of a function in radical form, including even root, is determined by considering the radicand equal and greater than zero. Moreover, the value of those x that make the denominator zero must be excluded from the domain. Thus: x0 pffiffiffi x x þ 1 6¼ 0 pffiffiffi Note that x x þ 1 6¼ 0 is true for any x. Hence: )x0 Choice (3) is the answer. 8.18. Based on the information given in the problem, we have: f ð xÞ ¼

1x 4x þ x3

The value of those x that make the denominator zero are not in the domain. Thus:   4x þ x3 6¼ 0 ) x 4 þ x2 6¼ 0 Note that 4 + x2 6¼ 0 for any x. Hence: )x¼0 Choice (4) is the answer. 8.19. Based on the information given in the problem, we have:

Therefore:

f ðxÞ ¼ 2x

ð1Þ

f ðgðxÞÞ ¼ 2x þ 2

ð2Þ

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Solutions of Problems: Functions, Algebra of Functions, and Inverse Functions

f ðgðxÞÞ ¼ 2gðxÞ

ð3Þ

Solving (2) and (3): 2gðxÞ ¼ 2x þ 2 ) gðxÞ ¼ x þ 1 Choice (3) is the answer. 8.20. Based on the information given in the problem, we have: f ð xÞ ¼ x  1

ð1Þ

f ð gð x Þ Þ ¼ x

ð2Þ

f ð gð x Þ Þ ¼ gð x Þ  1

ð3Þ

Thus:

Solving (2) and (3): gð x Þ  1 ¼ x ) gð x Þ ¼ x þ 1 Choice (1) is the answer. 8.21. Based on the information given in the problem, we have: f ðxÞ ¼

pffiffiffiffiffiffiffiffiffiffiffi 1x

To determine the inverse function of a function, it is convenient to use y instead of f(x). Then, we need to determine x based on y. After that, we must replace x with y and vice versa. Note that the domain of f 1(x) is the same as the range of f(x). Therefore: y¼

pffiffiffiffiffiffiffiffiffiffiffi or 1  x ) y2 ¼ 1  x ) x ¼ 1  y2 ) y ¼ 1  x2 ) f 1 ðxÞ ¼ 1  x2

Since the domain of f 1(x) is the same as the range of f(x), which is [0, 1), we need to add x  0 to f 1(x) as its domain. Thus: f

1

ð xÞ ¼ 1  x2 , x  0

Choice (1) is the answer. 8.22. Based on the information given in the problem, we have: f ðxÞ ¼ sin ðxÞ  2 To determine the inverse function of a function, it is convenient to use y instead of f(x). Then, we need to determine x based on y. After that, we must replace x with y and vice versa. Note that the domain of f 1(x) is the same as the range of f(x). Therefore:

8

Solutions of Problems: Functions, Algebra of Functions, and Inverse Functions

63 or

y ¼ sin ðxÞ  2 ) y þ 2 ¼ sin ðxÞ ) x ¼ arcð sin ðy þ 2ÞÞ ) y ¼ arcð sin ðx þ 2ÞÞ ) f 1 ðxÞ ¼ arcð sin ðx þ 2ÞÞ Choice (4) is the answer.

8.23. Based on the information given in the problem, we have: f ðxÞ ¼ 3x  2 gðxÞ ¼ 2 þ x First, we need to determine fog(x) as follows: fogðxÞ ¼ f ðgðxÞÞ ¼ f ð2 þ xÞ ¼ 3ð2 þ xÞ  2 ¼ 3x þ 4 To determine the inverse function of a function, it is convenient to use y instead of f(x). Then, we need to determine x based on y. After that, we must replace x with y and vice versa. Note that the domain of f 1(x) is the same as the range of f(x). Therefore: 1 4 1 4 y ¼ 3x þ 4 ) 3x ¼ y  4 ) x ¼ y  ) y ¼ x  3 3 3 3 Choice (1) is the answer. 8.24. Based on the information given in the problem, we have: f ðxÞ ¼ x3 þ 3x2 þ 3x þ 2 First, we need to define the function in a cube form as follows:   ) f ðxÞ ¼ x3 þ 3x2 þ 3x þ 1 þ 1 ¼ ðx þ 1Þ3 þ 1 To determine the inverse function of a function, it is convenient to use y instead of f(x). Then, we need to determine x based on y. After that, we must replace x with y and vice versa. Note that the domain of f 1(x) is the same as the range of f(x). Therefore: 1

1

1

) y ¼ ðx þ 1Þ3 þ 1 ) y  1 ¼ ðx þ 1Þ3 ) ðy  1Þ3 ¼ x þ 1 ) x ¼ ðy  1Þ3  1 ) y ¼ ðx  1Þ3  1 or

) f 1 ðxÞ ¼  1 þ

p ffiffiffiffiffiffiffiffiffiffiffi 3 x1

Choice (3) is the answer. 8.25. Based on the information given in the problem, we have: f ðxÞ þ xf ðxÞ ¼ x2 þ 1 The problem can be solved as follows:

64

8

x¼2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) x ¼ 2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )

(

Solutions of Problems: Functions, Algebra of Functions, and Inverse Functions

8 < f ð2Þ þ 2f ð2Þ ¼ 5

) )  ð 2Þ : f ð2Þ  2f ð2Þ ¼ 5 f ð2Þ þ ð2Þf ð2Þ ¼ ð2Þ2 þ 1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) f ð2Þ þ 2f ð2Þ ¼ 22 þ 1

(

f ð2Þ þ 2f ð2Þ ¼ 5 2f ð2Þ þ 4f ð2Þ ¼ 10

þ

)5f ð2Þ ¼ 5 ) f ð2Þ ¼  1 Choice (1) is the answer.

8.26. From trigonometry, we know that: 1 þ cos ð2xÞ ¼ 2 cos 2 ðxÞ Based on the information given in the problem, we have: f ð xÞ ¼ x2  2 The problem can be solved as follows: ) f ð2 cos ðxÞÞ ¼ ð2 cos ðxÞÞ2  2 ¼ 4 cos 2 ðxÞ  2 ¼ 2ð1 þ cos ð2xÞÞ  2 ¼ 2 cos ð2xÞ ) f ð f ð2 cos ðxÞÞÞ ¼ ð2 cos ð2xÞÞ2  2 ¼ 4 cos 2 ð2xÞ  2 ¼ 2ð1 þ cos ð4xÞÞ  2 ¼ 2 cos ð4xÞ ) f ð f ð f ð2 cos ðxÞÞÞÞ ¼ ð2 cos ð4xÞÞ2  2 ¼ 4 cos 2 ð4xÞ  2 ¼ 2ð1 þ cos ð8xÞÞ  2 ¼ 2 cos ð8xÞ Choice (4) is the answer. 8.27. Based on the information given in the problem, we have: f


 pffiffiffiffiffiffiffiffiffiffiffiffiffi x1 ¼ 2x  1 x

First, we need to determine the f(x) as follows: x1 1 t)x¼ ) f ðt Þ ¼ x 1t

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffi 1 1þt t !x 1þx 2 1¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) f ð xÞ ¼ 1t 1t 1x

The domain of a function in radical form, including even root, is determined by considering the radicand equal and greater than zero. Therefore: 1þx xþ1 0) 0) 1x 0 f ð xÞ ¼ 1 x0 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 ðf ðxÞÞ ¼ 1 Choice (1) is the answer. 8.38. Based on the information given in the problem, we have: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi 5x  x2 f ðxÞ ¼ log 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 can be determined by considering its argument greater than zero. Therefore:   8   8 8 5x  x2 > 2 > 2 0 < log < 5x  x2 < log 5x  x  log ð1Þ x  5x þ 4  0 4  1 4 ) ) ) 4 2 > : : x ð x  5Þ < 0 > 5x  x xðx  5Þ < 0 : x2  5x < 0 >0 4 ( 1x4 \ ðx  4Þðx  1Þ  0 ) ) )1x4 x ð x  5Þ < 0 0 > < log x ðx þ 9Þ  0 < log x ðx þ 9Þ  log x ð1Þ 0 x þ9>0 x þ9>0 > > > : : : x > 0, x 6¼ 1 x > 0, x 6¼ 1 x > 0, x 6¼ 1 Note that x2 + 9 > 0 and x2 + 8  0 are true for any x. Hence: \

) x > 0 , x 6¼ 1 Choice (4) is the answer.

8.40. Based on the information given in the problem, we have: f ðxÞ ¼ 2x  2½x þ 1 Based on the definition, we know that:  ½ x 2 þ1 ½ x  x < ½ x þ 1 ¼ ¼ ¼ ¼ ¼ ¼ ) 0  x  ½x < 1)0  2x  2½x < 2 ) 1  2x  2½x þ 1 < 3 ) 1  f ðxÞ < 3 ) Rf ¼ ½1, 3Þ Choice (2) is the answer. 8.41. Based on the information given in the problem, we have: f ð xÞ ¼ x2 þ 1 gð x Þ ¼ ) fogðxÞ ¼ f ðgðxÞÞ ¼ f

pffiffiffiffiffiffiffiffiffiffiffi x1


pffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffi2 x1 ¼ x1 þ1¼x1þ1¼x

Next, we need to determine the domain of the function. As we know, the domain of a function in radical form, including even root, is determined by considering the radicand equal and greater than zero: x  1  0 ) x  1 ) Dfog ¼ ½1, 1Þ Now, we can determine the range of the function based its domain as follows: Dfog ¼ ½1, 1Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) Rfog ¼ ½1, 1Þ fogðxÞ ¼ x ¼ Choice (2) is the answer. 8.42. Based on the information given in the problem, we have:

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8

Solutions of Problems: Functions, Algebra of Functions, and Inverse Functions

f ð xÞ ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2  2x þ 3

The problem can be solved as follows: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ) f ð x Þ ¼ ð x  1Þ 2 þ 2 As we know: pffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi hpffiffiffi  pffiffiffi pffiffiffi þ2 2, 1 ðx  1Þ2  0)ðx  1Þ2 þ 2  2) ðx  1Þ2 þ 2  2 ) f ðxÞ  2 ) R f ðxÞ ¼

Choice (1) is the answer. 8.43. Based on the information given in the problem, we have: f ðxÞ ¼ jx  2j ) D f ¼ ℝ

Based on the definition, two functions are equivalent if they are equal and their domains are the same. Choice (1):



2



x  3x þ 2 ðx  1Þðx  2Þ



¼ j x  2j



¼ g1 ð x Þ ¼

x1 x1

) D g 1 ¼ ℝ  f1g

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. Choice (2):



2



x  4 ðx  2Þðx þ 2Þ



¼ j x  2j



¼ g2 ð x Þ ¼

xþ2 xþ2

) Dg2 ¼ ℝ  f2g

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

ð x  2Þ 2 j x  2j 2 ¼ ¼ j x  2j j x  2j jx  2j ) Dg3 ¼ ℝ  f2g

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 ð x Þ ¼

j6x  12j 6jx  2j ¼ ¼ jx  2j 6 6

9

Problems: Factorization of Polynomials

Abstract

In this chapter, the basic and advanced problems of factorization of polynomials 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. 9.1. Which one of the following terms is not a factor of x3
7x2 + 6x. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) x 2) x
1 3) x
6 4) x + 6 9.2. Calculate the value of the following term: x2 Difficulty level Calculation amount 1) x + 2 2) x
2 3) x + 1 4) x
1

● Easy ● Small

○ Normal ○ Normal

x3 þ 8
2x þ 4

○ Hard ○ Large

9.3. Which one of the choices is a factor of x3 + y3? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) x
y 2) x2 + xy + y2 3) x2
xy + y2 4) x + y + xy

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72

9

9.4. Which one of the choices is a factor of x3
y3? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) x2 + xy + y2 2) x2
xy + y2 3) x + y 4) x + y + xy 9.5. Which one of the choices is not a factor of x3 + 2x2
3x? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) x + 1 2) x 3) x
1 4) x + 3 9.6. Which one of the choices is a factor of 6x3 + 7x2 + 2x? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 3x + 2 2) 2x
1 3) 3x2
2x 4) 2x2
x 9.7. Which one of the choices is a factor of x3 + x
10? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) x2
2x + 5 2) x2 + 2x + 5 3) x + 2 4) x2
1 9.8. Which one of the choices is a factor of 3a2 + 2ab
b2? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) a
2b 2) 3a
b 3) a
b 4) 2a
b 9.9. Which one of the choices is not a factor of x5
x4
4x + 4? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) x2
2 2) x + 1 3) x2 + 2 4) x
1 9.10. Calculate the value of x2 þ x12 if x þ 1x ¼ 2. Difficulty level ○ Easy ● Normal Calculation amount ● Small ○ Normal 1) 8 2) 6 3) 4 4) 2

○ Hard ○ Large

Problems: Factorization of Polynomials

9

Problems: Factorization of Polynomials

73

9.11. Calculate the value of the following term: ðx þ 1Þ3
3xðx þ 1Þ x3 þ 1 Difficulty level Calculation amount 3 2
3x 1) x þ1
3x x3 þ1 2) 1 3 3) xx3
3x þ1 4)

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

x3
3x2 x3 þ1

9.12. Calculate the value of (50.01)2
(49.99)2. Difficulty level ○ Easy ● Normal Calculation amount ● Small ○ Normal 1) 0.2 2) 0.02 3) 2 4) 0

○ Hard ○ Large

9.13. Which one of the following terms is not a factor of x4 + 2x3
x
2. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) x2
x + 1 2) x2 + x + 1 3) x + 2 4) x
1 9.14. Which one of the following terms shows the correct factorization of x4 + x2
2? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) (x2
2)(x2 + 1) 2) (x2 + 2)(x2 + 1) 3) (x2 + 2)(x + 1)(x
1) 4) (x2
2)(x + 1)(x
1) 9.15. Calculate the value of the term below: 3

x3
2x2 þ 1
2 3 x2
1 Difficulty level Calculation amount 1) 2 2) x
2 3) 1 4) x

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

9.16. Calculate the value of a3
b3 if a
b ¼ 1 and a2 + b2 ¼ 5. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 10 2) 9 3) 7

74

9

4) 6

9.17. Calculate the value of the following term: ða þ bÞ2 þ ðb þ cÞ2 þ ða þ cÞ2
ða þ b þ cÞ2 a2 þ b2 þ c 2 Difficulty level Calculation amount 1) 1 2) abþbcþac a2 þb2 þc2 3) 4)

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

ðaþbþcÞ2 a2 þb2 þc2 2ðabþbcþacÞ a2 þb2 þc2

9.18. Calculate the value of x3 + y3 if xy ¼ 5 and x + y ¼ 7. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 216 2) 238 3) 244 4) 264 9.19. Calculate the value of the term below: 2x4 þ x3
4x
2 x3
2 Difficulty level Calculation amount 1) 2x 2) 2x
1 3) 2x + 1 4) x + 1

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

9.20. Calculate the value of the term below: x4 þ x2 þ 1 x2 þ x þ 1 Difficulty level Calculation amount 1) x2 + x + 1 2) x2 + x 3) x2
x + 1 4) x2 + x
1

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

9.21. Calculate the value of the following term: x6 þ 4x2 þ 5 x2 þ 1 Difficulty level Calculation amount 1) x4 + x2 + 5 2) x4
x2 + 5

○ Easy ● Small

○ Normal ○ Normal

● Hard ○ Large

Problems: Factorization of Polynomials

9

Problems: Factorization of Polynomials

75

3) x4
2x2 + 5 4) x4 + 2x2 + 5

9.22. Calculate the value of the following term: 9

x2
1 3 3 x þ x2 þ 1 Difficulty level Calculation amount 3 1) x2 þ 1 3 2) x2
1 3 3) x2 þ 2 3 4) x2
2

○ Easy ● Small

○ Normal ○ Normal

● Hard ○ Large

9.23. Calculate the value of x2 + y2 if x4 + y4 + 4 ¼ 2(2x2 + 2y2
x2y2). Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 2) 1 3) 0 4) 8 9.24. Calculate the value of the following term:


ð2 þ 1Þ 22 þ 1 24 þ 1 . . . 264 þ 1 Difficulty level Calculation amount 1) 2256 + 1 2) 2128 + 1 3) 2128
1 4) 2256
1

○ Easy ○ Small

○ Normal ● Normal

● Hard ○ Large

Solutions of Problems: Factorization of Polynomials

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. The problem can be solved as follows:
x3
7x2 þ 6x ¼ x x2
7x þ 6 ¼ xðx
1Þðx
6Þ As can be seen, x + 6 is not a factor of the expression. Choice (4) is the answer. 10.2. Based on the rule of sum of two cubes, we know that: a3 þ b3 ¼ ða þ bÞ a2
ab þ b2




Therefore: x2

ðx þ 2Þðx2
2x þ 4Þ x3 þ 8 ¼ ¼xþ2
2x þ 4 x2
2x þ 4

Choice (1) is the answer. 10.3. Based on the rule of sum of two cubes, we know that: x3 þ y3 ¼ ðx þ yÞ x2
xy þ y2




Choice (3) is the answer. 10.4. Based on the rule of difference of two cubes, we know that: x3
y3 ¼ ðx
yÞ x2 þ xy þ y2




Choice (1) is the answer. 10.5. The problem can be solved as follows:
x3 þ 2x2
3x ¼ x x2 þ 2x
3 ¼ xðx þ 3Þðx
1Þ Choice (1) is the answer.

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Solutions of Problems: Factorization of Polynomials

10.6. The problem can be solved as follows:
6x3 þ 7x2 þ 2x ¼ x 6x2 þ 7x þ 2 ¼ xð3x þ 2Þð2x þ 1Þ Choice (1) is the answer. 10.7. Based on the rule of difference of two cubes, we know that: x3
y3 ¼ ðx
yÞ x2 þ xy þ y2




Therefore:


x3 þ x
10 ¼ x3
8 þ ðx
2Þ ¼ ðx
2Þ x2 þ 2x þ 4 þ ðx
2Þ ¼ ðx
2Þ x2 þ 2x þ 5 Choice (2) is the answer. 10.8. Based on the rule of difference of two squares, we know that: ð a þ bÞ ð a
bÞ ¼ a2
b2 The problem can be solved as follows: 3a2 þ 2ab
b2 ¼ 2a2 þ 2ab þ a2
b2 ¼ 2aða þ bÞ þ ða þ bÞða
bÞ ¼ ða þ bÞð2a þ a
bÞ ¼ ða þ bÞð3a
bÞ Choice (2) is the answer. 10.9. From the rule of difference of two squares, we know that: ð a þ bÞ ð a
bÞ ¼ a 2
b2 The problem can be solved as follows:





x5
x4
4x þ 4 ¼ x5
4x
x4
4 ¼ x x4
4
x4
4 ¼ x4
4 ðx
1Þ ¼ x2 þ 2 x2
2 ðx
1Þ Choice (2) is the answer. 10.10. Based on the information given in the problem, we have: 1 ¼2 x

ð1Þ

  1 1 1 2 2 ¼ x þ þ 2
2 ¼ x þ
2 x x2 x2

ð2Þ

xþ The problem can be solved as follows: x2 þ Solving (1) and (2): x2 þ Choice (4) is the answer. 10.11. As we know:

1 ¼ ð 2Þ 2
2 ¼ 2 x2

10

Solutions of Problems: Factorization of Polynomials

79

ða þ bÞ3 ¼ a3 þ 3a2 b þ 3ab2 þ b3 Therefore: ðx þ 1Þ3
3xðx þ 1Þ x3 þ 3x2 þ 3x þ 1
3x2
3x x3 þ 1 ¼ ¼ 3 ¼1 x3 þ 1 x þ1 x3 þ 1 Choice (2) is the answer. 10.12. Based on the rule of difference of two squares, we know that: ð a þ bÞ ð a
bÞ ¼ a2
b2 The problem can be solved as follows: ð50:01Þ2
ð49:99Þ2 ¼ ð50:01 þ 49:99Þð50:01
49:99Þ ¼ ð100Þð0:02Þ ¼ 2 Choice (3) is the answer. 10.13. Based on the rule of difference of two cubes, we know that: a3
b3 ¼ ða
bÞ a2 þ ab þ b2




The problem can be solved as follows:

x4 þ 2x3
x
2 ¼ x3 ðx þ 2Þ
ðx þ 2Þ ¼ ðx þ 2Þ x3
1 ¼ ðx þ 2Þðx
1Þ x2 þ x þ 1 Choice (1) is the answer. 10.14. Based on the rule of difference of two squares, we know that: ð a þ bÞ ð a
bÞ ¼ a2
b2 The problem can be solved as follows:




x4 þ x2
2 ¼ x4
1 þ x2
1 ¼ x2
1 x2 þ 1 þ x2
1 ¼ x2
1 x2 þ 2 ¼ ðx
1Þðx þ 1Þ x2 þ 2 Choice (3) is the answer. 10.15. Based on perfect square binomial formula, we have: a2 þ b2
2ab ¼ ða
bÞ2 The problem can be solved as follows:

 3 2  3 2 3 2 2 þ 1 2
1 x
2x x x
2x þ 1 ¼ 3
2 ¼
2
2 ¼ 1 3 3 x2
1 x2
1 x2
1 3

3 2

Choice (3) is the answer. 10.16. From the rule of difference of two cubes, we know that:

80

10

Solutions of Problems: Factorization of Polynomials

a3
b3 ¼ ða
bÞ a2 þ b2 þ ab




ð1Þ

Based on the information given in the problem, we have: a
b¼1

ð2Þ

a2 þ b 2 ¼ 5

ð3Þ

a
b ¼ 1) ða
bÞ2 ¼ 1 ) a2 þ b2
2ab ¼ 1

ð4Þ

5
2ab ¼ 1 ) ab ¼ 2

ð5Þ

From (2), we have: ð Þ2

Solving (3) and (4):

Solving (1), (2), (3), and (5): a3
b 3 ¼ 1  ð 5 þ 2 Þ ¼ 7 Choice (3) is the answer. 10.17. As we know: ða þ b þ cÞ2 ¼ a2 þ b2 þ c2 þ 2ab þ 2bc þ 2ac Therefore: ða þ bÞ2 þ ðb þ cÞ2 þ ða þ cÞ2
ða þ b þ cÞ2 a2 þ b2 þ c 2 ¼

a2 þ b2 þ 2ab þ b2 þ c2 þ 2bc þ a2 þ c2 þ 2ac
a2 þ b2 þ c2 þ 2ab þ 2bc þ 2ac a2 þ b2 þ c 2 ¼




a2 þ b2 þ c 2 ¼1 a2 þ b2 þ c 2

Choice (1) is the answer. 10.18. Based on the information given in the problem, we have: xy ¼ 5 xþy¼7 Based on the rule of sum of two cubes, we know that: x3 þ y3 ¼ ðx þ yÞ x2
xy þ y2




Therefore:  

x3 þ y3 ¼ ðx þ yÞ x2 þ 2xy þ y2
3xy ¼ ðx þ yÞ ðx þ yÞ2
3xy ¼ 7  72
3  5 ¼ 7  34 ¼ 238 Choice (2) is the answer. 10.19. The problem can be solved as follows:

10

Solutions of Problems: Factorization of Polynomials

81

2x4 þ x3
4x
2 ð2x4
4xÞ þ ðx3
2Þ 2xðx3
2Þ þ ðx3
2Þ ðx3
2Þð2x þ 1Þ ¼ ¼ ¼ ¼ 2x þ 1 x3
2 x3
2 x3
2 x3
2 Choice (3) is the answer.

10.20. Based on the rule of difference of two squares and perfect square binomial formula, we know that: ð a þ bÞ ð a
bÞ ¼ a2
b2 a2 þ b2 þ 2ab ¼ ða þ bÞ2 The problem can be solved as follows: 2

x4 þ x2 þ 1 x4 þ x2 þ 1 þ x2
x2 x4 þ 2x2 þ 1
x2 ðx2 þ 1Þ
x2 ¼ ¼ ¼ x2 þ x þ 1 x2 þ x þ 1 x2 þ x þ 1 x2 þ x þ 1 ¼

ð x2 þ 1 þ xÞ ð x2 þ 1
xÞ ¼ x2
x þ 1 x2 þ x þ 1

Choice (3) is the answer. 10.21. Based on the information given in the problem, the answer of the following term is requested: x6 þ 4x2 þ 5 x2 þ 1 The problem can be solved as follows:

Choice (2) is the answer. 10.22. Based on the rule of difference of two cubes, we know that: a3
b3 ¼ ða
bÞ a2 þ ab þ b2 Therefore: 9

x2
1 ¼ 3 3 x þ x2 þ 1 Choice (2) is the answer.




  3  3 2 3 x2
1 x2 þ x2 þ 1 x

3 2


2

þx þ1 3 2

3

¼ x2
1

82

10

Solutions of Problems: Factorization of Polynomials

10.23. As we know: ða þ b þ cÞ2 ¼ a2 þ b2 þ c2 þ 2ðab þ bc þ acÞ

ð1Þ

Therefore:

x4 þ y4 þ 4 ¼ 2 2x2 þ 2y2
x2 y2 ) x4 þ y4 þ 4
2 2x2 þ 2y2
x2 y2 ¼ 0 )
x2


2

þ
y2


2






þ 22 þ 2 2
x2 þ 2
y2 þ
x2
y2 ¼ 0

Solving (1) and (2):


2
x2 þ
y2 þ 2 ¼ 0 )
x2
y2 þ 2 ¼ 0 ) x2 þ y2 ¼ 2 Choice (1) is the answer.

10.24. Based on the rule of difference of two squares, we know that: ð a þ bÞ ð a
bÞ ¼ a2
b2 Now, the problem can be solved as follows:


ð2 þ 1Þ 22 þ 1 24 þ 1 . . . 264 þ 1  ð 2
1Þ






¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ )ð2
1Þð2 þ 1Þ 22 þ 1 24 þ 1 . . . 264 þ 1 ¼ 22
1 22 þ 1 24 þ 1 . . . 264 þ 1






¼ 24
1 24 þ 1 . . . 264 þ 1 ¼ 28
1 . . . 264 þ 1 ¼ . . . ¼ 264
1 264 þ 1 ¼ 2128
1 Choice (3) is the answer.

ð2Þ

Problems: Trigonometric and Inverse Trigonometric Functions

11

Abstract

In this chapter, the basic and advanced problems of trigonometric and inverse trigonometric functions 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. 11.1. Calculate the value of 2 cos (x) if sin ðxÞ ¼ 12 and the end of arc x is in the second quadrant. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large pffiffiffi 1)  3 pffiffiffi 2)  2 pffiffiffi 3) 2 pffiffiffi 4) 3 11.2. If cos(α)(1 + tan2(α)) > 0, then determine the quadrant that the end of arc α is located in that. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) First 2) Second 3) First or third 4) First or fourth 



11.3. Calculate the value of cos(10 ) + cos (190 ). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0  2) 2 cos (10 )  3) 2 sin (10 )  4) cos(200 ) 







11.4. Calculate the value of sin(200 ) + sin (10 ) + sin (20 ) + sin (190 ). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2 sin (10 )  2) 2 cos (10 ) 3) 0  4)  sin (10 )

# Springer Nature Switzerland AG 2021 M. Rahmani-Andebili, Precalculus, https://doi.org/10.1007/978-3-030-65056-8_11

83

84

11



Problems: Trigonometric and Inverse Trigonometric Functions



11.5. Calculate the value of sin(1860 ) + cos (1860 ) Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large pffiffi 1þ 3 1) 2 pffiffi 2) 321 pffiffi 3) 12 3 4) 0 



11.6. Calculate the value of tan(135 )sin2(60 ). Difficulty level ● Easy ○ Normal Calculation amount ● Small ○ Normal 1)  34 2)  12 3) 14 4) 34   7π  11.7. Calculate the value of cos 2 5π 4 þ sin 6 . Difficulty level ● Easy ○ Normal Calculation amount ● Small ○ Normal 1) 1 2) 12 3) 14 4) 0 



○ Hard ○ Large

○ Hard ○ Large





11.8. Calculate the value of (sin(60 )  sin (45 ))(cos(30 ) + cos (45 )). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 34 2) 14 3) 12 4) 1 11.9. If cos ðθÞ ¼  23 and tan(θ) cos (θ) > 0, which quadrant is the location of the end of arc θ? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) First 2) Second 3) Third 4) Fourth 11.10. In Figure 11.1, in triangle ABC, determine the value of sin(θ). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 35 2) 34 3) 45 4) 54

11

Problems: Trigonometric and Inverse Trigonometric Functions

85

Figure 11.1 The graph of problem 11.10

11.11. In Figure 11.2, determine the value of tan(α) + tan (β). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 56 2) 45

Figure 11.2 The graph of problem 11.11

3) 4)

6 5 5 4









11.12. Calculate the value of cos(60 ) cos (30 ) + sin (60 ) sin (30 ). Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large  1) tan(30 )  2) cot(45 )  3) sin(60 )  4) cos(60 ) 11.13. Which one of the statements below is correct? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large   1) sin(50 ) < sin (40 )   2) cos(50 ) < cos (40 )   3) tan(50 ) < tan (40 )   4) cot(40 ) < cot (50 ) 11.14. Simplify and calculate the final answer of the term below:             sin 60 cos 60 tan 60 csc 60 sec 60 cot 60       sin ð40 Þ cos ð40 Þ tan ð40 Þcscð40 Þ sec ð40 Þ cot ð40 Þ Difficulty level Calculation amount 1) 0 2) 12 3) 1

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

86

11

Problems: Trigonometric and Inverse Trigonometric Functions

4) 2 







11.15. Calculate the value of sin(135 ) + cos (45 ) + tan (225 ) + cot (315 ). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large pffiffiffi 1) 2 þ 2 pffiffiffi 2) 2 pffiffiffi 3) 2  2 4) 2 

11.16. Which one of the choices is equal to tan(10 )? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) tan(10 )  2) cot(100 )  3) tan(170 )  4) tan(190 ) pffiffi 11.17. Calculate the value of tan(θ) if sin ðθÞ ¼  55 and the end of arc θ is in the third quadrant. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 2)  12 3) 12 4) 2 

11.18. Which one of the choices is equivalent to the angle of 27 ? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large  1) 127  2) 127  3) 333  4) 53 11.19. Calculate the range of m if: cos ð3xÞ ¼ Difficulty level Calculation amount 1) (1, 2] 2) (0, 2) 3) (2, 3] 4) (2, 3)

○ Easy ● Small

○ Normal ○ Normal

m1 , 2



π π 22

○ Easy ● Small

○ Normal ○ Normal

m , 2

● Hard ○ Large

π 3π 0   cos ðαÞ 1 þ tan 2 ðαÞ > 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ) cos ðαÞ > 0

Figure 12.2 The graph of the solution of problem 12.2

Therefore, the end of arc α is in the first or fourth quadrant, as can be noticed from Figure 12.2. Choice (4) is the answer. 12.3. From trigonometry, we know that for an acute angle of θ:    cos 180 þ θ ¼  cos ðθÞ Therefore:              cos 10 þ cos 190 ¼ cos 10 þ cos 180 þ 10 ¼ cos 10  cos 10 ¼ 0 Choice (1) is the answer. 12.4. From trigonometry, we know that for an acute angle of θ:    sin 180 þ θ ¼  sin ðθÞ Therefore:

        sin 200 þ sin 10 þ sin 20 þ sin 190

                  ¼ sin 180 þ 20 þ sin 10 þ sin 20 þ sin 180 þ 10 ¼  sin 20 þ sin 10 þ sin 20  sin 10 ¼ 0 Choice (3) is the answer. 12.5. From trigonometry, we know that for an acute angle of θ:    sin n  360 þ θ ¼ sin ðθÞ, n 2 ℤ    cos n  360 þ θ ¼ cos ðθÞ, n 2 ℤ Moreover, we know that: 

sin 60





pffiffiffi 3 ¼ 2

12

Solutions of Problems: Trigonometric and Inverse Trigonometric Functions

91

  1 cos 60 ¼ 2

Therefore:

          sin 1860 þ cos 1860 ¼ sin 5  360 þ 60 þ cos 5  360 þ 60     ¼ sin 60 þ cos 60 ¼

pffiffiffi pffiffiffi 3 1 3þ1 þ ¼ 2 2 2

Choice (1) is the answer. 12.6. From trigonometry, we know that for an acute angle of θ:    tan 180  θ ¼  tan ðθÞ In addition, we know that:

  tan 45 ¼ 1 

sin 60





Therefore:

pffiffiffi 3 ¼ 2

                 tan 135 sin 2 60 ¼ tan 135 sin 2 60 ¼ tan 180  45 sin 2 60 ¼  tan 45 sin 2 60 ¼ 1  ¼ 

pffiffiffi2 3 2

3 4

Choice (1) is the answer. 12.7. From trigonometry, we know that for an acute angle of θ: cos ðπ þ θÞ ¼  cos ðθÞ sin ðπ þ θÞ ¼  sin ðθÞ Moreover, we know that:

  pffiffiffi 2 π cos ¼ 2 4 sin

  π 1 ¼ 6 2

Therefore: cos 2

          2    pffiffiffi2 2 5π 7π π π π π 1 1 1 þ sin ¼ cos 2 π þ þ sin π þ ¼  cos ¼   sin  ¼  ¼0 2 4 6 4 6 4 6 2 2 2

Choice (4) is the answer. 12.8. From trigonometry, we know that: 

sin 60





pffiffiffi 3 ¼ 2

92

12

Solutions of Problems: Trigonometric and Inverse Trigonometric Functions

  sin 45 ¼

pffiffiffi 2 2

  cos 30 ¼

pffiffiffi 3 2



cos 45





pffiffiffi 2 ¼ 2

In addition, based on the rule of difference of two squares, we know that: ð a þ bÞ ð a  bÞ ¼ a2  b2 The problem can be solved as follows: 

          sin 60  sin 45 cos 30 þ cos 45 ¼

pffiffiffi pffiffiffipffiffiffi pffiffiffi pffiffiffi2 pffiffiffi2 3 3 3 2 2 2 3 2 1  þ ¼  ¼  ¼ 2 2 2 2 2 2 4 4 4

Choice (2) is the answer. 12.9. Based on the information given in the problem, we know that: cos ðθÞ ¼ 

2 ) cos ðθÞ < 0 3

ð1Þ

tan ðθÞ cos ðθÞ > 0

ð2Þ

tan ðθÞ < 0

ð3Þ

Solving (1) and (2):

Figure 12.3 The graph of the solution of problem 12.9

By considering Figure 12.3 as well as (1) and (3), we can conclude that the location of the end of arc θ is in the second quadrant. Choice (2) is the answer.

12.10. From trigonometry, we know that:

12

Solutions of Problems: Trigonometric and Inverse Trigonometric Functions

Side opposite to the angle Hypotenuse

sin ðangleÞ ¼

Moreover, by using Pythagorean formula for the triangle shown in Figure 12.4, we can write: BC 2 ¼ AB2 þ AC 2 ) AB ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi BC 2  AC 2 ¼ 52  42 ¼ 3

Figure 12.4 The graph of the solution of problem 12.10

Therefore: sin ðθÞ ¼

AB 3 ¼ BC 5

Choice (1) is the answer.

12.11. From trigonometry, we know that: tan ðangleÞ ¼

Side opposite to the angle Side adjacent to the angle

Figure 12.5 The graph of the solution of problem 12.11

Therefore: tan ðαÞ þ tan ðβÞ ¼

CB CB 1 1 1 1 5 þ ¼ þ ¼ þ ¼ AB DB 1 þ 1 þ 1 1 þ 1 3 2 6

Choice (1) is the answer. 12.12. From trigonometry, we know that:   1 cos 60 ¼ 2

93

94

12

Solutions of Problems: Trigonometric and Inverse Trigonometric Functions

  cos 30 ¼ 

sin 60





pffiffiffi 3 2

pffiffiffi 3 ¼ 2

  1 sin 30 ¼ 2

Therefore: pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi         1   3 3 1 3 3 3 þ  ¼ þ ¼ ¼ sin 60 cos 60 cos 30 þ sin 60 sin 30 ¼  2 2 4 4 2 2 2 Choice (3) is the answer. 12.13. We know that for an acute angle, sin(α) and tan(α) have ascending trends. Therefore:     sin 40 < sin 50     tan 40 < tan 50 Moreover, for an acute angle, cos(α) and cot(α) are descending. Therefore:     cos 50 < cos 40     cot 50 < cot 40 Choice (2) is the answer. 12.14. From trigonometry, we know that: cot ðαÞ ¼

1 tan ðαÞ

sec ðαÞ ¼

1 cos ðαÞ

cscðαÞ ¼

1 sin ðαÞ

The problem can be solved as follows:             sin 60 cos 60 tan 60 csc 60 sec 60 cot 60       sin ð40 Þ cos ð40 Þ tan ð40 Þcscð40 Þ sec ð40 Þ cot ð40 Þ             sin 60 csc 60 cos 60 sec 60 tan 60 cot 60 ¼ ¼       sin ð40 Þcscð40 Þ cos ð40 Þ sec ð40 Þ tan ð40 Þ cot ð40 Þ   sin 60 sin

    cos 60 cos 160 tan 60 tan 160 ð Þ ð Þ 111 ¼ ¼ ¼1    1 1 111 sin ð40 Þ sin 40 cos ð40 Þ cos 40 tan ð40 Þ tan 140 ð Þ ð Þ ð Þ Choice (3) is the answer.

1



ð60 Þ

12

Solutions of Problems: Trigonometric and Inverse Trigonometric Functions

95

12.15. From trigonometry, we know that for an acute angle of θ:    sin 180  θ ¼ sin ðθÞ    tan 180 þ θ ¼ tan ðθÞ    cot 360  θ ¼  cot ðθÞ

In addition, we know that: 





pffiffiffi 2 ¼ 2







pffiffiffi 2 ¼ 2

sin 45

cos 45

  tan 45 ¼ 1   cot 45 ¼ 1 The problem can be solved as follows:                 sin 135 þ cos 45 þ tan 225 þ cot 315 ¼ sin 180  45 þ cos 45 þ tan 180 þ 45    þ cot 360  45 

¼ sin 45







þ cos 45





    þ tan 45  cot 45 ¼

pffiffiffi pffiffiffi pffiffiffi 2 2 þ þ11¼ 2 2 2

Choice (2) is the answer. 12.16. From trigonometry, we know that for an acute angle of θ: tan ðθÞ ¼  tan ðθÞ    cot 90 þ θ ¼  tan ðθÞ    tan 180  θ ¼  tan ðθÞ    tan 180 þ θ ¼ tan ðθÞ Choice (1):

Choice (2):

Choice (3):

Choice (4):

    tan 10 ¼  tan 10        cot 100 ¼ cot 90 þ 10 ¼  tan 10        tan 170 ¼ tan 180  10 ¼  tan 10

96

12

Solutions of Problems: Trigonometric and Inverse Trigonometric Functions

       tan 190 ¼ tan 180 þ 10 ¼ tan 10 Choice (4) is the answer. 12.17. Based on the information given in the problem, we know that the end of arc θ is in the third quadrant and: pffiffiffi 5 sin ðθÞ ¼  5

ð1Þ

sin 2 ðθÞ þ cos 2 ðθÞ ¼ 1

ð2Þ

tan ðθÞ ¼

sin ðθÞ cos ðθÞ

ð3Þ

The problem can be solved as follows: Solving (1) and (2):

Figure 12.6 The graph of the solution of problem 12.17

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cos ðθÞ ¼ 1  sin 2 ðθÞ ¼

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffi pffiffiffi2 5 1 2 1 ¼ 1  ¼  pffiffiffi 5 5 5

In (4), the negative value is acceptable because the end of arc θ is in the third quadrant (see Figure 12.6). Solving (1), (3), and (4):

pffiffi sin ðθÞ  55 1 tan ðθÞ ¼ ¼ ¼ cos ðθÞ  p2ffiffi 2 5

Choice (3) is the answer.

12.18. As we know from trigonometry, in a unit circle, the relation below holds for any pair of angles: 

θ2  θ1 ¼ n  360 , n 2 ℤ ) θ2  θ1 The problem can be solved as follows: Choice (1):

ð4Þ

12

Solutions of Problems: Trigonometric and Inverse Trigonometric Functions

97

      127  27 ¼ 154 ) 127 ≢  27 Choice (2):

Figure 12.7 The graph of the solution of problem 12.19

      127  27 ¼ 100 )  127 ≢  27 Choice (3):

      333  27 ¼ 360 ) 333   27

Choice (4):

      53  27 ¼ 80 ) 53 ≢  27

Choice (3) is the answer. 12.19. Calculate the range of m if: cos ð3xÞ ¼

m1 , 2



π π