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Printed at: Print Vision, Navi Mumbai
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P.O. No. 179827
Challenger
MathematicsVol. I
JEE (Main)
For all Engineering Entrance Examinations held across India.
Exhaustive coverage of MCQs under each sub-topic.
2049 Questions of the level of various competitive exams.
Includes solved MCQs of JEE (Main) from year 2014 to 2018.
Includes JEE (Main) - 2019 [8th April (Set - I)] Question Paper and Answer Keywith solutions provided in Q.R. Code form.
Concise theory for every topic. Model Test papers for thorough revision and practice.
Important inclusions: Shortcuts, Important Notes, Problems to Ponder.
Salient Features
TEID: 13620
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Target’s ‘Challenger Maths: Vol-I’ is a compact guidebook, extremely handy for preparation of various competitive exams like JEE (Main). Features of each chapter: Coverage of Theoretical Concepts that form a vital part of any competitive examination. Multiple Choice Questions segregated into two sections. Concept Building Problems – Contains questions of
various difficulty range and pattern. Practice Problems – Contains ample questions for thorough revision. ‘Important Note’ highlights the unique points about the topic. Shortcuts to help students save time while dealing with lengthy questions. Problems to Ponder: Various types of questions of different pattern created with the primary objective of
helping students to understand the application of various concepts of Maths. Two Model Test Papers are included to assess the level of preparation of the student on a competitive level. MCQs have been created and compiled with the following objective in mind – to help students solve complex problems which require strenuous effort and understanding of multiple-concepts. The level of difficulty of the questions is at par with that of various competitive examinations like CBSE, JEE (Main), AIEEE, TS EAMCET (Engg.), BCECE, Assam CEE, AP EAMCET (Engg.) and the likes. Also to keep students updated, questions from the most recent examinations of JEE (Main), of years 2014, 2015, 2016, 2017 and 2018 are covered exclusively. JEE (Main) - 2019 [8th April (Set - I)] Question Paper and Answer Key have been provided. Students can scan the Q.R. Code to access the solutions of the given Question Paper. The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you. Please write to us on : [email protected] A book affects eternity; one can never tell where its influence stops. From, Publisher Edition: Second
Disclaimer This reference book is based on the JEE (Main) syllabus prescribed by Central Board of Secondary Education (CBSE). We the publishers are making this reference book which constitutes as fair use of textual contents which are transformed by adding and elaborating, with a view to simplify the same to enable the students to understand, memorize and reproduce the same in examinations. This work is purely inspired upon the course work as prescribed by the National Council of Educational Research and Training (NCERT). Every care has been taken in the publication of this reference book by the Authors while creating the contents. The Authors and the Publishers shall not be responsible for any loss or damages caused to any person on account of errors or omissions which might have crept in or disagreement of any third party on the point of view expressed in the reference book. © reserved with the Publisher for all the contents created by our Authors. No copyright is claimed in the textual contents which are presented as part of fair dealing with a view to provide best supplementary study material for the benefit of students.
PREFACE
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No. Topic Name Page No.
1 Basic Algebra 1
2 Sets, Relations and Functions 19
3 Quadratic Equations 72
4 Trigonometry - I 104
5 Trigonometry - II 172
6 Basic coordinate geometry and Straight lines 238
7 Complex Numbers 306
8 Sequences and Series 340
9 Circles 383
10 Conics 430
11 Permutations and Combinations 500
12 Mathematical Induction and Binomial Theorem 527
13 Probability 563
14 Statistics 598
Model Test Paper - I 619
Model Test Paper - II 622
JEE (Main) - 2019: Question Paper & Answer Key 8th APRIL (Set – I) 625
CONTENT
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Why Challenger Series? Gradually, every year the nature of competitive entrance exams is inching towards conceptual
understanding of topics. Moreover, it is time to bid adieu to the stereotypical approach of solving a problem using a single conventional method. To be able to successfully crack the JEE (Main) examination, it is imperative to develop skills such as data interpretation, appropriate time management, knowing various methods to solve a problem, etc. With Challenger Series, we are sure, you’d develop all the aforementioned skills and take a more holistic approach towards problem solving. The way you’d tackle advanced level MCQs with the help of hints, tips, shortcuts and necessary practice would be a game changer in your preparation for the competitive entrance examinations.
What is the intention behind the launch of Challenger Series? The sole objective behind the introduction of Challenger Series is to severely test the student’s
preparedness to take competitive entrance examinations. With an eclectic range of critical and advanced level MCQs, we intend to test a student’s MCQ solving skills within a stipulated time period.
What do I gain out of Challenger Series? After using Challenger Series, students would be able to: a. assimilate the given data and apply relevant concepts with utmost ease. b. tackle MCQs of different pattern such as match the columns, diagram based questions, multiple
concepts and assertion-reason efficiently. c. garner the much needed confidence to appear for various competitive exams. Can the Questions presented in Problems to Ponder section be a part of the JEE (Main)
Examination? No, the questions would not appear as it is in the JEE (Main) Examination. However, there are fair chances that these questions could be covered in parts or with a novel question construction.
Why is then Problems to Ponder a part of this book? The whole idea behind introducing Problems to Ponder was to cover an entire concept in one question.
With this approach, students would get more variety and less repetition in the book.
Best of luck to all the aspirants!
Frequently Asked Questions
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Chapter 1: Basic Algebra
Algebraic Identities and Polynomials: 1. Algebraic Identities: i. (a + b)2 = a2 + 2ab + b2 ii. (a b)2 = a2 2ab + b2 iii. (a + b)2 (a b)2 = 4ab (a + b)2 = (a b)2 +4ab (a b)2= (a + b)2 4ab iv. a2 b2 = (a + b) (a b) v. (a + b)3 = a3 + b3 + 3ab (a + b) or (a + b)3 = a3 + 3a2b + 3ab2 + b3 vi. (a b)3 = a3 b3 3ab(a b) or (a b)3 = a3 3a2b + 3ab2 b3 vii. a3 + b3 = (a + b)3 3ab(a + b) or a3 + b3 = (a + b) (a2 ab + b2) viii. a3 b3 = (a b)3 + 3ab(a b) or a3 b3 = (a b) (a2 + ab + b2 ) ix. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) x. a3 + b3 + c3 3abc
= (a + b + c) (a2 + b2 + c2 ab bc ca) Illustrations:
1. If x + 1x
= 103
, x > 1, then evaluate
i. x 1x
ii. x2 2
1x
iii. x3 + 3
1x
iv. x3 – 31x
Solution:
i. 21x
x
= 21x
x
4(x) 1x
= 1009
4 = 649
x 1x
= 83
ii. x2 2
1x
= 1 1x xx x
= 10 83 3
= 809
iii. x3 + 3
1x
= 31x
x
3(x)
1x
1xx
= 100027
3 103
= 73027
iv. x3 3
1x
= 31 1 13x x x
x x x
= 38 83
3 3
= 51227
+ 8
= 72827
2. a, b, c are positive real numbers satisfying a3 + b3 + c3 = 3abc. Then show that a = b = c. Solution: a3 + b3 + c3 = 3abc a3 + b3 + c3 3abc = 0 (a +b + c) (a2 + b2 + c2 ab bc ca) = 0
…(i) Since, a, b, c are positive real numbers. (i) a2 + b2 + c2 ab bc ca= 0
12
(2a2 + 2b2 + 2c2 2ab 2bc 2ca) = 0
12
[( a2 2ab + b2) + (b2 2bc +c2)
+ (c2 2ca + a2) ]= 0
12
[(a b)2+ (b c)2 + (c a)2]= 0
a b = 0 and b c = 0 and c a = 0 a = b = c 2. Polynomials: A polynomial P(x ) in variable x is defined as P(x) = a0 + a1x + a2x2+…+ anxn …(i) where a0, a1, a2, … , an are real numbers
(constants) and an ( 0) is called the leading coefficient.
Degree of a polynomial is the index of the term of highest power.
(i) represents a polynomial of degree n. A polynomial P(x) = ax + b is a linear
polynomial.
1.1 Algebraic Identities and Polynomials 1.2 Absolute Value
1.3 Intervals and Inequations 1.4 Rational and Irrational Inequations
Basic Algebra 1
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Challenger Maths Vol - I (Engg.)
A polynomial P(x) = ax2 + bx + c is a 2nd degree polynomial, called a quadratic polynomial.
A polynomial P(x) = ax3 + bx2 + cx + d is a
cubic polynomial. If a polynomial equation P(x) = 0 is satisfied
by x = k, then k is a zero of the polynomial P(x).
Division Algorithm: If a polynomial P(x) is divided by a
polynomial D(x), giving quotient Q(x) and remainder R(x), then
P(x) = D(x) Q(x) + R(x), where R(x) = 0 or degree of R(x) < degree of D(x) e.g. If P(x) = x4 + 4 is divided by x + 2, then
quotient is a cubic polynomial. P(x) = x4 + 4 = (x + 2) (ax3 + bx2 + cx + d)+ r where r is the remainder (r is a constant as
divisor is of degree 1) Clearly a = 1 a = 1 gives a term 2x3. As there is no cubic
term in P(x), to eliminate 2x3 , b needs to be 2.
b= 2 gives a term 4x2. As there is no term of x2 in P(x), to eliminate 4x2, c needs to be 4.
c = 4 gives a term 8x. As there is no term of x in P(x), to eliminate 8x, d needs to be 8.
x4+ 4 = (x + 2) (x3 2x2 + 4x 8) + 20 r = 20 Remainder Theorem: If a polynomial P(x) is divided by x ,
then the remainder is given by P(). Factor theorem: For a polynomial P(x), x is a factor of
P(x) iff P() = 0 Illustration: Factorize the polynomial P(x) = 2 x3 11 x2 23 x + 14 Solution: To find one zero of a polynomial. Substitute x = 1, 1, 2, 2,… We observe P(2) = 0 x + 2 is a
factor. P(x) = 2 x3 11 x2 23 x + 14 = (x + 2) (2x2 15 x + 7) = (x + 2) (2x 1) (x 7)
Absolute Value: 1. Definition: For a real number x , absolute
value of x is denoted as | x | and defined as | x | = x , x < 0 = x, x 0 e.g. |3| = 3, |7| = 7 2. Graph of y = | x | y = x, x < 0 and
y = x , x 0 x = 0 is the critical
point. To get the graph,
some points like (1, 1), (1, 1), (2, 2), (2, 2) , (0,0) etc. are to be plotted.
x and x are linear polynomials and graphs of linear polynomials are lines.
A drastic change can be seen around the critical point.
3. Graph of y = | x a |, a > 0 y = a x, x < a y = x a, x a x = a is the critical point. 4. Properties of absolute value: For two real numbers a and b , i. | ab | = | a | | b | ii. Triangle Inequality: | a + b | | a | + | b | Equality occurs only when a and b
have same sign or one of them is (or both are) 0.
Equality occurs only when ab 0 Similar notation | | would be seen in the
chapter Complex Numbers with name modulus.
y
O x
y = –x y = x
a O x
y = x ay = a x
y
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Chapter 1: Basic Algebra
Intervals and Inequations: 1. Intervals: i. Closed Interval [a,b], a, b are real numbers. A closed interval [a,b] is the collection of all such real numbers which are greater than or equal to a, and less than or equal to b. [a,b] = { x : x R, a x b} ii. Open Interval (a, b), a,b are real numbers. An open interval (a, b) is the collection of
all such real numbers which are greater than a and less than b. (a,b) = { x : x R, a x b} iii. Other type of intervals: [a, b) = { x : x R, a x b} (a, b] = { x : x R, a x b} (a, ) ={ x : x R, x > a} [a, ) = { x : x R, x a } (, a) ={ x : x R, x < a} (, a] ={ x : x R, x a} 2. SETS: A well defined collection is known as
a SET in mathematics. A member of a set is called an element. e.g. A closed interval is a set and a number appearing in the interval is an element.
If an element x is present in a set A, it is denoted as x A
e.g. 3[2,7] If an element x is not present in a set
A, it is denoted as x A e.g. 2 (2,1) Two operations called union and
intersection are defined as follows: Union of two sets AB: AB= { x : x A or x B}
Intersection of two sets AB: AB = { x : x A and x B} e.g. A = [3, 5] , B = [2, 7] then AB= [3, 7] and AB = [2, 5] A set having only one element, say a,
is known as a singleton set. It is denoted as {a}
A set having no element is called a null set. It is denoted as { } or . e.g. A = the set containing all even integers.
B = the set containing all odd integers Then AB = More on sets can be found in the
chapter SETS, RELATIONS and FUNCTIONS.
3. Inequations: Let X be a given set. Consider two expressions in x , say p(x) and q(x), x X. p(x) > q(x) is an inequation. All the values of x for which the
expression p(x) is greater than the expression q(x) are called the solutions of the inequation p(x) > q(x).
The set of all such values is called the solution set.
i. Types of inequations: p(x) > q(x) is called a strict
inequation. p(x) q(x) is called a nonstrict
inequation. ii. Equivalence of inequations: A given inequation can undergo successive transformations and can be reduced to a simpler inequation known as equivalent inequation. Assertions used for equivalence are as
follows: p(x) < q(x) p(x) > q(x), xR Let p(x) > 0 and q(x) > 0, then
p(x) < q(x) 1( )p x
> 1( )q x
p(x) < q(x) p(x) + r(x) < q(x) + r(x) Let r(x) > 0, then p(x) < q(x) p(x) r(x) < q(x) r(x) Let p(x) 0 and q(x) 0, then p(x) < q(x) (p(x))2 < (q(x))2 ‘’ sign is used to show the equivalence . i.e., ‘’ is used for ‘ is same as ’.
Square brackets [ ] are used to include the end points of the interval. Round brackets ( ) are used to exclude the end points of the interval. and are symbolic representations of left and right ends of the number line, they are not actual values. So they are always shown with round brackets.
Important Notes
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4
Challenger Maths Vol - I (Engg.)
iii. Fundamental Inequations and their
solution sets. Inequations Solution Sets i. | x | < a (a, a) ii. | x | a [a, a] iii. | x | > a (, –a) (a, ) iv. | x | a (, –a] [a, )
Illustrations: Solve: 1. | x 7| < 3 2. |3x +5| 8 3. |6x 2| > 4 4. |2x 5| 3 Solution: 1. | x 7| < 3 3 < x 7 < 3 Adding 7 throughout, we get 4 < x < 10 Solution set = (4,10) 2. |3x +5| 8 8 3x +5 8 Subtracting 5 throughout, we get 13 3x 3
133 x 1
Solution set = 13 ,13
3. |6x 2| > 4 6x 2 < 4 or 6x 2 > 4 6x < 2 or 6x > 6
x < 13 or x > 1
Solution set = 1, 1,3
4. |2x 5| 3 2x 5 –3 or 2x 5 3 2x 5 3 or 2x 5 + 3 x 1 or x 4 Solution set = (,1] [4, ) Rational and Irrational Inequations: 1. Rational inequations: i. Rational function: Let Pn(x) and Qm(x) be polynomials of
degree n and m respectively. Then
R(x) = P ( )Q ( )
n
m
xx
is called a rational function.
R(x) 0 or R(x) < 0 are rational inequations. A rational inequation has only rational functions.
The zeroes of Pn(x) and Qm(x) are called the critical points. ii. Method of intervals for solving a rational
inequation: The method is based on the fact that “ a rational function retains its sign in the
interval between its two neighbouring critical points”.
If the rational inequation is a strict
inequation, then the critical points of the rational function are not solutions.
If the rational inequation is a non strict
inequation, then the zeroes of Pn(x) are solutions(and the zeroes of Qm(x) are not solutions) provided they are not zeroes of Qm(x).
Method: a. Mark the critical points on the number
line to partition the number line into finite number of sub-intervals.
b. On each sub-interval the L.H.S. of the given inequation has a particular sign.
c. Pick up any point of a sub-interval and check the sign of L.H.S of the given inequation. (At all the points of the subinterval L.H.S. has same sign as at this point.)
d. If the sign is same as the required sign, then the considered sub-interval is an interval of the solution set otherwise solution lies in the adjacent intervals.
Important Notes Multiplying an inequation by a negative
number reverses the inequality sign. Same quantity can be added on two sides of an
inequation without changing the inequalitysign.
Multiplying by a positive number to aninequation does not change the inequality sign.
If both sides of an inequation are non negative, then squaring both sides of the inequation does not change the inequality sign.
While solving ( )( )
p xq x
< k, k is a constant or
variable, never multiply q(x) to the R.H.S assign of q(x) is unknown.
e.g. (2x + 3) (x + 1)2 > (x + 1) is equivalent to(2x + 3) (x + 1) > 1 only if x + 1 0.
(If x + 1 is negative, then multiplying by
11x
would reverse the inequality sign.)
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Chapter 1: Basic Algebra
iii. Standard rational inequations and their solutions (a < b)
Inequations Solution Sets i. (x a) (x b) < 0 (a, b) ii. (x a) (x b) 0 [a, b] iii. (x a) (x b) > 0 (, a) (b, ) iv. (x a) (x b) 0 (, a] [b, ) Inequations Pictorial
Representation i. (x a) (x b) < 0
ii. (x a) (x b) 0
iii. (x a) (x b ) > 0
iv. (x a) (x b) 0
Illustrations:
1. The largest integer that satisfies ( 5)2
x xx
4 is
(A) 2 (B) 7 (C) 8 (D) 1
Solution: ( 5)2
x xx
4
( 5)2
x xx
4 0
2 9 8
2x x
x
0
( 1)( 8)2
x xx
0 …(i)
(A rational inequation) x = 1, 2, 8 are the critical points. Since, the given inequation is a non strict
inequation. x = 1, 8 are the solutions. …(ii) Mark 1, 2, 8 on number line to partition it. Select an interval, say (2,8) and a point from it, say x = 3. Substituting x = 3 in L.H.S. of (i), we get
(3 1)(3 8)(3 2)
= 10 which is less than 0.
The interval (2,8) is a part of the required solution set. …(iii)
By the method of intervals (8,) and (1,2) do not belong to the solution set.
(,1) belongs to the required solution set. …(iv)
(ii), (iii) and (iv) the solution set = (, 1] (2, 8]
option (C) is the correct answer. 2. For the inequation
2( 1)( 1)x xx
0, which
of the following is the correct statement? (A) All negative real numbers belong to
the solution set. (B) More than two positive integers
satisfy the inequation. (C) Exactly two positive integers satisfy
the inequation. (D) An isolated integer satisfies the inequation. Solution:
2( 1)( 1)x x
x
0
x = 1, 0, 1 are the critical points. Since, the given inequation is a non strict
inequation. x = 1,1 are the solutions. …(i)
Given inequation 1xx 0 …(ii)
(as (x + 1)2 always does not contribute in fixing sign.)
(ii) (0,1) belongs to the required solution set. …(iii)
By the method of intervals (,0) and (1,) do not belong to the solution set.
(i) and (iii) the solution set = (0,1] {1} option (D) is the correct answer. 3. The solution set of the inequation |x2 x 2| > 4 is (A) (2, 3) (B) (, 2) (C) (3, ) (D) (, 2) (3, ) Solution: |x2 x 2| > 4 |(x 2) (x +1)| > 4 |x 2| |x + 1| > 4 Critical points are 1, 2. |x2 x 2| = x2 x 2, x < 1 = – (x2 x 2), 1 x < 2 = (x2 x 2), x 2 For x –1 or x 2, Given inequation x2 x 2 4 x2 x 6 0 (x + 2) ( x 3) 0 x –2 or x 3 x (,2) (3,) is a solution … (i) For –1 x 2,
1 2 8
1 2 8
1 0 1
b a
[ ]
a b [ ]
) b a (
b a
) (
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Challenger Maths Vol - I (Engg.)
Given inequation – (x2 x – 2) 4 x2 x + 2 < 0
21
2
x + 74
0
(Note this step!) which is a meaningless statement. There is no solution in [–1, 2) (i) solution set = (, 2) (3, ) Aliter: Given inequation is of the form |x| a So, x2 – x – 2 – 4 or x2 – x – 2 4 x2 – x + 2 0 or x2 – x – 6 0 (x – 3) ( x + 2) 0 x –2 or x 3 x2 – x + 2 < 0 is not meaningful as shown
before. The solution set is (, 2) (3, ) option (D) is the correct answer. 2. Irrational inequations: Square root: A square root is denoted by and defined, for real x, as 2x = | x | …(i) So, 2x = x, x < 0 = x, x 0 (i) shows that we can find the square root of
a non negative real number i.e., the quantity whose square root we find has to be non negative and the value of square root is also non negative.
e.g. 27 = |7| = 7 Illustrations: 1. Solve 7 6x < x for real x. Solution: Given inequation is an irrational inequation. Before solving any irrational inequation, we find feasible region. Feasible region is not the solution set but
a set of those values for which the inequation is a meaningful statement.
L.H.S. of given inequation is nonnegative R.H.S. of given inequation must be
positive. (as the inequation is a strict inequation.)
i.e., x > 0 …(i)
Also, 7x 6 0 x 67
…(ii) (i) and (ii)
the feasible region = 6 ,7
… (iii)
In the feasible region, both sides of the given inequation are non negative and hence, we can square them.
Squaring the given inequation, we get 7x 6 < x2 x2 7x + 6 > 0 (x 1) (x 6) > 0 x < 1 or x > 6 …(iv) (iii) and (iv) the solution set
= 6 , 17
(6,)
2. For the inequation x 1 < 2 3 10x x , which of the following statements is true? (A) The greatest negative integer
satisfying the inequation is 2. (B) The least positive integer satisfying
the inequation is 2. (C) The number of integers that do not
satisfy the inequation are 7.
(D) The solution set is (,1) 11,5
Solution: First we find the feasible region. x2 + 3x – 10 0 (x + 5) (x 2) 0 x 5 or x 2 (Feasible region)…(i) R.H.S. of given inequation is non negative. If L.H.S. of given inequation is negative(and that happens when x < 1), then the given
inequation turns out to be a true statement. But (5,1) does not belong to the feasible
region. x 5 is a solution …(ii) Let x 1. Note that [1,2) does not belong to the
feasible region. So let x 2. Squaring the given inequation, we get x2 2x + 1 < 2 3 10x x
5x > 11 x > 115
…(iii)
(i), (ii) and (iii) the solution set
= (, 5] 11,5
Integers that do not satisfy the given inequation are –4, –3, –2, –1, 0, 1, 2.
option (C) is the correct answer.
67
1 6
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Chapter 1: Basic Algebra
Some important results:
i. For x > 0, 1xx
2
Equality occurs only when x = 1x
= 1
ii. For x < 0, 1xx
2
Equality occurs only when x = 1x
= 1
Illustrations on these can be seen in the following chapters.
1. If (p2 + q2)3 = (p3 + q3)2 and pq ≠ 0, then the
value of p qq p is
(A) 32
(B) 23
(C) 13
(D) 12
2. If a = 32 33 , b = 33 34 and c = 3–4 – 3–2,
then a3 + b3 + c3 equals
(A) 8
13
(B) 10
323
(C) 9
13 (D) 20
1043
3. Three distinct numbers a, b, c R. If a3 + b3 + c3 = 3abc, then
(A) a2 + b2 + c2 = ab + bc + ca (B) a + b + c = 0 (C) (a b)2 + (b c)2 + (c a)2 = 0 (D) (a + b + c)3 0 4. If x300 < 3500, then the greatest integer value
of x is (A) 4 (B) 5 (C) 6 (D) 7 5. If cba = 6561, then maximum value of abc, where a,b,c are positive integers is (A) 162 (B) 144 (C) 18 (D) 9 6. If 2x = 3y = 18z, then 1
x+ 1
y+ 1
z equals
(A) 2 1z y (B) 2 1
z y
(C) 0 (D) 2y
7. The remainder on dividing x100 2x51 + 1 by x2 1 is Ax + B. Then A + 2B is (A) 0 (B) 2 (C) 2 (D) 4 8. P(x) is a polynomial with integer
coefficients and has the lowest possible degree. If one of the zeroes of P(x) is 3 11 + 3 121 , then the product of all zeroes is
(A) 11 (B) 121 (C) 1331 (D) 132 9. If the expression
3 3 1( 1)
x xx x
remains
unaltered by the substitution x 11 x
and
x 1 1x
, then the roots of the equation 3 3 1
( 1)x x
x x
= 3 23 1
(1 )c c
c c
are
(A) c, 11 c
, 1cc
(B) 1 , ,1
cc c
1 c
(C) 1 + c, 1,1
cc c
(D) c 1, 1 ,2 1
cc c
Algebraic Identities and Polynomials 1.1
Concept Building Problems
Important Notes
Before solving an irrational inequation(or an irrational equation), feasible region must be obtained. The zeroes of Qm(x), while solving R(x ) < 0 or R(x) 0 or R(x) > 0 or R(x) 0, where
R(x) = P ( )Q ( )
n
m
xx
, never belong to the solution set.
The zeroes of Pn(x) and Qm(x) while solvingR(x) < 0 or
R(x) > 0, where R(x) = P ( )Q ( )
n
m
xx
, never belong to
the solution set. The zeroes of Pn(x), while solving R(x) 0 or
R(x) 0, where R(x) = P ( )Q ( )
n
m
xx
, belong to the
solution set.
SAMPLE C
ONTENT
8
8
Challenger Maths Vol - I (Engg.) 10. The graph of y = |x + 3| meets the xaxis at x = (A) 3 (B) 3 (C) 0 (D) 3 and 3 11. The sum of the values of x that satisfy
|2x 3| 4 = 3 is (A) 5 (B) 5 (C) 3 (D) 2 12. How many positive integer solutions does
the equation |2x 1| = 5 have? (A) 2 (B) 1 (C) 0 (D) 3 13. Let x = a satisfy |10x 3| = 0. Then the
value(s) of x satisfying 35 xa
= 20 is/are
(A) 2 only (B) 6 only (C) 2 and 6 (D) 2 and 6
14. If |x| + 1
xx
= 2
1x
x, then
(A) x > 1 (B) x < 1 (C) x < 0 (D) x > 0 15. If |x 1| + |5 2x| = |3x 6|, then x belongs
to the interval
(A) ( , 1] 5 ,2
(B) 5 ,2
(C) 51,2
(D) (, 1]
16. How many integer solutions does the
inequation |5 2x| < 1 have? (A) 1 (B) 2 (C) 3 (D) 0 17. If a real number x > 0 is such that 5
5
xx
> 0,
then the correct statement is (A) 3x + 10 < 25 (B) x2 4x 5 > 0 (C) (x 5) (x + 3)2 < 0 (D) |2x 5| < 5 18. If 53
2x 2, then the value of x which
satisfies the inequation is
(A) 47
(B) 25
(C) 38
(D) 110
19. The solution set of the inequation 12
xx
< 1 is
(A) 12,2
(B) 1,2
(C) 1, 2 ,2
(D) 1 ,2
20. The sum of integer values of x satisfying
|2x 4| < x 1 is (A) 2 (B) 3 (C) 1 (D) 6 21. The least positive and greatest negative
integer values of x that satisfy 2|x + 1| > x + 6 are a and b respectively. Then a b equals
(A) 6 (B) 7 (C) 8 (D) 9 22. The sum of all those integers which do not
satisfy the inequation ( 1)( 3)( 2)( 1)x xx x
0 is
(A) 1 (B) 5 (C) 3 (D) 2 23. The number of integers satisfying the
inequation 2 51x
x
> 2 is
(A) 3 (B) 2 (C) 1 (D) 0 24. If the solution set of the inequation
1 12
x x 2
2xis (a,b] (c,d) [f,),
then a +b + c + d + f is (A) 3 (B) 3 + 17 (C) 3 17 (D) 4 25. How many integers do not satisfy the
inequation 3|x 1| + x2 7 > 0? (A) 3 (B) 4 (C) 2 (D) 1 26. The solution set of the inequation
3 10x > 6 x is
(A) 10 , 63
(B) 10 , 43
(C) (4, 6] (D) (4, ) 27. The solution set of the inequation
(x 1) 2 2x x 0 is (A) ( ] {2} (B) [2, ) {1, 1} (C) [1, ) (D) ( ] [2, ) 28. The solution set of the inequation
18x < 2 x is (A) [18, 2) (B) [18, 2) (C) (, 2) (7, ) (D) [18, 2) (7, )
Absolute Value 1.2
Rational and Irrational Inequations1.4
Intervals and Inequations 1.3
SAMPLE C
ONTENT
9
Chapter 1: Basic Algebra
29. The solution set of the inequation 2 3 2x x > 2x 5 is (, a] , c db
f
,
then a + b + ( )c df is
(A) 8 (B) 6
(C) 113
(D) 2 1. If 4 x 6 and 3 y 1, then (A) 1 x y 5 (B) 1 x y 4 (C) 3 x y 6 (D) 5 x y 9 2. Simplified form of the expression
32 3
5 5a b
32 35 5a b
is
(A) 2 26 (4 3 )25
b a b (B) 2 26 (4 3 )
125b a b
(C) 2 218 (4 3 )
125b a b (D)
2 218 (4 9 )25
b a b 3. The product (a + b + c) [(a b)2 + (b c)2 + (c a)2] is (A) a3 + b3 + c3 (B) (a + b + c)3 (C) a3 (b + c) + b3 (c + a) + c3 (a + b) (D) 2(a3 + b3 + c3 3abc) 4. p3 + (q p)3 q3 = (A) 3(p q) (q 1) (1 p) (B) 3p3q3 (q p) (C) 3pq(q p) (D) 3q2 p2 (q p) 5. If p2 + 2
1p
= 83, then the value of p 1p
is
(A) 9 (B) 9 (C) 9 or 9 (D) none of these 6. If x 1
x= 6, then 4
4
1xx
is
(A) 1442 (B) 1158 (C) 1154 (D) 1296 7. If m + 1
m= 2, then 4
4
1mm
equals
(A) 4 (B) 2 (C) 8 (D) 16 8. If x =
4 83 22 3
, then x2 is
(A) 242
3
(B) 122
3
(C) 123
2
(D) 243
2
9. The remainder when polynomial x3 5x2 + x + 1 is divided by x 2, is
(A) 9 (B) 9 (C) 10 (D) 10 10. The number which divides (12)1 to give
quotient 12
3
and remainder zero, is
(A) 112
(B) 16
(C) 118
(D) 18 11. The solution set of the inequation
2 51
xx
1 is
(A) [3, 1) [ 2, ) (B) (, 3] (1, 2] (C) [3, 1) (1, 2] (D) (, 5 ) (1, 5 ] 12. The solution set of the inequation
(x 3) (x 3) (x 4) is (A) (5, ) (B) [5, ) (C) (, 3] (D) (, 3] [5, ) 13. The set of all values of x satisfying the
inequation (x 4) (x 10) 0 and (x 9) (2x + 1) 0 is
(A) 1 , 42
(B) [4, 9]
(C) 1 , 102
(D) [9, 10]
14. The solution set of the inequation
(x 2)3 (x 4)2 (x 10)17 0 is (A) ( (B) (, 2) (4, 10) (C) (, 2] [10, ) {4} (D) [2, 4] [10, ) 15. Consider the inequation 2
(2 1)( 4)x xx
0.
Which of the following is correct? (A) All negative integers satisfy the inequation. (B) There is an isolated integer that does
not satisfy the inequation. (C) Not all positive integers satisfy the
inequation. (D) The greatest negative integer
satisfying the inequation is 2. 16. The solution set of the inequation
4 21 1x x
< 1 is
(A) (, 1) (1, ) (B) (1, 1) (C) (0, 4) (D) (1, 0)
Practice Problems
SAMPLE C
ONTENT
10
10
Challenger Maths Vol - I (Engg.)
17. How many integers satisfy x2 3x < 4? (A) 4 (B) 5 (C) 6 (D) 7 18. The least integer satisfying |x 2| |x + 4| is (A) 0 (B) 1 (C) 2 (D) 1 19. How many integers do not satisfy the
inequation 13x
< 12
?
(A) 4 (B) 5 (C) 3 (D) 1 20. How many values of x satisfy x2 3x + 5 > 0 ? (A) One (B) Two (C) more than two but less than five (D) infinitely many 21. The number of integer values of x that
satisfy the inequation x2 5|x| + 6 < 0 is (A) 0 (B) 4 (C) 1 (D) infinitely many 22. The solution set of the inequation
|x2 2x 8| > 2x is (, a) (b, ). Then (A) a, b both are rational. (B) a, b both are irrational. (C) a is rational and b is irrational. (D) a is irrational and b is rational. 23. The solution set of the inequation
| |x 1| x | 4 is
(A) 3 5,2 2
(B) 3,
2
(C) 3 , 12
(D) 3 ,
2
24. How many integer values of x satisfy the
inequation 2
35 6
xx x
2 ?
(A) 3 (B) 2 (C) 1 (D) 0 25. The solution set of the inequation
2
2
11
x
x x< 1 is (a ,b) (c, ),
then ab + bc + ca is (A) 0 (B) 1
(C) 52
(D) 2 26. The number of integers for which
x4 3x3 x + 3 < 0 is (A) 3 (B) 2 (C) 1 (D) 0
27. The solution set of the inequation (x 3) < 2 4 5x x is
(A) (, 5] (B) [5, 1] (C) (, 5] [1, 3) (D) (, 5] [1, ) 28. The greatest integer x such that
2 1 3 13 2
x x > 1 is
(A) 3 (B) 2 (C) 1 (D) 1 29. The least positive integer satisfying
|x + 1| + |x + 4| > 7 is (A) 5 (B) 4 (C) 3 (D) 2 30. If the solution set of the inequation
2 3 2x x < 1 + 2 1x x is (, a][b, c), then a + b + c is
(A) not defined. (B) an irrational number. (C) an odd integer. (D) an even integer. 31. The sum of all integral values of x satisfying
the equation 2 2 8x x = x2 + 2x + 8 is (A) 7 (B) 5 (C) 0 (D) 6 32. The largest integral value of x satisfying the
inequation (x 1) (x + 2)2 (x 3) (x + 4)2 (x 5) < 0 is (A) 5 (B) 3 (C) 4 (D) 4 1. The least value of
|x 1| + |x 2| + |x 3| + |x 4|, x R is (A) 3 (B) 8 (C) 10 (D) 4 2. The equation x5 209x + 56 = 0 has two
roots whose product is unity. These two roots are
(A) 2+ 3 , 2 3 (B) 4+ 15 , 4 15 (C) 8 + 63 , 8 63 (D) 6 + 35 , 6 35 3. Given that a ≠ 0 R is a parameter. Find
the complete solution set of the inequation
x 2 1aa
2 ( 1)3
xa
Problems To Ponder
SAMPLE C
ONTENT
11
Chapter 1: Basic Algebra
1. (B) 2. (B) 3. (B) 4. (C) 5. (A) 6. (A) 7. (C) 8. (D) 9. (B) 10. (B) 11. (C) 12. (B) 13. (C) 14. (A) 15. (A) 16. (D) 17. (B) 18. (D) 19. (C) 20. (A) 21. (C) 22. (A) 23. (D) 24. (A) 25. (B) 26. (C) 27. (B) 28. (A) 29. (A)
1. (D) 2. (C) 3. (D) 4. (C) 5. (C) 6. (A) 7. (B) 8. (A) 9. (B) 10. (C) 11. (A) 12. (D) 13. (A) 14. (C) 15. (B) 16. (A) 17. (A) 18. (D) 19. (B) 20. (D) 21. (A) 22. (B) 23. (D) 24. (D) 25. (B) 26. (C) 27. (D) 28. (C) 29. (D) 30. (B) 31. (B) 32. (D)
1. (D) 2. (A) 3. For a < 0, x (, 2]
For 0 < a < 23
, x [2, )
For a > 23
, x (, 2]
For a = 23
, x R
1. (p2 + q2)3 = (p3 + q3)2
3(p2 + q2) p2q2 = 2p3q3
2 2p qpq = 2
3
p qq p = 2
3 2. a + b+ c = 0 a3 + b3 + c3 = 3abc
a = 19 1
27= 2
27, b = 1
27 1
81= 2
81,
c = 181
19
= 881
3abc = 10
323 = a3 + b3 + c3
3. a3 + b3 + c3 3abc = (a +b + c) (a2 + b2 + c2 ab bc ca) (a + b + c) (a2 + b2 + c2 ab bc ca) = 0 (a + b + c)
1
2
[(a – b)2 + (b – c)2 + (c – a)2] = 0 …(i)
a, b, c are distinct a ≠ b ≠ c (i) a + b + c = 0 4. x300 < 3500 x3 < 35 (=243) The greatest integer value of x is 6. 5. 6561 = 38 = 323 , (32)4 =
229 , (34)2 = 1281
Three possible triplets of (a,b,c) are (3,2,3), (9,2,2) and (81,2,1) Maximum value of abc = (81)(2) = 162
Answer Key
Concept Building Problems
Practice Problems
Problems To Ponder
Hints
Concept Building Problems
SAMPLE C
ONTENT
12
12
Challenger Maths Vol - I (Engg.)
6. 2x = 3y 3yx = 2
18z = 2z 32z 3y = 3zy
x
. 32z
2
3yz zx
= 3y yz + 2zx = xy
1x
+ 2y
= 1z
1x
+ 1y
+ 1z
= 1 2z y
+ 1
y+ 1
z
= 2 1z y
7. Let P(x) = x100 2x51 + 1 and Q(x) = x2 1 If S(x) is the quotient, then P(x) = Q(x) S(x) + Ax + B x100 2x51 + 1= (x2 1) S(x) + Ax + B P(1) = 0 0= (0) S(1) + A(1) + B A + B = 0 …(i) P(1) = 4 4 = (0) S(–1) + A (1) + B 4 = A + B …(ii) Solving (i) and (ii) , we get B = 2, A = 2 A + 2B = 2 8. Let a = 3 11 + 3 121 a + 3 11 + 3 121 = 0
a3 + 33 11 + 3
3 121 = 3a( 3 11 ) ( 3 121 )
…[ a + b + c = 0 a3 + b3 + c3 = 3abc] a3 11 121 = 3a(11) a3 33a 132 = 0 ‘a’ satisfies the cubic equation x3 33x 132 = 0 If the other two roots are b and c, then x3 33x 132 = (x a) (x b) (x c) abc = 132 9. x = 1
c satisfies the given equation.
Substituting x = 1c
in 11 x
and 11x
,
we get the other roots as 1
cc
and 1 – c. 10. x = 3 is the critical point where the graph
meets the x axis. 11. |2x 3| = 7 2x 3 = ± 7 2x = 10 or 4 x = 5 or 2 Sum of the values = 3
12. 2x 1 = ± 5 2x = 6, 4 x = 3,2 13. |10x 3| = 0 x = 3
10 = a
35 xa
= 20 |5x + 10| = 20
5x + 10 = ± 20 5x = 30, 10 x = 6, 2
14. |x| + 1
xx
= 2
1x
x |x| +
1xx
= 2
1x
x
Let a = x, b = 1
xx
.
Then, a + b = 2
1x x x
x
= 2
1x
x
Given equation is of the form | a | + | b | = | a + b | which shows that equality has occurred in the inequality | a | + | b | | a + b |
ab 0
2
1x
x 0 x = 0 or 1
1 x
> 0
x > 1 15. |x 1| + |5 2x| = |3x 6| |x 1| + |2x 5| = |3x 6|
Equality has occurred in the inequality |x 1| + |2x 5| | 3x 6|
(x 1) (2x 5) 0
x 1 or x 52
16. |5 2x| < 1 1 < 5 2x < 1 6 < 2x < 4 3 > x > 2 There is no integer in (2, 3). 17. 5
5
xx
> 0
( 5)5
xx
> 0 if x 5< 0 i.e, if 0 < x < 5
–1 > 0 (absurd)
55
xx
> 0
( 5)5
xx
> 0 if x 5 > 0 i.e., if x > 5
1 > 0
55
xx
> 0 is true for all x > 5.
Now, consider all the options, 3x + 10 < 25 3x < 15 x < 5 (x 5) (x + 3)2 < 0
SAMPLE C
ONTENT
13
Chapter 1: Basic Algebra
x 5 < 0 as (x + 3)2 > 0 for all x except x = 3
x < 5 |2x 5| < 5 5 < 2x 5 < 5 0 < 2x < 10 x < 5 x2 4x 5 > 0 (x 5) > 0 x > 5 18. 53
2x 2 53
2x 2 or 53
2x 2
3x 12
or 3x 94
x 16
or x 34
Solution set = 1 3, ,6 4
Only x = 110
lies in the solution set. 19. 1
2
xx
< 1
| x 1| > x + 2, if x + 2 < 0 …(i) and |x 1| < x + 2, if x + 2 > 0 …(ii) (i) is a true statement as | x 1 | is non
negative and x + 2 is negative (i.e, x + 2 < 0) x < 2 is a solution …(iii) (ii) (x + 2) < x 1 < (x + 2), if x + 2 > 0
x 2 < x 1 and x 1 < x + 2, if x > 2
2x > 1 and 1 < 2 x > 1
2 …(iv)
(iii) and (iv) Solution set
= 1, 2 ,2
20. |2x 4| < x 1 L.H.S. is non negative R.H.S. must be positive x > 1 (feasible region) …(i) Given inequation 1 x < 2x 4 < x 1 1 x < 2x 4 and 2x 4 < x 1 5 < 3x and x < 3
x > 53
and x < 3 …(ii)
(i) and (ii) solution set = 5 , 33
The only integer in solution set = 2 The required sum = 2 21. 2|x + 1| > x + 6 …(i) If x + 6 < 0 (i.e., x < 6) then (i) is a true
statement. x < 6 is a solution …(ii)
Let x 6 then (ii) 2(x +1) < x 6 or 2(x + 1) > x + 6 3x < 8 or x > 4
x 86, 4,3
…(iii)
(ii) and (iii) solution set
= 83
,
(4, )
a = 5, b = –3 a – b = 8 22. The critical points are 3, 1, 1, 2. x = 1, x = 3 are solutions. Method of intervals gives the solution set
(, 3] (1, 1] (2, ) The integers, which are not solutions, are
2,1,2. The required sum = 2+(1) + (2) = –1 23. 2 5
1x
x
2 > 0 71x
x
> 0
1
xx
< 0
The critical points are 1, 0. The method of intervals gives solution set
(1,0).
24. 1 12x x
2
2x
2 2( 2) 2
x xx x x
0 22( 2) 2 4
( 2) ( 2)x x xx x x
0
22 6 4
( 2)( 2)x x
x x x
0
2 3 2( 2) ( 2)x x
x x x
0
2 3 17The zeroes of 3 2 are 2
x x
The solution set is
3 17 3 172, (0, 2) ,2 2
a + b + c + d + f = 3 32 2 = 3
25. 3|x 1| + x2 7 > 0 …(i) Case I: x < 1 (i) x2 3x 4 > 0 (x 4) (x + 1) > 0 x < 1 or x > 4 x (, 1) is a solution …(ii)
(x + 1)
> 0 for x > 0
(true)
1 1
2
3
1
0
2 3 172
0 2 3 17
2
SAMPLE C
ONTENT
14
14
Challenger Maths Vol - I (Engg.)
Case II: x 1 (i) x2 + 3x 10 > 0 (x + 5) (x 2) > 0 x < 5 or x > 2 x (2,) is a solution …(iii) (ii) and (iii) solution set = () (2, ) 26. 3 10x > 6 x First we find feasible region: 3x 10 > 0 and 6 x 0
103
< x 6 …(feasible region)
Squaring given inequation, 3x 10 > 6 x 4x > 16 x > 4 The solution set = (4,6] 27. (x 1) 2 2x x 0 x = –1 and 1 make two sides of inequation
equal, hence they are solutions. Feasible region: x2 x 2 0 (x 2) (x + 1) 0 x ( ] [2, ) Also given inequation x 1 0 x 1 Solution set = [2,) {1, –1} 28. 18x < 2 x To get feasible region, x + 18 0 and 2 x > 0 18 x < 2 …(Feasible region) Squaring given inequation x + 18 < x2 4x + 4 x2 5x 14 > 0 (x 7) (x +2) > 0 x < 2 or x > 7 The solution set = [18, 2) 29. 2 3 2x x > 2x 5 … (i) Feasible region: x2 3x + 2 0 (x 1) (x 2) 0 x 1 or x 2
If x < 52
then 2x 5 < 0 (i.e. negative)
L.H.S of (i) is nonnegative and
R.H.S. is negative for x < 52
(i) is a true statement for x < 52
x (, 1] 52 ,2
is a solution …(ii)
Let x 52
, squaring (i), we get
x2 3x +2 > 4x2 20x + 25
3x2 17x + 23 < 0
3217
6x
> 23 + 289
12 3
2176
x
< 1312
136
< x 176
< 136
17 136 < x < 17 13
6
x 5 17 13,2 6
is a solution … (iii)
(ii) and (iii) give solution set
= (, 1) 17 132,6
a = 1, b = 2, c = 17, d = 13, f = 6
a + b + +c df
= 1 + 2 + 17 136 = 8
1. 3 y –1 1 y 3 … (i) Also 4 x 6 …(ii) From (i) and (ii), we get 5 x y 9
2. 32 3
5 5
a b 32 3
5 5
a b
= 2 3 2 35 5 5 5a b a b
22 35 5a b
22 3
5 5a b
+ 2 3 2 3
5 5 5 5a b a b
= 2 2 2 26 4 9 4 92
5 25 25 25 25b a b a b
= 2 26 12 9
5 25 25b a b
= 2 218 4 3125
b a b
3. (a + b + c) [(a b)2 + (b c)2 + (c a)2] = (a + b + c) [2(a2 + b2 + c2 ab bc ca)] = 2(a3 + b3 + c3 3abc) 4. p3 + (q p)3 q3 = [p +(q p)]3 3p (q p) (p + q p) q3 = q3 3pq (q p) q3 = 3pq(q p)
5. 2
1 pp
= p2 + 2
1p
2
= 83 2 = 81
p 1p
= ± 9
Practice Problems
SAMPLE C
ONTENT
15
Chapter 1: Basic Algebra
6. 44
1xx
= 2
22
1xx
2
= 221 2x
x
2
= [(6)2 + 2]2 2 = 1444 2 = 1442 7. 1
mm
= 2
Equality in 1m
m 2
m = 1m
= 1
m4 + 41
m = 1 + 1 = 2
8. x = 4 83 2
2 3
= 4 82 2
3 3
= 122
3
x2 = 2122
3
= 242
3
9. Let P(x) = x3 5x2 + x + 1 P(2) = Remainder = 8 20 + 2 + 1 = 9 10. Let x be the number. Then
(12)1 = 12
3
(x)
112
= 32x x = 1
18 11.
2 51
xx
+ 1 0 2 6
1x x
x
0
( 3)( 2)1
x xx
0 x = 3, 1, 2 are critical points out of which
x = 3, 2 belong to the solution set. Solution set = [3, 1) [ 2, ) 12. Let x 3 < 0 i.e. x < 3 Then given inequation 1 x 4 x 5 x < 3 is a solution …(i) Also x = 3 is a solution …(ii) Let x > 3. Then given inequation 1 x 4 x 5 x 5 is a solution …(iii) (i), (ii) and (iii) Solution set = (, 3] [5, )
13. (x 4) (x 10) 0 x 4 or x 10 …(i)
(x 9) (2x + 1) 0 12 x 9 …(ii)
From (i) and (ii), we get x 1 , 4
2
14. (x 2)3 (x 4)2 (x 10)17 0 … (i) (i) x = 2, 4, 10 are solutions … (ii) Solving (i) Solving (x 2) (x 10) 0. (as even powers of x 2 , x – 4 and x 10
do not contribute to the solution set) x 2 or x 10 … (iii) (ii) and (iii) Solution set = (, 2] [10, ) {4} 15. 2
(2 1)( 4)x xx
0
(2x + 1) (x + 4) 0, x ≠ 0
x 4 or x 12
and x ≠ 0
Solution set = (, 4] 1 ,2
{0}
16. 4 2
1 1x x
1 < 0
22 6 1
( 1)( 1)x xx x
< 0 2 2 5
( 1)( 1)x xx x
> 0 …(i)
Observe that x2 2x + 5 = (x 1)2 + 4 > 0
(i) 1( 1)( 1)x x
> 0 x < 1 or x > 1
Solution set = (, 1) (1,) 17. x2 3x 4 < 0 (x – 4) (x + 1) < 0 1 < x < 4 18. |x 2| |x + 4| Squaring x2 4x + 4 x2 + 8x + 16 12x 12 x 1 19. 1
3x < 1
2 … (i)
Case I: x < 0
(i) 13x
< 12 1
2 + 1
3x > 0
42( 3)
xx
> 0 x < 4 or x > 3
x(,4) (3,0) is a solution …(ii) Case II: x 0
(i) 13x 1
2< 0 5
2( 3)x
x
< 0
3 1 2
4 9 10 12
SAMPLE C
ONTENT
16
16
Challenger Maths Vol - I (Engg.)
52( 3)
xx
> 0
x < 3 or x > 5 x [0, 3) (5, ) is a solution … (iii) (ii) and (iii) give Solution set = (, 4) (3, 3) (5, ) Integers which do not satisfy the inequation
are 4, 3, 3, 4, 5. 20. x2 3x + 5 > 0
23
2x
+ 5 9
4 > 0
23
2x
+ 11
9 > 0 (a true statement)
x R is a solution. 21. x2 5|x| + 6 < 0 |x|2 5|x| + 6 < 0 (|x| 2) (|x| 3) < 0 2 < |x| < 3 |x| > 2 and |x| < 3 (x < 2 or x > 2) and (3 < x < 3) 3 < x < 2 or 2 < x < 3 There is no integer in (–3, –2) (2, 3) 22. |x2 2x 8| > 2x … (i) All x 0 are solutions … (ii) (i) |(x 4) (x + 2)| > 2x Let 0 < x < 4 (i) (4 –x) (x + 2) 2x x2 8 –2 2 x 2 2 0 x 2 2 is a solution … (iii) Let x ≥ 4 (i) (4 –x) (x + 2) > 2x x2 4x 8 > 0 (x 2)2 > 12 x 2 < 2 3 or x 2 > 2 3 x < 2 2 3 or x > 2 + 2 3 x > 2 + 2 3 is a solution …(iv) (ii), (iii) and (iv) give solution set = (, 2 2 ) (2 + 2 3 ,) 23. | |x 1| x | 4 …(i) Case I: x < 1 (i) |1 x x| 4 |2x 1| 4 4 2x 1 4
32
x 52
32
x < 1 is a solution …(ii)
Case II: x 1 (i) |x 1 x| 4 1 4 (true) x 1 is a solution …(iii)
(ii) and (iii) Solution set = 3 ,2
24.
2
35 6
xx x
2 0 …(i)
Case I: x 3 < 0 x < 3
(i) reduces to ( 3)( 3)( 2)
xx x
2 0
12x
+ 2 0 2 32
xx
0
32 x < 2
32 x < 2 is a solution
Case II: x 3 > 0 x > 3
(i) reduces to 12x 2 0
2 52
xx
0 2 5
2x
x
0
2 < x 52
(rejected)
The solution set is 3 , 22
There is no integer in 3 , 22
25.
2
21
1
x
x x< 1 …(i)
Observe that x2 + x + 1 = 21 3
2 4
x 34
x2 +x + 1 is positive for all x (i) | x2 – 1| < x2 + x + 1 x2 x 1 < x2 1 < x2 + x + 1 2x2 + x > 0 and x > 2
1 or 02
x x and x > 2
Solution set = 12,2
(0, )
a = 2, b = 12
, c = 0 ab + bc + ca = 1 26. Given inequation x3 (x 3) (x 3) < 0 (x 3) (x3 1) < 0 (x 3) (x 1) (x2 + x + 1) < 0 1 < x < 3 as x2 + x + 1 is positive for x. The only integer in the solution set is 2.
SAMPLE C
ONTENT
17
Chapter 1: Basic Algebra
27. (x 3) < 2 4 5x x … (i) Feasible region: x2 + 4x 5 0 (x + 5) (x 1) 0 x 5 or x 1
…[Feasible region] If x < 3 then L.H.S of (i) < 0 and R.H.S 0 x < 3 makes (i) a meaningful statement. x 5 or 1 x < 3 is a solution … (ii) Let x 3. Then squaring (i) we get x2 6x + 9 < x2 + 4x 5 14 < 10x
x > 75
x ≥ 3 is a solution … (iii) (ii) and (iii) Solution set = (, 5] [1, ) 28. 2 1 3 1
3 2x x
> 1
4x + 2 9x + 3 > 6
5x < 1 x < 15
the required greatest integer is 1. 29. |x + 1| + |x + 4| > 7 …(i) Case I: x < 4 (i) 2x 5 > 7 2x > 12 x < 6 x < 6 is a solution. Case II: 4 x < 1 (i) x + 4 x 1 > 7 3 > 7 (absurd)
No solution in [4,1) Case III: x 1 (i) 2x + 5 > 7 x > 1 x > 1 is a solution Solution set = (,6) (1,) Required least positive integer = 2 30. 2 3 2x x < 1 + 2 1x x Feasible region:
x2 x + 1 0 21 3 3
2 4 4x
x2 + 3x + 2 0 (x + 1) (x + 2) 0 x 2 or x 1 Feasible region = (,2] [1,) Squaring given inequation,
x2 + 3x + 2 < 1 + x2 x + 1 + 2 2 1x x
4x < 2 2 1x x … (i)
x < 0 is a solution as x < 0 makes L.H.S. negative and R.H.S. is non negative.
x (,2] [–1, 0) is a solution … (ii) .
Let x 0. Squaring (i), 4x2 x2 – x + 1 3x2 + x – 1 0
321
6
x 1 + 112
21
6
x 1336
1 136
x 1 136
x 1 130,6
is a solution … (iii)
(ii) and (iii)
Solution set = (, 2) 1 131,6
31. 2 2 8x x = x2 + 2x + 8 = (x2 2x – 8)
…(i) | a | = –a a is negative (i) x2 2x – 8 < 0 (x 4) (x + 2) < 0 2 < x < 4 The integral values of x satisfying
2 < x < 4 are 1, 0, 1, 2, 3 Required sum = 1 + 0 +1 + 2 + 3 = 5 32. For (x + 2)2 (x + 4)2 > 0 except for x = 2, 4
…(i) So the inequation will be equivalent to
(x 1) (x 3) (x 5) < 0 x (, 1) (3, 5) …(ii) From (i) and (ii), x (, 4) (4, 2) ( 2, 1) (3, 5) Largest integral value must be 4. 1. Given expression = |x 1| + | x 2| + |3 x | + |4 x |
= (|x 1| + | 4 x |) +(|x 2| + |3 x|) |x 1 + 4 x| + |x 2 + 3 x |
= 3 + 1 = 4 Least value is 4 provided equality occurs in | x 1| + | 4 x | 3 and | x 2| + |3 x | 1 i.e. (x 1) (4 x) 0 and (x 2) (x 3) 0
Problems To Ponder
SAMPLE C
ONTENT
18
18
Challenger Maths Vol - I (Engg.)
(x 1) (x 4) 0 and (x 2) (x 3) 0 1 x 4 and 2 x 3 2 x 3 Aliter:
By drawing graph of y = |x 1| + |x 2| + |x 3| + |x 4|, we can find the least value. y = 10 4x, x < 1 = 8 2x, 1 x < 2 = 4, 2 x < 3 = 2x 2, 3 x < 4 = 4x 10, x > 4 For values of x between 2 and 3, y takes
value 4 which is minimum. 2. The required two roots are roots
x2 + ax + 1 = 0 where sum of the roots = a. Then x5 209x + 56
= (x2 + ax + 1)[x3 ax2 + (a2 1) x+ 2a a3] 2a a3 = 56 …(i) and a4 3a2 208 = 0 …(ii) (a2 16) (a2 + 13) = 0 a2 16 = 0 a = ± 4 a = 4 satisfies (i) The required roots are the roots of the
equation x2 4x + 1 = 0
The required roots are 2 3 , 2+ 3 3. x 2 1a
a
2 ( 1)3
xa
213a
x 2 1aa
+ 23a
213a
x 6 43a
a = 2 4
3a
213a
x 2 213a
… (i)
If 1 23a
< 0 then (i) x 2
i.e., If 3 23a
a < 0 then x 2
For 0 < a < 23
, x [2, ) … (ii)
If 1 23a
> 0 then (i) x 2
i.e., If 3 23a
a > 0 then x 2
For a < 0 or a > 23
, x (, 2] … (iii)
(ii) and (iii) give solution set as For a < 0, x (, 2]
For 0 < a < 23
, x [2, )
For a > 23
, x (, 2]
For a = 23
, x R
Remark: Given inequation is an inequation
with a parameter. For every value of a
(except a = 0), there is an inequation. So
solving given inequation is equivalent to
solving a class of inequations resulting from
every value of a. One must remember that
an equation with a parameter must be
considered for all values of the parameter.
Even if one value of the parameter is
neglected, the solution set is not complete.
y
1 O
2
10
4
6 8
2 x
3 4
