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Std. XI Sci.

Perfect Mathematics - II

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storage and retrieval system without permission in writing from the Publisher.

Edition: July 2014

Mr. Vinodkumar J. Pandey B.Sc. (Mathematics)

G. N. Khalsa College, Mumbai

Dr. Sidheshwar S. Bellale M.Sc., B.Ed., PhD. (Maths) Department of Mathematics,

Dayanand Science College, Latur.

Mr. Vinod Singh M.Sc. (Mathematics) Mumbai University

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Std. XI Sci.

Perfect Mathematics - II

Written according to the New Text book (2012-2013) published by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune.

TEID : 752

Salient Features :

• Exhaustive coverage of entire syllabus.

• Covers answers to all textual and miscellaneous exercises.

• Precise theory for every topic.

• Neat, labelled and authentic diagrams.

• Written in systematic manner.

• Self evaluative in nature.

• Practice problems and multiple choice questions for effective

preparation.

PREFACE In the case of good books, the point is not how many of them you can get through, but rather how many can get through to you.

“Std. XI Sci. : PERFECT MATHEMATICS - II” is a complete and thorough guide critically analysed and extensively drafted to boost the students confidence. The book is prepared as per the Maharashtra State board syllabus and provides answers to all textual questions. At the beginning of every chapter, topic – wise distribution of all textual questions including practice problems has been provided for simpler understanding of different types of questions. Neatly labelled diagrams have been provided wherever required.

Practice Problems and Multiple Choice Questions help the students to test their range of preparation and the amount of knowledge of each topic. Important theories and formulae are the highlights of this book. The steps are written in systematic manner for easy and effective understanding.

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]

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Yours faithfully, Publisher

No. Topic Name Page No.

1 Sets, Relations and Functions 1 2 Logarithms 39 3 Complex Numbers 67 4 Sequence and Series 119 5 Permutations and Combinations 172 6 Method of Induction and Binomial Theorem 213 7 Limits 250 8 Differentiation 302 9 Integration 329

10 Statistics (Measures of Dispersion) 345 11 Probability 389

Log Tables Logarithms 439 Antilogarithms 441

Target Publications Pvt. Ltd. Chapter 01: Sets, Relations and Functions

1 1

01 Sets, relations and functions

Type of Problems Exercise Q. Nos.

To describe sets in Roster form 1.1 Q.1 (i., ii., iii.)

Practice Problems (Based on Exercise 1.1)

Q.1 (i., ii., iii.)

To describe sets in Set-Builder form

1.1 Q.2 (i., ii., iii.), Q.12(i. to iv.)


Q.2 (i., ii., iii.) Q.11 (i. to iv.)

Miscellaneous Q.1 (i., ii., iii.)

Practice Problems (Based on Miscellaneous)

Q.1(i., ii.)

Operations on Sets

1.1 Q.3 to Q.11 Q.13 (i. to iv.)


Q.3 to Q.10 Q.12 (i., ii.)

Miscellaneous Q.2, 3, 4


Q.2, 3, 4

Ordered Pairs

1.2 Q.1, 2, 6, 11


Q.1, 2, 5, 9

Cartesian product of two Sets

1.2 Q.3, 4, 5


Q.3, 4

Miscellaneous Q.5 (i., ii.)


Q.5

To find domain and range of a given relation

1.2 Q.7, 8, 9, 10


Q.6, 7, 8

Miscellaneous Q.6, 7


Q.6, 7

Target Publications Pvt. Ltd. Std. XI Sci.: Perfect Maths - II

2

Types of Functions

1.3 Q.9, 10


Q.9, 10



Q.9, 10, 21

To find values of the given function

1.3 Q.3, 5, 6, 7


Q.3, 4, 6, 7

Miscellaneous Q.13 to Q.17


Q.12 to Q.16

Operations on functions

1.3 Q.11 (i. to iii.)


Q.11 (i. to iii.)

Composite function

1.3 Q.12 to Q.15


Q.12 to Q.15

Miscellaneous Q.18 to Q.21


Q.17 to Q.20

Inverse function

1.3 Q.16


Q.16

Miscellaneous Q.11, 12


Q.11

To find domain and range of a given function

1.3 Q.1, 2, 4, 8


Q.1, 2, 5, 8



Q.8


3 3

Syllabus: Sets, Subset, Intervals, Types of sets, Power set, Ordered pair, Cartesian products of two sets, Definition of Relation, Domain, Co-domain and Range of Relation, Types of Relation, Definition of function, Types of functions, Operations on functions, Composite function, Inverse function, Binary operation, Real valued functions of the real variable. Introduction All basic concepts of modern mathematics are based on set theory. The concepts involving logic can be explained more easily with the help of set theory. It plays a crucial role in the study of relations, functions, probability and is used extensively in various other branches of mathematics. We shall briefly revise and study some more concepts about sets. Sets A set is a well-defined collection of objects. These objects may be actually listed or may be specified by a rule. A set is usually denoted by the capital letters A, B, C, N, R, etc. Each object in a set is called an element or a member of the set and is denoted by the small letters a, b, c, etc. If x is an element of set A, then we write it as x ∈ A and read it as ‘x belongs to A’ and if y is not an element of set A, then we write it as y ∉ A and read it as ‘y does not belong to A’. Example: If A = {2, 4, 6, 8}, then 4 ∈ A, 7 ∉ A, 8 ∈ A, 10 ∉ A The set of natural numbers, whole numbers, integers, rational numbers and real numbers are denoted by N, W, I, Q and R respectively. Methods of Representation of Sets There are two methods of representing a set which are as follows: i. Roster method (Listing method): In this method all the elements are listed or

tabulated. The elements are separated by commas and are enclosed within two braces(curly brackets).

Example: The set A of all positive even integers less than 9 can be written as A = {2, 4, 6, 8}.

ii. Set-Builder method: In this method, the set is described by the

characteristic property of its elements. In general, if all the elements of set A satisfy

some property P, then write A in set-builder notation as A = {x/x has property P} and read it as ‘A is the set of all x such that x has the property P’.

Example: Let B = {3, 4, 5, 6, 7, 8} Using the set-builder method, B can be written

as B = {x/x ∈ N, 3 ≤ x ≤ 8} Since B = {3, 4, 5, 6, 7, 8} can also be stated

as the set of natural numbers from 3 to 8 including 3 and 8.

Some standard sets are as follows: N = set of all natural numbers = {1, 2, 3,…..} Z or I = set of all integers = { ….−3, −2, −1, 0, 1, 2, 3 …} Q = set of all rational numbers

= p / p,q Z,q 0q

⎧ ⎫∈ ≠⎨ ⎬

⎩ ⎭

Subset: Set A is called a subset of set B, if every element of set A is also an element of set B i.e., if x ∈ A, then x ∈ B. We denote this relation as A ⊆ B and read it as ‘A is a subset for B’. It s clear that i. Every set is a subset of itself i.e., A ⊆ A. ii. An empty set φ is a subset of every set. Example: If A = {2, 4, 6, 8} and B = {2, 4, 6, 8, 10, 12}, then A ⊆ B. If A ⊆ B, then B is called a superset of A, denoted by B ⊇ A. Proper subset: If every element of set A is an element of set B and B contains at least one element which is not in A, then A is said to be a proper subset of B and it is denoted as A ⊂ B.


4

Example: If A = {2, 3, 5, 6}, B = {1, 2, 3, 4, 5, 6, 7} Here every element of set A i.e., 2, 3, 5, 6 is an element of set B. But B contains elements 1, 4, 7 which are not in A. Hence in the case we say that A is a proper subset of B and is denoted by A ⊂ B. Intervals Open interval If p, q ∈ R and p < q, then the set {x/x∈R, p < x < q} is called open interval and is denoted by (p, q). Here all the numbers between p and q ∈ (p, q) except p and q. ∴ (p, q) = {x/x∈R, p < x < q} Closed interval If p, q ∈ R and p < q, then the set {x/x∈R, p ≤ x ≤ q} is called closed interval and is denoted by [p,q]. Here all the numbers between p and q ∈ [p, q] including p, q. ∴ [p, q] = {x/x∈R, p ≤ x ≤ q} Semi-closed interval

If p, q ∈ R and p < q, then the set {x/x∈R, p ≤ x < q} is called semi-closed interval and is denoted by [p, q). ∴ [p, q) = {x/x∈R, p ≤ x < q} [p, q) includes p but excludes q. Semi-open interval

If p, q ∈ R and p < q, then the set {x/x∈R, p < x ≤ q} is called semi-open interval and is denoted by (p, q]. ∴ (p, q] = {x/x ∈ R, p < x ≤ q} (p, q] excludes p but includes q. Remarks: i. Set of all real numbers > p i.e., (p, ∞) = [x/x∈R, x > p}

Set of all real numbers ≥ p i.e., [p, ∞) = {x/x∈R, x ≥ p} ii. Set of real numbers < q i.e., (−∞, q) {x/x∈R, x < q} Set of real numbers ≤ q i.e., (−∞, q] = {x/x∈R, x ≤ q} iii. Set of all real numbers R is (−∞, ∞) R = (−∞, ∞) = {x/x ∈ R, − ∞ < x < ∞} Types of sets Empty set: A set which does not contain any element is called an empty set and it is denoted by φ or { }. It is also called null set or void set. Example: A = {x/x ∈ N, 3 < x < 4} B = {x/x is a positive integer < 1} Note: The set {0} and {φ} are not empty sets as they contain one element, namely 0 and φ respectively. Singleton set: A set which contains only one element is called a singleton set. Example: A = {5}, B = {3}, X = {x/x ∈ N, 1 < x < 3} The set A = set of all integers which are neither positive nor negative is a singleton set since A = {0} Finite set: A set which contains countable number of elements is called a finite set. Example: A = {a, b, c} B = {1, 2, 3, 4, 5} C = {a, e, i, o, u}

p q x R

Rp q x

q R

x p

x p q R

R

pR

q R

q R

R p


5 5

Infinite set: A set which contains uncountable number of elements is called an infinite set. Example: N = {1, 2, 3, 4……} Z = {…−3, −2, −1, 0, 1, 2, 3,…..} Note: i. An empty set is a finite set. ii. N, W, I, Q and R are infinite sets. Equal sets: Two sets A and B are said to be equal if they have the same elements and we denote this as A = B. From this definition it follows that “two sets A and B are equal if and only if A ⊆ B and B ⊆ A” Example: If A = {1, 2, 3, 4}, B = {2, 4, 1, 3}, then A = B. Equivalent sets: Two sets A and B are said to be equivalent, if they contain the same number of elements and we denote it as A ≈ B. Example: If A = {1, 2, 3, 4, 5}, B = {a, b, c, d, e}, then n(A) = n(B) ∴ A and B are equivalent sets. Note: Equal sets are always equivalent but equivalent sets need not be equal. Universal set: A non-empty set of which all the sets under consideration are subsets, is called a universal set. It is usually denoted by X or U. Example: If A = {1, 2, 3, 4}, B = {2, 8, 13, 15} and C = {1, 2, 3, …, 50} are sets under consideration, then the set N of all natural numbers can be taken as the universal set. Venn diagram: A set is represented by any closed figure such as circle, rectangle, triangle, etc. The diagrams representing sets are called venn diagrams. Example: i. A = {4, 6, 9} ii. A = {a, b, c, d, e, f} B = {b, e, f} B ⊂ A

Operations on sets Complement of a set: Let A be a subset of a universal set X then the set of all those elements of X which do not belong to A is called the complement of set A and it is denoted by A′ or Ac. Thus, A′ = {x/x ∈ X, x ∉ A} The shaded region in the above figure represents A′. Example: Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9} be an universal set and A = {1, 3, 5, 6, 8}. Then A′ = {2, 4, 7, 9} Properties: If X is the universal set and A, B ⊆ X, then i. (A′)′ = A ii. X′ = φ iii. φ′ = X iv. A ∩ A′ = φ v. A ∪ A′ = X vi. If A ⊆ B, then B′ ⊆ A′. Union of sets: If A and B are two sets, then the set of those elements which belong to A or to B or to both A and B is called the union of the sets A and B and is denoted by A ∪ B. i.e., A ∪ B = {x/x ∈ A or x ∈ B} The shaded portion in the below venn diagram represents A ∪ B. Example: i. If A = {1, 2, 3, 4}, B = {2, 4, 6, 8}, then A ∪ B = {1, 2, 3, 4, 6, 8} ii. If A is the set of all odd integers and B is the

set of all even integers, then A ∪ B is the set of all integers.

A.4.6 .9

.b.e .f

.aB

.d.c

A

A′

X

A

A BX

A ∪ B


6

Properties: If A, B, C are any three sets, then i. A ∪ φ = A ii. A ∪ X = X iii. A ∪ B = B ∪ A (Commutative law) iv. (A ∪ B) ∪ C = A ∪ (B ∪ C) (Associative law) v. A ∪ A = A (Idempotent law) vi. If A ⊂ B, then A ∪ B = B vii. A ⊂ A ∪ B, B ⊂ A ∪ B Intersection of sets If A and B are two sets, then the set of those elements which belong to both A and B i.e., which are common to both A and B is called the intersection of the sets A and B and is denoted by A ∩ B. Thus, A ∩ B = { x/x ∈ A and x ∈ B} The shaded portion in the below venn diagram represents A ∩ B. Example: If A = {1, 2, 3, 4, 5}, B = {1, 3, 5, 7, 9}, then A ∩ B = {1, 3, 5} Properties: If A, B, C are any three sets, then i. A ∩ φ = φ ii. A ∩ X = A iii. A ∩ B = B ∩ A (Commutative law) iv. (A ∩ B) ∩ C = A ∩ (B ∩ C) (Associative law) v. A ∩ A = A (Idempotent law) vi. If A ⊂ B, then A ∩ B = A vii. A ∩ B ⊂ A, A ∩ B ⊂ B Disjoint sets: Two sets A and B are said to be disjoint, if they have no element in common i.e., A ∩ B = φ. Example: If A = {2, 4, 6} and B = {3, 5, 7}, then A ∩ B = φ ∴ A and B are disjoint sets. The venn diagram of the disjoint sets A and B is shown below:

Difference of sets: If A and B are two sets then the set of all the elements of A which are not in B is called difference of sets A and B and is denoted by A − B. Thus, A − B = {x/x ∈ A and x ∉ B} Similarly, B − A = {x/x ∉ A and x ∈ B} In the below venn diagrams shaded region represents A − B and B − A . Example: If A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 6, 8}, then A − B = {1, 3, 5} and B − A = {8} Distributive Properties of union and intersection If a, b, c ∈ R, then a × (b + c) = (a × b) + (a × c) This is known as distributive property of multiplication over addition. In set theory, the operation of union and intersection of sets are both distributive over each other i.e., If A, B, C are any three sets, then i. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) ii. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) We verify these distributive laws using Venn diagrams shown below. The shaded portion in each figure shows the set obtained by performing the operation given below the figure. i. = ii. =

A ∩ B

A B

X

A − B

B

X

A

B − A

A

X

B

A ∩ (B ∪ C) (A ∩ B) ∪ (A ∩ C)

B

A

C

X

B

A

C

X

(A ∪ B) ∩ (A ∪ C)

B

A

C

A ∪ (B ∩ C)

X

B

A

C

X

A ∩ B = φ

A B

X


7 7

De-Morgan’s laws If A and B are two subsets of a universal set X, then i. (A ∪ B)′ = A′ ∩ B′ ii. (A ∩ B)′ = A′ ∪ B′ We verify these laws using Venn diagrams shown below. The shaded portion in each figure shows the set obtained by performing the operation below the figure: i. ii. Number of elements in a set Let A be a set. Then the total number of elements in it is denoted by n(A). Example: Let A = {8, 9, 10, 11, 12} ∴ n (A) = 5 The number of elements in the empty set φ is zero. i.e., n (φ) = 0 Results: For given sets A, B, C i. n(A ∪ B) = n(A) + n(B) − n(A ∩ B) ii. When A and B are disjoint sets, then n(A ∪ B) = n(A) + n(B) iii. n(A ∩ B′) + n(A ∩ B) = n(A) iv. n(A′ ∩ B) + n(A ∩ B) = n(B) v. n(A ∩ B′) + n(A ∩ B) + n(A′ ∩ B) = n(A ∪ B) vi. n(A∪ B∪ C) = n(A) + n(B) + n(C) − n(A∩B)

− n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C) Power set The set of all subsets of set A is called the power set of A and it is denoted by P(A). Example: If A = {a, b, c}, then P(A) ={φ, {a}, {b}, {c}, {a, b}, {b, c}, {a, c}, {a, b, c}} Note: If A contains n elements, then the power set of A i.e., P(A) contains 2n elements.

Exercise 1.1 1. Describe the following sets in Roster form: i. {x/x is a letter of the word

‘MARRIAGE’}

ii. 1 9/ isanintegerand2 2

⎧ ⎫− < <⎨ ⎬⎩ ⎭

x x x

iii. {x/x = 2n, n ∈ N} Solution: i. Let A = {x/x is a letter of the word

‘MARRIAGE’} ∴ A = {M, A, R, I, G, E}

ii. Let B = 1 9/ isan integer and2 2

⎧ ⎫− < <⎨ ⎬⎩ ⎭

x x x

∴ B = {0, 1, 2, 3, 4} iii. Let C = {x/x = 2n, n ∈ N} ∴ C = {2, 4, 6, 8, ….} 2. Describe the following sets in Set-Builder

form: i. {0} ii. {0, ± 1, ± 2, ± 3}

iii. 1 2 3 4 5 6 7, , , , , ,2 5 10 17 26 37 50

⎧ ⎫⎨ ⎬⎩ ⎭

Solution: i. Let A = {0} 0 is a whole number but it is not a natural

number ∴ A = {x / x ∈ W, x∉ N} ii. Let B = {0, ± 1, ± 2, ± 3} B is the set of elements which belongs to Z

from −3 to 3 ∴ B = {x / x ∈ Z, −3 ≤ x ≤ 3} iii. Let C = 1 2 3 4 5 6 7, , , , , ,

2 5 10 17 26 37 50⎧ ⎫⎨ ⎬⎩ ⎭

∴ C = 2n ,n N,n 7

n 1⎧ ⎫= ∈ ≤⎨ ⎬

+⎩ ⎭x / x

3. If A = {x / 6x2 + x − 15 = 0} B = {x / 2x2 − 5x − 3 = 0} C = {x / 2x2 − x − 3 = 0}, then find i. A ∪ B ∪ C ii. A ∩ B ∩ C Solution: A = { }2/ 6 15 0+ − =x x x

∴ 6x2 + x − 15 = 0 ∴ 6x2 + 10x − 9x − 15 = 0

X

A′ ∩ B′

A BAX

(A ∪ B)′

A BA =

(A ∩ B)′

A B

X A B

X

A′ ∪ B′

=


8

∴ 2x(3x + 5) − 3(3x + 5) = 0 ∴ (3x + 5) (2x − 3) = 0

∴ x = 53

− or x = 32

∴ A = 5 3,3 2

−⎧ ⎫⎨ ⎬⎩ ⎭

B = {x/2x2 − 5x − 3 = 0} ∴ 2x2 − 5x − 3 = 0 ∴ 2x2 − 6x + x − 3 = 0 ∴ 2x(x − 3) + 1(x − 3) = 0 ∴ (x − 3)(2x + 1) = 0

∴ x = 3 or x = 12−

∴ B = 1, 32−⎧ ⎫

⎨ ⎬⎩ ⎭

C = {x/2x2 − x − 3 = 0} ∴ 2x2 − x − 3 = 0 ∴ 2x2 − 3x + 2x − 3 = 0 ∴ x(2x − 3) + 1(2x − 3) = 0 ∴ (2x − 3) (x + 1) = 0

∴ x = 32

or x = −1

∴ C = 31,2

⎧ ⎫−⎨ ⎬⎩ ⎭

Thus,

i. A ∪ B ∪ C = 5 3 1 3, ,3 1,3 2 2 2

−⎧ ⎫ ⎧ ⎫ ⎧ ⎫− ∪ ∪ −⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭ ⎩ ⎭

= 5 1 3, 1, , , 33 2 2

− −⎧ ⎫−⎨ ⎬⎩ ⎭

ii. A ∩ B ∩ C = { } 4. If A, B, C are the sets of the letters in the

words ‘college’, ‘marriage’ and ‘luggage’ respectively, then verify that

[A − (B ∪ C)] = [(A − B) ∩ (A − C)]. Solution: A = {c, o, l, g, e} B = {m, a, r, i, g, e,} C = {l, u, g, a, e} B ∪ C = {m, a, r, i, g, e, l, u} A − (B ∪ C) = {c, o} A − B = {c, o, l} A − C = {c, o} ∴ [(A − B) ∩ (A − C)] = {c, o} = A − (B ∪ C) ∴ [A − (B ∪ C)] = [(A − B) ∩ (A − C)]

5. If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8} and universal set X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then verify

the following: i. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) ii. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) iii. (A ∪ B)′ = A′ ∩ B′ iv. (A ∩ B)′ = A′ ∪ B′ v. A = (A ∩ B) ∪ (A ∩ B′) vi. B = (A ∩ B) ∪ (A′ ∩ B) vii. n(A ∪ B) = n(A) + n(B) − n(A ∩ B) Solution: A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8} X ={1, 2, 3, 4, 5, 6, 7, 8, 9, 10} i. (B ∩ C) = {4, 5, 6} ∴ A ∪ (B ∩ C)= {1, 2, 3, 4, 5, 6} (A ∪ B) = {1, 2, 3, 4, 5, 6} (A ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8} ∴ (A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6} ∴ A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) ii. (B ∪ C) = {3, 4, 5, 6, 7, 8} ∴ A ∩ (B ∪ C) = {3, 4} A ∩ B = {3, 4} A ∩ C = {4} ∴ (A ∩ B) ∪ (A ∩ C) = {3, 4} ∴ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) iii. A ∪ B = {1, 2, 3, 4, 5, 6} ∴ (A ∪ B)′ ={7, 8, 9, 10} A′ = {5, 6, 7, 8, 9, 10}, B′ = {1, 2, 7, 8, 9, 10} ∴ A′ ∩ B′ = {7, 8, 9, 10} ∴ (A ∪ B)′ = (A′ ∩ B′) iv. A ∩ B = {3, 4} (A ∩ B)′ = {1, 2, 5, 6, 7, 8, 9, 10} A′ = {5, 6, 7, 8, 9, 10} B′ = {1, 2, 7, 8, 9, 10} ∴ A′ ∪ B′ = {1, 2, 5, 6, 7, 8, 9, 10} ∴ (A ∩ B)′ = A′ ∪ B′ v. A ∩ B = {3, 4} A ∩ B′ = {1, 2} ∴ (A ∩ B) ∪ (A ∩ B′) = {1, 2, 3, 4} ∴ A = (A ∩ B) ∪ (A ∩ B′) vi. A ∩ B = {3, 4} A′ ∩ B = {5, 6} ∴ (A ∩ B) ∪ (A′ ∩ B) = {3, 4, 5, 6} ∴ B = (A ∩ B) ∪ (A′ ∩ B)


9 9

vii. A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, A ∩ B = {3, 4}, A ∪ B = {1, 2, 3, 4, 5, 6} ∴ n(A) = 4, n(B) = 4, n(A ∩ B) = 2, n(A ∪ B) = 6 ∴ n(A) + n(B) − n(A ∩ B) = 4 + 4 − 2 = 6 ∴ n(A ∪ B) = n(A) + n(B) − n(A ∩ B) 6. If A and B are subsets of the universal set X

and n(X) = 50, n(A) = 35, n(B) = 20, n(A′ ∩ B′) = 5, find i. n(A ∪ B) ii. n(A ∩ B) iii. n(A′ ∩ B) iv. n(A ∩ B′). Solution: n(X) = 50, n(A) = 35, n(B) = 20, n(A′ ∩ B′) = 5 i. n(A ∪ B) = n(X) − [n(A ∪ B)′] = n(X) − n(A′ ∩ B′) = 50 − 5 = 45 ii. n(A ∩ B) = n(A) + n(B) − n(A ∪ B) = 35 + 20 − 45 = 10 iii. n(A′ ∩ B) = n(B) − n(A ∩ B) = 20 − 10 = 10 iv. n(A ∩ B′) = n(A) − n(A ∩ B) = 35 − 10 = 25 7. In a class of 200 students who appeared

certain examinations, 35 students failed in MHT-CET, 40 in AIEEE and 40 in IIT entrance, 20 failed in MHT-CET and AIEEE, 17 in AIEEE and IIT entrance, 15 in MHT-CET and IIT entrance and 5 failed in all three examinations. Find how many students

i. did not fail in any examination. ii. failed in AIEEE or IIT entrance. Solution: Let A = set of students who failed in MHT-CET B = set of students who failed in AIEEE C = set of students who failed in IIT entrance X = set of all students ∴ n(X) = 200, n(A) = 35, n(B) = 40, n(C) = 40, n(A ∩ B)= 20, n(B ∩ C) = 17, n(A ∩ C) = 15, n(A ∩ B ∩ C) = 5 i. n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C)

− n(A ∩ C) + n(A ∩ B ∩ C) = 35 + 40 + 40 − 20 − 17 − 15 + 5 = 68 ∴ No. of students who did not fail in any exam = n(X) − n(A ∪ B ∪ C) = 200 − 68 = 132

ii. No. of students who failed in AIEEE or IIT entrance = n(B ∪ C) = n(B) + n(C) − n(B ∩ C) = 40 + 40 − 17 = 63 8. From amongst 2000 literate individuals of a

town, 70% read Marathi newspapers, 50% read English newspapers and 32.5% read both Marathi and English newspapers. Find the number of individuals who read

i. at least one of the newspapers. ii. neither Marathi nor English

newspaper. iii. only one of the newspapers. Solution: Let M = set of individuals who read Marathi

newspapers E = set of individuals who read English

newspapers X = set of all literate individuals

∴ n(X) = 2000, n(M) = 70100

× 2000 = 1400

n(E) = 50100

× 2000 = 1000

n(M ∩ E) = 32.5100

× 2000 = 650

n(M ∪ E) = n(M) + n(E) − n(M ∩ E) = 1400 + 1000 − 650 = 1750 i. No. of individuals who read at least one of the

newspapers = n(M ∪ E) = 1750. ii. No. of individuals who read neither Marathi

nor English newspaper = n(M′ ∩ E′) = n(M ∪ E)′ = n(X) − n(M ∪ E) = 2000 − 1750 = 250 iii. No. of individuals who read only one of the newspaper = n(M ∩ E′) + n(M′ ∩ E) = n(M ∪ E) – n(M ∩ E) = 1750 – 650 = 1100

Perfect Mathematics - II - Target Publications · Publications Pvt. Ltd. Target Chapter 01: Sets,...

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