n(B)

A ⊂ B ⇒ A ∪ B = B

⇒ n(A ∪ B) = n(B)

Answer is A ∪ Β′

Answer is A ∩ B′

(C) The Answer is φ

the Answer is Q

The set of all the subsets of a given set Q is called the power set of Q and is denoted by P(Q) 

Ex. If S = {1} , then P (S) = { φ , s} 

If T = {a, b}, then P (T) = { φ , {a}, {b}, T}

A = { \(x\) : \(x\) = 4n, n∈ N and n ≤ 3} 

= { 4, 8, 12 } 

B = { \(x\) : \(x\) = 2n, n∈ N and n ≤ 6} 

= { 2, 4, 6, 8, 10, 12} 

Here A ⊂ B 

⇒ A and B are comparable.

Two sets A and B are said to be comparable, if one of them is a subset of the other, i.e., A and B are comparable if either A ⊆ B or B ⊆ A. 

Ex. A = {set of vowels}, B = {letters of english alphabet} are comparable as A ⊆ B

The superset of all the sets for a particular discussion is called the universal set. It is denoted by ∪ or ξ. 

Ex. The set of integers is the universal set for the set of positive numbers and negative numbers, also for prime numbers and composite numbers.

The set of elements of universal set, which are not in a given set (say P) is the complement of P, denoted by P′. 

P′ = {x ∈ ξ ; x ∉ P) 

Ex. ξ = {set of natural numbers}, P = { set of even numbers}. 

Then P′ = {set of odd numbers}

Here correct answer is D. none of this as A⋂(A⋂B) = (A⋂A)⋂B) 

= A⋂B 

Or we have to give one condition that A⊂B then correct answer is A because A⋂(A⋂B) = (A⋂A)⋂B) 

= A⋂B 

= A (∵ A⊂B⇒A⋂B = A)

n(A⋃B) = n(A) + n(B) - n(A⋂B) 

∴ n(A⋂B) = n(A) + n(B) - n(A⋃B) 

∴ n(A⋂B) = 20 + 25 - 40 = 5 

∴ n(A⋂B) = 5

Here, 

n(A) = 4 and n(B) = 7 

Now, n(A⋃B) = n(A)+n(B) - n(A⋂B) 

= 4 + 7 - n(A⋂B) 

= 11 - n(A⋂B) 

So, 

n(A⋃B) is maximum whenever n(A⋂B) is minimum and it is possible only when A⋂B = ϕ 

Now, 

A⋂B = ϕ then min(n(A⋂B)) = 0. 

∴ min(n(A⋃B)) =11 - 0 =11

Here,

A = {\((x,y):y = \frac{1}x,0≠ x∊R\)}and 

B = {(x,y):y = - x, xϵR 

A⋂B = {\((x,y):y = \frac{1}x,0≠ x∊R\)} and {(x,y):y = - x,xϵR}

= {\((x,y):-x = \frac{1}x, x∊R\)}

= {(x, y):-x² =1,x ϵ R} 

= {(x, y):x² =-1,x ϵ R} 

= ϕ (∵ there is no such that x² = - 1)

Overlapping sets: Two sets are called overlapping if they have at least one element in common. 

Ex. The sets {2, 3, 4} and {3, 6, 9} are overlapping as the element 3 is common to both of them.

But,

Disjoint sets: If two sets A and B have no elements in common, they are called disjoint sets. 

Ex. A ={ set of vowels} B = { set of consonants} are disjoint as they have no letter in common.

(i) Disjoint sets; as no girl can be of age below 15 years and also above 15 years

(ii) Overlapping sets; as boys above 27 years are also above 20 years.

(iii) Overlapping sets; as natural numbers from 50 to 59 are common to both the sets.

(iv) Overlapping sets; as students of class IX studying in I.C.S.E. board are common.

(v) Overlapping sets; as natural number 24 is common to both the sets.

(vi) Disjoint sets; as no letter is common to both the sets.

(i) A = {0, 1, 2, 4} i.e. n (A) = 4

(ii) B = {-3, -1, 1, 3, 5, 7} i.e. n(B) = 6

(iii) C = { } i.e. n(C) = 0

(iv) D = {3, 2, 2, 1,3, 1, 2} => D = {3, 2, 1,} i.e. n(D) = 3

(v) E = {16, 17, 18 19} i.e. n(E) = 4

(vi) F = {8, 9, 10, 11, 12, 13, 14} i.e. n(F) = 7

(i) [-4, 0] = {x : x ∈ R, -4 < x < 0} 

(ii) [6, 8] = {x : x ∈ R, 6 ≤ x ≤ 8} 

(iii) [- 3, 7] = {x : x ∈ R, -3 ≤ x < 7} 

(iv) [3, 10] = {x : x ∈ R, 3 < x ≤ 10}

(i) {x : x ∈ R, -3 < x < 6} 

-3 < x < 6 is open interval where -3 and 6 are not included. 

Hence, {x : x ∈ R, -3 < x < 6} has interval (-3, 6). 

(ii) {x : x ∈ R, -4 ≤ x ≤ 8} 

-4 ≤ x ≤ 8 is closed interval in which (-4 and 8) are included. 

Hence, interval is [-4, 8] 

(iii) {x : x ∈ R, 4 < x ≤ 9} 

4 < x ≤ 9 is semi closed interval where the interval is less than 4 and equal to 9. 

Hence, {x : x ∈ R, 4 < x ≤ 9} has interval [4, 9]. 

(iv) {x : x ∈ R, -6 ≤ x < -1} 

-6 ≤ x < -1 is semi closed interval where the interval is equal to -6 and less than -1. 

Hence, {x : x ∈ R, – 6 ≤ x < -1} has interval [-6, -1]

(i) (2, 5) = {x : x ∈ R, 2 < x < 5} 

(ii) [0, 7] = {x : x ∈ R, 0 ≤ x < 7} 

(iii) [2, 10] = {x : x ∈ R, 2 ≤ x < 10} 

(iv) [-5, 0] = {x : x ∈ R, -5 < x ≤ 0}

A = {x : x ∈ N, 2 ≤ x ≤ 9} = {2, 3, 4, 5, 6, 7, 8, 9} 

B = {x : x two digit natural number, sum of whose digits is 8} 

= {17, 26, 35, 44, 53, 62, 71} 

(i) A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9} ∪ {17, 26, 35, 44, 53, 62, 71} 

A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9, 17, 26, 35, 44, 53, 62, 71} 

(ii) A ∩ B = {2, 3, 4, 5, 6, 7, 8, 9} ∩ {17, 26, 35, 44, 53, 62, 71} = Φ 

(iii) A – B = {2, 3, 4, 5, 6, 7, 8,9} – {17, 26, 35, 44, 53, 62, 71} 

= All elements of A which arc not present in B. 

= {2, 3, 4, 5, 6, 7, 8, 9} = A 

(iv) (A – B) ∪ (B – A) 

A – B = {2, 3, 4, 5, 6, 7, 8, 9} 

B – A = {17, 26, 35, 44, 53, 62, 71} 

(A – B) ∪ (B – A) = {2, 3, 4, 5, 6, 7, 8, 9, 17, 26, 35, 44, 53, 62, 71} = A ∪ B

Answer is (A)

A = {2, 4, 6, 8} and B = {1, 4, 7, 8}

A – B= {2, 6} and B – A = {1, 7}

Answer is (B)

A ∩ B = Φ 

Φ, is common in A and B both. 

So, other elements of A is not present in B. 

Thus A – B = A

Answer is (A) A ∩ B = Φ ⇒ A = Φ and B = Φ

Because in A ∩ B = Φ, Φ is common in A and B.

(i) Even natural numbers less than 8 are 2, 4, 6.

Hence, {2, 4, 6} is a set. 

(ii) The collection of big cities in India is not a set because there is no scale to measure it. 

(iii) The collection of various geometrical figures is not a set because these figures are not in particular shape. 

(iv) The numbers which divide the numbers 46 are 1, 2, and 23.

Hence, {1, 2, 23} is a set. 

(v) The collection of the best 20 cricket batsman of the world is not a set because there is no scale to measure their qualities. 

(vi) The collection of all even integer is a set because all even integers are 2, 4, 6, 8, 10,……..

(vii) The collection of literature written by the poet Kali das is a set because it is renowned literature.

(viii) The collection of Great men contributed in Indian culture is not a set because we cannot give proper credit for their contribution.

Hence, in above option (i), (iv), (vi) and (vii) are sets.

Answer is (D)

As in the tabular form the set C = {5} and set E = {5}

So, C = E 

Pair of equal sets is C and E.

Answer is 

(D) {x : x ∈ N and x is a prime no.}.

Answer is (D)

Given, x² + x – 2 = 0 

x² + 2x – x – 2 = 0 

x(x + 2) – 1(x + 2) = 0 

(x + 2)(x – 1) = 0 

when x + 2 = 0 then x = -2 

when x – 1 = 0 then x = 1 

Hence, roster form of equation is {-2, 1} or {1, -2}

(i) 3 ∈ {1, 2, 3, 4, 5} 

(ii) 2.5 ∉ N 

(iii) 0 ∈ Q

Answer is (C)

Because in the two sets there are no common elements.

(i) {a, b} ⊂ {b, a, c} 

Elements of first set (a, b) is present in second set.

Hence, the statement is true. 

(ii) {a, e} ⊂ {x : x is vowel of English alphabet}

Tabular form of seconds set is {a, e, i, o, u}

Elements of first set (a, e is present is second set.

Hence, the statement is true. 

(iii) {1, 2, 3} ⊄ {1, 3, 2, 5}

Elements of first set is also the elements of seconds set.

Hence, the statement is false. 

(iv) {x : x is an even natural number less then 6} ⊂ (x : x is a natural number which divide 36} 

{2, 4} ⊂ {1, 2, 3, 4, 6, 9, 12, 18, 36}

Hence, the statement is true.

(i) {a} 

Subsets of {a} are Φ and {a}. 

Hence P{a} = {Φ, {a}} 

(ii) {1, 2, 3} 

Subsets of {1, 2, 3} are Φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3} and {1, 2, 3}. 

Hence, P{1, 2, 3} = {Φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} 

(iii) {a, b} 

Subsets of {a, b} are Φ, {a}, {b}, {a, b} 

Hence, P{a, b} = {Φ, {a}, {b}, {a, b}} 

(iv) Φ 

Φ is null set, then P(Φ) = {Φ}

Answer is (B)

Given, A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8} 

A – B means A contains the element which is not present in B. 

Thus, A – B = { 1, 3, 5}

Answer is (B)

B – A means B contains the element which is not present in A. 

Thus, B – A = {4, 5}

Answer is  (D)

Vowels of English alphabet are a, e, i, o, u. 

Thus, the set of vowels is {a, e, i, o, u}.