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If A is a finite set containing n element, then number of subsets of A is 2ⁿ 

U = {1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {2, 4, 6, 8}, B = {2, 3, 5, 7}

(i) (AUB)' = {2,3,4,5,6,7,8}' = {1,9}

A'∩B' = {1,3,5,7,9}∩{1,4,6,8,9} = {1,9}

∴ (AUB)'= A'∩B'

(ii) (A∩B)' = {2}' = {1,3,4,5,6,7,8,9}

A'UB' = {1,3,5,7,9}U{1,4,6,8,9}

={1,3,4,5,6,7,8,9}

(A∩B)'=A'UB'

f(x) = 1. 23x where x is number of American dollars. 

g(y) = 50.50y where y is number of Singapore dollars. 

gof(x) = g(f(x)) 

= g(1. 23x)

= 50.50 (1.23x) 

= 62.115 x

A(x) = 30, 000 + 0.04x, where x is merchandise rupee value 

S(x) = 25000 + 0.05 x 

(A + S) (x) = A(x) + S(x) 

= 30000 + 0.04x + 25000 + 0.05 x 

= 55000 + 0.09x 

(A + S) (x) = 55000+ 0.09x

They each sell x = 1,50,00,000 worth of merchandise 

(A + S) x = 55000 + 0.09 (1,50,00,000)

= 55000 + 13,50,000 

∴ Total income of family = Rs 14,05,000

Given, Sets are A = {1, 3, 5, 6, 7} and B = {3, 7, 8, 9} 

Now, A – B = {1, 3, 5, 6, 7} – {3, 7, 8, 9} = {1, 5, 6} 

And B – A = {3, 7, 8, 9} – { 1, 3, 5, 6, 7} = {8, 9} 

∴ Required Symmetric difference, 

A ∆ B = (A – B) ∪ (B – A) = {1, 5, 6} ∪ {8,9} = {1, 5, 6, 8, 9}

(i) Given set = {1, 2}. Hence, it is finite. 

(ii) Given set = {2}. Hence, it is finite. 

(iii) Given set = φ. Hence, it is finite. 

(iv) We know that there are infinitely many primes. Hence, given set is infinite. 

(v) There are infinitely many odd numbers. Hence, given set is infinite set. 

(vi) There are infinitely many parallel lines to the X-axis. Hence, given set is infinite set. 

(vii) Finite sets 

(viii) Infinite set 

(ix) Finite set 

(x) Infinite set.

Two sets A and B are said to be equal if they have exactly the same elements and we write A = B.

Finally, now we need to show 4 ⇒ 1. 

So, assume A ∩ B = A. 

To show: A ⊂ B 

∵A ∩ B = A, therefore A⊂B, and so (4) 

⇒ (1) is true.

Let x ϵ A ∪ (B –A) ⇒ x ϵ A or x ϵ (B – A) 

⇒ x ϵ A or x ϵ B and x ∉ A 

⇒ x ϵ B 

⇒ x ϵ (A ∪ B) [∵ B ⊂ (A ∪ B)] 

This is true for all x ϵ A ∪ (B–A) 

∴ A∪(B–A)⊂(A∪B)……(1) 

Conversely, 

Let x ϵ (A ∪ B) ⇒ x ϵ A or x ϵ B 

⇒ x ϵ A or x ϵ (B–A) [∵ B ⊂ (A ∪ B)] 

⇒ x ϵ A ∪ (B–A) 

∴ (A∪B)⊂ A∪(B–A)……(2) 

From 1 and 2 we get…

 A ∪ (B – A) = A ∪ B

We need to show that 3 ⇒ 4 

Assume that A ∪ B = B 

To show: A ∩ B = A. 

∵A ∪ B = B 

∴ A⊂ B and so A ∩ B = A. 

So (3) ⇒ (4) is true.

Let x ϵ A–(A–B) ⬄ x ϵ A and x ∉ (A–B) 

⇔ x ϵ A and x ∉ (A ∩ B) 

⇔ x ϵ A∩(A ∩ B) 

⇔ x ϵ (A ∩ B) 

∴ A–(A–B) = (A ∩ B)

Here, x ∈ Q 

i.e. x is a rational number 

We know that,

If a and b are two rational numbers, then \(\frac{a+b}{2}\) is a rational number between a and b such that a<\(\frac{a+b}{2}\) <b

So, the rational number 1 and 2 is 

\(\frac{1+2}{2}\) = \(\frac{3}{2}\)

∴ the set is not empty 

∴ It is not a null set.

We need to show that 2 ⇒ 3 

So assume that A–B = ϕ 

To show: A∪B = B. 

∵ A – B = ϕ 

∴ Every element of A is an element of B. 

So A ⊂ B and therefore A∪B = B 

So (2) ⇒ (3) is true.

= (A–A) ∩(A–B) 

= ϕ ∩ (A – B) 

= A – B.

Since, 0 ∈ G

∴ the set is not empty 

∴ It is not a null set

= (A ∩ A) ∪ (A ∩ B) [∵intersection is distributive over union] 

= A ∪ (A ∩ B) [∵ A ∩ A = A] 

= A

= (A–B) ∪ (B–B) 

= (A–B) ∪ ϕ 

= A–B

(i) Integers = …-3, -2, -1, 0, 1, 2, 3, … 

Given equation: 

x² = 4 

⇒ x = √4 

⇒ x = ± 2 

If x = -2, then x² = (-2)² = 4 

If x = 2, then x² = (2)² = 4 

So, there are two elements in a set. 

∴ It is not a singleton set. 

(ii) Integers = -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, … 

Given equations: 

x + 5 = 0 

⇒ x + 5 – 5 = 0 – 5 

⇒ x = -5 

So, there is only 1 element in a given set. 

∴ It is a singleton set. 

(iii) Integers = …, -2, -1, 0, 1, 2, …

Given equation: |x| = 1 

If x = -1, then |x| = |-1| = 1 

If x = 1, then |x| = |1| = 1 

So, there are 2 elements in a given set 

∴ It is not a singleton set. 

(iv) Natural Numbers = 1, 2, 3, … 

Given equation: x 2 = 16 

⇒ x = √16 ⇒ x = ± 4 

⇒ x = -4, 4 

but x = -4 not possible because x ∈ N

So, there is only 1 element in a set. 

∴ It is a singleton set. 

(v) Prime number = 2, 3, 5, 7, 11, … 

Even Prime number = 2 

∴ It is a singleton set.

A ∪ (A ∩ B) [∵ union is distributive over intersection] 

(A ∪ A) ∩ (A ∪ B)[∵ A ∪ A = A] 

= A ∩ (A ∪ B) 

= A.

Equal Sets = Two sets A and B are said to be equal if they have exactly the same elements & we write A = B 

We have, 

A = set of letters in the word, ALLOY

A = {A, L, O, Y} 

and B = set of letters in the word, LOYAL 

B = {L, O, Y, A} 

Here, A = B because the elements in both the sets are equal. The repetition of elements in a set does not change a set. 

Thus, A and B are equal sets.

To show that the following four statements are equivalent, we need to show that 1⇒2, 2⇒3, 3⇒ 4, 4⇒1 

We first show that 1⇒ 2 

Now A–B = {xϵ A: x ∉ B} .As A ⊂ B, 

∴ Each element of A is an element of B, 

∴ A–B = ϕ 

Hence we proved 1⇒ 2.

True

R and T can be represented in roster form as follows:

R = {…,-8, -6, -4, -2, 0, 2, 4, 6, 8,….}

T = {…,-18, -12, -6, 0, 6, 12, 18,….}

Since, every element of T is present in R

T ⊂ R

So, the given statement is true.

Equal Sets = Two sets A and B are said to be equal if they have exactly the same elements & we write A = B 

We have, 

C = set of letters in the word, ‘CATARACT.’ 

C = {C, A, T, R} 

and D = set of letters in the word, ‘TRACT.’ 

D = {T, R, A, C} 

Here, C = D because the elements in both the sets are equal. The repetition of elements in a set does not change a set. 

Thus, C and D are equal sets.

We have, ACB. 

To show: C – B ⊂ C – A 

Let, x ϵ C–B 

⇒ x ϵ C and x∉ B 

⇒ x ϵ C and x∉ A 

⇒ x ϵ C – A 

Thus, x ϵ C–B ⇒ x ϵ C – A 

This is true for all x ϵ C–B 

∴ C – B ⊂ C – A

Given: 

A ⊂ B 

To show: C–B ⊂ C–A 

Let x ϵ C– B 

⇒ x ϵ C and x ∉ B [by definition C–B] 

⇒ x ϵ C and x ∉ A 

⇒ x ϵ C–A 

Thus x ϵ C–B ⇒ x ϵ C–A. This is true for all x ϵ C–B. 

C – B ⊂ C – A

The set is said to be equal with another set if all elements of both the sets are equal and same.

Now,

A = {1, 2, 3}

B = {x ∈ R: x² – 2x+1=0}

x²–2x+1 = 0

(x–1)² = 0

∴ x = 1.

B = {1}

C= {1, 2, 2, 3}

In sets we do not repeat elements thus C can be written as {1, 2, 3}

D = {x ∈ R: x³ – 6x²+11x – 6 = 0}

For the x = 1, x²–2x+1=0

= (1)³ – 6(1)² + 11(1) – 6

= 1–6+11–6

= 0

For the x =2,

= (2)³–6(2)²+11(2)–6

= 8–24+22–6

= 0

For the x =3,

= (3)³–6(3)²+11(3)–6

= 27–54+33–6

= 0

∴ D = {1, 2, 3}

Thus, the set A, C and D are equal.

(A⋂B)∖(A⋂C)

= (A⋂B)∖A⋃(A⋂B)∖C (∵ De Morgen’s law with difference and intersection.) 

= ϕ⋃(A⋂B)∖C 

= (A⋂B)∖C 

= A⋂(B∖C) (∵ intersection with difference is difference with intersection)

Let x be some element in set A – B that is x ∈ (A – B) 

Now if we prove that x ∈ (A ∩ B’) then (A – B) = (A ∩ B’) 

x ∈ (A – B) means x ∈ A and x ∉ B 

Now x ∉ B means x ∈ B.’ 

Hence we can say that x ∈ A and x ∈ B'.

Hence x ∈ A ∩ B.’ 

And as x ∈ A ∩ B’ and also x ∈ A – B we can conclude that A – B = A ∩ B.’

A is true statement because 

A∖B is the set of elements which are in A but not in B 

⇒A∖B is the set of elements which are in A and in B' 

⇒A∖B is the set of elements of A⋂B' 

∴A∖B = A⋂B' 

B is true statement because 

A∖B is the set of elements which are in A but not in B 

⇒A∖B is the set of elements which are in A but not in A⋂B' 

⇒ A∖B = A∖(A⋂B') 

C is false statement because 

A∖B = A⋂B' and A∖B^'= A⋂B 

∴ A⋂B'≠A⋂B

⇒ A∖B≠B∖A.

A∖B=A⋂B' 

≠(A∖B)⋂(B∖A)

Here,

A = {1,2,3} and B = {3,4,5} 

A∖B={1,2,3}∖{3,4,5} 

={1,2,3}∖{3} 

={1,2} 

B∖A = {3,4,5}∖{1,2,3,4} 

= {3,4,5}∖{3} 

= {4,5} 

AΔB = (A∖B)⋃(B∖A) 

= {1,2}⋃{4,5} 

= {1,2,4,5}

(A∖B)⋃(B∖A)=[A⋂B' ]⋃[B⋂A'] 

= [A⋃(B⋂A' )]⋂[B'⋃(B⋂A' )] 

= [(A⋂B)⋃(A⋂A' )]⋂[(B'⋂B)⋃(B'⋂A' )] 

= [(A⋂B)⋃ϕ]⋂[ϕ⋃(B'⋂A' )] 

= [(A⋂B)]⋂[(B'⋂A' ) 

= [(A⋂B)]⋂[(B⋃A)' ] 

= (B⋃A)'⋂(A⋂B) 

=(A⋃B)∖(A⋂B)

Given; U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4,}, B = {2, 4, 6, 8} and C = {1, 4, 5, 6} 

(i) A’ = {5, 6, 7, 8, 9} 

(ii) B’ = {1, 3, 5, 7, 9} 

(iii) C’ = {2, 3, 7, 8, 9}

(iv) (B’)’ = {2, 4, 6, 8} 

(v) (A ∪ B)’ = {5, 7, 9} 

(vi) (A ∩ C)’ = {2, 3, 5, 6, 7, 8, 9} 

(vi) (B – C)’ = {1, 3, 4, 5, 6, 7, 9}

Here, 

A⊂B ⇒ B'⊂A' 

∴ A'⋂B' = B’ 

B'∖A' = B'∖(A'⋂B') 

= B'∖B' 

= ϕ

Correct option is (A) A ⊂ C

Correct option is (C) C

Given; A = {2, 4, 6, 8, 10, 12}, B = {3, 4, 5, 6, 7, 8, 10} 

(i) (A – B) = {2, 12} 

(ii) (B – A) = {5, 7} 

(iii) (A – B) ∪ (B – A) = Φ or {}

Given; A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g} 

(i) A ∩ (B – C) = {a, c} 

(ii) A – (B ∪ C) = {d} 

(iii) A – (B ∩ C) = {a, b, c, d}