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# NCERT Books Solutions for Class 11 Mathematics english Medium 1. Sets

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## 1. Sets

#### Exercise 1.5

Exercise 1.5

Q1. Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } and C = { 3, 4, 5, 6 }. Find

(i) A′

(ii) B′

(iii) (A ∪ C)′

(iv) (A ∪ B)′

(v) (A′)′

(vi) (B – C)′

Solution: Given that

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

(i) A' = {5, 6, 7, 8, 9}

(ii) B' = {1, 3, 5, 7, 9}

(iii) A ∪ C = {1, 2, 3, 4, 5, 6}

Therefore, (A ∪ C)′ = {7, 8, 9}

(iv) A ∪ B = {1, 2, 3, 4, 6, 8}

Therefore, (A ∪ B)′ = {5, 7, 9}

(v) A' = {5, 6, 7, 8, 9}

(A')' = A = {1, 2, 3, 4}

(vi) B - C = {2, 8}

(B - C)' = 1, 3, 4, 5, 6, 7, 9}

Q2. If U = { a, b, c, d, e, f, g, h}, find the complements of the following sets :
(i) A = {a, b, c}

(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}

(iv) D = { f, g, h, a}

Solution: Given that

U = { a, b, c, d, e, f, g, h}

(i) A = {a, b, c}

A' = {d, e, f, g, h}

(ii) B = {d, e, f, g}

B' = {a, b, c, h}

(iii) C = {a, c, e, g}

C' = {b, d, f, h}

(iv) D = { f, g, h, a}

D' = {b, c, d e}

Q3. Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i) {x : x is an even natural number}

(ii) { x : x is an odd natural number }
(iii) {x : x is a positive multiple of 3}

(iv) { x : x is a prime number }
(v) {x : x is a natural number divisible by 3 and 5}
(vi) { x : x is a perfect square }

(vii) { x : x is a perfect cube}
(viii) { x : x + 5 = 8 }

(ix) { x : 2x + 5 = 9}
(x) { x : x ≥ 7 }

(xi) { x : x ∈ N and 2x + 1 > 10 }

Solution: Given that U = { 1, 2, 3, 4, 5, 6, 7 ....}

(i) Let A = {x : x is an even natural number}

Or A = {2, 4, 6, 8 .....}

A' = { 1, 3, 5, 7 .....}

= {x : x is an odd natural number}

(ii) Let B = { x : x is an odd natural number }

Or     B = { 1, 3, 5, 7 .....}

B' = {2, 4, 6, 8 .....}

= {x : x is an even natural number}

(iii) Let C = {x : x is a positive multiple of 3}

Or     C = {3, 6, 9 ....}

C' = {1, 2, 4, 5, 7, 8, 10 .....}

= {x: x N and x is not a multiple of 3}

(iv) Let D = { x : x is a prime number }

Or     D = {2, 3, 5, 7, 11 ... }

D' = {1, 4, 6, 8, 9, 10 ...... }

= {x: x is a positive composite number and x = 1}

(v) Let E = {x : x is a natural number divisible by 3 and 5}

Or     E = {15, 30, 45 .....}

E' = {x: x is a natural number that is not divisible by 3 or 5}

(vi) Let F = { x : x is a perfect square }

F' = {x: x N and x is not a perfect square}

(vii) Let G = {x: x is a perfect cube}

G' = {x: x N and x is not a perfect cube}

(viii) Let H = {x: x + 5 = 8}

H' = {x: x N and x ≠ 3}

(ix) Let I = {x: 2x + 5 = 9}

I' = {x: x N and x ≠ 2}

(x) Let J = {x: x ≥ 7}

J' = {x: x N and x < 7}

(xi) Let K = {x: x N and 2x + 1 > 10}

K = {x: x N and x ≤ 9/2}

Q4. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}. Verify that
(i) (A ∪ B)′ = A′ ∩ B′

(ii) (A ∩ B)′ = A′ ∪ B′

Solution:

(i) U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}.

(A ∪ B)′ = A′ ∩ B′

A ∪ B = {2, 3, 4, 5, 6, 7, 8}

LHS = (A ∪ B)′ = {1, 9} ...(i)

RHS = A′ ∩ B′

= {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9}

= {1, 9} .... (ii)

LHS = RHS

Hence Verified.

Solution:

(ii) U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}.

(A ∩ B)′ = A′ ∪ B′

A ∩ B = {2}

LHS = (A ∩ B)′ = {1, 3, 4, 5, 6, 7, 8, 9 }

RHS = A′ ∪ B′

{1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9}

= {1, 3, 4, 5, 6, 7, 8, 9 }

LHS = RHS

Hence Verified

Q5. Draw appropriate Venn diagram for each of the following :
(i) (A ∪ B)′,

(ii) A′ ∩ B′,

(iii) (A ∩ B)′,

(iv) A′ ∪ B′

Solution:

(i) (A ∪ B)′

Venn diagram of (A ∪ B)′

(ii) A′ ∩ B′,

Venn diagram of A′ ∩ B′

Note: Venn diagram of A′ ∩ B′ will be same as (A ∪ B)′

Because (A ∪ B)′ = A′ ∩ B′

(iii) (A ∩ B)′

Venn diagram of (A ∩ B)′

(iv) A′ ∪ B′

Venn diagram of A′ ∪ B′

Q6. Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A′?

Solution:

A = {the set of all triangles with at least one angle different from 60°}

A' = {the set of all equilateral triangles}

Q7. Fill in the blanks to make each of the following a true statement :
(i) A ∪ A′ = . . .

(ii) φ′ ∩ A = . . .

(iii) A ∩ A′ = . . .

(iv) U′ ∩ A = . . .

Solution:

(i) A ∪ A′ = U

(ii) φ′ = U

Therefore φ′ ∩ A = U ∩ A = A

so, φ′ ∩ A = A

(iii) A ∩ A′ = φ

(iv) U′ ∩ A = φ

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##### NCERT Solutions⇒Class 11th ⇒ Mathematics ⇒ Chapter 1. Sets

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