NCERT Solutions for Class 11 Maths Chapter 1 Sets Exercise 1.5

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NCERT Solutions for Class 11 Maths Chapter 1 Sets Exercise 1.5

1. 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)'

Answer

Here, 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' = U – A = {5, 6, 7, 8, 9}

(ii) B' = U – B = {1, 3, 5, 7, 9}

(iii) A∪ C = {1, 2, 3, 4}∪ {3, 4, 5,6}
 = {1, 2, 3, 4, 5, 6}
Hence (A ∪ C)' = U – (A ∪ C) = {7, 8, 9}

(iv) A ∪ B = {1, 2, 3, 4}∪ {2, 4, 6, 8}
= {1, 2, 3, 4, 6, 8}
Hence (A ∪ B)' = U – (A ∪ B) = {5, 7, 9}

(v) A' = U – A = {5, 6, 7, 8, 9}
Hence (A')' = U – A'= {1, 2, 3, 4}

(vi) B – C = {2, 4, 6, 8} – {3, 4, 5, 6} = {2, 8}
Hence (B – C)' = U – (B – C) = {1, 3, 4, 5, 6, 7, 9}

2. 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}

Answer

Here, U – {a, b, c, d, e, f, g, h}
(i) A' = U – A = {a, b, c, d,e, f, g, h} – {a, b, c}
= {d, e, g, g, h}

(ii) B' = U – B = {a, b, c, d,e, f, g, h} – {d, e, f, g}
= {a, b, c, h}

(iii) C' = U – C = {a, b, c, d,e, f, g, h} – {a, c, e, g}
= {b, d, f, h}

(iv) D' = U – D = {a, b, c, d,e, f, g, h} – {f, g, h,a}
= {b, c, d,e}.

3. 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 ε N and 2x + 1 > 10}

Answer

(i) {x: x is an odd natural number}

(ii) {x: x is an even natural number}

(iii) {x: x ε N and x is not a multiple of 3}

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

(v) {x: x ε n N and x is neither a multiples of 3 nor of 5}.

(vi) {x: x ε N and x is not a perfect square}

(vii) {x: x ε N and x is not a perfect cube}

(viii) {x: x ε N and x ≠ 3}
[x + 5 = 8 ⇒ x = 8 – 5 = 3]

(ix) {x: x ε N and x ≠ 2}
[2x + 5 = 9 ⇒ 2x = 9 – 5 ⇒ 2x = 4 ⇒ x = 2]

(x) {x: x ε N and x < 7} = {1, 2, 3, 4, 5, 6},

(xi) { x : x ε N and x ≤ 9/2}
[2x + 1 > 10 ⇒ 2x > 10 – 1 ⇒ 2x = 9
⇒ x > 9/2]

4. 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'

Answer

(i) A ∪ B = {2, 4, 6, 8} ∪{2, 3, 5, 7}
= {2, 3, 4, 5, 6, 7, 8}
L.H.S. = (A ∪ B)¢ = U – (A ∪ B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 4, 5, 6, 7, 8}
= {1, 9}
Now, A' = U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}
and B' = U – B
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 5, 7}
= {1, 4, 6, 8, 9}
R.H.S. = A'∩ B'
= {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1,9}
L.H.S. = R.H.S.
Hence (A∪B)' = A'∩ B' is verified.

(ii) A ∩B = {2, 4, 6, 8} ∩ {2, 3, 5, 7} = {2}
L.H.S. = (A∩ B)' = U – (A∩ B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2}
= {1, 3, 4, 5, 6, 7, 8, 9}
R.H.S. = A' ∪ B'
= {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9}
= {1, 3, 4, 5, 6, 7, 8, 9}
L.H.S = R.H.S.
Hence, (A ∩ B) = A' ∪ B' is verified.

5. Draw appropriate Venn diagram for each of the following:
(i) (A ∪ B)'
(ii) A' ∩ B'
(iii) (A ∩ B)'
(iv) A' ∪ B'.

Answer

(i) (A ∪ B)'

The shaded region indicates (A ∪ B)'
(ii) A' ∩ B'
The shaded region indicates A' ∩ B'
(iii) (A ∩ B)'
The shaded region indicates (A ∩ B)'
(iv) A' ∪ B'
The shaded region indicates A' ∪ B' 
All shaded region formed by all horizontal and vertical lines is A' ∪ B'.

6. 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'?

Answer

A is the set of triangles in which no triangle is equilateral.
Hence A' is the set of all equilateral triangles.

7. Fill in the blanks to make each of the following a true statement: 
(i) A ∪ A' = .............. 
(ii) Ф' ∩ A = .............. 
(iii) A∩ A' = .............. 
(iv) U' ∩ A = .............. 

Answer

(i) A ∪ A' = U
(ii) Ф' ∩ A = U∩ A = A
(iii) A ∩ A'= Ф
(iv) U' ∩ A = Ф ∩ A = Ф
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