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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.5

**Exercise 1.5 Class 9 Maths Samacheer Question 1.**

Using the adjacent venn diagram, find the following sets:

(i) A – B

(ii) B – C

(iii) A’ ∪ B’

(vi) A’ ∩ B’

(v) (B ∪ C)’

(vi) A – (B ∪ C)

(vii) A – (B ∩ C)

Solution:

(i) A – B = {3, 4, 6}

(ii) B – C = {-1, 5, 7}

(iii) A’ ∪ B’

A’ = {1, 2, 0, -3, 5, 7, 8}

B’ = {-3, 0, 1, 2, 3, 4, 6)

A’ ∪ B’ = {-3, 0, 1, 2, 3, 4, 5, 6, 7, 8)

(iv) A’ ∩ B’

A’ ∩ B’ = {-3, 0, 1, 2}

(v) B ∪ C = {-3, -2, -1, 0, 3, 5, 7, 8}

(B ∪ C)’ = U – (B ∪ C)

= {-3, -2, -1, 0, 1,2, 3, 4, 5, 6, 7, 8} – {-3, -2, -1, 0, 3, 5, 7, 8}

(B ∪ C)’ = {1, 2, 4, 6}

(vi) A – (B ∪ C) = {-2, -1, 3, 4,6} – {-3, -2, -1, 0, 3, 5, 7, 8} = {4, 6}

A – (B ∩ C)

B ∩ C = {-2, 8}

A- (B ∩ C) = {-2, -1, 3, 4, 6} – {-2, 8} = {-1, 3, 4, 6}

**9th Maths Exercise 1.5 Samacheer Kalvi Question 2.**

If K = {a, b, d, e,f}, L = {b, c, d, g} and M {a, b, c, d, h} then find the following:

(i) K ∪ (L ∩ M)

(ii) K ∩ (L ∪ M)

(iii) (K ∪ L) ∩ (K ∪ M)

(iv) (K ∩ L) ∪ (K ∩ M) and verify distributive laws.

Solution:

K = {a, b, d, e, f}, L = {b, c, d, g} and M {a, b, c, d, h}

(i) K ∪ (L ∩ M)

L ∩ M = {b, c, d, g} ∩ {a, b, c, d, h} = {b, c, d}

K ∪ (L ∩ M) = {a, b, d, e, f } ∪ {b, c, d) = {a, b, c, d, e, f}

(ii) K ∩(L ∪ M)

L ∪ M = {a, b, c, d, g, h}

K ∩ (L ∪ M) = {a, b, d, e, f} ∩ {a, b, c, d, g, h} = {a, b, d}

(iii) (K ∪ L) ∩ (K ∪ M)

K ∪ L = {a, b, c, d, e, f, g}

K ∪ M = {a, b, c, d, e, f, h}

(K ∪ L) ∩ (K ∪ M) = {a, b, c, d, e,f}

(iv) (K ∩ L) ∪ (K ∩ M)

(K ∩ L) = {b, d)

(K ∩ M) = {a,b,d}

(K ∩ L) ∪ (K ∩ M) = {b, d} ∪ {a, b, d} = {a, b, d}

Distributive laws

K ∪ (L ∩ M) = (K ∪ L) ∩ (K ∪ M)

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

= {a, b, c, d, e, f}

Thus Verified.

K ∩ (L ∪ M) = (K ∩ L) ∪ (K ∩ M)

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

= {a, b, d}

Thus Verified.

**9th Maths Set Language Exercise 1.5 Question 3.**

If A = {x : x ∈ Z, -2 < x ≤ 4}, B = {x : x ∈ W, x ≤ 5}, C ={-4, -1, 0, 2, 3, 4}, then verify A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

Solution:

A = {x : x ∈ Z, -2 < x ≤ 4} = {-1, 0, 1, 2, 3, 4}

B = {x : x ∈ W, x ≤ 5} = {0, 1, 2, 3, 4, 5}

C = {-4, -1, 0, 2, 3, 4}

A ∪ (B ∩ C)

B ∩ C = {0, 1, 2, 3, 4, 5} ∩ {-4, -1, 0, 2, 3, 4} = {0, 2, 3, 4}

A ∪ (B ∩ C) = {-1, 0, 1, 2, 3, 4} ∪ (0, 2, 3, 4} ={-1, 0, 1, 2, 3, 4} …………. (1)

(A ∪ B) ∩ (A ∪ C)

A ∩ B = {0, 1, 2, 3, 4}

A ∩ C = {-1, 0, 2, 3, 4}

(A ∩ B) ∪ (A ∩ C) = {0, 1, 2, 3, 4} ∪ {-1, 0, 2, 3, 4}= {-1, 0, 1, 2, 3, 4} …………. (2)

From (1) and (2), it is verified that

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

**9th Maths Exercise 1.5 Question 4.**

Verify A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) using Venn diagrams.

Solution:

L.H.S A ∪ (B ∩ C)

From (2) and (5), it is verified that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

**Class 9 Maths Chapter 1 Samacheer Kalvi Question 5.**

If A = {b, c, e, g, h}, B = {a, c, d, g, i} and C = {a, d, e, g, h}, then show that A – (B ∩ C) = (A – B) ∪ (A – C).

Solution:

A = {b, c, e, g, h}

B = {a, c, d, g, i}

C = {a, d, e, g, h}

B ∩ C = {a, d, g}

A – (B ∩ C) = {b, c, e, g, h} – {a, d, g} = {b, c, e, h} ……..… (1)

A- B = {b, c, e, g, h} – {a, c, d, g, i} = {b, e, h}

A – C = {b, c, e, g, h} – {a, d, e, g, h} = {b, c}

(A – B) ∪ (A – C) = {b, c, e, h} ………..… (2)

From (1) and (2) it is verified that

A – (B ∩ C) = (A – B) ∪ (A – C)

**Samacheerkalvi.Guru 9th Maths Question 6.**

If A = {x : x = 6 n ∈ W and n < 6}, B = {x : x = 2n, n ∈ N and 2 < n ≤ 9} and C = {x : x = 3n, n ∈ N and 4 ≤ n < 10}, then show that A – (B ∩ C) = (A – B) ∪ (A – C)

Solution:

A = {x : x = 6n, n ∈ W, n < 6}

x = 6n

n = {0, 1, 2, 3, 4, 5}

⇒ x = 6 × 0 = 0

x = 6 × 1= 6

x = 6 × 2 = 12

x = 6 × 3 = 18

x = 6 × 4 = 24

x = 6 × 5 = 30

∴ A = {0, 6, 12, 18, 24, 30}

B = { x : x = 2n, n ∈ N, 2 < n ≤ 9}

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

x = 2 n

⇒ x = 2 × 3 = 6

2 × 4 = 8

2 × 5 = 10

2 × 6 = 12

2 × 7 = 14

2 × 8 = 16

2 × 9 = 18

∴ B {6, 8, 10, 12, 14, 16, 18}

C = { x : x = 3n, n ∈ N, 4 ≤ n < 10}

N = { 4, 5, 6, 7, 8, 9}

x = 3 × 4 = 12

⇒ x = 3 × 5 = 15

x = 3 × 6 = 18

x = 3 × 7 = 21

x = 3 × 8 = 24

x = 3 × 9 = 27

x = 2 × 9 = 18

∴ C = {12, 15, 18, 21, 24, 27}

A – (B ∩ C) = (A – B) ∪ (A – C)

L.H.S R.H.S

B ∩ C = {12,18}

A – (B ∩ C) = {0, 6, 12, 18, 24, 30} – {12, 18} = {0, 6, 24, 30} ……….…. (1)

(A – B) = {0, 24, 30}

(A – C) = {0, 6, 30}

(A – B) ∪ (A – C) = {0, 6, 24, 30} …………. (2)

From (1) and (2), it is verified that

A – (B ∩ C) = (A – B) ∪ (A – C).

**Samacheer Kalvi Guru 9th Maths Question 7.**

If A = {-2, 0, 1, 3, 5}, B = {-1, 0, 2, 5, 6} and C = {-1, 2, 5, 6, 7}, then show that A – (B ∪ C) = (A – B) ∩ (A – C).

Solution:

A = {-2, 0, 1, 3, 5},

B = {-1, 0, 2, 5, 6}

C ={-1, 2, 5, 6, 7}

B ∪ C = {-1, 0, 2, 5, 6, 7}

A – (B ∪ C) = {-2, 1, 3} …………. (1)

(A – B) = {-2, 1, 3}

(A – C) = {-2, 0, 1, 3}

(A – B) ∩ (A – C) = {-2, 1, 3} ………..… (2)

From (1) and (2), it is verified that . A – (B ∪ C) = (A – B) ∩ (A – C)

**9th Standard Maths Exercise 1.5 Question 8.**

if A={y: y = \(\frac{a+1}{2}\), a ∈ W and a ≤ 5},B = {y: y=\(\frac{2 n-1}{2}\),n ∈ W and n < 5} and C={−1,\(-\frac{1}{2}\), 1, \(\frac{3}{2}\), 2} then show that A – (B ∪ C) = (A – B) ∩ (A – C).

Solution:

(A – B) ∩ (A – C) = {3} …………. (2)

From (1) and (2), it is verified that A – (B ∪ C) = (A – B) ∩ (A – C).

**9th Maths Exercise 1.5 In Tamil Question 9.**

Verify A – (B ∩ C) = (A – B) ∪ (A – C) using Venn diagrams.

Solution:

∴ A – (B ∩ C) = (A – B) ∪ (A – C)

Hence it is proved.

**Kalvi Guru 9th Maths Question 10.**

If U = {4, 7, 8, 10, 11, 12, 15, 16}, A = {7, 8, 11, 12} and B = {4, 8, 12, 15}, then verify De Morgan’s Laws for complementation.

U = {4, 7, 8, 10, 11, 12, 15, 16}

A = {7, 8, 11, 12}, B = {4, 8, 12, 15}

De Morgan’s Laws for complementation.

(A ∪ B)’ = A’ ∩ B’

A ∪ B = {4, 7, 8, 11, 12, 15}

(A ∪ B)’ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 7, 8, 11, 12, 15}

= {10, 16} ……………. (1)

A’ = {4, 10, 15, 16}

B’ = {7, 10, 11, 16}

A’ ∩ B’ = {10, 16} ………………(2)

From (1) and (2) it is verified that (A ∪ B)’ = A’ ∩ B’.

**9th Class Math Exercise 1.5 Solution Question 11.**

Verify (A ∩ B)’ = A’ ∪ B’ using Venn diagrams.

Solution:

(A ∩ B)’ = A’ ∪ B’

(2) = (5)

∴ (A ∩ B)’ = A’ ∪ B’