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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1

Question 1.

Write the following in roster form.

(i) {x ∈ N : x^{2} < 121 and x is a prime}.

(ii) the set of all positive roots of the equation (x – 1)(x + 1)(x^{2} – 1) = 0.

(iii) {x ∈ N : 4x + 9 < 52}.

(iv) {x : \(\frac{x-4}{x+2}\) = 3, x ∈ R – {-2}}

Solution:

(i) Let A = { x ∈ N : x^{2} < 121 and x is a prime }

A = {2, 3, 5, 7}

(ii) The set of positive roots of the equations

(x – 1) (x + 1) (x^{2} – 1) = 0

(x – 1 ) (x + 1 ) (x + 1) (x – 1) = 0

(x + 1 )^{2} (x – 1)^{2} = 0

(x + 1)^{2} = 0 or (x – 1)^{2} = 0

x + 1 = 0 or x – 1 = 0

x = -1 or x = 1

A = { 1 }

(iii) Let A = { x ∈ N : 4x + 9 < 52 }

When x = 1, (4) × (1 ) + 9 = 4 + 9 = 13

When x = 2, (4) × (2) + 9 = 8 + 9 = 17

When x = 3, (4) × (3) + 9 = 12 + 9 = 21

When x = 4, (4) × (4) + 9 = 16 + 9 = 25

When x = 5, (4) × (5) + 9 = 20 + 9 = 29

When x = 6, (4) × (6) + 9 = 24 + 9 = 33

When x = 7, (4) × (7) + 9 = 28 + 9 = 37

When x = 8, (4) × (8) + 9 = 32 + 9 = 41

When x = 9, (4) × (9) + 9 = 36 + 9 = 45

When x = 10, (4) × (10) + 9 = 40 + 9 = 49

∴ A = { 1, 2, 3, 4, 5, 6 ,7, 8, 9, 10 }

(i.e.) x – 4 = 3(x + 2)

x – 4 = 3x + 6

– 4 – 6 = 3x – x

2x = -10 ⇒ x = -5

A = {-5}

Question 2.

Write the set {-1, 1} in set builder form.

Solution:

A = {x : x^{2} – 1 = 0, x ∈ R}

Question 3.

State whether the following sets are finite or infinite.

- {x ∈ N : x is an even prime number}
- {x ∈ N : x is an odd prime number}
- {x ∈ Z : x is even and less than 10}
- {x ∈ R : x is a rational number}
- {x ∈ N : x is a rational number}

Solution:

- Finite set
- Infinite set
- Infinite
- Infinite
- Infinite

Question 4.

By taking suitable sets A, B, C, verify the following results:

(i) A × (B ∩ C) = (A × B) ∩ (A × C).

(if) A × (B ∪ C) = (A × B) ∪ (A × C).

(iii) (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).

(iv) C – (B – A) = (C ∩ A) ∪ (C ∩ B).

(v) (B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A).

Solution:

To prove the following results let us take U = {1, 2, 5, 7, 8, 9, 10}

A = {1, 2, 5, 7}

B = {2, 7, 8, 9}

C = {1, 5, 8, 7}

(i) Let A = {1, 2}, B = {3, 4}, C = {4, 5}

B ∩ C = {3, 4} ∩ {4, 5}

B ∩ C = {4}

A × (B ∩ C) = {1, 2} × {4}

A × (B ∩ C) = { (1,4), (2,4) } —– (1)

A × B = {1, 2} × {3, 4}

A × B = { (1,3), (1, 4), (2, 3), (2, 4)}

A × C = {1, 2} × { 4, 5 }

A × C = {(1, 4), (1, 5), (2, 4), (2, 5)}

(A × B) ∩ (A × C) = {(1, 3), (1, 4), (2, 3), (2, 4)} ∩ { (1, 4), (1, 5), (2, 4), (2, 5)}

(A × B) ∩ (A × C) = {(1, 4), (2, 4)} —- (2)

From equations (1) and (2)

A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) To prove A × (B ∪ C) = (A × B) (A × C)

B = {2, 7, 8, 9}, C = {1, 5, 8, 10)

B ∪ C = {1, 2, 5, 7, 8, 9, 10}

A = {1, 2, 5, 7}

A × (B ∪ C) = {(1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1, 9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)) …. (1)

A × B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9),

(7, 2), (7, 7), (7, 8), (7, 9)}

A × C = {(1, 1), (1, 5), (1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)}

(A × B) ∪ (A × C) = (1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1,9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)} …… (2)

(1) = (2) ⇒ A × (B ∪ C) = (A × B) ∪ (A × C)

(iii) Let A = {1, 2}, B = {2, 3}

A × B = {1, 2} × {2, 3}

A × B = {(1, 2), (1, 3), (2, 2), (2, 3)}

B × A = {2, 3} × {1, 2}

B × A = {(2, 1), (2, 2), (3, 1), (3,2)}

(A × B) ∩ (B × A) = {(1, 2), (1, 3),(2, 2), (2, 3)} ∩ {(2, 1), (2, 2), (3, 1),(3, 2)}

(A × B) ∩ (B × A) = {(2, 2)} ——- (1)

A ∩ B = {1, 2} ∩ {2, 3}

A ∩ B = {2}

B ∩ A = {2, 3} ∩ {1, 2}

B ∩ A = {2}

(A ∩ B) × (B ∩ A) = {2} × {2}

(A ∩ B) × (B ∩ A) = {(2,2)} ———- (2)

From equations (1) and (2)

(A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A)

(iv) To prove C – (B – A) = (C ∩ A) ∪ (C ∩ B)

B – A = {8, 9}

C = {1, 5, 8, 10}

∴ LHS = C – (B – A) = {1, 5, 10} …… (1)

C ∩ A = {1}

U = {1, 2, 5, 7, 8, 9, 10}

B = {2, 7, 8, 9} ∴ B’ = {1, 5, 10}

C ∩ B = {1, 5, 10}

R.H.S. (C ∩ A) ∪ (C ∩ B) = {1} ∪ {1, 5, 10}

= {1, 5, 10} ……. (2)

(1) = (2) ⇒ LHS = RHS

(v) Let A = {1, 2, 3, 4} , B = {3, 4, 5, 6}, C = { 5, 6, 7, 8 )

B – A = {3, 4, 5, 6} – {1, 2, 3, 4}

B – A = {5, 6}

(B – A) ∩ C = {5, 6} ∩ {5, 6, 7, 8}

(B – A) ∩ C = {5, 6} ——– (1)

(B ∩ C) = {3, 4, 5, 6} ∩ {5, 6, 7, 8}

B ∩ C = {5, 6}

(B ∩ C) – A = {5, 6} – {1,2,3,4}

(B ∩ C) – A = {5, 6} ——- (2)

C – A = {5, 6, 7, 8} – {1, 2, 3, 4}

C – A = {5, 6, 7, 8}

B ∩ (C – A) = {3, 4, 5, 6} ∩ {5, 6, 7, 8}

B ∩ (C – A) = {5, 6} ——– (3)

From equations (1) , (2) and (3)

(B – A) ∩ C = (B ∩ C) – A = B ∩(C – A)

(vi) To prove (B – A) ∪ C ={1, 5, 8, 9, 10}

B – A = {8, 9},

C = {1, 5, 8, 10}

(B – A) ∪ C = {1, 5, 8, 9, 10} ……. (1)

B ∪ C = {1, 2, 5, 7, 8, 9, 10}

A – C = {2, 7}

(B ∪ C) – (A – C) = {1, 5, 8, 9, 10} ……… (2)

(1) = (2)

⇒ (B – A) ∪ C = (B ∪ C) – (A – C)

Question 5.

Justify the trueness of the statement.

“An element of a set can never be a subset of itself.”

Solution:

“An element of a set can never be a subset of itself ”

The statement is correct

Let A = {a, b, c, d} for a ∈ A

‘a’ cannot be a subset of ‘a’

Question 6.

If n(P(A)) = 1024, n(A ∪ B) = 15 and n(P(B)) = 32, then find n(A ∩ B).

Solution:

n(P( A)) = 1024 = 2^{10} ⇒ n( A) = 10

n(A ∪ B) = 15

n(P(B)) = 32 = 2^{5} ⇒ n(B) = 5

We know n(A ∪ B) = n{A) + n(B) – n(A ∩ B)

(i.e.) 15 = 10 + 5 – n(A ∩ B)

⇒ n(A ∩ B) = 15 – 15 = 0

Question 7.

If n(A ∩ B) = 3 and n(A ∪ B) = 10, then find n(P(A(A ∆ B)).

Solution:

Given n(A ∩ B) = 3 and n(A ∪ B) = 10

A ∆ B = (A – B) ∪ (B – A)

n(A ∆ B) = n [ (A – B ) ∪ (B – A)]

n(A ∆ B) = n(A – B) + n(B – A) —— (1)

(Since A – B and B – A are disjoint sets)

A ∪ B = (A – B) ∪ (B – A) ∪ (A ∩ B)

n(A ∪ B) = n[(A – B) ∪ (B – A) ∪ (A ∩ B)]

n(A ∪ B) = n (A – B) + n (B – A) + n (A ∩ B)

(Since A – B, B – A and A ∩ B are disjoint sets)

n(A ∪ B) = n(A ∆ B) + n(A ∩ B)

10 = n(A ∆ B) + 3

n(A ∆ B) = 10 – 3 = 7

∴ n(P(A ∆ B)) = 2^{7} = 128

Question 8.

For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A.

Solution:

A × A = 16 elements = 4 × 4

⇒ A has 4 elements

∴ A = {0, 1, 2, 3}

Question 9.

Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y, z are distinct elements.

Solution:

Given A and B be two sets such that n (A) = 3 and n(B) = 2.

Also given (x, 1), (y, 2), (z, 1) ∈ A × B

A = { x, y, z }, B = {1, 2}

Question 10.

If A × A has 16 elements, S = {(a, b) ∈ A × A : a < b} ; (-1, 2) and (0, 1) are two elements of S, then find the remaining elements of S.

Solution:

n(A × A) = 16 ⇒ n( A) = 4

S ={(-1, 0), (-1, 1), (0, 2), (1, 2)}

### Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1 Additional Questions

Question 1.

Write the following sets in roster form

(a) {x ∈ N; x^{3} < 1000}

(b) {The set of positive roots of the equation (x^{2} – 4) (x^{3} – 27) = 0}

Solution:

(a) A = {1, 2, 3, 4, 5, 6, 7, 8, 9}

(b) B = {2, 3}

Question 2.

By taking suitable sets A, B, C verify the following results

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

(ii) (B – A) ∪ C = (B ∪ C) – (A – C)

Solution:

Prove by yourself

Question 3.

Given n(A) = 7; n(B) = 8 and n(A ∪ B) = 10 find n[P(A ∩ B)].

Solution:

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

(i.e.,) 10 = 7 + 8 – n(A ∩ B)

⇒ n(A ∩ B) = 7 + 8 – 10 = 5

So n[P(A ∩ B)] = 25 = 32