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

**11th Maths Exercise 1.1 Answers 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) A = {2, 3, 5, 7}

(ii) B = {1}

(iii) 4x + 9 < 52

4x + 9 – 9 < 52 – 9

4x < 43

x < \(\frac{43}{4}\) (i.e.) x < 10.75 4

But x ∈ N

∴ 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}

**11th Std Maths Exercise 1.1 Answers Question 2.**

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

Solution:

A = {x: x^{2} = 1}

**11th Maths Book Exercise 1.1 Answers Question 3.**

State whether the following sets are finite or infinite.

(i) {x ∈ N : x is an even prime number}

(ii) {x ∈ N : x is an odd prime number}

(iii) {x ∈ Z : x is even and less than 10}

(iv) {x ∈ R : x is a rational number}

(v) {x ∈ N : x is a rational number}

Solution:

(i) Finite set

(ii) Infinite set

(iii) Infinite

(iv) and

(v) infinite

**Exercise 1.1 Class 11 Maths State Board 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) To prove: A × (B ∩ C) = (A × B) ∩ (A × C)

B ∩ C = {8}; A = {1, 2, 5, 7}

So A × (B ∩ C) = {1, 2, 5, 7} × {8}

= {(1, 8), (2. 8), (5, 8), (7, 8)}

Now A x 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)} …. ( 1)

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, 8), (2, 8), (5, 8), (7, 8)} ……… (2)

(1) = (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) 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)}

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

L.H.S. (A × B) ∩ (B × A) = {(2, 2), (2, 7), (7, 2), (7, 7)} …. (1)

R.H.S. A ∩ B = {2, 7}

B ∩ A = {2, 7}

(A ∩ B) × (B ∩ A) = {2, 7} × {2, 7}

= {(2, 2), (2, 7), (7, 2), (7, 7)} ……… (2)

(1) = (2) ⇒ LHS = RHS

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

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

Now B – A = {8, 9}

(B – A) ∩ C = {8} ……. (1)

B ∩ C = {8}

A = {1, 2, 5, 7}

So (B ∩ C) – A = {8} …… (2)

C – A = {8, 10}

B = {2, 7, 8, 9}

B ∩ (C – A) = {8} …. (3)

(1) = (2) = (3)

(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)

**11th Maths Exercise 1.1 Question 5.**

Justify the trueness of the statement.

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

Solution:

A set itself can be a subset of itself (i.e.) A ⊆ A. But it cannot be a proper subset.

**11th Maths Chapter 1 Exercise 1.1 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

**11 Maths Exercise 1.1 Question 7.**

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

Solution:

n(A ∪ B) = 10; n(A ∩ B) = 3

n(A ∆ B) = 10 – 3 = 7

and n(P(A ∆ B)) = 27 = 128

**11th Maths 1.1 Exercise 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}

**11th Maths Exercise 1.1 Answers In Tamil 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:

n(A) = 3 ⇒ set A contains 3 elements

n(B) = 2 ⇒ set B contains 2 elements –

we are given (x, 1), (y, 2), (z, 1) are elements in A × B ⇒ A = {x, y, z} and B = {1, 2}

**11th Maths Exercise 1.1 Answers State Board 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

**Samacheer Kalvi 11th Maths Example Sums 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}

**11th Maths Exercise 1.1 4th Sum 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

**11th Maths 1st Chapter Exercise 1.1 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