Tamilnadu Samacheer Book 11th Maths Solutions Chapter 1 Sets Ex 1.1 - Book Exercise Answers

 Tamilnadu Samacheer Book 11th Maths Solutions Chapter 1 Sets Ex 1.1 - Book Exercise Answers - Reduced syllabus 2021

• 11th Maths Book guide Ex 1.1 : Question 1.

Write the following in roster form.

(i) {x ∈ N : x² < 121 and x is a prime}.

(ii) the set of all positive roots of the equation (x – 1)(x + 1)(x² – 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}

• Question 2.

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

• Solution:

A = {x: x² = 1}

• 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

• 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)

• 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.

• 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 = 210 ⇒ n( A) = 10

n(A ∪ B) = 15

n(P(B)) = 32 = 25 ⇒ 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:

• n(A ∪ B) = 10; n(A ∩ B) = 3

n(A ∆ B) = 10 – 3 = 7

and n(P(A ∆ B)) = 27 = 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:

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}

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)}