Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 1 Sets, Relations and Functions Ex 1.1 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets, Relations and Functions Ex 1.1

Question 1.

Write the following in roaster form.

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

Answer:

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 equation (x – 1) ( x + 1) (x – 1 ) = 0

Answer:

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) {x ∈ N : 4x + 9 < 52}

Answer:

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}

(iv)

Answer:

Let A =

⇒ \(\frac{x-4}{x+2}\) = 3

⇒ x – 4 = 3(x + 2)

⇒ x – 4 = 3x + 6

⇒ 3x – x = – 4 – 6

2x = – 10

⇒ x = \(-\frac{10}{2}\) = -5

A = { -5 }

Question 2.

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

Answer:

A = {x : x^{2} – 1 = 0, x ∈ R}

Question 3.

State whether the following sets are finite or infinite.

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

Answer:

Let A = { x ∈ N : x is an even prime number )

A = {2}

A is a finite set.

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

Answer:

Let B = {x ∈ N : x is an odd prime number}

B = {1, 3, 5, 7, 11, …………….. }

B is an infinite set.

(iii) {x ∈ Z : x is even and < 10 }

Answer:

C = {x ∈ Z : x is even and< 10}

C = { ……….. -8, -6, -4, -2, 0, 2, 4, 6, 8}

C is an infinite set.

(iv) {x ∈ R : x is a rational number }

Answer:

D = { x ∈ R : x is a rational number }

D is an infinite set.

(v) {x ∈ N: x is a rational number }

Answer:

E = { x ∈ N : x is a rational number )

E = {1, 2, 3, 4, 5, 6, …………..)

Every integer is a rational number.

∴ E is an infinite set.

Question 4.

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

(i) A × (B ∩ C) = (A × B) ∩ (A × C)

Answer:

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) A × (B ∪ C) = (A × B) ∪ (A × C)

Answer:

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

B ∪ C = {3, 4} ∪ {4, 5}

B ∪ C = {3, 4, 5)

A × (B ∪ C) = {1, 2} × {3, 4, 5}

A × (B ∪ C) = { (1, 3),( 1, 4),(1, 5),(2, 3), (2, 4),(2,5)} ——– (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, 3) (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)} —— (2)

From equations (1) and (2)

A × (B U C) = (A × B) U (A × C)

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

Answer:

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) C – (B – A) = (C ∩ A) ∪ (C ∩ B’)

Answer:

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

B – A = { 2, 3, 4 ) – {1, 2, 3}

B – A = {4}

C – (B – A) = {3, 4, 5} – {4}

C – (B – A) = {3, 5} —- (1)

C ∩ A = {3, 4, 5} ∩ { 1, 2, 3)

C ∩ A = {3}

B’ = {1, 5}

C ∩ B’ = {3, 4, 5} ∩ {1, 5}

C ∩ B’ = {5}

(C ∩ A) ∪ (C∩B’) = {3} ∪ {5}

(C∩A) ∪ (C ∩ B’) = {3, 5} —— (2)

From equations (1) and (2)

C – (B – A) = (C ∩ A) u (C ∩ B’)

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

Answer:

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) (B – A) ∪ C = (B ∪ C) – (A – C)

Answer:

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 , 7 , 8 } ——- (1)

B ∪ C = { 3, 4, 5, 6 } ∪ { 5, 6, 7,8 }

B ∪ C = { 3, 4, 5, 6, 7, 8 }

A – C = { 1, 2, 3, 4 } – { 5, 6, 7, 8 }

A – C = { 1 , 2, 3 , 4 }

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

(B ∪ C) – (A – C) = { 5, 6, 7, 8 } —-(2)

From equations (1) and (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”.

Answer:

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, find n(A ∩ B).

Answer:

Given n(P(A)) = 1024 , n(A ∪ B) = 15, n(P(B)) = 32

n(P(A)) = 1024 = 2^{10} n(A) = 10

n(P(B)) = 32 = 2^{5} = n(B) = 5

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

15 = 10 + 5 – n(A ∩ B)

15 = 15 – n (A ∩ B)

n(A ∩ B) = 0

Question 7.

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

Answer:

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.

Answer:

Given A × A contains 16 elements.

∴ A contains 4 elements.

Also, (1, 3) and (0, 2) are two elements of A × A.

∴ 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.

Answer:

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.

Answer:

Given A × A has 16 elements.

∴ A has 4 elements.

Also S = {(a, b) ∈ A × A; a < b}

Given (-1, 2) and (0, 1) ∈ S

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

The elements of S are

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

∴ The other elements of the sets are

(-1, 0), (-1, 1) , (0, 2), (1, 2)