Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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 : 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}

Exercise 1.1 Class 11 Maths State Board Question 2.
Write the set {-1,1} in set builder form.
Solution:
A = {x: x2 = 1}

11th Std Maths 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

11th Maths Book Exercise 1.1 Answers 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 Answers In Tamil 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.

Samacheer Kalvi Guru 11th Maths 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

11th 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 Exercise 1.1 Answers State Board 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 Samacheer Solutions 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}

11 Maths Exercise 1.1 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 11th Maths Solution Question 1.
Write the following sets in roster form
(a) {x ∈ N; x3 < 1000}
(b) {The set of positive roots of the equation (x2 – 4) (x3 – 27) = 0}
Solution:
(a) A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
(b) B = {2, 3}

11th Maths Chapter 1 Exercise 1.1 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 Samacheer Maths Solutions 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