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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 : 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 : x−4x+2 = 3, x ∈ R – {-2}}
Solution:
(i) Let A = { x ∈ N : x2 < 121 and x is a prime }
A = {2, 3, 5, 7}
(ii) The set of positive roots of the equations
(x – 1) (x + 1) (x2 – 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 : x2 – 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 = 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:
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)) = 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:
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; 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}
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