Assertion (A): Roster form of the set A = {x : x ∈ I, x2 < 49} is A = {−6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6}.
Reason (R): Squares of integers 0, ±1, ±2, ±3, ±4, ±5 and ±6 are less than 49.
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Answer
The squares of the integers 0, ±1, ±2, ±3, ±4, ±5 and ±6 are
0, 1, 4, 9, 16, 25 and 36
and each one of these is less than 49, while (±7)2 = 49 which is not less than 49.
∴ Reason (R) is true.
As x ∈ I and x2 < 49, the only integers x which satisfy the condition are the ones listed above i.e. 0, ±1, ±2, ±3, ±4, ±5 and ±6.
∴ A = {−6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6}
∴ Assertion (A) is true.
The roster form of A is obtained exactly by picking out those integers whose squares are less than 49, which is what Reason (R) states.
Hence, Assertion (A) and Reason (R) are both true and Reason (R) is the correct explanation of Assertion (A).
Assertion (A): If A = {1, 2, 3} then the cardinal number of the power set of A is 9.
Reason (R): Number of subsets of a set containing n elements = 2n.
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Answer
We know that the number of subsets of a set containing n elements is 2n.
∴ Reason (R) is true.
Given,
A = {1, 2, 3}, so n(A) = 3.
The power set P(A) is the set formed by all the subsets of A, so the cardinal number of P(A) is the number of subsets of A.
∴ n(P(A)) = 2n(A) = 23 = 8, and not 9.
∴ Assertion (A) is false.
Hence, Assertion (A) is false, but Reason (R) is true.
Assertion (A): If A ⊂ B then A ∩ B = A.
Reason (R): If A and B are two sets, then A ⊂ B means every element of A is also an element of B.
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Answer
Given,
A ⊂ B, thus every member of A is also a member of B.
∴ Reason (R) is true.
So, the elements common to A and B (i.e. A ∩ B) will be equal to the members of A.
∴ A ∩ B = A
∴ Assertion (A) is true.
Hence, Assertion (A) and Reason (R) are both true and Reason (R) is the correct explanation of Assertion (A).
Assertion (A): A − B = A − (A ∩ B)
Reason (R): A − B = (A ∪ B) − B.
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Answer
Assertion (A): A − B = A − (A ∩ B)
Solving R.H.S.,
A − (A ∩ B) = A ∩ (A ∩ B)' (∵ X − Y = X ∩ Y')
= A ∩ (A' ∪ B') (De Morgan law)
= (A ∩ A') ∪ (A ∩ B') (Distributive law)
= φ ∪ (A ∩ B') (∵ A ∩ A' = φ)
= A ∩ B' = A − B
∴ Assertion (A) is true.
Reason (R): A − B = (A ∪ B) − B
Solving R.H.S.,
(A ∪ B) − B = (A ∪ B) ∩ B' (∵ X − Y = X ∩ Y')
= (A ∩ B') ∪ (B ∩ B') (Distributive law)
= (A ∩ B') ∪ φ (∵ B ∩ B' = φ)
= A ∩ B' = A − B
∴ Reason (R) is true.
Both are correct results about A − B, but Reason (R) removes B from the bigger set A ∪ B, whereas Assertion (A) removes A ∩ B from A itself. So Reason (R) is not used anywhere in proving Assertion (A).
Hence, Assertion (A) and Reason (R) are both true, but Reason (R) is not the correct explanation of Assertion (A).
Assertion (A): If A ⊂ B then A − B = φ.
Reason (R): If A and B are disjoint sets, then n(A ∪ B) = n(A) + n(B).
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Answer
Given,
A ⊂ B, thus every element of A is also an element of B.
So, there is no element which belongs to A but does not belong to B.
∴ A − B = φ
∴ Assertion (A) is true.
We know that,
n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
If A and B are disjoint sets, then A ∩ B = φ so that n(A ∩ B) = 0.
∴ n(A ∪ B) = n(A) + n(B)
∴ Reason (R) is true.
Reason (R) is a result about the cardinal number of the union of two disjoint sets, while Assertion (A) is about the difference of two sets one of which is contained in the other. So Reason (R) is not used in proving Assertion (A).
Hence, Assertion (A) and Reason (R) are both true, but Reason (R) is not the correct explanation of Assertion (A).