(1) Write the following in roster form.
(i) {x ∈ N : x2 < 121 and x is a prime}. Solution
(ii) the set of all positive roots of the equation (x − 1)(x + 1)(x2 − 1) = 0. Solution
(iii) {x ∈ N : 4x + 9 < 52}. Solution
(iv) {x : (x−4)/(x+2) = 3, x ∈ R − {−2}}. Solution
(2) Write the set {−1, 1} in set builder form. Solution
(3) State whether the following sets are finite or infinite.
(i) {x ∈ N : x is an even prime number}. Solution
(ii) {x ∈ N : x is an odd prime number}. Solution
(iii) {x ∈ Z : x is even and less than 10}. Solution
(iv) {x ∈ R : x is a rational number}. Solution
(v) {x ∈ N : x is a rational number}. Solution
(4) By taking suitable sets A,B,C, verify the following results:
(i) A × (B ∩ C) = (A × B) ∩ (A × C). Solution
(ii) A × (B ∪ C) = (A × B) ∪ (A × C) Solution
(iii) (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A) Solution
(iv) C − (B − A) = (C ∩ A) ∪ (C ∩ B'). Solution
(v) (B − A) ∩ C = (B ∩ C) − A = B ∩ (C − A). Solution
(vi) (B − A) ∪ C = (B ∪ C) − (A − C) Solution
(5) Justify the trueness of the statement:
“An element of a set can never be a subset of itself.”
(6) If n(P(A)) = 1024, n(A ∪ B) = 15 and n(P(B)) = 32, then find n(A ∩ B). Solution
(7) If n(A ∩ B) = 3 and n(A ∪ B) = 10, then find n(P(AΔB)). Solution
(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.
(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
(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
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