Let A = {x : x ∈ R, x > 4} and B = {x ∈ R : x < 5}. Then, A ∩ B = (a) \((4,5]\) (b) \((4,5)\) (c) \([4,5)\) (d) \([4,5]\) Solution \[ A=\{x:x>4\}=(4,\infty) \] \[ B=\{x:x

For any three sets A, B and C (a) \(A\cap(B-C)=(A\cap B)-(A\cap C)\) (b) \(A\cap(B-C)=(A\cap B)-C\) (c) \(A\cup(B-C)=(A\cup B)\cap(A\cup C’)\) (d) \(A\cup(B-C)=(A\cup B)-(A\cup C)\) Solution \[ B-C=B\cap C’ \] Therefore, \[ A\cap(B-C) \] \[ =A\cap(B\cap C’) \] \[ =(A\cap B)\cap C’ \] \[ =(A\cap B)-C \] Hence, option (b) is correct. Answer \[ \boxed{A\cap(B-C)=(A\cap B)-C} \]

Which of the following statement is false : (a) \(A-B=A\cap B’\) (b) \(A-B=A-(A\cap B)\) (c) \(A-B=A-B’\) (d) \(A-B=(A\cup B)-B\) Solution Using set identities, \[ A-B=A\cap B’ \] So, option (a) is true. Also, \[ A-(A\cap B)=A\cap(A\cap B)’ \] \[ = A\cap(A’\cup B’) \] \[ = A\cap B’ \] \[ = A-B \] Hence, option (b)

For any two sets A and B, (A − B) ∪ (B − A) = (a) \((A-B)\cup A\) (b) \((B-A)\cup B\) (c) \((A\cup B)-(A\cap B)\) (d) \((A\cup B)\cap(A\cap B)\) Solution By definition, \[ (A-B)\cup(B-A)=A\Delta B \] Also, \[ A\Delta B=(A\cup B)-(A\cap B) \] Therefore, \[ (A-B)\cup(B-A)=(A\cup B)-(A\cap B) \] Answer \[ \boxed{(A\cup B)-(A\cap B)} \]

The symmetric difference of A = {1, 2, 3} and B = {3, 4, 5} is (a) \(\{1,2\}\) (b) \(\{1,2,4,5\}\) (c) \(\{4,3\}\) (d) \(\{2,5,1,4,3\}\) Solution \[ A=\{1,2,3\}, \qquad B=\{3,4,5\} \] \[ A-B=\{1,2\} \] \[ B-A=\{4,5\} \] Symmetric difference: \[ A\Delta B=(A-B)\cup(B-A) \] \[ =\{1,2\}\cup\{4,5\} \] \[ =\{1,2,4,5\} \] Answer \[ \boxed{\{1,2,4,5\}} \] Correct option: (b)

“`html The symmetric difference of A and B is not equal to (a) \((A-B) \cap (B-A)\) (b) \((A-B) \cup (B-A)\) (c) \((A \cup B)-(A \cap B)\) (d) \(\{(A \cup B)-A\} \cup \{(A \cup B)-B\}\) Solution Symmetric difference is defined as \[ A \Delta B = (A-B)\cup(B-A) \] Also, \[ A \Delta B=(A\cup B)-(A\cap B) \]

If A = {1, 3, 5, B} and B = {2, 4}, then (a) \(4 \in A\) (b) \(\{4\} \subset A\) (c) \(B \subset A\) (d) none of these Solution Given, \[ A=\{1,3,5,B\} \] and \[ B=\{2,4\} \] Here, the set \(B\) itself is an element of \(A\), not the elements \(2\) and \(4\). Therefore,

For any two sets A and B, A ∩ (A ∪ B) = (a) \(A\) (b) \(B\) (c) \(\phi\) (d) none of these Solution Using absorption law, \[ A \cap (A \cup B)=A \] Answer \[ \boxed{A} \] Correct option: (a) Next Question / Full Exercise

The number of subsets of a set containing n elements is (a) \(n\) (b) \(2^n-1\) (c) \(n^2\) (d) \(2^n\) Solution If a set contains \(n\) elements, then each element has two choices: either included in a subset or not included in a subset Therefore, total number of subsets \[ =2 \times 2 \times 2 \cdots

Let A and B be two sets in the same universal set. Then, A − B = (a) \(A \cap B\) (b) \(A’ \cap B\) (c) \(A \cap B’\) (d) none of these Solution By definition, \[ A-B \] means the elements which belong to \(A\) but do not belong to \(B\). Therefore, \[ A-B=A