Educational

For two sets \(A \cup B = A\) iff (a) \(B\subseteq A\) (b) \(A\subseteq B\) (c) \(A\ne B\) (d) \(A=B\) Solution If \[ A\cup B=A \] then every element of \(B\) is already present in \(A\). Therefore, \[ B\subseteq A \] Answer \[ \boxed{B\subseteq A} \] Correct option: (a) Next Question / Full Exercise

If A and B are two disjoint sets, then \(n(A \cup B)\) is equal to (a) \(n(A)+n(B)\) (b) \(n(A)+n(B)-n(A\cap B)\) (c) \(n(A)+n(B)+n(A\cap B)\) (d) \(n(A)n(B)\) Solution For disjoint sets, \[ A\cap B=\Phi \] Therefore, \[ n(A\cap B)=0 \] Using the formula, \[ n(A\cup B)=n(A)+n(B)-n(A\cap B) \] \[ =n(A)+n(B)-0 \] \[ =n(A)+n(B) \] Answer \[ \boxed{n(A)+n(B)}

In set-builder method the null set is represented by (a) \(\{\}\) (b) \(\Phi\) (c) \(\{x:x\ne x\}\) (d) \(\{x:x=x\}\) Solution A null set contains no element. In set-builder form, \[ \{x:x\ne x\} \] means the set of all \(x\) such that \(x\) is not equal to itself. This is impossible for any element. Therefore, the set

Let A and B be two sets such that \[ n(A)=16,\quad n(B)=14,\quad n(A\cup B)=25 \] Then, \(n(A\cap B)\) is equal to (a) 30 (b) 50 (c) 5 (d) none of these Solution Using the formula, \[ n(A\cup B)=n(A)+n(B)-n(A\cap B) \] Substituting the values, \[ 25=16+14-n(A\cap B) \] \[ 25=30-n(A\cap B) \] \[ n(A\cap B)=30-25 \]

Let A and B be two sets such that \[ n(A)=16,\quad n(B)=14,\quad n(A\cup B)=25 \] Then, \(n(A\cap B)\) is equal to (a) 30 (b) 50 (c) 5 (d) none of these Solution Using the formula, \[ n(A\cup B)=n(A)+n(B)-n(A\cap B) \] Substituting the values, \[ 25=16+14-n(A\cap B) \] \[ 25=30-n(A\cap B) \] \[ n(A\cap B)=30-25 \]

Let U be the universal set containing 700 elements. If A, B are sub-sets of U such that \[ n(A)=200,\quad n(B)=300,\quad n(A\cap B)=100 \] Then, \[ n(A’ \cap B’)= \] (a) 400 (b) 600 (c) 300 (d) none of these Solution Using De Morgan’s law, \[ A’\cap B’=(A\cup B)’ \] Therefore, \[ n(A’\cap B’)=n(U)-n(A\cup B)

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 \[ \boxed{(A\cup B)-(A\cap B)} \]