Question

Given $A=\{x\in N:x<6\},B=\{3,6,9\}$ and $C=\{x\in N:2x-5\le 8\}$

• A. $A\cup$ (B $\cap$C)=(A $\cap$B) $\cap$(A $\cap$ C)
• B. (A $\cup$B)'=A'$\cap$B'
• C. A $\cup$B=null set
• D. None of the above

Answer 1

The correct option is B (A $\cup$B)'=A'$\cap$B'

$A=\{x\in N:x<6\}$

$\therefore A=\{1,2,3,4,5\}$

$B=\{3,6,9\}$

$C=\{x\in N:2x-5\le 8\}$

$\Rightarrow C=\{1,2,3,4,5,6\}$

Option A. $A\cup (B\cap C)=A\cup \{3,6\}=\{1,2,3,4,5,6\}$

RHS $=(A\cap B)\cap (A\cup C)=\left\{3\right\}\cap \{1,2,3,4,5,6\}=\left\{3\right\}$

LHS $\ne$ RHS

Option B. $(A\cup B{)}^{\prime }=A^{\prime }\cap B^{\prime }$ is always true by De Morgan's Law .

$A\cup B=\{1,2,3,4,5,6,9\}$

$(A\cup B{)}^{\prime }=\{7,8,10,11,...\}$

$A^{\prime }=\{x\in N:x\ge 7\}$ and $B^{\prime }=\{1,2,4,5,7,8,10,11,\mathrm{12...}\}$

$A^{\prime }\cap B^{\prime }=\{7,8,10,11,\mathrm{12...}\}$

$\therefore (A\cup B{)}^{\prime }=A^{\prime }\cap B^{\prime }$

Option C. $A\cap B=\left\{3\right\}\ne$ null set

Hence, only B is correct.