Question

Let $A=\{y:y={\log }_{2}x,x<16,x,y\in N\}$ and $B=\{x:x^2-7x+12=0\}.$ Then $(A\cup B)\times (A\cap B)$ is

• A. $\{3,3\}$
• B. $\left\{\right(1,3),(2,3),(3,3),(4,3\left)\right\}$
• C. $\left\{\right(1,3),(2,3),(3,3\left)\right\}$
• D. $\left\{\right(1,3),(2,3),(3,3),(1,4),(2,4),(3,4\left)\right\}$

Answer 1

The correct option is B $\left\{\right(1,3),(2,3),(3,3),(4,3\left)\right\}$

$A=\{y:y={\log }_{2}x,x<16,x,y\in N\}$
For $x<16,{\log }_{2}x<4$
$\therefore y\in (-\infty ,4)$ and $y\in N$
$\therefore A=\{1,2,3\}$
$B=\{x:x^2-7x+12=0\}\Rightarrow B=\{3,4\}$
$A\cup B=\{1,2,3,4\}A\cap B=\left\{3\right\}\therefore (A\cup B)\times (A\cap B)=\left\{\right(1,3),(2,3),(3,3),(4,3\left)\right\}$