Question

If $A=\{x\in N:{\log }_{2}x+\frac{3}{{\log }_{2}x}<4,x>1\}$ and $B=\{x\in Z:(4-x^2\left)\right(x^2-8x+15)\ge 0\},$ then the number of subsets of the set $(A\cup B)-(A\cap B)$ is

• A. $32$
• B. $64$
• C. $128$
• D. $256$

Answer 1

The correct option is C $128$

$A=\{x\in Z:{\log }_{2}x+\frac{3}{{\log }_{2}x}<4,x>1\}$
Let ${\log }_{2}x=t$
$\Rightarrow t+\frac{3}{t}<4$
$\Rightarrow t^2-4t+3<0[\because t>0]$
$\Rightarrow (t-1)(t-3)<0$
$\Rightarrow t\in (1,3)$
$\Rightarrow {\log }_{2}x\in (1,3)$
$\Rightarrow x\in (2,8)$
$\therefore A=\{3,4,5,6,7\}$

$B=\{x\in Z:(4-x^2\left)\right(x^2-8x+15)\ge 0\}$
$(2-x)(2+x)(x-3)(x-5)\ge 0$
$\Rightarrow x\in [-2,2]\cup [3,5]$
$\therefore B=\{-2,-1,0,1,2,3,4,5\}$

$A\cup B=\{-2,-1,0,1,2,3,4,5,6,7\}$
$A\cap B=\{3,4,5\}$
$n\left[\right(A\cup B)-(A\cap B\left)\right]=7$
Then, the number of subsets is $2^7=128.$