Question

Let $Z$ be the set of integers. If $A=\{x\in Z:{\log }_{10}(x^2-7x+13)=0\}$ and $B=\{x\in Z:(x+2\left)\right(x-3\left)\right(x^2-1)\le 0\}$ then the number of subsets of set $A\times B$ is

• A. $2^{10}-1$
• B. $2^8$
• C. $2^{10}$
• D. $2^8+1$

Answer 1

The correct option is C $2^{10}$

If ${\log }_{10}(x^2-7x+13)=0$
$\Rightarrow x^2-7x+13=1$
$\Rightarrow x^2-7x+12=0$
$\Rightarrow x^2-3x-4x+12=0$
$\Rightarrow (x-3)(x-4)=0$
$A=\{3,4\}$
So $A$ has $2$ elements.

Now,
$(x+2)(x-3)(x^2-1)\le 0$
$(x+2)(x-3)(x-1)(x+1)\le 0$
$B=\{-2,-1,1,2,3\}$
So $B$ has $5$ elements.

$A\times B$ has 10 elements
So the number of subsets is $2^{10}$