Question

If $A=\{x,x\in Z\text{and}\frac{x^2-4x+3}{x^2-8x+15}\le 0\}$, then $n\left(A\right)=$

Answer 1

$\frac{x^2-4x+3}{x^2-8x+15}\le 0\Rightarrow \frac{(x-1)(x-3)}{(x-3)(x-5)}\le 0\Rightarrow \frac{(x-1)}{(x-5)}\le 0,\text{where}x\ne 3\Rightarrow x\in [1,5)-\left\{3\right\}$
Since, $x\in Z$
$\Rightarrow x\in \{1,2,4\}$
$\therefore n\left(A\right)=3$