Question

The number of integers in solution set of $\sqrt{x}-3\le \frac{2}{\sqrt{x}-2}$ is

Answer 1

Given $\sqrt{x}-3\le \frac{2}{\sqrt{x}-2}$ will be valid
if $x\ge 0,x\ne 4$

Let $\sqrt{x}=t,t\ge 0$
$t-3\le \frac{2}{t-2}$
$\Rightarrow (t-3)-\frac{2}{t-2}\le 0$
$\frac{(t-3)(t-2)-2}{t-2}\le 0$
$\frac{t^2-5t+4}{t-2}\le 0$
$\frac{(t-1)(t-4)}{t-2}\le 0,(t\ne 2)$
$\Rightarrow t\in [0,1]\cup (2,4]\therefore x\in [0,1]\cup (4,16]$