Question

The number of integral solution of $\sqrt{x-1}>\sqrt{3-x}$, is

Answer 1

$\sqrt{x-1}>\sqrt{3-x}$ is meaningful if
$x-1\ge 0\text{and}3-x\ge 0$
$\text{or},1\le x\le 3$
$\text{Also},\sqrt{x-1}>\sqrt{3-x}\cdots \left(1\right)$
$\because$ both the sides are $+ive$
Hence squaring both sides will not change the inequality sign
$\therefore x-1>3-x\Rightarrow x>2\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ we have
$2<x\le 3$
So, number of integral solution is $1.$