Question

If $x>\sqrt{1-x},$ then greatest value of $x$ is

Answer 1

Given $x>\sqrt{1-x}$

For $\sqrt{1-x}$ to exists,
$1-x\ge 0\Rightarrow x\le 1\dots \left(i\right)$

As $\sqrt{1-x}$ is always positive, $x$ which is greater than $\sqrt{1-x},$ will also be a positive quantity
$\Rightarrow x>0\dots \left(ii\right)$

Squaring inequality both the sides, we get
$\Rightarrow x^2>1-x\Rightarrow x^2+x-1>0$
$\Rightarrow (x-\frac{\sqrt{5}-1}{2})(x+\frac{\sqrt{5}+1}{2})>0$
$\Rightarrow x\in (-\infty ,\frac{-\sqrt{5}-1}{2})\cup (\frac{\sqrt{5}-1}{2},\infty )\dots \left(iii\right)$

Hence, final solution is
$\left(i\right)\cap \left(ii\right)\cap \left(iii\right)$
$\Rightarrow x\in (\frac{\sqrt{5}-1}{2},1]$