Question

The domain of $f\left(x\right)=\sqrt{(\frac{1}{\sqrt{x}}-\sqrt{(x+1)})}$ is

• A. $x>-1$
• B. $(-1,\infty )$ \ $\left\{0\right\}$
• C. $(0,\frac{\sqrt{5}-1}{2}]$
• D. $[\frac{1-\sqrt{5}}{2},0)$

Answer 1

The correct option is C $(0,\frac{\sqrt{5}-1}{2}]$

$f\left(x\right)=\sqrt{\frac{1}{\sqrt{x}}-\sqrt{x+1}}x>0,x+1\ge 0\Rightarrow x\ge -1\frac{1}{\sqrt{x}}-\sqrt{x+1}\ge 0\Rightarrow \frac{1}{x}\ge x+1\Rightarrow x^2+x-1\le 0\Rightarrow x=\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2}\therefore x\in (0,\frac{\sqrt{5}-1}{2}]$