Question

The domain of $f\left(x\right)=\frac{1}{\sqrt{2x-1}}-\sqrt{1-x^2}$ is :

• A. $[\frac{1}{2},1]$
• B. [- 1, $\infty$]
• C. [ 1, $\infty$]
• D. none of these

Answer 1

Answer: (A)

Given, $f\left(x\right)=\frac{1}{\sqrt{2x-1}}-\sqrt{1-x^2}$
$=p\left(x\right)-q\left(x\right)$
where $p\left(x\right)=\frac{1}{\sqrt{2x-1}}$ and $q\left(x\right)=\sqrt{1+x^2}$
Now, Domain of p(x) exist when
$2x-1\ne 0$
$\Rightarrow x=\frac{1}{2}$ and $2x-1>0$
$\Rightarrow x=\frac{1}{2}$ and $x>\frac{1}{2}$
$\therefore x\in \left(\frac{1}{2},\infty \right)$
and domain of q(x) exists when
$\Rightarrow 1-x^2\ge 0\Rightarrow x^2\le 1\Rightarrow \left|x\right|\le 1$
$\therefore -1\le x\le 1$
$\therefore$ Common domain is $]\frac{1}{2},1[$