Question

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

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

Answer 1

Answer: (A)

$\bigg[ \frac{1}{2} , 1 \bigg]$

Answer 2

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 $\Rightarrow 1 -x^{2} \ge0 \Rightarrow x^{2 } \le1 \Rightarrow \left|x\right| \le1$ $\therefore -1 \le x \le1$ $\therefore$ Common domain is $\bigg] \frac{1}{2} , 1 \bigg[$