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

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