Q47 / 100 · BITSAT 2023 (Memory Based Paper 2) · Mathematics
The domain of the function $f(x)=\sqrt{x-\sqrt{1-x^2}}$ is
- A$\left[-1,-\frac{1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, 1\right]$
- B$[-1,1]$
- C$\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{1}{\sqrt{2}},+\infty\right)$
- Correct answerD$\left[\frac{1}{\sqrt{2}}, 1\right]$
Solution
For $f(x)$ to be defined, we must have
$x-\sqrt{1-x^2} \geq 0 \text { or } x \geq \sqrt{1-x^2}>0 \text { or } x^2 \geq 1-x^2$
or $x^2 \geq \frac{1}{2}$.
Now, $x^2 \geq \frac{1}{2} \Rightarrow\left(x-\frac{1}{\sqrt{2}}\right)\left(x+\frac{1}{\sqrt{2}}\right) \geq 0$
$\Rightarrow x \leq-\frac{1}{\sqrt{2}} \quad \text { or } \quad x \geq \frac{1}{\sqrt{2}}$
Also, $1-x^2 \geq 0$ or $x^2 \leq 1$.
Also, $x^2 \leq 1 \Rightarrow(x-1)(x+1) \leq 0 \Rightarrow-1 \leq x \leq 1$
Thus, $x>0, x^2 \geq \frac{1}{2}$ and $x^2 \leq 1 \Rightarrow x \in\left[\frac{1}{\sqrt{2}}, 1\right]$
$x-\sqrt{1-x^2} \geq 0 \text { or } x \geq \sqrt{1-x^2}>0 \text { or } x^2 \geq 1-x^2$
or $x^2 \geq \frac{1}{2}$.
Now, $x^2 \geq \frac{1}{2} \Rightarrow\left(x-\frac{1}{\sqrt{2}}\right)\left(x+\frac{1}{\sqrt{2}}\right) \geq 0$
$\Rightarrow x \leq-\frac{1}{\sqrt{2}} \quad \text { or } \quad x \geq \frac{1}{\sqrt{2}}$
Also, $1-x^2 \geq 0$ or $x^2 \leq 1$.
Also, $x^2 \leq 1 \Rightarrow(x-1)(x+1) \leq 0 \Rightarrow-1 \leq x \leq 1$
Thus, $x>0, x^2 \geq \frac{1}{2}$ and $x^2 \leq 1 \Rightarrow x \in\left[\frac{1}{\sqrt{2}}, 1\right]$
Hence, the correct answer is option: 'D'.