Question

The function $f\left(x\right)=sin^{- 1}\left(2 x - x^{2}\right)+\sqrt{2 - \frac{1}{\left|x\right|}}+\frac{1}{\left[x^{2}\right]}$ is defined in the interval (where $\left[\cdot \right]$ is the greatest integer function)

• A. $x\in \left(1 - \sqrt{2} , 1\right)$
• B. $x\in \left[1,1 + \sqrt{2}\right]$
• C. $x\in \left[1 - \sqrt{2} , 1 + \sqrt{2}\right]$
• D. $x\in \left[1 - \sqrt{2} , 2\right]$

Answer 1

Answer: (B)

(i) $-1 \leq 2 x - x ^{2} \leq 1$ (for $\sin ^{-1}$ to be defined) $\Rightarrow -1 \leq x^{2}-2 x \leq 1$
i.e. $x^{2}-2 x+1 \geq 0$ and $x^{2}-2 x-1 \leq 0$
$(x-1)^{2} \geq 0$ and $(x-1)^{2}-(\sqrt{2})^{2} \leq 0$
$x \in R$ and $(x-1-\sqrt{2})(x-1+\sqrt{2}) \leq 0$
$\Rightarrow x \in[1-\sqrt{2}, 1+\sqrt{2} \mid \ldots$
(ii) $2-\frac{1}{|x|} \geq 0 \Rightarrow \frac{1}{|x|} \leq 2 \Rightarrow |x| \geq \frac{1}{2} \Rightarrow x \in\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{1}{2}, \infty\right) \ldots$
(iii) $\left[ x ^{2}\right] \neq 0 \Rightarrow x ^{2} \notin[0,1)$
$\Rightarrow x \notin(-1,1) \Rightarrow x \in(-\infty,-1] \cup[1, \infty) \ldots(3)$
Hence, (1)$\cap(2) \cap(3)$
$\Rightarrow x \in[1,1+\sqrt{2}]$