Question

The function $f\left(x\right)=si{n}^{-1}\left(2x-x^2\right)+\sqrt{2-\frac{1}{|x|}}+\frac{1}{\left[x^2\right]}$ 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) $-1\le 2x-x^2\le 1$ (for ${\sin }^{-1}$ to be defined) $\Rightarrow -1\le x^2-2x\le 1$
i.e. $x^2-2x+1\ge 0$ and $x^2-2x-1\le 0$
$(x-1{)}^2\ge 0$ and $(x-1{)}^2-(\sqrt{2}{)}^2\le 0$
$x\in R$ and $(x-1-\sqrt{2})(x-1+\sqrt{2})\le 0$
$\Rightarrow x\in [1-\sqrt{2},1+\sqrt{2}\mid \dots$
(ii) $2-\frac{1}{|x|}\ge 0\Rightarrow \frac{1}{|x|}\le 2\Rightarrow |x|\ge \frac{1}{2}\Rightarrow x\in \left(-\infty ,-\frac{1}{2}\right]\cup \left[\frac{1}{2},\infty \right)\dots$
(iii) $\left[x^2\right]\ne 0\Rightarrow x^2\notin [0,1)$
$\Rightarrow x\notin (-1,1)\Rightarrow x\in (-\infty ,-1]\cup [1,\infty )\dots (3)$
Hence, (1)$\cap (2)\cap (3)$
$\Rightarrow x\in [1,1+\sqrt{2}]$