Question

Let $f(x)={\tan }^{-1}(\sqrt{[x]+[-x]})+\sqrt{2-|x|}+\frac{1}{x^2}$. Number of integers in the range of $f(x)$, is [Note: [y] denotes greatest integer function less than or equal to $y$.]

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

$2-|x|\ge 0$ or $|x|\le 2\Rightarrow -2\le x\le 2$
because of $\frac{1}{x^2}\Rightarrow x\ne 0$

Hence $x$ can only be integer
$\therefore x\in -2,-1,1,2$
$f(1)=2;f(2)=\frac{1}{4}$
Also $f$ is even. Hence range is $\left\{\frac{1}{4},2\right\}$
$\Rightarrow \text{ number of integers is }1$