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 | \geq 0 $ or $| x | \leq 2 \Rightarrow -2 \leq x \leq 2$
because of $\frac{1}{x^2} \Rightarrow x \neq 0$

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 $