Question

Let $f:R\to (0,\pi )$ be defined as $f\left(x\right)={\cot }^{-1}\left(\frac{2-|x|}{2+|x|}\right).$ Then which of the following statements is(are) TRUE?

• A. $f\left(x\right)$ is neither injective nor surjective.
• B. $f\left(x\right)$ is continuous and differentiable on $R$.
• C. $f\left(x\right)$ is both even function and non-periodic function.
• D. $\underset{x\to \frac{-2}{\sqrt{3}}}{lim}f\left(x\right)=\frac{5\pi }{12}$.

Answer 1

Answer: (A), (C), (D)

$\underset{x\to \frac{-2}{\sqrt{3}}}{lim}f\left(x\right)=\frac{5\pi }{12}$.

$f\left(x\right)$ is an even function.
Hence, $f$ is not injective.
We have $f\left(x\right)={\cot }^{-1}\left(\frac{2-|x|}{2+|x|}\right)$
$=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{2-|x|}{2+|x|}\right)=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1-\frac{|x|}{2}}{1+\frac{|x|}{2}}\right)$
$=\frac{\pi }{2}-({\tan }^{-1}1-{\tan }^{-1}\frac{|x|}{2})f\left(x\right)=\frac{\pi }{4}+{\tan }^{-1}\frac{|x|}{2}$
Since ${\tan }^{-1}\frac{|x|}{2}\in [0,\frac{\pi }{2}),$
$\therefore f\left(x\right)\in [\frac{\pi }{4},\frac{3\pi }{4})$
Hence, $f\left(x\right)$ is not surjective.

$f\left(x\right)=\{\begin{array}{cc}\frac{\pi }{4}-{\tan }^{-1}\frac{x}{2},& x\le 0\\ \frac{\pi }{4}+{\tan }^{-1}\frac{x}{2},& x>0\end{array}{f}^{\prime }\left(x\right)=\{\begin{array}{cc}-\frac{1}{1+(x/2{)}^2}\cdot \frac{1}{2},& x\le 0\\ \frac{1}{1+(x/2{)}^2}\cdot \frac{1}{2},& x>0\end{array}$

$f^{\prime }\left(0^{+}\right)=\frac{1}{2}$ and $f^{\prime }\left(0^{-}\right)=\frac{-1}{2}$
Clearly, $f\left(x\right)$ is non-differentiable at $x=0$

Also, $f$ is non-periodic.

$\underset{x\to \frac{-2}{\sqrt{3}}}{lim}f\left(x\right)=\underset{x\to \frac{-2}{\sqrt{3}}}{lim}(\frac{\pi }{4}-{\tan }^{-1}\frac{x}{2})=\frac{5\pi }{12}$