Question

The domain of the derivative of the function $f\left(x\right)={\tan }^{-1}x$. if $\left|x\right|\le 1=\left(\frac{1}{2}\right)\left[\left|x\right|-1\right]$ if $\left|x\right|>1$ is

• A. $R-\left\{0\right\}$
• B. $R-\left\{1\right\}$
• C. $R-\left\{-1\right\}$
• D. $R-\left\{-1,1\right\}$

Answer 1

The correct option is D

$R-\left\{-1,1\right\}$

Find the domain of the function $f\left(x\right)$:

Consider the given function as,

$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{ll}{\tan }^{-1}x& \text{if}\left|x\right|\le 1\\ \frac{1}{2}\left(\left|x\right|-1\right)& \text{if}\left|x\right|>1\end{array}\right. \\ f\left(x\right)=\left\{\begin{array}{ll}-\frac{1}{2}\left(x+1\right)& \text{if}x\le -1\\ {\tan }^{-1}x& \text{if}-1\le x\le 1\\ \frac{1}{2}\left(x-1\right)& x>1\end{array}\right. \end{gathered}$$

At, $x=-1$

LHL, $-\frac{1}{2}\left(-1+1\right)=0$

RHL, ${\tan }^{-1}\left(-1\right)=-\frac{\mathrm{\pi }}{4}$

Thus, $x=-1$is a discontinues function so that it is not differentiable.

At, $x=1$

LHL, ${\tan }^{-1}\left(1\right)=\frac{\mathrm{\pi }}{4}$

RHL, $\frac{1}{2}\left(1-1\right)=0$

Thus, $x=1$is a discontinues function so that it is not differentiable.

Hence, the domain of the function is $R-\left\{-1,1\right\}$

Therefore, the correct answer is option (D).