Question

The function $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{\mathrm{\pi }}{4}+{\tan }^{-1}\mathrm{x},\left|\mathrm{x}\right|\le 1\\ \frac{1}{2}\left(\left|\mathrm{x}\right|-1\right),\left|\mathrm{x}\right|>1\end{array}\right.$is

• A. both continuous and differentiable on $\mathrm{R}-\left\{-1\right\}$
• B. continuous on$\mathrm{R}-\left\{-1\right\}$ and differentiable on $\mathrm{R}-\left\{-1,1\right\}$
• C. continuous on $\mathrm{R}-\left\{1\right\}$ and differentiable on $\mathrm{R}-\left\{-1,1\right\}$
• D. both continuous and differentiable on $\mathrm{R}-\left\{1\right\}$

Answer 1

The correct option is C

continuous on $\mathrm{R}-\left\{1\right\}$ and differentiable on $\mathrm{R}-\left\{-1,1\right\}$

Explanation for correct option

Step 1. Check the function for continuity.

It is given that the function,

$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{\mathrm{\pi }}{4}+{\tan }^{-1}\mathrm{x},\left|\mathrm{x}\right|\le 1\\ \frac{1}{2}\left(\left|\mathrm{x}\right|-1\right),\left|\mathrm{x}\right|>1\end{array}\right.$

For continuity at $\mathrm{x}=-1$.

Left Hand Limit (L.H.L.) $=\left(\frac{\mathrm{\pi }}{4}\right)-\left(\frac{\mathrm{\pi }}{4}\right)$

$=0$

Right Hand Limit (R.H.L.) $=0$

So, the limits are continuous at $\mathrm{x}=-1$.

Now, we will check for continuity at $\mathrm{x}=1$.

L.H.L. $=0$

R.H.L. $=\left(\frac{\mathrm{\pi }}{4}\right)+\left(\frac{\mathrm{\pi }}{4}\right)$

$=\left(\frac{\mathrm{\pi }}{2}\right)$

So, the limits are not continuous at $\mathrm{x}=1$.

Step 2. Check the function for differentiability

For differentiability at $\mathrm{x}=-1$.

Left Hand Derivative (L.H.D.) $=\frac{1}{\left(1+1\right)}$

$=\frac{1}{2}$

Right Hand Derivative (R.H.D.) $=-\frac{1}{2}$

So, both derivative are not differentiable at $\mathrm{x}=-1$.

Thus,

$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{\mathrm{\pi }}{4}+{\tan }^{-1}\mathrm{x},\left|\mathrm{x}\right|\le 1\\ \frac{1}{2}\left(\left|\mathrm{x}\right|-1\right),\left|\mathrm{x}\right|>1\end{array}\right.=\left\{\begin{array}{l}\frac{\mathrm{\pi }}{4}+{\tan }^{-1}\mathrm{x},\mathrm{x}\in (-\infty ,-1]\cup [1,\infty )\\ \begin{array}{l}-\frac{\left(\mathrm{x}+1\right)}{2},\mathrm{x}\in (-1,0]\\ \frac{\mathrm{x}-1}{2},\mathrm{x}\in \left(0,1\right)\end{array}\end{array}\right.$

So, the given function is continuous on $\mathrm{R}-\left\{1\right\}$ and differentiable on $\mathrm{R}-\left\{-1,1\right\}$.

Hence, Option $\left(\mathrm{C}\right)$ is correct .