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Question:

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

  1. continuous on $\mathrm{R}-\{1\}$ and differentiable on

    $\mathrm{R}-\{-1,1\}$.

  2. both continuous and differentiable on

    $R-\{-1\}$.

  3. continuous on $R-\{-1\}$ and differentiable on

    $\mathrm{R}-\{-1,1\}$.

  4. both continuous and differentiable on

    $\mathrm{R}-\{1\}$


Correct Option: 1

Solution:

for continuity at $\mathrm{x}=-1$

L.H.L. $=\frac{\pi}{4}-\frac{\pi}{4}=0$

R.H.L. $=0$

so, continuous at $x=-1$

for continuity at $x=1$

L.H.L. $=0$

R.H.L. $=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}$

so, not continuous at $x=1$

For differentiability at $x=-1$

L.H.D. $=\frac{1}{1+1}=\frac{1}{2}$

R.H.D. $=-\frac{1}{2}$

so, non differentiable at $x=-1$

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