The function f(x)= l 4 + ^ -1 x,|x| 1 1 2 (|x|-1),|x|>1 . - Vidyadip

MCQ

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

A
continuous on $R -\{1\}$ and differentiable on $R-\{-1,1\}$
B
both continuous and differentiable on $R-\{-1\}$
C
continuous on $R -\{-1\}$ and differentiable on $R -\{-1,1\}$
D
both continuous and differentiable on $R -\{1\}$

✓

Answer

$f(x)=\left\{\begin{array}{ccc}\frac{\pi}{4}+\tan ^{-1} x & , & x \in(-\infty,-1] \cup[1, \infty) \\ -\frac{(x+1)}{2} & , & x \in(-1,0] \\ \frac{x-1}{2} & , & x \in(0,1)\end{array}\right.$for continuity at $x=-1$
$\text{L.H.L.} =\frac{\pi}{4}-\frac{\pi}{4}=0$
$\text{R.H.L.} =0$
so, continuous at $x=-1$ for continuity at $x=1$
$\text{L.H.L.} =0$
$\text{R.H.L.} =\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}$
so, not continuous at $x=1$ For differentiability at $x =-1$
$\text{L.H.D.} =\frac{1}{1+1}=\frac{1}{2}$
$\text{R.H.D.} =-\frac{1}{2}$
so, non differentiable at $x=-1$

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