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# Engineering

The function f(x)=\left\{\begin{aligned} \frac{\pi}{4}+\tan ^{-1} x,|x| \leq 1 \\ \frac{1}{2}(|x|-1),|x|>1 \end{aligned}\right. is:
Option: 1 continuous on R -{1} and differentiable on R- {-1,1}.
Option: 2 both continous and differentiable on R - {-1}
Option: 3 continous on R - {-1} and differntiable on R - {-1,-1}
Option: 4 both continous and differntiable on R - {-1}

Answers (1)
H Himanshu Meshram

 

for continuity at x = – 1

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

so, continuous at x = –1

for continuity at x = 1

L.H.L. = 0

\mathrm{R} . \mathrm{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

f(x)=\left\{\begin{aligned}{} \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{aligned}\right.

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