Functions Question 196
Question: The function $ y={e^{-|x|}} $ is
[AMU 2000]
Options:
A) Continuous and differentiable at $ x=0 $
B) Neither continuous nor differentiable at $ x=0 $
C) Continuous but not differentiable at $ x=0 $
D) Not continuous but differentiable at $ x=0 $
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Answer:
Correct Answer: C
Solution:
We have, $ f(x)= \begin{cases} {e^{-x}}, & x\ge 0 \\ e^{x}, & x<0 \\ \end{cases} . $ Clearly, $ f(x) $ is continuous and differentiable for all non zero x. Now $ \underset{x\to 0-}{\mathop{\lim }},f(x)=\underset{x\to 0}{\mathop{\lim }},e^{x}=1 $ , $ \underset{x\to {0^{+}}}{\mathop{\lim }},f(x)=\underset{x\to {0^{+}}}{\mathop{\lim }},f(x){e^{-x}}=1 $ Also, $ f(0)=e^{0}=1 $ . So, $ f(x) $ is continuous for all x. (LHD at $ x=0) $ $ ={{( \frac{d}{dx}(e^{x}) )}_{x=0}}=1 $ (RHD at $ x=0) $ $ f(x) $ So, $ \underset{x\to 7}{\mathop{\lim }},\frac{2-\sqrt{x-3}}{x^{2}-49} $ is not differentiable at $ L,{f}’,(1)=\underset{h\to 0}{\mathop{\lim }},\frac{f(1-h)-f(1)}{-h} $ . Hence $ f(x)={e^{-,|,x,|}} $ is everywhere continuous but not differentiable at $ x=0 $ .
BETA
