Functions Question 19
Question: If $ f(x)= \begin{cases} & x\frac{{e^{(1/x)}}-{e^{(-1/x)}}}{{e^{(1/x)}}+{e^{(-1/x)}}},,x\ne 0 \\ & ,0,,x=0 \\ \end{cases} . $ then which of the following is true
[Kurukshetra CEE 1998]
Options:
A) f is continuous and differentiable at every point
B) f is continuous at every point but is not differentiable
C) f is differentiable at every point
D) f is differentiable only at the origin
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(0+0)=\underset{h\to 0}{\mathop{\lim }},f(x)=\underset{h\to 0}{\mathop{\lim }},f(0+h) $ $ =\underset{h\to 0}{\mathop{\lim }},(0+h),\frac{{e^{1/0+h}}-{e^{-1/0+h}}}{{e^{1/0+h}}+{e^{-1/0+h}}} $ $ =\underset{h\to 0}{\mathop{\lim }},h\frac{{e^{1/h}}-{e^{-1/h}}}{{e^{1/h}}+{e^{-1/h}}} $ =0 and $ f(0-0)=\underset{h\to 0}{\mathop{\lim }},f(0-h)=\underset{h\to 0}{\mathop{\lim }},-h\frac{{e^{-1/h}}-{e^{1/h}}}{{e^{-1/h}}+{e^{1/h}}}=0 $ and $ f(0)=0 $ ;
$ \therefore ,f(0+0)=f(0-0)=f(0) $ Hence f is continuous at $ x=0. $ At remaining points $ f(x) $ is obviously continuous. Thus it is everywhere continuous. Again, $ L,{f}’(0)=\underset{h\to 0}{\mathop{\lim }},\frac{f(0-h)-f(0)}{-h} $ $ =\underset{h\to 0}{\mathop{\lim }}\frac{h,.,\frac{{e^{-1/h}}-{e^{1/h}}}{{e^{-1/h}}+{e^{1/h}}}-0}{-h}=-1 $ $ R,{f}’(0)=\underset{h\to 0}{\mathop{\lim }}\frac{f,(0+h)-f,(0)}{h} $ $ =\underset{h\to 0}{\mathop{\lim }},\frac{h\frac{{e^{1/h}}-{e^{-1/h}}}{{e^{1/h}}+{e^{-1/h}}}}{h}=1 $ $ \because L{f}’(0)\ne R,{f}’(0) $
$ \therefore f $ is not differentiable at $ x=0 $ .
BETA
