Functions Question 79
Question: If $ f(x),=, \begin{cases} x{e^{-,( \frac{1}{|,x,|},+,\frac{1}{x} )}}, & x\ne 0 \\ 0,, & x=0 \\ \end{cases} . $ , then $ f(x), $ is
[AIEEE 2003]
Options:
A) Continuous as well as differentiable for all x
B) Continuous for all x but not differentiable at $ x=0 $
C) Neither differentiable nor continuous at $ x=0 $
D) Discontinuous every where
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(0)=0 $ and $ f(x)=x{e^{-( \frac{1}{|x|}+\frac{1}{x} )}} $
R.H.L. = $ \underset{h\to 0}{\mathop{\lim }},(0+h){e^{-2/h}}=\underset{h\to 0}{\mathop{\lim }},\frac{h}{{e^{2/h}}}=0 $
L.H.L. = $ \underset{h\to 0}{\mathop{\lim }},(0-h){e^{-( \frac{1}{h},-,\frac{1}{h} )}}=0 $ ; \ $ f(x) $ is continuous.
$ R{f}’,(x)=\underset{h\to 0}{\mathop{\lim }},\frac{(0+h){e^{-( \frac{1}{h}+\frac{1}{h} )}}-h{e^{-( \frac{1}{h}+\frac{1}{h} )}}}{h}=0 $
$ L{f}’(x)=\underset{h\to 0}{\mathop{\lim }},\frac{(0-h){e^{-( \frac{1}{h}-\frac{1}{h} )}}-h{e^{-( \frac{1}{h}+\frac{1}{h} )}}}{-h}=1 $
Þ $ L{f}’(x)\ne R{f}’(x) $ . $ f(x) $ is not differentiable at $ x=0. $
BETA
