Functions Question 346
Question: Let $ g(x)=x.f(x), $ where $ f(x)= \begin{cases} & x\sin \frac{1}{x},x\ne 0 \\ & 0,x=0 \\ \end{cases} $ at $ x=0 $
[IIT Screening 1994; UPSEAT 2004]
Options:
A) g is differentiable but g’ is not continuous.
B) Both g and f are not differentiable.
C) Both f and g are differentiable.
D) g is differentiable and g’ is continuous.
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)= \begin{cases} x\sin \frac{1}{x}, & x\ne 0 \\ 0, & x=0 \\ \end{cases} . $ , $ g(x)= \begin{cases} x^{2}\sin \frac{1}{x}, & x\ne 0 \\ 0, & x=0 \\ \end{cases} . $
$ L{f}’(0)=\underset{h\to 0}{\mathop{\lim }}\frac{f(0-h)-f(0)}{-h} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{(0-h)\sin (-\frac{1}{h})-(0)}{-h}=\underset{h\to 0}{\mathop{\lim }}-\sin ( \frac{1}{h} ) $
= a quantity which lies between ? 1 and 1
$ R{f}’(0)=\underset{h\to 0}{\mathop{\lim }}\frac{f(0+h)-f(0)}{h} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{(0+h)\sin \frac{1}{h}-0}{h}=\underset{h\to 0}{\mathop{\lim }}\sin \frac{1}{h} $
= a quantity which lies between ? 1 and 1
Hence $ L{f}’(0)\ne R{f}’(0) $ \ $ f(x) $ is not differentiable at $ x=0 $
Now $ L{g}’(0)=\underset{h\to 0}{\mathop{\lim }}\frac{f(0-h)-f(0)}{0-h} $
$ L{g}’(0)=\underset{h\to 0}{\mathop{\lim }}\frac{{{(0-h)}^{2}}\sin (-\frac{1}{h})-0}{-h}=\underset{h\to 0}{\mathop{\lim }}h\sin ( \frac{1}{h} ) $
$ L{g}’(0)=0\times ( -1\le \sin \frac{1}{h}\le 1 ) $
Þ $ L{g}’(0)=0 $
and $ R{g}’(0)=\underset{h\to 0}{\mathop{\lim }}\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\frac{{{(0+h)}^{2}}\sin ( \frac{1}{h} )-0}{h} $
$ =\underset{h\to 0}{\mathop{\lim }}h\sin ( \frac{1}{h} )=0\times ( -1\le \sin ( \frac{1}{h} )\le 1 )=0 $
$ \because L{g}’(0)=R{g}’(0) $ then $ g(x) $ is differentiable at $ x=0 $
Now $ g(x)=x^{2}\sin \frac{1}{x} $ $ {g}’(x)=2x\sin \frac{1}{x}+x^{2}\cos \frac{1}{x}\times -\frac{1}{x^{2}} $
$ {g}’(x)=2x\sin \frac{1}{x}-\cos \frac{1}{x} $
Þ $ {g}’(x)=2f(x)-\cos \frac{1}{x} $
So, $ {g}’(x) $ is not differentiable at $ x=0 $ .
BETA
