Question

Let g(x) = x.f(x), where f(x) = $f\left(x\right)=\{\begin{array}{cc}xsin\frac{1}{x},& x\ne 0\\ 0,& x=0.\end{array}$ at x = 0

• A. g is differentiable but g' is not continuous
• B.
  g is differentiable while f is not
• C. Both f and g are differentiable
• D. g is differentiable and g' is continuous

Answer 1

Answer: (A), (B)

g is differentiable while f is not

$f\left(x\right)=\{\begin{array}{cc}xsin\frac{1}{x},& x\ne 0\\ 0,& x=0.\end{array}g\left(x\right)=f\left(x\right)=\{\begin{array}{cc}{x}^{2}sin\frac{1}{x},& x\ne 0\\ 0,& x=0.\end{array}$
L f'(0) = $\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{f(0-h)sin(-\frac{1}{h})-\left(0\right)}{-h}=\underset{h\to 0}{lim}-sin\left(\frac{1}{h}\right)$
= a quantity which lies between - 1 and 1
R f'(0) = $\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$
$=\underset{h\to 0}{lim}\frac{(0+h)sin\frac{1}{h}-0}{h}=\underset{h\to 0}{lim}sin\frac{1}{h}$
= a quantity which lies between - 1 and 1
Hence L f'(0) $\ne$ R f'(0)
$\therefore$ f(x) is not differentiable at x = 0
Now L g'(0) = $\underset{h\to 0}{lim}\frac{f(0-h)-f\left(\right(0)}{0-h}$
$L{g}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{(0-h{)}^{2}sin(-\frac{1}{h})-0}{-h}=\underset{h\to 0}{lim}hsin\left(\frac{1}{h}\right)$
L g'(0) = $0\times (-1\le sin\frac{1}{h}\le 1)\Rightarrow L{g}^{\prime }\left(0\right)=0$ and
R g'(0) = $=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{(0+h{)}^{2}sin\frac{1}{h}-0}{h}=\underset{h\to 0}{lim}hsin\left(\frac{1}{h}\right)=0\times (-1\le sin\left(\frac{1}{h}\right)\le 1)=0$
$\because$ L g'(0) = R g'(0) then g(x) is differentiable at x = 0
Now $g\left(x\right)=x^{2}sin\frac{1}{x}{g}^{\prime }\left(x\right)=2xsin\frac{1}{x}+x^{2}cos\frac{1}{x}\times \frac{1}{x^2}$
$g^{\prime }\left(x\right)=2xsin\frac{1}{x}-cos\frac{1}{x}\Rightarrow g^{\prime }\left(x\right)=2f\left(x\right)-cos\frac{1}{x}$
So, g'(x) is not differentiable at x = 0.