Let the function f, g and h be defined as follows : f(x) , = * 20… - Vidyadip

MCQ

Let the function $f, g$ and $h$ be defined as follows :

$f(x)\, = \left\{ {\begin{array}{*{20}{c}}{x\,\sin \,\left( {\frac{1}{x}}\right)\,\,\,\,\,\,\,for\,\, - 1 \le x \le 1\,\,and\,\,x \ne \,0}\\
{0\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\, = \,0}
\end{array}} \right.$

$g(x)\, = \left\{ {\begin{array}{*{20}{c}}{{x^2}\,\sin \,\left( {\frac{1}{x}} \right)\,\,\,\,\,\,\,for\,\, - 1 \le x \le 1\,\,and\,\,x \ne \,0}\\{0\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\, = \,0}\end{array}} \right.$ $h (x) = | x |^3$ for $- 1 \le x \le 1$ Which of these functions are differentiable at $x = 0$ ?

A
$f $ and $g$ only
B
$f$ and $h$ only
✓
$g$ and $h$ only
D
none

✓

Answer

Correct option: C.

$g$ and $h$ only

c
(1) $f(x)=x \sin \left(\frac{1}{x}\right)$ For $-1 \leq x \leq 1$ and $x \neq 0 \quad 0$ For $x=0$

$f(x)$ is not differentiable at $x=0$

$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{f(h)-0}{h}$

$=\lim _{h \rightarrow 0} \frac{h \sin \left(\frac{1}{h}\right)}{h}=\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right)$

which does not exist.

(2) $g(x)=x^{2} \sin \left(\frac{1}{x}\right)$ For $-1 \leq x \leq 1$ and $x \neq 0$ 0 For $x=0$

$R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{(0+h)^{2} \sin \left(\frac{1}{0+h}\right)-0}{h}$

$=\lim _{h \rightarrow 0} h \sin \left(\frac{1}{h}\right)=0$

Similarly $L f^{\prime}(0)=0$

Hence, $g(x)$ is differentiable at $x=0$.

(3) $h(x)=|x|^{3}$ For $-1 \leq x \leq 1$

$R H D=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{|h|^{3}-0}{h}$

$=\lim _{h \rightarrow 0} h^{2}=0$

$L H D=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{|-h|^{3}-0}{-h}$

$=\lim _{h \rightarrow 0}-h^{2}=0$

since $f^{\prime}(0)=R H D=L H D=0, h(x)$ is differentiable at $x=0$

Hence, only $g$ and $h$ are differentiable.

Hence, option $C$ is correct.

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