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.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 5. continuity and differentiation→STD 12 - 5. continuity and differentiation — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

If in the interval $[0,3]$

$f\left( x \right) = \left\{ \begin{gathered} x{\left\{ x \right\}^2},x \notin I \hfill \\ x\,\,\,\,\,\,\,\,\,\,,x \in I \hfill \\ \end{gathered} \right.,$

then which of the following statement is correct?

(where $\{.\}$ denotes fractional part function)

Consider the function $f(x)=\left\{\begin{array}{cc}\frac{x+5}{x-2}, & \text { if } x \neq 2 \\1, & \text { if } x=2\end{array}\right.$ Then, $f(f(x))$ is discontinuous

If $\int\frac{\cos8\text{x}+1}{\tan2\text{x}-\cot2\text{x}}\text{ dx}=\text{a}\cos8\text{x}+\text{C},$ then a =

Choose the correct answer from the given four options.The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}-\text{y}}+\text{x}^2\text{e}^{-\text{y}}$ is:

If $f\left( x \right) = a\left| {\sin \,x} \right| + b{e^{\left| x \right|}} + c{\left| x \right|^3}\,$, where $a, b, c \in R$ , is differentiable at $x = 0$, then

If ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{\pi }{2}$, then the value of ${x^2} + {y^2} + {z^2} + 2xyz$ is equal to

Let $y=y(x)$ be the solution of the differential equation $\sec \mathrm{x} d \mathrm{y}+\{2(1-\mathrm{x}) \tan \mathrm{x}+\mathrm{x}(2-\mathrm{x})\}$ $\mathrm{dx}=0$ such that $\mathrm{y}(0)=2$. Then $\mathrm{y}(2)$ is equal to :

The direction ratios of the line x - y + z - 5 = 0 = x - 3y - 6 are proportional to:

The points $A(4,\,5,\,1),B(0, - 1, - 1),C(3,\,9,\,4)$ and $D( - 4,\,4,\,4)$ are

$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)=$