Let the functions f, g and h be defined as fo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To check for differentiability at $x = 0$,we calculate the derivative using the limit definition $f'(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)}{h}$.

Vedclass · Accepted Answer

$g$ and $h$ only

Vedclass · Answer

(C) To check for differentiability at $x = 0$,we calculate the derivative using the limit definition $f'(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)}{h}$.$(1)$ For $f(x)$:$f'(0) = \lim_{h 	o 0} \frac{h \sin(1/h) - 0}{h} = \lim_{h 	o 0} \sin(1/h)$.Since $\lim_{h 	o 0} \sin(1/h)$ oscillates between $-1$ and $1$,the limit does not exist. Thus,$f(x)$ is not differentiable at $x = 0$.$(2)$ For $g(x)$:$g'(0) = \lim_{h 	o 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h 	o 0} h \sin(1/h)$.Since $|h \sin(1/h)| \le |h|$,by the Squeeze Theorem,the limit is $0$. Thus,$g(x)$ is differentiable at $x = 0$.$(3)$ For $h(x) = |x|^3$:$h'(0) = \lim_{h 	o 0} \frac{|h|^3 - 0}{h} = \lim_{h 	o 0} \frac{|h|^3}{h}$.For $h > 0$,$\frac{h^3}{h} = h^2 	o 0$. For $h Since the limit exists and is $0$,$h(x)$ is differentiable at $x = 0$.Therefore,$g$ and $h$ are differentiable at $x = 0$. The correct option is $C$.

Similar Questions

Identify the correct statement about the function $f(x) = \max(x^2 - 1, 7 - x^2, 5)$.

The number of points,at which the function $f(x) = |2x+1| - 3|x+2| + |x^2+x-2|$,$x \in R$ is not differentiable,is ............

If $f(x)$ is a differentiable function such that $f: R \to R$ and $f\left( \frac{1}{n} \right) = 0$ for all $n \ge 1, n \in I$,then:

The set of points where $f(x) = \frac{x}{4+|x|}$ is differentiable is

Assertion $(A)$: If $f(x)$ is not continuous at $x=a$,then it is not differentiable at $x=a$.
Reason $(R)$: If $f(x)$ is differentiable at a point,then it is continuous at that point.

Vedclass Test Series

Exam Paper Generator

Online Exam Module