Let $f(x) = \begin{cases} g(x) \cos(\frac{1}{x}) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$ where $g(x)$ is an even function differentiable at $x = 0$,passing through the origin. Then $f'(0)$:

  • A
    is equal to $1$
  • B
    is equal to $0$
  • C
    is equal to $2$
  • D
    does not exist

Explore More

Similar Questions

The number of points in the interval $(0,2)$ at which $f(x)=|x-0.5|+|x-1|+\tan x$ is not differentiable is

If $f(x) = \begin{cases} x^2 \left| \cos \frac{\pi}{x} \right|, & x \neq 0 \\ 0, & x = 0 \end{cases}$, then at $x = 2$, $f(x)$ is

If $f(x) = \text{sgn}(x^3)$,then

The number of points,where the function $f: R \rightarrow R, f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x^2-5x+4|$ is $NOT$ differentiable,is:

Let the functions $f, g$ and $h$ be defined as follows:
$f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right) & \text{for } -1 \le x \le 1, x \ne 0 \\ 0 & \text{for } x = 0 \end{cases}$
$g(x) = \begin{cases} x^2 \sin \left( \frac{1}{x} \right) & \text{for } -1 \le x \le 1, x \ne 0 \\ 0 & \text{for } x = 0 \end{cases}$
$h(x) = |x|^3$ for $-1 \le x \le 1$.
Which of these functions are differentiable at $x = 0$?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo