Let $f(x) = \begin{cases} x^{3/5} & \text{if } x \le 1 \\ -(x - 2)^3 & \text{if } x > 1 \end{cases}$. Then the number of critical points on the graph of the function is:

  • A
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$

Explore More

Similar Questions

If $f(x) = \begin{cases} 3x^2 + 12x - 1, & -1 \le x \le 2 \\ 37 - x, & 2 < x \le 3 \end{cases}$,then:

Let $f:(-2,2) \rightarrow \mathbb{R}$ be defined by $f(x) = \begin{cases} x[x] & , -2 < x < 0 \\ (x-1)[x] & , 0 \leq x < 2 \end{cases}$ where $[x]$ denotes the greatest integer function. If $m$ and $n$ respectively are the number of points in $(-2,2)$ at which $y = |f(x)|$ is not continuous and not differentiable,then $m + n$ is equal to $...........$.

Match the items given in List-$I$ with those of the items of List-$II$:
List-$I$List-$II$
$a$. If $y=|x|+|x-2|$, then at $x=2, \frac{dy}{dx}=$$i$. $2$
$b$. If $f(x)=|\cos 2x|$, then $f^{\prime}(\frac{\pi}{4}+)=$$ii$. $0$
$c$. If $f(x)=\sin(\pi[x])$, where $[\cdot]$ denotes the greatest integer function, then $f^{\prime}(1-)=$$iii$. $-2$
$d$. If $f(x)=\log|x-1|, x \neq 1$, then $f^{\prime}(\frac{1}{2})=$$iv$. Does not exist

If $f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x$,then $f'\left( \frac{\pi}{4} \right)$ is

Difficult
View Solution

Let $f(x)$ be a quadratic expression which is positive for all real $x$. If $g(x) = f(x) + f'(x) + f''(x)$,then for any real $x$,which one is correct?

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