Explore More

Similar Questions

Let $f: R \rightarrow R$ be defined as $f(x)=\begin{cases} \frac{a-b \cos 2 x}{x^2} & ; x<0 \\ x^2+c x+2 & ; 0 \leq x \leq 1 \\ 2 x+1 & ; x>1 \end{cases}$. If $f$ is continuous everywhere in $R$ and $m$ is the number of points where $f$ is $NOT$ differentiable,then $m+a+b+c$ equals:

If $f(x) = \begin{cases} x^{3}-3x+2, & x < 2 \\ x^{3}-6x^{2}+9x+2, & x \geq 2 \end{cases}$, then:

If a function $f(x)$ defined by $f(x)=\begin{cases} a e^{x}+b e^{-x}, & -1 \leq x<1 \\ c x^{2}, & 1 \leq x \leq 3 \\ a x^{2}+2 c x, & 3 < x \leq 4 \end{cases}$ is continuous for some $a, b, c \in R$ and $f'(0)+f'(2)=e$,then the value of $a$ is:

In the usual notation,the value of $\Delta \nabla$ is equal to

If $f$ is defined by $f(x) = \begin{cases} x & \text{for } 0 \leq x < 1 \\ 2-x & \text{for } x \geq 1 \end{cases}$,then at $x=1$,$f(x)$ is

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