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.
Reason $(R)$: If $f(x)$ is differentiable at a point,then it is continuous at that point.
- A$(A)$ and $(R)$ are both true,$(R)$ is the correct explanation of $(A)$
- B$(A)$ and $(R)$ are both true,$(R)$ is not the correct explanation of $(A)$
- C$(A)$ is true,$(R)$ is false
- D$(A)$ is false,$(R)$ is true
Explore More
Similar Questions
Let $K$ be the set of all real values of $x$,where the function $f(x) = \sin |x| - |x| + 2(x - \pi) \cos |x|$ is not differentiable. Then the set $K$ is
DifficultMHT CET 2024
View SolutionIf $f(x) = \begin{cases} x \left(1 + \frac{1}{2} \sin (\log x^2) \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$,then find the value of $\lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x}$.
The set of all those points,where the function $f(x) = \frac{x}{1 + |x|}$ is differentiable,is
Easy
View SolutionLet $f(x) = \begin{cases} \sin x, & \text{for } x \ge 0 \\ 1 - \cos x, & \text{for } x \le 0 \end{cases}$ and $g(x) = e^x$. Then $(g \circ f)'(0)$ is
Easy
View SolutionThe function $f(x) = |x| + |x - 1|$ is
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo