Let [x] denote the greatest integer less than | vedclass

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

Question

(D) The function is defined as $f(x) = [x] + |1 - x|$ on the interval $[-1, 3]$.Points of discontinuity for the greatest integer function $[x]$ in the

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) The function is defined as $f(x) = [x] + |1 - x|$ on the interval $[-1, 3]$.Points of discontinuity for the greatest integer function $[x]$ in the interval $[-1, 3]$ are $x = 0, 1, 2, 3$.At $x = 1$,the function is $f(x) = [x] + |1 - x|$. Since $|1 - x|$ is continuous everywhere,the non-differentiability arises from the jump discontinuities of $[x]$ at $x = 0, 1, 2, 3$ and the corner point of $|1 - x|$ at $x = 1$.Checking the points:$1$. At $x = 0$: $[x]$ is discontinuous,so $f(x)$ is not differentiable.$2$. At $x = 1$: $[x]$ is discontinuous and $|1 - x|$ has a corner,so $f(x)$ is not differentiable.$3$. At $x = 2$: $[x]$ is discontinuous,so $f(x)$ is not differentiable.$4$. At $x = 3$: $[x]$ is discontinuous,so $f(x)$ is not differentiable.Thus,the function is not differentiable at $x = 0, 1, 2, 3$. There are $4$ such points.

Let $[x]$ denote the greatest integer less than or equal to $x$. Then the number of points where the function $y = [x] + |1 - x|$ for $-1 \leq x \leq 3$ is not differentiable,is

Similar Questions

If $f(x) = \begin{cases} \frac{x-1}{2x^2-7x+5}, & \text{for } x \neq 1 \\ -\frac{1}{3}, & \text{for } x=1 \end{cases}$,then $f^{\prime}(1)$ is equal to:

$\text{The domain of the derivative of the function } f(x) = \begin{cases} \tan^{-1} x, & \text{if } |x| \le 1 \\ \frac{1}{2}(|x|-1), & \text{if } |x| > 1 \end{cases} \text{ is given by:}$

The function $f(x) = e^{-|x|}$ is

If $f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$,then $f(x)$ is differentiable on

Vedclass Test Series

Exam Paper Generator

Online Exam Module