If f(x) = begin{cases} 1 + x, & text{when } x | vedclass

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

Question

(A) To check the continuity of $f(x)$ at $x = 2$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 2$.$1$. Left-h

Vedclass · Accepted Answer

$f(x)$ is continuous at $x = 2$

Vedclass · Answer

(A) To check the continuity of $f(x)$ at $x = 2$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 2$.$1$. Left-hand limit: $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^-} (1 + x) = 1 + 2 = 3$.$2$. Right-hand limit: $\lim_{x 	o 2^+} f(x) = \lim_{x 	o 2^+} (5 - x) = 5 - 2 = 3$.$3$. Value of the function: $f(2) = 1 + 2 = 3$.Since $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^+} f(x) = f(2) = 3$,the function $f(x)$ is continuous at $x = 2$.

If $f(x) = \begin{cases} 1 + x, & \text{when } x \le 2 \\ 5 - x, & \text{when } x > 2 \end{cases}$,then which of the following is true?

Similar Questions

Let $f(x) = \begin{cases} x^2 - a & x < 3 \\ b\sqrt{x - 2} + a & 3 \leqslant x < 6 \\ 2x + b & x \geqslant 6 \end{cases}$. If $f(x)$ is continuous $\forall x \in R$,then find the value of $\frac{f(1) - f(3)}{4}$.

If $f(x) = \begin{cases} \frac{1 - \cos x}{x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then $k = $

Let $f(x) = \frac{1-\tan x}{4x-\pi}$,where $x \neq \frac{\pi}{4}$ and $x \in [0, \frac{\pi}{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,then $f(\frac{\pi}{4})$ is:

Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} 2x + 3, & \text{if } x \le 2 \\ 2x - 3, & \text{if } x > 2 \end{cases}$

If $f(x) = \begin{cases} \frac{x}{e^{1/x} + 1}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module