If f(x) = begin{cases} 1, & 0

Question

(C) To check the continuity at $x = \frac{3\pi}{4}$,we evaluate the left-hand limit,right-hand limit,and the value of the function at that point.First

Vedclass · Accepted Answer

$f(x)$ is continuous at $x = \frac{3\pi}{4}$

Vedclass · Answer

(C) To check the continuity at $x = \frac{3\pi}{4}$,we evaluate the left-hand limit,right-hand limit,and the value of the function at that point.First,the value of the function at $x = \frac{3\pi}{4}$ is $f\left(\frac{3\pi}{4}ight) = 1$.Next,the left-hand limit is $\lim_{x 	o \frac{3\pi}{4}^-} f(x) = 1$.Then,the right-hand limit is $\lim_{x 	o \frac{3\pi}{4}^+} f(x) = \lim_{h 	o 0} 2\sin \left(\frac{2}{9} \left(\frac{3\pi}{4} + hight)ight) = 2\sin \left(\frac{2}{9} \cdot \frac{3\pi}{4}ight) = 2\sin \left(\frac{\pi}{6}ight) = 2 \cdot \frac{1}{2} = 1$.Since $\lim_{x 	o \frac{3\pi}{4}^-} f(x) = \lim_{x 	o \frac{3\pi}{4}^+} f(x) = f\left(\frac{3\pi}{4}ight) = 1$,the function $f(x)$ is continuous at $x = \frac{3\pi}{4}$.

If $f(x) = \begin{cases} 1, & 0 < x \le \frac{3\pi}{4} \\ 2\sin \frac{2}{9}x, & \frac{3\pi}{4} < x < \pi \end{cases}$,then

Similar Questions

If $f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x > 0 \end{cases}$ is continuous at $x = 0$,then the value of $a$ will be

Let $f: R \rightarrow R$ be a continuous function satisfying $f(0)=1$ and $f(2x)-f(x)=x$ for all $x \in R$. If $\lim_{n \rightarrow \infty} \{f(x)-f(\frac{x}{2^n})\} = G(x)$,then $\sum_{r=1}^{10} G(r^2)$ is equal to

If the function $f(x)$ is continuous on its domain $[-2, 2]$,where $f(x) = \begin{cases} \frac{\sin ax}{x} + 3, & -2 \leq x < 0 \\ x + 5, & 0 \leq x \leq 1 \\ \sqrt{x^2 + 8} - b, & 1 < x \leq 2 \end{cases}$,then $7a + b + 1$ is equal to:

If $f(x) = \begin{cases} \frac{\sqrt{1+mx} - \sqrt{1-mx}}{x}, & -1 \le x < 0 \\ \frac{2x+1}{x-2}, & 0 \le x \le 1 \end{cases}$ is continuous in the interval $[-1, 1]$,then $m$ is equal to:

Let $f(x) = \begin{cases} \frac{\sin \pi x}{5x}, & x \ne 0 \\ k, & x = 0 \end{cases}$. If $f(x)$ is continuous at $x = 0$,then $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module