If f(x) = begin{cases} frac{1-sqrt{2} sin x}{ | vedclass

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

Question

(D) Since $f(x)$ is continuous at $x = \frac{\pi}{4}$,we must have $\lim_{x \rightarrow \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}\right)$.$\lim_{x \r

Vedclass · Accepted Answer

$1/4$

Vedclass · Answer

(D) Since $f(x)$ is continuous at $x = \frac{\pi}{4}$,we must have $\lim_{x \rightarrow \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}\right)$.$\lim_{x \rightarrow \frac{\pi}{4}} f(x) = \lim_{x \rightarrow \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4x}$.This is a $\frac{0}{0}$ form,so we apply $L$'Hospital's rule:$\lim_{x \rightarrow \frac{\pi}{4}} \frac{\frac{d}{dx}(1-\sqrt{2} \sin x)}{\frac{d}{dx}(\pi-4x)} = \lim_{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4}$.Substituting $x = \frac{\pi}{4}$:$= \frac{-\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{-4} = \frac{-1}{-4} = \frac{1}{4}$.Since $f\left(\frac{\pi}{4}\right) = a$,we have $a = \frac{1}{4}$.

If $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4x} & \text{if } x \neq \frac{\pi}{4} \\ a & \text{if } x = \frac{\pi}{4} \end{cases}$ is continuous at $x = \frac{\pi}{4}$,then $a$ is equal to

Similar Questions

$f(x) = \begin{cases} \frac{\log x}{x-1}, & \text{if } x \neq 1 \\ k, & \text{if } x=1 \end{cases}$ is continuous at $x=1$,then the value of $k$ is

The function $f(x) = \begin{cases} \frac{2}{5-x}, & x < 3 \\ 5-x, & x \geq 3 \end{cases}$ is

The function $f(x) = \begin{cases} \frac{x - |x|}{x}, & x \neq 0 \\ 2, & x = 0 \end{cases}$

If $[x]$ denotes the greatest integer not exceeding $x$ and if the function $f$ defined by $f(x)= \begin{cases} \frac{a+2 \cos x}{x^2} & , x < 0 \\ b \tan \frac{\pi}{[x+4]} & , x \geq 0 \end{cases}$ is continuous at $x=0$, then the ordered pair $(a, b)$ is equal to

The function $f(x) = p[x + 1] + q[x - 1],$ where $[x]$ is the greatest integer function,is continuous at $x = 1$ if:

Vedclass Test Series

Exam Paper Generator

Online Exam Module