If a function f is defined by f(x) = begin{ca | vedclass

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

Question

(A) Given that the function $f(x)$ is continuous at $x = \frac{\pi}{4}$,we must have $f(\frac{\pi}{4}) = \lim_{x 	o \frac{\pi}{4}} f(x)$.Since $f(\fr

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(A) Given that the function $f(x)$ is continuous at $x = \frac{\pi}{4}$,we must have $f(\frac{\pi}{4}) = \lim_{x 	o \frac{\pi}{4}} f(x)$.Since $f(\frac{\pi}{4}) = k$,we evaluate the limit:$k = \lim_{x 	o \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x}$.This is a $\frac{0}{0}$ indeterminate form. Applying $L$'$H$ôpital's Rule by differentiating the numerator and denominator with respect to $x$:$k = \lim_{x 	o \frac{\pi}{4}} \frac{\frac{d}{dx}(1-\sqrt{2} \sin x)}{\frac{d}{dx}(\pi-4 x)}$$k = \lim_{x 	o \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4}$Substituting $x = \frac{\pi}{4}$:$k = \frac{-\sqrt{2} \cos(\frac{\pi}{4})}{-4} = \frac{-\sqrt{2} 	imes \frac{1}{\sqrt{2}}}{-4} = \frac{-1}{-4} = \frac{1}{4}$.

If a function $f$ is defined by $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4 x}, & x \neq \frac{\pi}{4} \\ k, & x = \frac{\pi}{4} \end{cases}$ and is continuous at $x = \frac{\pi}{4}$,then $k = $

Similar Questions

If $f(x) = \begin{cases} \frac{5}{2} - x, & x < 2 \\ 1, & x = 2 \\ x - \frac{3}{2}, & x > 2 \end{cases}$,then:

If $f(x) = \begin{cases} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}, & -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & 0 \leq x \leq 1 \end{cases}$ is continuous in $[-1, 1]$, then $p = $

If $f(x) = \frac{e^{2x} - (1 + 4x)^{1/2}}{\ln(1 - x^2)}$ for $x \neq 0$,then $f$ has

If $f: R \rightarrow R$ is defined by $f(x) = x - [x]$, where $[x]$ is the greatest integer not exceeding $x$, then the set of points of discontinuity of $f$ is

If $f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ for $x \neq \pi$ is continuous at $x = \pi$,then $f(\pi) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module