If the function f defined on left(frac{pi}{6} | vedclass

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

Question

(A) Since $f(x)$ is continuous on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$,it must be continuous at $x=\frac{\pi}{4}$.Therefore,$f\left(\frac{\pi}{

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Since $f(x)$ is continuous on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$,it must be continuous at $x=\frac{\pi}{4}$.Therefore,$f\left(\frac{\pi}{4}\right) = \lim_{x \rightarrow \frac{\pi}{4}} f(x) = \lim_{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}$.This is a $\frac{0}{0}$ indeterminate form.Applying $L$'Hospital's rule,we differentiate the numerator and denominator with respect to $x$:$k = \lim_{x \rightarrow \frac{\pi}{4}} \frac{\frac{d}{dx}(\sqrt{2} \cos x - 1)}{\frac{d}{dx}(\cot x - 1)} = \lim_{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \sin x}{-\operatorname{cosec}^2 x} = \lim_{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \sin x}{\operatorname{cosec}^2 x}$.Substituting $x = \frac{\pi}{4}$:$k = \frac{\sqrt{2} \sin(\frac{\pi}{4})}{\operatorname{cosec}^2(\frac{\pi}{4})} = \frac{\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{(\sqrt{2})^2} = \frac{1}{2}$.

If the function $f$ defined on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$ by $f(x)=\begin{cases} \frac{\sqrt{2} \cos x-1}{\cot x-1}, & x \neq \frac{\pi}{4} \\ k, & x=\frac{\pi}{4} \end{cases}$ is continuous,then $k$ is equal to

Similar Questions

The function $f(x) = 2x - |x - x^2|$ is

The function $f(x) = [x]$,where $[x]$ denotes the greatest integer not greater than $x$,is

If a function defined by $f(x) = \frac{(3^x - 1)^2}{\sin x \log(1 + x)}$,$x \neq 0$,is continuous at $x = 0$,then $f(0) =$

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

If $f(x) = \begin{cases} \frac{x^2}{a} - a, & x < a \\ 0, & x = a \\ a - \frac{x^2}{a}, & x > a \end{cases}$ then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module