Search any question & find its solution
Question:
Answered & Verified by Expert
$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}$ equals
Options:
Solution:
1670 Upvotes
Verified Answer
The correct answer is:
$\frac{8}{\pi} f(2)$
$\frac{8}{\pi} f(2)$
$\lim _{x \rightarrow \frac{x}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}$ $\left[\therefore \frac{0}{0}\right.$ form $]$
Let
$$
\begin{array}{ll}
\text { Let } & L=\lim _{x \rightarrow \frac{x}{4}} \frac{f\left(\sec ^2 x\right) 2 \sec x \sec x \tan x}{2 x} \\
\therefore & L=\frac{2 f(2)}{\pi / 4}=\frac{8 f(2)}{\pi}
\end{array}
$$
Let
$$
\begin{array}{ll}
\text { Let } & L=\lim _{x \rightarrow \frac{x}{4}} \frac{f\left(\sec ^2 x\right) 2 \sec x \sec x \tan x}{2 x} \\
\therefore & L=\frac{2 f(2)}{\pi / 4}=\frac{8 f(2)}{\pi}
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.