Verified Solution • 2025–26 Session moderate

QJEEM22Ma57ce: JEE Main 2022 Mathematics - Class 12 JEEM22Ma57ce Solution: Let f : R → R be a differentiable function such that f ( π 4 )…

Let f : R $\to$ R be a differentiable function such that $f (\frac{π}{4}) = \sqrt{2}, f (\frac{π}{2}) = 0$ and $f^{'} (\frac{π}{2}) = 1$ and

let $g (x) = \int_{x}^{π / 4} (f^{'} (t) \sec t + \tan t \sec t f (t)) d t$ for $x \in [\frac{π}{4}, \frac{π}{2})$ . Then $lim_{x \to {(\frac{π}{2})}^{-}} g (x)$ is equal to :

Multiple Choice Options

A 2
B 3
C 4
D −3

Step-by-Step Verified Solution

Given :

f\left(\frac{\pi}{4}\right)=\sqrt{2}, f\left(\frac{\pi}{2}\right)=0

and

f^{\prime}\left(\frac{\pi}{2}\right)=1

\begin{aligned} g (x) = \int_{x}^{\frac{π}{4}} (f^{'} (t) \sec t + \tan t \sec t f (t)) d t \\ = [\sec t + f (t)]_{x}^{\frac{π}{4}} = 2 - \sec x f (x) \end{aligned}

Now,

\lim \limits_{x \rightarrow \frac{\pi^{-}}{2}} g(x)=\lim \limits_{h \rightarrow 0} g\left(\frac{\pi}{2}-h\right)

= lim_{h \to 0} 2 - (cosec h) f (\frac{π}{2} - h)

=\lim \limits_{h \rightarrow 0}\left[2-\frac{f\left(\frac{\pi}{2}-h\right)}{\sin h}\right]

=\lim \limits_{h \rightarrow 0}\left[2+\frac{f^{\prime}\left(\frac{\pi}{2}-h\right)}{\cos h}\right]

= 3

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