Let \(\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}\) be a differentiable function such that \(\mathrm{f}\left(\frac{\pi}{4}\right)=\sqrt{2}, \mathrm{f}\left(\frac{\pi}{2}\right)=0\) and \(\mathrm{f}^{\prime}\left(\frac{\pi}{2}\right)=1\) and let \(g(x)=\int_{x}^{\pi / 4}\left(f^{\prime}(t) \sec t+\tan t \sec t f(t)\right) d t\) for \(\mathrm{x} \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right)\). Then \(\lim _{\mathrm{x} \rightarrow\left(\frac{\pi}{2}\right)^{-}} \mathrm{g}(\mathrm{x})\) is equal to

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Definite Integrals

Let

f : R \to R

be a differentiable function such that

f (\frac{π}{4}) = 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

Answer: (B)

The integrand can be written as

\frac{d}{d t} (f (t) sec t)

.

Therefore:

g (x) = \int_{x}^{π /4} d (f (t) sec t) = [f (t) sec t]_{x}^{π /4} = f (\frac{π}{4}) sec \frac{π}{4} - f (x) sec x

g (x) = 2 \cdot 2 - f (x) sec x = 2 - \frac{f ( x )}{c o s x}

x \to (π /2)^{-}

, both

f (x) \to 0

and

cos x \to 0

, so we apply L'Hôpital's rule:

lim_{x \to (π /2)^{-}} \frac{f ( x )}{c o s x} = lim_{x \to (π /2)^{-}} \frac{f ^{'} ( x )}{- s i n x} = \frac{1}{- 1} = - 1

Therefore:

lim_{x \to (π /2)^{-}} g (x) = 2 - (- 1) = 3

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option