Let f: R → R be a differentiable function such that f((π)/4) = 2, f((π)/2) = 0 and f'((π)/2) = 1 and let g(x) = ∫ x ^(π /4) (f'(t) sec t + tan t sec t f(t)) dt for x [ (π)/4,(π)/2). Then lim x → ((π)/2)^(- g)(x) is equal to:

Integral Calculus: JEE Main 2022 Mathematics question

JEE Main 2022 · 28 Jun, Evening shift · Q3428

JEE Main 202228 Jun, Evening shiftMathematicsMediumIntegral Calculus369× since 2002

Let f : R

\to

R be a differentiable function such that

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

and

f'\left( {{\pi \over 2}} \right) = 1

and let

g(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt}

for

x \in \left[ {{\pi \over 4},{\pi \over 2}} \right)

. Then

\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)

is equal to :

JEE Main 2022 Integral Calculus Question with Solution (28 Jun Evening shift, Q3428) | JEE Arc