Let f be a differentiable function defined on [ 0,(π)/2 ] such that f(x) > 0 and f(x) + ∫₀ˣ f(t)√(1 - (log e f(t))²) dt = e, x [ 0,(π)/2 ]. Then (6 log e f((π)/6))² is equal to _ _ _ _ _.

Integral Calculus: JEE Main 2023 Mathematics question

JEE Main 2023 · 24 Jan, Evening shift · Q28

JEE Main 202324 Jan, Evening shiftMathematicsMediumIntegral Calculus369× since 2002

Let

f

a

differentiable function defined on

\left[ {0,{\pi \over 2}} \right]

such that

f(x) > 0

and

f(x) + \int_0^x {f(t)\sqrt {1 - {{({{\log }_e}f(t))}^2}} dt = e,\forall x \in \left[ {0,{\pi \over 2}} \right]}

. Then

\left( {6{{\log }_e}f\left( {{\pi \over 6}} \right)} \right)^2

is equal to __________.

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2023 Integral Calculus Question with Answer (24 Jan Evening shift, Q28) | JEE Arc