QJEEM21Ma9122: JEE Main 2021 Mathematics - Class 12 JEEM21Ma9122 Solution: Let , where . Then which one of the following is correct?

Question

Verified Solution • 2025–26 Session moderate

Let

g(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi \over 4}t + f(x)} \right)} dx

, where

f(x) = {\log _e}\left( {x + \sqrt {{x^2} + 1} } \right),x \in R

. Then which one of the following is correct?

Accepted Answer

\because

f(x) = \ln \left( {x + \sqrt {{x^2} + 1} } \right)

\therefore

f(x) + f( - x) = \ln \left( {\sqrt {{x^2} + 1} + x} \right) + \ln \left( {\sqrt {{x^2} + 1} - x} \right)

\therefore

f(x) + f( - x) = 0

.... (i)

\because

g(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi \over 4}t + f(x)} \right)} dx

= \int_0^{\pi /2} {\left\{ {\cos \left( {{\pi \over 4}t + f(x)} \right) + \cos \left( {{\pi \over 4}t + f( - x)} \right)} \right\}} dx

= \int_0^{\pi /2} {\left\{ {\cos \left( {{{\pi t} \over 4} + f(x)} \right) + \cos \left( {{{\pi t} \over 4} - f(x)} \right)} \right\}dx}

g(t) = 2\int_0^{\pi /2} {\cos {{\pi t} \over 4}.\cos (f(x))dx}

\therefore

g(1) = \sqrt 2 \int_0^{\pi /2} {\cos \left( {f(x)} \right)} dx

and

g(0) = 2\int_0^{\pi /2} {\cos \left( {f(x)} \right)} dx

\therefore

\sqrt 2 g(1) = g(0)

Step-by-Step Verified Solution