Let $g(t)=\int_{-\pi / 2}^{\pi / 2} \cos \left(\frac{\pi}{4} t+f(x)\right) d x$, where $f(x)=\log _{e}\left(x+\sqrt{x^{2}+1}\right), x \in R$. Then which one of the following is correct?

(D) $\sqrt{2} \mathrm{~g}(1)=\mathrm{g}(0)$

(A) $\mathrm{g}(1)=\mathrm{g}(0)$

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Definite Integrals

Let

g (t) = \int_{- π /2}^{π /2} cos (\frac{π}{4} t + f (x)) d x

, where

f (x) = lo g_{e} (x + x^{2} + 1), x \in R

. Then which one of the following is correct?

Answer: (D)

The function

f (x) = lo g_{e} (x + x^{2} + 1)

satisfies

f (- x) = - f (x)

, because

(x + x^{2} + 1) (- x + x^{2} + 1) = 1

. Expanding the integrand,

cos (\frac{π}{4} t + f (x)) = cos \frac{π t}{4} cos f (x) - sin \frac{π t}{4} sin f (x)

Over the symmetric interval

[- \frac{π}{2}, \frac{π}{2}]

the even part

cos f (x)

survives while the odd part

sin f (x)

integrates to zero, so

g (t) = C cos \frac{π t}{4}, C = \int_{- π /2}^{π /2} cos f (x) d x

Therefore

g (0) = C

and

g (1) = C cos \frac{π}{4} = \frac{C}{2}

, which means

2 g (1) = g (0)

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option