Let g(t) = ∫(cos(πt/4) + f(x)) dx | Definite Integration JEE Main 2021

JEE MAIN 2021

20-07-2021 S2

Question

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

Select the correct option:

g(1) = g(0)

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

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

g(1)+ g(0) = 0

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Mathematics

Easy

Start Preparing for JEE with Competishun

Report Issue