Solve the Following Questions


Let $\mathrm{g}(\mathrm{t})=\int_{-\pi / 2}^{\pi / 2} \cos \left(\frac{\pi}{4} \mathrm{t}+f(\mathrm{x})\right) \mathrm{dx}$, where

$f(x)=\log _{e}\left(x+\sqrt{x^{2}+1}\right), x \in \mathbf{R}$. Then which one of the following is correct?

  1. $g(1)=g(0)$

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

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

  4. $\mathrm{g}(1)+\mathrm{g}(0)=0$

Correct Option: , 2


$\mathrm{g}(\mathrm{t})=\int_{-\pi / 2}^{\pi / 2}\left(\cos \frac{\pi}{4} \mathrm{t}+\mathrm{f}(\mathrm{x})\right) \mathrm{d} \mathrm{x}$

$\mathrm{g}(\mathrm{t})=\pi \cos \frac{\pi}{4} \mathrm{t}+\int_{-\pi / 2}^{\pi / 2} \mathrm{f}(\mathrm{x}) \mathrm{dx}$

$\mathrm{g}(\mathrm{t})=\pi \cos \frac{\pi}{4} \mathrm{t}$

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

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now