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

We have,

$\begin{array}{l}\mathrm{g}\left(\mathrm{t}\right)={\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} \cos \left(\frac{\mathrm{\pi }}{4}\mathrm{t}+\mathrm{f}\left(\mathrm{x}\right)\right)\mathrm{dx}\\ \mathrm{g}\left(\mathrm{t}\right)={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \left[\cos \left(\frac{\mathrm{\pi }}{4}\right)\mathrm{tcos}\left(\mathrm{f}\right(\mathrm{x}\left)\right)-\sin \frac{\mathrm{\pi }}{4}\mathrm{tsin}\mathrm{f}\left(\mathrm{x}\right)\right]\mathrm{dx}\end{array}$

Given, $\mathrm{f}\left(\mathrm{x}\right)=\log \left(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\right)$

$\begin{array}{l}\mathrm{f}(-\mathrm{x})=\log \left(-\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\right)=-\log \left(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\right)\\ \mathrm{f}(-\mathrm{x})=-\mathrm{f}\left(\mathrm{x}\right)\end{array}$

f(x) is an odd function.

$\therefore \mathrm{g}\left(\mathrm{t}\right)={\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} \cos \frac{\mathrm{\pi }}{4}\mathrm{tcos}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} [\because \sin \mathrm{f}(\mathrm{x}\left)\text{ is an odd function }\right]$

$\Rightarrow \mathrm{g}\left(1\right)={\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} \frac{1}{\sqrt{2}}\cos \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\Rightarrow \sqrt{2}\mathrm{g}\left(1\right)={\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} \cos \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$

and $\mathrm{g}\left(0\right)={\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} \cos \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$

$\therefore \sqrt{2}\mathrm{g}\left(1\right)=\mathrm{g}\left(0\right)$