Question

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

• A. $g\left(1\right)+g\left(0\right)=0$
• B. $g\left(1\right)=\sqrt{2}g\left(0\right)$
• C. $\sqrt{2}g\left(1\right)=g\left(0\right)$
• D. $g\left(1\right)=g\left(0\right)$

Answer 1

The correct option is C $\sqrt{2}g\left(1\right)=g\left(0\right)$

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

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

$\therefore f\left(x\right)+f(-x)=0\cdots \left(i\right)$
Now $g\left(t\right)=\underset{-\pi /2}{\overset{\pi /2}{\int }}\cos (\frac{\pi }{4}t+f(x\left)\right)dx.$
$\Rightarrow g\left(t\right)=2\underset{0}{\overset{\pi /2}{\int }}\cos \frac{\pi t}{4}.\cos \left(f\right(x\left)\right)dx.$$\because \text{from}\left(i\right)$
$\therefore g\left(1\right)=\sqrt{2}\underset{0}{\overset{\pi /2}{\int }}\cos \left(f\right(x\left)\right)dx$
and $g\left(0\right)=2{\int }_0^{\pi /2}\cos \left(f\right(x\left)\right)dx.$
$\therefore \sqrt{2}g\left(1\right)=g\left(0\right)$