Question

A function $f$ is continuous and differentiable for all $x>0,$ such that $f^2\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)\frac{\cos t}{2+\sin t}dt$ and $f\left(x\right)\ne 0,f\left(\pi \right)=\ln 2,$ then $f\left(x\right)$ is

• A. $\frac{1}{2}\ln \left(\frac{x+5\cos x}{4}\right)$
• B. $\frac{1}{2}\ln \left(\frac{4}{2+\cos x}\right)$
• C. $\frac{1}{2}\ln (4+2\sin x)$
• D. $\frac{2+\cos x+\sin x}{2+\sin x}$

Answer 1

The correct option is C $\frac{1}{2}\ln (4+2\sin x)$

$f^2\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)\frac{\cos t}{2+\sin t}dt$
$\Rightarrow 2f\left(x\right)f^{\prime }\left(x\right)=f\left(x\right)\frac{\cos x}{2+\sin x}$ Differentiating w.r.t. $x$ using Leibnitz rule,
$\Rightarrow 2f^{\prime }\left(x\right)=\frac{\cos x}{2+\sin x}$
[as $f\left(x\right)\ne 0$ ]
$\Rightarrow 2\int f^{\prime }\left(x\right)dx=\int \frac{\cos x}{2+\sin x}dx\Rightarrow 2f\left(x\right)={\log }_e(2+\sin x)+{\log }_{e}C$
Put $x=\pi ;$ we have $f\left(\pi \right)=\ln 2$
$\therefore 2{\log }_{e}2={\log }_{e}2+{\log }_{e}C\Rightarrow {\log }_{e}C={\log }_{e}2\Rightarrow f\left(x\right)=\frac{1}{2}\ln (4+2\sin x)$