Question

Let $f\left(x\right)$ be a continuous and not a constant function of all $x$ in its domain, such that
$\left(f\right(x){)}^2=\underset{0}{\overset{x}{\int }}f\left(t\right)\frac{4}{\sin 2t-4{\sin }^{2}t+4}dt$ and $f\left(0\right)=0,$ then

• A. $f\left(\frac{3\pi }{4}\right)=\log \left(\frac{1}{2}\right)$
• B. $f\left(\frac{\pi }{4}\right)=\log \left(\frac{3}{2}\right)$
• C. $f\left(\frac{\pi }{4}\right)=\log \left(\frac{5}{4}\right)$
• D. $f\left(\frac{\pi }{2}\right)=2$

Answer 1

Answer: (A), (B)

$f\left(\frac{\pi }{4}\right)=\log \left(\frac{3}{2}\right)$

$\left(f\right(x){)}^2={\int }_0^{x}f\left(t\right)\frac{4}{\sin 2t-4si{n}^{2}t+4}dt,f\left(0\right)=0$

$\sin 2t=\frac{2\tan t}{1+{\tan }^{2}t}-4{\sin }^{2}t+4=4(1-{\sin }^{2}t)=4{\cos }^{2}t=\frac{4}{{\sec }^{2}t}$
$\left(f\right(x){)}^2={\int }_0^{x}f\left(t\right)\frac{4}{\frac{2\tan t}{{\sec }^{2}t}+\frac{4}{{\sec }^{2}t}}dt$
$\left(f\right(x){)}^2={\int }_0^{x}f\left(t\right)\frac{2{\sec }^{2}t}{2+\tan t}dt$
Taking derivative
$2f\left(x\right)\times f^{\prime }\left(x\right)=f\left(x\right)\times \frac{2{\sec }^{2}x}{2+\tan x}$
$f^{\prime }\left(x\right)=\frac{{\sec }^{2}x}{2+\tan x}dx$
On integrating both the sides, we have -
$\int f^{\prime }\left(x\right)dx=\int \frac{{\sec }^{2}x}{2+\tan x}dx$
Let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$\Rightarrow f\left(x\right)=\log |2+\tan x|+c$
$\because f\left(0\right)=0\Rightarrow c=-\log 2.$
$\therefore f\left(x\right)=\log |2+\tan x|-\log 2$
$f\left(\frac{\pi }{4}\right)=\log 3-\log 2=\log \frac{3}{2}$
$f\left(\frac{\pi }{2}\right)\ne 2$
$f\left(\frac{3\pi }{4}\right)=\log |2-1|-\log 2=\log \frac{1}{2}$