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

The correct option is A $f\left(\frac{3\pi }{4}\right)=log\left(\frac{1}{2}\right)$

$\left(f\right(x){)}^2={\int }_0^{x}f\left(t\right)\frac{4}{\underset{}{sin2t}-\underset{}{4si{n}^{2}t+4}}dt,f\left(0\right)=0$
$\begin{array}{c}\frac{2tant}{1+ta{n}^{2}t}\\ =\frac{2tant}{se{c}^{2}t}\end{array}\{\begin{array}{cc}4(1-si{n}^{2}t)=4co{s}^{2}t=\frac{4}{se{c}^{2}t}& \\ & \end{array}$
$\left(f\right(x){)}^2={\int }_0^{x}f\left(t\right)\frac{4}{\frac{2tant}{se{c}^{2}t}+\frac{4}{se{c}^{2}t}}dt$
$\left(f\right(x){)}^2={\int }_0^{x}f\left(t\right)\frac{2se{c}^{2}t}{2ttant}dt$
Taking derivative
$2f\left(x\right)\times f^{\prime }\left(x\right)=f\left(x\right)\times \frac{2se{c}^{2}x}{2ttanx}$
$f^{\prime }\left(x\right)=\frac{se{c}^{2}x}{2+tanx}dx$
Integrating
$\int f^{\prime }\left(x\right)dx=\int \frac{se{c}^{2}x}{2+tanx}dx$
Let $tanx=t\Rightarrow se{c}^{2}xdx=dt$
$\Rightarrow f\left(x\right)=log|2+tanx|+c$
$\because f\left(0\right)=0\Rightarrow C=-log2.$
$\therefore f\left(x\right)=log|2+tanx|-log2.)$
$f\left(\frac{\pi }{4}\right)=log3-log2=log\frac{3}{2}$
$f\left(\frac{\pi }{2}\right)\ne 2$
$f\left(\frac{3\pi }{4}\right)=log|2-1|-log2=logdfrac12$