Question

If $f\left(x\right)=\int (\cot \frac{x}{2}-\tan \frac{x}{2})dx,$ where $f\left(\frac{\pi }{2}\right)=0,$ then which of the following statements is (are) CORRECT ?

$($Note : $\text{sgn}\left(y\right)$ denotes the signum function of $y.)$

• A. $\underset{0}{\overset{\pi }{\int }}f\left(x\right)dx=-2\pi \ln 2$
• B. $\underset{0}{\overset{\pi }{\int }}f\left(x\right)dx=-\pi \ln 2$
• C. $\text{sgn}\left(f\left(\frac{2\pi }{3}\right)\right)=-1$
• D. $\text{sgn}\left(f\left(\frac{2\pi }{3}\right)\right)=1$

Answer 1

Answer: (A), (C)

The correct options are
A $\underset{0}{\overset{\pi }{\int }}f\left(x\right)dx=-2\pi \ln 2$
C $\text{sgn}\left(f\left(\frac{2\pi }{3}\right)\right)=-1$

$f\left(x\right)=\int (\cot \frac{x}{2}-\tan \frac{x}{2})dx$

$\Rightarrow f\left(x\right)=\int (\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}-\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}})dx$

$\Rightarrow f\left(x\right)=\int \left(\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{\sin \frac{x}{2}\cdot \cos \frac{x}{2}}\right)dx$

$\Rightarrow f\left(x\right)=2\int \left(\frac{\cos x}{2\sin \frac{x}{2}\cdot \cos \frac{x}{2}}\right)dx$
$=2\int \cot xdx$
$=2\ln |\sin x|+C$

At $x=\frac{\pi }{2},f\left(\frac{\pi }{2}\right)=0$
$\Rightarrow C=0$
$\therefore f\left(x\right)=2\ln |\sin x|=\ln {\sin }^{2}x$

$\underset{0}{\overset{\pi }{\int }}f\left(x\right)dx$
$=\underset{0}{\overset{\pi }{\int }}2\ln \left(\sin x\right)dx$
$=2\cdot 2\underset{0}{\overset{\pi /2}{\int }}\ln \left(\sin x\right)dx$
$=4\left(\frac{-\pi }{2}\ln 2\right)=-2\pi \ln 2$

Now$,f\left(\frac{2\pi }{3}\right)=\ln \frac{3}{4}=-ve$
$\Rightarrow \text{sgn}\left(f\left(\frac{2\pi }{3}\right)\right)=-1$