Question

Let $\text{sgn}\left(y\right)$ denote the signum function of $y.$ If $f\left(x\right)=\int (\cot \frac{x}{2}-\tan \frac{x}{2})dx$ and $f\left(\frac{\pi }{2}\right)=0,$ then which of the following statements is CORRECT?

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

Answer 1

The correct option is B $\text{sgn}\left(f\left(\frac{2\pi }{3}\right)\right)=-1$

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

$=2\ln \left|\sin \frac{x}{2}\right|+2\ln \left|\cos \frac{x}{2}\right|+C$

$f\left(x\right)=2\ln \left|\frac{\sin x}{2}\right|+C$
$\Rightarrow f\left(x\right)=2\ln |\sin x|+K$

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

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