Question

The point of discontinuity of $f\left(x\right)=\tan \left(\frac{\pi x}{x+1}\right)$ other than x = -1 are :

• A. $x=0$
• B. $x=\pi$
• C. $x=\frac{2m+1}{1-2m}$
• D. $x=\frac{2m-1}{2m+1}$

Answer 1

Answer: (C)

We have, function $f\left(x\right)=\tan \left(\frac{\pi x}{x+1}\right)$ and we know that function f (x) is discontinuous at those points, where $\tan \left(\frac{\pi x}{x+1}\right)=\tan \frac{\pi }{2}$
($\because \tan \frac{\pi }{2}$ is not defined)
By using $\tan \theta =\tan \alpha ,$ we have $\theta =m\pi +\alpha$
$\Rightarrow \frac{\pi x}{x+1}=m\pi +\frac{\pi }{2}$
$\Rightarrow \pi \left(\frac{x}{x+1}\right)=\pi \left(m+\frac{1}{2}\right)$
$\Rightarrow \left(\frac{x}{x+1}\right)=m+\frac{1}{2}$
$\Rightarrow \left(\frac{x}{x+1}\right)=\frac{2m+1}{2}$
$\Rightarrow \frac{x}{x+1}=\frac{2m+1}{2}$
$\Rightarrow 2x=(2m+1)x+(2m+1)$
$\Rightarrow (2-2m-1)x=2m+1\Rightarrow x=\frac{2m+1}{1-2m}$