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  4. The point of discontinuity of f(x) = tan((π x/x+1)) other than x = -1 are :

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

Continuity and Differentiability

Solution:

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} $
$\Rightarrow \:\: 2x = (2m +1) x + (2m +1)$
$\Rightarrow \:\: (2 - 2m - 1) x = 2 m + 1 \: \Rightarrow \: x = \frac{2m + 1}{1 - 2m}$