The point of discontinuity of $f (x) = \tan (\frac{π x}{x + 1})$ other then $x = - 1$ are.

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$x = \frac{2 m + 1}{1 - 2 m}$

$x = 0$

$x = π$

$x = \frac{2 m - 1}{2 m + 1}$

answer is C.

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Detailed Solution

We have, function $f (x) = \tan (\frac{π x}{x + 1})$ and we know that function f (x) is discontinuous at those points.

Where $\tan (\frac{π x}{x + 1}) = \tan \frac{π}{2} (∵ \tan \frac{π}{2}$ is not defined)

By using $\tan θ = \tan α, we have θ = m π + α$

$\begin{array}{l} \Rightarrow \frac{π x}{x + 1} = m π + \frac{π}{2} \\ \Rightarrow π (\frac{x}{x + 1}) = π (m + \frac{1}{2}) \\ \Rightarrow (\frac{x}{x + 1}) = m + \frac{1}{2} \\ \Rightarrow \frac{x}{x + 1} = \frac{2 m + 1}{2} \\ \Rightarrow 2 x = (2 m + 1) x + (2 m + 1) \\ \Rightarrow (2 - 2 m - 1) x = 2 m + 1 \Rightarrow x = \frac{2 m + 1}{1 - 2 m} \end{array}$