Discussion
The set of all points where the function \(f(x) = \frac{x}{x^{2}-4}, x \in \mathbb{R}\) , is discontinuous, is
Answer: E
The function is discontinuous at {-2, 2}.
A rational function is continuous wherever its denominator is non-zero. Here \(x^{2} - 4 = 0\) at x = 2 and x = -2, where f is not defined, so these are the points of discontinuity.
At x = 0 the numerator is zero but the denominator is -4, so f(0) = 0 and f is continuous there; any option including 0 is wrong. {2, 4} and {0, 2} miss the root -2.
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