$f (x) = \frac{4 - x^{2}}{4 x - x^{3}}$ . Find the value of $x$ at which $f (x)$ is discontinuous

x = 0, 1, - 2.

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x = 1, 2, - 2.

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x = 1, 1, - 2.

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x = 0, 2, - 2.

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Solution

The correct option is C $x = 0, 2, - 2.$
The function will be dicontinuous if the denominator doesnt exist
i.e., when,
$4 x - x^{3} = 0$
$x (x^{2} - 4) = 0$
$x = 0, 2, - 2$

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