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Number of points of discontinuity of $f (x) = [2 x^{3} - 5]$ in $[1, 2),$ is equal to: (where [x] denotes greatest integer less than or equal to $x$ )

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Detailed Solution

$f (x) = [2 x^{3} - 5] = [2 x^{3}] - 5$

Let $2 x^{3} = t$

$∴ g (t) = [t]$ where $t \in [2, 16)$

$g$ is continuous at $t = 2$ and discontinuous $t = 3, 4, ......., 15 = 13$ points.

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