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$f (x) = [x^{2}] - {[x]}^{2}$ , then (where [.] denotes greatest integer function)

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f is not continuous $x = 0 a n d x = 1$

f is continuous at $x = 0$ but not at $x = 1$

f is not continuous at $x = 0$ but continuous at $x = 1$

f is continuous at $x = 0$ and $x = 1$

answer is C.

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Detailed Solution

$f (0 -) = 0 - {(- 1)}^{2} = - 1 a n d f (0) = 0.$ Hence $f$ is not continuous at $x = 0$

(1) $f (1 -) = 0 - 0 = 0$ , $f (1 +) = 1 - 1 = 0$ $f (1) = 0$ and Thus $f$ is continuous at $x = 1$

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