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$f (x) = [x]^{2} - {x}^{2}$ where $[.]$ and $\{.\}$ denote the greatest integer function and the fractional part respectively, is

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continuous at $x = 1, - 1$

continuous at $x = - 1$ , but not at $x = 1$

continuous at $x = - 1$ but not at $x = - 1$

discontinuous at $x = 1 a n d x = - 1$

answer is D.

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Detailed Solution

$f (x) = [x^{2}] - {x}^{2} \Rightarrow f (- 1) = 1, f (- 1^{-}) = 1 - 1 = 0$

$\Rightarrow f (1) = 1, f (1^{+}) = 1 - 0 = 1$

$f (1^{-}) = 0 - 1 = - 1 Thus, f (x) is discontinuous at x = 1, - 1 .$

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