Let f, $g : R \to R$ be functions defined by $f (x) = \{\begin{cases} [x], x < 0 \\ | 1 - x |, x ⩾ 0 \end{cases}$ and $g (x) = \{\begin{cases} e^{x} - x, x < 0 \\ (x - 1)^{2} - 1, x ⩾ 0 \end{cases}$
where [x] denote the greatest integer less than or equal to x. Then, the function fog is discontinuous at exactly :

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answer is B.

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Detailed Solution

Check continuity at x = 0 and also check continuity at those x where g(x) = 0

at x = 0, 2

${\begin{array}{l} fog (0^{+}) = f (0) = 1 \\ fog (0^{-}) = f (1) = |1 - 1| = 0 \end{array}}^{}$

Hence, discontinuous at x = 0

$\begin{array}{l} fog (2^{+}) = f (0) = 1 \\ fog (2^{-}) = f (0) = - 1 \end{array}$

Hence, discontinuous at x = 2

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