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

Given:

$$f(x) = \begin{cases} [x], & x < 0 \\ |1 - x|, & x \geq 0 \end{cases}$$

$$g(x) = \begin{cases} e^x - x, & x < 0 \\ (x - 1)^2 - 1, & x \geq 0 \end{cases}$$

where $$[x]$$ is the greatest integer function.

Analyze $$g(x)$$ on different intervals:

• For $$x < 0$$: $$g(x) = e^x - x$$. Since $$g'(x) = e^x - 1 < 0$$, $$g$$ is decreasing. As $$x \to -\infty$$, $$g \to +\infty$$; at $$x = 0^-$$, $$g \to 1$$. So $$g(x) > 1$$ for all $$x < 0$$.
• For $$x \geq 0$$: $$g(x) = (x-1)^2 - 1 = x^2 - 2x = x(x - 2)$$. Minimum at $$x = 1$$: $$g(1) = -1$$. Also $$g(0) = 0$$ and $$g(2) = 0$$.

Compute $$f(g(x))$$ on each interval:

For $$x < 0$$: $$g(x) > 1 \geq 0$$, so $$f(g(x)) = |1 - g(x)| = g(x) - 1 = e^x - x - 1$$. This is continuous on $$(-\infty, 0)$$.

For $$x \geq 0$$: We need to split based on the sign of $$g(x)$$:

• $$g(x) < 0$$ for $$x \in (0, 2)$$: Here $$g(x) \in [-1, 0)$$, so $$f(g(x)) = [g(x)]$$. Since $$g(x) \in [-1, 0)$$, we get $$[g(x)] = -1$$. So $$f(g(x)) = -1$$ on $$(0, 2)$$.
• $$g(x) = 0$$ at $$x = 0$$ and $$x = 2$$: $$f(0) = |1 - 0| = 1$$.
• $$g(x) > 0$$ for $$x > 2$$: $$f(g(x)) = |1 - g(x)| = |1 - x^2 + 2x|$$.

Check continuity at critical points:

At $$x = 0$$:

• Left limit: $$\lim_{x \to 0^-} (e^x - x - 1) = 1 - 0 - 1 = 0$$
• Right limit: $$\lim_{x \to 0^+} [g(x)] = [g(0^+)]$$. For $$x$$ slightly greater than 0, $$g(x) = x(x-2) \approx -2x < 0$$, so $$[g(x)] = -1$$.
• Value: $$f(g(0)) = f(0) = |1 - 0| = 1$$
• Left limit $$= 0$$, right limit $$= -1$$, value $$= 1$$: Discontinuous $$\boldsymbol{\times}$$

At $$x = 2$$:

• Left limit: $$\lim_{x \to 2^-} (-1) = -1$$
• Right limit: $$\lim_{x \to 2^+} |1 - x^2 + 2x| = |1 - 4 + 4| = 1$$
• Value: $$f(g(2)) = f(0) = 1$$
• Left limit $$= -1 \neq 1$$: Discontinuous $$\boldsymbol{\times}$$

For $$x > 2$$: $$f(g(x)) = |1 - x^2 + 2x| = |-(x^2 - 2x - 1)| = |-(x - 1)^2 + 2|$$

This equals $$2 - (x-1)^2$$ when $$(x-1)^2 < 2$$, i.e., $$x < 1 + \sqrt{2} \approx 2.414$$, and $$(x-1)^2 - 2$$ otherwise. At $$x = 1 + \sqrt{2}$$, the expression inside the absolute value is 0, and the function is continuous.

No other points of discontinuity exist (the function on $$(-\infty, 0)$$ and $$(0, 2)$$ and $$(2, \infty)$$ are each continuous within their intervals).

Therefore, $$f \circ g$$ is discontinuous at exactly two points: $$x = 0$$ and $$x = 2$$.

The correct answer is Option B: two points.