Let $$f : \mathbb{R} \to \mathbb{R}$$ be defined as $$f(x) = \begin{cases} [e^x], & x < 0 \\ ae^x + [x-1], & 0 \leq x < 1 \\ b + [\sin(\pi x)], & 1 \leq x < 2 \\ [e^{-x}] - c, & x \geq 2 \end{cases}$$
where $$a, b, c \in \mathbb{R}$$ and $$[t]$$ denotes greatest integer less than or equal to $$t$$. Then, which of the following statements is true?

We analyze the piecewise function for continuity at the boundary points $$x = 0, 1, 2$$. For $$x < 0$$ we have $$f(x) = [e^x]$$, and since $$0 < e^x < 1$$ for $$x < 0$$, $$[e^x] = 0$$. For $$0 \le x < 1$$, $$f(x) = ae^x + [x-1]$$, and since $$x - 1 \in [-1, 0)$$, $$[x-1] = -1$$, so $$f(x) = ae^x - 1$$. For $$1 \le x < 2$$, $$f(x) = b + [\sin(\pi x)]$$; at $$x = 1$$, $$\sin(\pi) = 0$$ gives $$[\sin(\pi)] = 0$$ and $$f(1) = b$$, while for $$x \in (1, 2)$$, $$\sin(\pi x) < 0$$ since $$\pi x \in (\pi, 2\pi)$$ and $$-1 \le \sin(\pi x) < 0$$, so $$[\sin(\pi x)] = -1$$ and $$f(x) = b - 1$$. For $$x \ge 2$$, $$f(x) = [e^{-x}] - c$$, and since $$0 < e^{-x} < 1$$ for $$x \ge 2$$, $$[e^{-x}] = 0$$, hence $$f(x) = -c$$.

The left limit at $$x \to 0^-$$ is $$\lim_{x \to 0^-} [e^x] = 0$$, and the right value is $$f(0) = a\cdot1 - 1 = a - 1$$, so continuity at $$x = 0$$ gives $$a - 1 = 0$$ and thus $$a = 1$$.

At $$x = 1$$, $$f(1) = b$$ but the right-hand limit is $$\lim_{x \to 1^+} f(x) = b - 1$$, which never equals $$b$$ for any $$b$$, so $$f$$ is always discontinuous at $$x = 1$$, making Option A false.

The left limit at $$x \to 2^-$$ is $$b - 1$$ and $$f(2) = -c$$, so continuity at $$x = 2$$ requires $$b - 1 = -c$$, giving $$c = 1 - b$$.

For $$f$$ to be discontinuous at exactly one point (namely $$x = 1$$), it must be continuous at $$x = 0$$ and $$x = 2$$, so $$a = 1$$ and $$c = 1 - b$$. Since $$b$$ can be any real number, we have $$a + b + c = 1 + b + (1 - b) = 2 \neq 1$$.

Option B is false, and Option C is true. Therefore, the correct answer is Option C.