Consider the function : $$f(x) = [x] + |1 - x|$$, $$-1 \leq x \leq 3$$ where [x] is the greatest integer function.
Statement 1: $$f$$ is not continuous at $$x = 0, 1, 2$$ and 3.
Statement 2: f(x) =$$\begin{cases}-x, & -1 \le x < 0 \\1 - x, & 0 \le x < 1 \\1 + x, & 1 \le x < 2 \\2 + x, & 2 \le x \le 3\end{cases}$$

Breaking down $$f(x) = [x] + |1 - x|$$ across the intervals:

For $$-1 \leq x < 0$$: $$f(x) = -1 + (1 - x) = -x$$

For $$0 \leq x < 1$$: $$f(x) = 0 + (1 - x) = 1 - x$$

For $$1 \leq x < 2$$: $$f(x) = 1 + (x - 1) = x$$

For $$2 \leq x < 3$$: $$f(x) = 2 + (x - 1) = 1 + x$$

At $$x = 3$$: $$f(3) = 3 + |1 - 3| = 5$$

Comparing this with Statement-2, the interval $$1 \leq x < 2$$ should yield $$x$$, not $$1+x$$. Thus, Statement-2 is False.
The function is discontinuous at $$x = 0, 1, 2,$$ and $$3$$ (As G.I.F. is discontinuous at integral values). Thus, Statement-1 is True.