Let $$f(x) = -1 + |x - 2|$$, and $$g(x) = 1 - |x|$$; then the set of all points where $$f \circ g$$ is discontinuous is :

We have two real-valued functions defined on the entire real line:

$$f(x)= -1 + |x-2| \qquad\text{and}\qquad g(x)=1 - |x|.$$

First recall a basic fact: the absolute-value function $$|x|$$ is continuous for every real number. Adding or subtracting continuous functions (or constants) preserves continuity. Therefore both $$f(x)$$ and $$g(x)$$ are individually continuous at every real point. We now investigate the composition $$\bigl(f\circ g\bigr)(x)=f\!\bigl(g(x)\bigr).$$

To do that, we explicitly substitute $$g(x)$$ into $$f$$:

$$\bigl(f\circ g\bigr)(x)=f\!\bigl(g(x)\bigr)= -1 + \left|\,g(x)-2\,\right|.$$

First compute the inner difference:

$$g(x)-2 \;=\; \bigl(1-|x|\bigr)-2 \;=\; -1 - |x| \;=\; -\bigl(1+|x|\bigr).$$

Now apply the absolute value. Remember that $$1+|x| \ge 0$$ for all real $$x$$, so

$$\left|\, -\bigl(1+|x|\bigr)\,\right| \;=\; 1+|x|.$$

Substituting this back gives

$$\bigl(f\circ g\bigr)(x)= -1 + \bigl(1+|x|\bigr) = |x|.$$

Thus the composition simplifies beautifully to the simple absolute-value function $$|x|$$.

Because $$|x|$$ is continuous everywhere on the real line, the composition $$f\circ g$$ is also continuous everywhere. Hence there is no point of discontinuity.

Therefore the set of all points where $$f\circ g$$ is discontinuous is the empty set.

Hence, the correct answer is Option D.