Let $$f(x) = 15 - |x - 10|$$; $$x \in R$$. Then the set of all values of $$x$$, at which the function $$g(x) = f(f(x))$$ is not differentiable, is:

We have the first‐level function defined as $$f(x)=15-|x-10|,\;x\in\mathbb R.$$

To discuss the differentiability of the composite $$g(x)=f\!\bigl(f(x)\bigr),$$ we must recall a fact about compositions: a composition $$F(G(x))$$ fails to be differentiable at a point $$x=a$$ if (i) the inner function $$G(x)$$ is not differentiable at $$x=a$$, or (ii) the outer function $$F(u)$$ is not differentiable at the value $$u=G(a).$$ We shall therefore find:

1. The point(s) where $$f(x)$$ itself is not differentiable.
2. The point(s) where the argument of the outer $$f$$ inside $$g(x)$$ equals the non-differentiability point of $$f$$.

Step 1 : Differentiability of $$f(x)$$

The expression $$f(x)=15-|x-10|$$ contains an absolute value. We know the standard result:

For $$h(x)=|x-c|,$$ the function $$h(x)$$ is not differentiable at $$x=c.$$ Everywhere else its derivative exists.

Here $$|x-10|$$ has a corner at $$x=10,$$ so $$f(x)$$ is not differentiable only at

$$x=10.$$

Step 2 : Values of $$x$$ that make the argument of the outer $$f$$ equal to 10

Inside the composite we have $$f(x)$$ fed into another $$f$$. The outer $$f$$ will have a corner whenever its argument equals 10. Hence we must solve

$$f(x)=10.$$

Substituting the definition of $$f(x),$$ we get

$$15-|x-10|=10.$$

Transposing terms gives

$$|x-10| = 15-10 =5.$$

Using the property $$|y|=k \Longrightarrow y=\pm k,$$ we obtain

$$x-10 = 5 \quad\text{or}\quad x-10 = -5.$$

Thus

$$x = 15 \quad\text{or}\quad x = 5.$$

Step 3 : Collecting all points of non-differentiability for $$g(x)$$

As explained, $$g(x)=f(f(x))$$ fails to be differentiable at

• $$x=10$$ (inner $$f$$ is not differentiable), and
• $$x=5,\;15$$ (outer $$f$$ is hit precisely at its corner because $$f(x)=10$$ there).

No other values of $$x$$ cause either type of failure, so the complete set is

$$\{5,\;10,\;15\}.$$

Hence, the correct answer is Option A.