The number of points, where the function $$f: \mathbb{R} \to \mathbb{R}$$, $$f(x) = |x - 1|\cos|x - 2|\sin|x - 1| + (x - 3)|x^2 - 5x + 4|$$, is NOT differentiable, is

We have $$f(x) = |x - 1|\cos|x - 2|\sin|x - 1| + (x - 3)|x^2 - 5x + 4|$$. We note that $$x^2 - 5x + 4 = (x - 1)(x - 4)$$, so the critical points to examine for differentiability are $$x = 1, 2, 3,$$ and $$4$$.

At $$x = 2$$: Near $$x = 2$$, we have $$|x - 1| = x - 1$$ (smooth) and $$\sin|x - 1| = \sin(x - 1)$$ (smooth). Also, $$\cos|x - 2| = \cos(|x - 2|)$$, and since cosine is an even function, $$\cos|x - 2| = \cos(x - 2)$$, which is differentiable. For the second term, $$x^2 - 5x + 4 < 0$$ near $$x = 2$$, so $$|x^2 - 5x + 4| = -(x^2 - 5x + 4)$$, a polynomial. Hence $$f$$ is differentiable at $$x = 2$$.

At $$x = 3$$: Near $$x = 3$$, $$|x - 1| = x - 1$$ and $$|x - 2| = x - 2$$ are both smooth. Also $$x^2 - 5x + 4 < 0$$ near $$x = 3$$ (since $$9 - 15 + 4 = -2$$), so $$|x^2 - 5x + 4| = -(x^2 - 5x + 4)$$, again a polynomial. Hence $$f$$ is differentiable at $$x = 3$$.

At $$x = 1$$: The first term is $$|x - 1|\cos|x - 2|\sin|x - 1|$$. Near $$x = 1$$, $$|x - 1|\sin|x - 1| \approx (x - 1)^2$$ (since $$\sin|x-1|/|x-1| \to 1$$), which is smooth with derivative 0 at $$x = 1$$. So the first term is differentiable at $$x = 1$$.

The second term is $$(x - 3)|(x - 1)(x - 4)| = (x - 3)(4 - x)|x - 1|$$ near $$x = 1$$ (since $$x - 4 < 0$$). For $$x > 1$$: this equals $$(x - 3)(4 - x)(x - 1)$$, with derivative at $$x = 1$$: evaluating from the expansion, we get $$(-2)(3)(1) = -6$$... more precisely, we compute the derivative of $$g(x) = (x-3)(4-x)(x-1)$$ at $$x = 1$$: $$g'(x)$$ at $$x = 1$$ equals $$(4-x)(x-1) + (x-3)(-(x-1)) + (x-3)(4-x)$$ evaluated at $$x = 1$$, giving $$0 + 0 + (-2)(3) = -6$$.

For $$x < 1$$: this equals $$(x - 3)(4 - x)(1 - x)$$, and by the same method the derivative at $$x = 1$$ is $$0 + 0 + (-2)(3)(-1) = 6$$.

Since the left derivative ($$6$$) differs from the right derivative ($$-6$$), $$f$$ is not differentiable at $$x = 1$$.

At $$x = 4$$: The first term is differentiable ($$|x - 1|$$ and $$|x - 2|$$ are smooth near $$x = 4$$). The second term is $$(x - 3)(x - 1)|x - 4|$$.

For $$x > 4$$: $$(x - 3)(x - 1)(x - 4)$$. The derivative at $$x = 4$$ is: expanding as $$h(x) = (x^2 - 4x + 3)(x - 4)$$, we get $$h'(x) = (2x - 4)(x - 4) + (x^2 - 4x + 3)$$, so $$h'(4) = 0 + 3 = 3$$.

For $$x < 4$$: $$(x - 3)(x - 1)(4 - x)$$. The derivative at $$x = 4$$ is: expanding as $$h(x) = (x^2 - 4x + 3)(4 - x) = -x^3 + 8x^2 - 19x + 12$$, we get $$h'(x) = -3x^2 + 16x - 19$$, so $$h'(4) = -48 + 64 - 19 = -3$$.

Since the left derivative ($$-3$$) differs from the right derivative ($$3$$), $$f$$ is not differentiable at $$x = 4$$.

In total, $$f$$ is not differentiable at exactly 2 points: $$x = 1$$ and $$x = 4$$.

Hence, the correct answer is Option B.