1. Given function

$f(x)=|x-1|\cos|x-2|\sin|x-1|+(x-3)|x^2-5x+4|$

We need the number of points where $f$ is not differentiable.

2. Identify possible non-differentiable points

A function involving modulus can fail to be differentiable where the expression inside modulus becomes zero.

So check:

• $|x-1| \Rightarrow x=1$
• $|x-2| \Rightarrow x=2$
• $|x^2-5x+4| \Rightarrow x^2-5x+4=0$

Now, $x^2-5x+4=(x-1)(x-4)$ so possible points are $x=1,2,4$

These are the only candidates.

3. Examine the first term

Let $f_1(x)=|x-1|\cos|x-2|\sin|x-1|$

Use the fact that $\sin|x-1|$ is actually smoothable with $|x-1|$:

Since $\sin t$ is odd, but here $t=|x-1|\ge 0$, it is better to combine: $|x-1|\sin|x-1|$

Let $u=|x-1|$. Then consider $u\sin u$. As a function of $u$, this is differentiable at $u=0$ because $u\sin u \sim u^2 \quad (u\to 0)$ So no issue comes from this factor at $x=1$.

Also, $\cos|x-2|$ is differentiable everywhere because $\cos|x-2|=\cos(x-2)$ (since cosine is even).

Hence $f_1(x)=|x-1|\sin|x-1|\cos(x-2)$

Now define $g(t)=|t|\sin|t|$ For $t>0$, $g(t)=t\sin t$; for $t<0$, $g(t)=-t\sin(-t)=t\sin t$. Thus actually $g(t)=t\sin t$ for all $t$.

So $|x-1|\sin|x-1|=(x-1)\sin(x-1)$

Therefore $f_1(x)=(x-1)\sin(x-1)\cos(x-2)$ which is differentiable for all real $x$.

So the first term gives no non-differentiable point.

4. Examine the second term

Let $f_2(x)=(x-3)|x^2-5x+4|=(x-3)|(x-1)(x-4)|$

Possible trouble points are $x=1$ and $x=4$.

We check whether multiplication by $(x-3)$ removes the cusp or not.

At $x=1$

Near $x=1$, write $x^2-5x+4=(x-1)(x-4)$ Since $x-4<0$ near $x=1$, we have $|x^2-5x+4|=|x-1||x-4|$ with $|x-4|$ smooth and nonzero near $1$. Thus locally this behaves like a nonzero smooth factor times $|x-1|$. Also $(x-3)$ at $x=1$ equals $-2\ne 0$. So near $x=1$, $f_2(x)\sim C|x-1| \quad (C\ne 0)$ which is not differentiable at $x=1$.

To verify by sign:

• For $x<1$: $(x-1)(x-4)>0$, so $f_2(x)=(x-3)(x^2-5x+4)$
• For $1<x<4$: $(x-1)(x-4)<0$, so $f_2(x)=-(x-3)(x^2-5x+4)$

Left and right derivatives at $x=1$ will differ by sign, so not differentiable.

At $x=4$

Similarly, $(x-3)$ at $x=4$ equals $1\ne 0$. Near $x=4$, $|x^2-5x+4|$ behaves like nonzero smooth factor times $|x-4|$. Hence $f_2(x)\sim C|x-4| \quad (C\ne 0)$ which is not differentiable at $x=4$.

5. Check $x=2$

The only modulus involving $x=2$ is in $\cos|x-2|$, but $\cos|x-2|=\cos(x-2)$ which is differentiable. So $x=2$ is not a non-differentiable point.

6. Conclusion

The function is not differentiable only at: $x=1,\;4$

Hence the number of non-differentiable points is $\boxed{2}$

So the correct option is B.

7. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.