Let $f(x)=|x-1|+|x-2|+|x-3|$. The number of points where $f(x)$ is not differentiable is:

Solution

1. The function $f(x)=|x-1|+|x-2|+|x-3|$ is a sum of three absolute value functions.
2. Each term $|x-a|$ is non-differentiable at $x=a$.
3. The sum of functions is non-differentiable at points where any of the individual terms are non-differentiable, provided they do not cancel out.
4. Here, the points of non-differentiability are $x=1$, $x=2$, and $x=3$.
5. There are 3 such points. Hence the answer is (D).