JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Continuity and Differentiability

For $x\in [n,n+1)$ with $n$ an integer, $[x]=n$ and $f(x)=\min \{x-n,n+1-x\}$, which is $x-n$ for $x\le n+\frac{1}{2}$ and $n+1-x$ for $x\ge n+\frac{1}{2}$. So $f(x)$ is the distance from $x$ to the nearest integer, rising from $0$ at an integer to $\frac{1}{2}$ at the next half-integer and falling back to $0$.
At every integer of $[0,3]$ both one-sided limits are $0$ and $f$ itself is $0$ there, so $f$ is continuous throughout $[0,3]$ and $P$ is empty.
The slope changes from $+1$ to $-1$ at $x=\frac{1}{2},\frac{3}{2},\frac{5}{2}$ and from $-1$ to $+1$ at $x=1,2$, and at no other point of $(0,3)$, so $Q=\left\{\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2}\right\}$.
Thus $P$ has $0$ elements and $Q$ has $5$, so the sum is $5$.