Let f:[0,3]R be defined by f(x)=x-[x],1+[x]-x where [x] is the greatest integer x.

Question

Let $f:[0,3]	o\mathbb{R}$ be defined by $f(x)=\min\{x-[x],1+[x]-x\}$ where $[x]$ is the greatest integer $\le x$. Let $P$ be the set of $x\in(0,3)$ where $f$ is discontinuous, and $Q$ the set of $x\in(0,3)$ where $f$ is not differentiable. Then the sum of the number of elements in $P$ and $Q$ is equal to .............

Answer: 5

Accepted Answer

$f(x)=\min\{\{x\},1-\{x\}\}$, which is continuous everywhere on $[0,3]$ but has corners at $x=\frac12,\frac32,\frac52$ (where the two pieces cross) and at $x=1,2$ (where $\{x\}$ resets). So $n(P)=0$ and $n(Q)=5$, giving $n(P)+n(Q)=5$.

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