Question

Let $f:[0,3]\to R$ be defined by $f\left(x\right)=min\{x–[x],1+[x]–x\}$ where $\left[x\right]$ is the greatest integer less than or equal to $x$. Let $P$ denote the set containing all $x\in [0,3]$ where $f$ is discontinuous, and $Q$ denote the set containing all $x\in (0,3)$ where $f$ is not differentiable. Then the sum of number of elements in $P$ and $Q$ is equal to

Answer 1

Figure of $\left\{x\right\}:$


and figure of $1-\left\{x\right\}:$


Now, figure of $f\left(x\right)=min\left\{\right\{x\},1+\{x\left\}\right\}:$

So, there is no discontinuity in the interval $[0,3]$
$\therefore n\left(P\right)=0$
and set $Q,$ where $f$ is not differentiable is
$Q=\{\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2}\}$
$\therefore n\left(Q\right)=5$
Hence, $n\left(P\right)+n\left(Q\right)=5$