Let $$f : [0, 3] \rightarrow R$$ be defined by $$f(x) = \min\{x - [x], 1 + [x] - x\}$$ where $$[x]$$ 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 _________.

We have the function $$f:[0,3]\to\mathbb R$$ given by

$$f(x)=\min\{x-[x],\;1+[x]-x\},$$

where $$[x]$$ denotes the greatest integer less than or equal to $$x$$. For each integer $$n$$ with $$0\le n\le2$$, every point $$x$$ in the sub-interval $$[n,n+1)$$ satisfies $$[x]=n$$, so we may write

$$x-[x]=x-n\quad\text{and}\quad 1+[x]-x = 1+n-x.$$

If we introduce the fractional part $$\{x\}=x-n\;(0\le\{x\}<1)$$, then inside the same sub-interval we obtain the much simpler expression

$$f(x)=\min\{\{x\},\;1-\{x\}\}.$$

Now $$\{x\}$$ increases linearly from $$0$$ to $$1$$ while $$1-\{x\}$$ decreases linearly from $$1$$ to $$0$$. The smaller of the two is therefore

$$\{x\}\quad\text{when}\quad\{x\}\le\frac12,\qquad\text{and}\qquad1-\{x\}\quad\text{when}\quad\{x\}\ge\frac12.$$

Hence on each open unit interval $$(n,n+1)$$ the function assumes the piece-wise linear form

$$ f(x)= \begin{cases} \{x\}, & 0\le\{x\}\le\dfrac12,\\\\ 1-\{x\}, & \dfrac12\le\{x\}<1. \end{cases} $$

Continuity is checked first. At a non-end-point $$x$$ lying strictly between two consecutive integers, $$f$$ is given by one of the linear formulas above, so it is continuous there. At an integer $$k\;(k=0,1,2,3)$$ we have

$$\lim_{x\to k^-}f(x)=0,\qquad\lim_{x\to k^+}f(x)=0,$$

while directly from the definition

$$f(k)=\min\{k-k,\;1+k-k\}=0.$$

Since the left limit, the right limit and the value of the function coincide, $$f$$ is continuous at every integer as well. Thus the set

$$P=\{x\in[0,3]\mid f\text{ is discontinuous at }x\}$$

is empty, so $$|P|=0.$$

We now investigate differentiability. Inside each open interval $$(n,n+1)$$ we already have linear pieces, so the derivative exists there and equals

$$f'(x)= \begin{cases} 1, & 0<\{x\}<\dfrac12,\\\\ -1, & \dfrac12<\{x\}<1. \end{cases} $$

Possible trouble points are where these two formulas meet, namely where $$\{x\}=\dfrac12$$, and also the integers where two adjacent unit intervals meet.

(i) Points with fractional part &frac12.
Solving $$\{x\}=\dfrac12$$ in each interval $$(n,n+1)$$ gives $$x=n+\dfrac12.$$ Within the domain $$[0,3]$$ this yields the three points

$$x=0.5,\;1.5,\;2.5.$$

At each of them the left derivative equals $$1$$ while the right derivative equals $$-1$$, so $$f$$ is not differentiable there.

(ii) Integer points.
At an integer $$k$$ the left derivative (coming from the slope $$-1$$ of the segment that ends at $$x=k$$) equals $$-1$$, whereas the right derivative (coming from the slope $$1$$ of the segment that starts at $$x=k$$) equals $$1$$. Therefore $$f$$ fails to be differentiable at every integer.

All integers in our closed interval are $$0,1,2,3$$, but the set

$$Q=\{x\in(0,3)\mid f\text{ is not differentiable at }x\}$$

excludes the endpoints $$0$$ and $$3$$. Consequently only $$1$$ and $$2$$ from the list above belong to $$Q$$.

Collecting everything lying strictly between $$0$$ and $$3$$, the non-differentiable points are

$$1,\;2,\;0.5,\;1.5,\;2.5,$$

so $$|Q|=5.$$

Finally, the required sum is

$$|P|+|Q| = 0 + 5 = 5.$$

Hence, the correct answer is Option 5.