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 _________.
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Functions involving $\min$, $\max$, absolute value, or integer/fractional parts often have points of non-differentiability at the boundaries where their definition changes. Sketching a rough graph of the function is an excellent way to visually identify these "sharp corner" points.
Step 1: Rewrite the function Let $\{x\}=x-[x]$ be the fractional part of $x$. Then, \[ f(x)=\min\{\{x\},\,1-\{x\}\}. \] Thus, on each interval $[n,n+1)$, the function forms a symmetric ``V-shape'' with: - minimum value $0$ at integers, - maximum value $\frac{1}{2}$ at the midpoint. Step 2: Analyze continuity Possible discontinuities occur only at integers due to $[x]$. At $x=1$: \[ \lim_{x\to1^-}f(x)=0,\quad f(1)=0,\quad \lim_{x\to1^+}f(x)=0 \] At $x=2$: \[ \lim_{x\to2^-}f(x)=0,\quad f(2)=0,\quad \lim_{x\to2^+}f(x)=0 \] Hence, $f$ is continuous at all points of $[0,3]$. \[ P=\varnothing \quad\Rightarrow\quad |P|=0 \] Step 3: Analyze differentiability The function is not differentiable at points where the minimum switches or where sharp corners occur. Midpoints of intervals: \[ x=\tfrac{1}{2},\ \tfrac{3}{2},\ \tfrac{5}{2} \] Integers (change in slope): \[ x=1,\ 2 \] At these points, left-hand and right-hand derivatives are unequal. \[ Q=\left\{\tfrac{1}{2},\,1,\,\tfrac{3}{2},\,2,\,\tfrac{5}{2}\right\} \quad\Rightarrow\quad |Q|=5 \] Step 4: Final answer \[ |P|+|Q|=0+5=5 \] Answer: \(\boxed{5}\)
