Key Concepts and Formulas

• Continuity of a function: A function $f(x)$ is continuous at a point $x=c$ if $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$. For a piecewise function, continuity needs to be checked at the points where the definition of the function changes.
• Differentiability of a function: A function $f(x)$ is differentiable at a point $x=c$ if the left-hand derivative and the right-hand derivative at $x=c$ are equal. That is, $\lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$. A function must be continuous at a point to be differentiable at that point.
• Greatest Integer Function $[x]$: The greatest integer function $[x]$ gives the largest integer less than or equal to $x$.

Step-by-Step Solution

Step 1: Simplify the definition of $f(x)$ for $0 \leq x \leq 2$. The function is defined as $f(x) = \min \{1+x+[x], x+2[x]\}$ for $0 \leq x \leq 2$. We need to analyze this in sub-intervals based on the value of $[x]$.

• Case 1: $0 \leq x < 1$. In this interval, $[x] = 0$. So, $f(x) = \min \{1+x+0, x+2(0)\} = \min \{1+x, x\}$. Since $1+x > x$ for all $x$, we have $f(x) = x$ for $0 \leq x < 1$.

• Case 2: $1 \leq x < 2$. In this interval, $[x] = 1$. So, $f(x) = \min \{1+x+1, x+2(1)\} = \min \{2+x, x+2\}$. Since $2+x = x+2$, we have $f(x) = x+2$ for $1 \leq x < 2$.

Step 2: Rewrite the piecewise function $f(x)$ with simplified intervals. Based on the analysis in Step 1, we can rewrite $f(x)$ as: $f(x)=\left\{\begin{array}{lc}3 x, & x<0 \\ x, & 0 \leq x<1 \\ x+2, & 1 \leq x<2 \\ 5, & x>2\end{array}\right.$

Step 3: Identify points of discontinuity ($\alpha$). We need to check for continuity at the points where the definition of $f(x)$ changes: $x=0$, $x=1$, and $x=2$.

• At $x=0$: Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 3x = 3(0) = 0$. Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$. Function value: $f(0) = 0$. Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$, $f(x)$ is continuous at $x=0$.

• At $x=1$: Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$. Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x+2) = 1+2 = 3$. Function value: $f(1) = 1+2 = 3$. Since $\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$, $f(x)$ is not continuous at $x=1$.

• At $x=2$: Left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x+2) = 2+2 = 4$. Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} 5 = 5$. Function value: $f(2) = 2+2 = 4$. Since $\lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x)$, $f(x)$ is not continuous at $x=2$.

The points where $f$ is not continuous are $x=1$ and $x=2$. Therefore, $\alpha = 2$.

Step 4: Identify points of non-differentiability ($\beta$). A function must be continuous at a point to be differentiable there. We already found that $f(x)$ is not continuous at $x=1$ and $x=2$, so it cannot be differentiable at these points. We also need to check the differentiability at $x=0$, where the function is continuous but the definition changes.

• At $x=0$: We need to compare the left-hand derivative and the right-hand derivative. Left-hand derivative: $f'_{-}(0) = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{3h - 0}{h} = \lim_{h \to 0^-} 3 = 3$. Right-hand derivative: $f'_{+}(0) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h - 0}{h} = \lim_{h \to 0^+} 1 = 1$. Since $f'_{-}(0) \neq f'_{+}(0)$, $f(x)$ is not differentiable at $x=0$.

• At $x=1$: Since $f(x)$ is not continuous at $x=1$, it is not differentiable at $x=1$.

• At $x=2$: Since $f(x)$ is not continuous at $x=2$, it is not differentiable at $x=2$.

The points where $f$ is not differentiable are $x=0$, $x=1$, and $x=2$. Therefore, $\beta = 3$.

Step 5: Calculate $\alpha + \beta$. We found $\alpha = 2$ (number of points of discontinuity) and $\beta = 3$ (number of points of non-differentiability). $\alpha + \beta = 2 + 3 = 5$.

Common Mistakes & Tips

• Incorrectly simplifying the $\min$ function: Carefully analyze the inequality $1+x > x$ and $2+x = x+2$ for the given intervals.
• Forgetting to check continuity before differentiability: A function must be continuous at a point to be differentiable at that point. If it's discontinuous, it's automatically non-differentiable.
• Errors in calculating limits or derivatives: Double-check the evaluation of limits at the boundary points and the calculation of left-hand and right-hand derivatives.

Summary

The problem requires us to find the number of points where the given piecewise function $f(x)$ is not continuous ($\alpha$) and not differentiable ($\beta$). First, we simplified the definition of $f(x)$ for the interval $0 \leq x \leq 2$ by considering the greatest integer function $[x]$. We then analyzed the continuity of $f(x)$ at the points where its definition changes ($x=0, 1, 2$). Subsequently, we determined the differentiability by comparing left-hand and right-hand derivatives at the points of continuity where the function definition changes, and by noting that discontinuity implies non-differentiability. Finally, we summed the number of points of discontinuity and non-differentiability.

The final answer is $\boxed{5}$.