Question:

Step 1: State the condition for continuity at a point. For a function to be continuous at $x=c$, the left-hand limit, the right-hand limit, and the function value at $c$ must all be equal. In this case, for $f(x)$ to be continuous at $x=2$, we must have: $$ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^-} f(x) = f(2) $$ Step 2: Evaluate the function value and the right-hand limit at $x=2$. For $x \geq 2$, the function is defined as $f(x) = 2x + \alpha^2$. We evaluate $f(2)$ by substituting $x=2$: $$ f(2) = 2(2) + \alpha^2 = 4 + \alpha^2 $$ The right-hand limit, $\lim_{x \to 2^+} f(x)$, will also be $4 + \alpha^2$. Step 3: Evaluate the left-hand limit at $x=2$. For $x < 2$, the function is defined as $f(x) = \frac{\alpha x}{2} + 10$. We find the left-hand limit by substituting $x=2$: $$ \lim_{x \to 2^-} f(x) = \frac{\alpha(2)}{2} + 10 = \alpha + 10 $$ Step 4: Equate the limits and the function value to satisfy the continuity condition. From Step 1, for continuity, $f(2)$ must be equal to $\lim_{x \to 2^-} f(x)$. Therefore, we set the expressions from Step 2 and Step 3 equal: $$ 4 + \alpha^2 = \alpha + 10 $$ Step 5: Rearrange the equation into a standard quadratic form. Subtract $\alpha + 10$ from both sides of the equation to get a quadratic equation: $$ \alpha^2 - \alpha - 6 = 0 $$ Step 6: Factor the quadratic equation. We look for two numbers that multiply to -6 and add to -1. These numbers are -3 and 2. So, we can factor the quadratic equation: $$ (\alpha - 3)(\alpha + 2) = 0 $$ Step 7: Solve for the possible values of $\alpha$. Setting each factor equal to zero gives the possible values for $\alpha$: $$ \alpha - 3 = 0 \implies \alpha = 3 $$ $$ \alpha + 2 = 0 \implies \alpha = -2 $$ Thus, the possible values for $\alpha$ are $3$ and $-2$. Step 8: Determine the number of possible values of $\alpha$. We found two distinct values for $\alpha$ that ensure the continuity of $f(x)$ at $x=2$. These values are $\alpha = 3$ and $\alpha = -2$. Therefore, there are 2 possible values for $\alpha$. The final answer is $\boxed{2}$.