Question:

Step 1: Analyze the function definition and identify the pieces. We have a piecewise function where: - For $x \leq 1$: $f(x) = x^2 - 2|x| + a$ - For $x > 1$: $f(x) = 6 + x$ We need to find when $f(x)$ has a local minimum at $x = 1$. Step 2: Evaluate $f(1)$ from the left piece. Since $x = 1 \leq 1$, we use the first piece. At $x = 1$, we have $|1| = 1$, so: $$f(1) = 1^2 - 2(1) + a = 1 - 2 + a = a - 1$$ Step 3: Evaluate the right-hand limit as $x$ approaches 1 from the right. For $x > 1$, we use $f(x) = 6 + x$. Therefore: $$\lim_{x \to 1^+} f(x) = 6 + 1 = 7$$ Step 4: Analyze the behavior of $f(x)$ for $x \leq 1$. We need to split this into two cases based on the sign of $x$: **Case 1:** For $0 \leq x \leq 1$, we have $|x| = x$, so: $$f(x) = x^2 - 2x + a$$ $$f'(x) = 2x - 2 < 0 \text{ for } x < 1$$ This means $f$ is strictly decreasing on $[0, 1)$. **Case 2:** For $x < 0$, we have $|x| = -x$, so: $$f(x) = x^2 + 2x + a$$ $$f'(x) = 2x + 2 = 0 \text{ at } x = -1$$ This shows there is a local minimum at $x = -1$ in the left piece. Step 5: Determine the condition for a local minimum at $x = 1$. For $f(x)$ to have a local minimum at $x = 1$, we need $f(1)$ to be less than or equal to the values of $f(x)$ for all $x$ in a neighborhood of 1. - From the left: Since $f$ is decreasing on $(0, 1)$, we have $f(1) < f(x)$ for $x \in (0, 1)$ near 1. ✓ - From the right: Since $f(x) = 6 + x$ is increasing for $x > 1$, we need: $$f(1) \leq \lim_{x \to 1^+} f(x)$$ $$a - 1 \leq 7$$ $$a \leq 8$$ Step 6: Determine if the inequality should be strict. For a strict local minimum (which is the standard definition), we require: $$f(1) < \lim_{x \to 1^+} f(x)$$ $$a - 1 < 7$$ $$a < 8$$ This gives us $a \leq 7$. Step 7: Count the positive integral values of $a$. The positive integers satisfying $a < 8$ are: $$a \in \{1, 2, 3, 4, 5, 6, 7\}$$ This gives us **7 positive integral values**. The answer is **Option 2: 7**.