Question:

Step 1: Understand the condition for strictly increasing function. For $f(x) = (x^2 + ax + 2a)e^x$ to be strictly increasing on $(-\infty, \infty)$, we need $f'(x) \geq 0$ for all $x \in \mathbb{R}$ (with equality only at isolated points). Step 2: Find the derivative of $f(x)$. Using the product rule: $$f'(x) = (2x + a)e^x + (x^2 + ax + 2a)e^x$$ Factoring out $e^x$: $$f'(x) = e^x[(2x + a) + (x^2 + ax + 2a)]$$ $$f'(x) = e^x[x^2 + (a+2)x + (a + 2a)]$$ $$f'(x) = e^x[x^2 + (a+2)x + 3a]$$ Step 3: Apply the condition for non-negativity. Since $e^x > 0$ for all $x$, the sign of $f'(x)$ depends entirely on $g(x) = x^2 + (a+2)x + 3a$. For $f'(x) \geq 0$ for all $x$, we need: $$g(x) = x^2 + (a+2)x + 3a \geq 0 \text{ for all } x \in \mathbb{R}$$ Step 4: Apply the discriminant condition. For a quadratic $g(x)$ to be non-negative for all real $x$, its discriminant must be non-positive: $$\Delta = (a+2)^2 - 4(1)(3a) \leq 0$$ Expanding: $$a^2 + 4a + 4 - 12a \leq 0$$ $$a^2 - 8a + 4 \leq 0$$ Step 5: Find the roots of the quadratic inequality. Using the quadratic formula: $$a = \frac{8 \pm \sqrt{64 - 16}}{2} = \frac{8 \pm \sqrt{48}}{2} = \frac{8 \pm 4\sqrt{3}}{2} = 4 \pm 2\sqrt{3}$$ Step 6: Determine the range of $a$. Since the coefficient of $a^2$ is positive, the quadratic $a^2 - 8a + 4 \leq 0$ is satisfied when: $$4 - 2\sqrt{3} \leq a \leq 4 + 2\sqrt{3}$$ Step 7: Calculate the numerical bounds. Since $\sqrt{3} \approx 1.732$, we have $2\sqrt{3} \approx 3.464$. Therefore: $$4 - 3.464 \leq a \leq 4 + 3.464$$ $$0.536 \leq a \leq 7.464$$ Step 8: Identify the integral values of $a$. The integers in the interval $[0.536, 7.464]$ are: $$a \in \{1, 2, 3, 4, 5, 6, 7\}$$ This gives us **7 integral values** of $a$. **Final Answer:** The number of integral values of $a$ is **7**, which corresponds to **Option 3**.