Question:

Step 1: Recognize the differential equation structure. We observe that the given equation $x^2 f'(x) + 2xf(x) = e^x$ can be rewritten as a derivative of a product. Notice that: $$\frac{d}{dx}\left[x^2 f(x)\right] = x^2 f'(x) + 2xf(x)$$ Therefore, our equation becomes: $$\frac{d}{dx}\left[x^2 f(x)\right] = e^x$$ Step 2: Integrate both sides to find the general solution. Integrating both sides with respect to $x$: $$x^2 f(x) = \int e^x \, dx = e^x + C$$ where $C$ is the constant of integration. Solving for $f(x)$: $$f(x) = \frac{e^x + C}{x^2}$$ Step 3: Apply the initial condition to determine the constant. We are given that $f(2) = \frac{e^2}{4}$. Substituting into our general solution: $$\frac{e^2 + C}{4} = \frac{e^2}{4}$$ This gives us $e^2 + C = e^2$, so $C = 0$. Therefore, the particular solution is: $$f(x) = \frac{e^x}{x^2}$$ Step 4: Find the derivative to locate critical points. Using the quotient rule: $$f'(x) = \frac{e^x \cdot x^2 - e^x \cdot 2x}{x^4} = \frac{e^x(x - 2)}{x^3}$$ Critical points occur where $f'(x) = 0$. Since $e^x > 0$ for all $x$, we need $x - 2 = 0$, giving us $x = 2$ as the only critical point. Step 5: Analyze the sign of $f'(x)$ to determine the nature of critical points. We examine the sign of $f'(x) = \frac{e^x(x-2)}{x^3}$ in different regions: **For $x > 0$:** Both $e^x > 0$ and $x^3 > 0$, so the sign depends on $(x-2)$: - When $0 < x < 2$: $(x-2) < 0$, so $f'(x) < 0$ (function is decreasing) - When $x > 2$: $(x-2) > 0$, so $f'(x) > 0$ (function is increasing) At $x = 2$, $f'(x)$ changes from negative to positive, indicating a local minimum. **For $x < 0$:** We have $e^x > 0$, $x^3 < 0$, and $(x-2) < 0$, so: $$f'(x) = \frac{(+)(-)}{(-)} = \frac{(-)}{(-)} > 0$$ The function is strictly increasing for all $x < 0$, with no critical points in this region. Step 6: State the conclusion. Based on our analysis, $f(x)$ has a local minimum at $x = 2$ and no local maxima. **The answer is Option 4: $f(x)$ has no local maxima but local minima.**