Question:

Step 1: Recognize the ODE. The equation $x^2 f'(x) + 2x f(x) = e^x$ can be written as $\frac{d}{dx}[x^2 f(x)] = e^x$. Step 2: Integrate both sides. $x^2 f(x) = e^x + C$, so $f(x) = \frac{e^x + C}{x^2}$. Step 3: Apply the initial condition $f(2) = \frac{e^2}{4}$. $\frac{e^2 + C}{4} = \frac{e^2}{4} \Rightarrow C = 0$. Therefore $f(x) = \frac{e^x}{x^2}$. Step 4: Analyze $f(x) = 1$, i.e., $e^x = x^2$. Let $h(x) = e^x - x^2$. $h'(x) = e^x - 2x$. $h''(x) = e^x - 2 = 0 \Rightarrow x = \ln 2$. At $x = \ln 2$: $h'(\ln 2) = 2 - 2\ln 2 > 0$ (since $\ln 2 < 1$). So $h'(x) > 0$ for all $x$ (checking: $h'(x) = e^x - 2x$, minimum at $x = \ln 2$ gives $2 - 2\ln 2 > 0$). Thus $h(x)$ is strictly increasing, meaning $e^x = x^2$ has exactly one real solution. Option (a) is TRUE. Step 5: Analyze $f(x) = 3$, i.e., $e^x = 3x^2$. Let $g(x) = e^x - 3x^2$. $g'(x) = e^x - 6x = 0$. This has two solutions (one negative, one positive near $x \approx 1.79$). Analyzing the behavior: as $x \to -\infty$, $g \to +\infty$; $g$ decreases to a local min, increases to a local max, then increases to $+\infty$. The equation $e^x = 3x^2$ has exactly three real solutions. Option (b) is TRUE. Step 6: Find critical points of $f(x) = \frac{e^x}{x^2}$. $f'(x) = \frac{e^x \cdot x^2 - 2x \cdot e^x}{x^4} = \frac{e^x(x-2)}{x^3}$. Setting $f'(x) = 0$: $x = 2$ (since $e^x > 0$ always). Step 7: Determine the nature of the critical point at $x = 2$. For $x < 2$ (and $x > 0$): $f'(x) < 0$ (decreasing). For $x > 2$: $f'(x) > 0$ (increasing). So $x = 2$ is a local minimum. There is no local maximum. Option (d) is TRUE, option (c) is FALSE. Step 8: The correct options are (a), (b), and (d), i.e., options 1, 2, and 4.