Question:

<div class="solution"> <p><strong>Step 1:</strong> We start by analyzing the given limits to extract information about the cubic polynomial \(y = f(x)\). The first limit, \(\lim_{x \to 0}(1+f(x))^{\frac{1}{x}} = e^{-1}\), suggests a relationship with the exponential function and implies that as \(x\) approaches 0, \(1 + f(x)\) approaches \(e^{-x}\), because \((e^{-1})^x = e^{-x}\) for small \(x\). This indicates \(f(x)\) approaches \(-1\) as \(x\) approaches 0, since \(e^{-x} \approx 1 - x\) for small \(x\), implying \(f(x) \approx -x\) for small \(x\).</p> <p><strong>Step 2:</strong> The second limit, \(\lim_{x \to 0}\left(x^3 f\!\left(\frac{1}{x}\right)\right)^{\frac{1}{x}} = e^2\), can be rewritten as \(\lim_{x \to 0} e^{\frac{1}{x} \ln\left(x^3 f\!\left(\frac{1}{x}\right)\right)} = e^2\). This implies \(\lim_{x \to 0} \frac{1}{x} \ln\left(x^3 f\!\left(\frac{1}{x}\right)\right) = 2\). Simplifying, we find \(\lim_{x \to 0} \ln\left(x^3 f\!\left(\frac{1}{x}\right)\right) = 2x\), which further simplifies to \(\lim_{x \to 0} x^3 f\!\left(\frac{1}{x}\right) = e^{2x}\). For small \(x\), \(f\!\left(\frac{1}{x}\right)\) must behave in such a way that this limit holds, suggesting \(f\!\left(\frac{1}{x}\right)\) approaches \(e^{2x}x^{-3}\) as \(x\) approaches 0.</p> <p><strong>Step 3:</strong> Considering \(f(x)\) is a cubic polynomial, we can express it as \(f(x) = ax^3 + bx^2 + cx + d\). From the first limit, we deduced \(f(x) \approx -x\) for small \(x\), implying \(d = 0\) and \(c = -1\) (since for small \(x\), terms with higher powers of \(x\) are negligible). The second limit implies that as \(x\) approaches infinity, \(f(x)\) behaves like \(2x^3\), because \(f\!\left(\frac{1}{x}\right)\) approaches \(2x^3\) as \(x\) approaches 0 (considering the substitution \(x \to \frac{1}{x}\)). This suggests \(a = 2\), since the leading term of \(f(x)\) dominates its behavior for large \(x\).</p> <p><strong>Step 4:</strong> With \(a = 2\), \(c = -1\), and \(d = 0\), we have \(f(x) = 2x^3 + bx^2 - x\). The sum of the roots of \(f(x) = 0\) is given by \(-\frac{b}{2}\) (from Vieta's formulas), and the product of the roots is 0 (since \(d = 0\), implying at least one root is 0). The limit \(\lim_{x \to \infty}\left(\frac{f(x)}{x^3}\right) = \lim_{x \to \infty}\left(2 + \frac{b}{x} - \frac{1}{x^2}\right) = 2\), confirming the behavior of \(f(x)\) for large \(x\).</p> <p><strong>Answer:</strong> The correct options are "Product of all real roots of \(f(x) = 0\) is 0" and "\(\lim_{x \to \infty}\left(\frac{f(x)}{x^3}\right) = 2\)".</p> <div class="key-concept"><strong>Key Concept:</strong> Understanding the behavior of cubic polynomials, applying limits to deduce coefficients, and using Vieta's formulas to relate coefficients to roots.</div> </div>