<p>Let \(f(x)\) be a polynomial of degree four having extreme values at \(x = 1\) and \(x = 2\). If \(\lim_{x \to 0}\left[1 + \dfrac{f(x)}{x^2}\right] = 3\), then \(f(2)\) is equal to</p>
Step-by-Step Solution
Key Concept: Since f(x) has extreme values at x=1 and x=2, both are critical points where f'(1)=0 and f'(2)=0. Combined with the limit condition that forces f(0)=0 and f'(0)=0 (making x=0 a double root of f), we can construct f(x)=ax²(x-1)(x-2) and use the limit to find the constant.
<p><strong>Step 1:</strong> Since f(x) has extreme values at x=1 and x=2, we have f'(1)=0 and f'(2)=0. Since f is degree 4, f'(x) is degree 3 with roots at x=1 and x=2.</p><p><strong>Step 2:</strong> Analyze the limit: $\lim_{x \to 0}\left[1+\frac{f(x)}{x^2}\right] = 3$ implies $\lim_{x \to 0}\frac{f(x)}{x^2} = 2$. This means f(0)=0 and f'(0)=0 (double root at x=0).</p><p><strong>Step 3:</strong> Therefore f'(x) has roots at x=0 (double), x=1, and x=2. So: $f'(x) = ax^2(x-1)(x-2) = ax^2(x^2-3x+2)$ for some constant a.</p><p><strong>Step 4:</strong> Integrate: $f(x) = a\int x^2(x^2-3x+2)dx = a\int(x^4-3x^3+2x^2)dx = a\left(\frac{x^5}{5}-\frac{3x^4}{4}+\frac{2x^3}{3}\right) + C$. Since f(0)=0, we have C=0.</p><p><strong>Step 5:</strong> Use the limit condition: $\lim_{x \to 0}\frac{f(x)}{x^2} = \lim_{x \to 0}a\left(\frac{x^3}{5}-\frac{3x^2}{4}+\frac{2x}{3}\right) = a\cdot\frac{2}{3} = 2$, so $a = 3$.</p><p><strong>Step 6:</strong> Calculate $f(2) = 3\left(\frac{32}{5}-\frac{3\cdot16}{4}+\frac{2\cdot8}{3}\right) = 3\left(\frac{32}{5}-12+\frac{16}{3}\right) = 3\left(\frac{96-180+80}{15}\right) = 3\cdot\frac{-4}{15} = -\frac{4}{5}$</p><p>∴ Answer: B</p>
Correct Answer: B