<p>Let <em>f</em>(<em>x</em>) be a polynomial of degree 4 having extreme values at <em>x</em> = 1 and <em>x</em> = 2. If \(\lim_{x \to 0}\left(\frac{f(x)}{x^2}+1\right)=3\), then <em>f</em>(−1) is equal to</p>
Step-by-Step Solution
Key Concept: Since f(x) has extreme values at x = 1 and x = 2, we know f'(1) = 0 and f'(2) = 0. This means f'(x) = k(x-1)(x-2) for some constant k, and we can integrate to find f(x). The limit condition determines the constant of integration.
<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.</p><p><strong>Step 2:</strong> For a degree 4 polynomial, f'(x) is degree 3. Since f'(x) has roots at x = 1 and x = 2, we write: f'(x) = a(x-1)(x-2)(x-r) for some constants a and r.</p><p><strong>Step 3:</strong> Evaluate the limit: $\lim_{x \to 0}\left(\frac{f(x)}{x^2}+1\right)=3$ implies $\lim_{x \to 0}\frac{f(x)}{x^2}=2$.</p><p>This means f(x) = 2x² + higher order terms, so f(0) = 0 and f'(0) = 0.</p><p><strong>Step 4:</strong> Since f'(0) = 0, we have: f'(x) = a(x-1)(x-2)·x, so f'(x) = ax(x-1)(x-2).</p><p><strong>Step 5:</strong> Integrate: $f(x) = a\int x(x-1)(x-2)dx = a\int (x^3-3x^2+2x)dx = a\left(\frac{x^4}{4}-x^3+x^2\right)+C$</p><p>Since f(0) = 0, we have C = 0.</p><p><strong>Step 6:</strong> From the limit condition, comparing f(x) ≈ 2x² for small x: the coefficient of x² is a(1) = 2, so a = 2.</p><p><strong>Step 7:</strong> Therefore: $f(x) = 2\left(\frac{x^4}{4}-x^3+x^2\right) = \frac{x^4}{2}-2x^3+2x^2$</p><p><strong>Step 8:</strong> $f(-1) = \frac{1}{2}-2(-1)+2(1) = \frac{1}{2}+2+2 = \frac{9}{2}$</p><p>∴ Answer: A</p>
Correct Answer: A