Limits, Continuity & Differentiability
Limits and Polynomial Functions
Grade 12
Question:
<p>Let \(f(x) = ax^4 + bx^3 + cx^2 + dx + e\). If \(\displaystyle\lim_{x \to 0}\left(\dfrac{f(x)}{x^2} + 1\right) = 3\), \(f'(1) = 0\) and \(f'(2) = 0\), then the value of \(a\) is:</p>
<p>\(\dfrac{1}{4}\)</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(\dfrac{1}{3}\)</p>
<p>\(\dfrac{1}{6}\)</p>
Step-by-Step Solution
Key Concept: Since the limit exists and equals 3, the numerator f(x) must be divisible by x² (meaning f(0) = 0 and f'(0) = 0), allowing us to extract that c = -e. Combined with two derivative conditions at x = 1 and x = 2, we can determine all coefficients.
<p><strong>Step 1:</strong> Analyze the limit condition.</p><p>For the limit to exist and equal 3: lim(x→0)[f(x)/x² + 1] = 3</p><p>This means lim(x→0)[f(x)/x²] = 2, which requires f(x) to be divisible by x².</p><p>Therefore: f(0) = 0 ⟹ <strong>e = 0</strong></p><p>And: f'(0) = 0 ⟹ <strong>d = 0</strong></p><p><strong>Step 2:</strong> Simplify f(x) with these constraints.</p><p>f(x) = ax⁴ + bx³ + cx²</p><p>The limit becomes: lim(x→0)[(ax⁴ + bx³ + cx²)/x²] = lim(x→0)[ax² + bx + c] = c = 2</p><p>So <strong>c = 2</strong></p><p><strong>Step 3:</strong> Apply derivative conditions.</p><p>f'(x) = 4ax³ + 3bx² + 2cx = 4ax³ + 3bx² + 4x</p><p>From f'(1) = 0: 4a + 3b + 4 = 0 ... (i)</p><p>From f'(2) = 0: 32a + 12b + 8 = 0 ... (ii)</p><p><strong>Step 4:</strong> Solve the system.</p><p>Equation (ii) ÷ 4: 8a + 3b + 2 = 0 ... (ii')</p><p>Subtracting (i) from (ii'): 4a - 2 = 0</p><p><strong>∴ a = 1/2 = Answer: B</strong></p>
Correct Answer: B