<p>If <em>f</em>(<em>x</em>) − <em>l</em>(<em>x</em>) has four zeroes, where <em>l</em>(<em>x</em>) is linear and <em>f</em>(<em>x</em>) = <em>x</em><sup>4</sup> + 2<em>x</em><sup>3</sup> + <em>c</em><em>x</em><sup>2</sup> + 9<em>x</em> + 4, then find 100<em>b</em> where <em>b</em> is the largest value of <em>c</em> for which the second derivative has at least two zeroes. (Hint: the second derivative <em>f</em>″(<em>x</em>) = 6<em>x</em><sup>2</sup> + 6<em>x</em> + <em>c</em> = 0 has two zeroes if and only if the discriminant 36 − 24<em>c</em> > 0.)</p>
Step-by-Step Solution
Key Concept: If f(x) - l(x) has four zeroes where l(x) is linear, then f(x) - l(x) must be a perfect square of a quadratic (since a degree 4 polynomial with 4 equal roots or two pairs of equal roots). This means f(x) = (ax² + bx + d)² + l(x), constraining the coefficients of f(x). The second derivative f''(x) = 12x² + 12x + 2c must have real roots, giving the condition on c.
<p><strong>Step 1:</strong> Since f(x) - l(x) is a degree 4 polynomial with exactly four zeroes, it must have the form f(x) - l(x) = (px² + qx + r)² for some p, q, r.</p><p><strong>Step 2:</strong> This means f(x) = (px² + qx + r)² + (mx + n) where l(x) = mx + n is linear. Expanding: f(x) = p²x⁴ + 2pqx³ + (q² + 2pr)x² + (2qr + m)x + (r² + n).</p><p><strong>Step 3:</strong> Comparing with f(x) = x⁴ + 2x³ + cx² + 9x + 4: we need p² = 1, so p = 1. Then 2pq = 2, giving q = 1. Also q² + 2pr = 1 + 2r = c, so c = 1 + 2r.</p><p><strong>Step 4:</strong> The second derivative is f''(x) = 12x² + 12x + 2c. For f''(x) to have at least two real zeroes (distinct or equal), the discriminant must satisfy: Δ = 144 - 96c ≥ 0, which gives 36 - 24c ≥ 0, so c ≤ 3/2.</p><p><strong>Step 5:</strong> From Step 3, c = 1 + 2r where r can vary. The largest value of c satisfying c ≤ 1.5 is b = 1.5 = 3/2.</p><p><strong>Step 6:</strong> Therefore, 100b = 100 × 1.5 = 150.</p><p>∴ Answer: <strong>150</strong></p>
Correct Answer: 150