<p><strong>Ex. 25 (B):</strong> If <p>a, b, c ∈ ℝ</p> such that <p>2a - 3b + 6c = 0</p>, then the equation <p>f(x) = (ax³/3) + (bx²/2) + cx + d</p> satisfies what condition?</p>
Step-by-Step Solution
Key Concept: Apply Rolle's Theorem: if f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), then f'(ξ) = 0 for some ξ ∈ (a,b).
<p><strong>Step 1:</strong> Let <p>f'(x) = ax² + bx + c</p></p><p><strong>Step 2:</strong> Then <p>f(0) = d</p></p><p><strong>Step 3:</strong> And <p>f(-1) = -(a/3) + (b/2) - c + d = (2a - 3b + 6c)/(-6) + d = 0 + d = d</p> [since <p>2a - 3b + 6c = 0</p>]</p><p><strong>Step 4:</strong> Since <p>f(0) = f(-1)</p>, by Rolle's Theorem, <p>f'(x) = 0</p> has at least one root in <p>(-1, 0)</p> (q)</p><p><strong>Step 5:</strong> Also, <p>f'(x) = 0</p> has roots in <p>(-1, 1)</p> (r)</p>
Correct Answer: q, r