<p><strong>Ex. 25 (C):</strong> Let <p>a, b, c</p> be non-zero real numbers such that <p>∫₀¹ (1 + cos 8x)(ax² + bx + c) dx = ∫₀² (1 + cos 8x)(ax² + bx + c) dx</p>. Then the equation <p>ax² + bx + c = 0</p> has at least one root in which interval?</p>
Step-by-Step Solution
Key Concept: The equality of integrals means the antiderivatives are equal at two different points, which by Rolle's Theorem implies the derivative is zero between those points.
<p><strong>Step 1:</strong> Given condition: <p>∫₀¹ (1 + cos 8x)(ax² + bx + c) dx = ∫₀² (1 + cos 8x)(ax² + bx + c) dx</p></p><p><strong>Step 2:</strong> This implies <p>f(1) = f(2)</p> where <p>f(x) = ∫(1 + cos 8x)(ax² + bx + c) dx</p></p><p><strong>Step 3:</strong> By Rolle's Theorem, <p>f'(x) = (1 + cos 8x)(ax² + bx + c) = 0</p> has at least one root in <p>(1, 2)</p></p><p><strong>Step 4:</strong> Since <p>1 + cos 8x > 0</p> for <p>x ∈ (0, 2)</p>, we need <p>ax² + bx + c = 0</p> to have a root in <p>(0, 1)</p> (s), <p>(0, 2)</p> (t), and <p>(-1, 1)</p> (r)</p>
Correct Answer: s, t, r