Definite Integration
Properties of definite integrals
Grade 12
Question:
<p>If <em>a</em>, <em>b</em>, <em>c</em> be nonzero real numbers such that<br>\(\int_0^1 (1+\cos^8 x)(ax^2+bx+c)dx = \int_0^2 (1+\cos^8 x)(ax^2+bx+c)dx\)<br>then the quadratic equation \(ax^2+bx+c=0\) has</p>
<p>(a) no root in (0, 2)</p>
<p>(b) at least one root in (1, 2)</p>
<p>(c) at least one root in (0,1)</p>
<p>(d) two imaginary roots</p>
Step-by-Step Solution
Key Concept: If a definite integral over [0,1] equals the integral over [0,2], then the polynomial (ax²+bx+c) must have a specific symmetry property about x=1. This means ∫₀¹(1+cos⁸x)(ax²+bx+c)dx = ∫₁²(1+cos⁸x)(ax²+bx+c)dx, implying the integrand must change sign symmetrically about some point.
<p><strong>Step 1:</strong> Given: ∫₀¹(1+cos⁸x)(ax²+bx+c)dx = ∫₀²(1+cos⁸x)(ax²+bx+c)dx</p><p><strong>Step 2:</strong> This implies: ∫₀¹(1+cos⁸x)(ax²+bx+c)dx = ∫₀¹(1+cos⁸x)(ax²+bx+c)dx + ∫₁²(1+cos⁸x)(ax²+bx+c)dx</p><p><strong>Step 3:</strong> Therefore: ∫₁²(1+cos⁸x)(ax²+bx+c)dx = 0</p><p><strong>Step 4:</strong> Since (1+cos⁸x) > 0 for all x ∈ [1,2], and the integral equals zero, the polynomial ax²+bx+c must change sign on [1,2]. Combined with Step 2 analysis, this means x=1 is a root of ax²+bx+c=0</p><p><strong>Step 5:</strong> Substituting x=1: a(1)²+b(1)+c = 0, giving a+b+c=0. This is the condition for one root being x=1</p><p><strong>Step 6:</strong> With a+b+c=0 and the integral structure, the quadratic has two real roots: one at x=1 and the other at x=c/a (by Vieta's formulas)</p><p>∴ Answer: C (The quadratic has one root at x=1 and roots are real)</p>
Correct Answer: C