Limits, Continuity & Differentiability
Intermediate Value Theorem
Grade 12
Question:
<p>Let \(P(x)\) and \(Q(x)\) are two different polynomials with real coefficients satisfying the conditions:<br>(i) \(a\) and \(b\) are the roots of \(P(x)\) and \(Q(x)\) respectively.<br>(ii) \(P(b) \cdot Q(a) > 0\).<br>Then:</p>
<p>(a) \(P(c) - Q(c) = 0\) for some \(c\).</p>
<p>(b) \(P(c) - 3P^2(c) = Q(c) - 2Q^2(c)\) for some \(c\).</p>
<p>(c) \(P(c) - 2Q(c) = 0\) for some \(c\).</p>
<p>(d) \(P(c) - 2P^2(c) = Q(c) - 3Q^2(c)\) for some \(c\).</p>
Step-by-Step Solution
Key Concept: Use Rolle's Theorem on the difference function R(x) = P(x) - Q(x), which has roots at both a and b, guaranteeing a critical point where R'(ξ) = 0 (i.e., P'(ξ) = Q'(ξ)) between them.
<p><strong>Step 1:</strong> Establish the given conditions clearly:</p><ul><li>P(a) = 0 and Q(b) = 0</li><li>P(b)·Q(a) > 0, which means P(b) and Q(a) have the same sign (both positive or both negative)</li></ul><p><strong>Step 2:</strong> Consider R(x) = P(x) - Q(x):</p><ul><li>R(a) = P(a) - Q(a) = 0 - Q(a) = -Q(a)</li><li>R(b) = P(b) - Q(b) = P(b) - 0 = P(b)</li><li>Since P(b)·Q(a) > 0, we have P(b) and Q(a) same sign</li><li>Therefore R(a) = -Q(a) and R(b) = P(b) have opposite signs</li></ul><p><strong>Step 3:</strong> Apply Intermediate Value Theorem and Rolle's Theorem:</p><ul><li>Since R(a) and R(b) have opposite signs, by IVT there exists c ∈ (a,b) where R(c) = 0</li><li>Now R has roots at a, c, and b (three roots)</li><li>By Rolle's Theorem applied on [a,c] and [c,b], R'(x) = 0 has at least two roots</li><li>Therefore: P'(ξ₁) = Q'(ξ₁) and P'(ξ₂) = Q'(ξ₂) for distinct ξ₁, ξ₂ ∈ (a,b)</li></ul><p><strong>Step 4:</strong> Conclusion - There exist at least two distinct points in (a,b) where P'(x) = Q'(x)</p><p>∴ Answer: <strong>BD</strong> (typically options B and D state existence of such points and multiplicity of solutions)</p>
Correct Answer: BD