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 the sign properties of polynomials at their roots combined with the constraint P(b)·Q(a) > 0 to determine the relative positions of roots a and b, then analyze continuity and differentiability of rational functions formed from P and Q.
<p><strong>Step 1:</strong> Since a is a root of P(x): P(a) = 0. Since b is a root of Q(x): Q(b) = 0.</p><p><strong>Step 2:</strong> Given P(b)·Q(a) > 0, both P(b) and Q(a) must have the same sign (both positive or both negative).</p><p><strong>Step 3:</strong> Consider the rational function R(x) = P(x)/Q(x). At x = a: R(a) = P(a)/Q(a) = 0/Q(a). At x = b: R(b) = P(b)/Q(b) = P(b)/0, which is undefined (vertical asymptote).</p><p><strong>Step 4:</strong> Since P(b)·Q(a) > 0, the point x = a is where the numerator is zero and denominator is non-zero (either positive or negative depending on the sign of Q(a)). The function is continuous at x = a but has a vertical asymptote at x = b.</p><p><strong>Step 5:</strong> The constraint P(b)·Q(a) > 0 ensures that between roots a and b, neither function changes sign in a way that creates removable discontinuities. This maintains differentiability on appropriate intervals.</p><p><strong>Step 6:</strong> Valid conclusions: P/Q is continuous at x = a (Option B), and depending on the polynomial structure, P/Q is differentiable at x = a (Option D).</p><p>∴ Answer: BD</p>
Correct Answer: BD