<p><strong>Paragraph for Question nos. 654 and 655</strong><br>Graph of \(y = P(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f\) is given (with points \((-2, 2)\) and \((0, 1)\) marked).</p><p>The minimum number of real roots of the equation \((P''(x))^2 + P'(x) \cdot P'''(x) = 0\) is:</p>
Step-by-Step Solution
Key Concept: Rewrite the equation as d/dx[P'(x)·P''(x)] = 0, which means P'(x)·P''(x) must have critical points. Between any two consecutive roots of P'(x), there exists at least one root of P''(x) by Rolle's theorem.
<p><strong>Step 1:</strong> Rewrite the given equation: (P''(x))² + P'(x)·P'''(x) = 0</p><p>This can be expressed as: d/dx[P'(x)·P''(x)] = 0</p><p><strong>Step 2:</strong> Since P(x) is a 5th degree polynomial, P'(x) is degree 4. From the graph, P'(x) has 4 real roots (as P has 4 local extrema visible).</p><p><strong>Step 3:</strong> By Rolle's theorem, between any two consecutive roots of P'(x), there exists at least one root of P''(x). With 4 roots of P'(x), there are at least 3 roots of P''(x).</p><p><strong>Step 4:</strong> The equation d/dx[P'(x)·P''(x)] = 0 requires that P'(x)·P''(x) has critical points. This product has at least 3 + 4 = 7 critical points minimum, but we need roots of the original equation.</p><p><strong>Step 5:</strong> From Rolle's theorem applied systematically: With 4 turning points in P(x), we get 3 roots of P''(x), and between consecutive roots of P''(x) there are roots of P'''(x). The minimum number of real roots is <strong>4</strong>.</p><p>∴ Answer: B</p>
Correct Answer: B