Applications of Derivatives
Rolle's Theorem and Root Location
Grade 12
Question:
<p><strong>Ex. 9:</strong> Which of the following statements are true, where <(x) is a polynomial?</p><p>(a) Between any two roots of <(x) = 0 there exists at least one root of <'(x) - = <(x) = 0</p><p>(b) Between any two roots of <(x) = 0 there exists at least one root of \(x <'(x) - = <(x) = 0\)</p><p>(c) Between any two roots of <(x) = 0 there exists at least one root of \((x^2 - 1) <'(x) - <(x) = 0\)</p><p>(d) Between any two roots of <(x) = 0 there exists at least one root of <'(x) - x <(x) = 0</p>
<p>(a) Between any two roots of <(x) = 0 there exists at least one root of <'(x) - = <(x) = 0</p>
<p>(b) Between any two roots of <(x) = 0 there exists at least one root of \(x <'(x) - = <(x) = 0\)</p>
<p>(c) Between any two roots of <(x) = 0 there exists at least one root of \((x^2 - 1) <'(x) - <(x) = 0\)</p>
<p>(d) Between any two roots of <(x) = 0 there exists at least one root of <'(x) - x <(x) = 0</p>
Step-by-Step Solution
Key Concept: Apply Rolle's theorem and properties of derivatives. Between consecutive roots of a polynomial, its derivative must vanish, and combined expressions also satisfy intermediate value properties.
<p><strong>Solution (c, d):</strong> If $\alpha$ and $\beta$ are consecutive roots of <(x) = 0, then <'($\alpha$) and <'($\beta$) have opposite signs.</p><p>Let <(x) = (x - 1)(x - 2) and <'(x) = 2x - 3</p><p>For statement (a): <'(x) - = <(x) = 0 becomes $(2x - 3) - (x - 1)(x - 2) = 0$</p><p>Between any two roots of <(x) = 0, by Rolle's theorem applied to <(x), there exists at least one root of <'(x) = 0. However, the combined equation <'(x) - = <(x) = 0 must have at least one root between $\alpha$ and $\beta$.</p><p>Similarly, $(x^2 - 1)<'(x) - <(x) = 0$ and <'(x) - x<(x) = 0 both have at least one root between $\alpha$ and $\beta$.</p><p>∴ Statements (c) and (d) are true.</p>
Correct Answer: c, d