Applications of Derivatives
Monotonicity and roots
Grade 12
Question:
<p>When the given equation \(2x^3 + 3x + k = 0\) has two distinct real roots in \([0, 1]\), then \(f'(x)\) will change sign. But \(f'(x) = 6x^2 + 3 > 0\), for all values of \(x \in \mathbb{R}\). The number of values of \(k\) for which the equation \(2x^3 + 3x + k = 0\) has two distinct real roots in \([0,1]\) is:</p>
<p>Two</p>
<p>Three</p>
<p>Infinitely many</p>
<p>No value of \(k\) exists</p>
Step-by-Step Solution
Key Concept: A cubic function with always positive derivative (monotonically increasing) cannot have two distinct real roots in any interval. The question tests understanding that monotonicity contradicts the possibility of two distinct roots.
<p><strong>Step 1:</strong> Let f(x) = 2x³ + 3x + k. We need to analyze if this can have two distinct real roots in [0,1].</p><p><strong>Step 2:</strong> Find f'(x) = 6x² + 3. Since x² ≥ 0 for all x ∈ ℝ, we have f'(x) = 6x² + 3 ≥ 3 > 0 for all x ∈ ℝ.</p><p><strong>Step 3:</strong> Since f'(x) > 0 for all x, the function f(x) is strictly increasing on [0,1] and throughout ℝ.</p><p><strong>Step 4:</strong> A strictly monotonic function can intersect any horizontal line (including the x-axis where f(x) = 0) at most once. Therefore, f(x) = 0 can have at most one real root in [0,1].</p><p><strong>Step 5:</strong> It is impossible for f(x) = 0 to have two distinct real roots in [0,1], regardless of the value of k.</p><p><strong>Step 6:</strong> The number of values of k for which the equation has two distinct real roots in [0,1] is 0.</p><p>∴ Answer: D (0)</p>
Correct Answer: D