<p>The real number <em>k</em> for which the equation \(2x^3 + 3x + k = 0\) has two distinct real roots in \([0, 1]\)</p>
<p>lies between 2 and 3.</p>
<p>lies between \(-1\) and 0.</p>
<p>does not exist.</p>
<p>lies between 1 and 2.</p>
Step-by-Step Solution
Key Concept: For a cubic function to have two distinct real roots in [0,1], it must have a local extremum within the interval where the function changes sign. However, we need to check if this is actually possible given the constraints of the function f(x) = 2x³ + 3x + k.
<p><strong>Step 1: Find the derivative</strong></p><p>Let f(x) = 2x³ + 3x + k</p><p>f'(x) = 6x² + 3</p></p><p><strong>Step 2: Analyze the derivative</strong></p><p>f'(x) = 6x² + 3 = 3(2x² + 1)</p><p>Since x² ≥ 0 for all real x, we have 2x² + 1 ≥ 1 > 0 for all x ∈ ℝ</p><p>Therefore f'(x) > 0 for all x ∈ [0,1]</p></p><p><strong>Step 3: Determine the nature of f(x)</strong></p><p>Since f'(x) > 0 throughout [0,1], the function f(x) is strictly increasing on [0,1].</p></p><p><strong>Step 4: Analyze roots in [0,1]</strong></p><p>A strictly increasing continuous function can cross the x-axis at most once in any interval. Therefore, f(x) = 0 can have at most one real root in [0,1].</p></p><p><strong>Step 5: Conclusion</strong></p><p>For two distinct real roots to exist in [0,1], the function must have a local extremum (maximum or minimum) in the interior of [0,1] where f changes sign. But f'(x) > 0 everywhere, so no such extremum exists. Therefore, it is impossible for f(x) = 2x³ + 3x + k = 0 to have two distinct real roots in [0,1] for any value of k.</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C