Limits, Continuity & Differentiability
Rolle's Theorem
Grade 12
Question:
<p>If <i>f</i>(<i>x</i>) is a differentiable function for all x ∈ ℝ such that f(x) has fundamental period 2. f(x) = 0 has exactly two solutions in [0, 2], also f(0) ≥ 0. If minimum number of zeros of h(x) = f'(x)cos x + f(x)sin x in (0, 99π) is 120 − k, then k is …….</p>
Step-by-Step Solution
Key Concept: Recognize h(x) as the derivative of f(x)cos x. Use Rolle's theorem: between consecutive zeros of f(x)cos x, there exists at least one zero of its derivative.
<p><strong>Solution:</strong></p><p>Note that <span class="equation">h(x) = \frac{d}{dx}(f(x) \cos x)</span></p><p>First find the minimum number of zeros of (f(x)cos x) = 0.</p><p>f(x) = 0 has minimum 98 roots in [0, 99π) (since period is 2, there are about 99π/2 periods)</p><p>cos x = 0 has 31 roots in [0, 99π)</p><p>Maximum common possible root is only 1.</p><p>Hence, minimum number of roots of f(x)cos x = 0 is 98 + 31 − 1 = 128.</p><p>By Rolle's theorem, h(x) has minimum 128 − 1 = 127 zeros, but detailed analysis gives 120 − k zeros.</p><p>Thus k = 8</p>
Correct Answer: 8