Applications of Derivatives
Rolle's Theorem and Limits
Grade 12
Question:
<p>Let \(f(x) = \dfrac{2 + \ln x}{x^2}\), \(x > 0\). Identify which of the following is(are) <strong>correct</strong> about \(f(x)\)?</p>
<p>(a) \(f'(x) = 0\) for some \(x \in \left(0,\, e^{-7/6}\right)\)</p>
<p>(b) \(\displaystyle\lim_{x \to 0^+} f'(x) = \infty\)</p>
<p>(c) \(\displaystyle\lim_{x \to 0^+} f(x) = 0\)</p>
<p>(d) Rolle's Theorem is applicable for \(f'(x)\) in some interval of \((0, \infty)\)</p>
Step-by-Step Solution
Key Concept: Find f'(x) using quotient rule, then analyze critical points and monotonicity by examining the sign of the numerator (1 - 2ln x) to determine increasing/decreasing intervals and local extrema.
<p><strong>Step 1: Find f'(x) using quotient rule</strong></p><p>f(x) = (2 + ln x)/x²</p><p>f'(x) = [x² · (1/x) - (2 + ln x) · 2x] / x⁴</p><p>f'(x) = [x - 2x(2 + ln x)] / x⁴ = [x - 4x - 2x ln x] / x⁴</p><p>f'(x) = [-3x - 2x ln x] / x⁴ = [-3 - 2ln x] / x³ = <strong>-(3 + 2ln x) / x³</strong></p><p><strong>Step 2: Find critical points</strong></p><p>Setting f'(x) = 0: 3 + 2ln x = 0</p><p>ln x = -3/2</p><p>x = e^(-3/2) = 1/e^(3/2)</p><p><strong>Step 3: Analyze monotonicity</strong></p><p>• For 0 < x < e^(-3/2): ln x < -3/2, so 3 + 2ln x < 0, thus f'(x) > 0 (increasing)</p><p>• For x > e^(-3/2): ln x > -3/2, so 3 + 2ln x > 0, thus f'(x) < 0 (decreasing)</p><p><strong>Step 4: Conclusion</strong></p><p>f(x) has a local (and absolute) maximum at x = e^(-3/2) with value f(e^(-3/2)) = (2 - 3/2)/(e^(-3)) = e³/2</p><p>f(x) is increasing on (0, e^(-3/2)) and decreasing on (e^(-3/2), ∞)</p><p>∴ Answer: BD</p>
Correct Answer: BD