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 \( x \in \left(0, e^{\frac{-7}{6}}\right) \)</p>
<p>(b) \( \lim_{x \to 0^+} f'(x) = \infty \)</p>
<p>(c) \( \lim_{x \to 0^+} f(x) = 0 \)</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 the sign of f'(x) to determine monotonicity intervals and identify critical points. The numerator of f'(x) determines where f is increasing/decreasing.
<p><strong>Step 1: Find f'(x) using quotient rule</strong></p><p>f(x) = (2 + ln x)/x²</p><p>f'(x) = [(1/x)·x² - (2 + ln x)·2x]/x⁴ = [x - 2x(2 + ln x)]/x⁴</p><p>f'(x) = [x - 4x - 2x ln x]/x⁴ = [-3x - 2x ln x]/x⁴ = [-(3 + 2ln x)]/x³</p><p><strong>Step 2: Analyze the sign of f'(x)</strong></p><p>Since x > 0, we have x³ > 0. So sign of f'(x) depends on -(3 + 2ln x)</p><p>f'(x) = 0 when 3 + 2ln x = 0 ⟹ ln x = -3/2 ⟹ x = e^(-3/2)</p><p><strong>Step 3: Determine monotonicity</strong></p><p>• For 0 < x < e^(-3/2): ln x < -3/2, so (3 + 2ln x) < 0, thus -(3 + 2ln x) > 0 ⟹ f'(x) > 0 (increasing)</p><p>• For x > e^(-3/2): ln x > -3/2, so (3 + 2ln x) > 0, thus -(3 + 2ln x) < 0 ⟹ f'(x) < 0 (decreasing)</p><p><strong>Step 4: Identify conclusions</strong></p><p>• f(x) is increasing on (0, e^(-3/2))</p><p>• f(x) is decreasing on (e^(-3/2), ∞)</p><p>• f(x) has a maximum at x = e^(-3/2)</p><p>∴ Answer: BCD</p>
Correct Answer: BCD