Applications of Derivatives
Monotonicity and bounds
Grade 12
Question:
<p>Let <span class="math">g:[1, 6] \to [0, \infty)</span> be a real valued differentiable function satisfying <span class="math">g'(x) = \frac{2}{x + g(x)}</span> and <span class="math">g(1) = 0</span>, then the maximum value of <span class="math">g</span> cannot exceed</p>
<p>(a) <span class="math">\log 2</span></p>
<p>(b) <span class="math">\log 6</span></p>
<p>(c) <span class="math">6 \log 2</span></p>
<p>(d) <span class="math">2 \log 6</span></p>
Step-by-Step Solution
Key Concept: Use the positivity of the derivative to establish that g is increasing, then integrate the differential equation to find an upper bound.
<p><strong>Step 1:</strong> Since <span class="math">g'(x) = \frac{2}{x + g(x)} > 0</span> for all <span class="math">x \in [1, 6]</span>, we have that <span class="math">g(x)</span> is an increasing function on <span class="math">[1, 6]</span>.</p><p><strong>Step 2:</strong> Integrating both sides:</p><p><span class="math">\int_1^6 g'(x) \, dx \geq \int_1^6 \frac{2}{x} \, dx</span></p><p><strong>Step 3:</strong> <span class="math">g(6) - g(1) \geq 2(\log e x)|_1^6</span></p><p><span class="math">g(6) - 0 \geq 2 \log 6</span></p><p><strong>Step 4:</strong> Since <span class="math">g(1) = 0</span>, we get <span class="math">g(6) \geq 2 \log 6</span>, so <span class="math">g</span> cannot exceed <span class="math">2 \log 6</span>.</p><p>∴ Answer is (d).</p>
Correct Answer: D