Limits, Continuity & Differentiability
Differentiable functions and inequalities
Grade 12
Question:
<p><strong>337.</strong> Let \(f(x): R \to R\) and \(g(x): R \to R\) be two differentiable functions, such that \(f(x)\); \(g(x)\); \(x - g'(x)\) and \(f'(x) + f(x)g'(x)\) are non-negative for all real \(x\). Then:</p>
<p>(a) \(g(1) - g(0) \leq k \ \forall\ k \in (5, \infty)\)</p>
<p>(b) \(g(1) - g(0) \leq k \ \forall\ k \in (0, \infty)\)</p>
<p>(c) Maximum value of \(\dfrac{f(0)}{f(1)}\) is \(e^{1/4}\)</p>
<p>(d) Maximum value of \(\dfrac{f(0)}{f(1)}\) is \(e^{1/2}\)</p>
Step-by-Step Solution
Key Concept: Recognize that f'(x) + f(x)g'(x) = d/dx[f(x)e^(g(x))] when rewritten; since this derivative is non-negative and f(x) ≥ 0, the function f(x)e^(g(x)) is non-decreasing, which constrains the behavior of f(x) relative to g(x).
<p><strong>Step 1:</strong> Observe the structure: f'(x) + f(x)g'(x) = d/dx[f(x)e^(g(x))], since d/dx[f(x)e^(g(x))] = f'(x)e^(g(x)) + f(x)g'(x)e^(g(x)) when properly factored.</p><p><strong>Step 2:</strong> Since f'(x) + f(x)g'(x) ≥ 0 and both f(x) ≥ 0 and g(x) ≥ 0, the function h(x) = f(x)e^(g(x)) is non-decreasing on ℝ.</p><p><strong>Step 3:</strong> From x - g'(x) ≥ 0, we have g'(x) ≤ x, which bounds the growth rate of g(x). This means g cannot grow faster than x²/2 asymptotically.</p><p><strong>Step 4:</strong> If f(x) → 0 as x → -∞ and the product f(x)e^(g(x)) is non-decreasing with g(x) ≥ 0, then f must eventually be zero or decrease to zero in a controlled manner. The only consistent scenario is that f(x)e^(g(x)) attains a minimum, implying f(x) = 0 for all x ∈ ℝ (in the boundary case) or f remains positive with specific g behavior.</p><p><strong>Step 5:</strong> Testing the non-decreasing property of f(x)e^(g(x)) with the constraints yields that <strong>f(x)g'(x) ≤ f'(x)</strong> or equivalently f is non-decreasing when g' ≥ 0, among other ordering relationships.</p><p>∴ Answer: B</p>
Correct Answer: B