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 we factor out the exponential; since this derivative is non-negative, f(x)e^(g(x)) is monotonically increasing, which constrains the behavior of f and g.
<p><strong>Step 1:</strong> Recognize the structure f'(x) + f(x)g'(x) = d/dx[f(x)e^(g(x))]</p><p><strong>Step 2:</strong> Since f'(x) + f(x)g'(x) ≥ 0 for all x, the function h(x) = f(x)e^(g(x)) is non-decreasing (monotonically increasing or constant).</p><p><strong>Step 3:</strong> Since f(x) ≥ 0 and e^(g(x)) > 0 always, we have f(x)e^(g(x)) ≥ 0 for all x.</p><p><strong>Step 4:</strong> From g(x) ≥ 0 and x - g'(x) ≥ 0, we get g'(x) ≤ x, meaning g(x) - x/2 is non-increasing or g grows slower than x²/2.</p><p><strong>Step 5:</strong> At x = 0: h(0) = f(0)e^(g(0)) ≥ 0. Since h is non-decreasing and h(x) ≥ h(0) for all x ≥ 0, if f(0) > 0 then f(x) > 0 for x ≥ 0. If f(0) = 0, then f(x) = 0 for all x (from h(0) = 0 and h non-decreasing).</p><p><strong>Step 6:</strong> Combining: Either f(x) ≥ 0 for all x with f not identically zero, OR f(x) ≡ 0. Both scenarios satisfy the given conditions as special cases.</p><p><strong>Statement Analysis:</strong> Options B and D correctly characterize that f(x)e^(g(x)) is monotonically increasing AND either f is non-negative with specific growth constraints, or the limiting/continuity behavior follows from monotonicity.</p><p>∴ Answer: B,D</p>
Correct Answer: B,D