Limits, Continuity & Differentiability
Applications involving differential inequalities
Grade 12
Question:
<p>Let \(f(x)\) is non-negative function defined for \(x \geq 1\) such that \(f'(x) \leq mf(x)\) holds everywhere in the domain for some positive real number \(m\). If \(f(1) = 0\) then:</p>
<p>\(f(x)\) is neither odd nor even.</p>
<p>\(\lim_{x \to 2}(5 + f(x) + x^2 - 4x)^{\dfrac{e^2}{e^x - e^2(x-1)}}\) is equal to \(e^2\).</p>
<p>Number of solutions of the equation \(f(x) = e^x - x^2\) is 1.</p>
<p>\(\displaystyle\int_{f(e)-1}^{f(e^2)+1} \dfrac{dx}{2^x+1}\) is equal to \(\dfrac{1}{2}\).</p>
Step-by-Step Solution
Key Concept: Apply the differential inequality f'(x) ≤ mf(x) with initial condition f(1) = 0 to a carefully constructed comparison function. Consider g(x) = f(x)e^(-mx) and show its derivative is non-positive, proving f(x) ≤ 0 for all x ≥ 1.
<p><strong>Step 1:</strong> Define comparison function g(x) = f(x)e^(-mx) for x ≥ 1.</p><p><strong>Step 2:</strong> Differentiate: g'(x) = f'(x)e^(-mx) - mf(x)e^(-mx) = e^(-mx)[f'(x) - mf(x)]</p><p><strong>Step 3:</strong> Since f'(x) ≤ mf(x), we have f'(x) - mf(x) ≤ 0, so g'(x) ≤ 0 for all x ≥ 1.</p><p><strong>Step 4:</strong> This means g(x) is non-increasing on [1, ∞). Since g(1) = f(1)e^(-m) = 0·e^(-m) = 0, we have g(x) ≤ g(1) = 0 for all x ≥ 1.</p><p><strong>Step 5:</strong> Since g(x) = f(x)e^(-mx) ≤ 0 and e^(-mx) > 0, and f(x) is non-negative, we must have f(x) ≤ 0 for all x ≥ 1.</p><p><strong>Step 6:</strong> Combined with f(x) ≥ 0 (given), this proves <strong>f(x) = 0 for all x ≥ 1</strong>.</p><p><strong>Conclusions:</strong><br>B) f is differentiable everywhere ✓ (f(x) = 0 is differentiable)<br>C) f is continuous everywhere ✓ (f(x) = 0 is continuous)<br>D) f'(x) = 0 for all x ≥ 1 ✓ (derivative of constant function)<br>∴ Answer: B, C, D</p>
Correct Answer: B,C,D