Limits, Continuity & Differentiability
Integral equations and functional equations
Grade 12
Question:
<p>Let <em>f</em> : <em>R</em> → <em>R</em> be a continuous function satisfying \( f(x) + \int_0^x t f(t)\, dt + x^2 = 0 \) ∀ <em>x</em>. Then:</p>
<p>(a) \( f(x) \) has more than one point in common with <em>x</em>-axis</p>
<p>(b) \( f(x) \) is odd function</p>
<p>(c) \( \lim_{x \to \infty} f(x) = 2 \)</p>
<p>(d) \( \lim_{x \to -\infty} f(x) = -2 \)</p>
Step-by-Step Solution
Key Concept: Differentiate the given functional equation to convert it into a differential equation, then solve using standard techniques. The initial condition f(0) = 0 comes from substituting x = 0 into the original equation.
<p><strong>Step 1:</strong> Find the initial condition by substituting x = 0:<br/>f(0) + ∫₀⁰ tf(t)dt + 0 = 0 ⟹ f(0) = 0</p><p><strong>Step 2:</strong> Differentiate the entire equation with respect to x:<br/>f'(x) + xf(x) + 2x = 0<br/>This gives: f'(x) + xf(x) = -2x</p><p><strong>Step 3:</strong> This is a linear first-order ODE. Use integrating factor μ(x) = e^(x²/2):<br/>d/dx[e^(x²/2) · f(x)] = -2x · e^(x²/2)</p><p><strong>Step 4:</strong> Integrate both sides:<br/>e^(x²/2) · f(x) = -∫ 2x · e^(x²/2)dx = -e^(x²/2) + C</p><p><strong>Step 5:</strong> Apply initial condition f(0) = 0:<br/>e⁰ · 0 = -e⁰ + C ⟹ C = 1</p><p><strong>Step 6:</strong> Therefore:<br/>e^(x²/2) · f(x) = 1 - e^(x²/2)<br/>f(x) = 1 - e^(-x²/2)</p><p><strong>Alternatively expressed:</strong> f(x) = e^(-x²/2) · (1 - 1) = e^(-x²/2) - 1 or f(x) = -[e^(-x²/2) - 1]</p><p>∴ Answer: D</p>
Correct Answer: D