Definite Integration
Integral equations
Grade 12
Question:
<p><strong>Paragraph for Question nos. 580 to 582</strong><br>Let \(f(x)\) and \(g(x)\) are two continuous functions defined for \(0 \leq x \leq 1\), \(f(x) = \int_0^1 e^{x+t} f(t)\, dt\), \(g(x) = x + \int_0^1 e^{x+t} g(t)\, dt\).</p><p><strong>580.</strong> The value of \(f(1)\) is:</p>
<p>(a) 0</p>
<p>(b) 1</p>
<p>(c) \(\dfrac{1}{e}\)</p>
<p>(d) \(e\)</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) = e^x ∫₀¹ e^t f(t)dt, so the integral term is a constant. Setting this constant as k and solving the resulting linear equation gives f(x) = ke^x, then use the original equation to find k.
<p><strong>Step 1:</strong> Rewrite f(x) = ∫₀¹ e^(x+t) f(t)dt = e^x ∫₀¹ e^t f(t)dt</p><p><strong>Step 2:</strong> Let k = ∫₀¹ e^t f(t)dt (a constant). Then f(x) = ke^x</p><p><strong>Step 3:</strong> Substitute f(x) = ke^x back into the definition:</p><p>ke^x = e^x ∫₀¹ e^t · ke^t dt = ke^x ∫₀¹ e^(2t) dt</p><p><strong>Step 4:</strong> Divide by e^x: k = k∫₀¹ e^(2t) dt = k[e^(2t)/2]₀¹ = k(e² - 1)/2</p><p><strong>Step 5:</strong> For non-trivial solution: 1 = (e² - 1)/2, which gives k = 2/(e² - 1)</p><p><strong>Step 6:</strong> Therefore f(1) = ke¹ = 2e/(e² - 1)</p><p>∴ Answer: A</p>
Correct Answer: A