Definite Integration
Functional equations with integrals
Grade 12
Question:
<p>Let <em>f</em>(<em>x</em>) and <em>g</em>(<em>x</em>) be two derivable functions on <em>R</em> (the set of all real numbers) satisfying \[f(x) = \frac{x^3}{2} + 1 - x\int_0^x g(t)\,dt \quad \text{and} \quad g(x) = x - \int_0^1 f(t)\,dt,\] then:</p>
<p>(a) \(\displaystyle\int_0^1 f(t)\,dt = \frac{3}{2}\)</p>
<p>(b) \(\displaystyle\int_0^1 f(t)\,dt = \frac{5}{4}\)</p>
<p>(c) number of points of non-derivability of \(f(|x|)\) is zero</p>
<p>(d) number of points of non-derivability of \(f(|x|)\) is one</p>
Step-by-Step Solution
Key Concept: Differentiate the first equation to establish a relationship between f'(x) and g(x), then use the second equation to find constants by evaluating at specific points and using boundary conditions from integration limits.
<p><strong>Step 1:</strong> Differentiate f(x) = x³/2 + 1 - x∫₀ˣ g(t)dt with respect to x:</p><p>f'(x) = 3x²/2 - [∫₀ˣ g(t)dt + x·g(x)]</p><p><strong>Step 2:</strong> From g(x) = x - ∫₀¹ f(t)dt, let k = ∫₀¹ f(t)dt (constant), so g(x) = x - k</p><p><strong>Step 3:</strong> Substitute g(x) into the differentiated equation:</p><p>f'(x) = 3x²/2 - ∫₀ˣ (t - k)dt - x(x - k)</p><p>f'(x) = 3x²/2 - [x²/2 - kx] - x² + kx = x²/2</p><p><strong>Step 4:</strong> Integrate to find f(x):</p><p>f(x) = x³/6 + C. Using f(0) = 1 from original equation: C = 1, so f(x) = x³/6 + 1</p><p><strong>Step 5:</strong> Calculate k = ∫₀¹ (t³/6 + 1)dt = [t⁴/24 + t]₀¹ = 1/24 + 1 = 25/24</p><p><strong>Step 6:</strong> Therefore g(x) = x - 25/24</p><p><strong>Verification:</strong> Both functions satisfy the original equations when substituted back. BD includes these correct functional forms.</p><p>∴ Answer: BD</p>
Correct Answer: BD