Definite Integration
Integral equations
Grade 12
Question:
<p>If \(f(x) = 2 + \displaystyle\int_{-1}^{1}\left(\dfrac{tx^2}{2} + \dfrac{9x}{14}\right)f(t)\,dt\), then:</p>
<p>Rolle's Theorem is applicable for \(y = f(x)\) in \([-2, -1]\)</p>
<p>\(\lim_{x \to 0} f(x) = 0\)</p>
<p>\(f\) is continuous and derivable on \(R\)</p>
<p>maximum value of \(f(x)\) does not exist</p>
Step-by-Step Solution
Key Concept: Recognize that the integral ∫₋₁¹ produces constants (since it integrates over a fixed interval), so f(x) must be a polynomial. Set up the functional equation by equating coefficients after substituting the assumed form of f(x).
<p><strong>Step 1:</strong> Recognize that f(x) = 2 + ∫₋₁¹(tx²/2 + 9x/14)f(t)dt. Since the integral over [-1,1] is independent of x, let:</p><p>A = ∫₋₁¹ tf(t)dt and B = ∫₋₁¹ f(t)dt</p><p>Then: f(x) = 2 + (A/2)x² + (9B/14)x</p><p><strong>Step 2:</strong> This means f(x) must be quadratic: f(x) = 2 + αx² + βx where α = A/2 and β = 9B/14</p><p><strong>Step 3:</strong> Calculate A = ∫₋₁¹ t(2 + αt² + βt)dt = ∫₋₁¹(2t + αt³ + βt²)dt = 0 + 0 + 2β/3 = 2β/3</p><p>So: α = (1/2)(2β/3) = β/3</p><p><strong>Step 4:</strong> Calculate B = ∫₋₁¹(2 + αt² + βt)dt = 4 + 2α/3 + 0 = 4 + 2α/3</p><p>So: β = (9/14)(4 + 2α/3) = 36/14 + 6α/14 = 18/7 + 3α/7</p><p><strong>Step 5:</strong> From α = β/3 and β = 18/7 + 3α/7:</p><p>α = (1/3)(18/7 + 3α/7) = 6/7 + α/7</p><p>6α/7 = 6/7 → α = 1, therefore β = 3</p><p><strong>Step 6:</strong> Thus f(x) = 2 + x² + 3x = x² + 3x + 2 = (x+1)(x+2)</p><p>∴ Answer: A,C</p>
Correct Answer: A,C