<p><strong>Paragraph for Question nos. 583 to 584</strong><br>Let \(y = f(x)\) be a differentiable function passing through \((1, 0)\). Let slope of the tangent at the point \((x, f(x))\) be \(m_1\) and slope of line joining the point and origin be \(m_2\). Also \(\left|\dfrac{\log(m_1 + m_2)}{\log x}\right| = \dfrac{2}{1}\).</p><p>If \(f_1(x)\) and \(f_2(x)\) are 2 functions satisfying the above property where \(f_1(x)\) is an algebraic function and \(f_2(x)\) is a transcendental function.</p><p>The value of \(\displaystyle\int_{-1}^{1}\left(f_1(x) + \dfrac{1}{4x}\right)dx\) is equal to:</p>
Step-by-Step Solution
Key Concept: From the condition |log(m₁ + m₂)/log x| = 2, derive that m₁ + m₂ = x² or x⁻². Since m₁ = f'(x) and m₂ = f(x)/x, this gives a differential equation. For the algebraic function f₁(x), solving f'(x) + f(x)/x = x² yields f₁(x) = x²/3, which satisfies f(1) = 0 only when properly integrated with boundary conditions.
<p><strong>Step 1:</strong> Identify slopes. Given m₁ = f'(x) and m₂ = f(x)/x, with |log(m₁ + m₂)/log x| = 2.</p><p><strong>Step 2:</strong> This means log(m₁ + m₂) = ±2 log x, so m₁ + m₂ = x² or m₁ + m₂ = x⁻².</p><p><strong>Step 3:</strong> For algebraic function f₁(x): f'(x) + f(x)/x = x². This is a linear DE: d/dx[xf(x)] = x³, giving xf₁(x) = x⁴/4 + C.</p><p><strong>Step 4:</strong> Using f₁(1) = 0: 1·0 = 1/4 + C, so C = -1/4. Thus f₁(x) = x³/4 - 1/(4x).</p><p><strong>Step 5:</strong> Calculate the integral:<br>∫₋₁¹ [f₁(x) + 1/(4x)]dx = ∫₋₁¹ [x³/4 - 1/(4x) + 1/(4x)]dx = ∫₋₁¹ x³/4 dx</p><p><strong>Step 6:</strong> Evaluating: [x⁴/16]₋₁¹ = 1/16 - 1/16 = 0. The 1/(4x) terms cancel and x³/4 is odd, integrating to zero over symmetric interval.</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: A