<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 \(\left|\dfrac{\log(m_1 + m_2)}{\log x}\right| = \dfrac{2}{1}\).<br>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><strong>583.</strong> 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 given condition |log(m₁ + m₂)/log x| = 2, derive that m₁ + m₂ = x² or x⁻², then use f(1)=0 to determine the specific algebraic function f₁(x), and finally evaluate the definite integral by exploiting symmetry properties.
<p><strong>Step 1:</strong> Identify slopes: m₁ = f'(x) and m₂ = f(x)/x</p><p><strong>Step 2:</strong> From |log(m₁ + m₂)/log x| = 2, we get |log(m₁ + m₂)| = 2|log x|, so log(m₁ + m₂) = ±2log x</p><p><strong>Step 3:</strong> This gives m₁ + m₂ = x² or m₁ + m₂ = 1/x²</p><p><strong>Step 4:</strong> For algebraic function, take f'(x) + f(x)/x = x². This is a linear ODE: d/dx[xf(x)] = x³, so xf(x) = x⁴/4 + C</p><p><strong>Step 5:</strong> Using f(1) = 0: 1·0 = 1/4 + C, so C = -1/4. Therefore f₁(x) = (x⁴ - 1)/(4x) = x³/4 - 1/(4x)</p><p><strong>Step 6:</strong> Evaluate ∫₋₁¹ [f₁(x) + 1/(4x)]dx = ∫₋₁¹ [x³/4 - 1/(4x) + 1/(4x)]dx = ∫₋₁¹ x³/4 dx</p><p><strong>Step 7:</strong> Since x³/4 is an odd function, ∫₋₁¹ x³/4 dx = 0</p><p>∴ Answer: A (which equals 0)</p>
Correct Answer: A