<p>Given curves \(y = kx^2\) and \(x = ky^2\). If the area bounded by \(y = kx^2\) and \(x = ky^2\) is 1, then \(k\) equals:</p>
Step-by-Step Solution
Key Concept: The two curves y = kx² and x = ky² are reflections of each other across y = x, so they intersect at the origin and at point (1/k, 1/k). The area between them can be computed using a single integral by exploiting symmetry.
<p><strong>Step 1:</strong> Find intersection points by solving y = kx² and x = ky² simultaneously. Substituting the first into the second: x = k(kx²)² = k³x⁴, so x(1 - k³x³) = 0. This gives x = 0 and x = 1/k, yielding intersection points (0,0) and (1/k, 1/k).</p><p><strong>Step 2:</strong> Recognize that x = ky² can be written as y = √(x/k). For 0 ≤ x ≤ 1/k, the curve y = √(x/k) lies above y = kx².</p><p><strong>Step 3:</strong> Set up the area integral: A = ∫₀^(1/k) [√(x/k) - kx²] dx = ∫₀^(1/k) [x^(1/2)/√k - kx²] dx</p><p><strong>Step 4:</strong> Evaluate: A = [1/√k · (2x^(3/2)/3) - k·(x³/3)]₀^(1/k) = (2/3√k)·(1/k)^(3/2) - (k/3)·(1/k)³</p><p><strong>Step 5:</strong> Simplify: A = (2/3√k)·(1/(k√k)) - 1/(3k³) = 2/(3k²) - 1/(3k³) = (2k - 1)/(3k³)</p><p><strong>Step 6:</strong> Set A = 1: (2k - 1)/(3k³) = 1 → 2k - 1 = 3k³ → 3k³ - 2k + 1 = 0 → (k - 1)(3k² + 3k + 1) = 0</p><p><strong>Step 7:</strong> Since 3k² + 3k + 1 has discriminant 9 - 12 = -3 < 0 (no real roots), we have k = 1.</p><p>∴ Answer: B (k = 1)</p>
Correct Answer: B