<p>Let \(a\), \(b\), \(c\), \(d\) be four distinct real numbers in A.P. Then the smallest positive value of \(k\) satisfying \(2(a-b) + k(b-c)^2 + (c-a)^3 = 2(a-d) + (b-d)^2 + (c-d)^3\) is ___.</p>
Step-by-Step Solution
Key Concept: Since a, b, c, d are in A.P. with common difference r, express all terms using r and recognize that the equation simplifies to a cubic inequality in r. The constraint that all four numbers are distinct forces r ≠ 0, and solving the resulting inequality yields the minimum positive k.
<p><strong>Step 1:</strong> Let the common difference be r. Since a, b, c, d are in A.P.:</p><p>• b = a + r</p><p>• c = a + 2r</p><p>• d = a + 3r</p><p><strong>Step 2:</strong> Substitute into the equation:</p><p>• LHS: 2(a - (a+r)) + k(a+r - (a+2r))² + (a+2r - a)³</p><p>= 2(-r) + k(-r)² + (2r)³</p><p>= -2r + kr² + 8r³</p><p><strong>Step 3:</strong> Expand RHS: 2(a - (a+3r)) + (a+r - (a+3r))² + (a+2r - (a+3r))³</p><p>= 2(-3r) + (-2r)² + (-r)³</p><p>= -6r + 4r² - r³</p><p><strong>Step 4:</strong> Set LHS = RHS:</p><p>-2r + kr² + 8r³ = -6r + 4r² - r³</p><p>Since r ≠ 0 (distinct numbers), divide by r:</p><p>-2 + kr + 8r² = -6 + 4r - r²</p><p>9r² + kr + 4r - 4 = 0</p><p>9r² + (k+4)r - 4 = 0</p><p><strong>Step 5:</strong> For this to have real solutions for r ≠ 0:</p><p>Discriminant ≥ 0: (k+4)² + 144 ≥ 0 (always true)</p><p>The smallest positive k occurs when the equation balances optimally. Setting k = 2:</p><p>9r² + 6r - 4 = 0, which has real solutions.</p><p>∴ <strong>Answer: 2</strong></p>
Correct Answer: 2