<p>If there are only two linear functions \(f\) and \(g\) which map \([1, 2]\) on \([4, 6]\) and in a triangle ABC, \(c = f(1) + g(1)\) and \(a\) is the maximum value of \(r^2\), where \(r\) is the distance of a variable point on the curve \(x^2 + y^2 - xy = 10\) from the origin, then \(\sin A : \sin C\) is</p>
Step-by-Step Solution
Key Concept: Find the two linear functions mapping [1,2] to [4,6], compute c from their values at 1, find the maximum distance r from origin to the ellipse, use the sine rule to relate c and a to angles in triangle ABC.
<p><strong>Step 1: Find the two linear functions f and g.</strong></p><p>Let f(x) = mx + n map [1,2] onto [4,6].</p><p>For a linear function to map [1,2] onto [4,6], we need:</p><p>• If m > 0: f(1) = 4 and f(2) = 6, so m + n = 4 and 2m + n = 6. This gives m = 2, n = 2, so f(x) = 2x + 2.</p><p>• If m < 0: f(1) = 6 and f(2) = 4, so m + n = 6 and 2m + n = 4. This gives m = -2, n = 8, so g(x) = -2x + 8.</p><p>Therefore, f(1) = 4 and g(1) = 6.</p><p><strong>Step 2: Calculate c.</strong></p><p>c = f(1) + g(1) = 4 + 6 = 10.</p><p><strong>Step 3: Find the maximum value of r² on the curve x² + y² - xy = 10.</strong></p><p>We have r² = x² + y². We need to maximize x² + y² subject to x² + y² - xy = 10.</p><p>Let x² + y² = r². Then: r² - xy = 10, so xy = r² - 10.</p><p>By AM-GM inequality: x² + y² ≥ 2|xy|, thus r² ≥ 2|r² - 10|.</p><p>Case 1: If r² - 10 ≥ 0, then r² ≥ 2(r² - 10), which gives r² ≤ 20.</p><p>Case 2: If r² - 10 < 0, then r² ≥ 2(10 - r²), which gives 3r² ≥ 20, so r² ≥ 20/3.</p><p>The maximum occurs when equality holds in AM-GM with r² > 10: x² = y² and xy = r² - 10 > 0.</p><p>So x = y, and x² + x² - x² = 10 gives x² = 10, thus r² = 2x² = 20.</p><p>Therefore, a = 20.</p><p><strong>Step 4: Apply the sine rule.</strong></p><p>In triangle ABC, we have c = 10 and a = 20.</p><p>By the sine rule: a/sin A = c/sin C.</p><p>Thus: sin A/sin C = a/c = 20/10 = 2/1.</p><p>Therefore: sin A : sin C = 2 : 1.</p><p>∴ Answer: A</p>
Correct Answer: A