<p>Let maximum value of the expression \(y = |k - 3|\cos 2x + |t - 4|\sin 2x + 3\) where \(0 \leq k \leq 6\) and \(1 \leq t \leq 7\) is equal to 6, then find the minimum value of \((k^2 + t^2)\).</p>
Step-by-Step Solution
Key Concept: The maximum of a·cos(2x) + b·sin(2x) + c is c + √(a² + b²). Since this maximum equals 6, we get √(|k-3|² + |t-4|²) = 3, which means k and t lie on a circle of radius 3 centered at (3,4). Minimize k² + t² by finding the point on this circle closest to the origin.
<p><strong>Step 1:</strong> The maximum value of a·cos(2x) + b·sin(2x) is √(a² + b²). Here a = |k-3| and b = |t-4|.</p><p><strong>Step 2:</strong> Maximum of y = √(|k-3|² + |t-4|²) + 3 = 6, so √(|k-3|² + |t-4|²) = 3.</p><p><strong>Step 3:</strong> This means (k-3)² + (t-4)² = 9. The point (k,t) lies on a circle with center C(3,4) and radius 3.</p><p><strong>Step 4:</strong> We need to minimize k² + t² = distance² from origin O(0,0) to point (k,t) on this circle.</p><p><strong>Step 5:</strong> Minimum distance from O to the circle = |OC| - radius = √(3² + 4²) - 3 = 5 - 3 = 2.</p><p><strong>Step 6:</strong> The closest point lies on the line joining O(0,0) and C(3,4), at distance 2 from O. This point is (k,t) = 2·(3,4)/5 = (6/5, 8/5).</p><p><strong>Step 7:</strong> Verify: (6/5 - 3)² + (8/5 - 4)² = (-9/5)² + (-12/5)² = 81/25 + 144/25 = 225/25 = 9 ✓</p><p><strong>Step 8:</strong> Minimum value of k² + t² = (6/5)² + (8/5)² = 36/25 + 64/25 = 100/25 = 4.</p><p>∴ Answer: <strong>4</strong></p>
Correct Answer: 4