<p>82. If the equation \(\cot^4 x - 2\csc^2 x + a^2 = 0\) has at least one solution, then the sum of all possible integral values of \(a\) is equal to</p>
Step-by-Step Solution
Key Concept: Convert the equation to a single variable by using cot²x = csc²x - 1, then find the range of the resulting quadratic expression to determine which integral values of a permit real solutions.
<p><strong>Step 1:</strong> Use the identity cot²x = csc²x - 1. Let y = csc²x where y ≥ 1 (since csc²x ≥ 1 for all real x).</p><p><strong>Step 2:</strong> Rewrite cot⁴x = (csc²x - 1)² = (y - 1)². The equation becomes: (y - 1)² - 2y + a² = 0</p><p><strong>Step 3:</strong> Expand: y² - 2y + 1 - 2y + a² = 0, which gives y² - 4y + (1 + a²) = 0</p><p><strong>Step 4:</strong> For the original equation to have at least one solution, this quadratic in y must have at least one root y ≥ 1.</p><p><strong>Step 5:</strong> The roots are y = (4 ± √(16 - 4(1 + a²)))/2 = 2 ± √(3 - a²). For real roots: 3 - a² ≥ 0, so |a| ≤ √3 ≈ 1.732</p><p><strong>Step 6:</strong> Check if at least one root satisfies y ≥ 1. The larger root is y = 2 + √(3 - a²). For |a| ≤ √3, we have y ≥ 2 + 0 = 2 ≥ 1 ✓. The condition is always satisfied when real roots exist.</p><p><strong>Step 7:</strong> Therefore |a| ≤ √3, giving integral values: a ∈ {-1, 0, 1}</p><p>∴ Sum = -1 + 0 + 1 = <strong>0</strong></p>
Correct Answer: 4