If y = tan ⁻¹( sec x³ - tan x³ ),π/2 < x³ < 3π/2, then

Solution

Maths sir · 4 steps

Turn sec minus tan into one tangent, check the range, and y becomes a simple polynomial in x.

1. Half-angle form. Let t = x³. Write 1 - sin t = (cos t/2 - sin t/2)² and cos t = (cos t/2 - sin t/2)(cos t/2 + sin t/2). Cancel one common factor, then divide the top and bottom by cos t/2.

   sec t - tan t = (1 - sin t)/(cos t) = (1 - tan t/2)/(1 + tan t/2) = tan(π/4 - t/2)

2. Check the range. For π/2 < t < 3π/2, the angle π/4 - t/2 lies between -π/2 and 0. That is inside the principal range of tan⁻¹, so y = π/4 - x³/2.
3. Differentiate. y' = -3x²/2 and y'' = -3x. So x² y'' = -3x³, and -6y = -3π/2 + 3x³.
4. So. x² y'' - 6y = -3π/2, which means x² y'' - 6y + 3π/2 = 0. The answer is (b).

The idea to keep. When you see sec t ± tan t inside tan⁻¹, rewrite it as tan(π/4 ± t/2). Then check that this angle lies in (-π/2, π/2) before you cancel.

Answer from the NTA final answer key. Solution by PrepWiser.