Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of 16((sec ⁻¹ x)²+(cosec⁻¹ x)²) is :

Solution

Maths sir · 4 steps

Swap cosec⁻¹x for π/2 - sec⁻¹ x, then complete the square.

1. One variable. Let θ = sec⁻¹ x. For |x| ≥ 1, cosec⁻¹x = π/2 - θ. Here θ ∈ [0, π] with θ ≠ π/2.

   θ² + (π/2-θ)² = 2(θ - π/4)² + π²/8

2. Minimum. The square is smallest, zero, at θ = π/4, which is reached at x = √2. The expression is then 16 × π²/8 = 2π².
3. Maximum. The square is largest at the endpoint farthest from π/4, which is θ = π (at x = -1). Then π² + (-π/2)² = 5π²/4, and 16 × 5π²/4 = 20π².
4. So. Maximum plus minimum is 20π² + 2π² = 22π², option (c).

The idea to keep. When two inverse functions add to a constant, write the expression in one variable and complete the square. Then check both ends of the principal-value range for the maximum.

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