Let a , b ⊂ 0 , 2 π be the largest interval for which sin - 1 sin θ - cos - 1 sin θ > 0 , θ ∈ 0 , 2 π , holds . If α x 2 + β x + sin - 1 x 2 - 6 x + 10 + cos - 1 x 2 - 6 x + 10 = 0 and α - β = b - a , then α is equal to;

Step-by-step solution

Given, sin - 1 sin θ - cos - 1 sin θ > 0 , θ ∈ 0 , 2 π ∵ sin - 1 θ + cos - 1 θ = π 2 ∴ sin - 1 sin θ - π 2 - sin - 1 sin θ > 0 sin - 1 sin θ > π 4 ⇒ θ ∈ π 4 , 3 π 4 a , b = π 4 , 3 π 4 ⇒ b - a = π 2 Given b - a = α - β ∴   α - β = π 2     . . . . . ( 1 ) Now α x 2 + β x + sin - 1 x 2 - 6 x + 10 + cos - 1 x 2 - 6 x + 10 = 0 N