Let ( {\a}, {\b}) \subset(0,2 \pi) be the largest interval for which \sin ^{-1}( \sin \theta)-\cos ^{-1}( \sin \theta)>0, \theta \in(0,2 \pi) holds. If \alpha x^2+\beta x+ \sin ^{-1} (x^2-6 x+10) +\cos ^{-1} (x^2-6 x+10) =0 and \alpha-\beta=b-a, then \alpha is equal to : [31-Jan-2023 Shift 2]

Inverse Trigonometric Functions

Section: Mathematics

Question : 32 of 95

Marks: +1, -0

Let (a,b)⊂(0,2π) be the largest interval for which sin‌−1(sin‌θ)−cos−1(sin‌θ)>0,θ∈(0,2π) holds. If
‌αx2+βx+sin‌−1(x2−6x+10)+cos−1
‌(x2−6x+10)=0
and α−β=b−a, then α is equal to :

[31-Jan-2023 Shift 2]

‌
π
48
‌
π
16
‌
π
8
‌
π
12

Validate

Go to Question:

Prev Question

Next Question