MathematicsMedium66×since 2002Q4115Let (a, b) ⊂(0,2π)\subset(0,2 \pi)⊂(0,2π) be the largest interval for which sin−1(sinθ)−cos−1(sinθ)>0,θ∈(0,2π)\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi)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\alpha x^{2}+\beta x+\sin ^{-1}\left(x^{2}-6 x+10\right)+\cos ^{-1}\left(x^{2}-6 x+10\right)=0αx2+βx+sin−1(x2−6x+10)+cos−1(x2−6x+10)=0 and α−β=b−a\alpha-\beta=b-aα−β=b−a, then α\alphaα is equal to :Aπ16\frac{\pi}{16}16πBπ48\frac{\pi}{48}48πCπ8\frac{\pi}{8}8πDπ12\frac{\pi}{12}12πCheck answerSkip