Trigonometry & Inverse Trigonometry
Interval from inequality involving inverse trig and quadratic application
nta_pyq_2023_jan
Grade 12
Question:
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 - 6x + 10) + \cos^{-1}(x^2 - 6x + 10) = 0 and \alpha - \beta = b - a, then \alpha is equal to:
\frac{\pi}{48}
\frac{\pi}{16}
\frac{\pi}{8}
\frac{\pi}{12}
Step-by-Step Solution
Key Concept: \sin^{-1}(t) + \cos^{-1}(t) = \pi/2. Find the largest interval where \sin^{-1}(\sin\theta) > \cos^{-1}(\sin\theta), i.e., \sin\theta > 1/\sqrt{2}, giving (\pi/4, 3\pi/4).
Interval is (\pi/4, 3\pi/4), so b - a = \pi/2 = \alpha - \beta. With \sin^{-1}(t)+\cos^{-1}(t)=\pi/2 at t=1 (x=3), equation gives 9\alpha + 3\beta + \pi/2 = 0. With \beta = \alpha - \pi/2: 12\alpha = \pi, so \alpha = \pi/12.
Correct Answer: 4