Inverse Trigonometric Functions Ans 6
Q6 - 31 January - Shift 2
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 :
(1) $\frac{\pi}{48}$
(2) $\frac{\pi}{16}$
(3) $\frac{\pi}{8}$
(4) $\frac{\pi}{12}$
Show Answer
Answer: (4)
Solution:
Formula: Property of sine and cosine inverse function : $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$
$ \begin{aligned} & \sin ^{-1} \sin \theta-(\frac{\pi}{2}-\sin ^{-1} \sin \theta)>0 \\ & \Rightarrow \sin ^{-1} \sin \theta>\frac{\pi}{4} \\ & \Rightarrow \sin \theta>\frac{1}{\sqrt{2}} \\ & \text{ So, } \theta \in(\frac{\pi}{4}, \frac{3 \pi}{4}) \\ & \theta \in(\frac{\pi}{4}, \frac{3 \pi}{4})=(a, b) \\ & b-a=\frac{\pi}{2}=\alpha-\beta \\ & \Rightarrow \beta=\alpha-\frac{\pi}{2} \\ & \Rightarrow \alpha x^{2}+\beta x+\sin ^{-1}[(x-3)^{2}+1]+\cos ^{-1}[(x-3)^{2}+1]=0 \\ & x=3,9 \alpha+3 \beta+\frac{\pi}{2}+0=0 \\ & \Rightarrow 9 \alpha+3(\alpha-\frac{\pi}{2})+\frac{\pi}{2}=0 \\ & \Rightarrow 12 \alpha-\pi=0 \\ & \alpha=\frac{\pi}{12} \end{aligned} $
