Let (a, b) (0,2 ) be the largest interval for which ^-1( )- ^-1( )0, (0,2 ), holds.

Question

Let (a, b) $\subset (0,2\pi )$ be the largest interval for which ${\sin }^{-1}(\sin 	heta )-{\cos }^{-1}(\sin 	heta )>0,	heta \in (0,2\pi )$,

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 :

Accepted Answer

$\sin^{-1}(\sin\theta)-\left(\dfrac{\pi}{2}-\sin^{-1}(\sin\theta)\right)>0\Rightarrow\sin^{-1}(\sin\theta)>\dfrac{\pi}{4}\Rightarrow\sin\theta>\dfrac{1}{\sqrt2}$.

So $\theta\in\left(\dfrac{\pi}{4},\dfrac{3\pi}{4}\right)=(a,b)$, giving $b-a=\dfrac{\pi}{2}$, hence $\alpha-\beta=\dfrac{\pi}{2}\Rightarrow\beta=\alpha-\dfrac{\pi}{2}$.

Since $x^2-6x+10=(x-3)^2+1\ge1$, and $\sin^{-1}u+\cos^{-1}u=\dfrac{\pi}{2}$ is only defined at $u=1$, we need $(x-3)^2+1=1\Rightarrow x=3$. At $x=3$:

$$9\alpha+3\beta+\frac{\pi}{2}=0\ \Rightarrow\ 9\alpha+3\left(\alpha-\frac{\pi}{2}\right)+\frac{\pi}{2}=0\ \Rightarrow\ 12\alpha-\pi=0\ \Rightarrow\ \alpha=\frac{\pi}{12}.$$

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