Properties Of Inverse Trigonometric Functions

(Where )

${\sin }^{-1}\sin \theta -\left(\frac{\pi }{2}-{\sin }^{-1}\sin \theta \right)>0$
$\Rightarrow {\sin }^{-1}\sin \theta >\frac{\pi }{4}$
$\Rightarrow \sin \theta >\frac{1}{\sqrt{2}}$
$\text{So,}\theta \in \left(\frac{\pi }{4},\frac{3\pi }{4}\right)$
$\theta \in \left(\frac{\pi }{4},\frac{3\pi }{4}\right)=(a,b)$
$\Rightarrow b-a=\frac{\pi }{2}=\alpha -\beta$
$\Rightarrow \beta =\alpha -\frac{\pi }{2}$
$\Rightarrow \alpha x^2+\beta x+{\sin }^{-1}\left[(x-3{)}^2+1\right]+{\cos }^{-1}\left[(x-3{)}^2+1\right]=0$
$x=3,9\alpha +3\beta +\frac{\pi }{2}+0=0$
$\Rightarrow 9\alpha +3\left(\alpha -\frac{\pi }{2}\right)+\frac{\pi }{2}=0$
$\Rightarrow 12\alpha -\pi =0$
$\alpha =\frac{\pi }{12}$
So, the correct option is (D) :

Step 1: Analyze the quadratic expression in the denominator.
Let .
By completing the square: .
Since , the minimum value of is (at ).
Thus, .


Step 2: Find the range of the inner fraction.
Let .
As , (from the positive side).
As , .
So, .


Step 3: Find the range of the function .
The function is strictly increasing for .
Minimum value (limit as ): . Since never actually reaches 0, the lower bound is open: .
Maximum value (at ): .
So the range is .

Conclusion: The range is , which corresponds to option (3).