Let  $x=\frac{m}{n}$ (m, n are co-prime natural numbers) be a solution of the equation  $\cos \left(2{\sin }^{-1}x\right)=\frac{1}{9}$ and  $\alpha ,\beta (\alpha >\beta )$ be the roots of the equation $m{x}^2-nx-m+n=0$, then which of the following is are CORRECT?

$\cos \left(2{\sin }^{-1}x\right)=\frac{1}{9}\Rightarrow 1-2{\sin }^2\left({\sin }^{-1}x\right)=\frac{1}{9}\Rightarrow \frac{8}{9}=2x^2\Rightarrow x=\pm \frac{2}{3}$

$x=\frac{2}{3}=\frac{m}{n}asm,n\in N,$

The roots  $m{x}^2-nx-m+n=0$  and  $\alpha =1, \beta =\frac{1}{2}$

$$\begin{gathered} f\left(\theta \right)=\frac{{\sin }^4\theta +{\cos }^2\theta +2{\cos }^2\theta }{{\sin }^2\theta +{\cos }^2\theta }=1+\frac{2{\cos }^2\theta }{1+{\cos }^4\theta -{\cos }^2\theta } \\ =1+\frac{2}{{\sec }^2\theta +{\cos }^2\theta -1},\min =1,\max =1+\frac{2}{1}=3 \end{gathered}$$

$Range=[1,3]=[\alpha ,6\beta ]$