Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Given: $\cos \left(2{\sin }^{-1}x\right)=\frac{1}{9}$

Let, ${\sin }^{-1}x=\theta$

$\Rightarrow \cos \left(2\theta \right)=\frac{1}{9}$

$\Rightarrow 1-2{\sin }^2\theta =\frac{1}{9}$

$\Rightarrow \frac{8}{9}=2{\sin }^2\theta$

$\Rightarrow \sin \theta =\pm \frac{2}{3}$

$\Rightarrow x=\pm \frac{2}{3}$

It is given that $m$ and $n$ are co-prime natural numbers,

$\Rightarrow m=2, n=3$

Also, $m{x}^2-nx-m+n=0$.

$\Rightarrow 2x^2-3x-2+3=0$

$\Rightarrow 2x^2-3x+1=0$

$\Rightarrow 2x^2-2x-x+1=0$

$\Rightarrow 2x\left(x-1\right)-\left(x-1\right)=0$

$\Rightarrow \left(2x-1\right)\left(x-1\right)=0$

$\Rightarrow x=1,\frac{1}{2}$

$\Rightarrow \alpha =1, \beta =\frac{1}{2}$

Out of the given options, $\left(1,\frac{1}{2}\right)$ lines on $5x+8y=9$.