Let x=m/n (m,n are co-prime natural numbers) be a solution of the equation (2^-1x)=19 and …

Question

Let $x=m/n$ ($m,n$ are co-prime natural numbers) be a solution of the equation $\cos(2\sin^{-1}x)=\dfrac19$ and let $\alpha,\beta$ ($\alpha>\beta$) be the roots of the equation $mx^2-nx-m+n=0$. Then the point $(\alpha,\beta)$ lies on the line

Accepted Answer

$\cos(2\sin^{-1}x)=1-2x^2=\dfrac19\Rightarrow x^2=\dfrac{4}{9}\Rightarrow x=\dfrac23$ (taking the positive co-prime form), so $m=2,n=3$.

The quadratic becomes $2x^2-3x+1=0\Rightarrow(2x-1)(x-1)=0\Rightarrow x=1,\dfrac12$. So $\alpha=1,\ \beta=\dfrac12$ (since $\alpha>\beta$).

Checking $(\alpha,\beta)=\left(1,\dfrac12\right)$ against the options: $5(1)+8\left(\dfrac12\right)=5+4=9$, which satisfies $5x+8y=9$.

Solutions · 1