Explanation
Given: cos(2sin−1x)=91
Using cos2θ=1−2sin2θ:
1−2x2=91⟹2x2=98⟹x2=94
x=32=nm⟹m=2,n=3
Equation: mx2−nx−m+n=0
2x2−3x−2+3=0⟹2x2−3x+1=0
\text{Roots are } \alpha = 1, \beta = \frac{1}{2} \quad (\alpha > \beta)
Point (1,1/2) lies on 5x+8y=9
5(1)+8(21)=5+4=9
\begin{aligned}
& \text{一} & 2x^{2}-3x-2+3 & =0 \\
& \text{一} & 2x^{2}-3x+1 & =0 \\
& \text{一} & 2x^{2}-2x-x+1 & =0 \\
& \text{一} & 2x^{2}(x-1)-(x-1) & =0 \\
& & (2x-1)(x-1) & =0 \\
& \text{一} & & \mathrm{x} & & =1,\frac{1}{2} \\
& \mathrm{:} & & \mathrm{0} & & =1, \\
& & & \mathrm{B}=1 \\
& \text{一} & (\alpha,\beta) & =
\begin{pmatrix}
1,\frac{1}{2}
\end{pmatrix} \\
& \mathrm{(1)} & 5(1)-8\left({\frac{1}{2}}\right) & =-9 \\
& \Rightarrow & & \mathrm{1} & & =-9\mathrm{~false} \\
& \mathrm{(2)} & 3(1)-2\left({\frac{1}{2}}\right) & =-2 \\
& \Rightarrow & & \mathrm{2} & & =-2\mathrm{~false}
\end{aligned}
(3) 3(1)+2(21)=−2
⇒4 = 2 false
(1) 5(1)+8(21)=9
⇒9 = 9 true
