JEE Main 2024 — Mathematics Trigonometry

Given: $\mathrm{cos}(2{\mathrm{sin}}^{-1}x)=\frac{1}{9}$cos(2sin−1x)=91​

Let, ${\mathrm{sin}}^{-1}x=\theta$sin−1x=θ

$$\Rightarrow \mathrm{cos}(2\theta )=\frac{1}{9}$$⇒cos(2θ)=91​

$$\Rightarrow 1-2{\mathrm{sin}}^{2}\theta =\frac{1}{9}$$⇒1−2sin2θ=91​

$$\Rightarrow \frac{8}{9}=2{\mathrm{sin}}^{2}\theta$$⇒98​=2sin2θ

$$\Rightarrow \mathrm{sin}\theta =\pm \frac{2}{3}$$⇒sinθ=±32​

$$\Rightarrow x=\pm \frac{2}{3}$$⇒x=±32​

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

$$\Rightarrow m=2,n=3$$⇒m=2,n=3

Also, $m{x}^{2}-nx-m+n=0$mx2−nx−m+n=0.

$$\Rightarrow 2{x}^{2}-3x-2+3=0$$⇒2x2−3x−2+3=0

$$\Rightarrow 2{x}^{2}-3x+1=0$$⇒2x2−3x+1=0

$$\Rightarrow 2{x}^{2}-2x-x+1=0$$⇒2x2−2x−x+1=0

$$\Rightarrow 2x(x-1)-(x-1)=0$$⇒2x(x−1)−(x−1)=0

$$\Rightarrow (2x-1)(x-1)=0$$⇒(2x−1)(x−1)=0

$$\Rightarrow x=1,\frac{1}{2}$$⇒x=1,21​

$$\Rightarrow \alpha =1,\beta =\frac{1}{2}$$⇒α=1,β=21​

Out of the given options, $(1,\frac{1}{2})$(1,21​) lines on $5x+8y=9$5x+8y=9.