Trigonometry & Inverse Trigonometry
arccos Equation — Quadratic Roots on a Line
nta_pyq_2024_jan
Grade 12
Question:
Let $x=\dfrac{m}{n}$ ($m$, $n$ are co-prime natural numbers) be a solution of the equation $\cos(2\sin^{-1}x)=\dfrac{1}{9}$ 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
$3x+2y=2$
$5x-8y=-9$
$3x-2y=-2$
$5x+8y=9$
Step-by-Step Solution
Key Concept: Let $\sin^{-1}x=\theta$. $\cos2\theta=1-2\sin^2\theta=1/9\Rightarrow\sin^2\theta=4/9\Rightarrow x=\pm2/3$. Since $m,n$ are co-prime natural numbers, $x=2/3$, $m=2,n=3$. Solve $2x^2-3x-1+3=0$ i.e. $2x^2-3x+1=0$.
$x=2/3,m=2,n=3$. Equation: $2x^2-3x+1=0\Rightarrow\alpha=1,\beta=1/2$. $(1,1/2)$ lies on $5x+8y=9$.
Correct Answer: 4