Let x=m/n (m, n are co-prime natural numbers) be a solution of the equation cos (2 sin ⁻¹ x)=1/9 and let α, β(α >β) be the roots of the equation m x²-n x-m+ n=0. Then the point (α, β) lies on the line

Solution

Maths sir · 3 steps

Write cos(2θ) as 1-2 sin² θ and the inverse sine disappears, leaving a plain equation in x.

1. Remove the inverse. Let θ = sin⁻¹ x, so sin θ = x. Then cos 2θ = 1 - 2 sin² θ = 1 - 2x².

   1 - 2x² = 1/9 ⇒ x² = 4/9 ⇒ x = ±2/3

2. Find the quadratic. Since m and n are natural numbers, x = 2/3, so m = 2 and n = 3. The equation becomes 2x² - 3x + 1 = 0, that is (2x-1)(x-1) = 0. The roots are 1 and 1/2, so α = 1 and β = 1/2.

   Option	Left side at (1, 1/2)	Right side	On the line?
   (a) 3x - 2y = -2	3 - 1 = 2	-2	no
   (b) 3x + 2y = 2	3 + 1 = 4	2	no
   (c) 5x + 8y = 9	5 + 4 = 9	9	yes
   (d) 5x - 8y = -9	5 - 4 = 1	-9	no

3. So. The point (1, 1/2) satisfies 5x + 8y = 9, so the answer is (c).

The idea to keep. When a trig function acts on an inverse trig function, put θ for the inverse and use an identity in sin θ or cos θ. The equation turns algebraic straight away.

Answer from the NTA final answer key. Solution by PrepWiser.