If 2 sin ³ x+sin 2 x cos x+4 sin x-4=0 has exactly 3 solutions in the interval [0, (n π)/2], n ∈ N, then the roots of the equation x²+n x+(n-3)=0 belong to :

Solution

Maths sir · 4 steps

Expand sin 2x with cos² x = 1 - sin² x; the cubes cancel and sin x = 2/3.

1. Simplify. Write sin 2x cos x = 2 sin x cos² x = 2 sin x(1-sin² x) = 2 sin x - 2 sin³ x. Put this back into the equation and the sin³ x terms cancel.

   2 sin³ x + 2 sin x - 2 sin³ x + 4 sin x - 4 = 0 ⇒ 6 sin x = 4 ⇒ sin x = 2/3

2. Count solutions. Take α ∈ (0, π/2) with sin α = 2/3. The solutions in increasing order are α, π - α, 2π + α, 3π - α, and so on. The interval [0, 2π] (n = 4) holds only 2 of them and [0, 3π] (n = 6) holds 4, but [0, 5π/2] holds exactly 3 because 2π + α < 5π/2 < 3π - α, so n = 5.
3. Solve the quadratic. The quadratic becomes x² + 5x + 2 = 0, with roots x = (-5 ± √17)/2 ≈ -0.44 and -4.56. As a check, the sum is -5 < 0 and the product is 2 > 0, so both roots have the same sign and that sign is negative.
4. So. Both roots lie in (-∞, 0), which is option (d).

The idea to keep. To count solutions of sin x = k in [0, L], list them in order: α, π-α, 2π+α, …. To find the signs of the roots, check the sum and product before solving.

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