Question

For what value of θ, 1 lies between the roots of the quadratic equation 3x^2 – 3sinθ x – 2cos^2θ = 0?

(a) 2nπ + π/6 < θ < 2nπ + 5π/6

(b) 2nπ + π/3 < θ < 2nπ + 5π/3

(c) 2nπ + π/6 ≤ θ ≤ 2nπ + 5π/6

(d) 2nπ + π/3 ≤ θ ≤ 2nπ + 5π/3

This question was posed to me during an interview.

This interesting question is from Applications of Quadratic Equations topic in division Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer 1

Right choice is (a) 2nπ + π/6 < θ < 2nπ + 5π/6

The best explanation: Let, f(x) = 3x^2 – 3sinθ x – 2cos^2θ

The coefficient of x^2 > 0

 f(1) < 0

So, 3 – 3sinθ – 2cos^2θ < 0

    => 2sin^2θ – 3sinθ + 1 < 0

    => (2sinθ – 1)(sinθ – 1) < 0

    => ½  2nπ + π/6 < θ < 2nπ + 5π/6