Question

If the roots of the given equation (cosp-1)x^2+(cosp)x+sinp=0 are real.then
(a)p€(-π,0). (b)p€[-π/2,π/2]
(c)p€(0,π). (d)p€(0,2π)

Answer 1

The roots of the equation, (cosp - 1)x² + (cosp)x + sinp = 0, are real.

To check : which one of the following is correct option ?

(a) P ∈ (-π, 0) (b) P ∈ [-π/2, π/2]

(c) P ∈ (0, π) (d) P ∈ (0, 2π)

solution : for real roots,

Discriminant, D = b² - 4ac ≥ 0

⇒(cosp)² - 4(cosp - 1)(sinp) ≥ 0

⇒cos²p + 4sinp (1 - cosp) ≥ 0

here cos²p ≥ 0 for all real value of p.

1 - cosp ≥ 0 for all real value of p.

means, (+ve) + 4sinp(+ ve) ≥ 0

here we can see, inequality will be always correct when we take sinp ≥ 0

but sinp will be positive only when p ∈ (0, π)

Therefore the roots of equation will be real only if P ∈ (0, π).

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