Question

f(x) = sinx – sin2x in [0, π]

Answer 1

We have, f(x) = sinx – sin2x in [0, π]

We know that all trigonometric functions are continuous and differentiable their domain, given function is also continuous and differentiable

So, condition of mean value theorem are satisfied.

Hence, there exists atleast one c ∈ (0, π) such that,

f'(c) = `("f"(pi) - "f"(0))/(pi - 0)`

⇒ cos c – 2 cos 2c =  `(sin pi - sin 2pi - sin 0 + sin 0)/(pi - 0)`

⇒ 2 cos 2c – cos c = 0

⇒ 2(2 cos²c – 1) – cos c = 0

⇒ 4cos²c – cos c – 2 = 0

⇒ cos c = `(1 +- sqrt(1 + 32))/8`

= `(1 +- sqrt(33))/8`

⇒ c = `cos^-1 ((1 +- sqrt(33))/8) ∈ (0, π)`

Hence, mean value theorem has been verified.