Verify Rolle’s theorem for each of the following functions:f(x) = si

1.	Verify Rolle’s theorem for each of the following functions:f(x) = sin x - sin 2x in [0, 2π]
Answer» Condition (1): Since, f(x) = sinx - sin2x is a trigonometric function and we know every trigonometric function is continuous. ⇒ f(x) = sinx - sin2x is continuous on [0,2π]. Condition (2): Here, f’(x) = cosx - 2cos2x which exist in [0,2π]. So, f(x) = sinx - sin2x is differentiable on (0,2π) Condition (3): Here, f(0) = sin0 - sin0 = 0 And f(2π) = sin(2π) - sin(4π) = 0 i.e. f(0) = f(2π) Conditions of Rolle’s theorem are satisfied. Hence, there exist at least one c ϵ (0,2π) such that f’(c) = 0 i.e. cosx - 2cos2x = 0 i.e. cosx - 4cos²x+2 = 0 i.e. 4cos²x - cosx - 2 = 0 i.e. cos x = \(\frac{1\pm\sqrt{33}}{8}\) i.e. c = 32° 32’ or c = 126°23’ Value of c = 32°32’ ϵ (0,2π) Thus, Rolle’s theorem is satisfied.

Verify Rolle’s theorem for each of the following functions:f(x) = sin x - sin 2x in [0, 2π]

Answer»

Condition (1):

Since, f(x) = sinx - sin2x is a trigonometric function and we know every trigonometric function is continuous.

⇒ f(x) = sinx - sin2x is continuous on [0,2π].

Condition (2):

Here, f’(x) = cosx - 2cos2x which exist in [0,2π].

So, f(x) = sinx - sin2x is differentiable on (0,2π)

Condition (3):

Here, f(0) = sin0 - sin0 = 0

And f(2π) = sin(2π) - sin(4π) = 0

i.e. f(0) = f(2π)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one c ϵ (0,2π) such that f’(c) = 0

i.e. cosx - 2cos2x = 0

i.e. cosx - 4cos²x+2 = 0

i.e. 4cos²x - cosx - 2 = 0

i.e. cos x = \(\frac{1\pm\sqrt{33}}{8}\)

i.e. c = 32° 32’ or c = 126°23’

Value of c = 32°32’ ϵ (0,2π)

Thus, Rolle’s theorem is satisfied.