Q. 193.5( 2 Votes )

# Verify Rolle’s th

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 cϵ(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. i.e. c=32° 32’ or c=126°23’

Value of c=32°32’ϵ(0,2π)

Thus, Rolle’s theorem is satisfied.

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