Answer :

First, let us write the conditions for the applicability of Rolle’s theorem:


For a Real valued function ‘f’:


a) The function ‘f’ needs to be continuous in the closed interval [a,b].


b) The function ‘f’ needs differentiable on the open interval (a,b).


c) f(a) = f(b)


Then there exists at least one c in the open interval (a,b) such that f’(c) = 0.


Given function is:


f(x) = sinx – sin2x on [0,]


We know that sine function is continuous and differentiable over R.


Let’s check the values of the function ‘f’ at the extremums.


f(0) = sin(0)–sin2(0)


f(0) = 0 – sin(0)


f(0) = 0


f() = sin() – sin2()


f() = 0 – sin(2)


f() = 0


We got f(0) = f(). So, there exists a cϵ such that f’(c) = 0.


Let’s find the derivative of the function ‘f’





f’(x) = cosx – 2(2cos2x – 1)


f’(x) = cosx – 4cos2x + 2


We have f’(c) = 0


cosc – 4cos2c + 2 = 0





We can see that cϵ(0,)


Rolle’s theorem is verified.


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