Values of c of Rolle's theorem for `f(x)=sin x-sin 2x` on `[0,pi]`

Answer

N/a

Explanation

We have, `f(x) = sinx - sin2x "in" [0,pi]`
(i) Since , we know that sine functions are continuous functions hence `f(x) = sinx - sin2x` is a continuous function in `[0,pi]`
(ii) `f(x) = cosx - cos2x . 2 = cos x - 2 cos 2x ` , which exist in `(0,pi)`
So, `f(x)` is differentiable in `(0,pi)`. Conditions of mean value theorem are satisfied.
Hence `3c in (0,pi)` such that, `f'(c) = (f(pi)-f (0))/(pi-0)`
`rArr cos - 2 cos 2x = (sin pi - sin 2pi - sin 0 + sin 2.0)/(pi-0)`
`rArr 2cos 2c -cos c = (0)/( pi)`
`rArr 2.(2cos^(2)c-1)-cosc= (0)`
`rArr 4 cos^(2)-2-cosc=0`
`rArr 4 cos^(2)c-cosc-2 = 0`
`rArr cos c = (1+-sqrt(1+32))/(8) = (1+-sqrt(33))/(8)`
`:. c = cos^(-1)((1+-sqrt(33))/(8))`
Also, `cos^(-1) ((1+-sqrt(33))/(8)) in (0,pi)`
Hence, mean value ther em has been verified.