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

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Solution :We have,`F(x) = SINX - sin2x "in" [0,pi]`
(i) SINCE , we know thatsinefunctionsare continuous functionshence `f(x)=sinx - sin2x` isa continuous function in `[0,pi]`
(ii) `f(x) = cosx -COS2X. 2 = cos x - 2 cos 2X ` , which existin `(0,pi)`
So, `f(x)`is differentiable in `(0,pi)`.Conditionsof mean value theoremare 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 2cos2c -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 cosc = (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, meanvalue therem has been verified.


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