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Question
Find the absolute maximum and absolute minimum values of the function f given by f(x)=sin2x-cosx,x ∈ (0,π)
Solution
f(x)=sin2x-cosx
f'(x)=2 sinx.cosx+sinx
=sinx(2cosx+1)
Equating f’(x) to zero.
f'(x)=0
sin x(2cos x + 1) = 0
sin x = 0
∴ x = 0, π
`2cos x + 1 = 0`
`⇒cos x =-1/2`
`therefore x=(5pi)/6`
`f(0) = sin20 – cos 0 = − 1`
`f((5pi)/6)=sin^2(5pi/6)-cos((5pi)/6)`
`=sin^2(pi/6)+cos(pi/6)`
`=1/4-sqrt3/2`
`=((1-2sqrt3)/sqrt4)`
`f(pi) = sin^2pi – cospi = 1`
Of these values, the maximum value is 1, and the minimum value is −1.
Thus, the absolute maximum and absolute minimum values of f(x) are 1 and −1, which it attains at x = 0 and x = π.
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