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Find the absolute maximum and absolute minimum values of the function f given by f(x)=sin^2 x-cosx,x ∈ (0,π) - CBSE (Commerce) Class 12 - Mathematics

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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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Solution Find the absolute maximum and absolute minimum values of the function f given by f(x)=sin^2 x-cosx,x ∈ (0,π) Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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