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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

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### 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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