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प्रश्न
Find the absolute maximum and minimum values of the function f given by f (x) = cos2 x + sin x, x ∈ [0, π].
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उत्तर
Here, f(x) = cos2 x + sin x, x ϵ (0, π)
∴ f'(x) = 2 cos x (- sin x) + cos x
= cos x (- 2 sin x + 1)
For maximum / minimum values, f (x) = 0
⇒ cos x (- 2 sin x + 1) = 0
`=> sin x = 1/2 => x = pi/6` and cos x = 0 ⇒ x = `pi/2`
In the interval [0, π], the critical points are x = `pi/6` and x = `pi/2`.
∴ f(0) = cos2 0 + sin 0 = 1
`f(pi/6) = cos^2 pi/6 + sin pi/6`
`= (sqrt3/2)^2 + 1/2`
`= 3/4 + 1/2`
`= (3 + 2)/4 = 5/4`
`f(pi/2) = cos^2 pi/2 + sin pi/2`
= 0 + 1 = 1
Hence, the absolute maximum and minimum value are `5/4` and 1, respectively.
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