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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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प्रश्न

Find the absolute maximum and minimum values of the function f given by f (x) = cos2 x + sin x, x ∈ [0, π].

योग
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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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अध्याय 6: Application of Derivatives - Exercise 6.6 [पृष्ठ २४३]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 6 Application of Derivatives
Exercise 6.6 | Q 14 | पृष्ठ २४३

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