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

Solution:

Maxima and minima are known as the extrema of a function. Maxima and minima are the maximum or the minimum value of a function within the given set of ranges

We have f (x) = cos² x + sin x

Therefore,

f' (x) = 2 cos x (- sin x) + cos x

= - 2 sin x cos x + cos x

Now,

f' (x) = 0

⇒ - 2 sin x cos x + cos x = 0

⇒ cos x = 2 sin x cos x

⇒ cos x (2 sin x - 1) = 0

⇒ sin x = 1/2 or cos x = 0

⇒ x = π / 6 or π / 2

Since x ∈ [0, π]

Now, evaluating the value of f at critical points x = π/6, π/2 and at the end points of the interval [0, π] i.e.,

at x = 0 and x = π,

we have:

f (π / 6) = cos² (π / 6) + sin (π / 6)

= (√3 / 2)² + 1/2

= 5/4

f (0) = cos² (0) + sin (0)

= 1 + 0

= 1

f (π) = cos² (π) + sin (π)

= (- 1)² + 0

= 1

f (π / 2) = cos² (π / 2) + sin (π / 2)

= 0 + 1

= 1

Hence, the absolute maximum value of f is 5/4 occurring at π / 6 and the absolute minimum value of f is 1 occurring at x = 0, π / 2, π

NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 14