Show that f(x) = sin x is increasing on (0, π/2) and decreasing on (π/2, π) and neither increasing nor decreasing in (0, π).

Given:- Function f(x) = sin x

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = sin x

⇒

⇒ f’(x) = cosx

Taking different region from 0 to 2π

a) let

⇒ cos(x) > 0

⇒ f’(x) > 0

Thus f(x) is increasing in

b) let

⇒ cos(x) < 0

⇒ f’(x) < 0

Thus f(x) is decreasing in

Therefore, from above condition we find that

⇒ f(x) is increasing in and decreasing in

Hence, condition for f(x) neither increasing nor decreasing in (0,π)

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