Show that f(x) = log sin x is increasing on (0, π/2) and decreasing on (π/2, π).

Given:- Function f(x) = log 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) = log sin x

⇒

⇒

⇒ f’(x) = cot(x)

Taking different region from 0 to π

a) let

⇒ cot(x) > 0

⇒ f’(x) > 0

Thus f(x) is increasing in

b) let

⇒ cot(x) < 0

⇒ f’(x) < 0

Thus f(x) is decreasing in

Hence proved

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