Show that the function f(x) = cot–1 (sin x + cos x) is decreasing on (0, π/4) and increasing on (π/4, π/2).
Given:- Function f(x) = cot–1 (sin x + cos 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) = cot–1 (sin x + cos x)
⇒
Now, as given
⇒ Cosx – sinx< 0 ; as here cosine values are smaller than sine values for same angle
⇒
⇒ f’(x) < 0
hence, Condition for f(x) to be decreasing
Thus f(x) is decreasing on interval