Book: RD Sharma - Mathematics (Volume 1)
Chapter: 17. Increasing and Decreasing Functions
Subject: Maths - Class 12th
Q. No. 7 of Exercise 17.2
Listen NCERT Audio Books to boost your productivity and retention power by 2X.
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)
