Solution We have
f(x) = sin x + cos x,
or
f ′(x) = cos x – sin x
Now 
( )
0
f
′x =
 gives sin x = cos x which gives  that 
4
x
=π
, 5
4
π  as 0
2
≤x
≤ π
The points 
4
x
=π
 and 
5
4
x
=π
 divide the interval [0, 2π] into three disjoint intervals,
namely, 0, 4
π



 , π
π
4
5
,4


  and 5 ,2
π4
π



 Note that
5
( )
0 if
0,
,2
4
4
f
x
x
π
π




′
>
∈
∪
π








or
f  is increasing in the intervals 0, 4
π



  and 
π45 ,2
π




Also
′
<
∈


f
x
x
( )
,
0
4
5
4
if
π
π
or
 f  is decreasing in 
π
π
4
5
,4



Fig 6 6
Rationalised 2023-24
 MATHEMATICS
158
Interval
Sign of 
f( )
′x
Nature of function
0, 4
π




> 0
f  is increasing
π
π
4
5
,4


< 0
f  is decreasing
π45 ,2
π




> 0
f  is increasing
EXERCISE 6 2
1