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Find the intervals in which f(x) = sin 3x – cos 3x, 0 < x < π, is strictly increasing or strictly decreasing. - Mathematics

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Find the intervals in which f(x) = sin 3x – cos 3x, 0 < x < π, is strictly increasing or strictly decreasing.

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Solution

 

Consider the function f'(x) = sin3x - cos3x.

f'(x)= 3cos3x+3sin3x

=3(sin3x + cos3x)

`=3sqrt2{sin3xcos(pi/4)+cos3xsin(pi/4)}`

`=3sqrt(2){sin(3x+pi/4)}`

For the increasing interval f'(x)>0

`3sqrt2{sin(3x+pi/4)}>0`

`sin(3x+pi/4)>0`

0<3x+`π/4`<π

`=>0<3x<(3pi)/4`

⇒ 0 < x < π/4

Also

`sin(3x+pi/4)>0`

when, `2pi<3x+pi/4<3pi`

=>`(7pi)/4<3x<(11pi)/4`

Therefore, intervals in which function is strictly increasing in  0 < x < π/4 and 7π/12< x <11π/12.

Similarly, for the decreasing interval f'(x)< 0.

`3sqrt2{sin(3x+pi/4)}<0`

`sin(3x+pi/4)<0`

`=>pi<3x+pi/4<2pi`

`=>(3pi)/4<3x<(7pi)/4`

⇒ π/4 < x <7π/12

Also

`sin(3x+pi/4)<0`

When

 `3pi<3x+pi/4<4pi`

`=>(11pi)/4<3x<(15pi)/4`

`=>(11pi)/12

The function is strictly decreasing in `pi/4and  `(11pi)/12

π4<x<7π12 and 11π12<x<π" data-mce-style="position: relative;" data-mce-tabindex="0">π4<x<7π12 and 11π12<x<π

 
Concept: Increasing and Decreasing Functions
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