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