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Question
If `f'(x)=k(cosx-sinx), f'(0)=3 " and " f(pi/2)=15`, find f(x).
Solution
f'(x) = k(cos x - sin x) ….(given)
`f(x)=intf'(x)dx`
`=intk(cosx-sinx)dx`
`=kint(cosx-sinx)dx`
f(x)=k(sinx+cosx)+c .....(i)
f'(0)=3........................(given)
k(cos0-sin0)=3
k(1)=3
k=3.............................(ii)
`also,f(pi/2)=15`
`k[sin(pi/2)+cos(pi/2)]+c=15`
3(1+0)+c=15
c=12
Putting (ii) and (iii) in (i), we get
f(x)=(3sinx+cosx)+12
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