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Question
If `f'(x)=k(cosx-sinx), f'(0)=3 " and " f(pi/2)=15`, find f(x).
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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) = 3(sinx + cosx) + 12
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