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If f'(x)=k(cosx-sinx), f'(0)=3 and f(pi/2)=15, find f'(x). - Mathematics and Statistics

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प्रश्न

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

योग

उत्तर

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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2014-2015 (October)

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