A continuous function `f: R rarrR` satisfies the equation `f(x)=x+int_(0)^(1) f(t)dt.` Which of the following options is true ?

Options

A. `f(x+y)=f(x)+f(y)`
B. `f(x+y)=f(x)f(y)`
C. `f(x+y)=f(x)+f(y)+f(x)f(y)`
D. `f(x+y)=f(xy)`

Answer

C

Explanation

`f(x)=x+underset(0)overset(1)intf(t)dt`
`f'(x)=1+f(x)rArrf'(x)-f(x)=1`
`rArre^(-x)f'(x)-f(x)e^(-x)=e^(-x)`
`rArr(d)/(dx)(f(x)e^(-x))=e^(-x)`
`rArrf(x)e^(-x)=(e^(-x))/(-1)+e`
`rArrf(x)=-1+ce^(x)`
`f(0)=0-1+ce^(0)rArrc=1`
`f(x)=e^(x)-1`
`f(x)+f(y)+f(x)f(y)=e^(x)-1+e^(y)-1+(e^(x)-1)(e^(y)-1)`
`e^(x)-1+e^(y)-1+e^(x).e^(y)-e^(y)-e^(x)+1`
`=f(x+y)`