If `f'(x)=f(x)+int_(0)^(1)f(x)dx`,given `f(0)=1`, then the value of `f(log_(e)2)` is

Options

A. `1/(3+e)`
B. `(5-e)/(3-e)`
C. `(2+e)/(e-2)`
D. none of these

Answer

B

Explanation

Differentiating we get `f''(x)=f'(x)`
or `int(df'(x))/(f'(x))=int dx`
or In `f'(x)=x+c` or `f'(x)=Ae^(x)`……………..1
or `intf'(x)dx=intAe^(x)dx` or `f(x)=Ae^(x)+B`……………2
Now `f(0)=1` or `A+B=1`
`:. f'(x)=f(x)+int_(0)^(1)(Ae^(x)+1-A)dx`
`Ae^(x)=(Ae^(x)+1-A)+|(Ae^(x)+(1-A)x|_(0)^(1)`
or `Ae^(x)=(Ae^(x)+1-A)+(Ae+(1-A)-A)`
or `1=A+(Ae+1-A-A)=0`
or `A(e-3)=-2`
or `A=2/(3-e)` and `B=1-2/(3-e)=(1-e)/(3-e)`
or `f(log_(e)2)-4/(3-e)+(1-e)/(3-e)=(5-e)/(3-e)`