if f be a differentiable function such that `f(x) =x^(2)int_(0)^(x)e^(-t)f(x-t).` dt. Then f(x) =

Options

A. 0
B. `(x^(3))/(3)+x^(2)`
C. not possible
D. `5x^(2)`

Answer

B

Explanation

`f(x) =x^(2)+underset(0)overset(x)inte^(-t)f(x-t)dt`
`=" " x^(2)+underset(0)overset(x)inte^((x-t))f(x-(x-t))dt`
`rArrf(x)=x^(2)+e^(-x)underset(0)overset(x)inte^(t)f(t)dt`
`rArr f'(x)=2x+e^(-x)[e^(x)f(x)]-e^(-x)underset(0)overset(x)inte^(t)f(t)dt`
`rArrf(x)+f'(x)=x^(2)+2x+f(x)`
`rArrf'(x)=x^(2)+2x` Now integrate
`rArrf(x)=(x^(3))/(3)+x^(2)+c`
But `f(0)=0rArrc=0`
`:. f(x)=(x^(3))/(3)+x^(2)`
For Lagrange's mean value theorem f(x) must be continous in [0,2] and differentiable in (0,2)
Hence, a=3
`m+b=2" "....(i)`
f'(x) `[{:(-2x,0ltxle1),(m,1ltxle2):}`
`rArr m=-2` and hence b=4