Let `f:R to R` be a differentiable function such that `f(x)=x^(2)+int_(0)^(x)e^(-t)f(x-t)dt`. `f(x)` increases for

Options

A. `(-2)/(3)`
B. `(2)/(3)`
C. `(1)/(3)`
D. `-(1)/(3)`

Answer

B

Explanation

We have,
`f(x)=x^(2)+underset(0)overset(x)int e^(-t)(x-t)dt" "`……..(i)
`rArr f(x)=x^(2)+underset(0)overset(x)int e^(-(x-t))f(x-(x-t))dt`
`rArr f(x)=x^(2)+underset(0)overset(x)int e^(-x)e^(t)f(t) dt`
`rArr f(x)=x^(2)+e^(-x)underset(0)overset(x)int e^(t)f(l)dt`
`rArr f'(x)=2x+e^(-x){e^(x)f(x)}-e^(-x)underset(0)overset(x)e^(t)f(t) dt`
`f'(x)=2x+f(x)-underset(0)overset(x)int e^((-x-t))f(t) dt`
`f'(x)=2x+f(x)-underset(0)overset(x)int e^((-x-(x-t)))f(x-t)dt`
`rArr f'(x)=2x+f(x)-underset(0)overset(x)int e^(-t)f(x-t)dt`
`rArr f'(x)=2x+f(x)+x^(2)-f(x)" "`[Using (i)]
`f'(x)=2x+x^(2)`
`rArr f(x)=x^(2)+(x^(3))/(3)+C" "`......(ii)
From (i), we obtain f(0)=0.
Putting x=0, in (ii), we obtain C=0
`:. f(x)=x^(2)+(x^(3))/(3)`.
`rArr f(-1)=1-(1)/(3)=(2)/(3)`