Let f(x) be a quadratic function such that `f(0)=1` and `int(f(x))/(x^2(x+1)^3)dx` is a rational function, then the value of `f'(0)` is

Options

A. 0
B. 2
C. 3
D. 5

Answer

c

Explanation

Let g(x) `=int(f(x))/(x^(2)(x+1)^(3))dx` . it given that f(x) is a quadratic function . so , let
`(f(x))/(x^(2)(x+1)^(3))=(A)/(x)+(B)/(x^(2))+(C)/(x+1)+(D)/((x+1)^(2))+(E)/((x+1)^(3))`
`thereforeg(x)=int(f(x))/(x^(2)(x+1)^(3))dx`
`int={(A)/(x)+(B)/(x^(2))+(C)/(x+1)+(D)/((x+1)^(2))+(E)/((x+1)^(3))}` dx is . . . (i)
`rArrg(x)=AIn x-(B)/(x)+CIn(x+1)-(D)/((x+1))-(E)/(2(x+1)^(2))+K`
It is given that g (x) is a rational function.
`thereforeA=C=0`
Putting A= C =0 in (i) , we get
`g(x)=int{(B)/(x^(2))+(D)/((x+1)^(2))+(E)/((x+1)^(3))}` dx
`rArrg(x) =int{(B(x+1)^(3)+Dx^(2)(x+1)+Ex^(2))/(x^(2)(x+1)^(3))}` dx
`thereforef(x) =B(x+1)^(3)+Dx^(2)(x+1)+Ex^(2)[becauseg(x)=int(f(x))/(x^(2)(x+1)^(3))]`
`rArrf(x)=(B+D)x^(3)+(3B+D+E)x^(2)+3Bx+B`
But , f(x) is a quadratic function such that f(0) =1 . therefore , `B+D=0andB=1rArrB=1,D=-1`
`thereforef(x)=(2+E)x^(2)+3x+1`
`rArr f(x) =2(2+E)x+3`
`rArrf'(0)=3`