Let the `f (x) ` be differentiabe function on the interval `(0,oo) ` such that `f (1) =1 and lim _(t to x) ((t ^(2) f (x) -x ^(2) f (t))/(t ^(2) -x ^(2))) =1/2 AAx gt 0,` then `f (x)` is:

Options

A. `(1)/(3x)+(2)/(3)x^(2)`
B. `-(1)/(3x)+(4)/(3)x^(2)`
C. `-(1)/(x)+(2)/(x^(2))`
D. `(1)/(x)`

Answer

A

Explanation

We, have
`underset(trarrx)(lim)(t^(2)f(x)-x^(2)f(t))/(t-x)=1`
`rArr" "underset(trarrx)(lim)(t^(2)f(x)-x^(2)f(x)+x^(2)f(x)-x^(2)f(t))/(t-x)=1`
`rArr" "underset(trarrx)(lim)((t^(2)-x^(2))/(t-x))f(x)-x^(2) underset(trarrx)(lim)(f(t)-f(x))/(t-x)=1`
`rArr" "f(x)underset(trarrx)(lim)(t+x)-x^(2)underset(trarrx)(lim)(f(t)-f(x))/(t-x)=1`
`rArr" "2xf(x)-x^(2)f'(x)=1`
`rArr" "x^(2)f'(x)-2xf(x)=-1`
`rArr" "f'(x)-(2)/(x)f(x)=-(1)/(x^(2))" ...(i)"`
This is a linear differential equation I.F. `=e^(-int(2)/(x)dx)=(1)/(x^(2))`
Multiplying both sides of (i) by I.F. `=(1)/(x^(2))` and integrating with respect to x, we get
`(f(x))/(x^(2))=(1)/(3x^(3))+C" ...(ii)"`
It is given that f(1) = 1. So, putting x = 1 and f(1) = 1 in (ii), we get `C=2//3`.
Putting C= 2/3 in (i), we get
`(f(x))/(x^(2))=(1)/(3x^(3))+(2)/(3)rArrf(x)=(1)/(3x)+(2)/(3)x^(2)`