`"Let "f:(0.oo)rarrR" be a differentiable function such that "f'(x)=2-(f(x))/(x)" for all "x in (0,oo) and f(1) ne 1.` Then

Options

A. `underset(xrarr0^(+))(lim)f'((1)/(x))=a`
B. `underset(xrarr0^(+))(lim)xf((1)/(x))=2`
C. `underset(xrarr0^(+))(lim)x^(2)f'(x)=0`
D. `|f(x)|le2` for all `x in (0,2)`

Answer

A

Explanation

The function f(x) satisfies the differential equation
`(df(x))/(dx)+(1)/(x)f(x)=2" …(i)"`
This is a linear differential equation with I.F. = `e^(int(1)/(2)dx)=e^(logx)=x`
Multiplying (i) by I.F.= x and integrating, we obtain
`df(x)=x^(2)+C`
`rArr" "f(x)=x+(C)/(x)`
`rArr" "f'(x)=1-(C)/(x^(2))`
`rArr" "f'((1)/(x))=1-Cx^(2)`
It is given that `f(1) ne 1`. So, `C ne 0`.
`therefore" "underset(xrarr0^(+))(lim)((1)/(x))=underset(xrarr0^(+))(lim)(1-Cx^(2))=1,`
`underset(xrarr0^(+))(lim)x^(2)f'(x)=underset(xrarr0^(+))(lim)(x^(2)-C)=-C,`
`and, underset(xrarr0^(+))(lim)xf((1)/(x))=underset(xrarr0^(+))(lim)x((1)/(x)+Cx)=1`
Thus, option (a) is correct.
We observe that for `C=1, f(x)=x+(1)/(x) ge 2`.
So, `|f(x)|le2` is not true . This can also be observed from the fact that `underset(xrarr0^(+))(lim)f(x)rarroo" for "C gt0`.