The body must start from $x=1$.Then, the work done against friction must be equal to the initial kinetic energy given to the body in the form of initial velocity of the body

$\large\frac{1}{2}$$mv^2=\int \limits_1^{\infty} \mu mg \;dx$

$\large\frac{v^2}{2}$$=Ag \int \limits_1^{\infty} \large\frac{1}{x^2}$$dx$

$\large\frac{v^2}{2}$$=Ag \bigg[ \large\frac{-1}{x}\bigg] _1 ^{\infty}$

$v^2=2 Ag$

$v=\sqrt {2gA}$

Hence a is the correct answer.