If you use the linear approximation to $y=ln(x)$ at $x=1$ to approximate the value of $ln(1\mathrm{.}06),$ then $ln(1\mathrm{.}06)$ is approximately equal to

$($a$)1\mathrm{.}01.$

$($b$)0\mathrm{.}03.$

$($c$)0\mathrm{.}06$

$($d$)0\mathrm{.}12.$

$($e$)1\mathrm{.}06.$

correct anwser