Ans: D
I may not be true. Suppose $a=3$ and $b=-6$. It is obvious that $a > b$ but $a^2 < b^2$.
I may not be true. Suppose $a=3$ and $b=-6$. It is obvious that $a > b$ but $a^2 < b^2$.
II must be true. This is the addition law of inequality.
III must be true. For $k<0$, then $k^2>0$. Therefore we have
$\begin{array}{rcl}
a & > & b \\
\dfrac{a}{k^2} & > & \dfrac{b}{k^2}
\end{array}$
