Ans: D
$y=k_1x^2+\dfrac{k_2}{x}$, where $k_1$ and $k_2$ are non-zero constants.
$y=k_1x^2+\dfrac{k_2}{x}$, where $k_1$ and $k_2$ are non-zero constants.
For $x=1$ and $y=-4$, we have
$\begin{array}{rcl}
-4 & = & k_1(1)^2 + \dfrac{k_2}{1} \\
-4 & = & k_1+k_2~\ldots \unicode{x2460}
\end{array}$
For $x=2$ and $y=5$, we have
$\begin{array}{rcl}
5 & = & k_1(2)^2 + \dfrac{k_2}{2} \\
5 & = & 4k_1 +\dfrac{k_2}{2} \\
10 & = & 8 k_1 +k_2~\ldots \unicode{x2461}
\end{array}$
$\unicode{x2461} – \unicode{x2460}$, we have
$\begin{array}{rcl}
10- (-4) & = & 7 k_1 \\
k_1 & = & 2
\end{array}$
Sub. $k_1=2$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
-4 & = & 2 + k_2 \\
k_2 & = & -6
\end{array}$
Therefore, $y= 2x^2-\dfrac{6}{x}$.
For $x=-2$, we have
$\begin{array}{rcl}
y & = & 2(-2)^2 – \dfrac{6}{-2} \\
& = & 11
\end{array}$
