Ans: D
Let $y = k_1 + \dfrac{k_2}{x}$, where $k_1$ and $k_2$ are constants. For $x=1$ and $y=-1$, we have
Let $y = k_1 + \dfrac{k_2}{x}$, where $k_1$ and $k_2$ are constants. For $x=1$ and $y=-1$, we have
$\begin{array}{rcl}
k_1 + \dfrac{k_2}{1} & = & -1\\
k_1 + k_2 & = & -1~\ldots \unicode{x2460}
\end{array}$
For $x=2$ and $y=1$, we have
$\begin{array}{rcl}
k_1 + \dfrac{k_2}{2} & = & 1 \\
k_1 + \dfrac{k_2}{2} & = & 1~\ldots \unicode{x2461}
\end{array}$
$\unicode{x2461} – \unicode{x2460}$, we have
$\begin{array}{rcl}
\dfrac{-k_2}{2} & = & 2 \\
k_2 & = & -4
\end{array}$
Sub. $k_2=-4$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
k_1 + (-4) & = & -1 \\
k_1 & = & 3
\end{array}$
Therefore, $y=3 – \dfrac{4}{x} $. When $y=2$, we have
$\begin{array}{rcl}
2 & = & 3 – \dfrac{4}{x} \\
\dfrac{4}{x} & = & 1 \\
x & = & 4
\end{array}$
