2013-II-34

Ans: D
$\left\{\begin{array}{ll}
x-\log y = 2 & \ldots \unicode{x2460} \\
x^2 – \log y^2 -10 = 2 & \ldots \unicode{x2461}
\end{array}\right.$

From $\unicode{x2460}$, we have

$\begin{array}{rcl}
x-\log y & = & 2 \\
x & = & 2+ \log y \ldots \unicode{x2462}
\end{array}$

Sub. $\unicode{x2462}$ into $\unicode{x2461}$, we have

$\begin{array}{rcl}
(2+\log y)^2 – \log y^2 -10 & = & 2 \\
(\log y)^2 + 4 \log y + 4 -2\log y -12 & = & 0 \\
(\log y)^2 +2\log y -8 & = & 0 \\
(\log y +4)(\log y -2) & = & 0
\end{array}$

Therefore, $\log y = -4$ or $\log y =2$.

Hence, we have $y = 10^{-4} = \dfrac{1}{10~000}$ or $y= 10^2 = 100$.