Q45 / 90 · KVPY 2015 (SB/SX) · Mathematics
The number of ordered pairs $(x, y)$ of real numbers that satisfy the simultaneous equations.
$$
x+y^{2}=x^{2}+y=12
$$ is
- A0
- B1
- C2
- Correct answerD4
Solution
$$
\begin{array}{l}
\mathrm{x}+\mathrm{y}^{2}=\mathrm{x}^{2}+\mathrm{y}=12 \\
\mathrm{x}+\mathrm{y}^{2}=\mathrm{x}^{2}+\mathrm{y} \\
x-y=x^{2}-y^{2} \Rightarrow x=y, \mathrm{x}+\mathrm{y}=1 \\
\text { when } \mathrm{x}=\mathrm{y} \quad \mathrm{x}^{2}+\mathrm{x}=12 \\
\quad \mathrm{x}^{2}+\mathrm{x}-12=0 \\
\quad \mathrm{x}^{2}+4 \mathrm{x}-3 \mathrm{x}-12=0 \\
\quad(\mathrm{x}+4)(\mathrm{x}-3)=0 \\
\quad \mathrm{x}=-4,3 \\
\quad(3,3)(-4,-4) \\
\text { When } \mathrm{y}=1-\mathrm{x} \\
\quad \mathrm{x}+(1-\mathrm{x})^{2}=12 \\
\mathrm{x}^{2}-\mathrm{x}-11=0 \\
\quad x=\frac{1 \pm \sqrt{1+44}}{2}
\end{array}
$$
So four pair.
Hence, the correct answer is option: 'D'.