Oxford University MAT 2018 Solved Papers

If $\left(\mathit{x},\mathit{y}\right)$ is a solution to this equation, then so is $\left(\mathit{y},\mathit{x}\right)$. So the graph must have reflectional symmetry in the line $\mathit{x}=\mathit{y}$. So A cannot be a sketch of this curve. The answer is (c).
As a check, note that $\mathit{C}$ could be a sketch with $\mathit{p}\left(\mathit{x}\right)={\mathit{x}}^2-1∕2$ and $\mathit{D}$ could be a sketch with $\mathit{p}\left(\mathit{x}\right)=\mathit{x}$. For $\mathit{B}$, look at the part of the sketch with $\mathit{y}$ constant. Along this line segment, $\mathit{p}\left(\mathit{x}\right)+\mathit{p}\left(\mathit{y}\right)=0$ and $\mathit{p}\left(\mathit{y}\right)$ is a constant. But if this polynomial is zero in some interval, it must be zero for all $\mathit{x}$, so $\mathit{p}\left(\mathit{x}\right)$ must be constant! But then either all points should be in the sketch (if the constant is zero) or none should (otherwise). This sketch is neither, and therefore impossible.