Oxford University MAT 2020 Solved Papers

For and $\mathit{B}=4-\mathit{\pi }{\mathit{r}}^2$. For $\mathit{r}>\sqrt{2},\mathit{A}=\mathit{\pi }{\mathit{r}}^2-4$ and $\mathit{B}=0$, which eliminates options (a) and (c). Now options (b) and (e) both have $\mathit{A}+\mathit{B}=0$ at some value of $\mathit{r}$, which would require $\mathit{A}=0$ and $\mathit{B}=0$, that is; no area inside the square that's not also inside the circle and vice versa. This is impossible, and the only remaining option is the $\left(\mathit{d}\right)$.