UPSC CDS II 2024 Math Paper

Concept:The diagonals of concentric squares increase uniformly based on the distance between corresponding corners.Explanation:The innermost square has area $1$, so its side $a_1 = 1$ and diagonal $d_1 = a_1\sqrt{2} = \sqrt{2}$.The distance between corresponding corners of consecutive squares is $1$ unit. Since corners lie on the same radial line from the center, the distance from the center to a corner increases by $1$ each time. The diagonal is twice that distance, so the diagonal increases by $2$ each step.Thus, $d_n = d_1 + 2(n-1) = \sqrt{2} + 2n - 2 = 2n + \sqrt{2} - 2$. Therefore, statement I is correct.For statement II, the area of a square is $\frac{d_n^2}{2}$. The area between the $n$th and $(n-1)$th square is:$\frac{d_n^2}{2} - \frac{d_{n-1}^2}{2} = \frac{(d_n - d_{n-1})(d_n + d_{n-1})}{2}$.$d_n - d_{n-1} = 2$, and $d_n + d_{n-1} = (2n+\sqrt{2}-2) + (2n+\sqrt{2}-4) = 4n + 2\sqrt{2} - 6$.So the difference $= \frac{2(4n+2\sqrt{2}-6)}{2} = 4n + 2\sqrt{2} - 6$, which depends on $n$. Hence, statement II is false.Answer:Only statement I is correct.