UPSC NDA I 2025 Mathematics Paper

Concept:For any non‑zero real number $k$, the sum $\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right)$ equals $\frac{\pi}{2}$. This follows from the identity $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ when $xy < 1$, and from the complementary angle property when $k>0$.Explanation:Given that $k$ is a root of $x^{2}-4x+1=0$, solve the equation:$x = \frac{4 \pm \sqrt{16-4}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$.So $k = 2+\sqrt{3}$ or $k = 2-\sqrt{3}$. Both are positive.We need $\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right)$.Using the sum formula: let $a = k$, $b = \frac{1}{k}$. Then$\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) = \tan^{-1}\left(\frac{k + \frac{1}{k}}{1 - k \cdot \frac{1}{k}}\right) = \tan^{-1}\left(\frac{k + \frac{1}{k}}{1-1}\right) = \tan^{-1}\left(\frac{k + \frac{1}{k}}{0}\right)$.The denominator is zero, so the expression is undefined in the usual arctan range. However, for any positive $k$, the sum $\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right)$ is known to be $\frac{\pi}{2}$. This can also be verified by noting that $\tan^{-1}(k) + \cot^{-1}(k) = \frac{\pi}{2}$ and $\cot^{-1}(k) = \tan^{-1}(1/k)$ for $k>0$.Answer:$\frac{\pi}{2}$ (Option D)