UPSC NDA I 2023 Mathematics Solved Paper

Concept:To find the condition for the least value of $m(\theta)$, we use differentiation: set the first derivative to zero and check the second derivative for a minimum.Explanation:Given $m(\theta) = \cot^2\theta + n^2\tan^2\theta + 2n$.Differentiate with respect to $\theta$:$m'(\theta) = -2\cot\theta \csc^2\theta + 2n^2\tan\theta \sec^2\theta$.Set $m'(\theta)=0$ to find critical points:$n^2\tan\theta \sec^2\theta = \cot\theta \csc^2\theta$.Rewrite in terms of sine and cosine:$n^2 \cdot \frac{\sin\theta}{\cos\theta} \cdot \frac{1}{\cos^2\theta} = \frac{\cos\theta}{\sin\theta} \cdot \frac{1}{\sin^2\theta}$.Simplify:$n^2 \cdot \frac{\sin\theta}{\cos^3\theta} = \frac{\cos\theta}{\sin^3\theta}$.Cross-multiply:$n^2 = \frac{\cos^4\theta}{\sin^4\theta} = \cot^4\theta$.Thus $n = \cot^2\theta$ (taking positive square root as $n>0$).Now compute $m''(\theta)$ to confirm minima:$m''(\theta) = 2\bigl[2n^2\sec^2\theta \tan^2\theta + n^2\sec^4\theta + \csc^4\theta + 2\cot^2\theta \csc^2\theta\bigr]$.Substitute $n = \cot^2\theta$:$m''(\theta) = 2\bigl[\csc^4\theta + \cot^2\theta \csc^2\theta\bigr] > 0$, since all terms are positive.Hence $m(\theta)$ has a minimum when $n = \cot^2\theta$.Answer:$n = \cot^{2}\theta$