UPSC NDA I 2023 Mathematics Solved Paper

Concept:Use differentiation to find the minimum value of a function.Set the first derivative to zero to locate critical points, then check the second derivative for minima.Explanation:Given $m(\theta) = \cot^2 \theta + n^2 \tan^2 \theta + 2n$.Differentiate with respect to $\theta$: $m'(\theta) = -2\cot\theta\csc^2\theta + 2n^2\tan\theta\sec^2\theta = 0$.Simplify: $n^2 \tan\theta \sec^2\theta = \cot\theta \csc^2\theta$.Using identities, this reduces to $n^2 = \cot^4 \theta$, so $n = \cot^2 \theta$.Second derivative test: $m''(\theta) = 2[2n^2\sec^2\theta\tan^2\theta + n^2\sec^4\theta + \csc^4\theta + 2\cot^2\theta\csc^2\theta] > 0$, confirming a minimum at $n = \cot^2 \theta$.Substitute $n = \cot^2 \theta$ into $m(\theta)$:$m(\theta) = \cot^2 \theta + (\cot^2 \theta)^2 \tan^2 \theta + 2\cot^2 \theta$$= \cot^2 \theta + \cot^4 \theta \tan^2 \theta + 2\cot^2 \theta$Since $\tan^2 \theta = \frac{1}{\cot^2 \theta}$, we get $\cot^4 \theta \cdot \frac{1}{\cot^2 \theta} = \cot^2 \theta$.Thus $m(\theta) = \cot^2 \theta + \cot^2 \theta + 2\cot^2 \theta = 4\cot^2 \theta = 4n$.Hence the least value is $4n$.Answer:$4n$ (Option D).