UPSC CDS 2 2021 Math Paper

Concept:For positive $a$ and $b$, minimum of $a\cos^3\theta + b\sec^3\theta$ in $0^{\circ} \le \theta < 90^{\circ}$ can be found by derivative or by AM–GM inequality.Explanation:Let $y = \cos^3\theta + \sec^3\theta$.Since $\theta$ is acute, both $\cos\theta$ and $\sec\theta$ are positive.$\sec\theta = \frac{1}{\cos\theta}$, so $y = \cos^3\theta + \frac{1}{\cos^3\theta}$.By AM–GM inequality: $\cos^3\theta + \frac{1}{\cos^3\theta} \ge 2\sqrt{\cos^3\theta \cdot \frac{1}{\cos^3\theta}} = 2$.Equality holds when $\cos^3\theta = \frac{1}{\cos^3\theta}$, i.e., $\cos^6\theta = 1 \Rightarrow \cos\theta = 1$.Thus $\theta = 0^{\circ}$ (within given range).So minimum value is $1^3 + 1^3 = 2$.Alternatively, using calculus:Differentiate: $\frac{dy}{d\theta} = -3\cos^2\theta \sin\theta + 3\sec^2\theta \cdot \sec\theta \tan\theta$.Simplify to $\frac{dy}{d\theta} = 3\tan\theta (-\cos^3\theta + \sec^3\theta)$.Set derivative to zero: $\tan\theta = 0$ or $\cos^3\theta = \sec^3\theta$.For $\theta \in [0,90^{\circ})$, $\tan\theta = 0$ gives $\theta = 0$; $\cos^3\theta = \sec^3\theta$ also gives $\cos\theta = 1$, so $\theta = 0$.Check endpoint: at $\theta = 0$, $y = 2$. For $\theta > 0$, $y > 2$ (e.g., $\theta = 60^{\circ}$ gives $0.125 + 8 = 8.125$).Hence minimum is $2$.Answer:The minimum value is $2$, which corresponds to option C.