UPSC CDS 2 2020 Math Paper

Concept:Use the identity $\sin^2 \theta + \cos^2 \theta = 1$ to express the given expression in one variable and find its minimum.Explanation:Given expression: $9 \sin^2 \theta + 16 \cos^2 \theta$.Replace $\cos^2 \theta$ with $1 - \sin^2 \theta$:$9 \sin^2 \theta + 16(1 - \sin^2 \theta) = 9 \sin^2 \theta + 16 - 16 \sin^2 \theta = 16 - 7 \sin^2 \theta$.Since $\sin^2 \theta$ lies between 0 and 1, the term $16 - 7 \sin^2 \theta$ is smallest when $\sin^2 \theta$ is largest, i.e., $\sin^2 \theta = 1$.Thus minimum value $= 16 - 7(1) = 9$.Alternatively, rewrite as $9 \sin^2 \theta + 16 \cos^2 \theta = 9(1 - \cos^2 \theta) + 16 \cos^2 \theta = 9 + 7 \cos^2 \theta$. Minimum occurs when $\cos^2 \theta = 0$, also giving $9$.Shortcut: For $a \sin^2 \theta + b \cos^2 \theta$ with $a, b > 0$, the minimum is the smaller of $a$ and $b$. Here $9 < 16$, so minimum is $9$.Answer:$9$ (Option B)