UPSC NDA I 2019 Mathematics Solved Paper

Concept:Use trigonometric identities to expand $p^2$, $q^2$, and $4pq$, then add them to match the given form and solve for $A$ and $B$.Explanation:Given $p = X \cos \theta - Y \sin \theta$ and $q = X \sin \theta + Y \cos \theta$.Compute $p^2 = X^2 \cos^2 \theta + Y^2 \sin^2 \theta - 2 XY \cos \theta \sin \theta$.Compute $q^2 = X^2 \sin^2 \theta + Y^2 \cos^2 \theta + 2 XY \cos \theta \sin \theta$.Compute $4pq = 4 \bigl[ X^2 \cos \theta \sin \theta + XY \cos^2 \theta - XY \sin^2 \theta - Y^2 \sin \theta \cos \theta \bigr]$.Add $p^2 + 4pq + q^2$:$p^2 + q^2$ gives $X^2(\cos^2 \theta + \sin^2 \theta) + Y^2(\sin^2 \theta + \cos^2 \theta) + \text{cross terms that cancel} = X^2 + Y^2$.The cross terms from $p^2$ and $q^2$ cancel; adding $4pq$ introduces new cross terms.Simplify using $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ and $2 \sin \theta \cos \theta = \sin 2\theta$:$p^2 + 4pq + q^2 = X^2(1 + 2 \sin 2\theta) + Y^2(1 - 2 \sin 2\theta) + 4 XY \cos 2\theta$.Given that this equals $AX^2 + BY^2$, we equate coefficients:$A = 1 + 2 \sin 2\theta$, $B = 1 - 2 \sin 2\theta$, and $\cos 2\theta = 0$.$\cos 2\theta = 0$ implies $2\theta = \frac{\pi}{2}$, so $\theta = \frac{\pi}{4}$. Then $\sin 2\theta = \sin \frac{\pi}{2} = 1$.Thus $A = 1 + 2(1) = 3$.Answer:$A = 3$ (Option B).