UPSC NDA I 2023 Mathematics Solved Paper

Concept:Use the Law of Cosines in triangle PQR to relate side lengths and the cosine of angle P.Explanation:Given $\cos P = \frac{1}{3}$ and side lengths $PQ = 2n+2$, $PR = 2n+4$, $QR = 2n+6$.Apply the Law of Cosines: $\cos P = \frac{PQ^2 + PR^2 - QR^2}{2 \cdot PQ \cdot PR}$.Substitute the given sides:$\frac{1}{3} = \frac{(2n+2)^2 + (2n+4)^2 - (2n+6)^2}{2(2n+2)(2n+4)}$.Expand and simplify the numerator:$(2n+2)^2 = 4n^2+8n+4$, $(2n+4)^2 = 4n^2+16n+16$, $(2n+6)^2 = 4n^2+24n+36$.So numerator becomes $(4n^2+8n+4)+(4n^2+16n+16)-(4n^2+24n+36) = 4n^2 -16$.Denominator: $2(2n+2)(2n+4) = 2(4n^2+12n+8) = 8n^2+24n+16$.Thus $\frac{1}{3} = \frac{4n^2-16}{8n^2+24n+16}$.Cross-multiply: $8n^2+24n+16 = 12n^2-48$.Bring all terms to one side: $0 = 4n^2 -24n -64$.Divide by 4: $n^2 -6n -16 = 0$.Factor: $(n-8)(n+2) = 0$.Since side lengths must be positive, $n = 8$ (the negative root is discarded).Answer:$n = 8$