UPSC NDA I 2018 Mathematics Solved Paper

Concept:Maximising area of trapezium using differentiation. The fourth side is the longer base, expressed as $6 + 2x$.Explanation:Let the non‑parallel sides be $6$ cm each. Drop perpendiculars to the longer base. The shorter base is $6$ cm.Let $x$ be the horizontal offset from each vertex to the point where perpendicular meets the base.Height $h = \sqrt{6^2 - x^2} = \sqrt{36 - x^2}$ by Pythagoras.Area $A = \frac{1}{2}(\text{sum of parallel sides})\times h = \frac{1}{2}\bigl(6 + (6+2x)\bigr)\sqrt{36-x^2} = (6+x)\sqrt{36-x^2}$.Differentiate $A$ with respect to $x$: $\frac{dA}{dx} = (6+x)\cdot\frac{-2x}{2\sqrt{36-x^2}} + \sqrt{36-x^2}$.Set $\frac{dA}{dx}=0$: $\frac{-(6+x)x}{\sqrt{36-x^2}} + \sqrt{36-x^2}=0 \Rightarrow -(6+x)x + (36-x^2) = 0 \Rightarrow x = 3$ (positive).Length of fourth side $= 6 + 2x = 6 + 2(3) = 12$ cm.Answer:$12$ cm (Option D).