UPSC NDA 2 2021 Mathematics Solved Paper

Concept:The diagonal of a rectangle inscribed in a circle equals the circle's diameter.For a fixed diagonal, the rectangle with maximum area is a square.Explanation:The circle has radius $2$ units, so its diameter is $4$ units.Let the rectangle have length $L$ and breadth $B$. The diagonal is $\sqrt{L^2 + B^2}$.Since the diagonal equals the diameter: $\sqrt{L^2 + B^2} = 4$.Squaring gives $L^2 + B^2 = 16$.To maximise area $A = L \times B$ under this constraint, use symmetry or calculus.By symmetry, maximum area occurs when $L = B$ (the rectangle becomes a square).Substituting $L = B$ into $L^2 + B^2 = 16$ gives $2L^2 = 16$, so $L^2 = 8$.Hence $L = B = \sqrt{8}$.Thus maximum area $= L \times B = \sqrt{8} \times \sqrt{8} = 8$ square units.Alternatively, using calculus: differentiate area $A = LB$ with respect to $L$, use constraint to find $L = B$, same result.Answer:$8$ square units (option C).