UPSC NDA I 2019 Mathematics Solved Paper

Concept:Use the AM-GM inequality to find the minimum value of a sum when the product of the terms is constant.Explanation:We have two terms $a^2x$ and $b^2y$.Apply the AM-GM inequality for two non-negative numbers:$\frac{a^2x + b^2y}{2} \ge \sqrt{a^2x \cdot b^2y}$.This simplifies to $a^2x + b^2y \ge 2 \sqrt{a^2 b^2 xy}$.Given $xy = c^2$, substitute: $\sqrt{a^2 b^2 \cdot c^2} = abc$.Thus, $a^2x + b^2y \ge 2abc$.Equality holds when $a^2x = b^2y$, which is possible for certain $x, y$ satisfying $xy = c^2$.Therefore, the minimum value of $a^2x + b^2y$ is $2abc$.Answer:2abc