UPSC CDS I 2019 Math Paper

Concept:For positive numbers, Arithmetic Mean ≥ Geometric Mean (AM ≥ GM). Use this to find the minimum of $x + \frac{1}{x}$ and add the constant.Explanation:Given expression: $x + \frac{x+2}{2x}$.Simplify: $x + \frac{x}{2x} + \frac{2}{2x} = x + \frac{1}{2} + \frac{1}{x}$.So we need the minimum of $x + \frac{1}{x} + \frac{1}{2}$ for $x > 0$.Consider $x$ and $\frac{1}{x}$ (both positive since $x > 0$).By AM ≥ GM: $\frac{x + \frac{1}{x}}{2} \ge \sqrt{x \cdot \frac{1}{x}} = \sqrt{1} = 1$.Hence $x + \frac{1}{x} \ge 2$.Equality when $x = \frac{1}{x}$, i.e., $x = 1$ (since $x > 0$).Thus the minimum value of $x + \frac{1}{x}$ is $2$.Now add the constant $\frac{1}{2}$: minimum of whole expression is $2 + \frac{1}{2} = 2\frac{1}{2}$.Answer:The minimum value is $2\frac{1}{2}$, which corresponds to option C.