UPSC CDS I 2019 Math Paper

Concept:Use componendo-dividendo to simplify each fraction, then add and simplify using algebraic manipulation.Explanation:Given $x = \frac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}}$.First compute $\frac{x}{2\sqrt{2}} = \frac{4\sqrt{6}}{2\sqrt{2}(\sqrt{2}+\sqrt{3})} = \frac{2\sqrt{3}}{\sqrt{2}+\sqrt{3}}$.Similarly, $\frac{x}{2\sqrt{3}} = \frac{4\sqrt{6}}{2\sqrt{3}(\sqrt{2}+\sqrt{3})} = \frac{2\sqrt{2}}{\sqrt{2}+\sqrt{3}}$.For the first term $\frac{x+2\sqrt{2}}{x-2\sqrt{2}}$: apply componendo-dividendo using $\frac{x}{2\sqrt{2}} = \frac{2\sqrt{3}}{\sqrt{2}+\sqrt{3}}$.$\frac{x+2\sqrt{2}}{x-2\sqrt{2}} = \frac{\frac{x}{2\sqrt{2}} + 1}{\frac{x}{2\sqrt{2}} - 1} = \frac{\frac{2\sqrt{3}}{\sqrt{2}+\sqrt{3}} + 1}{\frac{2\sqrt{3}}{\sqrt{2}+\sqrt{3}} - 1}$.Simplify numerator: $\frac{2\sqrt{3} + \sqrt{2} + \sqrt{3}}{\sqrt{2}+\sqrt{3}} = \frac{3\sqrt{3}+\sqrt{2}}{\sqrt{2}+\sqrt{3}}$.Denominator: $\frac{2\sqrt{3} - \sqrt{2} - \sqrt{3}}{\sqrt{2}+\sqrt{3}} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{2}+\sqrt{3}}$.Thus $\frac{x+2\sqrt{2}}{x-2\sqrt{2}} = \frac{3\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$. (Equation 1)For the second term $\frac{x+2\sqrt{3}}{x-2\sqrt{3}}$: similarly, using $\frac{x}{2\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{2}+\sqrt{3}}$,$\frac{x+2\sqrt{3}}{x-2\sqrt{3}} = \frac{\frac{2\sqrt{2}}{\sqrt{2}+\sqrt{3}} + 1}{\frac{2\sqrt{2}}{\sqrt{2}+\sqrt{3}} - 1} = \frac{2\sqrt{2} + \sqrt{2}+\sqrt{3}}{2\sqrt{2} - \sqrt{2} - \sqrt{3}} = \frac{3\sqrt{2}+\sqrt{3}}{\sqrt{2} - \sqrt{3}} = -\frac{3\sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}}$. (Equation 2)Add Equation 1 and Equation 2:$\frac{x+2\sqrt{2}}{x-2\sqrt{2}} + \frac{x+2\sqrt{3}}{x-2\sqrt{3}} = \frac{3\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} - \frac{3\sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}} = \frac{(3\sqrt{3}+\sqrt{2}) - (3\sqrt{2}+\sqrt{3})}{\sqrt{3}-\sqrt{2}} = \frac{2\sqrt{3} - 2\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{2(\sqrt{3}-\sqrt{2})}{\sqrt{3}-\sqrt{2}} = 2$.Answer:The value is $2$, which corresponds to option D.