UPSC CDS I 2019 Math Paper

Concept:The given infinite nested radical is a periodic expression; let the whole value be $x$ and form an equation by substituting the repeating part.Explanation:Let $x = 2 + \sqrt{2 + \sqrt{2 + \dots}}$. Then the part under the first square root is also the same infinite nested radical but starting with $2 + \dots$, so it equals $x - 2$. Thus, $x - 2 = \sqrt{2 + (x - 2)} = \sqrt{x}$. So we have the equation $x - 2 = \sqrt{x}$. Rearrange: $x - 2 = \sqrt{x}$. Square both sides: $(x - 2)^2 = x$. Expand: $x^2 - 4x + 4 = x$. Bring all terms: $x^2 - 5x + 4 = 0$. Factor: $(x - 4)(x - 1) = 0$. Thus $x = 4$ or $x = 1$. Since $x$ is clearly greater than $2$ (because $2$ plus a positive square root), $x = 1$ is impossible. Hence $x = 4$.Answer:$4$ (Option D)