UPSC NDA 2 2019 Mathematics Solved Paper

Concept:An infinite continued fraction can be expressed as a quadratic equation by recognizing the repeating pattern.Explanation:Let the given expression be equal to $y$. $y = 2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \cdots}}}$Since the pattern repeats indefinitely, the entire denominator after the first $2$ is also $y$. Thus, $y = 2 + \frac{1}{y}$. Multiply both sides by $y$: $y^2 = 2y + 1$. Rearranging: $y^2 - 2y - 1 = 0$. Solve the quadratic using the formula: $y = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$. The two solutions are $1 + \sqrt{2}$ and $1 - \sqrt{2}$. Now note that the original expression starts with $2$ and adds positive fractions, so $y$ must be greater than $2$. $1 - \sqrt{2} \approx -0.414$ is negative, so it is rejected. $1 + \sqrt{2} \approx 2.414$ is greater than $2$ and is valid.Hence the value is $1 + \sqrt{2}$.Answer:$1 + \sqrt{2}$, which corresponds to option B.