UPSC NDA 2 2016 Mathematics Solved Paper

Concept:For a quadratic equation $x^2 - (1 - 2a^2)x + (1 - 2a^2) = 0$, the sum of roots $\alpha + \beta = (1 - 2a^2)$ and product $\alpha\beta = (1 - 2a^2)$.Explanation:The given condition is $\frac{1}{\alpha^2} + \frac{1}{\beta^2} < 1$.Rewrite as $\frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} < 1$.Use identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.Substitute the values: $\frac{(1 - 2a^2)^2 - 2(1 - 2a^2)}{(1 - 2a^2)^2} < 1$.Factor numerator: $\frac{(1 - 2a^2)[(1 - 2a^2) - 2]}{(1 - 2a^2)^2} < 1$.Cancel one factor of $(1 - 2a^2)$ (provided $1 - 2a^2 \neq 0$): $\frac{(1 - 2a^2) - 2}{1 - 2a^2} < 1$.Simplify numerator: $\frac{-2a^2 - 1}{1 - 2a^2} < 1$.Subtract 1 from both sides: $\frac{-2a^2 - 1}{1 - 2a^2} - 1 < 0$.Combine into single fraction: $\frac{-2a^2 - 1 - (1 - 2a^2)}{1 - 2a^2} < 0$.Simplify numerator: $\frac{-2a^2 - 1 - 1 + 2a^2}{1 - 2a^2} = \frac{-2}{1 - 2a^2} < 0$.Multiply both sides by -1 (reverse inequality): $\frac{2}{1 - 2a^2} > 0$.Thus $1 - 2a^2 > 0$, since numerator 2 is positive.Therefore $1 > 2a^2$, i.e., $a^2 < \frac{1}{2}$.Answer:Option A: $a^2 < \frac{1}{2}$