UPSC NDA 2 2016 Mathematics Solved Paper

Concept:For a quadratic equation of the form $ax^2 + bx + c = 0$, real roots exist when the discriminant $D = b^2 - 4ac$ is greater than or equal to zero.Explanation:The given quadratic equation is $x^2 - (1 - 2a^2)x + (1 - 2a^2) = 0$.Here, $a = 1$, $b = -(1 - 2a^2)$, and $c = (1 - 2a^2)$.For real roots, $D \ge 0$.So $[-(1 - 2a^2)]^2 - 4 \cdot 1 \cdot (1 - 2a^2) \ge 0$.This simplifies to $(1 - 2a^2)^2 - 4(1 - 2a^2) \ge 0$.Factor $(1 - 2a^2)$: $(1 - 2a^2)[(1 - 2a^2) - 4] \ge 0$.Thus $(1 - 2a^2)(-2a^2 - 3) \ge 0$.Multiply both sides by $-1$ (reversing inequality): $(1 - 2a^2)(2a^2 + 3) \le 0$.Since $2a^2 + 3$ is always positive (as $a^2 \ge 0$), the product is $\le 0$ only when $1 - 2a^2 \le 0$.Hence $1 \le 2a^2$, i.e., $a^2 \ge \frac{1}{2}$.Answer:$a^2 \ge \frac{1}{2}$