UPSC NDA 2 2019 Mathematics Solved Paper

Concept:For any real $A$, $|A| > A$ is true only when $A < 0$.Explanation:Let $A = x^{2} - 3x + 2$.If $A \ge 0$, then $|A| = A$, so $|A| > A$ becomes $A > A$, which is false.If $A < 0$, then $|A| = -A$, and the inequality $|A| > A$ becomes $-A > A$, i.e., $2A < 0$, which is the same as $A < 0$.Thus, the given inequality $|x^{2} - 3x + 2| > x^{2} - 3x + 2$ simplifies directly to $x^{2} - 3x + 2 < 0$.Factor the quadratic: $(x-1)(x-2) < 0$.The product is negative when one factor is positive and the other is negative, which occurs strictly between the roots $1$ and $2$.Therefore, $1 < x < 2$.Answer:Option C: $1 < x < 2$