UPSC NDA I 2024 Mathematics Paper

Concept:The quadratic equation $ax^2 + bx + c = 0$, where $a$, $b$, $c$ are in geometric progression with $a>0$ and $c>0$, is solved using the relation $b^2 = ac$.Explanation:Since $a$, $b$, $c$ are in GP, we have $b^2 = ac$.Also, $a>0$ and $c>0$ imply $ac>0$.The discriminant of $ax^2 + bx + c = 0$ is $D = b^2 - 4ac = ac - 4ac = -3ac$.Because $ac>0$, $D = -3ac < 0$.Thus the roots are imaginary. Statement 1 is correct.Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-b \pm \sqrt{-3ac}}{2a}$.Let $b = \sqrt{ac}$ (taking positive value; negative yields same ratio).Then the two roots are $\frac{-\sqrt{ac} + i\sqrt{3ac}}{2a}$ and $\frac{-\sqrt{ac} - i\sqrt{3ac}}{2a}$.Their ratio (first divided by second):$= \frac{-\sqrt{ac} + i\sqrt{3ac}}{-\sqrt{ac} - i\sqrt{3ac}} = \frac{-1 + i\sqrt{3}}{-1 - i\sqrt{3}}$.Recognize that cube roots of unity: $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.Thus the ratio simplifies to $\frac{2\omega}{2\omega^2} = \frac{1}{\omega}$.This is equivalent to the ratio $1 : \omega$ (since $1/\omega = \omega^2$, but the statement "ratio of roots is $1:\omega$" holds because $\omega$ is a cube root of unity). Statement 2 is correct.Product of roots $= \frac{c}{a}$.From $b^2 = ac$, we get $c = \frac{b^2}{a}$.Hence product $= \frac{b^2/a}{a} = \frac{b^2}{a^2}$. Statement 3 is correct.Answer:Statements 1, 2, and 3 are all correct. The correct option is D (1, 2 and 3).