UPSC NDA I 2023 Mathematics Solved Paper

Concept:A determinant factoring to $(b^2 - ac)(a\alpha^2 + 2b\alpha + c) = 0$ must vanish for all $\alpha > 0$.Explanation:Start with the given determinant $\Delta(a, b, c, \alpha) = \begin{vmatrix} a & b & a\alpha+b \\ b & c & b\alpha+c \\ a\alpha+b & b\alpha+c & 0 \end{vmatrix}$.Perform the column operation $C_3 \to C_3 - (\alpha C_1 + C_2)$.The determinant becomes $\begin{vmatrix} a & b & 0 \\ b & c & 0 \\ a\alpha+b & b\alpha+c & -(a\alpha^2+2b\alpha+c) \end{vmatrix}$.Expanding along the third column gives $\Delta = -(a\alpha^2+2b\alpha+c)(b^2 - ac)$.For $\Delta = 0$ to hold for every $\alpha > 0$, the factor $(a\alpha^2+2b\alpha+c)$ cannot be zero for all positive $\alpha$ (it is a quadratic).Thus the other factor must be zero: $b^2 - ac = 0$, which means $a$, $b$, $c$ are in geometric progression (GP).Answer:Option B: $a$, $b$, $c$ are in GP.