UPSC NDA I 2023 Mathematics Solved Paper

Concept:The determinant $\Delta(a, b, c, \alpha)$ is defined in a specific form, and setting it to zero yields a quadratic equation in $\alpha$.Explanation:Given $\Delta(a, b, c, \alpha) = \begin{vmatrix} a & b & a\alpha + b \\ b & c & b\alpha + c \\ a\alpha + b & b\alpha + c & 0 \end{vmatrix}$.Apply the column operation $C_3 \to C_3 - (\alpha C_1 + C_2)$ to simplify the determinant.After the operation, we get: $\Delta(a, b, c, \alpha) = \begin{vmatrix} a & b & 0 \\ b & c & 0 \\ a\alpha + b & b\alpha + c & -(a\alpha^2 + 2b\alpha + c) \end{vmatrix}$.Now expand along the third column. The determinant becomes $-(a\alpha^2 + 2b\alpha + c)(b^2 - ac) = 0$.For the given case, $\Delta(7, 4, 2, \alpha) = 0$.Substitute $a = 7$, $b = 4$, $c = 2$ into the expression.First, $a\alpha^2 + 2b\alpha + c = 7\alpha^2 + 2\cdot4\cdot\alpha + 2 = 7\alpha^2 + 8\alpha + 2$.Second, $b^2 - ac = 4^2 - 7\cdot2 = 16 - 14 = 2$, which is nonzero.Therefore, $-(7\alpha^2 + 8\alpha + 2) \cdot 2 = 0$ implies $7\alpha^2 + 8\alpha + 2 = 0$.Hence $\alpha$ is a root of the quadratic equation $7x^2 + 8x + 2 = 0$.Answer:$\alpha$ is a root of $7x^2 + 8x + 2 = 0$, which corresponds to option C.