UPSC NDA I 2021 Mathematics Solved Paper

Concept:The function $f(x)$ is defined as a $3 \times 3$ determinant. For the given values of $x$, the determinant evaluates to zero because either a whole row or a whole column consists entirely of zeros.Explanation:First, evaluate $f(-1)$:$f(-1) = \begin{vmatrix} 1 & -1 & 0 \\ -2 & 2 & 0 \\ 6 & 12 & 0 \end{vmatrix}$.The third column has all entries equal to $0$, so the determinant is $0$.Thus, $f(-1) = 0$.Next, evaluate $f(0)$:$f(0) = \begin{vmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 4 & 0 \end{vmatrix}$.The second row is entirely zero, so the determinant is $0$.Thus, $f(0) = 0$.Finally, evaluate $f(1)$:$f(1) = \begin{vmatrix} 1 & 1 & 2 \\ 2 & 0 & 2 \\ 0 & 0 & 0 \end{vmatrix}$.The third row is entirely zero, so the determinant is $0$.Thus, $f(1) = 0$.Now add them:$f(-1) + f(0) + f(1) = 0 + 0 + 0 = 0$.Answer:$0$