JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

\text{Let } D_k = \begin{vmatrix} 1 & 2k & 2k-1 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{vmatrix}. \text{ If } \sum_{k=1}^{n} D_k = 96, \text{ then } n \text{ is equal to \_\_\_\_\_.}

Choose the correct answer:

Explanation

1. Determinant Solution

D_k = \begin{vmatrix} 1 & 2k & 2k-1 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{vmatrix}

\sum_{k=1}^{n} D_k = \begin{vmatrix} \sum 1 & \sum 2k & \sum (2k-1) \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{vmatrix}

\Rightarrow \begin{vmatrix} n & n(n+1) & n^2 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{vmatrix} = 96

Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ :

\begin{vmatrix} n & n^2+n & n^2 \\ 0 & 2 & 0 \\ 0 & 0 & n+2 \end{vmatrix} = 96

\Rightarrow n[2(n+2) - 0] = 96

\Rightarrow 2n^2 + 4n - 96 = 0

\Rightarrow n^2 + 2n - 48 = 0

\Rightarrow (n+8)(n-6) = 0

\mathbf{n = 6}