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JEE Main 06-Apr-2024 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 06-Apr-2024 Paper - Shift 14 Marks
MCQ
For $\alpha, \beta \in R$ and a natural number $n$, let
$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$. Then $2 A_{10}-A_8$ is
  • A
    $4 \alpha+2 \beta$
  • B
    $2 \alpha+4 \beta$
  • C
    2n
  • D
    $0$
✓

Answer

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$

$2 A_{10}- A _8=\left|\begin{array}{ccc}20 & 1 & \frac{ n ^2}{2}+\alpha \\ 40 & 2 & n ^2-\beta \\ 56 & 3 & \frac{ n (3 n -1)}{2}\end{array}\right|-\left|\begin{array}{ccc}8 & 1 & \frac{ n ^2}{2}+\alpha \\ 16 & 2 & n ^2-\beta \\ 22 & 3 & \frac{ n (3 n -1)}{2}\end{array}\right|$

$\begin{array}{l}\Rightarrow\left|\begin{array}{ccc}12 & 1 & \frac{n^2}{2}+\alpha \\ 24 & 2 & n ^2-\beta \\ 34 & 3 & \frac{ n (3 n -1)}{2}\end{array}\right| \\ \Rightarrow\left|\begin{array}{ccc}0 & 1 & \frac{ n ^2}{2}+\alpha \\ 0 & 2 & n ^2-\beta \\ -2 & 3 & \frac{ n (3 n -1)}{2}\end{array}\right|\end{array}$

$\begin{array}{l}\Rightarrow-2\left(\left( n ^2-\beta\right)-\left( n ^2+2 \alpha\right)\right) \\ \Rightarrow-2(-\beta-2 \alpha) \Rightarrow 4 \alpha+2 \beta\end{array}$

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