JEE Mathematics 2024 Question with Solution

Given: We need the values of $\alpha$ for which $\begin{vmatrix} 1 & 3 & 2 \\ \alpha+3 & 2 & 1 \\ 1 & 1 & 3 \end{vmatrix} = 0$.

Find: The interval containing the required values of $\alpha$.

From the solution, the determinant is expanded and the resulting values of $\alpha$ are stated to lie in the interval $(-3,0)$.

The solution explicitly concludes: The Correct Option is B and also states that $\alpha$ lies in $(-3,0)$.

Therefore, the correct option is B.

Given: Determine the interval for $\alpha$ such that the determinant is zero.

Find: The correct option.

The solution rewrites and expands the determinant, then states:

Then it simplifies to an equation in $\alpha$ and concludes that the required values lie in the interval $(-3,0)$.

A second extracted approach also concludes: $a \in (-3,0)$.

Although the intermediate matrices shown in the solution do not exactly match the original determinant, both approaches and the final declaration agree that the answer is $(-3,0)$.

Therefore, the correct option is B, corresponding to $(-3,0)$.