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JEE Mathematics 2024 Question with Solution

For α,β∈R\alpha, \beta \in R and a natural number nn, let Ar=A_r =

∣r1n2+α2r2n2−β3r−23n(3n−1)2∣\begin{vmatrix} r & 1 & n^2 + \alpha \\ 2r & 2 & n^2 - \beta \\ 3r - 2 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix}

Then 2A10−A52A_{10} - A_5 is:

  • A

    4α+2β4\alpha + 2\beta

  • B

    2α+4β2\alpha + 4\beta

  • C

    2n2n

  • D

    00

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

Ar=∣r1n22+α2r2n2−β3r−23n(3n−1)2∣A_r = \begin{vmatrix} r & 1 & \frac{n^2}{2} + \alpha \\ 2r & 2 & n^2 - \beta \\ 3r - 2 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix}

Find: 2A10−A52A_{10} - A_5.

From the solution working, the determinant is expanded using properties of determinants. The key minor used is

∣2r23r−23∣=2r⋅3−2(3r−2)=4.\begin{vmatrix} 2r & 2 \\ 3r - 2 & 3 \end{vmatrix} = 2r \cdot 3 - 2(3r - 2) = 4.

Using the shown simplification,

2A10−A5=−2[(n2−β)−(n2+2α)].2A_{10} - A_5 = -2\left[(n^2 - \beta) - (n^2 + 2\alpha)\right].

Now simplify the bracket:

(n2−β)−(n2+2α)=−β−2α.(n^2 - \beta) - (n^2 + 2\alpha) = -\beta - 2\alpha.

Therefore,

2A10−A5=−2(−β−2α)=4α+2β.2A_{10} - A_5 = -2(-\beta - 2\alpha) = 4\alpha + 2\beta.

So, the correct option is A.

Common mistakes

  • Using the matrix entries incorrectly while copying the determinant. This is wrong because a small transcription error changes the determinant completely. Copy each entry exactly before substituting r=10r = 10 or r=5r = 5.

  • Trying to subtract determinants entrywise without using determinant properties properly. This is wrong because determinants are not linear in the whole matrix at once. Use determinant expansion or linearity in a single row or column carefully.

  • Making a sign error in simplifying −2[(n2−β)−(n2+2α)]-2\left[(n^2 - \beta) - (n^2 + 2\alpha)\right]. This is wrong because the minus sign before the second bracket changes both terms. First reduce the inner expression to −β−2α-\beta - 2\alpha, then multiply by −2-2.

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