The values of $\alpha$, for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$, lie in the interval

(C) $\left(-\frac{3}{2}, \frac{3}{2}\right)$

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Determinants

The values of

α

, for which

11 2 α + 3 \frac{3}{2} \frac{1}{3} 3 α + 1 α + \frac{3}{2} α + \frac{1}{3} 0 = 0

, lie in the interval

Answer: (B)

Δ = 11 2 α + 3 \frac{3}{2} \frac{1}{3} 3 α + 1 α + \frac{3}{2} α + \frac{1}{3} 0 = 0.

Expand along the third row. The minor for

a_{31} = 2 α + 3

M_{31} = \frac{3}{2} \frac{1}{3} α + \frac{3}{2} α + \frac{1}{3} = \frac{3}{2} (α + \frac{1}{3}) - \frac{1}{3} (α + \frac{3}{2}) = \frac{3 α}{2} - \frac{α}{3} + \frac{1}{2} - \frac{1}{2} = \frac{7 α}{6} .

The minor for

a_{32} = 3 α + 1

(with cofactor sign

- 1

) is

- M_{32} = - 11 α + \frac{3}{2} α + \frac{1}{3} = - [(α + \frac{1}{3}) - (α + \frac{3}{2})] = - (- \frac{7}{6}) = \frac{7}{6} .

Δ = (2 α + 3) \cdot \frac{7 α}{6} + (3 α + 1) \cdot \frac{7}{6} = \frac{7}{6} [(2 α + 3) α + (3 α + 1)] = \frac{7}{6} (2 α^{2} + 6 α + 1) .

Setting

Δ = 0

2 α^{2} + 6 α + 1 = 0 \Rightarrow α = \frac{- 6 \pm 36 - 8}{4} = \frac{- 3 \pm 7}{2} .

Since

7 \approx 2.65

, both roots — approximately

- 0.18

and

- 2.82

— lie in

(- 3, 0)

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option