If $\left|\begin{array}{ccc}\mathrm{x}+1 & \mathrm{x} & \mathrm{x} \\ \mathrm{x} & \mathrm{x}+\lambda & \mathrm{x} \\ \mathrm{x} & \mathrm{x} & \mathrm{x}+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 \mathrm{x}+81)$, then $\lambda$, $\frac{\lambda}{3}$ are the roots of the equation

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Determinants

x + 1 x x x x + λ x x x x + λ^{2} = \frac{9}{8} (103 x + 81)

, then

λ

\frac{λ}{3}

are the roots of the equation

Answer: (B)

Substitute

x = 0

into the determinant:

100 0 λ 0 00 λ^{2} = \frac{9}{8} \times 81

This gives

λ^{3} = \frac{9 ^{3}}{8}

, so

λ = \frac{9}{2}

and

\frac{λ}{3} = \frac{3}{2}

.

The quadratic with roots

\frac{9}{2}

and

\frac{3}{2}

is:

x^{2} - (\frac{9}{2} + \frac{3}{2}) x + \frac{9}{2} \cdot \frac{3}{2} = 0

x^{2} - 6 x + \frac{27}{4} = 0

Multiplying by 4:

4 x^{2} - 24 x + 27 = 0

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option