JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If $x + 1 x x am p; x am p; x + λ am p; x am p; x am p; x am p; x + λ^{2} = \frac{9}{8} (103 x + 81)$ then $\lambda, \frac{\lambda}{3}$ are the roots of the equation:

Choose the correct answer:

Explanation

Solution

Determinant solve karein: $C_1 \to C_1 + C_2 + C_3$ apply karne par:

$\begin{vmatrix} 3x+1 & x & x \\ 3x+\lambda & x+\lambda & x \\ 3x+\lambda^2 & x & x+\lambda^2 \end{vmatrix}$

Iska expansion karne par $x$ ka linear term aur constant term nikalenge.
Comparison: Equation ko $\frac{9}{8}(103x + 81)$ se compare karne par $\lambda$ ki value nikalte hain.

Solve karne par $\lambda = \frac{9}{2}$ aati hai.
Roots: Roots honge $\lambda = \frac{9}{2}$ aur $\frac{\lambda}{3} = \frac{3}{2}$ .
Equation: $(x - \frac{9}{2})(x - \frac{3}{2}) = 0 \implies x^2 - 6x + \frac{27}{4} = 0 \implies 4x^2 - 24x + 27 = 0$ .

Correct Option: (3) $4x^2 - 24x + 27 = 0$