Aspire Faculty ID #18095 · Topic: JEE Main 2026 (28 January Morning Shift) · Just now
JEE Main 2026 (28 January Morning Shift)

If $ \alpha, \beta $, where $ \alpha < \beta $, are the roots of the equation $ \lambda x^2 - (\lambda + 3)x + 3 = 0 $ such that $ \frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3} $, then the sum of all possible values of $ \lambda $ is:

Solution

$ \frac{\beta - \alpha}{\alpha \beta} = \frac{1}{3}, \quad \alpha + \beta = \frac{\lambda + 3}{\lambda}, \quad \alpha \beta = \frac{3}{\lambda} $

$ \beta - \alpha = \frac{\alpha \beta}{3} = \frac{1}{\lambda} $

on squaring

$ \alpha^2 + \beta^2 - 2\alpha \beta = \frac{1}{\lambda^2} \quad ...(1)$

$ \alpha^2 + \beta^2 + 2\alpha \beta = \frac{(\lambda + 3)^2}{\lambda^2} \quad ...(2)$

(2) $-$ (1)

$ 4\alpha \beta = \frac{(\lambda + 3)^2 - 1}{\lambda^2} $

$ \Rightarrow \frac{12}{\lambda} = \frac{\lambda^2 + 6\lambda + 8}{\lambda^2} $

$ \Rightarrow \lambda^2 - 6\lambda + 8 = 0 $

$ \Rightarrow \lambda = 0, 2, 4 $

Sum of possible values of $ \lambda $ is $ = 6 $

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