Answer: (2)

Solution

Let roots of the quadratic equation are $\alpha, \beta$.

Given, $\lambda = \frac{\alpha}{\beta}$ and $\lambda + \frac{1}{\lambda} = 1 \Rightarrow \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = 1$

$\frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = 1$ …(1)

The quadratic equation is $3m^2x^2 + m(m-4)x + 2 = 0$

$\therefore \alpha + \beta = \frac{m(4-m)}{3m^2} = \frac{4-m}{3m}$ and $\alpha\beta = \frac{2}{3m^2}$

Put these values in eq (1),

$\frac{\left(\frac{4-m}{3m}\right)^2}{\frac{2}{3m^2}} = 3$

$\Rightarrow (m-4)^2 = 18$

$\Rightarrow m = 4 \pm \sqrt{18}$

Therefore, least value is $4 - \sqrt{18} = 4 - 3\sqrt{2}$