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(B) Let the roots of the equation be $\alpha$ and $\beta$. Given $\lambda = \frac{\alpha}{\beta}$.Given $\lambda + \frac{1}{\lambda} = 1$,we have $\fr

Vedclass · Accepted Answer

$4 - 3\sqrt{2}$

Vedclass · Answer

(B) Let the roots of the equation be $\alpha$ and $\beta$. Given $\lambda = \frac{\alpha}{\beta}$.Given $\lambda + \frac{1}{\lambda} = 1$,we have $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = 1$.This simplifies to $\frac{\alpha^2 + \beta^2}{\alpha\beta} = 1$,or $\alpha^2 + \beta^2 = \alpha\beta$.Adding $2\alpha\beta$ to both sides,we get $(\alpha + \beta)^2 = 3\alpha\beta$.From the quadratic equation $3m^2x^2 + m(m - 4)x + 2 = 0$,the sum of roots $\alpha + \beta = -\frac{m(m - 4)}{3m^2} = -\frac{m - 4}{3m}$ and the product of roots $\alpha\beta = \frac{2}{3m^2}$.Substituting these into the equation $(\alpha + \beta)^2 = 3\alpha\beta$:$\left(-\frac{m - 4}{3m}ight)^2 = 3 \left(\frac{2}{3m^2}ight)$.$\frac{(m - 4)^2}{9m^2} = \frac{2}{m^2}$.Since $m 
eq 0$,we multiply by $9m^2$:$(m - 4)^2 = 18$.$m^2 - 8m + 16 = 18 \Rightarrow m^2 - 8m - 2 = 0$.Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$m = \frac{8 \pm \sqrt{64 - 4(1)(-2)}}{2} = \frac{8 \pm \sqrt{72}}{2} = \frac{8 \pm 6\sqrt{2}}{2} = 4 \pm 3\sqrt{2}$.The least value is $4 - 3\sqrt{2}$.

If $\lambda$ is the ratio of the roots of the quadratic equation in $x$,$3m^2x^2 + m(m - 4)x + 2 = 0$,then the least value of $m$ for which $\lambda + \frac{1}{\lambda} = 1$ is:

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