Question

If $\alpha ,\beta$ are the roots of the equation
$\lambda (x^2-x)+x+5=0$ and if $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{5}$, then $\frac{{\lambda }_1}{{\lambda }_2^2}+\frac{{\lambda }_2}{{\lambda }_1^2}$ equals

• A. $4192$
• B. $4144$
• C. $4096$
• D. $4048$

Answer 1

Answer: (D)

$4048$

Given $\alpha ,\beta$ are the roots of

$\lambda (x^2-x)+x+5=0$

$\Rightarrow \lambda x^2-\lambda x+x+5=0$

$\lambda x^2+(1-\lambda )x+5=0$

$\therefore \alpha +\beta =\frac{-(1-\lambda )}{\lambda }=\frac{\lambda -1}{\lambda }$

$\alpha \beta =5/\lambda$

Now $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{5}$

$\Rightarrow \frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{4}{5}$

$\Rightarrow \frac{{(\alpha +\beta )}^2-2\alpha \beta }{2\alpha \beta }=\frac{4}{5}\Rightarrow \frac{{\left(\frac{\lambda -1}{\lambda }\right)}^2-\frac{2\times 5}{\lambda }}{5/\lambda }=\frac{4}{5}$

$\Rightarrow \frac{{\lambda }^2-2\lambda +1-10\lambda }{5\lambda }=\frac{4}{5}$

$\Rightarrow {\lambda }^2-16\lambda +1=0$

If ${\lambda }_1,{\lambda }_2$ are the roots then

${\lambda }_1+{\lambda }_2=16,{\lambda }_1{\lambda }_2=1$

$\therefore \frac{{\lambda }_1}{{\lambda }_2^2}+\frac{{\lambda }_2}{{\lambda }_1^2}=\frac{{\lambda }_1^3+{\lambda }_2^3}{{\left({\lambda }_1{\lambda }_2\right)}^2}=\frac{{\left(16\right)}^3-3\left(16\right)}{1^2}=4048$