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If $\lambda _{1} \: and\: \lambda _{2}$ are the two values of $\lambda$ such that the roots $\alpha \: and\: \beta$ of the quadratic equation

$\lambda \left ( x^{2} -x\right )+x+5=0$ satisfy $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+\frac{4}{5}=0$ then $\frac{\lambda _{1}}{\lambda _{2}^{2}}+\frac{\lambda _{2}}{\lambda _{1}^{2}}$ is equal to :

Option 1)
488

Option 2)
536

Option 3)
512

Option 4)
504

Answers (2)

Given that $\lambda(x^{2}-\lambda)+x+5=0$

$\therefore\ \;\lambda x^{2}+x(1-\lambda)+5=0$

$\therefore\ \;\alpha+\beta=\frac{\lambda-1}{\lambda}$ and $\therefore\ \;\alpha+\beta=\frac{5}{\lambda}$

$\therefore\ \;\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+\frac{4}{5}=0$

$\therefore\ \;\frac{\alpha^{2}+\beta^{2}}{\alpha\beta}+\frac{4}{5}=0$

$\therefore\ \;\frac{(\alpha+\beta)^{2}-2\alpha \beta}{\alpha\beta}+\frac{4}{5}=0$

$\therefore\ \; \frac{\left(\frac{\lambda-1}{\lambda} \right )^{2}-2.\frac{5}{\lambda}}{\frac{5}{\lambda}}+\frac{4}{5}=0$

$\Rightarrow\ \;(\lambda-1)^{2}-10\lambda+4=0$

$\Rightarrow\ \;\lambda^{2}-2\lambda+1-10\lambda+4=0$

$\Rightarrow\ \;\lambda^{2}+8\lambda+1=0$

$\therefore\ \;\lambda_{1}+\lambda_{2}=8$ , $\lambda_{1}.\lambda_{2}=1$

$\Rightarrow\ \;\frac{\lambda_{1}}{\lambda_{2}^{2}}+\frac{\lambda_{2}}{\lambda_{1}^{2}}$ $\Rightarrow\ \;\frac{\lambda_{1}^{3}+\lambda_{2}^{3}}{\lambda_{1}^{2}\lambda_{2}^{2}}$ $\Rightarrow\ \;\frac{(\lambda_{1}+\lambda_{2})^{3}-3\lambda_{1}\lambda_{2}(\lambda_{1}+\lambda_{2})}{(\lambda_{1}\lambda_{2})^{2}}$

$\Rightarrow\ \;\frac{8^{3}-3.1.8}{1^{2}}=8(8^{2}-3)=8\times 61=488$

Correct option is 1.

Option 1)

488

This is the correct option.

Option 2)

536

This is an incorrect option.

Option 3)

512

This is an incorrect option.

Option 4)

504

This is an incorrect option.

Posted by

Sabhrant Ambastha

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