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}=0then\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.

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