Question

If $\alpha$ and $\beta$ are the roots of the quadratic equation$\lambda (x^2-x)+x+5=0$ and ${\lambda }_1,{\lambda }_2$are two values $\lambda$obtained from$(ɑ/\beta )+(\beta /ɑ)=4/5$, then $\frac{{\lambda }_1}{{{\lambda }_1}^2}+\frac{{\lambda }_2}{{{\lambda }_2}^2}$are two values of $\lambda$

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

Answer 1

The correct option is D

$4048$

Explanation for correct option:

Step 1. Expressing the given the quadratic equation :

Given that $\alpha$ and $\beta$ are the roots of the quadratic equation$\lambda (x^2-x)+x+5=0$.

$$\begin{gathered} \lambda (x^2-x)+x+5=0 \\ \Rightarrow \lambda x^2-x(\lambda -1)+5=0 \end{gathered}$$

$\Rightarrow (\alpha +\beta )=\frac{(\lambda -1)}{\lambda }$and $\alpha \beta =\frac{5}{\lambda }$

Step 2. find the value of ${\mathit{\lambda }}_{\mathbf{1}}\mathbf{,}\mathbf{}{\mathit{\lambda }}_{\mathbf{2}}$:

Also given ${\lambda }_1,{\lambda }_2$are two values $\lambda$obtained from $(ɑ/\beta )+(\beta /ɑ)=4/5$

$$\begin{gathered} \frac{({\alpha }^2+{\beta }^2)}{\alpha \beta }=\frac{4}{5} \\ \Rightarrow \frac{{(\alpha +\beta )}^2-2\alpha \beta }{\alpha \beta }=\frac{4}{5} \\ \Rightarrow \frac{{\left({\displaystyle \frac{\lambda -1}{\lambda }}\right)}^2-2\left({\displaystyle \frac{5}{\lambda }}\right)}{\left(\frac{5}{\lambda }\right)}=\frac{4}{5} \\ \Rightarrow \frac{{(\lambda -1)}^2-10\lambda }{5\lambda }=\frac{4}{5} \\ \Rightarrow {(\lambda -1)}^2=14\lambda \\ \Rightarrow {\lambda }^2-2\lambda +1=14\lambda \\ \Rightarrow {\lambda }^2-16\lambda +1=0 \end{gathered}$$

$\Rightarrow {\lambda }_1+{\lambda }_2=16;{\lambda }_1.{\lambda }_2=1$

Step 3. Find the value of $\frac{{\mathbf{\lambda }}_{\mathbf{1}}}{{{\mathbf{\lambda }}_{\mathbf{2}}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\lambda }}_{\mathbf{2}}}{{{\mathbf{\lambda }}_{\mathbf{1}}}^{\mathbf{2}}}$:

$$\begin{gathered} \Rightarrow \frac{{\lambda }_1}{{{\lambda }_2}^2}+\frac{{\lambda }_2}{{{\lambda }_1}^2}=\frac{{{\lambda }_1}^3+{{\lambda }_2}^3}{{\left({\lambda }_1{\lambda }_2\right)}^2} \\ \Rightarrow \frac{{\lambda }_1}{{{\lambda }_2}^2}+\frac{{\lambda }_2}{{{\lambda }_1}^2}=\frac{({\lambda }_1+{\lambda }_2\left)\right({{\lambda }_1}^2+{{\lambda }_2}^2-{\lambda }_1{\lambda }_1)}{{\left({\lambda }_1{\lambda }_2\right)}^2}\mathbf{}\mathbf{\left[}\mathbf{\because }{\mathbf{a}}^{\mathbf{3}}\mathbf{+}{\mathbf{b}}^{\mathbf{3}}\mathbf{=}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{a}\mathbf{b}\mathbf{)}\mathbf{\right]} \\ \Rightarrow \frac{{\lambda }_1}{{{\lambda }_2}^2}+\frac{{\lambda }_2}{{{\lambda }_1}^2}=\frac{({\lambda }_1+{\lambda }_2\left)\right[({\lambda }_1+{\lambda }_2{)}^2-3{\lambda }_1{\lambda }_1]}{{\left({\lambda }_1{\lambda }_2\right)}^2} \\ \Rightarrow \frac{{\lambda }_1}{{{\lambda }_2}^2}+\frac{{\lambda }_2}{{{\lambda }_1}^2}=16({16}^2-3) \\ \Rightarrow \frac{{\lambda }_1}{{{\lambda }_2}^2}+\frac{{\lambda }_2}{{{\lambda }_1}^2}=16\times 253 \\ \Rightarrow \frac{{\lambda }_1}{{{\lambda }_2}^2}+\frac{{\lambda }_2}{{{\lambda }_1}^2}=4048 \end{gathered}$$

Hence, the correct option is (D).