Question

If $\alpha_{1}, \beta_{1}, \gamma_{1}, \delta_{1}$ are the roots of the equation $a x^{2}+b x^{3}+c x^{2}+d x+e=0$ and $\alpha_{2}, \beta_{2}, \gamma_{2}, \delta_{2}$ are the roots of the equation $e x^{4}+d x^{3}+c x^{2}+b x+a=0$ such that $0<\alpha_{1}

• A. 10
• B. 12
• C. 6
• D. 8

Answer 1

Answer: (D)

Given
$\alpha_{1}, \beta_{1}, \gamma_{1}, \delta_{1}$ are the roots of equation
$a x^{4}+b x^{3}+c x^{2}+d x+e=0$ and
$\alpha_{2}, \beta_{2}, \gamma_{2}, \delta_{2}$ are the roots of equation
$e x^{4}+d x^{3}+c x^{2}+b x+a=0$
$\therefore $ Clearly,
$\alpha_{1}, \beta_{1}, \gamma_{1}, \delta_{1}$ are the reciprocal of roots of
$a x^{4}+b x^{3}+c x^{2}+d x+e=0$
Also, $\delta_{1}>\gamma_{1}>\beta_{1}>\alpha_{1}>0$
and $\delta_{2}>\gamma_{2}>\beta_{2}>\alpha_{2}>0$
$\therefore \delta_{1}$ and $\alpha_{2}$ are reciprocal
Similarly $\gamma_{1}$ and $\beta_{2}, \gamma_{2}$ and $\beta_{1}$ and $\alpha_{1}$ and $\delta_{2}$ are reciprocal $\therefore \alpha_{1} \delta_{2}=1, \beta_{1} \gamma_{2}=1, \beta_{2} \gamma_{1}=1, \alpha_{2} \delta_{1}=1$
$\alpha_{1}-\delta_{2}=2$
$\therefore \alpha_{1}-\frac{1}{\alpha_{1}}=2$
$ \Rightarrow \alpha_{1}^{2}-2 \alpha_{1}-1=0$
and $\delta_{1}-\alpha_{2}=4$
$\delta_{1}-\frac{1}{\delta_{1}}=4=\delta_{1}^{2}-4 \delta_{1}-1=0$
$\therefore \alpha_{1}$ and $\delta_{1}$ are roots of
$a x^{4}+b x^{3}+c x^{2}+d x+e=0$
$\therefore \left(x^{2}-2 x-1\right)\left(x^{2}-4 x-1\right)$
$=a x^{4}+b x^{3}+c x^{2}+d x+e$
Put $x=1$,
$(1-2-1)(1-4-1)$
$=a+b+c+d+c$
$8=a+b+c+d+e$