Question

If ${\alpha }_1,{\beta }_1,{\gamma }_1,{\delta }_1$ are the roots of the equation $a{x}^2+b{x}^3+c{x}^2+dx+e=0$ and ${\alpha }_2,{\beta }_2,{\gamma }_2,{\delta }_2$ are the roots of the equation $e{x}^4+d{x}^3+c{x}^2+bx+a=0$ such that $0<{\alpha }_1<b_1<{\gamma }_1<{\delta }_1$, $0<{\alpha }_2<{\beta }_2<{\gamma }_2<{\delta }_2,{\alpha }_1-{\delta }_2=2={\beta }_1-{\gamma }_2;{\gamma }_1-{\beta }_2={\delta }_1-{\alpha }_2=4$ , then $a+b+c+d+e=$

• 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
$a{x}^4+b{x}^3+c{x}^2+dx+e=0$ and
${\alpha }_2,{\beta }_2,{\gamma }_2,{\delta }_2$ are the roots of equation
$e{x}^4+d{x}^3+c{x}^2+bx+a=0$
$\therefore$ Clearly,
${\alpha }_1,{\beta }_1,{\gamma }_1,{\delta }_1$ are the reciprocal of roots of
$a{x}^4+b{x}^3+c{x}^2+dx+e=0$
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
$a{x}^4+b{x}^3+c{x}^2+dx+e=0$
$\therefore \left(x^2-2x-1\right)\left(x^2-4x-1\right)$
$=a{x}^4+b{x}^3+c{x}^2+dx+e$
Put $x=1$,
$(1-2-1)(1-4-1)$
$=a+b+c+d+c$
$8=a+b+c+d+e$