Question

If $\alpha ,\beta ,\gamma ,\delta$ are the roots of equation $2x^4+4x^3-3x^2+3x+1=0$, then value of $\frac{1}{2\alpha -1}+\frac{1}{2\beta -1}+\frac{1}{2\gamma -1}+\frac{1}{2\delta -1}$ is

• A. $\frac{16}{19}$
• B. $-\frac{16}{19}$
• C. $\frac{19}{16}$
• D. $-\frac{19}{16}$

Answer 1

The correct option is B $-\frac{16}{19}$

$\alpha ,\beta ,\gamma ,\delta$ are the roots of equation $2x^4+4x^3-3x^2+3x+1=0$
Let $y=\frac{1}{2\alpha -1}$
$\Rightarrow \alpha =\frac{y+1}{2y}$
Putting it in the equation, we get
$2{\left(\frac{1+y}{2y}\right)}^4+4{\left(\frac{1+y}{2y}\right)}^3-3\left(\frac{1+y}{2y}\right)+3\left(\frac{1+y}{2y}\right)+1=0\Rightarrow 2(1+y{)}^4+8y(1+y{)}^3-12y^2(1+y{)}^2+24y^3(1+y)+16y^4=0\Rightarrow 2(y^4+4y^3+\dots )+8y(y^3+3y^2+\dots )-12y^2(y^2+2y+1)+24y^3(1+y)+16y^4=0$

This equation has roots as $\frac{1}{2\alpha -1},\frac{1}{2\beta -1},\frac{1}{2\gamma -1},\frac{1}{2\delta -1}$
So,
$\frac{1}{2\alpha -1}+\frac{1}{2\beta -1}+\frac{1}{2\gamma -1}+\frac{1}{2\delta -1}=-\frac{\text{Coefficient of}{y}^3}{\text{Coefficient of}{y}^4}=-\frac{(8+24-24+24)}{2+8-12+24+16}=-\frac{32}{38}=-\frac{16}{19}$