Question

Let $\alpha$ and $\beta$ are the roots of equation $a{x}^2+bx+c=0\left(a\ne 0\right).$ If $1,\alpha +\beta ,\alpha \beta$ are in arithmetic progression and $\alpha ,2,\beta$ are in harmonic progression, then the value of $\frac{{\left(\alpha \right)}^2+{\left(\beta \right)}^2-2{\left(\alpha \right)}^2{\left(\beta \right)}^2}{2\left({\left(\alpha \right)}^2+{\left(\beta \right)}^2\right)}$ is equal to

• A. 0
• B. 0.5
• C. 1
• D. 1.5

Answer 1

Answer: (D)

$1,\alpha +\beta ,\alpha \beta$ are in A.P. $\Rightarrow 1,\frac{-b}{a},\frac{c}{a}$ are in A.P.
$\Rightarrow 1+\frac{c}{a}=\frac{-2b}{a}\Rightarrow a+c+2b=0$.... (1)
$\frac{1}{\alpha },\frac{1}{2},\frac{1}{\beta }$ are in A.P.
$\Rightarrow \frac{1}{\alpha }+\frac{1}{\beta }=1$
$\Rightarrow \alpha +\beta =\alpha \beta$
$\Rightarrow \frac{-b}{a}=\frac{c}{a}\Rightarrow b+c=\mathrm{0....}$ (2)
From 1 & 2 we get,
$a=-b=c$
$\Rightarrow \alpha ,\beta$ are roots of equation $x^2-x+1=0$
Now, $\frac{{\alpha }^2+{\beta }^2-2{\alpha }^2{\beta }^2}{2{\alpha }^2+{\beta }^2}=\frac{1}{2}-\frac{\alpha {\beta }^2}{\alpha +{\beta }^2-2\alpha \beta }$
$=\frac{1}{2}-\frac{1^2}{1^2-21}=\frac{1}{2}+1=1.5$