Question

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