KVPY SA Exam 2016 Question Paper

$\mathit{P}\left(\mathit{x}\right)=\mathit{a}{\mathit{x}}^2+\mathit{b}\mathit{x}+\mathit{c}$ $=\mathit{a}\left(\mathit{x}-\mathit{\alpha }\right)\left(\mathit{x}-\mathit{\beta }\right)$
and α + β + αβ + 1 – 1 = (α + 1) (β + 1) – 1 $=\frac{\left(\mathit{a}-\mathit{b}+\mathit{c}\right)}{\mathit{a}}-1$
$\Rightarrow \mathit{\alpha }+\mathit{\beta }+\mathit{\alpha }\mathit{\beta }$ $=\frac{\mathit{b}}{\mathit{a}}-1={\mathit{\lambda }}_1-1$
i.e. $\frac{\mathit{b}}{\mathit{a}}$ is integer = ${\mathit{\lambda }}_1$
If $\mathit{b}=\mathit{a}{\mathit{\lambda }}_1$
then, $\mathit{c}=\mathit{a}\left(2{\mathit{\lambda }}_1-1\right)$ {because a, b, c are in A.P.}
$\therefore \mathit{P}\left(\mathit{x}\right)=\mathit{a}{\mathit{x}}^2+\mathit{a}{\mathit{\lambda }}_1\mathit{x}+\mathit{a}\left(2{\mathit{\lambda }}_1-1\right)$
$=\mathit{a}\left[{\mathit{x}}^2+{\mathit{\lambda }}_1\mathit{x}+\left(2{\mathit{\lambda }}_1-1\right)\right]$
$\mathit{D}={\mathit{\lambda }}_1^2-4\left(2{\mathit{\lambda }}_1-1\right)$ is perfect square for integral roots
$\Rightarrow \mathit{D}={\mathit{\lambda }}_1^2-8{\mathit{\lambda }}_1+4$ is perfect square
Let $\mathit{D}={\left({\mathit{\lambda }}_1-4\right)}^2-12={\mathit{k}}^2$ {where k ∈ I}
$\Rightarrow \left({\mathit{\lambda }}_1-4-\mathit{k}\right)\left({\mathit{\lambda }}_1-4+\mathit{k}\right)=12$
This gives ${\mathit{\lambda }}_1-4-\mathit{k}=2$ & ${\mathit{\lambda }}_1-4+\mathit{k}=6$
$\Rightarrow {\mathit{\lambda }}_1=8$ & $\mathit{k}=1$
$\therefore \mathit{\alpha }+\mathit{\beta }+\mathit{\alpha }\mathit{\beta }=8-1=7$