KVPY 2020-21 SB SX Exam Previous Papers

${\mathit{x}}^3+\mathit{a}{\mathit{x}}^2+\mathit{b}\mathit{x}+\mathit{c}=0$ $=\left(\mathit{x}-\mathit{a}\right)\left(\mathit{x}-\mathit{b}\right)\left(\mathit{x}-\mathit{c}\right)$
$\mathit{a}+\mathit{b}+\mathit{c}=-\mathit{a}$
$\Rightarrow 2\mathit{a}+\mathit{b}+\mathit{c}=0$ ...(i)
$\mathit{a}\mathit{b}+\mathit{b}\mathit{c}+\mathit{c}\mathit{a}=\mathit{b}$...(ii)
$\mathit{a}\mathit{b}\mathit{c}=-\mathit{c}\Rightarrow \mathit{a}\mathit{b}=-1\left[\because \mathit{c}\ne 0\right]$ ...(iii)
Also a is a root of equation
$\Rightarrow 2{\mathit{a}}^3+\mathit{a}\mathit{b}+\mathit{c}=0\mathrm{}$ $\Rightarrow 2{\mathit{a}}^3-1+\mathit{c}=0$
$\Rightarrow \mathit{c}=1-2{\mathit{a}}^3$
from (1)
$2{\mathit{a}}^2+\mathit{a}\mathit{b}+\mathit{a}\mathit{c}=0$
$2{\mathit{a}}^2-1+\mathit{a}\left(1-2{\mathit{a}}^3\right)=0$
$2{\mathit{a}}^2-2{\mathit{a}}^4+\mathit{a}-1=0$
$2{\mathit{a}}^2\left(1-\mathit{a}\right)\left(1+\mathit{a}\right)+\left(\mathit{a}-1\right)=0$
$\Rightarrow \left(1-\mathit{a}\right)\left[2{\mathit{a}}^2\left(\mathit{a}+1\right)-1\right]=0$
$\Rightarrow \mathit{a}=1$ or $2{\mathit{a}}^3+2{\mathit{a}}^2-1=0$
when $\mathit{a}=1,\mathit{b}=\frac{-1}{\mathit{a}}=-1$ and $\mathit{c}=1-2{\mathit{a}}^3=-1$
when $2{\mathit{a}}^3+2{\mathit{a}}^2-1=0$
There will be only one real solution of
$\mathit{f}\left(\mathit{x}\right)=2{\mathit{x}}^3+2{\mathit{x}}^2-1=0$
as $\mathit{f}\prime \left(\mathit{x}\right)=6{\mathit{x}}^2+4\mathit{x}=0\mathrm{}$ $\Rightarrow \mathit{x}=0,\frac{-2}{3}$

$\therefore$ corresponding to this real value of a one triplet is possible
$\therefore$ Exactly two triplets $\left(\mathit{a},\mathit{b},\mathit{c}\right)$ are possible