CAT 2022 Slot 3 Paper

${\mathit{x}}^2-6\mathit{x}+10=3\mathit{p}-\mathit{p}\mathit{x}$
${\mathit{x}}^2-\left(6-\mathit{p}\right)\mathit{x}+10-3\mathit{p}=0$
Since the equation will have real roots,
${\left(6-\mathit{p}\right)}^2-4\times \left(10-3\mathit{p}\right)\ge 0$
${\mathit{p}}^2-12\mathit{p}+12\mathit{p}+36-40\ge 0$
${\mathit{p}}^2\ge 4$
$\mathit{p}\ge 2,\mathit{p}\le -2$
Now, when $\mathit{p}=-2,\mathit{x}=4$. Since it is given that , thus this value will be discarded.
Now, $\frac{1}{2}$ and $-\frac{1}{2}$ do not come in the mentioned range.
when $\mathit{p}=2,\mathit{x}=2$
Thus, the minimum possible value of $\mathit{p}$ will be 2 .
Thus, the correct option is $\mathrm{B}$.

Alternate explanation:
Since
$3-\mathit{x}>0$
Let $3-\mathit{x}=\mathit{y}$. So, $\mathit{y}>0$.
Now, $\frac{{\mathit{x}}^2-6\mathit{x}+10}{3-\mathit{x}}=\frac{{\mathit{x}}^2-6\mathit{x}+9+1}{3-\mathit{x}}$
$\Rightarrow \frac{{\left(3-\mathit{x}\right)}^2+1}{3-\mathit{x}}$
Since $3-\mathit{x}=\mathit{y}$, the equation will transform to $\frac{{\mathit{y}}^2+1}{\mathit{y}}$ or $\mathit{y}+\frac{1}{\mathit{y}}$
The minimum value of the expression $\mathit{y}+\frac{1}{\mathit{y}}$ for $\mathit{y}>0$ will at $\mathit{y}=1$ i.e., Minimum value $=1+1=2$
Thus, the correct option is B.