CAT 2022

Slot 3

QA

Question & Solution

Let $\frac{x^2-6x+10}{3-x}=p$

$x^2-6x+10=3p-px$

$x^2-\left(6-p\right)x+10-3p=0$

Since the equation will have real roots,

$\left(6-p\right)^2-4\times\left(10-3p\right)\ge0$

$p^2-12p+12p+36-40\ge0$

$p^2\ge4$

$p\ge2\ ,\ p\le-2$

Now, when $p=-2$, $x = 4$.  Since it is given that $x<3$, thus this value will be discarded.

Now, $\frac{1}{2}$ and $-\frac{1}{2}$ do not come in the mentioned range.

when $p=2$, $x = 2$

Thus, the minimum possible value of p will be 2.

Thus, the correct option is B.

Alternate explanation:

Since $x<3$,

$3-x>0$

Let $3-x=y$. So, $y>0$.

Now, $\frac{x^2-6x+10}{3-x}=\frac{x^2-6x+9+1}{3-x}$

=> $\frac{\left(3-x\right)^2+1}{3-x}$

Since $3-x=y$, the equation will transform to $\frac{y^2+1}{y}$ or $y+\frac{1}{y}$

The minimum value of the expression $y+\frac{1}{y}$ for $y>0$ will at $y=1$

i.e., Minimum value = $1+1=2$

Thus, the correct option is B.