CAT 2022 Slot 3 · Quant · Question 2

The denominator suggests substituting t=3-x, which is positive because x<3.

Then x=3-t, and the numerator becomes {\left( 3-t \right)}^{2}-6\left( 3-t \right)+10.

Expanding, 9-6t+{t}^{2}-18+6t+10={t}^{2}+1.

A quicker route to the same place: {x}^{2}-6x+10={\left( x-3 \right)}^{2}+1={t}^{2}+1.

So the whole expression is \frac{{t}^{2}+1}{t}=t+\frac{1}{t}.

Both terms are positive, so AM-GM applies: t+\frac{1}{t}\ge 2\sqrt{t\times \frac{1}{t}}.

The product inside the root is 1, so the bound is 2\times 1=2.

Equality holds when t=\frac{1}{t}, that is t=1, which corresponds to x=2 — comfortably inside the allowed range.

So the minimum value is 2 — the third option.

Shortcut. Completing the square in the numerator is what reveals the structure: the expression is always t+\frac{1}{t} in disguise. Since x<3 keeps t positive, the answer must be positive, which eliminates both negative options immediately.