CAT 2022 Slot 3 QA Q2 Inequalities & Modulus

Question 2 of the Quantitative Ability (QA) section, CAT 2022 Slot 3, with CAT's official answer and a worked solution.

The minimum possible value of \(\frac{x^2-6 x+10}{3-x}\), for \(x\lt3\), is

A
\(\frac{1}{2}\)
B
\(-\frac{1}{2}\)
C
\(2\)
D
\(-2\)

Answer: (C) 2 Inequalities & Modulus MCQ

Solution

Put \(t=3-x\gt0\). Then \(x^2-6x+10=(x-3)^2+1=t^2+1\).
The expression is \(\frac{t^2+1}{t}=t+\frac1t\ge2\) by AM-GM, with equality at \(t=1\) (i.e. \(x=2\)).
Minimum value \(=2\).

Shortcut. Denominator is positive, so the value is positive: only \(\frac12\) or 2 survive, and \(t+\frac1t\ge2\).