CAT 2023 Slot 2 QA Q7 Inequalities & Modulus

Question 7 of the Quantitative Ability (QA) section, CAT 2023 Slot 2, with CAT's official answer and a worked solution.

Let \(k\) be the largest integer such that the equation \((x-1)^2+2 k x+11=0\) has no real roots. If \(y\) is a positive real number, then the least possible value of \(\frac{k}{4 y}+9 y\) is

TITA: type in the answer. There are no options and no negative marking.

Answer: 6 Inequalities & Modulus TITA

Solution

Expand: \(x^2 + (2k-2)x + 12 = 0\).
No real roots: \((2k-2)^2 < 48 \Rightarrow (k-1)^2 < 12 \Rightarrow k - 1 \le 3\), so the largest integer is \(k = 4\).
Then \(\frac{k}{4y} + 9y = \frac1y + 9y \ge 2\sqrt{\frac1y\cdot 9y} = 6\) by AM-GM, with equality at \(y = \frac13\).
Least value \(= 6\).