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CAT 2023 Slot 2 — QA Question 5

Discriminant and Roots of Quadratic EquationEasy

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

Answer & solution

Answer: 6

Solution

Easy

Two-part problem. First, expand and use the "no real roots" condition (discriminant <0\lt 0) to find the largest integer kk. Then substitute kk and minimise k4y+9y\dfrac{k}{4y}+9y with AM–GM.

1

Expand the quadratic:

(x−1)2+2kx+11=0⇒ x2−2x+1+2kx+11=0⇒ x2+(2k−2)x+12=0\begin{aligned} &(x-1)^2+2kx+11=0\\ &\Rightarrow\ x^2-2x+1+2kx+11=0\\ &\Rightarrow\ x^2+(2k-2)x+12=0 \end{aligned}
2

No real roots ⇒\Rightarrow discriminant <0\lt 0:

(2k−2)2−4(1)(12)<0⇒ 4(k−1)2−48<0⇒ (k−1)2<12\begin{aligned} &(2k-2)^2-4(1)(12)\lt 0\\ &\Rightarrow\ 4(k-1)^2-48\lt 0\\ &\Rightarrow\ (k-1)^2\lt 12 \end{aligned}

So k−1<12≈3.46k-1\lt\sqrt{12}\approx 3.46, giving k−1≤3k-1\le 3, i.e. the largest integer is k=4k=4.

3

Substitute k=4k=4 into the target expression:

k4y+9y=44y+9y=1y+9y\frac{k}{4y}+9y=\frac{4}{4y}+9y=\frac{1}{y}+9y
4

Minimise with AM–GM (valid since y>0y\gt 0):

1y+9y≥21y⋅9y=29=6\begin{aligned} &\frac{1}{y}+9y\ge 2\sqrt{\frac{1}{y}\cdot 9y}=2\sqrt{9}=6 \end{aligned}

Equality when 1y=9y⇒y=13\dfrac{1}{y}=9y\Rightarrow y=\tfrac13.

Least value=6\text{Least value}=\mathbf{6}
CAT 2023 Slot 2 QA Q5: Let k be the largest integer such that the equation (x - 1) 2 + 2kx + 11 = 0 has no real roots. If y is a posi — Solution | TheCATExam