CAT 2023 Slot 2 QA Question 7: 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 $\dfrac{k}{4y} + 9y$ is

Question

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 $\dfrac{k}{4y} + 9y$ is

Accepted Answer

First find the largest integer value of $k$ for which the quadratic has no real roots.

$(x - 1)^{2} + 2kx + 11 = 0$
$x^{2} + (2k - 2)x + 12 = 0$

For no real roots, the discriminant must be negative.
$(2k - 2)^{2} - 4 	imes 1 	imes 12 
$4(k - 1)^{2} 
$(k - 1)^{2}

Hence, the largest possible int

Let kkk be the largest integer such that the equation (x−1)2+2kx+11=0(x - 1)^{2} + 2kx + 11 = 0(x−1)2+2kx+11=0 has no real roots. If yyy is a positive real number, then the least possible value of k4y+9y\dfrac{k}{4y} + 9y4yk​+9y is

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 $\dfrac{k}{4y} + 9y$ is