CAT 2023

Slot 2

QA

Question & Solution

It is given that $\left(x-1\right)^2+2kx+11=0$ has no real roots. (Where k is the largest integer)

$\left(x-1\right)^2+2kx+11=0$, which can be written as:

=> $x^2-2x+1+2kx+11=0$

=> $x^2-2\left(k-1\right)x+12=0$

We know that for no real roots, D < 0 => b^2 -4ac < 0

Hence, $\left\{2\left(k-1\right)\right\}^2-4\cdot1\cdot12\ <0$

=> $4\left(k-1\right)^2<48$

=> $\left(k-1\right)^2<12$

Since k is an integer, it implies (k-1) is also an integer. 

Therefore, from the above inequality, we can say that the largest possible value of (k-1) = 3 => The largest possible value of k is 4.

Now we need to calculate the least possible value of $\frac{k}{4y}+9y$. 

$\frac{k}{4y}+9y$ can be written as $\frac{4}{4y}+9y\ =\ \frac{1}{y}+9y$

The least possible value of $9y\ +\frac{1}{y}$ can be calculated using A.M-G.M inequality.

Using A.M-G.M inequality, we get:

$\ \frac{\ 9y+\frac{1}{y}}{2}\ge\ \sqrt{\ 9y\times\ \frac{1}{y}}$

=> $\ \frac{\ 9y+\frac{1}{y}}{2}\ge\ \sqrt{\ 9}$

=> $\ \frac{\ 9y+\frac{1}{y}}{2}\ge3$

=> $\ \ 9y+\frac{1}{y}\ge6$

Hence, the least possible value is 6