CAT 2023 Slot 2 Solved Paper

It is given that ${\left(\mathit{x}-1\right)}^2+2\mathit{k}\mathit{x}+11=0$ has no real roots. (Where $\mathrm{k}$ is the largest integer) ${\left(\mathit{x}-1\right)}^2+2\mathit{k}\mathit{x}+11=0$, which can be written as:
$\Rightarrow {\mathit{x}}^2-2\mathit{x}+1+2\mathit{k}\mathit{x}+11=0$
$\Rightarrow {\mathit{x}}^2-2\left(\mathit{k}-1\right)\mathit{x}+12=0$
We know that for no real roots,
Hence,


Since $\mathit{k}$ is an integer, it implies ( $\mathit{k}-1$ ) is also an integer.
Therefore, from the above inequality, we can say that the largest possible value of $\left(\mathit{k}-1\right)=3\Rightarrow$ The largest possible value of $\mathrm{k}$ is 4 .
Now we need to calculate the least possible value of $\frac{\mathit{k}}{4\mathit{y}}+9\mathit{y}$.
$\frac{\mathit{k}}{4\mathit{y}}+9\mathit{y}$ can be written as $\frac{4}{4\mathit{y}}+9\mathit{y}=\frac{1}{\mathit{y}}+9\mathit{y}$
The least possible value of $9\mathit{y}+\frac{1}{\mathit{y}}$ can be calculated using A.M-G.M inequality.
Using A.M-G.M inequality, we get:
$\frac{9\mathit{y}+\frac{1}{\mathit{v}}}{2}\ge \sqrt{9\mathit{y}\times \frac{1}{\mathit{y}}}$
$\Rightarrow \frac{9\mathit{y}+\frac{1}{\mathit{y}}}{2}\ge \sqrt{9}$
$\Rightarrow \frac{9\mathit{y}+\frac{1}{\mathit{y}}}{2}\ge 3$
$\Rightarrow 9\mathit{y}+\frac{1}{\mathit{y}}\ge 6$
Hence, the least possible value is 6