CAT 2023 Slot 2 Solved Paper

It is given that $(x-1)^2+2 k x+11=0$ has no real roots. (Where $k$ is the largest integer) $(x-1)^2+2 k x+11=0$, which can be written as:$\;\Rightarrow x^2-2x+1+2kx+11=0$$\;\Rightarrow x^2-2(k-1)x+12=0$We know that for no real roots, $D<0\Rightarrow b^2-4ac<0$Hence, $\{2(k-1)\}^2-4\cdot 1\cdot 12<0$$\;\Rightarrow 4(k-1)^2<48$$\;\Rightarrow (k-1)^2<12$Since $k$ is an integer, it implies ( $k-1$ ) is also an integer.Therefore, from the above inequality, we can say that the largest possible value of $(k-1)=3 \Rightarrow$ The largest possible value of $k$ is 4 .Now we need to calculate the least possible value of $\;\frac{k}{4y}+9y$.$\;\frac{k}{4y}+9y$ can be written as $\frac{4}{4y}+9y=\;\frac{1}{y}+9y$ The least possible value of $9y+\;\frac{1}{y}$ can be calculated using A.M-G.M inequality.Using A.M-G.M inequality, we get:$\;\;\frac{9y+\frac{1}{v}}{2} \ge \sqrt{9y \times \;\frac{1}{y}}$$\Rightarrow \;\frac{9y+\frac{1}{y}}{2} \ge \sqrt{9}$$\Rightarrow \;\frac{9y+\frac{1}{y}}{2} \ge 3$$\Rightarrow 9y+\;\frac{1}{y} \ge 6$Hence, the least possible value is 6