It is given that $ (x-1)^2 + 2kx + 11 = 0 $ has no real roots. (Where k is the largest integer)
$ (x-1)^2 + 2kx + 11 = 0 $, which can be written as:
=> $ x^2 - 2x + 1 + 2kx + 11 = 0 $
=> $ x^2 - 2(k-1)x + 12 = 0 $
We know that for no real roots, $ D $ < $ 0 $ => $ b^2 - 4ac $ < $ 0 $
Hence, $ {(2(k-1))}^2 - 4 \cdot 1 \cdot 12$ < $ 0 $
=> $ 4(k-1)^2 $ < $ 48 $
=> $ (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
=> 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} \geq \sqrt{9y \times \frac{1}{y}} $
=> $ \frac{9y + \frac{1}{y}}{2} \geq \sqrt{9} $
=> $ \frac{9y + \frac{1}{y}}{2} \geq 3 $
=> $ 9y + \frac{1}{y} \geq 6 $
Hence, the least possible value is 6
