Question 7.

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