Range of gain K for stability by the Routh-Hurwitz criterion

Step-by-step derivation

1. Construct the Routh array. Row $s^3$ takes the coefficients of $s^3$ and $s^1$; row $s^2$ takes those of $s^2$ and $s^0$: $$\begin{array}{c|cc} s^3 & 1 & 2 \\ s^2 & 3 & K \end{array}$$
2. Compute the $s^1$ row from the two rows above it, using the standard determinant pattern: $$b_1 = \frac{(3)(2) - (1)(K)}{3} = \frac{6-K}{3}.$$
3. The $s^0$ row is simply the last entry carried down: $$\begin{array}{c|cc} s^3 & 1 & 2 \\ s^2 & 3 & K \\ s^1 & \frac{6-K}{3} & 0 \\ s^0 & K & \end{array}$$
4. The Routh-Hurwitz criterion: the system is stable if and only if every entry in the first column has the same sign. The first two are $1$ and $3$, both positive, so all of them must be positive.
5. From the $s^1$ row: $$\frac{6-K}{3} > 0 \;\Longrightarrow\; K < 6.$$
6. From the $s^0$ row: $$K > 0.$$
7. Combining the two conditions gives $$\mathbf{0 < K < 6},$$ which is option (A).
8. What happens at the edges. At $K = 6$ the entire $s^1$ row becomes zero - a pair of poles sits exactly on the imaginary axis and the system oscillates. The frequency comes from the auxiliary equation $3s^2 + 6 = 0 \Rightarrow s = \pm j\sqrt{2}$, so it rings at $\sqrt{2}$ rad/s.