The problem requires finding the values of the gain $K$ and the parameter $a$ such that the system is marginally stable (oscillates) at the frequency $\omega = 2 \text{ rad/s}$.

1. Form the Closed-Loop Characteristic Equation

The open-loop transfer function is $G(s) = \frac{K(s + 1)}{s^3 + a s^2 + 2s + 1}$. The characteristic equation is $1 + G(s) = 0$:

$$s^3 + a s^2 + 2s + 1 + K(s + 1) = 0$$

Grouping powers of $s$:

$$s^3 + a s^2 + (2 + K)s + (1 + K) = 0 \quad (\text{Eq. 1})$$

2. Apply Routh-Hurwitz Criterion for Marginal Stability

For marginal stability, the Routh array must have a row of zeros, and all preceding rows must be non-zero and positive. We construct the Routh array:

The coefficient $b_1$ is:

$$b_1 = \frac{a(2 + K) - 1(1 + K)}{a}$$

For oscillation, we set $b_1 = 0$ (assuming $a \neq 0$):

$$a(2 + K) - (1 + K) = 0$$ $$2a + aK - 1 - K = 0 \quad (\text{Eq. 2})$$

3. Use the Oscillation Frequency ($\omega = 2 \text{ rad/s}$)

If $b_1 = 0$, the auxiliary equation $A(s)$ is formed from the $s^2$ row coefficients, and its roots must be $\pm j\omega$:

$$A(s) = a s^2 + (1 + K) = 0$$

Substitute $s = j\omega = j2$:

$$a (j2)^2 + (1 + K) = 0$$ $$-4a + 1 + K = 0$$ $$K = 4a - 1 \quad (\text{Eq. 3})$$

4. Solve for $a$ and $K$

Substitute $K$ from Eq. 3 into Eq. 2:

$$2a + a(4a - 1) - 1 - (4a - 1) = 0$$ $$2a + 4a^2 - a - 1 - 4a + 1 = 0$$ $$4a^2 - 3a = 0$$ $$a(4a - 3) = 0$$

Since $a=0$ makes the characteristic equation $s^3 + (2+K)s + (1+K) = 0$ unstable (missing $s^2$ term), we choose the non-zero root:

$$4a = 3 \quad \Rightarrow \quad a = 0.75$$

Substitute $a = 0.75$ back into Eq. 3 to find $K$:

$$K = 4(0.75) - 1 = 3 - 1$$ $$K = 2$$

5. Conclusion

The system oscillates at $2 \text{ rad/s}$ when $K = 2$ and $a = 0.75$.

The correct option is Option 1.