To solve this problem, we need to find the values of \(K\) and \(\alpha\) that make the closed-loop system have poles at \((-3 \pm j\sqrt{5})\).

The open-loop transfer function is given as:

\(L(s) = C(s)G(s) = K(s + \alpha)\frac{1}{(s+1)(s+2)}\)

The closed-loop transfer function \(T(s)\) is:

\(T(s) = \frac{L(s)}{1 + L(s)} = \frac{K(s + \alpha)}{(s + 1)(s + 2) + K(s + \alpha)}\)

To have poles at \(s = -3 \pm j\sqrt{5}\), the characteristic equation of the closed-loop system should be:

\((s - (-3 + j\sqrt{5}))(s - (-3 - j\sqrt{5}))\)

Since these poles are complex conjugates, their product gives:

\((s + 3)^2 + 5 = s^2 + 6s + 14\)

This means the denominator of \(T(s)\) must be:

\((s+1)(s+2) + K(s+\alpha) = s^2 + 6s + 14\)

Expanding and equating coefficients,

• Constant term: \(2 + K\alpha = 14 \to K\alpha = 12\)
• Coefficient of \(s: 3 + K = 6 \to K = 3\)

Substituting \(K = 3\) into \(K\alpha = 12\) gives:

\(\alpha = \frac{12}{3} = 4\)

Therefore, the correct pair is \((K, \alpha) = (3, 4)\).