 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-307/4/ii/solution

b_{0} = \frac{3}{2} 2 (N_{c} + 1) - \frac{1}{2} (2 N_{f}) = 3 (N_{c} + 1) - N_{f} .

N_{f} = 3 (N_{c} + 1),

and the theory is infrared free for larger

N_{f}

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