 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/discrete-mathematics-probability-and-continuous-mathematics/4/c/i/solution

Differentiating on the given domain yields

j^{'} (x) = \frac{1}{x l n x}, j^{''} (x) = - \frac{l n x + 1}{x ^{2} ( l n x ) ^{2}} < 0 (x > 1) .

ln x + 1

are positive there. Thus $j$ is strictly concave, and is not convex.

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