 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ib/paper-1/4a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-07

g (t x + (1 - t) y) \leq t g (x) + (1 - t) g (y)

for all

x, y

in its domain and

0 \leq t \leq 1

. If the second derivative exists on an interval, this is equivalent to

g^{''} \geq 0

there. The equivalence follows from the nondecreasing first derivative, or from the supporting-line inequalities obtained by integration. Here

f^{'} (x) = lo g x + 1, f^{''} (x) = 1/ x > 0 (x > 0) .

Thus it is a strictly convex function, and in particular convex.

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