 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-mathematics-continuous-mathematics-and-probability-resit/6/c/solution

g (x) = \frac{1}{2} (1 - \frac{1}{x - 1}) .

x > 2

, then

x - 1 > 1

, so

0 < 1/ (x - 1) < 1

. Subtracting from

1

and dividing by

2

proves

0 < g (x) < \frac{1}{2} (x > 2) .

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