 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2013/f/1/a/ii/solution

The denominator is positive on

B

because

x_{3} > - 1

\nabla g (x) = (\frac{1}{x _{3} + 3}, \frac{1}{x _{3} + 3}, - \frac{x _{1} + x _{2}}{( x _{3} + 3 ) ^{2}}) .

In particular

g

is continuously differentiable throughout

B

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