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

Differentiating on the given domain yieldsThe denominator and $\ln x+1$ are positive there. Thus $j$ is strictly concave, and is not convex.

The denominator is positive throughout the domain. Holding $y$ fixed, differentiate with respect to $x$:At the interior point $(1,1/2)$ this second partial derivative is $-2/9<0$, so the restriction to a horizontal line is locally strictly concave, ruling out convexity. At the interior point $(1,2)$ it is $4/9>0$, ruling out concavity. Restrictions of a convex function or a concave function to every line segment must retain that property. Consequently $k$ is neither convex nor concave on its stated domain.

Since $\mathbf{a}\ne 0$, the line $\mathbf{x}=t\mathbf{a}/\Vert \mathbf{a}{\Vert }^2$ lies in the domain and satisfies ${\mathbf{a}}^{\mathsf{T}}\mathbf{x}=t$. The restriction of $l$ to this line is $t^3$, whose second derivative is $6t$: negative for $t<0$ and positive for $t>0$. It can be neither globally convex nor globally concave. Hence $l$ is neither. Equivalently, its Hessian matrix is $6({\mathbf{a}}^{\mathsf{T}}\mathbf{x})\mathbf{a}{\mathbf{a}}^{\mathsf{T}}$, changing the sign of its nonzero directional curvature across ${\mathbf{a}}^{\mathsf{T}}\mathbf{x}=0$.