Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 1 / b

Not injective; surjective. Indeed, $f(0)=f(-1)=-9$ although $0\ne -1$. This polynomial is continuous, with $f(x)\to -\infty$ as $x\to -\infty$ and $f(x)\to \infty$ as $x\to \infty$. For any prescribed real $y$, choose $a<b$ with $f(a)<y<f(b)$; the intermediate value theorem supplies a preimage between $a$ and $b$.

Both injective and surjective. The natural logarithm is strictly increasing on the positive real numbers, so it is injective. For every $y\in \mathbb{R}$, $x=e^y>0$ satisfies $\ln x=y$; hence the function is surjective, with the exponential function as its inverse function.

Not injective; its image of a function is exactly $\{1,2,3,\dots \}$. For example, both $x=e^{-1}$ and $x=e^{-1/2}$ give the ceiling function value $1$. Since $0<x<1$, $(\ln x{)}^2>0$, so the value $0$ is never attained. For every positive integer $j$,Each such interval is nonempty and contained in $(0,1)$, with the upper endpoint $1$ excluded when $j=1$. Thus the answer concerning surjectivity depends on the convention for the natural numbers: onto if $\mathbb{N}=\{1,2,\dots \}$, and not onto if $0\in \mathbb{N}$. The source does not settle that convention.

Not injective; surjective. The equal values $f(0)=f(2)=1$ disprove injectivity. For any $y>0$, take $x=1+y^{-1/2}>1$. This lies in the stated domain of a function, andConsequently every member of the codomain has a preimage.