Mathematics - Functions Question with Solution | TestHub

$x^2-8x+18$ is not zero for any real number because $x^2-8x+18$ can be written as $(x-4{)}^2+2$, and the numerator $x^2+4x+30$ is also positive because $(x+2{)}^2+26\equiv x^2+4x+30$. Since $f$ takes values that are only positive for real '$x$', the range of $f$ is a subset of $(0,\infty )$. Therefore, the range $f\ne R\Rightarrow f$ is not onto. Also:

$$\begin{array}{rl}& f(0)=\frac{5}{3}\Rightarrow \frac{x^2+4x+30}{x^2-8x+18}=\frac{5}{3}\Rightarrow 2x^2=52x\Rightarrow x=26\text{ if }x\ne 0\\ & \therefore f(10)=\frac{5}{3}=26\therefore f\text{ is not injective. }\end{array}$$