Question

Let $f:R\to R$ be a function defined by $f\left(x\right)=\frac{x^2-8}{x^2+2}$. Then, f is
(a) one-one but not onto
(b) one-one and onto
(c) onto but not one-one
(d) neither one-one nor onto

Answer 1

Injectivity:
Let x and y be two elements in the domain (R), such that
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow \frac{x^2-8}{x^2+2}=\frac{y^2-8}{y^2+2} \\ \Rightarrow \left(x^2-8\right)\left(y^2+2\right)=\left(x^2+2\right)\left(y^2-8\right) \\ \Rightarrow x^2{y}^2+2x^2-8y^2-16=x^2{y}^2-8x^2+2y^2-16 \\ \Rightarrow 10x^2=10y^2 \\ \Rightarrow x^2=y^2 \\ \Rightarrow x=\pm y \end{gathered}$$
So, f is not one-one.

Surjectivity:
$$\begin{gathered} f\left(-1\right)=\frac{{\left(-1\right)}^2-8}{{\left(-1\right)}^2+2}=\frac{1-8}{1+2}=\frac{-7}{3} \\ \mathrm{and}f\left(1\right)=\frac{{\left(1\right)}^2-8}{{\left(1\right)}^2+2}=\frac{1-8}{1+2}=\frac{-7}{3} \\ \Rightarrow f\left(-1\right)=f\left(1\right)=\frac{-7}{3} \end{gathered}$$
$\Rightarrow$f is not onto.
The correct answer is (d).