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

The correct option is D neither one-one nor onto

We have, $f\left(x\right)=\frac{x^2-8}{x^2+2}$
Clearly $f\left(1\right)=f(-1)=-\frac{7}{3}\Rightarrow$ f is not one-one
Now, Let $y=\frac{x^2-8}{x^2+2}$
$\Rightarrow y(x^2+2)=x^2-8$
$\Rightarrow (y-1)x^2=-8-2y=-2(y+4)$
$\Rightarrow x^2=-2\times \frac{y+4}{y-1}$
Now since, $x\in R$
$\Rightarrow \frac{y+4}{y-1}\le 0\Rightarrow y\in [-4,1)\ne$ codomain of f
Hence f is neither one-one nor onto.