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

Since $f\left(x\right)=f(-x)\therefore f$ is not one-one.

Let $y\in R$.

Then $f\left(x\right)=y\Rightarrow y=\frac{x^2-8}{x^2+2}\Rightarrow x^2=\frac{8+2y}{1-y}$

For $x$ to be real, $(8+2y)(1-y)\ge 0$

and $1-y\ne 0\Rightarrow (y+4)(y-1)\le 0$ and $y\ne 1\Rightarrow -4\le y<1$

$\therefore$ Range of $f=[-4,1)\subset R$

$\therefore f$ in not onto.