Let f : R→R be a function defined by f(x)

Solution:

Injectivity:
Let x and y be two elements in the domain (R), such that

$f(x)=f(y)$

$\Rightarrow \frac{x^{2}-8}{x^{2}+2}=\frac{y^{2}-8}{y^{2}+2}$

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

$\Rightarrow x^{2} y^{2}+2 x^{2}-8 y^{2}-16=x^{2} y^{2}-8 x^{2}+2 y^{2}-16$

$\Rightarrow 10 x^{2}=10 y^{2}$

$\Rightarrow x^{2}=y^{2}$

 

$\Rightarrow x=\pm y$

So, f is not one-one.

Surjectivity:

$f(-1)=\frac{(-1)^{2}-8}{(-1)^{2}+2}=\frac{1-8}{1+2}=\frac{-7}{3}$

and $f(1)=\frac{(1)^{2}-8}{(1)^{2}+2}=\frac{1-8}{1+2}=\frac{-7}{3}$

$\Rightarrow f(-1)=f(1)=\frac{-7}{3}$

$\Rightarrow f$ is not onto. The correct answer is $(d)$.