Question

Let $f:R\to R$ is defined by $f\left(x\right)=\frac{|x|-1}{|x|+1}$ then $f$ is :

• A. Both one - one and onto
• B. One - one but not 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.

As given $f\left(x\right)=\frac{(|x|-1)}{(|x|+1)}$.

We know that the definition if if distinct elements in the domain of a function f have distinct images in the co-domain, then f is said to be one-one.

Here for $x=1$ and $x=1.5$ both f has same image in co-domain so it is not one-one function.

Now, definition of onto function is that if each element in the co-domain have at least one preimage in the domain.

Here I had taken co-domain of f as R.

Method to find if function is onto is that first find its inverse function and then find domain of inverse function.

If domain of inverse function is equal to co-domain of given function then given function is onto function.

Here $\left(f^{-1}\right)\left(x\right)=\frac{(1-x)}{(x-1)}+\frac{(1-x)}{(x-1)}$. {} is fractional part.

Now our inverse function is not defined for $x=1$ so our function is not onto.