JEE Main 2014 — Functions Question with Solution

$f: R\to R, f\left(x\right)=\frac{\left|x\right|-1}{\left|x\right|+1}$

For $x=0, f\left(x\right)=-1$

And $f\left(x\right)=\frac{1-{\displaystyle \frac{1}{\left|x\right|}}}{1+{\displaystyle \frac{1}{\left|x\right|}}}$

If $x\to \pm \infty , f\left(x\right)\to 1$

Thus, the graph of the function $f\left(x\right)$ is given by

If a line parallel to the $x$-axis intersects the graph of a function at more than one point, then the function is many-one.

And, it is clear from the graph that, the given function is many-one.

Also, the range of the function is $[-1, 1)$ which is not same as the co-domain which is $R$ and if the co-domain of a function and range are not same then the function is not onto.

Hence, the given function is neither one-one nor onto.