Let f : R → R be a function defined by $f(x)=\frac{e^{[x]}-e^{-x}}{e^x+e^{-x}}$. Then

f is neither one-one nor onto

 $f : R → R$

$f(x)=\frac{e^{|x|}-e^{-x}}{e^x+e^{-x}}$

$f(-2)=\frac{e^{|-2|}-e^{2}}{e^{-2}+e^{2}}=\frac{e^{2}-e^{2}}{e^{-2}+e^{2}}=0$

$f(-3)=\frac{e^{|-3|}-e^{3}}{e^{-3}+e^{3}}=\frac{e^{3}-e^{3}}{e^{-3}+e^{3}}=0$

Hence, we can see for different values of x we are getting same values of f(x). That means, the given function is many one.

∴  f is not injective.

For $x<0$

$f(x)=0$

For $x>0$

$f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$

$⇒f(x)=\frac{e^x-e^{-x}-2e^{-x}}{e^x+e^{-x}}=1-\frac{2e^{-x}}{e^x+e^{-x}}$

The value of $\frac{2e^{-x}}{e^x+e^{-x}}$ is always positive.

∴  The value of f(x) is always less than 1.

Numbers more than 1 are not included in the range but they are included in co-domain.