The given function f:R→{ x∈R:−1<x<1 } is defined by f( x )= x 1+| x | .
Let f( x )=f( y ), then,
x 1+| x | = y 1+| y |
If x is positive and y is negative, then,
x 1+x = y 1−y 2xy=x−y
Here, x is positive and y is negative, so,
x>y and x−y>0
But 2xy is negative.
So, it is not possible that 2xy=x−y.
If x is negative and y is positive, then,
x 1−x = y 1+y 2xy=y−x
Which cannot be possible.
If both x and y are positive, then,
x 1+x = y 1+y x+xy=y+xy x=y
If both x and y are negative, then,
x 1−x = y 1−y x−xy=y−xy x=y
Thus, f is one-one.
Let y∈R such that −1<y<1.
If y is negative, then there exists x= y 1+y ∈R in such a way that,
f( x )=f( y 1+y ) = y 1+y 1+| y 1+y | = y 1+y 1+ −y 1+y =y
If y is positive, then there exists x= y 1−y ∈R in such a way that,
f( x )=f( y 1−y ) = y 1−y 1+| y 1−y | = y 1−y 1+ y 1−y =y
Thus, f is onto.
Therefore, f is one-one and onto.