Question

Show that function f : R → { x ∈ R : −1 < x < 1} defined by f ( x ) = , x ∈ R is one-one and onto function.

Answer 1

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.