solution:

 

The capacity f : R → {x ∈ R : – 1 < x < 1} characterized by f(x) =     ???? / 1+|????|

 

,x ∈ R

 

For one-one:

 

Say x, y ∈ R

 

According to meaning of |x|;

 

|????| = {−????, ???? < 0

 

????, ???? ≥ 0

 

????

 

So f(x) = { 1−????

 

1+????

 

, ???? < 0

 

, ???? ≥ 0

 

For x ≥ 0

 

\[f\left( x \right)\text{ }=\text{ }x/\left( 1+x \right)\]

\[f\left( y \right)\text{ }=\text{ }y/\left( 1+y \right)\]

In the event that \[f\left( x \right)\text{ }=\text{ }f\left( y \right),\text{ }x/\left( 1+x \right)\text{ }=\text{ }y/\left( 1+y \right)\text{ }x\left( 1\text{ }+\text{ }y \right)\text{ }=\text{ }y\text{ }\left( \text{ }1+x \right)\]

 

 x= y

 

For x < 0 f(x) = x/(1-x)

 

\[f\left( y \right)\text{ }=\text{ }y/\left( 1-y \right)\]

In the event that f(x) = f(y),

 

\[x/\left( 1-x \right)\text{ }=\text{ }y/\left( 1-y \right)\]

 

\[x\left( 1\text{ }-\text{ }y \right)\text{ }=\text{ }y\text{ }\left( \text{ }1-x \right)\]

 x= y

 

In both the conditions, x = y. Thusly, f is one-one.

 

Again for onto:

 

????

 

f(x) = { 1−????

 

1+????

 

For x < 0

 

, ???? < 0

 

, ???? ≥ 0

 

\[y\text{ }=\text{ }f\left( x \right)\text{ }=\text{ }x/\left( 1-x \right)\text{ }y\left( 1-x \right)\text{ }=\text{ }x~\] or on the other hand x(1+y) = y

 

or on the other hand x = y/(1+y) … (1)

 

For x ≥ 0

 

\[y\text{ }=\text{ }f\left( x \right)\text{ }=\text{ }x/\left( 1+x \right)\text{ }y\left( 1+x \right)\text{ }=\text{ }x~\] or on the other hand x = y/(1-y) … (2)

 

Presently we have two distinct upsides of x from both the case.

 

Since y ∈ {x ∈ R : – 1 < x < 1}

 

The worth of y lies between – 1 to 1.

 

In the event that y = 1

 

x = y/(1-y) (not characterized)

 

In the event that y = – 1

 

x = y/(1+y) (not characterized)

 

So x is characterized for every one of the upsides of y, and x ∈ R This shows that, f is onto.

 

Reply: f is one-one and onto.