Q. 4 Show that the function f : R \rightarrow \{x \in R : - 1 < x < 1\} defined byf(x) = \frac{x}{1 + |x|}   x \in R   is one one and onto function.

Answers (1)

The function   f : R \rightarrow \{x \in R : - 1 < x < 1\} defined by 

f(x) = \frac{x}{1 + |x|}  , x \in R

One- one:

         Let   f(x)=f(y)               , x,y \in R

                 \frac{x}{1+\left | x \right |}=\frac{y}{1+\left | y \right |}

  It is observed that if x is positive and y is negative.

          \frac{x}{1+x}= \frac{y}{1+y}

Since x is positive and y is negative.

x> y\Rightarrow x-y> 0    but 2xy is negative.

x-y\neq 2xy

Thus, the case of x is positive and y is negative is removed.

Same happens in the case  of y is positive and x is negative so this case is also removed.

When x and y both are positive:

                                        f(x)=f(y)

                                      \frac{x}{1+x}= \frac{y}{1+y}                   

                                   x(1+y)=y(1+x) 

                                    x+xy=y+xy

                                           x=y

When x and y both are negativef(x)=f(y)

                                                    \frac{x}{1-x}= \frac{y}{1-y}

                                                    x(1-y)=y(1-x)

                                                      x-xy=y-xy

                                                             x=y

                          \therefore  f is one-one.

Onto:

Let  y \in R  such that  -1< y< 1

If y is negative, then x= \frac{y}{y+1} \in R

f(x)=f(\frac{y}{y+1} )= \frac{\frac{y}{1+y}}{1+ | \frac{y}{1+y}|}= \frac{\frac{y}{1+y}}{1+ \frac{-y}{1+y}}=\frac{y}{1+y-y}=y 

If y is positive, then x= \frac{y}{1-y} \in R

f(x)=f(\frac{y}{1-y} )= \frac{\frac{y}{1-y}}{1+ | \frac{y}{1-y}|}= \frac{\frac{y}{1-y}}{1+ \frac{-y}{1-y}}=\frac{y}{1-y+y}=y

Thus, f is onto.

Hence, f is one-one and onto.

 

 

           

   

                                        

                       

 

                 

 

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