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

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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 negative: $f(x)=f(y)$

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

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

$x-xy=y-xy$

$x=y$

$\therefore$ 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.

Posted by

seema garhwal

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