Question

Let f : R $\to$ R be defined by f(x) = $\frac{|x|-1}{|x|+1}$ then f is :

• A. both one - one and onto
• B. one - one but not onto
• C. onto but not one - one
• D. neither one - one nor onto.

Answer 1

Answer: (D)

f(x) = $\frac{|x|-1}{|x|+1}$
for one-one function if $f(x_1)=f(x_2)$ then $x_1$ must be equal to $x_2$
Let $f(x_1)=f(x_2)$
$\frac{|x_1|-1}{|x_2|+1}=\frac{|x_2|-1}{|x_2|+1}$
$\left|x_1\right|\left|x_2\right|-\left|x_1\right|+\left|x_2\right|-1=\left|x_1\right|\left|x_2\right|+\left|x_1\right|-\left|x_2\right|-1$
$\Rightarrow \left|x_1\right|-\left|x_2\right|=\left|x_2\right|-\left|x_1\right|$
$2\left|x_1\right|=2\left|x_2\right|$
$\left|x_1\right|=\left|x_2\right|$
$x_1=x_2,x_1=-x_2$
here $x_1$ has two values therefore function is many one function.
For onto : $f\left(x\right)=\frac{|x|-1}{|x|+1}$
for every value of $f\left(x\right)$ there is a value of x in domain set.
If $f\left(x\right)$ is negative then $x=0$
for all positive value of $f\left(x\right)$, domain contain atleast one element. Hence $f\left(x\right)$ is onto function.