Question

Let f : R $\to$ R be defined by f(x) = $\frac{\left|x\right|-1}{\left|x\right|+1}$ 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) = $\frac{\left|x\right|-1}{\left|x\right|+1}$
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{\left|x_{1}\right|-1}{\left|x_{2}\right|+1} = \frac{\left|x_{2}\right|-1}{\left|x_{2}\right|+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 : $f\left(x\right) = \frac{\left|x\right|-1}{\left|x\right|+1}$
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.