Question

If $f:R-\left\{1\right\}\to R$ is defined by $f\left(x\right)=\frac{x^{1/3}}{x^{1/3}-1},$ then $f$ is

• A. an onto but not a one-one function.
• B. a bijective function.
• C. a one-one but not an onto function
• D. a many-one and an into function.

Answer 1

The correct option is C a one-one but not an onto function

For a one-one function,
Let $f\left(a\right)=f\left(b\right)$
$\Rightarrow \frac{a^{1/3}}{a^{1/3}-1}=\frac{b^{1/3}}{b^{1/3}-1}\Rightarrow a^{1/3}{b}^{1/3}-a^{1/3}=a^{1/3}{b}^{1/3}-b^{1/3}\Rightarrow a=b$
$\therefore f$ is a one-one function.

For onto, let $f\left(x\right)=y$
$\Rightarrow \frac{x^{1/3}}{x^{1/3}-1}=y\Rightarrow x^{1/3}=y{x}^{1/3}-y\Rightarrow x^{1/3}=\frac{y}{y-1}\Rightarrow x=\frac{y^3}{(y-1{)}^3}$
So, the range is $R-\left\{1\right\}$ which is not equal to codomain. Hence, $f$ is not an onto function.