Question

Let $f:R\to R$ be defined as $f\left(x\right)=x^4$.
Choose the correct answer.

• A. $f$ is one-one onto
• B. $f$ is many-one onto
• C. $f$ is one-one but not onto
• D. $f$ is neither one-one nor onto

Answer 1

Answer: (D)

$f$ is neither one-one nor onto

$f:R\to R$ is defined as $f\left(x\right)=x^4$.

Let $x,y\in R$ such that $f\left(x\right)=f\left(y\right)$
$\Rightarrow x^4=y^4$
$\Rightarrow x=\pm y$
$\therefore f\left(x_1\right)=f\left(x_2\right)$ does not imply that $x_1=x_2$
For instance $f\left(1\right)=f(-1)=1$
$\therefore f$ is not one-one.

Consider an element $-2$ in co-domain $R$.

It is clear that there does not exist any $x$ in domain $R$ such that $f\left(x\right)=-2$

$\therefore f$ is not onto.
Hence, function $f$ is neither one-one nor onto.
The correct answer is $D$.