Question:
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as $f(x)=x^{4}$. 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
Solution:
$f: \mathbf{R} \rightarrow \mathbf{R}$ is defined as $f(x)=x^{4}$.
Let $x, y \in \mathbf{R}$ such that $f(x)=f(y)$.
$\Rightarrow x^{4}=y^{4}$
$\Rightarrow x=\pm y$
$\therefore f\left(x_{1}\right)=f\left(x_{2}\right)$ does not imply that $x_{1}=x_{2}$.
For instance,
$f(1)=f(-1)=1$
∴ 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(x) = 2.
∴ f is not onto.
Hence, function f is neither one-one nor onto.
The correct answer is D.