Let f: R → R be defined as


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


$f: \mathbf{R} \rightarrow \mathbf{R}$ 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 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.



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