Let $f:R \to R$ 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.

Option D is correct

$f:R \to R$ defined as $f(x) = {x^4}.$

Injectivity

Let ${x_1},\;{x_2} \in R,\;\;\;f({x_1}) = f({x_2})$
$\Rightarrow x_1^4 = x_2^4 \Rightarrow x_1^2 = x_2^2 \Rightarrow {x_1} = \pm {x_2}$
Therefore, f is not one-one.

Surjectivity

$f:R \to R.$ Let, $y \in R$ (co-domain of f ) be any element, then there exist $x \in R$ (domain of f) such that
$f(x) = y \Rightarrow y = {x^4} \Rightarrow x = \pm \;{y^{1/4}}$

Now, $f({y^{1/4}}) = {({y^{1/4}})^4} = y$ and $f( - {y^{1/4}}) = {( - {y^{1/4}})^4} = - y$
Therefore, f is not onto.

Thus, f is neither one-one nor onto.