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

Solve for one-one.
$f\left(x\right)=x^4$
Now, $f\left(x_1\right)=(x_1{)}^4,f(x_2)=(x_2{)}^4$
For one-one,
$f\left(x_1\right)=f\left(x_2\right)$
$(x_1{)}^4=(x_2{)}^4$
$x_1=x_2$ or $x_1=-x_2$
Thus, $f\left(x_1\right)=f\left(x_2\right)$ does not only imply that $x_1=x_2$ (additionally, $x_1=-x_2$ also)
So, $f$ is not one-one.

Solve for onto.
Let $f\left(x\right)=y,$ such that $y\in R$
$x^4=y$
$x=\pm y^{\frac{1}{4}}$
Here, $y$ is a real number, it can be negative also, which is not possible as root of negative number is not real.
Hence, $x$ is not real.
$\therefore f$ is not onto.
Hence, $f$ is neither one-one nor onto,
Option D is correct.