Let f : R R be defined as f ( x ) = x 4 . Choose the correct answer:

Question

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

Smart Achievers · Accepted Answer

(4)

Ans. $f$ is neither one-one nor onto

Explanation:

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

Let $x, y \in R$ such that

$f (x) = f (y)$

$\Rightarrow$ $x^{4} = y^{4}$

$\Rightarrow$ $x^{4} - y^{4} = 0$

$\Rightarrow$ $(x^{2} + y^{2}) (x^{2} - y^{2}) = 0$

$\Rightarrow$ $x^{2} = y^{2}$

$\Rightarrow$ $x = \pm y$

$∴$ $f (x) = f (y)$ does not imply that $x = y$

For instance,

$f (1) = f (- 1) = 1$

Here $f (1) = f (- 1)$ but $- 1 \neq 1$ . Hence, $f$ is not one-one.

To check onto,

$f (x) = x^{4}$

Let $f (x) = y$ such that $y \in R$

$\Rightarrow$ $x^{4} = y$

$\Rightarrow$ $x = \pm y^{1 / 4}$

Note that $y$ is a real number, but it can also be negative.

For example, put $y = - 3$

$\Rightarrow$ $x = {(\pm 3)}^{1 / 4}$

$\Rightarrow$ $x = \pm {(- \sqrt{3})}^{1 / 2}$

which is not possible, as root of a negative number is not a real i.e., $f (x) = x^{4}$ is not a onto function.

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