#### Need solution for RD Sharma maths class 12 chapter Functions exercise 2.1 question 5 sub question (v)

Neither an injection nor a surjection.

Given:

$f:R \rightarrow R$, defined by $f(x)=\left | x \right |$.

Hint:

Injective  function means every element in the domain has a distinct image in the  co-domain.

Surjective

function means every element in the co-domain has at least one  pre image in the domain of function.

Bijection ⇒ Function should fulfill the injection, surjection condition.

Solution:

Let us check if the given function is injective, surjective, or bijective.

Injection:

Let x and y be any two elements in the domain (R) such that$f(x)=f(y)$

\begin{aligned} &f(x)=|x|, f(y)=|y| \\ &x=\pm y \end{aligned}

Therefore f is not an injection.

Surjection test:

Let y be any element in the codomain (R) , such that$f(x)=y$  for some element x in R .

\begin{aligned} &f(x)=y \\ &|x|=y \\ &x=\pm y \in Z \end{aligned}

Therefore, f  is surjective.

Since f  is not an injective , but surjective. So the f  is not bijective.