**Toolbox:**

- A function $f: A \rightarrow B$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one function.
- A function$ f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
- |x| is always non negative

Given $f:R \rightarrow R$ defined by $f(x)= |x| = x$ if $x \geq 0$ and $-x$ if $x <0$:

Step1: Injective or One-One function:

We see that $f(-1)= |-1| = -1$, as $-1 <0$ and $f(1) = |1| = 1$ as $1 \geq 0$.

$\Rightarrow f(-1) = f(1) \rightarrow -1 \neq 1$.

Therefore, $f:R \rightarrow R$ defined by $f(x)= |x|$ is not one-one.

Step 2: Surjective or On-to function:

Given $-1 \in R$ for $f$ to be onto, there must exist an element $x$ in R such that $f(x)=|x|=-1$

However, we know that $f(x) = |x|$ is always non-negative.

Therefore, there cannot be any element $x \in R$ such that $f(x) = |x| = -1$.

Hence $f:R \rightarrow R$ defined by $f(x)= |x|$ is not onto.

Solution: The modulus function $f:R \rightarrow R$ defined by $f(x)= |x|$ is neither one-one nor onto