Q4:

Show that the Modulus Function $f : R \to R$, given by $f(x) = |x|$, is neither one-one nor onto, where $|x|$ is $x$, if $x$ is positive or 0 and $|x|$ is $– x$, if $x$ is negative.

Step 1: Formal Definition of the Modulus Function

The function $f: \mathbb{R} \to \mathbb{R}$ is defined as $f(x) = |x|$, where the domain is the set of real numbers $\mathbb{R}$ and the codomain is the set of real numbers $\mathbb{R}$. The rule of assignment is defined piecewise as follows:

$f(x) = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}$

Step 2: Analysis of Injectivity (One-One)

A function $f: A \to B$ is injective (one-one) if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$. [Definition of Injective Mapping]

Consider two distinct real numbers $x_1 = 1$ and $x_2 = -1$.

• $f(1) = |1| = 1$
• $f(-1) = |-1| = 1$

Since $f(1) = f(-1) = 1$, but $1 \neq -1$, there exist distinct elements in the domain that map to the same image in the codomain. [Violation of the definition of one-one function]

Conclusion: The function $f$ is not one-one.

Step 3: Analysis of Surjectivity (Onto)

A function $f: A \to B$ is surjective (onto) if for every $y \in B$, there exists at least one $x \in A$ such that $f(x) = y$. [Definition of Surjective Mapping]

The codomain of the function is $\mathbb{R}$. Consider a negative value in the codomain, for instance, $y = -2$.

By the definition of the modulus function, $f(x) = |x| \geq 0$ for all $x \in \mathbb{R}$. There is no real number $x$ such that $|x| = -2$, because the absolute value of any real number is non-negative by definition. [Property of Absolute Value]

Conclusion: Since there exists at least one element (all negative real numbers) in the codomain that has no pre-image in the domain, the function $f$ is not onto.

Step 4: Graphical Representation

Step 5: Theoretical Synthesis

Final Solution: The function $f(x) = |x|$ is neither one-one nor onto, as it fails the horizontal line test (multiple domain elements map to one codomain element) and the range does not cover the negative subset of the codomain.