If $f: R \rightarrow R$ is defined by $f(x) = x + 2|x + 1| + 2|x - 1|$, then the element in the co-domain, which has a unique pre-image in the domain is

$A$ function $f$ from the set of natural numbers $\mathbb{N}$ to the set of integers $\mathbb{Z}$ is defined by $f(n) = \begin{cases} \frac{n-1}{2}, & \text{if } n \text{ is odd} \\ -\frac{n}{2}, & \text{if } n \text{ is even} \end{cases}$. The function $f$ is:

Let $Z$ denote the set of integers. Define $f: Z \rightarrow Z$ by $f(x) = \begin{cases} \frac{x}{2}, & x \text{ is even} \\ 0, & x \text{ is odd} \end{cases}$. Then $f$ is:

Let $S = \{1, 2, 3, 4, 5, 6\}$. Then the probability that a randomly chosen onto function $g: S \to S$ satisfies $g(3) = 2g(1)$ is:

Let $f: N \to N$ be defined by $f(x) = x^2 + x + 1$,where $x \in N$. Then $f$ is:

If $f(x) = \begin{cases} x, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q} \end{cases}$ and $g(x) = \begin{cases} x, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q} \end{cases}$,then the function $(f - g)$ is: