If $f: Z \rightarrow Z$, $f(x) = \begin{cases} \frac{x}{2}, & \text{if } x \text{ is even} \\ 0, & \text{if } x \text{ is odd} \end{cases}$, then $f$ is

Show that the function $f: N \rightarrow N$,given by $f(1)=f(2)=1$ and $f(x)=x-1$ for every $x>2$,is onto but not one-one.

The number of bijective functions $f: Z \rightarrow Z$ such that $f(x+y)=f(x)+f(y)$ for all $x, y \in Z$ is:

If $f(x) = \begin{cases} x, & \text{when } x \text{ is rational} \\ 0, & \text{when } x \text{ is irrational} \end{cases}$ and $g(x) = \begin{cases} 0, & \text{when } x \text{ is rational} \\ x, & \text{when } x \text{ is irrational} \end{cases}$,then $(f - g)$ is:

The number of functions $f$ from the set $A = \{x \in N: x^{2}-10x+9 \leq 0\}$ to the set $B = \{n^{2}: n \in N\}$ such that $f(x) \leq (x-3)^{2}+1$ for every $x \in A$ is:

Let the functions $f$ and $g$ be $f: [0, \frac{\pi}{2}] \rightarrow R$ given by $f(x) = \sin x$ and $g: [0, \frac{\pi}{2}] \rightarrow R$ given by $g(x) = \cos x$,where $R$ is the set of real numbers. Consider the following statements:
Statement $(I)$: $f$ and $g$ are one-one.
Statement $(II)$: $f+g$ is one-one.
Which of the following is correct?