If $f: R \rightarrow R$ is defined by $f(x) = x^2 + 3x + 4$,then the function $f$ is . . . . . . .

The function $f: C \rightarrow C$ defined by $f(x) = \frac{ax + b}{cx + d}$ for $x \in C$, where $bd \neq 0$, reduces to a constant function if:

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} x+4 & \text{for } x < -4 \\ 3x+2 & \text{for } -4 \leq x < 4 \\ x-4 & \text{for } x \geq 4 \end{cases}$ then the correct matching of List-$I$ from List-$II$ is:

List-$I$	List-$II$
$(A)$ $f(-5) + f(-4)$	$(i)$ $14$
$(B)$ $f(|f(-8)|)$	$(ii)$ $4$
$(C)$ $f(f(-7) + f(3))$	$(iii)$ $-11$
$(D)$ $f(f(f(f(0)))) + 1$	$(iv)$ $-1$
	$(v)$ $1$
	$(vi)$ $0$

Let $f(x) = \begin{cases} -a & \text{if } -a \leq x \leq 0 \\ x+a & \text{if } 0 < x \leq a \end{cases}$ where $a > 0$ and $g(x) = \frac{f(|x|) - |f(x)|}{2}$. Then the function $g: [-a, a] \rightarrow [-a, a]$ is

The function $f: N \rightarrow N$ defined by $f(x) = \begin{cases} x+1, & x \text{ is odd} \\ x-1, & x \text{ is even} \end{cases}$ is . . . . . . .

Show that the function $f : R \rightarrow \{ x \in R : -1 < x < 1 \}$ defined by $f(x) = \frac{x}{1+|x|}$ for all $x \in R$ is a one-one and onto function.