If $f: N \rightarrow R$ is defined by $f(1)=-1$ and $f(n+1)=3f(n)+2$ for $n \geq 1$, then $f$ is

Let $f :(0,1) \rightarrow R$ be a function defined by $f(x)=\frac{1}{1-e^{-x}}$,and $g(x)=(f(-x)-f(x))$. Consider two statements:
$(I)$ $g$ is an increasing function in $(0,1)$
$(II)$ $g$ is one-one in $(0,1)$
Then,

Consider a function $f: [0, \frac{\pi}{2}] \rightarrow \mathbb{R}$ given by $f(x) = \sin x$ and $g: [0, \frac{\pi}{2}] \rightarrow \mathbb{R}$ given by $g(x) = \cos x$. Show that $f$ and $g$ are one-one,but $f + g$ is not one-one.

In each of the following cases,state whether the function is one-one,onto or bijective. Justify your answer. $f : R \rightarrow R$ defined by $f(x) = 1 + x^2$.

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

The function $f : R \rightarrow (-1, 1)$ defined by $f(x) = \frac{e^x - 1}{e^x + 1}$ is: