Let $f : R \to R$ be defined by $f(x) = \frac{|x| - 1}{|x| + 1}$. Then $f$ is

Let $R$ be the set of all real numbers. Statement $I$: The function $f: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R$ defined by $f(x) = \sec x + \tan x$ is a one-one function. Statement $II$: The function $f: [0, \infty) \rightarrow R$ defined by $f(x) = x^2$ is a one-one function. Which of the above statements is(are) true?

Let $f: N \rightarrow N$ be defined by $f(n) = \begin{cases} \frac{n+1}{2}; & \text{if } n \text{ is odd} \\ \frac{n}{2}; & \text{if } n \text{ is even} \end{cases}$. Then $f$ is:

The function $f:[0, \infty) \rightarrow [0, \infty)$ defined by $f(x) = \frac{x}{1+x}$ is

The function $f: R \rightarrow R$ is defined by $f(x)=3^{-x}$. Observe the following statements about it:
$I$. $f$ is one-one
$II$. $f$ is onto
$III$. $f$ is a decreasing function
Out of these, the true statements are:

The real-valued function $f: R \rightarrow [ \frac{5}{2}, \infty )$ defined by $f(x) = | 2x + 1 | + | x - 2 |$ is