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

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

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

If $f(x) = (\frac{3}{5})^x + (\frac{4}{5})^x - 1$,$x \in R$,then the equation $f(x) = 0$ has

Let $A = \{0, 1, 2, 3, 4, 5, 6, 7\}$. Then the number of bijective functions $f: A \rightarrow A$ such that $f(1) + f(2) = 3 - f(3)$ is equal to $.....$

Let $f, g: N - \{1\} \rightarrow N$ be functions defined by $f(a) = \alpha$,where $\alpha$ is the maximum of the powers of those primes $p$ such that $p^{\alpha}$ divides $a$,and $g(a) = a + 1$,for all $a \in N - \{1\}$. Then,the function $f + g$ is.