$f(x) = x + \sqrt{x^2}$ is a function from $R \to R$,then $f(x)$ is

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

Show that the function $f: R_* \rightarrow R_*$ defined by $f(x) = \frac{1}{x}$ is one-one and onto,where $R_*$ is the set of all non-zero real numbers. Is the result true if the domain $R_*$ is replaced by $N$ with the co-domain remaining the same as $R_*$?

The mapping $f: R \to R$ defined as $f(x) = \cos x, x \in R$ is:

If the function $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} 2x-3, & \text{if } x < -2 \\ x^2-1, & \text{if } -2 \leq x \leq 2 \\ 3x+2, & \text{if } x > 2 \end{cases}$ then $f$ is

Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases}$. Then for all $x$,$f(g(x))$ is equal to