The function $f: N-\{1\} \rightarrow N$ defined by $f(n) = \text{the highest prime factor of } n$ is:

The function $f: R \to R$ defined by $f(x) = x^2$ for all $x \in R$ is

If $f(x) = \sin([\pi^2]x) - \sin([-\pi^2]x)$,where $[x]$ denotes the greatest integer function $\leq x$,then which of the following is not true?

Let $f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right) & \text{when } x \neq 0 \\ 1 & \text{when } x = 0 \end{cases}$ and $A = \{x \in \mathbb{R} : f(x) = 1\}$. Then,$A$ has

Consider the identity function $I_{N}: N \rightarrow N$ defined as $I_{N}(x) = x$ for all $x \in N$. Show that although $I_{N}$ is onto,the function $I_{N} + I_{N}: N \rightarrow N$ defined as $(I_{N} + I_{N})(x) = I_{N}(x) + I_{N}(x) = x + x = 2x$ is not onto.

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