For each $n \in \mathbf{N}$, let $A_n=\{(n+1) k / k \in \mathbf{N}\}$ and $X=\bigcup_{n \in \mathbf{N}} A_n \cdot A$ mapping $f: X \rightarrow N$ defined by $f(x)=x$, $\forall x \in \mathbf{X}$, is

If $f:[-3,2] \rightarrow[0, \sqrt[3]{x}]$ is an onto function defined by $f(n)=\left\{\begin{array}{cc}2+\sqrt[3]{n}, & -3 \leq n \leq-1 \\ n^{2 / 3}, & -1 \leq n \leq 2\end{array}\right.$, then $x=$

Let $[x]$ denote the greatest integer not more than $x$. If $A$ and $B$ are the domains of the functions $f(x)=\frac{x-[x]}{\sqrt{|x|-x}}$ and $g(x)=\frac{x-[x]}{\sqrt{|x|+x}}$ respectively, then

If $\operatorname{sech}^{-1}(1 / 2)-\operatorname{cosech}^{-1}(3 / 4)=\log _e k$, then