If $f(x)=-|x|$, then $(f \circ f \circ f)(x) + (f \circ f \circ f)(-x) =$

If $g(f(x))=|\sin x|$ and $f(g(x))=(\sin \sqrt{x})^2$, then

Let $S, T, U$ be three non-void sets and $f: S \rightarrow T, g: T \rightarrow U$ be functions such that $g \circ f: S \rightarrow U$ is surjective. Then,

If $f: A \rightarrow B$ and $g: B \rightarrow C$ are functions such that $g \circ f: A \rightarrow C$ is onto,then a necessary condition is:

Let $f(x) = \begin{cases} x, & x < 0 \\ 1 + x^2, & x \geq 0 \end{cases}$ and $g(x) = 1 + x - [x]$,then the range of $f(g(x))$ is (where $[.]$ denotes the greatest integer function).

If $f(x)=x^2+1$ and $g(x)=\frac{1}{x}$,then the value of $f(g(g(f(x))))$ at $x=1$ is