Let $g(x)=1+x-[x]$ and ${ }^{\prime}$

$$ f(x)= \begin{cases}-1, & x<0 \\ 0, & x=0,[x] \text { denotes the greatest integer less } \\ 1, & x>0\end{cases} $$

than or equal to $x$. Then for all $x, f(g(x))=$

The remainder obtained when $(2 m+1)^{2 n}(m, n \in N)$ is divided by 8 is

A value of $\theta$ lying between 0 and $\pi / 2$ and satisfying $\left|\begin{array}{ccc}1+\sin ^2 \theta & \cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & 1+\cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta\end{array}\right|=0$ is

If the system of equations $2 x+p y+6 z=8$, $x+2 y+q z=5$ and $x+y+3 z=4$ has infinitely many solutions, then $p=$