Question:
Let $g(x)=1+x-[x]$ and $f(x)=\left\{\begin{array}{ll}-1, & x<0_{\square} \\ 0, & x=0, \\ 1, & x>0\end{array}\right.$, where $[x]$ denotes the greatest integer less than or equal to $x .$ Then for all $\left.x, f(g)\right)$ is equal to
(a) $x$
(b) 1
(c) $f(x)$
(d) $g(x)$
Solution:
(b) 1
When, $-1
Then, $g(x)=1+x-[x]$
$\quad=1+x-(-1)=2+x$
$\therefore f(g(x))=1$
When, $x=0$
Then, $g(x)=1+x-[x]$
$\therefore f(g(x))=1$
When, $x>1$
Then, $g(x)=1+x-[x]$
$\therefore f(g(x))=1+x-1=x$
Therefore, for each interval f(g(x))=1