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# If $g(x)=1+x-[x]$ where $[x]$ is greatest integer $x$ and $f(x)=\left\{\begin{array}{111}-1,\:\:\:\:x<0\\0,\:\:\:\:\:x=0\\1,\:\:\:\:\:x>0\end{array}\right.$ then $fog(x)=?$

$\begin{array}{1 1} -1 \\ 0 \\ 1 \\ f(x) \end{array}$

Toolbox:
• $x-[x]$ is fractional function =a, where $a\in [0,1)$
Let,$g(x)=b$ where $b\in [1,2)$
$fog(x)=f(b) = 1$ since b is >0