Let $g(x)=1+x-[x]$ and $f(x)=\left\{\begin{array}{cc}-1 & x<0 \\ 0 & x=0 \\ 1 & x>1\end{array}\right.$. Then for all x, f {g(x)} is equal to :
Answer & explanation
Correct answer: option 2
$g(x) =1+x-[x]$ ($[x]=x-\{x\}$ (fractional part function))
$⇒g(x)=1+[x]$
so, $0≤\{x\}≤1$
$⇒g(x)≥1$
so $f(g(x))=1$ as $g(x)≥1$
for input ≥ 1, $f = 1$