Let $g(x)=1+x-[x]$ and $ f(x)=$ $\begin{Bmatrix} -1 & x<0, \\ 0 & x=0,\\ 1 & x>0 \end{Bmatrix}$ then for all $x, f[g(x)]$ is equal to
(a) $x$
(b) 1
(c) $f(x)$
(d) $g(x)$
(2001, 1M)
Correct Answer: 9. (b)
Solution:
since $x-[x]>0$
$f[g(x)]=1$, since $f(x)=1$ for all $x>0$
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