Let g(x)=1+x−[x] and f(x)


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) $g(x)$


(b) 1

When, $-1

Then, $g(x)=1+x-[x]$


$\therefore f(g(x))=1$

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

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