Let g(x)=1+x−[x] and f(x)=−1,x 0 then for all x,f(g(x)) is equal to

g(x)=1+n−n=1,x=n∈zf(g(x))=−1g(x) 0

Let g(x)=1+x−[x] and f(x)=−1,x 0 then for all x,f(g(x)) is equal to

g(x)=1+n−n=1,x=n∈zf(g(x))=−1g(x) 0

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$Let g (x) = 1 + x - [x] and$ $f (x) = \{\begin{array}{ccc} - 1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{array}$

then for all $x, f (g (x))$ is equal to

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$f (x)$

$g (x)$

answer is B.

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Detailed Solution

$g (x) = 1 + n - n = 1, x = n \in z$

$f (g (x)) = \{\begin{cases} - 1 & \begin{matrix} g (x) < & 0 \end{matrix} \\ 0 & \begin{matrix} g (x) = & 0 \end{matrix} \\ 1 & \begin{matrix} g (x) > & 0 \end{matrix} \end{cases}$

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