Let g(x) = 1 + x - [x] and f(x) = begin{cases | vedclass

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(B) Given $g(x) = 1 + x - [x]$.We know that the fractional part function is defined as $\{x\} = x - [x]$.Therefore,$g(x) = 1 + \{x\}$.Since $0 \le \{x

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(B) Given $g(x) = 1 + x - [x]$.We know that the fractional part function is defined as $\{x\} = x - [x]$.Therefore,$g(x) = 1 + \{x\}$.Since $0 \le \{x\} Thus,$1 \le g(x) Now,we evaluate $f(g(x))$. Since $g(x) > 0$ for all real values of $x$,we look at the definition of $f(x)$ for $x > 0$.According to the definition,$f(x) = 1$ for all $x > 0$.Since $g(x)$ is always in the interval $[1, 2)$,which is strictly greater than $0$,$f(g(x)) = 1$ for all $x$.

Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \\ 1, & \text{if } x > 0 \end{cases}$. Then for all values of $x$,the value of $f(g(x))$ is:

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