Question

Let \( g(x) = 1 + x - \lfloor x \rfloor \) and \[ f(x) = \begin{cases} -1, & x<0\\ 0, & x = 0 \\ 1, & x>0 \end{cases} \] where \( \lfloor x \rfloor \) denotes the greatest integer less than or equal to \( x \). Then for all \( x \), \( f(g(x)) = \)

• A. \( 1 \)
• B. \( x \)
• C. \( f(x) \)
• D. \( g(x) \)

Hint:

For expressions involving greatest integer \( \lfloor x \rfloor \), note that \( x - \lfloor x \rfloor \) always lies in \( [0, 1) \). Use this to bound the function output.

Answer 1

The Correct Option is A

Step 1: Consider \( g(x) = 1 + x - \lfloor x \rfloor \). This is always a number in the interval \( (1, 2) \) for any real \( x \), since \( x - \lfloor x \rfloor \in [0, 1) \).
Step 2: So, \( g(x) \in (1, 2) \Rightarrow g(x)>0 \ \forall x \in \mathbb{R} \).
Step 3: Given \( f(x) \) is: \[ f(x) = \begin{cases} -1, & x<0
0, & x = 0
1, & x>0 \end{cases} \]
Step 4: Since \( g(x)>0 \Rightarrow f(g(x)) = 1 \) \[ \boxed{f(g(x)) = 1 \text{ for all } x} \]