Question

Let $g(x) = 1 + x - [x]$ and

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

Then for all $x \in R$, $f(g(x))$ is equal to (where, $[x]$ denotes the greatest integer less than or equal to $x$)

• A. x
• B. 1
• C. f(x)
• D. g(x)

Answer 1

Answer: (B)

1

Answer 2

To find $f(g(x))$, we first need to understand the function $g(x)$.

Step 1: Analyze $g(x)$

The function $g(x)$ is defined as $g(x) = 1 + x - [x]$, where $[x]$ denotes the greatest integer less than or equal to $x$. We know that the fractional part of $x$, denoted by $\{x\}$, is given by $\{x\} = x - [x]$. So, we can rewrite $g(x)$ as:

$g(x) = 1 + \{x\}$

Step 2: Determine the range of $g(x)$

The fractional part $\{x\}$ always satisfies the inequality:

$0 \le \{x\} < 1$

Adding 1 to all parts of the inequality, we get:

$1 + 0 \le 1 + \{x\} < 1 + 1$

$1 \le g(x) < 2$

This means that for any real number $x$, the value of $g(x)$ will always be in the interval $[1, 2)$.

Step 3: Analyze $f(x)$

The function $f(x)$ is defined as:

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

Step 4: Evaluate $f(g(x))$

From Step 2, we know that $1 \le g(x) < 2$. This implies that $g(x)$ is always strictly positive ($g(x) > 0$) for all $x \in R$. Now, we use the definition of $f(x)$ from Step 3. Since the input to $f$ (which is $g(x)$) is always greater than 0, $f(g(x))$ will always be 1.

Therefore, for all $x \in R$, $f(g(x)) = 1$.