Question

Let $\mathrm{g}\left(\mathrm{x}\right)=1+\mathrm{x}–\left[\mathrm{x}\right]$and $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}-1\mathrm{x}<0\\ 0\mathrm{x}=0\\ 1\mathrm{x}>0\end{array}\right.$Then for all$\mathrm{x}$, $\mathrm{f}\left(\mathrm{g}\right(\mathrm{x}\left)\right)$ is equal to

• A. $\mathrm{x}$
• B. $1$
• C. $\mathrm{f}\left(\mathrm{x}\right)$
• D. $\mathrm{g}\left(\mathrm{x}\right)$

Answer 1

The correct option is B

$1$

Determine the value of $\mathrm{f}\left(\mathrm{g}\right(\mathrm{x}\left)\right)$

We know that:

$$\begin{gathered} \mathrm{x}=\left[\mathrm{x}\right]+\left\{\mathrm{x}\right\} \\ \Rightarrow \mathrm{x}-\left[\mathrm{x}\right]=\left\{\mathrm{x}\right\}\left\{\mathrm{x}\right\}\in [0,1) \end{gathered}$$

Here $\left\{\mathrm{x}\right\}$ is the fractional part of $\mathrm{x}$. Now we have:

$$\begin{gathered} \mathrm{f}\left(\mathrm{g}\right(\mathrm{x}\left)\right)=\mathrm{f}(1+\mathrm{x}-\left[\mathrm{x}\right]) \\ =\mathrm{f}(1+\left\{\mathrm{x}\right\}) \end{gathered}$$

Here, $\left(1+\left\{\mathrm{x}\right\}\right)>0$. So, for $\mathrm{x}>0,\mathrm{f}\left(\mathrm{x}\right)=1$.

Therefore, $\mathrm{f}\left(\mathrm{g}\right(\mathrm{x}\left)\right)=1$

Hence, option B is the correct answer.