Question

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

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

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Finding the value of $f\left(g\left(x\right)\right)$:

Given that,

$g\left(x\right)=1+x-\left[x\right]$,

Any whole number can be written as

$x=\left[x\right]+\left\{x\right\}$

$\Rightarrow x-\left[x\right]=\left\{x\right\}...\left(1\right)$

$\begin{array}{rcl}f\left(g\left(x\right)\right)& =& \left\{\begin{array}{l}-1,g\left(x\right)<0\\ 0,g\left(x\right)=0\\ 1,g\left(x\right)>0\end{array}\right.\\ & =& \left\{\begin{array}{l}-1,1+x-\left[x\right]<0\\ 0,1+x-\left[x\right]=0\\ 1,1+x-\left[x\right]>0\end{array}\right.\\ & =& \left\{\begin{array}{l}-1,1+\left\{x\right\}<0\\ 0,1+\left\{x\right\}=0\\ 1,1+\left\{x\right\}>0\end{array}\right.\left[\mathrm{from}1\right]\end{array}$

We know that,

$\left\{x\right\}\in [0,1)$

So,

$\begin{array}{rcl}f\left(g\left(x\right)\right)& =& \left\{\begin{array}{l}-1,\left\{x\right\}<-1\\ 0,\left\{x\right\}=-1\\ 1,\left\{x\right\}>-1\end{array}\right.\end{array}$

Here, only $\left\{x\right\}>-1$$\in [0,1)$ .

Therefore, the value of $f\left(g\left(x\right)\right)$ is $1$.

Hence, the correct option is B.