Question

Let $g\left(x\right)=1+x-\left[x\right]$ and $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}-1& x<0\end{array}\\ \begin{array}{cc}0& x=0\end{array}\\ \begin{array}{cc}1& x>0\end{array}\end{array}$. 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 $1$

We have,
$g\left(x\right)=\{\begin{array}{c}1+n-n=1,x=n\in Z\\ 1+n+k-n=1+k,x=n+k\end{array}$
where $n\in Z,0<k<1$
Now $f\left[g\left(x\right)\right]=\{\begin{array}{c}\begin{array}{cc}-1& g\left(x\right)<0\end{array}\\ \begin{array}{cc}0& g\left(x\right)=0\end{array}\\ \begin{array}{cc}1& g\left(x\right)>1\end{array}\end{array}$
Clearly, $g\left(x\right)>0$ for all $x$. So $f\left[g\left(x\right)\right]=1$, for all $x$.