Question

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

Answer: (B)

$1$

Let $g\left(x\right)=1+x-\left[x\right]=1+\left\{x\right\}>0$

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

Hence $f\left\{g\right(x\left)\right\}=1$