Question

Let $g\left(x\right)=1+x-\left[x\right]\mathrm{and}f\left(x\right)=\left\{\begin{array}{ll}-1,& x<0\boxed{}\\ 0,& x=0,\\ 1,& x>0\end{array}\right.$, where [x] denotes the greatest integer less than or equal to x. Then for all $x,f\left(g\left(x\right)\right)$ is equal to
(a) x
(b) 1
(c) f(x)
(d) g(x)

Answer 1

(b) 1

$$\begin{gathered} \mathrm{When},-1<x<0 \\ \mathrm{Then},g\left(x\right)=1+x-\left[x\right] \\ =1+x-\left(-1\right)=2+x \\ \therefore f\left(g\left(x\right)\right)=1 \\ \mathrm{When},x=0 \\ \mathrm{Then},g\left(x\right)=1+x-\left[x\right] \\ =1+x-0=1+x \\ \therefore f\left(g\left(x\right)\right)=1 \\ \mathrm{When},x>1 \\ \mathrm{Then},g\left(x\right)=1+x-\left[x\right] \\ =1+x-1=x \\ \therefore f\left(g\left(x\right)\right)=1 \end{gathered}$$

Therefore, for each interval f(g(x))=1