Question

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

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

Answer 1

Answer: (B)

$1$

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

$=1+\left\{x\right\}$

$\left\{x\right\}$ donates the fractional part of x

so, g(x) is always positive

so, input to $f\left(g\right(x\left)\right)$ is always positive and

Hence,

$fog\left(x\right)=1$