Question

Let $\left[x\right]$ denote the integral part of $x\in R,g\left(x\right)=x-\left[x\right]$. Let $f\left(x\right)$ be any continuous function with $f\left(0\right)=f\left(1\right)$, then the function $h\left(x\right)=f\left(g\right(x\left)\right)$:

• A. has finitely many discontinuities
• B. is discontinuous at some $x=c$, $c\in I$
  
• C. is continuous on $R$
• D. is a constant function

Answer 1

Answer: (C)

is continuous on $R$

Let $c$ be an integer,
${lim}_{x\to c^{+}}f\left(g\right(x\left)\right)={lim}_{x\to c^{+}}f(x-[x\left]\right)={lim}_{h\to 0}f(c+h-[c+h\left]\right)=f\left(0\right)=f\left(1\right)$
${lim}_{x\to c^{-}}f\left(g\right(x\left)\right)={lim}_{x\to c^{-}}f(x-[x\left]\right)={lim}_{h\to 0}f(c-h-[c-h\left]\right)=f\left(1\right)=f\left(0\right)$
$f\left(g\right(c\left)\right)=f(c-[c\left]\right)=f\left(0\right)=f\left(1\right)$
$\therefore f\left(g\right(x\left)\right)$ is continuous on $I$
Lets assume $c$ is not an integer (say $c=2.5$)
${lim}_{x\to {2.5}^{+}}f\left(g\right(x\left)\right)={lim}_{x\to {2.5}^{+}}f(x-[x\left]\right)={lim}_{h\to 0}f(2.5+h-[2.5+h\left]\right)=f\left(0.5\right)$
${lim}_{x\to {2.5}^{-}}f\left(g\right(x\left)\right)={lim}_{x\to {2.5}^{-}}f(x-[x\left]\right)={lim}_{h\to 0}f(2.5-h-[2.5-h\left]\right)=f\left(0.5\right)$
$f\left(g\right(2.5\left)\right)=f(2.5-[2.5\left]\right)=f\left(0.5\right)$
$\therefore f\left(g\right(x\left)\right)$ is clearly a continuous function on $R$, but we are not able to say that its a constant function.