Define f: R rightarrow R by f(x) = [x] + sqrt | vedclass

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(B) Let $x = n + f$,where $n = [x]$ is an integer and $0 \leq f < 1$ is the fractional part of $x$.Then $f(x) = n + \sqrt{f}$.For any integer $n$,cons

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(B) Let $x = n + f$,where $n = [x]$ is an integer and $0 \leq f Then $f(x) = n + \sqrt{f}$.For any integer $n$,consider the limit as $x 	o n^+$. Here $x = n + h$ where $h 	o 0^+$,so $[x] = n$ and $x - [x] = h$. Thus,$\lim_{x 	o n^+} f(x) = n + \sqrt{0} = n$.For the limit as $x 	o n^-$,consider $x = n - h$ where $h 	o 0^+$. Then $[x] = n - 1$ and $x - [x] = 1 - h$. Thus,$\lim_{x 	o n^-} f(x) = (n - 1) + \sqrt{1 - 0} = n - 1 + 1 = n$.Since $\lim_{x 	o n^+} f(x) = \lim_{x 	o n^-} f(x) = f(n) = n$,the function is continuous at all integers $n \in Z$.Since the function is continuous at all integers and also continuous between any two consecutive integers (where it behaves as $n + \sqrt{x-n}$),the function $f(x)$ is continuous for all $x \in R$.

$A$. $\|x\| + \|x - 2\|$	$I$. Right hand limit does not exist at $x = 2$.
$B$. $\text{cosech } x$	$II$. Continuous only for non-zero real values of $x$.
$C$. $x - [x]$	$III$. Limit is zero for all real $x$.
$D$. $\sqrt{2 - x}$	$IV$. Continuous for all real value of $x$.
	$V$. Discontinuous at all integral values of $x$.

Define $f: R \rightarrow R$ by $f(x) = [x] + \sqrt{x - [x]}$ for $x \in R$,where $[x]$ denotes the greatest integer function. Then the set of points at which $f$ is continuous is

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