If $\left[ x \right]$ is an integer function and $\left\{ x \right\}=x-\left[ x \right]$ then $f\left( x \right)=\left[ x \right]+\sum\limits_{r=1}^{100}{\dfrac{\left\{ x+r \right\}}{100}}$
A. $4x$
B. $2x$
C. $4\left[ x \right]+100\left\{ x \right\}$
D. $x$

Hint: We first try to simplify the summation part where we break the fraction part with respect to the constants and the integer part. We find that the summation becomes independent of $r$. We get the rest added and find the final solution.

Complete step by step answer:
We first simplify the expression $\sum\limits_{r=1}^{100}{\dfrac{\left\{ x+r \right\}}{100}}$.
As $\dfrac{1}{100}$ is constant, we get $\sum\limits_{r=1}^{100}{\dfrac{\left\{ x+r \right\}}{100}}=\dfrac{1}{100}\sum\limits_{r=1}^{100}{\left\{ x+r \right\}}$.
As $\left\{ x \right\}=x-\left[ x \right]$, we can write $\left\{ x+r \right\}=\left( x+r \right)-\left[ x+r \right]$.
The values of $r$ are integer, therefore,
$\left\{ x+r \right\}=x+r-\left[ x \right]-r=x-\left[ x \right]$
The summation becomes independent of $r$.So,
\[\sum\limits_{r=1}^{100}{\dfrac{\left\{ x+r \right\}}{100}}=\dfrac{1}{100}\sum\limits_{r=1}^{100}{\left\{ x+r \right\}} \\
\Rightarrow \sum\limits_{r=1}^{100}{\dfrac{\left\{ x+r \right\}}{100}}=\dfrac{100\left( x-\left[ x \right] \right)}{100}=x-\left[ x \right]\]
So, $f\left( x \right)=\left[ x \right]+\sum\limits_{r=1}^{100}{\dfrac{\left\{ x+r \right\}}{100}}$
$\Rightarrow f\left( x \right)=\left[ x \right]+x-\left[ x \right] \\
\therefore f\left( x \right)=x$

Hence, the correct option is D.

Note: The addition of integer for the box function omits the integer as the integral part of the given number is equal to itself. Therefore, we can always take away the maximum possible integer from the main number to find the solution.