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If $f(x) = {2^{{{\sin }^3}\pi x + x - [x]}}$, where $[x]$ denotes the integral part of $x$ is a periodic function with period:

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Hint: A function that repeats itself after every constant interval of time is known as the periodic function, and the concept of modelling the periodic function using trigonometric identities are known as modelling of periodic behaviour functions. In this question, we need to determine the time period of the function which could be segregated into two defined functions. The net time period of the function will be the LCM of the time period of each individual function.

Complete step by step solution: The given function $f(x) = {2^{{{\sin }^3}\pi x + x - [x]}}$ can be segregated into two parts such as $f(x) = {2^{{{\sin }^3}\pi x}} \times {2^{x - [x]}}$ so, we need to determine the time period of two functions, i.e., ${\sin ^3}\pi x$ and $x - [x]$ such that the overall time period will be the LCM of the individual time periods of the functions.
Now, calculating the time period for the function ${\sin ^3}\pi x$ by the concept of trigonometric identity $\sin (2\pi + \theta ) = \sin \theta $ as:
$
  {\sin ^3}\pi x = {\sin ^3}\pi (2 + x) \\
   = {\sin ^3}(2\pi + \pi x) \\
   = {\sin ^3}\pi x \\
 $ So, the time period of ${\sin ^3}\pi x$ is 2.
Again, calculating the time period of the function $x - [x]$ by using the concept of an integral part $[x + 1] = [x] + [1] = [x] + 1$ as:
$
  x - [x] = x + 1 - [x + 1] \\
   = x + 1 - [x] - 1 \\
   = x - [x] \\
 $ So, the time period of $x - [x]$ is 1.
Now, the overall time period of the given function $f(x) = {2^{{{\sin }^3}\pi x + x - [x]}}$ is the LCM of the time period of the functions ${\sin ^3}\pi x$ and $x - [x]$ which is 2 and 1, respectively.
$
  T = LCM(2,1) \\
   = 2 \\
 $ Hence, the time period of the function $f(x) = {2^{{{\sin }^3}\pi x + x - [x]}}$ is 2.

Note: One complete pattern is known as a cycle. The horizontal length of the cycle is known as the period of the function. Trigonometric identities should be well known by the candidates to solve this type of question.