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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:

Last updated date: 13th Jun 2024
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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.