Question

The period of the function $3^{({\sin }^2\pi x+x-\left[x\right]+{\sin }^4\pi x)}$, where $[.]$ denote the greatest integer function, is

Answer 1

As ${\sin }^2\pi x=\frac{1-\cos 2\pi x}{2}$
Since $\cos 2\pi x$ is a periodic function with period $\frac{2\pi }{2\pi }=1$, ...(1)
therefore ${\sin }^2\pi x$ is periodic with period $1$.
And $x-\left[x\right]$ is a periodic function with period $1$ ...(2)
${\sin }^4\pi x={\left({\sin }^2\pi x\right)}^2=\frac{1}{4}{(1-\cos 2\pi x)}^2$
$=\frac{1}{4}(1+{\cos }^{2}2\pi x-2\cos 2\pi x)=\frac{1}{8}(3+\cos 4\pi x-4\cos 2\pi x)$
Since $\cos 4\pi x$ is a periodic function with period $\frac{2\pi }{4\pi }=\frac{1}{2}$
and $\cos 2\pi x$ is a periodic function with period $\frac{2\pi }{2\pi }=1$,
therefore , period of ${\sin }^4\pi x$ is equal to
$L.C.M(1,\frac{1}{2})=\frac{L.C.M(1,1)}{H.C.F(1,2)}=\frac{1}{1}=1$ ...(3)
From (1),(2) and (3)
Period of $3^{({\sin }^2\pi x+x-\left[x\right]+{\sin }^4\pi x)}=1$