Question

Which of the following functions are aperiodic ($\left[x\right]$ denotes greatest integer function.)

• A. $f\left(x\right)=2+(-1{)}^{\left[x\right]}$
• B. $f\left(x\right)=\frac{1}{x}\cos x$
• C. $f\left(x\right)=\tan \left(\frac{\pi }{4}(x-\left[x\right])\right)$
• D. $f\left(x\right)=x+\sin x$

Answer 1

Answer: (B), (D)

The correct options are
B $f\left(x\right)=x+\sin x$
D $f\left(x\right)=\frac{1}{x}\cos x$

We know that if $f\left(x\right)$ is a periodic then $f(x+T)=f\left(x\right)$.

Consider, $f\left(x\right)=2+(-1{)}^{\left[x\right]}$

Let $f\left(x\right)$ is periodic,
$f(x+T)=f\left(x\right)$
$\Rightarrow 2+(-1{)}^{[x+T]}=2+(-1{)}^{\left[x\right]}$
$\Rightarrow (-1{)}^{[x+T]}=(-1{)}^{\left[x\right]}$
$\Rightarrow [x+T]=\left[x\right]$
Also, we know that $\frac{1}{x}$ is aperiodic, so $f\left(x\right)$ is aperiodic function.
Thus, $f\left(x\right)=\frac{1}{x}\cos x$ is aperiodic function.

Consider, $f\left(x\right)=\tan \left(\frac{\pi }{4}\right(x-\left[x\right]\left)\right)$
For periodic function, we have

$f(x+T)=f\left(x\right)$
$\Rightarrow \tan \left(\frac{\pi }{4}\right(x+T-[x+T]\left)\right)=\tan \left(\frac{\pi }{4}\right(x-\left[x\right]\left)\right)$
$\Rightarrow x+T-[x+T]-(x-[x\left]\right)=2n$
$\Rightarrow T-[x+T]+\left[x\right]=2n$
If $T$ is an integer, then only above equation can give a independent value of $x.$
$\Rightarrow T-\left[x\right]+T+\left[x\right]=2n$
$\Rightarrow T=n$
So, for smallest value of $n,$ period of $f\left(x\right)$ is $1$
Now consider, $f\left(x\right)=x+\sin x$
Let $f\left(x\right)$ is periodic function,
$f(x+T)=f\left(x\right)$
$\Rightarrow x+T+\sin (x+T)=x+\sin x$
$\Rightarrow -T=\sin (x+T)-\sin \left(x\right)$
$\Rightarrow -T=\cos (x+\frac{T}{2})\sin \frac{T}{2}$
Here, $T$ is not independent of $x,$ hence $f\left(x\right)$ is aperiodic.