Question

Which of the following function is periodic?

• A. $f(x)=x-[x]$, where $[x]$ denotes the greatest integer less than or equal to the real number $x$
• B. f(x)=\sin \frac{1}{x} \text { for } x \neq 0, f(0)=0
• C. $f(x)=x \cos x$
• D. None of these

Answer 1

Answer: (A)

Clearly, $f(x)=x-[x]=\{x\}$
which has period $1$ .
Let $\sin \left(\frac{1}{x}\right)$ be periodic with period $T$.
Then, $\sin \left(\frac{1}{x+T}\right)=\sin \left(\frac{1}{x}\right)$
$\Rightarrow \frac{1}{x+T}=n \pi+(-1)^{n} \frac{1}{x} $
$\Rightarrow x=\left(x^{2}+T x\right) n \pi+(-1)^{n}(x+T)$
$\Rightarrow T=\frac{x\left(1-(-1)^{n}\right)-x^{2} n \pi}{n \pi x+(-1)^{n}}$
Now for a variable $x$ and constant $T$, the given relation cannot hold $\forall$ allowable $x$.
Hence, $\sin \frac{1}{x}$ is not periodic.
Similarly, for $f(x)=x \cos x$, let $T$ be the period.
Hence, $(x+T) \cos (x+T)=x \cos x$
$\Rightarrow T \cos (x+T)=x\{\cos x-\cos (x+T)\}$
$\Rightarrow T \cos (x+T)=2 x \sin \left(x+\frac{T}{2}\right) \sin \frac{T}{2}$
$\Rightarrow \frac{T}{\sin \frac{T}{2}}=\frac{2 x \sin \left(x+\frac{T}{2}\right)}{\cos (x+T)}$
Note that $L H S$ is a constant while $R H S$ varies as $x$ varies for allowable values of $x$.
Hence, no such $T$ is possible, so $x \cos x$ is also non-periodic.