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, $\sin \left(\frac{1}{x+T}\right)=\sin \left(\frac{1}{x}\right)$
$\Rightarrow \frac{1}{x+T}=n\pi +(-1{)}^n\frac{1}{x}$
$\Rightarrow x=\left(x^2+Tx\right)n\pi +(-1{)}^n(x+T)$
$\Rightarrow T=\frac{x\left(1-(-1{)}^n\right)-x^{2}n\pi }{n\pi x+(-1{)}^n}$
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, $(x+T)\cos (x+T)=x\cos x$
$\Rightarrow T\cos (x+T)=x\{\cos x-\cos (x+T)\}$
$\Rightarrow T\cos (x+T)=2x\sin \left(x+\frac{T}{2}\right)\sin \frac{T}{2}$
$\Rightarrow \frac{T}{\sin \frac{T}{2}}=\frac{2x\sin \left(x+\frac{T}{2}\right)}{\cos (x+T)}$
Note that $LHS$ is a constant while $RHS$ varies as $x$ varies for allowable values of $x$.
Hence, no such $T$ is possible, so $x\cos x$ is also non-periodic.