Question

Which of the following functions is periodic

• A. $\text{sgn}\left(e^{-x}\right)$
• B. $\sin x+|\sin x|$
• C. $min.\{\sin x,|x|\}$
• D. $[x+\frac{1}{2}]+[x-\frac{1}{2}]+2[-x]$
  (where [.] denotes greatest integer function)

Answer 1

Answer: (A), (B), (C), (D)

$[x+\frac{1}{2}]+[x-\frac{1}{2}]+2[-x]$

(where [.] denotes greatest integer function)
$\left(A\right)\text{sgn}\left(e^{-x}\right)$
We know that,
$e^{-x}>0\Rightarrow \text{sgn}\left(e^{-x}\right)=1$
Constant function is periodic.

$\left(B\right)\sin x+|\sin x|$
As $\sin x$ and $|\sin x|$ both are periodic with period $2\pi$ and $\pi$ respectively, so
The period of $\sin x+|\sin x|$ is $2\pi$

$\left(C\right)min.\{\sin x,|x|\}$
As $|x|>\sin x$, so
$min.\{\sin x,|x|\}=\sin x$
Hence, this is a periodic function

$\left(D\right)[x+\frac{1}{2}]+[x-\frac{1}{2}]+2[-x]$
$=(x+\frac{1}{2})-\{x+\frac{1}{2}\}+(x-\frac{1}{2})-\{x-\frac{1}{2}\}+2(-x-\{-x\left\}\right)=-\{x+\frac{1}{2}\}-\{x-\frac{1}{2}\}-2\{-x\}$
We know that $\left\{x\right\}$ is periodic, so the function is also periodic.