Question

Which of the following functions are periodic?

• A. $f\left(x\right)=\log \left(x\right),x>0$
• B. $f\left(x\right)=e^x,x\in R$
• C. $f\left(x\right)=x–\left[x\right],x\in R$
• D. $f\left(x\right)=x+\left[x\right],x\in R$

Answer 1

The correct option is C

$f\left(x\right)=x–\left[x\right],x\in R$

A periodic function is a function that repeats itself at regular intervals.

Explanation for the correct option:

$\mathrm{Option}\left(\mathrm{C}\right):f\left(x\right)=x–\left[x\right],x\in R$

$\left[x\right]$ is the greatest integer less than or equal to the real number $x$.

$\left[x\right]=x-\left\{x\right\}$ where $\left\{x\right\}$ is the fractional part.

Now putting $x+1$ in place $x$ in the given function.

$f\left(x+1\right)=x+1–\left[x+1\right],x\in R$

$\Rightarrow f\left(x+1\right)=x+1–\left(\left[x\right]+1\right)$

$\Rightarrow f\left(x+1\right)=x+1–\left[x\right]-1$

$\Rightarrow f\left(x+1\right)=x–\left[x\right]$

$\Rightarrow f\left(x+1\right)=f\left(x\right)$

So the function is periodic. The period of the function is $1$.

Hence, the correct option is $C)f\left(x\right)=x–\left[x\right],x\in R$.

Explanation for the incorrect options:

$A)f\left(x\right)=\log \left(x\right),x>0$

Putting $x=x+1$ in the function:

$f\left(x+1\right)=\log \left(x+1\right)$

$\Rightarrow f\left(x+1\right)\ne f\left(x\right)$

Hence, this function is not periodic.

$B)f\left(x\right)=e^x,x\in R$

Putting $x=x+1$ in the function:

$f\left(x+1\right)=e^{x+1}$

$\Rightarrow f\left(x+1\right)=e·e^x$

$\Rightarrow f\left(x+1\right)=e·f\left(x\right)$

$\Rightarrow f\left(x+1\right)\ne f\left(x\right)$

Hence, this function is not periodic.

$D)f\left(x\right)=x+\left[x\right],x\in R$

Now putting $x+1$ in place $x$ in the given function.

$f\left(x+1\right)=x+1+\left[x+1\right],x\in R$

$\Rightarrow f\left(x+1\right)=x+1+\left(\left[x\right]+1\right)$

$\Rightarrow f\left(x+1\right)=x+1+\left[x\right]+1$

$\Rightarrow f\left(x+1\right)=x+\left[x\right]+2$

$\Rightarrow f\left(x+1\right)\ne f\left(x\right)$

Hence, this function is not periodic.

Hence, the correct option is $C)f\left(x\right)=x–\left[x\right],x\in R$.