Question

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

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

Answer 1

Answer: (A), (C)

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

Step 1. Explanation for the correct option.

Option A. $f\left(x\right)=\frac{1}{x}\cos x$

We know that $\frac{1}{x}$ and $\cos x$ is periodic, so $f\left(x\right)$ is a periodic function.

Thus, $f\left(x\right)=\frac{1}{x}\cos x$ is a periodic function.

Also, $f\left(x\right)$ is a periodic function if $f\left(x+T\right)=f\left(x\right)$.

So, Option A. is the correct option

Option C. $f\left(x\right)=2+{\left(-1\right)}^{\left[x\right]}$

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

Now,

$$\begin{gathered} 2+{\left(-1\right)}^{\left[x+T\right]}=2+{\left(-1\right)}^{\left[x\right]} \\ \Rightarrow {\left(-1\right)}^{\left[x+T\right]}={\left(-1\right)}^{\left[x\right]} \\ \therefore \left[x+T\right]=\left[x\right] \end{gathered}$$

So, this is a periodic function.

So, Option C is the correct option

Step 2. Explanation for the incorrect options.

Option B. $f\left(x\right)=x+\sin x$

$f\left(x+T\right)=x+T+\sin \left(x+T\right)$

Now,

$$\begin{gathered} x+T+\sin \left(x+T\right)=x+\sin x \\ \Rightarrow -T=\sin \left(x+T\right)-\sin x \\ \Rightarrow -T=2\cos \left(x+\frac{T}{2}\right)\sin \frac{T}{2}\left[\because \sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)\right] \end{gathered}$$

Here $T$ is dependent of $x$.

Therefore, $f\left(x\right)=x+\sin x$ is not a periodic function.

Hence, the correct options are A and C.