$f(x)= \begin{cases} -1+\sin(k_1\pi x) & x\;\text{is rational} \\ 1+\cos(k_2 \pi x) & x\;\text{is irrational} \end{cases}$

If $f(x)$ is periodic function, then

(A) Either $k_1, k_2\in\text{rational} \;\text{or}\;k_1, k_2\in\text{irrational}$

(B) $k_1, k_2\in\text{rational only}$

(C) $k_1, k_2\in\text{irrational only}$

(D) $k_1, k_2\in\text{irrational such that }\dfrac{k_1}{k_2}\; \text{rational}$

My Approach:

Case $(1)$: Assuming Period $T_1$ to be rational

Let $x$ is rational

Then $f(x)=f\left(x+T_1\right)\implies -1+\sin(k_1\pi x)=-1+\sin\left(k_1 \pi(x+T_1)\right)$

$\implies \sin(k_1\pi x)=\sin\left(k_1 \pi(x+T_1)\right)$

$\implies k_1\pi (x+T_1)=n\pi+k_1\pi x\implies k_1(x+T_1)=n+k_1x\implies k_1=\dfrac{n}{T_1}$

Hence $k_1$ should be rational.


Assuming Period $T_1$ to be irrational.

Let $x$ is rational and using the same steps as Case$(1)$ I got $k_1=\dfrac{n}{T_1}$

Hence $k_1$ should be irrational.

I am not getting any option and correct answer given is option (B)

Also, Did I make any error in Case $(2)$?

  • $\begingroup$ You have given Multiple Choices , I assume it is some test where quick thinking can save time. Intuitively : We can easily get the Correct Answer (B) : All other Choices allow irrational numbers. Intuitively , we can see that $ k_1 \in \{ e^e , \sqrt{2} , \log(2) \} $ & $ k_2 \in \{ e^\pi , \sqrt{3} , \pi^{\pi} \} $ will give some weird function which can not have a period !! $\endgroup$
    – Prem
    Commented May 1, 2023 at 10:49

1 Answer 1


What you have shown are the following:

  • If $T_1$ is rational, then $k_1$ is rational.
  • If $T_1$ is irrational, then $k_1$ is irrational.

You do not know if $T_1$ can be either rational or irrational.

In fact, you can show that $T_1$ cannot be irrational. Indeed, if $T_1$ is the period of $f$, then we have $f(0) = f(T_1)$. However, if $T_1$ were to be irrational, then by definition of $f$, we would get $f(0) = -1 + \sin(0) = -1$ and $f(T_1) = 1 + \cos(k_2 T_1 \pi) \in [0,2]$. Therefore, they cannot be equal.

  • $\begingroup$ Other than what you suggested Did i do everything correct? $\endgroup$
    – mathophile
    Commented May 28, 2023 at 14:08
  • $\begingroup$ @mathophile Yes, I think so. $\endgroup$
    – VTand
    Commented May 29, 2023 at 3:10

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