Question

: If $$f\left(x\right)=\text{sin root}\left(\text{x}\right)$$ then period of $$f\left(x\right)$$ is
(1) $$pi$$
(2) $${\displaystyle \frac{pi}{2}}$$
(3) $$2pi$$
(4) None of these

Answer 1

Hint: In order to solve this question, we will prove that the function $$\sin \sqrt{x}$$ is not periodic which means option (4) is the correct answer. So, to solve this we first assume that the function is periodic and hence we get $$\sin \sqrt{x+T}=\sin \sqrt{x}$$ .Now substituting $$x=0$$ and $$x=T$$ we will get two equations. Dividing the two equations we will find an equation which will be a contradictory statement. Hence, we prove that the function is not periodic.

Complete step-by-step answer:
Now let us assume that the function $$\sin \sqrt{x}$$ is periodic and the period is $$T$$
As we know that if the function is periodic, then
$$f\left(x+T\right)=f\left(x\right)$$
Hence, we can say that
$$\sin \sqrt{x+T}=\sin \sqrt{x}$$
Now substituting $$x=0$$ we get
$$\sin \sqrt{T}=\sin \sqrt{0}$$
We know that $$\sin 0=0$$
Therefore, we get $$\sin \sqrt{T}=0---\left(1\right)$$
Now we know that if $$\sin x=0$$ then $$x=2n\pi$$
Hence, we get $$\sqrt{T}=2n\pi ---\left(2\right)$$
Now again consider $$\sin \sqrt{x+T}=\sin \sqrt{x}$$
Now let us substitute $$x=T$$
Hence, we get
$$\sin \sqrt{T+T}=\sin \sqrt{T}$$
From equation $$\left(1\right)$$ we have $$\sin \sqrt{T}=0$$
Therefore, on substituting the value in the above equation we get
$$\sin \sqrt{2T}=0$$
Again, using the concept that if $$\sin x=0$$ then $$x=2m\pi$$ we get
$$\sqrt{2T}=2m\pi ---\left(3\right)$$
Now on dividing equation $$\left(3\right)$$ by equation $$\left(2\right)$$ we get,
$${\displaystyle \frac{\sqrt{2T}}{\sqrt{T}}}={\displaystyle \frac{2m\pi }{2n\pi }}$$
$$\Rightarrow \sqrt{2}={\displaystyle \frac{m}{n}}$$
As we know that $$\sqrt{2}$$ is an irrational number and hence cannot be expressed in the form of $${\displaystyle \frac{m}{n}}$$
Therefore, we arrive at a contradiction
Thus, our assumption is wrong that the function $$\sin \sqrt{x}$$ is periodic.
Hence, we can say that the function $$\sin \sqrt{x}$$ is not periodic
which means it does not have any period.
So, the correct answer is “Option 4”.

Note: Note that the domain of periodic function is always $$\left(-\mathrm{\infty },\mathrm{\infty }\right)$$ .But in our case we have the domain of function as $$\left(0,\mathrm{\infty }\right)$$ .Hence we can directly say that the function is not periodic, and it does not have any period. Also, remember that the converse of the statement is not true: that every function with domain $$\left(-\mathrm{\infty },\mathrm{\infty }\right)$$ is not periodic. For example, let $$y=x$$ the function has domain $$\left(-\mathrm{\infty },\mathrm{\infty }\right)$$ but is not periodic.