Question

Period of the function $\sin \sqrt{x}+\cos \sqrt{x}$ is
$\begin{align}
  & a)4{{\pi }^{2}} \\
 & b)\sqrt{2\pi } \\
 & c)1 \\
 & d)\text{does not exist} \\
\end{align}$

Answer 1

Hint: Now we will first consider the function $\sin \sqrt{x}$ . let us assume 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 is a contradictory statement. Hence we prove that the function $\sin \sqrt{x}$ is not periodic. Hence $\sin \sqrt{x}+\cos \sqrt{x}$ is also not periodic.

Complete step-by-step answer:
Now let us first consider the function $\sin \sqrt{x}$ .
Let us say that the function is periodic and the period is T.
Hence we can say that $\sin \sqrt{x+T}=\sin \sqrt{x}$
Now substituting x = 0 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,
$\sin \sqrt{T+T}=\sin \sqrt{T}$
Now from equation (1) we have $\sin \sqrt{T}=0$ hence substituting this value in the equation we get,
$\sin \sqrt{2T}=0$
Now again we know that if $\sin x=0$ then $x=2n\pi $
Hence using this we can say that $\sqrt{2T}=2m\pi $
$\sqrt{2T}=2m\pi ............\left( 3 \right)$
Now let us divide equation (3) by equation (2). Hence we get,
$\dfrac{\sqrt{2T}}{\sqrt{T}}=\dfrac{2m\pi }{2n\pi }$
$\Rightarrow \sqrt{2}=\dfrac{m}{n}$
Now we know that $\sqrt{2}$ is irrational and hence cannot be written in the form of $\dfrac{p}{q}$ .
Hence we arrive at a contradiction.
The contradiction arises because of our wrong assumption that $\sin \sqrt{x}$ is Periodic.
Hence we can say that the function $\sin \sqrt{x}$ is non periodic.
Now addition of any function to a non-periodic function is not periodic.
Hence we can say that $\sin \sqrt{x}+\cos \sqrt{x}$ is not a periodic function.

So, the correct answer is “Option d”.

Note: Now note that the domain of periodic function is always $\left( -\infty ,\infty \right)$ . In our case we have the domain of function is $\left( 0,\infty \right)$ . Hence we can directly say that the function is not periodic. Now note that the converse of the statement is not true which means every function with domain $\left( -\infty ,\infty \right)$ is not periodic. Take y = x for example. The function has domain $\left( -\infty ,\infty \right)$ but is not periodic.