Question

: If $f\left( x \right) = {\text{sin root}}\left( {\text{x}} \right)$ then period of $f\left( x \right)$ is
(1) $pi$
(2) $\dfrac{{pi}}{2}$
(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 $\sin \sqrt T = 0{\text{ }} - - - \left( 1 \right)$
Now we know that if $\sin x = 0$ then $x = 2n\pi$
Hence, we get $\sqrt T = 2n\pi {\text{ }} - - - \left( 2 \right)$
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 $\sin \sqrt T = 0{\text{ }}$
Therefore, on substituting the value in the above equation we get
$\sin \sqrt {2T} = 0{\text{ }}$
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,
$\dfrac{{\sqrt {2T} }}{{\sqrt T }} = \dfrac{{2m\pi }}{{2n\pi }}$
$\Rightarrow \sqrt 2 = \dfrac{m}{n}$
As we know that $\sqrt 2$ is an irrational number and hence cannot be expressed in the form of $\dfrac{m}{n}$
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 $\left( { - \infty ,\infty } \right)$ .But in our case we have the domain of function as $\left( {0,\infty } \right)$ .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 $\left( { - \infty ,\infty } \right)$ is not periodic. For example, let $y = x$ the function has domain $\left( { - \infty ,\infty } \right)$ but is not periodic.