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.