
Period of the function is
Answer
501.3k+ views
Hint: Now we will first consider the function . let us assume the function is periodic and hence we get . 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 is not periodic. Hence is also not periodic.
Complete step-by-step answer:
Now let us first consider the function .
Let us say that the function is periodic and the period is T.
Hence we can say that
Now substituting x = 0 we get
Now we know that if then .
Hence we get
Now again consider .
Now let us substitute x = T . Hence we get,
Now from equation (1) we have hence substituting this value in the equation we get,
Now again we know that if then
Hence using this we can say that
Now let us divide equation (3) by equation (2). Hence we get,
Now we know that is irrational and hence cannot be written in the form of .
Hence we arrive at a contradiction.
The contradiction arises because of our wrong assumption that is Periodic.
Hence we can say that the function is non periodic.
Now addition of any function to a non-periodic function is not periodic.
Hence we can say that is not a periodic function.
So, the correct answer is “Option d”.
Note: Now note that the domain of periodic function is always . In our case we have the domain of function is . 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 is not periodic. Take y = x for example. The function has domain but is not periodic.
Complete step-by-step answer:
Now let us first consider the function
Let us say that the function is periodic and the period is T.
Hence we can say that
Now substituting x = 0 we get
Now we know that if
Hence we get
Now again consider
Now let us substitute x = T . Hence we get,
Now from equation (1) we have
Now again we know that if
Hence using this we can say that
Now let us divide equation (3) by equation (2). Hence we get,
Now we know that
Hence we arrive at a contradiction.
The contradiction arises because of our wrong assumption that
Hence we can say that the function
Now addition of any function to a non-periodic function is not periodic.
Hence we can say that
So, the correct answer is “Option d”.
Note: Now note that the domain of periodic function is always
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