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Period of the function sinx+cosx is
a)4π2b)2πc)1d)does not exist

Answer
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Hint: Now we will first consider the function sinx . let us assume the function is periodic and hence we get sinx+T=sinx . 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 sinx is not periodic. Hence sinx+cosx is also not periodic.

Complete step-by-step answer:
Now let us first consider the function sinx .
Let us say that the function is periodic and the period is T.
Hence we can say that sinx+T=sinx
Now substituting x = 0 we get
sinT=0............(1)
Now we know that if sinx=0 then x=2nπ .
Hence we get T=2nπ..........(2)
Now again consider sinx+T=sinx .
Now let us substitute x = T . Hence we get,
sinT+T=sinT
Now from equation (1) we have sinT=0 hence substituting this value in the equation we get,
sin2T=0
Now again we know that if sinx=0 then x=2nπ
Hence using this we can say that 2T=2mπ
2T=2mπ............(3)
Now let us divide equation (3) by equation (2). Hence we get,
2TT=2mπ2nπ
2=mn
Now we know that 2 is irrational and hence cannot be written in the form of pq .
Hence we arrive at a contradiction.
The contradiction arises because of our wrong assumption that sinx is Periodic.
Hence we can say that the function sinx is non periodic.
Now addition of any function to a non-periodic function is not periodic.
Hence we can say that sinx+cosx 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 (0,) . 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.