Which of the following functions have period 2?

(A) The period of $\cos \mathrm{\pi x} \mathrm{is} \frac{2\mathrm{\pi }}{\mathrm{\pi }}$ = 2, and period of {x} is 1 Hence, period of the given function is L.C.M. of (1, 2) = 2
(B) Solving $\tan \left(\frac{\mathrm{\pi }}{2}[\mathrm{x}+\mathrm{T}]\right)=\tan \left(\frac{\mathrm{\pi }}{2}\left[\mathrm{x}\right]\right)$ i.e., [x + T] – [x] = 2n gives a value of T independent of x only if T is an integer.
In that case, the above equation reduces to
[x] + T – [x] = 2n
i.e., T = 2n
Hence, period of f (x), is the smallest positive value of T, i.e., 2.
(C) We have period of sin x = 2$\mathrm{\pi }$ and period of {x} = 1 Hence, period of the given functions is L.C.M. of (2p, 1) which does not exist since 2$\mathrm{\pi }$ is an irrational number. Hence, the function is not periodic
(D) Let us solve sin{cos (x + T)} = sin{cos x} i.e., cos (x + T) = np + (– 1) n cos x, n ∈ I Putting n = 0, gives cos (x + T) = cos x, which gives T = 2$\mathrm{\pi }$ as the smallest positive value. For no other value of n can a value of T be found independent of x.
Hence, the required fundamental period is 2p. The correct option is (1) and (2)