Correct Option: A

Given (A) x[n] = 4 sin [2n]
Fundamental period,

N =	2π.m
	Ω

Here, Ω = 2

So, N =	2π	·m = π
	2

which is non-periodic.
(B) x[n] = 4 sin [0·2πn]
Ω = 0·2π

N =	2π	· m =	2	· m = 10 × 1 = 10
	Ω		0·2π

i.e., periodic with period = 10.
(C) x[n] = 4 cos [6πn]
Ω = 6

N =	2π	· m =	2π	· m = 2 × 3 6 = 1
	Ω		6π

i.e., periodic with period = 1
(D)

x[n] = 4 cos			6πn
			35

Ω =	6π
	35

N =	2π	· m =	2π	· m =	2 × 35	· m
	Ω		6π/35		6

= 35 with m = 3
i.e., periodic with period = 35.
Hence, alternative (A) is the correct choice.