Let [x] be the greatest integer less than or equals to x. Then, at which of the following point(s) the function f(x) = xcos(π(x+[x])) is discontinuous?

The correct options are A, C and D.

f(x) = xcos(π[x] + πx)

At x = 2
lim_{x→2⁻} [xcos(π[x] + πx)] = 2cos(π + 2π) = 2cos(3π) = -2
lim_{x→2⁺} [xcos(π[x] + πx)] = 2cos(2π + 2π) = 2cos(4π) = 2
Hence, f(x) is discontinuous at x = 2

At x = 0
lim_{x→0⁻} xcos(π[x] + πx) = 0cos(-π + 0) = 0
lim_{x→0⁺} xcos(π[x] + πx) = 0cos(0 + 0) = 0
Hence, f(x) is continuous at x = 0

At x = 1
lim_{x→1⁻} xcos(π[x] + πx) = 1cos(0 + π) = cos(π) = -1
lim_{x→1⁺} xcos(π[x] + πx) = 1cos(π + π) = cos(2π) = 1
Hence, f(x) is discontinuous at x = 1

At x = -1
lim_{x→-1⁻} xcos(π[x] + πx) = -1cos(-2π + π) = -cos(-π) = 1
lim_{x→-1⁺} xcos(π[x] + πx) = -1cos(-π - π) = -cos(-2π) = -1
Hence, f(x) is discontinuous at x = -1

Hence, options A, C and D are correct.