We need to find where $f(x)=x\cos\big(\pi(x+[x])\big)$ is discontinuous among the given points.

A function involving $[x]$ can only have possible discontinuities at integers, because $[x]$ is continuous on each interval $(n,n+1)$ and jumps at integers.

So we only need to test the given integer points: $x=-1,1,0,2$.

1. Simplify the function on intervals

Let $x=n+t$ where $n=[x]$ and $0\le t<1$. Then $[x]=n,$ so $x+[x]=(n+t)+n=2n+t.$ Hence $\cos\big(\pi(x+[x])\big)=\cos\big(\pi(2n+t)\big).$ Using periodicity, $\cos(\pi(2n+t))=\cos(2n\pi+\pi t)=\cos(\pi t).$ But since $t=x-n$, $\cos(\pi t)=\cos(\pi(x-n))=\cos(\pi x- n\pi)=(-1)^n\cos(\pi x).$ A cleaner way near an integer is to directly use one-sided values.

2. Check continuity at a general integer $x=n$

At $x=n$, $f(n)=n\cos(\pi(n+[n]))=n\cos(\pi(n+n))=n\cos(2n\pi)=n.$ So the function value at integer $n$ is $f(n)=n.$

Left-hand limit at $x=n$

For $x\to n^-$, we have $[x]=n-1$. Thus $f(x)=x\cos\big(\pi(x+n-1)\big).$ Taking limit as $x\to n$, $\lim_{x\to n^-} f(x)=n\cos\big(\pi(n+n-1)\big)=n\cos((2n-1)\pi).$ Since $\cos((2n-1)\pi)=-1$, $\lim_{x\to n^-} f(x)=-n.$

Right-hand limit at $x=n$

For $x\to n^+$, we have $[x]=n$. Thus $f(x)=x\cos\big(\pi(x+n)\big).$ So $\lim_{x\to n^+} f(x)=n\cos\big(\pi(n+n)\big)=n\cos(2n\pi)=n.$

Thus at integer $n$: $\text{LHL}=-n,\qquad \text{RHL}=n,\qquad f(n)=n.$

For continuity, we need $-n=n,$ which gives $n=0.$ So the function is continuous at $x=0$ and discontinuous at every nonzero integer.

3. Test the given options

A: $x=-1$

Here $n=-1\ne 0$. So discontinuous.

B: $x=1$

Here $n=1\ne 0$. So discontinuous.

C: $x=0$

Here $n=0$. Then $\text{LHL}=0,\quad \text{RHL}=0,\quad f(0)=0,$ so continuous.

D: $x=2$

Here $n=2\ne 0$. So discontinuous.

4. Final answer

The function is discontinuous at $x=-1,\;1,\;2.$ So the correct options are: $\boxed{\text{A, B, D}}$

5. Comparison with stored correct answer

Stored correct answer: A, B, D

Our derived answer matches it exactly.