Let $$f(x)= [x]^{2}-[x+3]-3, x\in \mathbb R$$, where $$[\cdot]$$ is the greatest integer funtion. Then

Let $$[\,\cdot\,]$$ denote the greatest-integer (floor) function. Write $$n=[x]$$. Then $$x\in[n,n+1)$$ with $$n\in\mathbb Z$$.

Since $$x+3\in[n+3,n+4)$$, we have $$[x+3]=n+3$$.

Hence for every $$x\in[n,n+1)$$,

$$\begin{aligned} f(x)&=[x]^{2}-[x+3]-3 \\[2mm] &=n^{2}-(n+3)-3 \\[2mm] &=n^{2}-n-6 \\[2mm] &=(n-3)(n+2)\,.\end{aligned}$$

Thus $$f(x)$$ depends only on the integer $$n=[x]$$ and is constant over each unit interval $$[n,n+1)$$.

Sign analysis of $$g(n)=n^{2}-n-6=(n-3)(n+2)$$

• $$g(n)=0$$ at $$n=-2,\,3$$.
• $$g(n)\gt0$$ for $$n\le-3$$ and for $$n\ge4$$.
• $$g(n)\lt0$$ for $$-1\le n\le2$$.

Translating back to $$x$$ (remember $$x\in[n,n+1)$$):

• $$f(x)\gt0$$ for $$x\in(-\infty,-2)\cup[4,\infty)$$.
• $$f(x)=0$$ for $$x\in[-2,-1)\cup[3,4)$$.
• $$f(x)\lt0$$ for $$x\in[-1,3)$$.

Checking the options

Option A: “$$f(x)\gt0$$ only for $$x\in[4,\infty)$$”. There is another positive region $$(-\infty,-2)$$, so Option A is incorrect.

Option B: “$$f(x)\lt0$$ only for $$x\in[-1,3]$$”. Indeed whenever $$f(x)\lt0$$, $$x\in[-1,3)$$, which is contained in $$[-1,3]$$, and there are no negative values outside this interval. Therefore Option B is correct.

Option C: “$$f(x)=0$$ for finitely many values of $$x$$”. The zeros occur throughout the full intervals $$[-2,-1)$$ and $$[3,4)$$, giving infinitely many points. Hence Option C is wrong.

Option D: Evaluate $$\displaystyle\int_{0}^{2}f(x)\,dx$$.

Between $$0\le x&lt1$$: $$[x]=0\Rightarrow f(x)=0^{2}-3-3=-6$$.
Between $$1\le x&lt2$$: $$[x]=1\Rightarrow f(x)=1^{2}-4-3=-6$$.

Therefore $$\int_{0}^{2}f(x)\,dx=\int_{0}^{1}(-6)\,dx+\int_{1}^{2}(-6)\,dx=-6(1)-6(1)=-12\neq-6$$. So Option D is also wrong.

Hence the only correct choice is Option B.