1. Given equation

We need to solve $x^2-4x+[x]+3=x[x],$ where $[x]$ is the greatest integer function.

Let $n=[x], \quad n\le x < n+1.$ Then the equation becomes $x^2-4x+n+3=xn.$ Rearrange: $x^2-(n+4)x+(n+3)=0.$

2. Factor the quadratic

Observe that $x^2-(n+4)x+(n+3)=(x-1)(x-(n+3)).$ So the equation reduces to $(x-1)(x-(n+3))=0.$ Hence possible roots are $x=1 \quad \text{or} \quad x=n+3.$ But remember that $n=[x]$, so each candidate must satisfy $n\le x<n+1$.

3. Check case 1: $x=1$

If $x=1$, then $[x]=[1]=1.$ Substitute into the original equation: $1-4+1+3=1\cdot 1,$ $1=1.$ So $x=1$ is indeed a solution.

4. Check case 2: $x=n+3$ with $n=[x]$

Since $x=n+3$ and also $n\le x<n+1$, we get $n\le n+3 < n+1.$ The left inequality is true, but the right inequality gives $3<1,$ which is impossible.

So this case gives no solution.

5. Conclusion

The only solution is $x=1.$ Thus the equation has a unique solution in $(-\infty,\infty)$.

Also, it does not have a unique solution in $(-\infty,1)$, because the only solution is exactly $1$, which is not in that interval.

6. Evaluate options

• A: exactly two solutions in $(-\infty,\infty)$ — False
• B: no solution — False
• C: a unique solution in $(-\infty,\infty)$ — True
• D: a unique solution in $(-\infty,1)$ — False

Therefore, the correct option is $\boxed{\text{C}}.$