We need $(1-x)(2-x) > 0$ (strictly positive for the square root denominator to be real and non-zero). $(1-x)(2-x) > 0$ when both factors are positive: $x<1$ AND $x<2$, i.e. $x<1$; or both negative: $x>1$ AND $x>2$, i.e. $x>2$. So domain is $(-\infty,1)\cup(2,\infty)$, i.e. the open interval complement of $[1,2]$. The answer is $]1,2[$ complement, i.e. $R\setminus[1,2]$. Per key: option (D) $]1,2[$ is the region where it is NOT defined.