1. Let set $M=\{-2,0,1\}, N=\{1,2,3,4,5\}$, mapping $f: M \rightarrow N$ such that for any $x \in M$, $x+f(x)+x f(x)$ is an odd number, then the number of such mappings $f$ is ( ).
A. 45
B. 27
C. 15
D. 11
1. $\mathrm{A}$

Since when $x=-2$, $x+f(x)+x f(x)=-2-f(-2)$ is odd, so $f(-2)$ can take 1,3,5, which gives 3 possible values;

When $x=0$, $x+f(x)+x f(x)=f(0)$ is odd, so $f(0)$ can take $1,3,5$, which gives 3 possible values;

When $x=1$, $x+f(x)+x f(x)=1+2 f(1)$ is odd, so $f(1)$ can take 1,2, $3,4,5$, which gives 5 possible values;
Therefore, by the multiplication principle, there are $3 \times 3 \times 5=45$ mappings.