If $R(x,y)$ is a recursive relation, then $\exists y\leq 2$ such that $R(x,y)$ is recursive
Fixing $y$ in $R(x,y)$ results in a special case of the problem "whether $\langle x,y \rangle \in R$", and hence the latter should be no more harder than the general problem of deciding if $\langle x,y \rangle \in R$. So, intuitively if $R(x,y)$ is a recursive relation then $R(x,y_0)$ should also be recursive for some fixed $y_0$.  

Formally, recall that a relation $R(x,y)$ is recursive if its characteristic function 

$$ c_R(x,y) =  \begin{cases} 
      1  &amp; \text{ if } \langle x,y \rangle \in R \\
      0 &amp; \text{ if } \langle x,y \rangle \notin R 
\end{cases}
$$
is recursive. 

Now assume that $R(x,y)$ is recursive. Then $c_R(x,y)$ is also recursive. Let $M(x,y)$ be a TM computing $c_R(x,y)$. For a fixed $y_0$ our new TM $M'(x)$ will take $x$, call $M(x, y_0)$ and return its value. Since $M(x,y)$ computes a recursive function (always returns $0$ or $1$), $M'(x)$ also computes a recursive function, namely $c_R(x,y_0)$. Thus $R(x,y_0)$ is recursive.