Significance of adjoint relationship with Ext instead of Hom For categories $\mathcal{C}$ and $\mathcal{D}$, a pair of functors $\mathcal{C} \xrightarrow{\;R\;} \mathcal{D}$ and $\mathcal{C} \xleftarrow{\;L\;} \mathcal{D}\,$ are an adjoint pair if for any objects $X$ of $\mathcal{C}$ and $Y$ of $\mathcal{D}$ we have a bijection 
$$
\operatorname{Hom}_\mathcal{C}(LY,X) 
\;\simeq\; 
\operatorname{Hom}_\mathcal{D}(Y,RX)
\,.
$$
Recently, I've come across an exercise* asking to prove that a pair of functors are an adjoint pair, and then additionally to prove that
$$
\operatorname{Ext}^1_\mathcal{C}(X,LY) 
\;\simeq\; 
\operatorname{Ext}^1_\mathcal{D}(RX,Y)
\,.
$$
What is the significance of this adjoint-like relationship with $\operatorname{Ext}$ instead of $\operatorname{Hom}$? The exercise I'm looking at didn't provide any motivation for proving this.

* In the exercise, $X$ and $Y$ are categories of quiver representations, and $R$ and $L$ are reflection functors across a sink and source vertex respectively. This category is hereditary, so $\operatorname{Ext}^i$ vanish for $i > 1$.
 A: $\operatorname{Ext}$ is the derived functor of $\operatorname{Hom}$. So you're basically looking at a "derived" adjunction. Or rather an adjunction which is compatible with the dg-structure, I guess.
A: The isomorphism is not true for an arbitrary choice of adjoint functors $L$ and $R$, even when the abelian categories $\mathcal{C}$ and $\mathcal{D}$ are hereditary. Thus, a possible motivation for the the authors might be to show that the functors they are considering (reflection functors) are particularly nice.
To see a counter-example, take $\mathcal{C}$ to be the category of modules over the path algebra $A$ of the quiver $1\to 2$ over a field $k$, and let $\mathcal{D}$ be the category of vector spaces over $k$.  Take $L = ?\otimes_k A$ and $R= \operatorname{Hom}_A(A, ?)$ to be the adjoint pair.  Finally, take $X$ to be the representation $k\to 0$ and $Y=k$.
Then $LY = A$, so $\operatorname{Ext}_A^1 (X, LY) \cong k$, but $\operatorname{Ext}_k^1 (RX, Y) = 0$, since all $\operatorname{Ext}^1$ groups vanish in the category of vector spaces.  Therefore, the two extension groups are not isomorphic.  
