A delicate question about derived functors Let $A\subseteq B \subseteq C$ be three triangulated categories, such that $A$ is a full triangulated sub-category of $B$, and $B$ is a full triangulated sub-category of $C = K(R)$.
Let $F: C \to D$ be a triangulated functor, where $D = K(S)$ is a fourth triangulated category. 
Here, $K(R)$ and $K(S)$ are the homotopy categories over some rings $R,S$.
Assume that:


*

*For each $b \in B$, there is some $a\in A$, and a quasi-isomorphism $a\to b$.

*If $a\in A$ is acyclic, then $F(a)$ is also acyclic.

*The derived functor $LF : D(R) \to D(S)$ exists. 


My question is: Can we compute $LF$ on objects from $B$ by using $A$-resolutions?
 A: I have found the following counterexample. I'll use homological terminology, i.e. degree $-1$ differentials. This means that you have to exchange 'above' and 'below' if you want to think cohomologically, etc.
Let $R=S=k[\epsilon]/\epsilon^2$ be the ring of dual numbers over a field $k$ and $F=-\otimes_Rk$ the plain tensor product, where $k=R/\epsilon$. Moreover, let $A$ be the subcategory of bounded complexes and $B$ the subcategory of bounded above complexes with bounded homology. Any complex $X$ in $B$ is quasi-isomorphic to a complex obtained by truncating $t_{\geq n}X$ below the minimum degree where its homology vanishes. This complex $t_{\geq n}X$ is in $A$ and comes equipped with a quasi-isomorphism $t_{n\geq 0}X\rightarrow X$. Hence 1 holds. Moreover, acyclic bounded complexes over $R$ are contractible, since the only non-contractible acyclic complex in $C=K(R)$ is
\[\cdots \rightarrow R\stackrel{\epsilon}\rightarrow R\stackrel{\epsilon}\rightarrow R\rightarrow\cdots\]
and those which contain it as a direct summand. Therefore 2 holds since $F$ preserves contractible complexes. The complex $b=k$ concentrated in degree zero is in $A$ so its $A$-resolution is $a=k$ itself, but $F(k)=k$ and $LF(k)=\bigoplus_{n\geq 0} \operatorname{Tor}_n^R(k,k)[n]=\bigoplus_{n\geq 0} k[n]$.
EDIT: OK, I notice I've been silly. You can just take $B=A$. I think you believe that, as a consequence of 2, $LF=F$ on $A$, but it doesn't as I show above.
