Define the function $V : [0,1] \rightarrow \mathbb{R}$ with parameter $a>0$ as
$$V(x) = (1-x^a)C_1(a) + (1-x)^a C_2(a)$$
where
$$C_1(a)-C_2(a) \begin{cases} <0 & a< 1\\ =0 & a=1\\ >0 & a>1, \end{cases} $$ with $C_1'(a) >0$ and $C_2'(a) <0$. Thanks to numerical analysis in Mathematica I believe that this function is quasiconcave in $x$. Indeed, it seems that it's either strictly concave, monotonically increasing or monotonically decreasing. I've played around with the different ways of defining quasiconcavity to prove it, but I'm making virtually no headway. I'd be grateful for any pointers or ideas!
I got it. We have
\begin{align*} V'(x) &= -a\left[x^{a-1} C_1(a)+(1-x)^{a-1}C_2(a)\right],\\ V''(x) &= -a(a-1)\left[x^{a-2} C_1(a)-(1-x)^{a-2}C_2(a)\right], \end{align*} Note that for $C_1(a), C_2(a)\geq0$ and $C_1(a), C_2(a)\leq 0$ the function $V$ is monotonically decreasing and increasing, respectively, and thus quasiconcave. Further, for $a\geq 1$, $C_1(a) >0$ and $C_2(a)<0$ it is concave and thus quasiconcave. Conversely, fox $a<1$ and $C_1(a) <0$ and $C_2(a) >0$ it is concave and thus quasiconcave. Together with the observation \begin{align*} C_1(a)-C_2(a) \begin{cases} <0 & a< 1\\ =0 & a=1\\ >0 & a>1, \end{cases} \end{align*} this completes the proof, as the function is either concave or monotonically increasing or decreasing and thus quasiconcave.