Define $$f_K(x)=\sum_{i=K+1}^{2K} \binom{2K}{i}x^{i-1}(1-x)^{2K-i}.$$ How to show that $qf_K(x)-f_K(1-x)$ has exactly one real root in $(0,1)$ for any $q > 0$ and $K \geq 1$. The proof for $q=1$ goes as follows:
\begin{align} f_k(x)-f_K(1-x)&= \sum_{i=K+1}^{2K} \binom{2K}{i}\left[x^{i-1}(1-x)^{2K-i}-(1-x)^{i-1}x^{2K-i}\right]\\ &= \sum_{i=K+1}^{2K} \binom{2K}{i} x^{2K-i}(1-x)^{2K-i}\left[x^{2(i-K)-1}-(1-x)^{2(i-K)-1}\right]\\ &= (2x-1)\sum_{i=K+1}^{2K} \binom{2K}{i}x^{2K-i}(1-x)^{2K-i} \sum_{r=0}^{2(i-K)-2} x^r (1-x)^{2(i-K)-2-r} \end{align}
Clearly, the above expression is $>0$ if $x > 1/2$, $<0$ if $x < 1/2$, and zero at $x=1/2$. So it has a unique root at $x=1/2$. But this argument does not generalize for other values of $q$.
Let $L\left(x\right):=q\cdot f_{K}\left(x\right)-f_{K}\left(1-x\right)$. We have $L\left(0\right)=-1<0$ and $L\left(1\right)=q>0$, so by continuity, $L\left(x\right)=0$ has at least one solution (existence). Towards uniqueness, you have shown that there exists a unique $x$ such that $f_{K}\left(x\right)-f_{K}\left(1-x\right)=0$, so we write $L\left(x\right)=\left(q-1\right)\cdot f_{K}\left(x\right)$. Show that has a unique maximum $x^{*}$ by taking the derivative wrt $x$. Show then that $L\left(x^{*}\right)>0.$ Since $L\left(0\right)<0$, $L\left(x^{*}\right)>0$ and $L\left(\cdot\right)$ is strictly increasing in $\left[0,x^{*}\right]$ we have that it has a unique solution in that interval.