Constructing Jacobians of rank 1

Let $K$ be a number field, let $g\ge 1$ be an integer and let $f(x)=(x-a_1)\cdots (x-a_{2g+1})\in O_K[x]$ be a polynomial that splits into $2g+1$ distinct linear factors. Write $C$ for the hyperelliptic curve given by $C:y^2=f(x)$ and write $J=\mathrm{J}\mathrm{a}\mathrm{c}(C)$ for its Jacobian. Under mild technical assumptions on $f$ that are satisfied almost always, we prove that there exists some $d\in K^{\times }$ such that the quadratic twist $J^d$ has rank exactly equal to $1$. As a consequence, we deduce that for any positive integer $g$, there exists an absolutely simple abelian variety over $K$ with dimension equal to $g$ and rank equal to $1$.