Hamilton completion and the path cover number of sparse random graphs

We prove that for every $\varepsilon > 0$ there is $c_0$ such that if $G\sim G(n,c/n)$, $c\ge c_0$, then with high probability $G$ can be covered by at most $(1+\varepsilon)\cdot \frac{1}{2}ce^{-c} \cdot n$ vertex disjoint paths, which is essentially tight. This is equivalent to showing that, with high probability, at most $(1+\varepsilon)\cdot \frac{1}{2}ce^{-c} \cdot n$ edges can be added to $G$ to create a Hamiltonian graph.