 Solution

Both positive-density components have nonnegative pressure, so the Friedmann acceleration equation gives $\ddot{a}<0$ whenever $a>0$. Suppose $a(t_1)=a_1>0$ and $\dot{a}(t_1)=v_1>0$. Going backward while the solution is regular, $\dot{a}(t)\ge v_1$. Integrating,and thereforeA positive scale factor cannot persist indefinitely into the past: the right side vanishes at the finite time $t_1-a_1/v_1$.

For completeness, in this model the past endpoint must actually have $a\to 0$, rather than another singularity at positive $a$. From part (d) write ${\rho }_m=C_m/a^3$ and ${\rho }_r=C_r/a^4$ with $C_m,C_r>0$. ThenThe right side strictly decreases with $a$ and is positive at $a_1$, so it remains positive throughout $0<a<a_1$. There is no earlier bounce. It is smooth at every positive $a$, permitting continuation there. The lookback integral isNear $a=0$ its integrand is $a/\sqrt{K_r}$, which is integrable. ThusThe early asymptotic behavior is $a\simeq [2\sqrt{K_r}(t-t_0){]}^{1/2}$. Both densities diverge at this Big Bang boundary. Writing $a(t_0)=0$ denotes the limiting singular endpoint of the classical solution, not an ordinary regular spacetime slice.