Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 53 / 1 / ii / c / Solution

The Hubble parameter is $H=An{t}^{n-1}>0$, andFor $n=1$ this is positive at every $t>0$, so expansion is exponential and inflationary throughout. For $0<n<1$, acceleration begins only afterThe interval $0<t<t_{\ast }$ is decelerating, while $t=t_{\ast }$ is the transition. Thus the model admits inflation, but not arbitrarily near its specified time origin when $n<1$; choosing parameters can place the transition early. Equivalently ${ϵ}_H=(1-n)/(An{t}^n)$ becomes smaller than one at the same time.