高一数学 追加！在线等

1.设：0<=t1<=t2
因为g(t)为常数，g(t1)=g(t2)
即[g(0)-p/r][e^(rt1/v)-e^(rt2/v)]=0
则g(0)=p/r
2.设0<t1<t2
g(t1)-g(t2)=[g(0)-p/r][e^(rt1/v)-e^(rt2/v)]
=[g(0)-p/r][[e^(rt2/v)-e^(rt1)/v]/[e^r(t1+t2)/v]
因为g(0)-p/r<0,0<t1<t2
则g(t1)<g(t2),污染越来越严重