18.设aaa为正实数,且a≠1a\ne 1a=1,已知函数f(x)={(2a−1)x+3a,x<0ax,x⩾0f(x)=\left\{{\left.\begin{array}{l}{({2a-1})x+3a,x<0}\\ {{a^x},x\geqslant 0}\end{array}\right.}\right.f(x)={(2a−1)x+3a,x<0ax,x⩾0,使得函数在RRR上单调递减成立的充分不必要条件是((( )))