8.已知函数,f(x)={2x−1,x<ax2,x⩾af(x)=\left\{\begin{array}{l}{{2}^{x-1},x<a}\\ {{x}^{2},x\geqslant a}\end{array}\right.f(x)={2x−1,x<ax2,x⩾a,则"a=la=la=l"是"f(x)f(x)f(x)是RRR上的增函数"的((( )))