8．已知函数f(x)={(12)∣x+1∣,x⩽0∣lnx∣,x>0f(x)=\left\{\begin{array}{l}{(\frac{1}{2})^{\vert x+1\vert },x\leqslant 0}\\ {\vert lnx\vert ,x>0}\end{array}\right.f(x)={(21)∣x+1∣,x⩽0∣lnx∣,x>0，若函数g(x)=4[f(x)]2−(4t+3)f(x)+3tg(x)=4[f(x)]^{2}-(4t+3)f(x)+3tg(x)=4[f(x)]2−(4t+3)f(x)+3t有七个不同的零点，则实数ttt的取值范围是((( )))