31．已知函数f(x)=e∣x−1∣−sin(π2x)f(x)={e^{\vert {x-1}\vert }}-\sin ({\frac{\pi }{2}x})f(x)=e∣x−1∣−sin(2πx)，则使得f(x)>f(2x)f(x)>f(2x)f(x)>f(2x)成立的xxx的取值范围是 ___(0,23)({0,\frac{2}{3}})(0,32)___．