(Ⅰ)a=-[1/2]时,函数f(x)=ex-
x2
2+
1
2x−1,求导数可得f′(x)=ex-x+[1/2]
∴f′(1)=e-[1/2],f(1)=e-1
∴曲线y=f(x)在点(1,f(1))处的切线方程为y-(e-1)=(e-[1/2])(x-1),即(e-[1/2])x-y-[1/2]=0;
(Ⅱ)由f(x)≥0得ax≤ex-[1/2]x2-1,因为x≥
1
2,所以a≤
ex−
1
2x2−1
x.
令g(x)=
ex−
1
2x2−1
x,则g′(x)=
ex(x−1)−
1
2x2+1
x2.
令h(x)=ex(x-1)-[1/2]x2+1,所以h′(x)=x(ex-1).
因为x≥
1
2,所以h′(x)>0,所以h(x)在[[1/2],+∞)上单调增
所以h(x)≥h([1/2])=[7/8]-
1
2
e>0
所以g′(x)>0
∴g(x)在[[1/2],+∞)上单调增
∴g(x)≥g(