f(x)=e^x+ax-1
f(x)≥x^2在(0,1)上恒成立
e^x+ax-1≥x^2在(0,1)上恒成立
e^x-x^2-1≥-ax
∵x>0
∴(e^x-x^2-1)/x≥-a
设g(x)=(e^x-x^2-1)/x
g'(x)=[(x-1)e^x-(x^2-1)]/x^2
=[(x-1)(e^x-(x+1))]/x^2
设h(x)=e^x-(x+1)
h'(x)=e^x-1
x∈(0,1)
∴h'(x)>0
∵h(0)=1-1=0
∴h(x)>0
即e^x-(x+1)>0
∵x-1
f(x)=e^x+ax-1
f(x)≥x^2在(0,1)上恒成立
e^x+ax-1≥x^2在(0,1)上恒成立
e^x-x^2-1≥-ax
∵x>0
∴(e^x-x^2-1)/x≥-a
设g(x)=(e^x-x^2-1)/x
g'(x)=[(x-1)e^x-(x^2-1)]/x^2
=[(x-1)(e^x-(x+1))]/x^2
设h(x)=e^x-(x+1)
h'(x)=e^x-1
x∈(0,1)
∴h'(x)>0
∵h(0)=1-1=0
∴h(x)>0
即e^x-(x+1)>0
∵x-1