设a为实属,函数f（x）=e^2-2x+2a,x属于R 求证：a>ln2-1且x>0时,e^2>x^2-2ax+1

f'(x)=e^x-2>0,x>ln2f(x)的极小值(也是最小值)是f(ln2)=2-2ln2+2a.因为a>ln2-1,即f(ln2)=2-2ln2+2a>0,f(x)=e^x-2x+2a>0恒成立.设F(x)=e^x-x^2+2ax-1,F'(x)=e^x-2x+2a=f(x)>0.所以,F(x)为增函数.当x>0时,F(x)>F(0)=0,...