已知函数f(x0=1\3x^3-a\2x^2+2x+1,且x1,x2是f(x)的两个极点,0<x1<1<x2<3,若|x1-x2|大于等于m^2-2bm-2,对b属于-1到1恒成立,求m范围急
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已知函数f(x0=1\3x^3-a\2x^2+2x+1,且x1,x2是f(x)的两个极点,0<x1<1<x2<3,若|x1-x2|大于等于m^2-2bm-2,对b属于-1到1恒成立,求m范围急
已知函数f(x0=1\3x^3-a\2x^2+2x+1,且x1,x2是f(x)的两个极点,0<x1<1<x2<3,
若|x1-x2|大于等于m^2-2bm-2,对b属于-1到1恒成立,求m范围
急
已知函数f(x0=1\3x^3-a\2x^2+2x+1,且x1,x2是f(x)的两个极点,0<x1<1<x2<3,若|x1-x2|大于等于m^2-2bm-2,对b属于-1到1恒成立,求m范围急
f(x)=1\3x^3-a\2x^2+2x+1,f‘(x)=x^2-ax^2+2,x=[a±√(a²-8)]/2,∵0<x1<1<x2<3,∴0<[a-√(a²-8)]/2<1,1<[a+√(a²-8)]/2<3,得3<a<11/3.∵|x1-x2|=√(a²-8)],∴1<|x1-x2|<7/3,
∵|x1-x2|大于等于m^2-2bm-2,∴1<m^2-2bm-2<7/3,当m^2-2bm-2>1时,
{m-[b+√(b²+3)]}{m-[b-√(b²+3)]}>0,b属于-1到1时,[b+√(b²+3)]>0,[b-√(b²+3)]<0,则取
m>[b+√(b²+3)]或m<[b-√(b²+3)];当m^2-2bm-2<7/3时,[b-√(b²+13/3)]<m<[b+√(b²+13/3)],
综合以上,[b-√(b²+13/3)]<m<[b-√(b²+3)];[b+√(b²+3)]m<[b+√(b²+13/3)],结合b属于-1到1,则m范围:-1-4√3/3<m<-1,1<m<1+4√3/3.