一道高考数学函数导数题已知函数f(x)=lnx,g(x)=k·(x-!)/(x+1)(Ⅱ)当x>1时,函数f(x)> g(x)恒成立,求实数k的取值范围;(Ⅲ)设正实数a1,a2,a3,an满足a1+a2+a3+...+an=1,求证:ln(1+1/a1²)+ln(1+a2²)+.+ln(1+1/an
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![一道高考数学函数导数题已知函数f(x)=lnx,g(x)=k·(x-!)/(x+1)(Ⅱ)当x>1时,函数f(x)> g(x)恒成立,求实数k的取值范围;(Ⅲ)设正实数a1,a2,a3,an满足a1+a2+a3+...+an=1,求证:ln(1+1/a1²)+ln(1+a2²)+.+ln(1+1/an](/uploads/image/z/2518054-70-4.jpg?t=%E4%B8%80%E9%81%93%E9%AB%98%E8%80%83%E6%95%B0%E5%AD%A6%E5%87%BD%E6%95%B0%E5%AF%BC%E6%95%B0%E9%A2%98%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dlnx%2Cg%28x%29%3Dk%C2%B7%EF%BC%88x-%21%29%2F%28x%2B1%29%28%E2%85%A1%29%E5%BD%93x%3E1%E6%97%B6%2C%E5%87%BD%E6%95%B0f%28x%29%3E+g%28x%29%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82%E5%AE%9E%E6%95%B0k%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%EF%BC%9B%28%E2%85%A2%29%E8%AE%BE%E6%AD%A3%E5%AE%9E%E6%95%B0a1%2Ca2%2Ca3%2Can%E6%BB%A1%E8%B6%B3a1%2Ba2%2Ba3%2B...%2Ban%3D1%2C%E6%B1%82%E8%AF%81%EF%BC%9Aln%281%2B1%2Fa1%26%23178%3B%29%2Bln%281%2Ba2%26%23178%3B%29%2B.%2Bln%281%2B1%2Fan%26%23)
一道高考数学函数导数题已知函数f(x)=lnx,g(x)=k·(x-!)/(x+1)(Ⅱ)当x>1时,函数f(x)> g(x)恒成立,求实数k的取值范围;(Ⅲ)设正实数a1,a2,a3,an满足a1+a2+a3+...+an=1,求证:ln(1+1/a1²)+ln(1+a2²)+.+ln(1+1/an
一道高考数学函数导数题
已知函数f(x)=lnx,g(x)=k·(x-!)/(x+1)
(Ⅱ)当x>1时,函数f(x)>
g(x)恒成立,求实数k的取值范围;
(Ⅲ)设正实数a1,a2,a3,an满足a1+a2+a3+...+an=1,
求证:ln(1+1/a1²)+ln(1+a2²)+.+ln(1+1/an²)>2n²/(n+2)
这题第(Ⅱ)答案是(﹣∞,2】.第(Ⅲ)答案有点看不懂,
由(2)知,在inx>2·(x-1)/(x+1)恒成立
令x=1+1/an²(0<an<1),则in( 1+1/an²)>2/(2an²+1)>2/(2an+1)
所以ln(1+1/a1²)+ln(1+1/a2²)+.+ln(1+1/an²)>2(1/(2a1+1)+1/(2a2+1)+...1/(2an+1))
又2(1/(2a1+1)+1/(2a2+1)+...1/(2an+1))[(2a1+1)+(2a2+1)+...(2an+1)]≥n²(请问这步怎么来,请详解,
所以2(1/(2a1+1)+1/(2a2+1)+...1/(2an+1))≥2n²/(n+2)
所以:ln(1+1/a1²)+ln(1+a2²)+.+ln(1+1/an²)>2n²/(n+2)
一道高考数学函数导数题已知函数f(x)=lnx,g(x)=k·(x-!)/(x+1)(Ⅱ)当x>1时,函数f(x)> g(x)恒成立,求实数k的取值范围;(Ⅲ)设正实数a1,a2,a3,an满足a1+a2+a3+...+an=1,求证:ln(1+1/a1²)+ln(1+a2²)+.+ln(1+1/an