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设函数(x)=ax^2lnx+b(x-1)(x>0),曲线y=f(x)过点(e,e^2-e+1)且在(1,0)处的切线方程为y=0(1)求a,b的值(2)证明:当x≥1时,f(x)≥(x-1)^2(3)若当x≥1时,f(x)≥m(x-1)^2恒成立,求实数m的取值范围多谢^^
题目详情
设函数(x)=ax^2lnx+b(x-1)(x>0),曲线y=f(x)过点(e,e^2-e+1)且在(1,0)处的切线方程为y=0
(1)求a,b的值
(2)证明:当x≥1时,f(x)≥(x-1)^2
(3)若当x≥1时,f(x)≥m(x-1)^2恒成立,求实数m的取值范围
多谢^_^
(1)求a,b的值
(2)证明:当x≥1时,f(x)≥(x-1)^2
(3)若当x≥1时,f(x)≥m(x-1)^2恒成立,求实数m的取值范围
多谢^_^
▼优质解答
答案和解析
(1) f(x)=ax^2lnx+b(x-1)
f(e)=a*e^2+b(e-1)=e^2-e+1
(a-1)e^2+(b+1)(e-1)=0 (1)
f'(x)=2axlnx+ax+b
f'(1)=a+b=0 得: a=-b
将a=-b代入(1):(-b-1)e^2+(b+1)(e-1)=0
(b+1)(e-1)=(b+1)e^2
b=-1 a=1
(2) f(x)=x^2lnx-(x-1)
设g(x)=x^2lnx-(x-1)-(x-1)^2
则 g(1)=0
∵ g'(x)=2xlnx+x-1-2(x-1)=2xlnx-x+1>=0 (x>=1)
∴ 当x>=1时,g(x)递增, g(x)>=0 即 f(x)≥(x-1)^2
(3) f(x)=x^2lnx-(x-1)>m(x-1)^2
设 F(x)=x^2lnx-(x-1)-m(x-1)^2
则 F(1)=0
F'(x)=2xlnx+x-1-2m(x-1)>=0
m <=(2xlnx+x-1)/[2(x-1)]
∵(2xlnx+x-1)/[2(x-1)]>3/2
∴m<=3/2
f(e)=a*e^2+b(e-1)=e^2-e+1
(a-1)e^2+(b+1)(e-1)=0 (1)
f'(x)=2axlnx+ax+b
f'(1)=a+b=0 得: a=-b
将a=-b代入(1):(-b-1)e^2+(b+1)(e-1)=0
(b+1)(e-1)=(b+1)e^2
b=-1 a=1
(2) f(x)=x^2lnx-(x-1)
设g(x)=x^2lnx-(x-1)-(x-1)^2
则 g(1)=0
∵ g'(x)=2xlnx+x-1-2(x-1)=2xlnx-x+1>=0 (x>=1)
∴ 当x>=1时,g(x)递增, g(x)>=0 即 f(x)≥(x-1)^2
(3) f(x)=x^2lnx-(x-1)>m(x-1)^2
设 F(x)=x^2lnx-(x-1)-m(x-1)^2
则 F(1)=0
F'(x)=2xlnx+x-1-2m(x-1)>=0
m <=(2xlnx+x-1)/[2(x-1)]
∵(2xlnx+x-1)/[2(x-1)]>3/2
∴m<=3/2
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