Question

已知函数f(x)=x^2+ax-lnx,a属于r.(1)若函数f(x)在[1,2]上是减函数,求实数a的取值范围.(2)令g(x)=f(x)-x^2

已知函数f(x)=x^2+ax-lnx,a属于r.(1)若函数f(x)在[1,2]上是减函数,求实数a的取值范围.(2)令g(x)=f(x)-x^2……是哪一个高考题？

Answer 1

已知函数f(x)=x^2+ax-lnx,a€R. ①若函数f(x)在[1,2]上是减函数，求实数a的取值范围。 ②令g(x)=f(x)-x^2,若x€(0,e]时，g(x)的最小值是3，求a值。

1. f'(x) = 2x+a+(-1/x)
=> 当x属于[1,2]时，f'(x)是增函数
=> f'(1)<=f'(x)<=f'(2)
=> a+1 <= f'(x) <= a+ 7/2
函数f(x)在[1,2]上是减函数
=> 当x属于[1,2]时,f'(x)<=0
=> a+ 7/2 <=0
=> a≤-7/2

2. g(x)=f(x)-x^2 = ax-lnx
=> g'(x)=a-(1/x)
=> 当x属于(0,e]时，g'(x)是增函数
=> g'(x)<=g'(e) = a- (1/e)
=> (1)、当a<=e时,g'(x)<=0
=> g(x)在(0,e]上是减函数
=> g(x)>=g(e)=ae-1 ,函数g(x)的最小值是3
=> ae-1 = 3
=> a = 4/e
(2)、当a>e时： 当x∈(0,1/a] =>g(x)是减函数
当x∈(1/a,e] =>g(x)是增函数
=> g(x)>=g(1/a)=1-ln(1/a) =3
=> ln(1/a) = -2
=> a = e^2