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)

Answer 1

1.f'(x) = 2x+a+(-1/x)

=> 当x属于[1,2]时,f'(x)是增函数

=> f'(1) a+1 g'(x)=a-(1/x)

=> 当x属于(0,e]时,g'(x)是增函数

=> g'(x) 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

3.当x属于(0,e]时,设 h(x) = e^2*x^2-5/2 - (x+1)lnx

根据单调性最值来做