Question

已知函数f(x)=loga(ax^2-x+1/2)

(1)当a=3/8时，求函数f(x)的单调递减区间 （2）当0小于a小于1时， f(x)在1,2的闭区间上恒大于0，求实数a的取值范围
求详解 多谢 急

Answer 1

1)
f(x) = log(a)[ax^2-x + 1/2 ]
当a = 3/8
f(x) = log(3/8)[3/8)x^2-x +1/2]
令g(x) =(3/8)x^2-x +1/2
因为0《3/8《1
所以log(3/8)^x是单调减函数
所以g'(x) = (3/4)x -1 <0
x < 4/3

又因为g(x) >0
=> (3/8)x^2-x +1/2 > 0
3x^2-8x +4 > 0
(3x-2)(x-2) >0
x>2 or x< 2/3

综上所述
x < 4/3 and (x>2 or x< 2/3)
=> x< 2/3
f(x)的单调递减区间 , x< 2/3
(2)
因为0<a<1上 f(x)> 0 在 x ∈[1,2]恒成立
令h(x) = ax^2-x + 1/2
h'(x) = 2ax - 1 =0
x = 1/(2a)
h''(x) = 2a > 0 ( min) (0<a<1)
minh(x) = h(1/(2a))
= a(1/(2a))^2 - 1/(2a) + 1/2
= -1/(4a) + 1/2
minf(x) = f(1/(2a))
= log(a)[-1/(4a) + 1/2] > 0
-1/(4a) + 1/2 > 1
1/2 > 1/(4a)
4a > 2
a > 1/2