Question

求实数a的取值范围

Answer 1

f(x)=ax�0�5+x-1+3a(a属于R)在区间 [-1,1]上有零点
即ax�0�5+x-1+3a=0在[-1,1]上有实数解
即a(x�0�5+3)=1-x
即 a=(1-x)/(x�0�5+3)有实数解
令 g(x)=(1-x)/(x�0�5+3)则
a的范围即是g(x)的值域
g'(x)=[-x�0�5-3-2x(1-x)]/(x�0�5+3)�0�5
=(x�0�5-2x-3)/(x�0�5+3)�0�5
=(x+1)(x-3)/(x�0�5+3)�0�5
∵-1≤x≤1∴ (x+1)(x-3) ≤0
∴ g'(x)≤0
∴g(x)是减函数
∴x=-1 g(x)max=1/2
x=1,g(x)min=0
∴g(x)值域为[0,1/2]
∴实数a的取值范围是[0,1/2]

Answer 2

我可以告诉你，画函数图就好！图象大约分三种

Answer 3

负1/2到1/2

Answer 4

大于五