(1)a>0,f(x)=x(ax+1),x1=0.x2=-1/a,根据|f(x)|的图像性质,要使|f(x)|≤1在[0,1]上成立,只要满足f(1)≤1即可。有a+1≤1,a≤0.所以a>0时,无解。
(2)0<-1/2a≤1,a≤-1/2.f(x)=x(ax+1),x1=0.x2=-1/a,根据|f(x)|的图像性质, 要使|f(x)|≤1在[0,1]上成立, 只要满足f(-1/2a)≤1和|f(1)|≤1,有a≤-1/4,-2≤a≤0。所以-2≤a≤-1/2.
(3)-1/2a>1,-1/2<a<0.f(x)=(ax+1),x1=0.x2=-1/a,根据|f(x)|的图像性质, 要使|f(x)|≤1在[0,1]上成立,只要满足f(1)≤1,,有a≤0.所以,-1/2<a<0
由(1),(2),(3)知,2≤a<0
温馨提示:答案为网友推荐,仅供参考