Question

已知函数f(x)=ax^2+2ax+4(a>3),若X1<X2.X1+X2=1-a,则f(x1)>f(x2)??/

Answer 1

f(x1)-f(x2)
=ax1^2+2ax1+4-(ax2^2+2ax2+4)
=a(x1²-x2²)+a(x1-x2)
=a(x1+x2)(x1-x2)+a(x1-x2)
=a(x1-x2)(x1+x2+1)
∵X1+X2=1-a,a>3
∴1-a<-2
∴x1+x2<-2
∴x1+x2+1<-1<0
∵a>3,x1-x2<0
∴f(x1)-f(x2)>0
∴f(x1)>f(x2)

Answer 2

f(x)=ax²+2ax+4(a>3),若X1<X2.X1+X2=1-a
f(x1)-f(x2)=ax1²+2ax1+4-ax2²-2ax2-4=a（x1²-x2²）+2a（x1-x2）=（x1-x2)[a（x1+x2）+2a]
=（x1-x2)[a（1-a）+2a]=（x1-x2)（3a-a²）
∵X1<X2∴x1-x2＜0
∵a>3∴3a-a²＜0
∴f(x1)-f(x2)＞0

Answer 3

令y=o得X1+X2=-2而题意则是X1+X2=1-aa＞3则X1+x2＜-2按照抛物线A＞3，开口向上，假设X2与与Y=0时的X2在同一点。那么X1在Y=0时X1的左边，所以Y（X1）大于Y（x2）

Answer 4

看不懂额