Question

14题怎么做?

Answer 1

以A为坐标原点,AD为y正半轴,AB为x正半轴
设AE=a 则BG=2a ,E(0,1),F(a,0),G(2,2a)
Kef=-1/a ,Kag=2a/2=a
AE方程:y=-1/a*x+1...............(1)
AG议程:y=ax.............................(2)
解(1)(2)得
x=a/(a^2+1),y=a^2/(a^2+1)
O[a/(a^2+1),a^2/(a^2+1)]
OB^2=[2-a/(a^2+1)]^2+a^4/(a^2+1)^2
=(2a^2-a+2)^2/(a^2+1)^2+[a^4/(a^2+1)^2]
=(5a^4-4a^3+9a^2-4a+4)/(a^2+1)^2
=[5(a^2+1)^2-(4a^3+a^2+4a^2+1)]/(a^2+1)^2
=5-[(4a^3+a^2+4a^2+1)/(a^2+1)^2]
=5-{[a^2(4a+1)+(4a+1)]/(a^2+1)^2}
=5-[(4a+1)(a^2+1)/(a^2+1)^2]
=5-[(4a+1)/(a^2+1)]
所以求 减数:(4a+1)/(a^2+1)最大值
令(4a+1)/(a^2+1)=t
ta^2+t=4a+1
ta^2-4a+t-1=0
判别式=16-4t(t-1)>=0
t^2-t-4<=2
t<=(√17-1)/2
(4a+1)/(a^2+1)最大值=(√17-1)/2

Answer 2

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