1:若不等式x^2+2+|x^3-2x|>=ax对x属于(0,4)恒成立,则实数a的取值范围是?2:设x>0,y>0,且x^2+y^2/2=1,求x跟下(1+y^2)的最大值~
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1:若不等式x^2+2+|x^3-2x|>=ax对x属于(0,4)恒成立,则实数a的取值范围是?2:设x>0,y>0,且x^2+y^2/2=1,求x跟下(1+y^2)的最大值~
1:若不等式x^2+2+|x^3-2x|>=ax对x属于(0,4)恒成立,则实数a的取值范围是?
2:设x>0,y>0,且x^2+y^2/2=1,求x跟下(1+y^2)的最大值~
1:若不等式x^2+2+|x^3-2x|>=ax对x属于(0,4)恒成立,则实数a的取值范围是?2:设x>0,y>0,且x^2+y^2/2=1,求x跟下(1+y^2)的最大值~
解(1)x^2+2+|x^3-2x|≥ax
因为x∈(0,4)
所以两边同时除以x
所以有:x+2/x+|x^2-2|≥a
对于x+2/x+|x^2-2|在x=√2时有最小值
所以a∈(-∞,2√2]
(2)设x=cosα,y=√2sinα,0<α<π/2
x√(1+y^2)=cosα√[1+2(sinα)^2]
=√[(cosα)^2+2(sinαcosα)^2]
=√{(cosα)^2+(1/2)[sin(2α)]^2}
=√{(1/2)[1+cos(2α)]+(1/2)[sin(2α)]^2}
=√(1/2)√{1+cos(2α)+[sin(2α)]^2}
=√(1/2)√{2+cos(2α)-[cos(2α)]^2}
=√(1/2)√{-[cos(2α)-1/2]^2+9/4}
0<α<π/2
0<2α<π
-1<cos(2α)<1
-3/2<cos(2α)-1/2<1/2
0≤[cos(2α)-1/2]^2<1/4或0≤[cos(2α)-1/2]^2<9/4
0≤[cos(2α)-1/2]^2<9/4
-9/4<-[cos(2α)-1/2]^2≤0
0<9/4-[cos(2α)-1/2]^2≤9/4
0<√{9/4-[cos(2α)-1/2]^2}≤3/2
0<x√(1+y^2)≤(3/2)√(1/2)
最大值(3/2)√(1/2)