Question

若x满足不等式2(log0.5x)∧2+7log0.5x+3≤0,则f(x)=log0.5(x/2)log0.5(x/4)的最值?

Answer 1

2(log0.5x)∧2+7log0.5x+3≤0
[2(log0.5x)+1]*[(log0.5x)+3]≤0
-3≤(log0.5x)≤-1/2
log0.5(x/2)log0.5(x/4)=[(log0.5x)+1]*[(log0.5x)+2]=(log0.5x)∧2+3log0.5x+2
1/4≤(log0.5x)∧2+3log0.5x+2≤2
f(x)=log0.5(x/2)log0.5(x/4)的最小值是1/4，最大值是2.

Answer 2

解：∵在圆柱形水管的截面圆O中，OF是圆O的半径，直线CP切圆O于点C、P
∴OF垂直并平分CP ∵连接OC、OP，OC=OP=R
∴则Rt△OCG的高OG=OF-GF=R-2 ∴R=√(R-2)²+(20/2)²=26
∵连接OA、OB，OA=OB=R ∴则Rt△OEA的高OE=OF-GF-EG=R-4=26-4=22
∴AE=√(OA²-OE²)=26²-22²=8√3 ∴AB=8√3×2=16√3