已知A={x|(2x+1)/(x-3)>=1},B={y|y=barctant,-1

Question

已知A={x|(2x+1)/(x-3)>=1},B={y|y=b arctan t,-1<=t<=√3/3,b<=0},A∩B=Φ,求b取值范围
RT,另注：
arctan t表示反正切,当t∈R时,arctant∈(-π/2,π/2),tan(arctant)=t,例如：arctan1=π/4,arctan(-√3)=-π/3,arctan(√3/3)=π/6,y=arctant在R上单调递增

Answer 1

已知A={x|(2x+1)/(x-3)≧1},B={y|y=b arctan t,-1≦t≦√3/3,b≦0},A∩B=Φ,求b取值范围
A={x|(2x+1)/(x-3)≧1}={x︱(2x+1)/(x-3)-1≧0}={x︱(x+4)/(x-3)≧0}={x︱x≦-4或x>3}
B={y|y=b arctan t,-1≦t≦√3/3,b≦0}={y|y/b=arctant,-1≦t≦√3/3,b≦0}
={y︱-π/4≦y/b≦π/6,b≦0}={y︱bπ/6≦y≦-bπ/4,b≦0}
A∩B=Φ,∴-4