∵函数y=lg(x2+2x+m)的值域为R,
∴f(x)可以取所有的正数可得,△≥0
∴△≥0,可得4-4m≥0,
解得m≤1,
故选:C.
函数y=lg(x2+2x+m)的值域是R,则m的取值范围是( ) A.m>1 B.m≥1 C.m≤1 D.m∈R
函数y=lg(x2+2x+m)的值域是R,则m的取值范围是( )
A. m>1
B. m≥1
C. m≤1
D. m∈R
A. m>1
B. m≥1
C. m≤1
D. m∈R
数学人气:950 ℃时间:2019-10-05 02:31:02
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