当x=2时,y=loga5>0,
∴a>1.由x2+2x-3>0⇒x<-3或x>1,
易见函数t=x2+2x-3在(-∞,-3)上递减,
故函数y=loga(x2+2x-3)(其中a>1)也在(-∞,-3)上递减.
故选A
函数y=loga(x2+2x-3),当x=2时,y>0,则此函数的单调递减区间是( ) A.(-∞,-3) B.(1,+∞) C.(-∞,-1) D.(-1,+∞)
函数y=loga(x2+2x-3),当x=2时,y>0,则此函数的单调递减区间是( )
A. (-∞,-3)
B. (1,+∞)
C. (-∞,-1)
D. (-1,+∞)
A. (-∞,-3)
B. (1,+∞)
C. (-∞,-1)
D. (-1,+∞)
数学人气:464 ℃时间:2020-03-30 23:24:05
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