已知f(x)=loga(a﹣x+1)+bx(a>0,a≠1)是偶函式,則( )A.b=且f(a)>f() B...
問題詳情:已知f(x)=loga(a﹣x+1)+bx(a>0,a≠1)是偶函式,則()A.b=且f(a)>f() B.b=﹣且f(a)<f()C.b=且f(a+)>f() D.b=﹣且f(a+)<f()【回答】C【解答】解:∵f(x)=loga(a﹣x+1)+bx(a>0,a≠1)是偶函式,∴f(﹣x)=f(x),即loga(ax+1)﹣bx=loga(a﹣x+1)+bx,∴loga(ax+1)﹣bx=loga(ax+1)+(b﹣1)x,∴﹣b=b﹣1,∴b=,...