设f(x)=|lgx|,若0< a< b< c,且f(a)> f(c> f(b),下列关系式正确的是 (A)ac+1< a+c (B)ac+1> a+c (C)ac+1=a+c (D)ac> 1

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答案:A∵0<a<b<c,则lga<lgb<lgc又∵|lga|>|lgc|>|lgb|∴lga<0,lgb>0,lgc>0∴c>b>1>a>0∴(ac+1)-(a+c)=ac+1-a-c=(c-1)(a-1)<0∴ac+1<a+c.