2019版数学人教B版选修4-5训练：1-5-2 综合法和分析法 WORD版含解析.docx

1、1.5.2综合法和分析法课时过关能力提升1.若1x10,则下面不等式正确的是()A.(lg x)2lg x2lg(lg x)B.lg x2(lg x)2lg(lg x)C.(lg x)2lg(lg x)lg x2D.lg(lg x)(lg x)2lg x2解析:1x10,0lg x1,0lg x22,0(lg x)21,lg(lg x)0.(lg x)2-lg x2=(lg x)2-2lg x=lg x(lg x-2)0,(lg x)2lg x2.lg(lg x)(lg x)2lg x2.答案:D2.已知0a10B.logab+logba-20C.logab+logba+20D.logab+l

2、ogba+20解析:0a1b,logab0.(-logab)+1-logab2,-logab+1-logab-2,即logab+1logab-2,当且仅当logab=-1,即b=1a时等号成立.logab+logba-2.logab+logba+20.答案:D3.设1313b13a1,则aa,ab,ba之间的大小关系是()A.aaabbaB.aabaabC.abaabaD.abbaaa解析:1313b13a1,0ab1,abaa.又aaba=aba,且0ab0,0aba1,aaba.abaa、”).答案:0)m,则宽为4xm,则水池总造价y=4120+2x2+8x280=480+320x+4x

3、480+1 280=1 760(元).当且仅当x=2时等号成立.答案:1 7607.给出下列命题:ab0ba1;ab0a-2b,cd,abcd0acbd;ab0|a+b|a|+|b|b0,cd0adbc.其中为真命题的是.(填所有真命题的序号)答案:8.设a,b,c均为正数,求证:12a+12b+12c1b+c+1c+a+1a+b.证明a,b,c均为正数, 1212a+12b12ab1a+b,当且仅当a=b时等号成立;同理1212b+12c12bc1b+c,当且仅当b=c时等号成立; 1212c+12a12ca1c+a,当且仅当a=c时等号成立.以上三个不等式相加,得12a+12b+12c1b+c+1c+a+1a+b,当且仅当a=b=c时等号成立.9.已知函数f(x)=lg1x-1,x0,12,若x1,x20,12,且x1x2,求证:12f(x1)+f(x2)fx1+x22.证明要证明原不等式成立,只需证明1x1-11x2-12x1+x2-12.x1,x20,12,x1x2,1x1-11x2-1-2x1+x2-12=1x1x2-1x1-1x2-4(x1+x2)2+4x1+x2=(x1-x2)2(1-x1-x2)x1x2(x1+x2)20.1x1-11x2-12x1+x2-12,lg1x1-11x2-1lg2x1+x2-12,即12f(x1)+f(x2)fx1+x22.