數(shù)學必修五課程小練習與單元統(tǒng)練模塊試題mah

班學姓成11ba(2)ab，cacbd；(3)ac2bc2ab (4)c2c班學姓成11ba(2)ab，cacbd；(3)ac2bc2ab (4)c2cc(5)acbd，且ab0cd00a2(1)a4b4ab(3)ab，cdc32bc2d2a(4)ab且algcblgc0c10cacb已知ba0c0，；acbd；11；abccabcabc0，.ab14.(0.4)若b2，則的取值范圍為.b5.(0.8)已知3ba2,abab的取值范圍.6.0.4分)abcddcabcdadbc.[下方區(qū)域為改錯區(qū)域，嚴禁用作草稿常規(guī)作 4(理由陳述，地方不夠，可以另附紙張，大小以試卷為準，標明題號1.(0.4)0<a12.②ax常規(guī)作 4(理由陳述，地方不夠，可以另附紙張，大小以試卷為準，標明題號1.(0.4)0<a12.②ax(1)2①log(1a)④(1a)nan,nN③sin(1a)sin(1a2.(0.4)已知3ab2，1ab4a的取值范圍為，2ab的取值范圍.13.(0.4)已知2x4y1x2.(”填空1log1.(”填空113225.(0.7)已知abcRabc0abc0111 6.(0.7)已知數(shù)列{a}a2n1,n為偶數(shù)nN*，求數(shù)列{a}S的公式nnnn2，nn1.2.3. (2)(3)(4)(1)≤(4)(,1]U(0,)24.[6,4]，(0,5.acd6.1.2.3. (2)(3)(4)(1)≤(4)(,1]U(0,)24.[6,4]，(0,5.acd6.1.(0.4)0<a121.ax(①loga(1a)x③sin(1a)sin(1④(1a)a,n *②2因為0a1，所以1a011a1，所以有1aa222ylogax在(0,上單調(diào)遞減，所以有l(wèi)oga(1alogaa1x0時不成立③因為11a3 11a1a，22又因為函數(shù)ysinx在 )上單調(diào)遞增，所以sin(1a)sin(12④因為1aa0，由不等式性質(zhì)可知(1a)nannN*2.(0.4)已知3ab2，1ab4，2ab的取值范圍.的取值范圍為1a302ab7（1）因為3ab2，1ab所以22a61a（2）設uabvab，則3u2，1v所以auvbuv22所以2ab1u3v3u133v 02ab7 13.(0.4)已知2x4y1x2.(”填空x14(5y11xy∴xy 251log1.(”填空11322lg 1131lg3lg2lg3lg22＝3(作差法1211213 3(lg3lg2)(lg313.(0.4)已知2x4y1x2.(”填空x14(5y11xy∴xy 251log1.(”填空11322lg 1131lg3lg2lg3lg22＝3(作差法1211213 3(lg3lg2)(lg3lg11＝∴l(xiāng)og13＞log123 ＞ 231111112∴l(xiāng)og13＞log＝1，log ＜log ∴l(xiāng)og13＞log223323111 5.(0.7)已知abcRabc0abc0∴a2b2c22ab2ac2bc0--abc∴abacbc ∴a2b2c2abbc111abc且abacbc ∵ 111∴ 0---- 6.(0.7)已知數(shù)列{a}a2n1,n為偶數(shù)nN*，求數(shù)列{a}n的公式nnnnn2，n解：依題意可知數(shù)列{an}的前幾項Sn(a1a3Lan2an)(a2a4Ln2(32(n1)1)233n23n 2，426（2）當n2n23(n1)2