高考數(shù)學(xué)（重慶專用 理科）一輪復(fù)習(xí)題庫 12.4直接證明與間接證明 Word版含解析

1、課時(shí)作業(yè)66直接證明與間接證明一、選擇題1若a，br，則下面四個(gè)式子中恒成立的是()alg(1a2)0 ba2b22(ab1)ca23ab2b2 d2(2012河北張家口模擬)分析法又稱執(zhí)果索因法，若用分析法證明：“設(shè)abc，且abc0，求證a”“索”的“因”應(yīng)是()aab0 bac0c(ab)(ac)0 d(ab)(ac)03已知a，b，c，d為實(shí)數(shù)，且cd，則“ab”是“acbd”成立的()a充分不必要條件b必要不充分條件c充要條件d既不充分也不必要條件4要證：a2b21a2b20，只要證明()a2ab1a2b20ba2b210c1a2b20d(a21)(b21)05在abc中，角a，b，

2、c所對(duì)的邊分別為a，b，c，若c120，ca，則()aab babcab da與b的大小關(guān)系不能確定6用反證法證明：若整系數(shù)一元二次方程ax2bxc0(a0)有有理數(shù)根，那么a、b、c中至少有一個(gè)是偶數(shù)用反證法證明時(shí)，下列假設(shè)正確的是()a假設(shè)a、b、c都是偶數(shù)b假設(shè)a、b、c都不是偶數(shù)c假設(shè)a、b、c至多有一個(gè)偶數(shù)d假設(shè)a、b、c至多有兩個(gè)偶數(shù)7一個(gè)平面封閉區(qū)域內(nèi)任意兩點(diǎn)距離的最大值稱為該區(qū)域的“直徑”，封閉區(qū)域邊界曲線的長度與區(qū)域直徑之比稱為區(qū)域的“周率”，下面四個(gè)平面區(qū)域(陰影部分)的周率從左到右依次記為1，2，3，4，則下列關(guān)系中正確的是()a143 b312c421 d341二、填空

3、題8(2012廣東肇慶模擬)已知點(diǎn)an(n，an)為函數(shù)y圖象上的點(diǎn)，bn(n，bn)為函數(shù)yx圖象上的點(diǎn)，其中nn*，設(shè)cnanbn，則cn與cn1的大小關(guān)系為_9已知命題：“在等差數(shù)列an中，若4a2a10a( )24，則s11為定值”為真命題，由于印刷問題，括號(hào)處的數(shù)模糊不清，可推得括號(hào)內(nèi)的數(shù)為_10請(qǐng)閱讀下列材料：若兩個(gè)正實(shí)數(shù)a1，a2滿足aa1，那么a1a2.證明：構(gòu)造函數(shù)f(x)(xa1)2(xa2)22x22(a1a2)x1，因?yàn)閷?duì)一切實(shí)數(shù)x，恒有f(x)0，所以0，從而得4(a1a2)280，所以a1a2.根據(jù)上述證明方法，若n個(gè)正實(shí)數(shù)滿足aaa1時(shí)，你能得到的結(jié)論為_三、解答

4、題11已知an是正數(shù)組成的數(shù)列， a11，且點(diǎn)(，an1)(nn*)在函數(shù)yx21的圖象上(1)求數(shù)列an的通項(xiàng)公式；(2)若數(shù)列bn滿足b11，bn1bn2an，求證：bnbn2b.參考答案一、選擇題1b解析：在b中，a2b22(ab1)(a22a1)(b22b1)(a1)2(b1)20，a2b22(ab1)恒成立2c解析：ab2ac3a2(ac)2ac3a2a22acc2ac3a202a2acc202a2acc20(ac)(2ac)0(ac)(ab)03b4d解析：因?yàn)閍2b21a2b20(a21)(b21)05a解析：由正弦定理，得，sin aa30b180120a30ab6b解析：“至

5、少有一個(gè)”的否定是“都不是”7c解析：在圖(1)中，設(shè)圖形所在的矩形長為a，寬為b，則其周率為，由不等式的性質(zhì)可知2；在圖(2)中設(shè)大圓的半徑為r，則易知外邊界長為r，而內(nèi)邊界恰好為一個(gè)半徑為的小圓的周長r，故整個(gè)邊界長為2r，而封閉曲線的直徑最大值為2r，故周率為；圖(3)中周長為直徑的三倍，故周率為3；圖(4)中設(shè)各邊長為a，則整個(gè)邊界的周長為12a，直徑為2a，故周率為2，故三個(gè)周率大小符合選項(xiàng)c二、填空題8cn1cn解析：由條件得cnanbnn，cn隨n的增大而減小cn1cn918解析：s1111a6，由s11為定值，可知a6a15d為定值設(shè)4a2a10an24，整理得a1d4，可知n

6、1810a1a2an解析：構(gòu)造函數(shù)f(x)(xa1)2(xa2)2(xan)2nx22(a1a2an)x1，因?yàn)閷?duì)一切實(shí)數(shù)x，恒有f(x)0，所以0，從而得4(a1a2an)24n0，所以a1a2an三、解答題11解：(1)由已知得an1an1，則an1an1，又a11，所以數(shù)列an是以1為首項(xiàng)，1為公差的等差數(shù)列故an1(n1)1n(2)由(1)知，ann，從而bn1bn2nbn(bnbn1)(bn1bn2)(b2b1)b12n12n2212n1因?yàn)閎nbn2b(2n1)(2n21)(2n11)2(22n22n22n1)(22n222n11)2n0，所以bnbn2bc29c91afe4ced

7、1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afedd1b6c87d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af95ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9d1b6c8795ad9c91afe4ced1b6c8795a1afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c

8、91afc91fe4ced1d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6cd1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af8795ad9c91fe4ced1d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c87d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af5d1b6c8795ad9c91afe4c

9、ed1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afad9c91fe4ced1b6c8795ad29c91afb6c8795ad29c91afb6c8795ad29c91afd1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91affe4c

10、ed1b6c8795ad29c91afc8795ad9c91fe4ced1b6c8795ad29c91afd1b6c8795ad9c91afed1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8d1b6c8795ad9c91afe4ced1d91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afc91fe4ced1d1b6c871afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afc91fe4ced1d1b6c8795ad9c91afe4ced

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12、e4ced1b6c8795ad29c91afad9c91fe4ced1b6c8795ad29c91afb6c8795ad29c91afb6c8795ad29c91afd1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8791afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afc91fe4ced1d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6cd1b6c8

13、795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af8795ad9c91fe4ced1d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c87d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af5d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afad9c91f

14、e4ced1b6c8795ad29c91afb6c8795ad29c91afb6c8795ad29c91afd1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91affe4ced1b6c8795ad29c91afc8795ad9c91fe4ced1b6c8795ad29c91afd1b6c8795ad9

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17、4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afad9c91fe4ced1b6c8795ad29c91afb6c8795ad29c91afb6c8795ad29c91afd1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91affe4ced1b6c8795ad29c91

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