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高一年级期末必修四必修五试题(数学)一.选择题:(每小题4分共40分)在下列每小题给出的四个选项中,只有一个符合要求,请选出并填入下列表中相应的位置题号12345678910代码1.不等式x-2y+60表示的平面区域在直线:x-2y+6=0的·························()1A.右上方B.右下方C.左上方D.左下方2.若A为△ABC内角,则下列函数中一定取正值的是:········································()2A.sinAB.cosAC.tanAD.sin2A3在△ABC中3,2ba.B=60那么角A等于:·······································()3A.135B.90C.45D.304.设0<b<a<1,则下列不等式成立的是:··························································()4A.ab<b2<1B.0loglog2121abC.a2<ab<1D.ba)21()21(215.设数列{an}是等差数列,若a2=3,a7=13.数列{an}的前8项和为:·······················()5A.128B.80C.64D.566.在△ABC中,若BbAacoscos,则△ABC的形状是:·····································()6A.等腰三角形B.直角三角形C.等腰或直角三角形D.等腰直角三角形7.数列{an}的通项公式为11nnan,前n项和Sn=9,则n等于:···············()7A.98B.99C.96D.978.不等式1||31xyxy表示区域的面积为:························································()8A.1B.21C.25D.239.若ab0,则下列不等式中一定成立的是……………………………………()9A.abba11B.abba11C.11ababD.bababa2210.已知数列{an}的通项公式an=n2+-11n-12,则此数列的前n项和取最小值时,项数n等于()10A.10或11B.12C.11或12D.12或13二.填空题:(每小题4分共20分)11.不等式125x的解集为:.12.在各项都为正项的等比数列{an}中a1=3,S3=21,则a3+a4+a5=.13.在△ABC中,角A.B.C.的对边分别为:a,b,c,若BsinCsin,bcba32322则角A=.14..若数列:12+22+32+42+••••••+n2=6)12)(1(nnn则:数列:1,2,2,3,3,3,4,4,4,4,•••••••••••••••的前100项的和是.15.x,y满足约束条件0,002063yxyxyx若目标函数z=ax+b(a0,b0)的是最大值为12.则ba32的最小值为三.解答题()16.(10分)已知:A.B.C为△ABC的三内角,且其对边分别为a,b,c,若21CsinBsinCcosBcos.(Ⅰ)求A.(Ⅱ)若432cb,a,求△ABC的面积.17.(10分)若不等式0252xax的解集是221xx,(1)求a的值;(2)求不等式01522axax的解集.18.(8分)若实数x,y满足:001xyx求:xy的范围19.(6分)设正数x;y满足x+2y=1求yx11的最小值20.(6分)已知数列{an}的首项12,3211nnnaaaanN*(Ⅰ)证明数列{11na}是等比数列.(Ⅱ)数列{nan}的前n项的和Sn2013年度高一年级期末考试试题(数学)答案:一选择题4.特殊值+筛选2141ab6.将ab分别换成sinAsinB7.再叠加分母有理化后nna18.用的方法:用23||21ADS9.强烈建议“逆证法”如:C、假abaabbababab11D、真22222222ababababbababa10.令an=0得n=12,∴S11=S12由开中向上的抛物线性质可知:当n≤12时an≤0,当n>0时an>0也就是an从第十三项开始大于零,S13=S12+正数S12。以后单调递增。二填空题11.(-,-2)∪(3,+)12.8413.30°解∵BCsin32sin∴bc32bababbbba776)32(322222令32,7,1cab则再由余弦定理即得14.954解:在相同的数n中,最后一个n是原数列的第(1+2+……+n)项,如:最后一个3是第1+2+3=6项,131002)1(nnn最大的由也就是最后一个13是数列的第91项945914)13.......21(222100S15.126411联立两直线得)6.4(A是目标函数z=ax+b的最优解12=4a+bbbaabababa446441261232变量分离后再用均值定理题号12345678910代码BACDCABDDCA063yx02yx)6.4(A23AB2,1)21,21(CD三解答题:16.解:Ⅰ)原式可化为:12021cos21AACBcos即:)(Ⅱ)由余弦定理可知:bc16bccbbccb120bc2cb32222222)(cos)(∴bc=4,323421120421Abc21Ssinsin17(1)2525221025221221aaxaxx,x的两根由韦达定理可是方程(2)ax2-5x+a2-10可化为:-2x2-5x+30即2x2+5x-30(2x-1)(x+3)0213x18.解:0,1xxy110111xyxxy即19.223223)22(3221221121yxxyyxxyyxxyyyxxyxyxyx证明:)22.2(22”时取“也就是即当且仅当yxyxyxxy223)11(minyx20.Ⅰ))(*nnnNnaaa1212112311112111212121111nnnnnnnaaaaaaa)而(为公比的等比数列、为首项}是以数列{2121112111111nnnaqaaⅡ)nnnnnnnnnanaa22111212121111)()()()(nnn.......n.......S212132122113213213221211n212211232121nnnn.......n.......S)(()(13221211212211232121nnnnn.......n.......S)(()(11142221141221121-1214121212123121214121nnnnnnnnnnnnnn.....nnS)(】)(【)()())((nnnnnnS2212121)(
本文标题:人教版高一数学必修四必修五期末测试题
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