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1、银川二中 2015-2016 学年第一学期高三年级统练一数 学 试 卷（理科）一、选择题（本大题共12 小题，每小题5 分，满分 60 分.在每小题给出的四个选项中，只有一项是符合题目要求的.）1设集合5,2,3,2,1,6,5,4,3,2,1BAU，则)(BCAU=（）A1，3B2C 2，3D3答案：A试题分析：因为1,3,4UC B，所以()1,2,31,3,41,3UAC B，选 A考点：集合运算2在等差数列na中，若,32a943aa，则61aa=（）A18 B14 C2 D27答案：B试题分析：34255161699612,714.aaaaadaaa a选 B 考点：等差数列通项3函
2、数 f（x）x34x 5的图象在x1 处的切线在x 轴上的截距为（）A10 B 5C 1 D37答案：D试 题 分 析：因 为2()34fxx，所 以(1)7kf，切 线 方 程 为：(1)7(1)107(1)yfxyx，令0y得37x，选 D考点：导数几何意义4等比数列na的前 n 项和为nS，已知12310aaS，95a，则1a=（）A31 B31 C91 D91答案：D试 题 分 析：232112321311 01 099Saaaaaaaaaq，因 此 由425111199999aa qaa，选 D考点：等比数列通项5将函数)62sin(xy图象向左平移4个单位，所得函数图象的一条对称轴
3、的方程是（）A12x B6x C 3x D 12x答案：A试 题 分 析：函 数)62sin(xy图 象 向 左 平 移4个 单 位，所 得 函 数 为sin(2()sin(2)463yxx，所 以 由2,()32xkkZ得 对 称 轴 方 程 为,()122kxkZ，从而一条对称轴的方程是12x，选 A考点：三角函数图像与性质6已知|a|1，|b|2，a与b的夹角为60，则ab在a上的投影为（）A1 B2 C772 D77答案：B试题分析：ab在a上的投影为()112cos602|aaba，选 B考点：向量投影7已知,0，22)3cos(，则2tan（）A33 B3或33 C 33 D 3答
4、案：C试题分析：40,333，因 此由22)3cos(得3553,2,tan2.341263选 C考点：特殊角三角函数值8在 ABC中，M为边 BC上任意一点，N为 AM的中点，ANABAC，则 的值为（）A12 B13 C14 D1答案：A试题分析：因为222AMANABAC，所以1221,2选 A考点：向量共线表示9已知命题p：函数2()21(0)fxaxxa在（0，1）内恰有一个零点；命题 q：函数2 ayx在(0,)上是减函数，若p 且q为真命题，则实数a的取值范围是（）A1aBa2 C 12答案：C试题分析：命题p 为真时：0,(1)2201afaa；命题q 为真时：文档编码:CY4
5、Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9
6、文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10
7、T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9
8、 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2
9、X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5
10、K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码
11、:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X
12、10G920,2aa，因为 p 且q为真命题，所以命题 p 为真，命题 q 为假，即12a，选 C考点：命题真假10ABC中，角,A B C成等差数列是sin(3cossin)cosCAAB成立的（）A充分不必要条件 B必要不充分条件C充要条件 D既不充分也不必要条件答案：A试题分析：由题意得：sin()(3cossin)coscossin3 coscoscos0tan3ABAABABABAB或即23AB或，而角,A B C成等差数列，则3B，因此角,A B C成等差数列是sin(3cossin)cosCAAB成立的充分不必要条件，选A考点：充要关系11在正项等比数列an中，存在两项nmaa,
13、，使得nmaa 41a，且5672aaa，则nm51的最小值是（）A47 B135 C625D352答案：A试 题 分 析：2765221(),2aaaqqqq舍；由mna a 21a得224m nq，即22222,6mnmn，因此nm511515151()(6)(62)(62 5)6666mnnmnmmnmnmn，但等于 号 取 不 到，从 而 逐 一 验 证12345,54321mmmmmnnnnn得24mn时nm51取最小值为74，选 A考点：等比数列性质12函数)(xfy为定义在R上的减函数，函数)1(xfy的图像关于点（1，0）对称，,x y满足不等式0)2()2(22yyfxxf，
14、(1,2),(,)MN x y，O为坐标原点，则当41x时，OM ON的取值范围为（）A,12B3,0 C12,3 D12,0文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9
15、HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q
16、8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文
17、档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T
18、10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9
19、ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X
20、1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K
21、4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9答案：D试题分析：由题意得函数()yf x的图像关于点（0，0）对称，即函数()yf x为奇函数，因此由0)2()2(22yyfxxf得22(2)(2)f xxfyy，2222,()(2)0 xxyyxyxy因为41x所以可行域为一个三角形ABC及其内部，其中(11)(4 4),(42)ABC，而2OMONxy，所以过点C时取最小值0，过点 B时取最大值12，选 D 考点：线性规划求最值二、填空题（本大题共4 小题，每小题5 分，共 20 分.）
22、13已知m，()yf x，x，且2ab与ab垂直，则实数的值为答案：29试题分析：由题意得：229(2)()02.2ababab考点：向量数量积14已知数列na的前 n 项的和nS满足nSn)1(log2，则na=答案：12n试题分析：2log(1)12nnnSnS，所以111222(2)nnnnnnaSSn，又11=211aS，因此na=12n考点：数列通项15已知函数()2sin(),(0)6f xx的图象与y 轴交于 P，与 x 轴的相邻两个交点记为 A，B，若 PAB的面积等于 ，则 _答案：12试题分析：由题意得：(01)2TPAB，因此111=.22，考点：三角函数性质16 ABC
23、为锐角三角形，内角CBA,的对边长分别为cba,，已知2c，且AABC2sin2)sin(sin，则a的取值范围是_答案：2 5 2 3(,)53试题分析：sinsin()2sin 2sin()sin()2sin 22sincos4sincosCBAAABBAABAAA文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10
24、I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D
25、7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9
26、HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q
27、8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文
28、档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T
29、10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9
30、ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9，因为ABC为锐角三角形，所以sin2sin2BAba，因为ABC为锐角三角形，所以2222220,0,abcacb即2254,34,aa解得a的取值范围是25 23(,)53考点：正弦定理，余弦定理应用三、解答题(本大题 6 小题,共 70 分,解答应写出文字说明、证明过程或演算步骤，并把解答写在答卷纸
31、的相应位置上)17（本题满分12 分）已知等差数列na中，,21a1053aa（1）求数列na的通项公式;（2）令1nnnaab，证明：2111121nbbb答案：（1）1nan（2）详见解析试题分析：（1）求等差数列通项公式，一般利用待定系数法，求出首项及公差，代入通项公式1(1)naand即可。本题已知首项，只需由1053aa解出公差即可；（2）证明数列和式不等式，一般方法为以算代证，先分析数列通项1nb的特征，可知利用裂项相消法求和，即由1111111()nnnnnbaadaa得其和为11111()nd aa1122n，因此只需再利用放缩即可得证试题解析：（1）联立解得：,21a,1d1
32、nan；（2）证明：由（1）知，)2)(1(nnbn;212121)2)(1(143132111121nnnbbbn考点：等差数列通项，裂项求和18（本小题满分12 分）f（x）=ab，其中向量a=（m，cos2x），b=（1+sin2x，1），xR，且函数()yf x的图象经过点(,2)4（）求实数m的值（）求函数()yf x的最小值及此时x值的集合。答案：（）1m（）最小值为12，3|,8x xkkZ文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4
33、Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9
34、文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10
35、T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9
36、 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2
37、X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5
38、K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码
39、:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9试题分析：（）由向量数量积坐标公式得函数()yfx(1sin 2)cos2mxx，再根据()4f2 得1m（）先利用配角公式将()yf x化为基本三角函数：()1sin
40、 2cos212 sin(2)4f xxxx，再根据正弦函数性质求最小值及对应x值：当si n(2)14x时，()yf x的 最 小 值 为12，x值 的 集 合 为3|,8x xkkZ试题解析：（）()fxa b(1sin 2)cos2mxx由已知()4f(1sin)cos222m，得1m（）由（）得()1sin 2cos212 sin(2)4f xxxx当sin(2)14x时，()yfx的最小值为12，由sin(2)14x，得x值的集合为3|,8x xkkZ考点：向量数量积，配角公式，三角函数性质19（本题满分12 分）如图，在ABC中，BC边上的中线AD长为 3，且10cos8B，1co
41、s4ADC（1）求sinBAD的值；（2）求AC边的长答案：（1）64（2）4试题分析：（1）利用角的关系BADADCABD，再结合两角差正弦公式展开就可求解（2）先在三角形ABD中，由正弦定理解出BD长，即 CD长：由正弦定理，得文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4
42、Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9
43、文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10
44、T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9
45、 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2
46、X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5
47、K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码
48、:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9sinsinADBDBBAD，即33 6684BD，解 得2BD 故2DC；再 在 三 角 形ADC 中 由 余 弦 定 理 解 出AC：2222cosACADDCAD DCADC221322 32()164；AC=4试题解析：（1）;863sin,810cosBB451sin,41cosADCADC;46)sin(sinBADCBAD（2）
49、在ABD中，由正弦定理，得sinsinADBDBBAD，即33 6684BD，解得2BD故2DC，从而在ADC中，由余弦定理，得2222cosACADDCAD DCADC221322 32()164；AC=4 ;考点：正余弦定理20（本小题满分12 分）已知等比数列na是递增数列，,3252aa1243aa，数列nb满足11b，且nnnabb221（Nn）（1）证明：数列nnab是等差数列；（2）若对任意Nn，不等式nnbbn1)2(总成立，求实数的最大值答案：（1）详见解析（2）12试题分析：（1）求等比数列通项公式，一般利用待定系数法，求出首项及公比，代入通项公式11nnaa q即可。本题
50、先利用等比数列性质转化条件25343232a aa a；再结合1243aa联立方程组解出34,a a，根据等比数列na递增性，舍去一解，最后根据43aqa求出公比及首项。证明数列为等差数列，一般利用定义进行证明，即证11nnnnbbaa为一个常数（2）不等式恒成立，先利用变量分离，转化为研究函数最值，即文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9 HJ2X1R9D7I9 ZS10I10T10X10G9文档编码:CY4Q8M5K4M9