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2012高考数学试题(全国卷)解析一选择题:(共12个小题,每小题5分,满分60分)1. 复数=(A) 2+i(B) 2-i(C) 1+2i(D)1-2i选(C)【解法一】:分母实数化后,直接得结果.【解法二】:在复平面内画出-1+3i和1+i对应的向量,易知两向量的夹角大于45o小于90o,四个选项中只有(C)满足。2.已知集合A=1,3,B =1,m,AB =A,则m =(A) 0或(B) 0或3(C) 1或 (D) 1或3选(B)【解法一】:由AB =A BA mA m =3或m =当m =时,m = 0或m = 1(舍去),综上:m = 0或3【解法二】:筛选法,将各选项一一验证知(B)满足条件3.椭圆的中心在原点,焦距为4,一条准线为x = - 4,则该椭圆的方程为(A) =1(B) =1(C) =1 (D) =1选(C)【解法一】:由条件得方程组解得:c =2, a2 = 8, b2 = 4 .【解法二】:筛选法,将各选项一一验证知(C)满足条件4.已知正四棱柱ABCD -A1B1C1D1中,AB = 2,CC1 = 2,E为CC1的中点,则直线AC1与平面BED的距离为:(A) 2(B) (C) (D) 1选(D)【解法一】:设ACBD = O,作CHOE交OE于H,易知:平面OCE平面BDE,则CH平面BDE,由AB = 2,CC1 = 2 CO = CE,则H为OE的中点 CH =CE =1显然A、C到平面BED的距离相等,所以AC1与平面BED的距离为1 .【解法二】:(等体积法)VC-BED = VE-BCD SBDEd = SBDCCE d =1【解法三】:(向量法)略5.已知等差数列an的前n项和为Sn,a5 = 5,S5 =15,则数列的前100项和为(A) (B) (C) (D) 选(A)【解】:由a5 = 5,S5 =15得a1 = 1,d = 1,则an = n,=-易得的前n项和为1-,所以的前100项和为1-=6.ABC中,AB边的高为CD,= a,= b,ab = 0,| a | = 1,| b | = 2,则=(A)a -b(B) a -b(C)a -b(D) a -b选(D)【解法一】:如图所示,A、B、D共线,可设= l+(1-l)= la +(1-l)bCDAB = 0la +(1-l)b) ( a-b) = 0la2 - (1-l)b2 =0l =+= -b +a+b =a -b【解法二】:DECDEAABC CE:DE = DE:AE = BC:AC = 1:2 CE:AE = 1:4,CE:AC = 1:5,= b,同理CF:CB = 4:5,= a=+=+= -b +a+b =a -b【解法三】:因为AD AB = AC 2, BD AB = BC 2所以AD:BD = AC 2: BC 2= 4:1 AD:AB = 4:5=( a - b)【解法四】:以CA为x轴、CB为y轴建立坐标系,求出D的坐标为(,),从而可得=(-,)a = (0,1),b = (2,0),经检验=( a - b)7.已知a 为第二象限的角,sina + cosa =,则cos2a =(A) -(B)-(C) (D) 选(A)【解法一】:设cosa = x,sina = y,则P(x,y)是圆弧C:x 2+ y2 = 1(x 0)与直线l:x + y =的交点联立以上两方程解得:x =,y =则cos2a = x 2-y2 = -【解法二】:设cosa = x,sina = y,则x 2+ y2 = 1(x y ) 且x + y = (x - y) 2 +(x + y)2 = 2(x 2+ y2 ) = 2 (x - y) 2 =x - y = - cos2a = x 2-y2 = (x - y) (x + y) = -【解法三】:由sina + cosa = sin(45o+a ) = cos(45o+a ) = -(45o+a 为第二象限的角) cos2a =sin(90o+2a) = 2sin(45o+a )cos(45o+a ) = -【解法四】:由sina + cosa = sin(45o+a ) = sin165o =sin(45o+a ) =sin135ok360 o+ 135o 45o+a k360 o+ 165o (kZ)k360 o+ 90o a k360 o+ 120o (kZ) k360 o+ 180o 2a k360 o+ 240o (kZ) cos2a -显然只有(A)成立.8.已知F1、F2为双曲线C:x 2-y2 =2的左、右焦点,点P在C上,|PF1|=2|PF2|,则cosF1PF2 = (A) (B) (C)(D) 选(C)【解】:不妨设P在右支上,由双曲线的定义知:|PF1| - |PF2| = 2,由|PF1|=2|PF2|得:|PF1|= 4,|PF2| =2,|F1F2| = 4,由余弦定理得:cosF1PF2 =9.已知x = lnp,y =log 5 2,z =,则(A) x y z(B) z x y(C) z y x(D) y z 1, y与z都小于1,= log 2 52,= yz y z x10.已知函数y =x 3-3x + c的图像与x轴恰有两个公共点,则c =(A) -2或2(B) -9或3(C) -1或1(D) -3或1选(A)【解】:y =3x 2-3=3(x -1) (x +1),x = -1、x = 1分别是函数的极大值点、极小值点.画出简图,可知恰有一个极值为0,即13 - 31+ c = 0或(-1)3 - 3(-1)+ c = 0 c = 211.将字母a, a, b, b, c, c排成三行两列,要求每行的字母互不相同,每列的字母也互不相同,则不同点排列方法共有(A) 12种(B) 18种(C) 24种(D) 36种选(A)【解】:先排第一列,排法有= 6种,再排第二列,排法有2种,由分步计数原理,共有12种排法.12.正方形ABCD的边长为1,点E在边AB上,点F在边BC上,AE = BF =,动点P从E出发沿直线向F运动,每当碰到正方形的边时反弹,反弹时反射角等于入射角,当点P第一次碰到E时,P与正方形的边碰撞的次数为(A) 16(B) 14(C) 12(D) 10选(B)【解】:二填空题:(共4个小题,每小题5分,满分20分)13.若x、y满足约束条件 x - y +10x + y -30 x + 3y -30,则z =3x - y的最小值为 -1 【解】:如图,A(0, 1),B(3, 0)是最优解,代入目标函数可得z =3x - y的最小值为-1.14.当函数y = sinx - cosx (0x 2p)取得最大值时,x =【解】:sinx -cosx =2sin(x -) (-x -)y = sinx -cosx (0x 2p)取得最大值时, x -= x =15.若(x +) n的展开式中第三项与第七项的二项式系数相等,则该展开式中的系数为 56 【解】:由条件知:=n = 8, 通项Tr+1 =,令8 -2r = -2,得r = 5,则展开式中的系数为=56 .16. 三棱柱ABC-A1B1C1中,底面边长和侧棱长都相等,BAA1=CAA1= 60o,则异面直线AB1与BC1所成的角的余弦值为 【解法一】:设= a,= b,= c,则= a + b, = b + c - a,由条件知:| a | = | b | = | c | = r , = = = 60o,则ab = bc = ac = r 2| a + b | 2 = | a | 2+ | b | 2 +2 ab = 3r 2| b + c - a | 2 = | a | 2+ | b | 2 + | c | 2 + 2bc -2ab -2ac = 2 r 2设=q,则cos q =【解法二】:如图,D、E、F、G分别为其所在棱的中点,则DEAB1,EFBC1易证AA1BCCC1BC,设AB=2,DE =AB1=,EF=BC1=,DF=C1G=,在DEF中,由余弦定理的cosDEF= -则AB1与BC1所成的角的余弦值为【解法三】:如图,D、E、H、G分别为其所在棱的中点,则DEAB1,DHBC1,解三角形DEH可求得cosEDH= 二解答题:(共6个小题,满分70分)17.(本小题满分10分)ABC的内角A、B、C的对边分别为a、b、c,已知cos(A - C ) + cosB = 1,a = 2c,求C .【解】:由cos(A - C ) + cosB = 1 cos(A - C ) - cos(A + C ) = 12sinAsinC = 1 由a = 2c及正弦定理 sinA = 2sinC 联立、得sin2C = sinC =(sinC 0),sinA =1 A = 90o, C =30o18. (本小题满分12分)如图,四棱锥PABCD中,底面ABCD为菱形,PA底面ABCD,AC = 2,PA = 2,E是PC上的一点,PE = 2EC .()证明:PC平面BED ;()设二面角A -PB - C为90o,求PD与平面PBC所成的角的大小.()【证法一】:如图,设ACBD = O,设= a,= b,则| a | = 2, | b | = 2,ab = 0= a - b,=-=-=( b - a) + a =( a +2b)=( a +2b )( b - a) =(2b 2 - a 2 - ab ) = 0 PCOE,ABCD为菱形 ACBD,AC是PC在平面ABCD上的射影,所以PCBD PC平面BED【证法二】:PC =2,CE =, CE CP =CO CA,又ECO =PCA,则ECO PCA CEO =PAC PCOE,下同证法一.【证法三】:建立如图所示的坐标系,则A(, 0,0),P(, 0,2)C(-, 0,0),=+=+=(-,0,)设B(0,m,0) ,则:=(2, 0,2)(-,0,) = 0=(2, 0,2)(0,m,0) = 0 CPOE且CPOB CP平面BED.() 【解法一】:作AFPB交PB于F,由条件知平面PAB平面PBC则AF平面PBC AFBC由PA底面ABCD PABC所以BC平面PABBCABABCD为正方形 AB = 2= PAF是PB的中点 AF =,ADBC AD平面PBC D到平面PBC的距离d = AF =设PD与平面PBC所成的角为q,则sinq =q = 30o .【解法二】:建立如图所示的坐标系,则A(, 0,0),P(, 0,2)C(-, 0,0),设B(0,m,0) ,因为平面PAB平面ABCD,所以平面PAB的法向量与平面ABCD平行可设p = (a, b, 0)为平面PAB的法向量p= 0 (a, b, 0) (-, m,0) = 0bm =a,取b =得p = (m, , 0)设q = (x, y, z )为平面PBC的法向量q= 0,q= 0 (x, y, z )(2, 0,2) = 0,(x, y, z )(, m,0) = 0x + z = 0,x + my = 0,取z =得q = (-1, , )由二面角A -PB - C为90o pq = 0 (m, , 0)(-1, , ) = 0 m =(,,2),q = (-1, 1, )设PD与平面PBC所成的角为q,则sinq =|cos| =q = 30o .19. (本小题满分12分)乒乓球比赛规则规定:一局比赛,双方比分在10平前,一方连续发球2次后,对方再连续发球2次,依次轮换. 每次发球,胜方得1分,负方得0分. 设在甲、乙的比赛中,每次发球,发球方得1分的概率为0.6,各次发球的胜负结果相互独立. 甲、乙的一局比赛中,甲先发球.()求开始第四次发球时,甲、乙的比分为1比2的概率 ;()x 表示开始第四次发球时乙的得分,求x 的期望.()【解】:记“发球者胜”为事件A,事件“开始第四次发球时,甲、乙的比分为1比2”为:A A + A A + 其概率p = 0.60.40.6 + 0.40.60.6 + 0.40.40.4 = 0.352 (即)() 【解】:x 所有可能的取值有0, 1, 2, 3 .p1 = P(x = 0) = P(AA A ) = 0.60.60.4 = 0.144 (即)p4 = P(x = 3) = P(A ) = 0.40.40.6 = 0.096 (即)由()知p3 = P(x = 2) = 0.352 (即)所以p2 = P(x = 1) =1- p1- p3 -p4 = 0.408 (即)Ex = 0 p1 + 1 p2 + 2 p3 + 3 p4 = 1.4 (即)20. (本小题满分12分)设函数f (x ) = ax + cosx , x 0, p .()讨论f (x )的单调性;()设f (x ) 1 + sinx ,求a的取值范围.()【解】: f (x ) = a - sinx , x 0, p .因为x 0, p时,0 sinx 1,则(1) a1时,f (x ) 0恒成立,f (x )在0, p上是增函数;(2) a0时,f (x ) 0恒成立,f (x )在0, p上是减函数;(3) 0 a 1时,f (x )的零点为x 1 = arcsina和x 1 = p-arcsina,此时f (x )的递减区间为arcsina,p-arcsina,递增区间为0,arcsina和p-arcsina,p.()【解】:设g (x ) = f (x ) - (1 + sinx),g (0 ) = 0不等式 f (x ) 1 + sinx g (x ) g (0 ) x = 0时,g (x )取得最大值0.g (x ) = a - sinx - cosx = a -因为,则-1(1) a时,g (x )0恒成立,g (x )在0, p上是增函数g (x ) g (0 ) f (x ) 1 + sinx ,与条件矛盾.(2) a-1时,g (x ) 0恒成立,g (x )在0, p上是减函数g (x ) g (0 ) = 0 f (x ) 1 + sinx恒成立.(3) -1 a 1时,g (x )有唯一零点x 0 = -arcsin,当x 0, x 0 )时,g (x )0g (x )的最大值为g (0 )或g (p )依题意,g (p ) g (0 ) a p -20 a -1 a(4) 1 a 0, g (x )在0, x 1上是增函数 g (x 1 ) g (0 )与g (x ) g (0 )相矛盾.综上a的取值范围为 (-,21. (本小题满分12分)已知抛物线C:y = (x +1) 2与圆M:(x -1) 2 +( y -) 2 = r 2 (r 0)有一个公共点A,且在A处两曲线的切线为同一直线l.()求r ;()设m、n是异于l且与C及M都相切的两条直线,m、n的交点为D,求D到l的距离.()【解法一】:由条件知:C上的动点P(x , (x +1) 2 )到M(1,)的距离的最小值,|PM|2 = (x -1) 2 + (x +1) 2 - 2 = (x +1) 4- 4x +,设f (x ) = (x +1) 4- 4x +,f (x ) = 4(x +1) 3 -1 , f (x )有唯一零点x = 0,则x = 0时,f (x )最小,此时r =【解法二】:设A(t,(t +1) 2 ),因为l是C的切线,则l的斜率k =2(t +1)因为l是M的切线,则l的斜率k = -=,则= 2(t +1) (t +1) 3 =1 t = 0,则A的坐标为(0,1)r = | AM | =()【解】:设m、n在C上的切点分别为P1(x 1 , y1)、P2 (x 2 , y2). 由切线公式得m的方程分别为:=(x 1 +1)( x +1)即y1+ y = 2(x 1 +1)( x +1) 同理n的方程分别为:y2+ y = 2(x 2 +1)( x +1)联立解得:x =,y = (x 1 +1) (x 2 +1)即D的坐标为( , (x 1 +1) (x 2 +1)由()知l的方程为:2x - y +1= 0D到l的距离d = =抛物线C在点(t,(t+1)2 )处的切线为:2(t+1)x - y +1- t 2 = 0圆M与该切线相切的条件为:= | 2t 2 - 4t +5 | 2 = 54(t+1)2 +1 t 4- 4t 3- 6t 2 = 0, 显然t0,那么t 2- 4t 2- 6 = 0则x 1,x 2是方程的两根 x 1x 2 = - 6 d =22. (本小题满分12分)函数f (x ) = x 2 - 2x - 3 .定义数列x n如下:x 1= 2,x n+1是过两点P(4, 5)、Qn(x n, f (x n )的直线PQn与x轴的交点的横坐标.()证明:2x n x n+13 ;()求数列x n的通项公式.()【解】:直线PQn的方程为:y-5 =(x - 4),令y = 0得:x = x n+1 = + 4 = + 4 = -+ 4 =x n+1 =先证明2x n 3(数学归纳法)(1)n =1时,不等式显然成立(2)假设n = k时,不等式成立,即2x k 3 , kN*,则2x k+1 3 23 2(x k + 2)4x k + 33(x k + 2)x k3(显然成立),所以2x k+1 3成立,即n = k +1时,不等式成立再证明x n x n+1不等式x n x n+1 x n +2x n 4x n +3 x n -2x n -3 0 -1x n 3(显然成立)综上:2x n x n+13()【解】:由x n+1 =是公比为的等比数列,首项为= -= -() n - 1 x n =
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