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2015·-全国卷1精校完整解析版

2015·-全国卷1精校完整解析版
2015·-全国卷1精校完整解析版

20152全国卷Ⅰ(文科数学)

1.A1[20152全国卷Ⅰ] 已知集合A={x|x=3n+2,n∈N},B={6,8,10,12,14},则集合A ∩B中元素的个数为( )

A.5 B.4 C.3 D.2

2.F1、F2[20152全国卷Ⅰ] 已知点A(0,1),B(3,2),向量AC→=(-4,-3),则向量BC→=( ) A.(-7,-4) B.(7,4) C.(-1,4) D.(1,4)

3.L4[20152全国卷Ⅰ] 已知复数z满足(z-1)i=1+i,则z=( )

A.-2-i B.-2+i C.2-i D.2+i

4.K2[20152全国卷Ⅰ] 如果3个正整数可作为一个直角三角形三条边的边长,则称这3个数为一组勾股数,从1,2,3,4,5中任取3个不同的数,则这3个数构成一组勾股数的概率为( )

A.

3

10

B.

1

5

C.

1

10

D.

1

20

5.H5、H7[20152全国卷Ⅰ] 已知椭圆E的中心为坐标原点,离心率为1

2

,E的右焦点与抛物线C:

y2=8x的焦点重合,A,B是C的准线与E的两个交点,则|AB|=( )

A.3 B.6 C.9 D.12

6.G12[20152全国卷Ⅰ] 《九章算术》是我国古代内容极为丰富的数学名著,书中有如下问题:“今有委米依垣内角,下周八尺,高五尺.问:积及为米几何?”其意思为:“在屋内墙角处堆放米(如图1-1,米堆为一个圆锥的四分之一),米堆底部的弧长为8尺,米堆的高为5尺,问米堆的体积和堆放的米各为多少?”已知1斛米的体积约为1.62立方尺,圆周率约为3,估算出堆放的米约有( )

图1-1

A.14斛 B.22斛 C.36斛 D.66斛

7.D2[20152全国卷Ⅰ] 已知{a n}是公差为1的等差数列,S n为{a n}的前n项和.若S8=4S4,则a

10

=( )

A.17

2

B.

19

2

C.10 D.12

8.C4[20152全国卷Ⅰ] 函数f(x)=cos(ωx+φ)的部分图像如图1-2所示,则f(x)的单调递

减区间为( )

图1-2

A.?

????k π-14,k π+34,k ∈Z B.? ????2k π-14,2k π+34,k ∈Z C.? ????k -14,k +34,k ∈Z D.?

?

???2k -14,2k +34,k ∈Z

9.L1[20152全国卷Ⅰ] 执行图1-3所示的程序框图,如果输入的t =0.01,则输出的n =( )

图1-3

A .5

B .6

C .7

D .8

10.B6、B7[20152全国卷Ⅰ] 已知函数f (x )=???2x -1

-2,x ≤1,

-log 2

(x +1),x >1,且f (a )=-3,则f (6

-a )=( )

A .-74

B .-54

C .-34

D .-1

4

11.G2[20152全国卷Ⅰ] 圆柱被一个平面截去一部分后与半球(半径为r )组成一个几何体,该几何体三视图中的正视图和俯视图如图1-4所示.若该几何体的表面积为16+20π,则r =( )

图1-4

A .1

B .2

C .4

D .8

12.B6、B7[20152全国卷Ⅰ] 设函数y =f (x )的图像与y =2

x +a

的图像关于直线y =-x 对称,且

f (-2)+f (-4)=1,则a =( )

A .-1

B .1

C .2

D .4

13.[20152全国卷Ⅰ] 在数列{a n }中,a 1=2,a n +1=2a n ,S n 为{a n }的前n 项和.若S n =126,则n =________.

14.B12[20152全国卷Ⅰ] 已知函数f (x )=ax 3+x +1的图像在点(1,f (1))处的切线过点(2,7),则 a =________.

15.E5[20152全国卷Ⅰ] 若x ,y 满足约束条件???x +y -2≤0,

x -2y +1≤0,2x -y +2≥0,

则z =3x +y 的最大值为________.

16.H6[20152全国卷Ⅰ] 已知F 是双曲线C :x 2

-y 2

8=1的右焦点,P 是C 的左支上一点,A (0,

66) ,当△APF 周长最小时,该三角形的面积为________.

17.C5、C8[20152全国卷Ⅰ] 已知a ,b ,c 分别是△ABC 内角A ,B ,C 的对边,sin 2B =2sin A sin

C .

(1)若a =b ,求cos B;

(2)若B =90°,且a =2, 求△ABC 的面积.

18.G5[20152全国卷Ⅰ] 如图1-5,四边形ABCD 为菱形,G 为AC 与BD 的交点,BE ⊥平面ABCD .

(1)证明:平面AEC⊥平面BED;

(2)若∠ABC=120°,AE⊥EC, 三棱锥E-ACD的体积为

6

3

,求该三棱锥的侧面积.

图1-5

19.I4[20152全国卷Ⅰ] 某公司为确定下一年度投入某种产品的宣传费,需了解年宣传费x(单位:千元)对年销售量y(单位:t)和年利润z(单位:千元)的影响.对近8年的年宣传费x i和年销售量y i(i=1,2,…,8)数据作了初步处理,得到下面的散点图及一些统计量的值.

图1-6

x y w 错误!(w i-w )2

(y i-y )

46.6 563 6.8 289.8 1.6 1469 108.8

其中w

i =x

i

,w=

1

8

i=1

8

w

i

.

(1)根据散点图判断,y=a+bx与y=c+d x哪一个适宜作为年销售量y关于年宣传费x的回归方程类型?(给出判断即可,不必说明理由)

(2)根据(1)的判断结果及表中数据,建立y关于x的回归方程.

(3)已知这种产品的年利润z与x,y的关系为z=0.2y-x.根据(2)的结果回答下列问题:

(i)年宣传费x=49时,年销售量及年利润的预报值是多少?

(ii)年宣传费x为何值时,年利润的预报值最大?

附:对于一组数据(u

1,v

1

),(u

2

,v

2

),…,(u

n

,v

n

),其回归直线v=α+βu的斜率和截距的最

小二乘估计分别为

β^=错误!,错误!=v-错误!u.

20.H1、H3、H4[20152全国卷Ⅰ] 已知过点A(0,1)且斜率为k的直线l与圆C:(x-2)2+(y -3)2=1交于M,N两点.

(1)求k的取值范围;

(2)OM→2ON→=12,其中O为坐标原点,求|MN|.

21.B9,B11,B12[20152全国卷Ⅰ] 设函数f(x)=e2x-a ln x.

(1)讨论f(x)的导函数f′(x)零点的个数;

(2)证明:当a>0时,f(x)≥2a+a ln 2 a .

22.N1[20152全国卷Ⅰ] 选修4-1:几何证明选讲

如图1-7,AB是⊙O的直径,AC是⊙O的切线,BC交⊙O于点E.

(1)若D为AC的中点,证明:DE是⊙O的切线;

(2)若OA=3CE,求∠ACB的大小.

23.N3[20152全国卷Ⅰ] 选修4-4:坐标系与参数方程

在直角坐标系xOy中,直线C1:x=-2,圆C2:(x-1)2+(y-2)2=1,以坐标原点为极点,x轴

的正半轴为极轴建立极坐标系.

(1)求C1,C2的极坐标方程;

(2)若直线C3的极坐标方程为θ=π

4

(ρ∈R),设C2与C3的交点为M,N,求△C2MN的面积.

24.N4[20152全国卷Ⅰ] 选修4-5:不等式选讲

已知函数f(x)=|x+1|-2|x-a|,a>0.

(1)当a=1时,求不等式f(x)>1的解集;

(2)若f(x)的图像与x轴围成的三角形面积大于6,求a的取值范围.

答案

1.D [解析] 集合A={2,5,8,11,14,17,…},所以A∩B={8,14},所以A∩B中有2个元素.

2.A [解析] AB →=(3,1),BC →=AC →-AB →=(-4,-3)-(3,1)=(-7,-4).

3.C [解析] 设复数z =a +b i(a ,b ∈R ),代入(z -1)i =1+i 得(a -1+b i)i =1+i ,即-b +(a -1)i =1+i.根据复数相等可得???-b =1,a -1=1,

得a =2,b =-1,所以复数z =2-i.

4.C [解析] 从1,2,3,4,5中任取3个不同的数有(1,2,3),(1,2,4),(1,2,5),(1,3,4),(1,3,5),(1,4,5),(2,3,4),(2,3,5),(2,4,5),(3,4,5),共10种取法,其中只有(3,4,5)是一组勾股数,所以构成勾股数的概率为

110

. 5.B [解析] 抛物线C :y 2=8x 的焦点坐标为(2,0),准线方程为x =-2,即椭圆的半焦距c

=2.又离心率e =c a =2a =12,所以a =4,于是b 2

=12,则椭圆的方程为x 216+y 212=1.A ,B 是C 的准线x

=-2与E 的两个交点,把x =-2代入椭圆方程得y =±3,所以|AB |=6.

6.B [解析] 米堆的体积即为四分之一的圆锥的体积,设圆锥底面半径为r ,则1

432πr =8,

得r =

16π,所以米堆的体积为13314πr 235≈3209(立方尺),3209

÷1.62≈22(斛). 7.B [解析] 由S 8=4S 4,得8a 1+837231=4? ?

?

??4a 1+433231,解得a 1=12,所以a 10=12+(10-1)31=19

2

.

8.D [解析] 由图知T 2=54-1

4=1,所以T =2,即2π

||ω=2,所以ω=±π.

因为函数f (x )的图像过点? ????

14,0,

所以当ω=π时,

ω4+φ=π

2

+2k π,k ∈Z , 解得φ=π

4+2k π,k ∈Z ;

当ω=-π时,

ω4+φ=-π

2

+2k π,k ∈Z , 解得φ=-π

4

+2k π,k ∈Z .

所以f (x )=cos ? ?

???πx +π4,由2k π<πx +π4<π+2k π解得2k -14

9.C [解析] 经推理分析可知,若程序能满足循环,则每循环一次,S 的值减少一半,循环6次后S 的值变为126=164>0.01,循环7次后S 的值变为127=1

128<0.01,此时不再满足循环的条件,所以

结束循环,于是输出的n =7.

10.A [解析] 因为2x -1-2>-2恒成立,所以可知a >1,于是由f (a )=-log 2(a +1)=-3得a =7,所以f (6-a )=f (-1)=2-1-1-2=-7

4

.

11.B [解析] 由三视图可知,此组合体的前半部分是一个底面半径为r ,高为2r 的半圆柱(水平放置),后半部分是一个半径为r 的半球,其中半圆柱的一个底面与半球的半个圆面重合,所以此几何体的表面积为2r 22r +12πr 2+1

2

πr 2+πr 22r +2πr 2=4r 2+5πr 2=16+20π,解得r =2.

12.C [解析] 在函数y =f (x )的图像上任设一点P (x ,y ),其关于直线y =-x 的对称点为P ′

(x ′,y ′),则有?????y ′-y

x ′-x =1,x +x ′2

+y +y ′2=0,解得???x ′=-y ,

y ′=-x .

由于点P ′(x ′,y ′)在函数y =2

x +a

的图

像上,于是有-x =2-y +a ,得-y +a =log 2(-x ),即y =f (x )=a -log 2(-x ),所以f (-2)+f (-4)=a -log 22+a -log 24=2a -3=1,所以a =2.

13.D36 [解析] 由a 1=2,a n +1=2a n 可知数列{a n }为等比数列,公比为2,所以S n =2(1-2n )

1-2=

126,得n =6.

14.1 [解析] 因为f ′(x )=3ax 2+1,所以函数在点(1,f (1)),即点(1,2+a )处的切线的斜率k =f ′(1)=3a +1.又切线过点(2,7),则经过点(1,2+a ),(2,7)的直线的斜率k =2+a -7

1-2

,所以3a +1=2+a -7

1-2

,解得a =1.

15.4 [解析] 作出约束条件表示的可行域如图所示,当目标函数线平移至经过可行域的顶点

A (1,1)时,目标函数z 取得最大值,故z max =331+1=4.

16.12 6 [解析] 由已知得a=1,c=3,则F(3,0),|AF|=15.设F1是双曲线的左焦点,根据双曲线的定义有|PF|-|PF1|=2,所以|PA|+|PF|=|PA|+|PF1|+2≥|AF1|+2=17,即点P是线段AF1与双曲线的交点时,|PA|+|PF|=|PA|+|PF1|+2最小,即△APF周长最小,此时,sin∠OAF

=1

5

,cos∠PAF=1-2sin2∠OAF=

23

25

,即有sin∠PAF=

46

25

.由余弦定理得|PF|2=|PA|2+|AF|2-

2|PA||AF|cos∠PAF,即(17-|PA|)2=|PA|2+152-2|PA|315323

25

,解得|PA|=10,于是S△APF=

1

2

|PA|2|AF|2sin∠PAF=1

2

3103153

46

25

=12 6.

17.解:(1)由题设及正弦定理可得b2=2ac. 又a=b,所以可得b=2c,a=2c.

由余弦定理可得cos B=a2+c2-b2

2ac

1

4

.

(2)由(1)知b2=2ac.

因为B=90°,所以由勾股定理得a2+c2=b2.

故a2+c2=2ac,得c=a=2,

所以△ABC的面积为1.

18.解:(1)证明:因为四边形ABCD为菱形,所以AC⊥BD. 因为BE⊥平面ABCD,所以AC⊥BE,故AC⊥平面BED.

又AC?平面AEC,所以平面AEC⊥平面BED.

(2)设AB=x,在菱形ABCD中,由∠ABC=120°,可得AG=GC=

3

2

x,GB=GD=

x

2

.

因为AE⊥EC,所以在Rt△AEC中,可得EG=

3

2

x.

由BE⊥平面ABCD,知△EBG为直角三角形,可得BE=

2

2

x.

由已知得,三棱锥E-ACD的体积V E-ACD=1

3

3

1

2

AC2GD2BE=

6

24

x3=

6

3

故x =2.从而可得AE =EC =ED =6,

所以△EAC 的面积为3,△EAD 的面积与△ECD 的面积均为 5. 故三棱锥E - ACD 的侧面积为3+2 5.

19.解:(1)由散点图可以判断,y =c +d x 适宜作为年销售量y 关于年宣传费x 的回归方程类型.

(2)令w =x ,先建立y 关于w 的线性回归方程.由于 d^=错误!=错误!=68,

c ^

=y -d^w =563-6836.8=100.6,

所以y 关于w 的线性回归方程为y ^=100.6+68w ,因此y 关于x 的回归方程为y ^=100.6+68x. (3)(i )由(2)知,当x =49时,年销售量y 的预报值 y ^

=100.6+6849=576.6, 年利润z 的预报值

z ^

=576.630.2-49=66.32.

(ii )根据(2)的结果知,年利润z 的预报值

z ^

=0.2(100.6+68x)-x =-x +13.6x +20.12, 所以当x =13.6

2=6.8,即x =46.24时,z ^取得最大值.

故年宣传费为46.24千元时,年利润的预报值最大. 20.解:(1)由题设,可知直线l 的方程为y =kx +1. 因为l 与C 交于两点,所以|2k -3+1|

1+k

2

<1, 解得4-73

所以k 的取值范围为? ????

4-73,

4+73. (2)设M (x 1,y 1),N (x 2,y 2).

将y =kx +1代入方程(x -2)2+(y -3)2=1,整理得(1+k 2)x 2-4(1+k )x +7=0, 所以x 1+x 2=

4(1+k )1+k 2

,x 1x 2=7

1+k 2

OM →2ON →=x 1x 2

+y 1y 2

=(1+k 2)x 1x 2+k (x 1+x 2)+1 =

4k (1+k )

1+k 2

+8.

由题设可得4k (1+k )

1+k 2+8=12,解得k =1,所以直线l 的方程为y =x +1.

故圆心C 在直线l 上,所以|MN |=2.

21.解:(1)f (x )的定义域为(0,+∞),f ′(x )=2e 2x -a x

(x >0). 当a ≤0时,f ′(x )>0,f ′(x )没有零点.

当a >0时,因为e 2x 单调递增,-a x

单调递增,所以f ′(x )在(0,+∞)上单调递增.又f ′(a )>0,

当b 满足0

4

时,f ′(b )<0,故当a >0时,f ′(x )存在唯一零点.

(2)证明:由(1)可设f ′(x )在(0,+∞)上的唯一零点为x 0.当x ∈(0,x 0)时,f ′(x )<0;当x ∈(x 0,+∞)时,f ′(x )>0.

故f (x )在(0,x 0)上单调递减,在(x 0,+∞)上单调递增,所以当x =x 0时,f (x )取得最小值,最小值为f (x 0).

由于2e2x 0-a x 0=0,所以f (x 0)=a 2x 0+2ax 0+a ln 2a ≥2a +a ln 2

a .

故当a >0时,f (x )≥2a +a ln 2

a

.

22.解:(1)证明:连接AE ,由已知得,AE ⊥BC ,AC ⊥AB . 在Rt △AEC 中,由已知得,DE =DC ,故∠DEC =∠DCE . 连接OE ,则∠OBE =∠OEB .

又∠ACB +∠ABC =90°,所以∠DEC +∠OEB =90°,故∠OED =90°,即DE 是⊙O 的切线.

(2)设CE =1,AE =x ,由已知得AB =23,BE =12-x 2.

由射影定理可得,AE 2=CE 2BE ,所以x 2=12-x 2,即x 4+x 2-12=0, 可得x =3,所以∠ACB =60°.

23.解:(1)因为x =ρcos θ,y =ρsin θ,所以C 1的极坐标方程为ρcos θ=-2,C 2的极坐标方程为ρ2-2ρcos θ-4ρsin θ+4=0.

(2)将θ=π

4

代入ρ2-2ρcos θ-4ρsin θ+4=0,得ρ2-32ρ+4=0,解得ρ1=22,ρ

2

=2,故ρ1-ρ2=2,即|MN |= 2.

由于圆C 2的半径为1,所以△C 2MN 的面积为1

2

.

24.解:(1)当a =1时,f (x )>1化为|x +1|-2|x -1|-1>0. 当x ≤-1时,不等式化为x -4>0,无解; 当-10,解得2

3

当x ≥1时,不等式化为-x +2>0,解得1≤x <2. 所以f (x )>1

的解集为??????

???

?x 23

(2)由题设可得,f (x )=???x -1-2a ,x <-1,

3x +1-2a ,-1≤x ≤a ,-x +1+2a ,x >a ,

所以函数f (x )的图像与x 轴围成的三角形的三个顶点分别为A ? ??

??

2a -13,0,B (2a +1,0),C (a ,a +1),△ABC 的面积为23

(a +1)2.

由题设得2

3(a +1)2>6,故a >2,

所以a 的取值范围为(2,+∞).

2015高考数学全国卷1(完美版)

2015高考数学全国卷1(完美版)

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