当前位置:文档之家› 2018-2019学年四川省遂宁中学外国语实验学校高一下学期期中考试试卷数学Word版含解析

2018-2019学年四川省遂宁中学外国语实验学校高一下学期期中考试试卷数学Word版含解析

2018-2019学年四川省遂宁中学外国语实验学校高一下学期期中考试试卷数学Word版含解析
2018-2019学年四川省遂宁中学外国语实验学校高一下学期期中考试试卷数学Word版含解析

2018-2019学年四川省遂宁中学外国语实验

学校高一下学期期中考试试卷

数学科试题

满分:150分 考试时间:120分钟

第Ⅰ卷(选择题,满分60分)

一、选择题 (本大题共12小题, 每小题5分,共60分. 在每小题给出的四个选项中, 有且只有一项是符合题目要求的) 1.sin 18°cos 27°+cos 18°sin 27°的值是( ) A.

22 B.12 C.32 D .-2

2

2.D 是△ABC 的边AB 上的中点,则向量CD

→等于( ) A .-BC

→+12BA → B .-BC →-12BA → C.BC →-12BA → D.BC →+12BA → 3.在△ABC 中,C =60°,AB =3,BC =2,那么A 等于( )

A .135°

B .105°

C .45°

D .75° 4.在数列{a n }中,a n +1-a n =2,a 2=5,则{a n }的前4项和为( ) A .9 B .22 C .24 D .32 5.已知向量a =(1,2),b =(1,0),c =(3,4).若λ为实数,(a +λb )∥c ,则λ等于( ) A.14 B.1

2 C .1 D .2 6.已知sin α=1010

,则

2sin 2α+sin 2αcos ? ????α-π4等于( )A.-255 B.-35

10

C.-310

10

D.255

7.△ABC 的内角A ,B ,C 的对边分别为a ,b ,c ,已知b =2,B =π6,C =π

4,则△ABC 的面积为( )

A .23+2 B.3+1 C .23-2 D.3-1 8. 已知sin x +3cos x =65,则cos ? ??

??

π6-x =( )

A.-3

5 B.35 C.-4

5 D.45

9.若O 为△ABC 所在平面内任一点,且满足(OB →-OC →)·(OB →+OC →-2OA →)=0,

则△ABC 的形状为( ) A .正三角形 B .直角三角形 C .等腰三角形

D .等腰直角三角形

10.已知|OA →|=1,|OB →|=3,OA →·OB →=0,点C 在∠AOB 内,且OC →与OA →的夹角为30°,设OC →=mOA →+nOB →(m ,n ∈R ),则m n 的值为( )

A .2 B.5

2 C .

3 D .4

11. 已知数列{},{}n n a b 满足*1121

,1,()21n n n n n

b a a b b n N a +=+==∈-,则2017b =( )

A.

20172016 B. 20182017 C. 2019

2018

D.1 12.设α、β都是锐角,且cos α=55,sin(α+β)=3

5,则cos β等于( ) A.2525 B.255 C.2525或255 D.55或525

第II 卷(非选择题 共90分)

二.填空题(本大题包括4个小题,每小题5分,共20分) 13.2

2cos sin 88

ππ

-= . 14.设x ∈R ,向量a =(x,1),b =(1,-2),且a ⊥b ,则|a +b |=________. 15. 已知数列{a n }满足a 1=1,a n =n -1

n ·a n -1(n ≥2且n ∈N *),则a n =________. 16.如图所示,为测一建筑物的高度,在地面上选取A ,B 两点,从A ,B 两点分别测得建筑物顶端的仰角为30°,45°,且A ,B 两点间的距离为60 m ,则该建筑物的高度为______m.

三、解答题(本大题共6小题,共70分.解答应写出文字说明,

证明过程或演算步骤.)

17.(本题满分10分)已知)0,1(=,)1,2(=.

(1)求3a b +; (2)当k 为何实数时,-ka b 与3a b +平行,平行时它们是同向还是反向?

18.(本题满分12分)在等差数列{a n }中,a 1=1,a 3=-3. (1)求数列{a n }的通项公式;

(2)若数列{a n }的前k 项和S k =-35,求k 的值.

19. (本题满分12分)在平面直角坐标系xOy 中,已知向量m =? ????22,-22,

n =(sin x ,cos x ),x ∈? ?

???0,π2.

(1)若m ⊥n ,求tan x 的值; (2)若m 与n 的夹角为π

3,求x 的值.

20. (本题满分12分)数列{a n }满足a 1=1,a 2=2,a n +2=2a n +1-a n +2. (1) 设b n =a n +1-a n ,证明{b n }是等差数列; (2) 求{a n }的通项公式.

21. (本小题满分12分)已知函数f(x)=sin ?

?

???x +π6+cosx.

(1) 求函数f(x)的最大值,并写出当f(x)取最大值时x 的取值集合;

(2) 若α∈?

?

???0,π2,f ? ????α+π6=335,求f(2α)的值.

22. (本小题满分12分)在△ABC 中,角A ,B ,C 的对边分别为a ,b ,c ,且a 2-(b -c )2=(2-3)bc ,sin A sin B =cos 2C

2,BC 边上的中线AM 的长为7. (1)求角A 和角B 的大小; (2)求△ABC 的面积.

数学科试题参考答案

2.答案 A 解析 如图,CD

→=CB →+BD →=CB →+12BA →=-BC →+12BA →.

3.答案 C 解析 由正弦定理知BC sin A =AB sin C ,即2

sin A =

3

sin 60°

, 所以sin A =2

2,又由题知,BC

5.答案 B 解析 ∵a +λb =(1+λ,2),c =(3,4),且(a +λb )∥c , ∴1+λ3=24,∴λ=12,

6.【答案】D 【解析】由sin α=10

10.

故2sin 2α+sin 2αcos ? ????α-π4=2sin α()sin α+cos α2

2

()sin α+cos α=22sin α=

255. 7.答案 B 解析 ∵b =2,B =π6,C =π

4.

由正弦定理b sin B =c sin C ,得c =b sin C sin B =2×22

12=22,A =π-(π6+π4)=7

12π,

∴sin A =sin(π4+π3)=sin π4cos π3+cos π4sin π

3=6

+24

.

则S △ABC =12bc ·sin A =1

2×2×22×6+24=3+1.

8.【答案】B 【解析】sin x + 3 cos x =2? ??

??12sin x +3

2cos x

=2? ????sin π6sin x +cos π6cos x =2cos ? ????π6-x =65,∴cos ? ??

??π6-x =3

5.

9.答案 C 解析 因为(OB →-OC →)·(OB

→+OC →-2OA →)=0,

即CB →·(AB →+AC →)=0,因为AB →-AC →=CB →,所以(AB →-AC →)·(AB →+AC →)=0, 即|AB

→|=|AC →|,所以△ABC 是等腰三角形,故选C. 10.答案 C 解析 ∵OA →·OB →=0,∴OA →⊥OB →,

以OA 为x 轴,OB 为y 轴建立直角坐标系(图略), OA

→=(1,0),OB →=(0,3),OC →=mOA →+nOB →=(m ,3n ). ∵tan 30°=3n m =33,∴m =3n ,即m

n =3,故选C. 11. 【答案】B

【解析】∵1n n a b +=,112a =,∴11

2b =,

∵121n n n b b a +=

-,∴1

12n n b b +=-,∴111

111

n n b b +-=---, 又∵112b =,∴1121b =--. ∴数列11n b ????-??

是以﹣2为首项,﹣1为公差的等差数列,

1

11n n b =---,∴1n n b n =+.则201720172018

b =.故答案为:B 12. 答案 B 解析:依题意得sin α=1-cos 2α=25

5,

cos(α+β)=±1-sin 2(α+β)=±

45.

又α,β均为锐角,所以0<α<α+β<π,cos α>cos(α+β). 因为45>55>-45,所以cos(α+β)=-45.

于是cos β=cos[(α+β)-α]=cos(α+β)cos α+sin(α+β)sin α=-45×55+35×25

5=2525.

13.【答案】2

【解析】[由二倍角公式得2

2

cos sin 8

8

π

π

-=cos

4

2

=

π

14.答案

10 解析 ∵a ⊥b ,∴a·b =0,即x -2=0, ∴x =2,∴a =(2,1),∴a 2=5,b 2=5, ∴|a +b |=

(a +b )2=

a 2+2a·

b +b 2=

5+5=10.

15. 答案a n =1

n . 【解析】∵a n =n -1n a n -1 (n ≥2),∴a n -1=n -2n -1a n -2,…,a 2=

12a 1.

以上(n -1)个式子相乘得a n =a 1·12·2

3·…·n -1n =a 1n =1n .

当n =1时也满足此等式,∴a n =1

n .

16. 答案:30+30 3 解析:在△PAB 中,∠PAB =30°,∠APB =15°,AB =60, sin 15°=sin(45°-30°)=sin 45°cos 30°-cos 45°sin 30°

=22×32-22×12=6-24.由正弦定理,得PB sin 30°=AB sin 15°

所以PB =1

2×606-24

=30(6+2).

所以建筑物的高度为PBsin 45°=30(6+2)×2

2=(30+303) m. 17.解:(1)因为a =(1,0),b =(2,1),

所以a +3b =(7,3),∴|a +3b |= 72+32=58. …………5分

(2)ka -b =(k -2,-1),a +3b =(7,3), …………7分

因为ka -b 与a +3b 平行,所以3(k -2)+7=0,即k =-1

3

. …………

8分

此时ka -b =(k -2,-1)=? ??

??

-73,-1,a +3b =(7,3), …………

9分

则a +3b =-3(ka -b ),即此时向量a +3b 与ka -b 方向相反. …………10分

18.解 (1)设等差数列{a n }的公差为d ,则a n =a 1+(n -1)d . 由a 1=1,a 3=-3,可得1+2d =-3,解得d =-2. 从而a n =1+(n -1)×(-2)=3-2n .

(2)由(1)可知a n =3-2n ,所以S n =n [1+(3-2n )]2

=2n -n 2

.

由S k =-35,可得2k -k 2=-35,即k 2-2k -35=0,解得k =7或k =-5.

又k ∈N *,故k =7.

19.解 (1)因为m =? ????

22

,-22,n =(sin x ,cos x ),m ⊥n .

所以m ·n =0,即22sin x -2

2cos x =0,所以sin x =cos x ,所以tan x =1.

(2)因为|m |=|n |=1,所以m ·n =cos π3=12,即22sin x -22cos x =12,所以sin ? ??

?

?

x -π4=12,

因为0

12. 20.(1)证明 由a n +2=2a n +1-a n +2, 得a n +2-a n +1=a n +1-a n +2, 即b n +1=b n +2. 又b 1=a 2-a 1=1,

所以{b n }是首项为1,公差为2的等差数列. (2)解 由①得b n =1+2(n -1)=2n -1, 即a n +1-a n =2n -1.

于是∑n

k =1 (a k +1-a k )=∑n

k =1 (2k -1), 所以a n +1-a 1=n 2,即a n +1=n 2+a 1.

又a 1=1,所以{a n }的通项公式为a n =n 2-2n +2.

21 解:(1) f(x)=sin ?

????

x +π6+cosx =32sinx +12cosx +cosx =32sinx +32cosx =3

sin ? ????x +π3.

当x +π3=2k π+π2(k ∈Z),即x =2k π+π

6(k ∈Z)时,f(x)取得最大值 3. 此时x 的取值集合为{x|x =2k π+π

6,k ∈Z}.

(2) 由(1)知,f(x)=3sin ? ????x +π3,又f ?

????α+π6=33

5,

所以3sin ? ????

α+π6+π3=3cos α=335,即cos α=35.

因为α∈?

????

0,π2,所以sin α=45,sin2α=2sin αcos α=2×45×35=2425,

cos2α=2cos 2α-1=-7

25,

所以f(2α)=3sin ? ????

2α+π3=32sin2α+32cos2α=32×2425-32×725=243-2150.

22. 解 (1)由a 2-(b -c )2=(2-3)bc ,得a 2-b 2-c 2=-3bc , ∴cos A =b 2+c 2-a 22bc =32,又0<A <π,∴A =π

6.

由sin A sin B =cos 2

C 2,得1

2sin B =1+cos C 2

,即sin B =1+cos C ,

则cos C <0,即C 为钝角,∴B 为锐角,且B +C =5π

6,

则sin(5π6-C )=1+cos C ,化简得cos(C +π3)=-1,解得C =2π3,∴B =π6.

(2)由(1)知,a =b ,由余弦定理得AM 2=b 2+(a 2)2-2b ·a 2·cos C =b 2+b 24+b 22

=(7)2

解得b =2,故S △ABC =12ab sin C =12×2×2×3

2= 3.

人教版高中英语必修一高一上学期英语期中考试题

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