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山东省广饶第一中学2016届高三10月阶段质量检测数学文试卷 Word版含答案

月考试题

第Ⅰ卷

一、选择题(本大题共10小题,每小题5分,满分50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.已知集合{}

2,0x M y y x ==>,{}

)2lg(2x x y x N -==,则M N 为( ) A.()2,1 B.()+∞,1 C.[)+∞,2 D.[)+∞,1 2. 若b a b a >是任意实数,且、,则下列不等式成立..

的是( ) A.2

2

b a > B.1-b a D.1133a b

????< ? ?????

3. 若ππ,42θ??

∈????

,sin 2θ,则sin θ=( )

A.35

B.45 D.3

4

4.若)0)(sin()(:;,2

:≠+=∈+=ω?ωππ

?x x f q Z k k P 是偶函数,则p 是q 的( )

A.充要条件

B.充分不必要条件

C.必要不充分条件

D.既不充分也不必要条件

5. 下列判断正确的是( )

A. 若命题p 为真命题,命题q 为假命题,则命题“p q ∧”为真命题

B. 命题“若0xy =,则0x =”的否命题为“若0xy =,则0x ≠”

C. “1sin 2α=

”是“ 6

π

α=”的充分不必要条件 D. 命题“,20x x ?∈>R ”的否定是“ 00,20x x ?∈≤R ”

6.函数21

()3cos

log 22

f x x x π

=--的零点个数为( ) A.2 B.3 C.4 D.5

7. 函数()sin()f x A x ω?=+(其中π

0,2

A ?><)的图象如图1所示,为了得到

x x g 2sin )(=的图象,则只需将()f x 的图象( )

A.向右平移π6个长度单位

B.向右平移π

12

个长度单位

C.向左平移

π6个长度单位 D.向左平移π

12

个长度单位 图1

8.在△ABC 中,若→

→→→→→→

?+?+?=CB CA BC BA AC AB AB 2

,则△ABC 是( ) A.等边三角形 B.锐角三角形 C.钝角三角形 D.直角三角形

9. 已知曲线ππ2sin +cos -44y x x ????

= ? ?????

与直线21=y 相交,若在y 轴右侧的交点自

左向右依次记为P 1, P 2, P 3,…,则|51P P |等于( ) A. π B. 2π C. 3π D. 4π

10. 已知函数()2,0,2,0,x f x x x ≥?=?

-+

32f x f x -<的x 的取值范围为( ) A. B. (-3,1) C. [-3,0) D. (-3,0)

第Ⅱ卷

二、填空题(本大题共5小题,每小题5分,共25分.将答案填在答题卷相应位置上.)

11. 已知向量b a ,夹角为45?

,且a =1,

a 2,则

b =________.

13. 设,x y 满足约束条件:,0,1,3,x y x y x y ≥??

-≥-??+≤?则2z x y =-的取值范围为 .

14. 在△ABC 中,角A 、B 、C 所对的边分别为

a 、

b 、

c ,且满足sin cos a B b A =,则

cos B C -的最大值是 . 15. 已知0,0x y >>,若2282y x m m x y

+>+恒成立,则实数m 的取值范围是 .

三、解答题:本大题共6小题,满分75分.解答须写出文字说明、证明过程和演算步骤.

16.(12分)已知P:,23

1

1≤--

x q:),0(01222>≤-+-m m x x 且p ?是q ?的必要不充分条件,求实数m 的取值范围

17.(12分)已知函数x

x

x x x f sin 2sin )cos (sin )(-=

(1)求)(x f 的定义域及最小正周期; (2)求)(x f 的单调递减区间.

18.(12分)已知a ,b ,c 分别为△ABC 三个内角A ,B ,C 的对边,c = 3a sin C -c cos A .

(1)求A ;

(2)若a =2,△ABC.的面积为3,求b ,c.

19. (12分)已知向量m (sin ,1),x =n cos ,cos 2)(0)2

A

x x A =>,函数

()f x =m ·n

的最大值为6. (1)求A ;

(2)将函数()y f x =的图象向左平移12

π

个单位,再将所得图象上各点的横坐标缩短为原来的

12倍,纵坐标不变,得到函数()y g x =的图象.求()g x 在50,24π??????

的值域.

20. (13分)如图2,建立平面直角坐标系xoy ,x 轴在地平面上,y 轴垂直于地平面,单位长度为1千米.某炮位于坐标原点.已知炮弹发射后的轨迹在方程

221

(1)(0)20

y kx k x k =-

+>表示的曲线上,其中k 与发射方向有关.炮的射程是指炮弹落地点的横坐标. (1)求炮的最大射程;

(2)设在第一象限有一飞行物(忽略其大小),其飞行高度为3.2千米,试问它的

横坐标a 不超过多少时, 炮弹可以击中它?请说明理由.

图2

21.(14分)设f (x )=a ln x +x -1,

(1)求)(x f 的单调区间

(2)证明:当,1=a x >1时,f (x )<3

2(x -1);

月考试题二答案

一选择题

1A 2D 3D 4A 5D 6B 7A 8D 9B 10D

11. 12. 1或2 13. ]3,3[- 14.1 15.()4,2-

16.解:由题意知:命题:若非p 是非q 的必要而不充分条件的等价命题即逆否命题为:p 是q 的充分不必要条件.

p:|1|≤2-2≤-1≤2-1≤≤3-2≤x≤10.

q:x 2-2x+1-m 2≤0

[x-(1-m)][x-(1+m)]≤0.(*)

∵p 是q 的充分不必要条件,

∴不等式|1

|≤2的解集是x 2-2x+1-m 2≤0(m >0)解集的真子集.

又∵m >0,∴不等式(*)的解集为1-m≤x≤1+m. 又∵1-m=-2与1+m=10不同时成立,

∴∴m≥9.

∴实数m 的取值范围是[9,+∞)

17.解:(1)由sin x ≠0得x ≠k π(k ∈Z),

故f (x )的定义域为{x ∈R|x ≠k π,k ∈Z}. …………………2分

因为f (x )=(sin x -cos x )sin2x

sin x

=2cos x (sin x -cos x ) =sin2x -cos2x -1

=2sin ???

?2x -π4-1, …………………………4分 所以f (x )的最小正周期T =2π

2=π. …………………………5分 (2)函数y =sin x 的单调递减区间为????2k π+π2,2k π+3π2(k ∈Z). …………………………6分

由2k π+π2≤2x -π4≤2k π+3π

2,x ≠k π(k ∈Z),

得k π+3π8≤x ≤k π+7π

8(k ∈Z).

f (x )

?

???

k π+3π8,kx +7π8(k ∈

Z). …………………………10分

18.解:(1)由c =3a sin C -c cos A 及正弦定理得

3sin A sin C -cos A sin C -sin C =0. …………………………3分

由于sin C ≠0,所以sin ????A -π6=12. 又0

3. …………………………6分

(2)△ABC 的面积S =1

2

bc sin A =3,故bc =4.

而a 2=b 2+c 2-2bc cos A ,故b 2+c 2=8. …………………………10分 解得b =c =2. …………………………12分 19.解:(1)f (x )=m ·n =3A sin x cos x +A 2cos2x =A ? ??

??32sin2x +12cos2x =A sin ? ????2x +π6. 因为A >0,由题意知,A =6.

(2)由(1)f (x )=6sin ?

????2x +π6. 将函数y =f (x )的图象向左平移π12个单位后得到y =6sin ????

??2? ????x +π12+π6=6sin ? ????2x +π3的图象;再将所得图象上各点横坐标缩短为原来的12倍,纵坐标不变,得到y =6sin ?

????4x +π3的图象.

因此,g (x )=6sin ?

????4x +π3. 因为x ∈?

?????0,5π24,所以4x +π3∈??????π3,7π6.

故g (x )在?

?????0,5π24上的值域为[-3,6].

20.解:(1)令y =0,得kx -1

20(1+k 2)x 2=0,…………………………2分 由实际意义和题设条件知x >0,k >0,

故x =20k 1+k

2=20k +1k ≤20

2=10,当且仅当k =1时取等号. …………………………4分 所以炮的最大射程为10 km. …………………………5分

(2)因为a >0,所以炮弹可击中目标?存在k >0,使3.2=ka -1

20

(1+k 2)a 2成立

?关于k 的方程a 2k 2-20ak +a 2+64=0有正根 …………………………7分 ?判别式Δ=(-20a )2-4a 2(a 2+64)≥0

?a ≤6. …………………………11分 所以当a 不超过6 km 时,可击中目标. …………………………12分

21(1)解:定义域为()∞+,

0 21

21)('f -+=x x a x =

x

x

a 22+ 当a ≥0时,0)('>x f

当a<0时,令0)('>x f 解得24x a >;令0)('

综上所述:当a ≥0时,f(x)的递增区间为),(∞+0

当a<0时,f(x)的递增区间为)

,(∞+2a 4,f(x)的递减区间为)(2a 4,0 (2)证明:(1)(证法一)记g (x )=ln x +x -1-32

(x -1).则当x >1时,

g ′(x )=1x +12x -3

2

<0,g (x )在(1,+∞)上单调递减.

又g (1)=0,有g (x )<0,即f (x )<3

2

(x -1).

(证法二)

由均值不等式,当x >1时,2x

2.①

令k (x )=ln x -x +1,则k (1)=0,k ′(x )=1

x

-1<0,

故k (x )<0,即ln x

由①②得,当x >1时,f (x )<3

2

(x -1).

话题10:春节-安徽省合肥市第一中学高考英语新题型读后续写、读写任务技巧突破专项课讲义

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合肥市2019届高三调研性检测数学试题-理科含答案

合肥市2019届高三调研性检测 数学试题(理科) (考试时间:120分钟 满分:150分) 第Ⅰ卷 一、选择题:本大题共12小题,每小题5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)已知集合{}12M x x =-<<,{}13N x x =≤≤,则M N = (A)(]1,3- (B)(]1,2- (C)[)1,2 (D)(]2,3 (2)已知复数122i z i -= -(i 为虚数单位),则||z = (A)15 (B)35 (C)4 5 (D)1 (3)右图是在北京召开的第24届国际数学家大会的会标,会标是根据我国古代数学家赵爽弦图设计的,颜色的明暗使它看上去像一个风车,代表中国人民热情好客.已知图中直角三角形两条直角边的长分别为2和3.若从右图内随机取一点,则该点取自 阴影区域的概率为 (A)23 (B)8 9 (C)1213 (D)2425 (4)已知实数x y ,满足条件00220x y x y x y -≤?? +≥??+-≤? ,则2z x y =-的取值范围是 (A)26 3??-????, (B)20 3?? ???? , (C)[)6 -+∞, (D)[)0 +∞, (5)已知直线:50l x y +-=与圆222 :(2)(1)(0)C x y r r -+-=>相交所得的 弦长 为C 的半径r = 222 (D)4 (6)执行右面的程序框图,若输出的结果为15,则判断框中的条件是 (A)4?i < (B)5?i < (C)6?i < (D)7?i < (7)已知t a n 3α=,则s in c o s 22ππαα???? -?+ ? ????? 的值为 (A)310 (B)310- (C)3 5 (D)35- (8)已知双曲线22 22:1(00)x y M a b a b -=>>,的焦距为4,两条渐近线 的夹角为60o ,则双曲线M 的标准方程是 (A)2213x y -= (B)2213x y -=或22 13y x -= (C)221124x y -= (D)221124x y -=或22 1412 x y -= (9)已知某几何体的三视图如图所示,其中正视图和侧视图都由半圆及矩形组成,俯视图由正方形及其内切圆组成,则该几何体的表面积等于 (A)488π+ (B)484π+ (C)648π+ (D)644π+ (10)若将函数()()()2c o s 1c o s 1c o s f x x x x =+-图象上所有点的横坐标伸长为原来的2倍(纵坐标不变),得到函数()y g x =的图象,则函数()y g x =的单调递减区间为 (A)()2k k k Z πππ??-+∈????, (B)() 2k k k Z πππ?? +∈???? ,

2018届合肥市高三一模试题-理科数学

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2018届安徽省合肥市高三第一次质量检测理科数学试题及答案

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