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2019届高三数学9月月考试题 文

2019届高三数学9月月考试题 文
2019届高三数学9月月考试题 文

1

3

x

的图象是(

<φ<)的最小正周期为π,将该函数的A.关于点(,0)对称 B.关于直线x=对称

2019届高三数学9月月考试题文

一、选择题(本题共12道小题,每小题5分,共60分)

1.复数z=

2-i

2+i(i为虚数单位)在复平面内对应的点所在象限为()

A.第一象限B.第二象限C.第三象限D.第四象限

2.已知集合A={x|x2﹣x﹣6≥0},B={x|﹣3≤x≤3},则A∩B等于()

A.[﹣3,﹣2]B.[2,3]C.[﹣3,﹣2]∪{3}D.[2,3]∪{﹣3}

3.已知命题p:?x∈[1,2],使e x-a≥0.若?p是假命题,则实数a的取值范围为()

A.(-∞,e2]

B.(-∞,e]

C.[e,+∞)

D.[e2,+∞)

4.下列命题正确的是()

A.命题“若α=β,则sinα=sinβ”的逆否命题为真命题

B.命题“若a

C.命题“?x>0,5x>0”的否定是“?x≤0,5x0≤0”

D.“x<-1”是“ln

(x+2)<0”的充分不必要条件

1

5.设a=log3,b=()0.2,c=23,则()

1

2

A.a

6.函数y=

x ln x

)

A B C D

7.已知函数f(x)=sin(ωx+φ)(w>0,-ππ

22

图象向左平移

π

6

个单位后,得到的图象对应的函数为奇函数,则f(x)的图象是()π5π

1212

,0)对称 D.关于直线 x = 对称

8.要想得到函数 y = sin 2 x + ? 的图象,只需将 y = sin 2 x + ? 的图像(

)

π ? C.向右平移 个单位

D.向右平移 个单位

,则 tan - 2θ ? = ( )

7

D. - ? 3 ?

3 3 ? 3 0, 3 ? ? 3 ? ) ? 0, ? ,0) ? ,+∞ ?? A. (-∞,

B . (- C. D . 3 ? 3 ? ? ? 3 3 3 ? 3 A . -

5

照此规律,第五个不等式为

.

2 ) 的部分图象如图所示,则

C.关于点(

12

π

12

? ? π ? ? 3 ? ?

6 ?

A.向左平移 π

6

个单位

B.向左平移

π

12

个单位

π

6

π

12

9.已知 tan θ =

1 ? π ?

2 ? 4 ?

A.7

B. -7

C. 1

1

7

10.函数 y = 3x 2 - 2 ln x 的单调增区间为(

? ? ? ,+∞ ?

?

? ?

?

4x + a

11. 已知函数 f ( x ) = 是奇函数,则 f (a) 的值为(

2x

5 3

3

B .

C . -

D .

2

2

2 2

12. 设函数 f ′(x )是奇函数 f (x )(x ∈R )的导数,f (-1)=0,当 x >0 时,xf ′(x )-f (x )

<0,则使得 f (x )>0 成立的 x 的取值范围是(

A.(-∞, -1)∪(0,1)

B.(-1,0)∪(1,+∞)

C.(-∞,-1)∪(-1,0)

D.(0,1)∪(1,+∞)

二、填空题(本题共 4 道小题,每小题 5 分,共 20 分)

13.观察下列不等式

1 + 1

22 <

3 2

1 +

1 + 1 1 5 + < , 2

2 3

3 3 1 1 1 5 + + <

22 32 42 3

……

...

14.若函数 y = A s in(ωx + ?)( A > 0,ω > 0, ? <

该函数解析式是 .

π

15.已知函数 f ( x ) = xe x + 2 ,则曲线 y = f ( x ) 在点 (0, f (0)) 处的切线方程为

16 关于函数 f (x )=4sin (2x+ ),(x ∈R )有下列结论:

①y=f(x )是以 π 为最小正周期的周期函数;

②y=f(x )可改写为 y=4cos (2x ﹣

③y=f(x )的最大值为 4;

);

④y=f(x )的图象关于直线 x=

对称;

则其中正确结论的序号为

三、解答题(本题共 2 道小题,第 1 题 0 分,第 2 题 0 分,共 0 分)

17 已知函数 f ( x ) =

3 sin 2 x - 2cos 2 x - 1, x ∈ R .

(Ⅰ) 求函数 f ( x) 的最小正周期和最小值;

(Ⅱ) 在 ?ABC 中, A, B, C 的对边分别为 a, b , c ,若 c =

求 a, b 的值.

3, f (C ) = 0,sin B = 2sin A ,

18.已知△ABC 的内角 A , B , C 的对边分别为 a , b , c ,且 2c ? cos B - b = 2a .

(Ⅰ)求角 C 的大小;

(Ⅱ)设角 A 的平分线交 BC 于 D ,且 AD = 3 ,若 b = 2 ,求△ABC 的面积 .

1 9.如图,在半径为 30cm 的半圆形铁皮上截取一块矩形材料 A (点 A ,B 在直径上,点 C ,D

在半圆周上),并将其卷成一个以 AD 为母线的圆柱体罐子的侧面(不计剪裁和拼接损耗)

(1)若要求圆柱体罐子的侧面积最 大,应如何截取?

(2)若要求圆柱体罐子的体积最大,应如何截取?

20.已知函数 f (x )= 2

3

ax 3 + x 2 + x - 1.

[

e 在直角坐标系 xOy 中,曲线 C 的参数方程为 ? ? y =

sin α 2

坐标方程为 ρcos θ + ? = -2 2 .

( I )当 a = -

的取值范围.

1

2 时,求 f (x )的单调区间; (Ⅱ)若函数 f (x )在 1,3]上单调递增,试求出 a

21.已知函数 f (x )= kx 2 x

(k >0).

(1)求函数 f (x )的单调区间;

(2)当 k=1 时,若存在 x >0,使 lnf (x )>ax 成立,求实数 a 的取值范围.

(二)选考题:共 10 分。请考生在第 22、23 题中任选一题作答。如果多做,则按所做的第

一题计分。

22.[选修 4-4:坐标系与参数方程](10 分)

1 ? x = 2cos α ? 3 ? 3

( α 为参数),将曲线 C 上各点的横

1

坐标都缩短为原来的 1

倍,纵坐标坐标都伸长为原来的 3 倍,得到曲线 C ,在极坐标系(与直

2

角坐标系 xOy 取相同的长度单位,且以原点 O 为极点,以 x 轴非负半轴为极轴)中,直线 l 的极

? π ? ?

4 ?

(1)求直线 l 和曲线 C 的直角坐标方程;(2)设点 Q 是曲线 C 上的一个动点,求它到直线 l 的距

2

2

离的最大值.

23.[选修 4-5:不等式选讲](10 分)

已知函数 f (x ) = 2x + 1 + 2x - 3 .

(Ⅰ)解不等式 f (x ) ≤ 6 ;

(Ⅱ)当 x ∈ R 时, f (x ) > a + 2 ,求实数 a 的取值范围.

.

14. y = 2sin(2 x + )

) - 2 = 0 ,所以 sin(2C -

试卷答案

一;选择题

1.D

2.C

3.B

4.A

5.A

6.B

7.B

8.B

9.D 10.D 11.C 12 A

二:填空题

13.1 + 1 1 1 1 1 11

+ + + + < 22 32 42 52 62 6

π

4

15. x - y + 2 = 0

16 故答案为:①②③④

三:解答题

17(Ⅰ) f ( x ) =

3 sin 2 x - 2 cos 2 x - 1 = 3 sin 2 x - (cos 2 x + 1) - 1 = 2sin(2 x -

π

6

) - 2

所以 f ( x ) 的最小正周期 T = 2π 2

= π ,最小值为 - 4

(Ⅱ)因为 f (C ) = 2sin(2C -

π π

6 6

) = 1

π π 11π π π

π 又 C ∈ (0, π ),2C - ∈ (- , ) ,所以 2C - = ,得: C =

6 6 6 6 2 3

因为 sin B = 2sin A ,由正弦定理得: b = 2a

由余弦定理得: c 2 = a 2 + b 2 - 2ab cos C = a 2 + 4a 2 - 2a 2 = 3a 2

又c=3,所以a=1,b=2

18.(Ⅰ)法一:由已知及余弦定理得2c?a2+c2-b2=2a+b,整理得a2+b2-c2=-ab.…2分

2ac

cos C=a2+b2-c2=-ab=-1,………………3分

2ab2ab2

又在△ABC中,0<C<,………………4分

∴C=2π,即角C的大小为2π..………………5分

33

法二:由已知及正弦定理得2sin C?cos B-sin B=2sin A,

又在△ABC中,sin A=sin(B+C)=sin B cos C+cos B sin C,.......……2分

∴2sin C cos B–sin B=2sin B cos C+2cos B sin C,

即2sin B cos C=–sin B,又sin B≠0,………………3分

∴cos C=-1,又0<C<,………………4分

2

∴C=2π,即角C的大小为2π..………………5分

33

(Ⅱ)由(Ⅰ)C=2π,依题意得如图,在△ADC中,AC=b=2,AD=3,

3

由正弦定理得sin∠CDA=AC?sin C=2?3=2,.………………7分

AD322

∵在△ADC中,0<∠CDA<,C为钝角,........………....………8分

∴∠CDA=π,故∠CAD=π-2π-π=π..………………9分43412

∵在△ABC中,AD是角A的平分线,∴∠CAB=π,.……….……10分

6

∴△ABC是等腰三角形,BC=AC=2..………………11分

故△ABC的面积S=1BC?AC sin2π=1?2?2?3=3..…………….…12分

23222

19【解答】解:(1)连接OC,设BC=x,则AB=2900-x2,(其中0<x<30),

∴S=2x900-x2=2x2(900-x2)≤x2+(900-x2)=900,

当且仅当x2=900﹣x2,即x=152时,S取最大值900;

π

,令 V′(x )=0,得 x=10 3 ;

π

时,函数 f (x ) = - x 3

+ x 2 + x - 1. f '(x ) = - x 2 + 2 x + 1,

1 (

)

单调递减区间是 - ∞,1 -

( ) (

)

[ [

?[在 x ∈ [1,3]

a ≥

2 x 2 因为 g '(x ) = - - ?, 1 ? [

∴取 BC=15 2 cm 时,矩形 ABCD 的面积最大,最大值为 900cm 2.

(2)设圆柱底面半径为 r ,高为 x ,

则 AB=2 900 - x 2 =2π r ,解得 r=

900 - x 2

π

∴V=π r 2h=

900x - x 3

π ,(其中 0<x <30);

∴V′= 900 - 3x 2

因此 V (x )=在(0,10 3 )上是增函数,在(10 3 ,30)上是减函数;

∴当 x=10 3 时,V (x )取得最大值 V (10)= 6000 3

∴取 BC=10 3 cm 时,做出的圆柱形罐子体积最大,最大值为

6000 3 π

cm 3.

20 (I )当 a = - 1 1 2 3

令 f '(x ) > 0, 即 x 2 - 2 x - 1 < 0, 解得1 - 2 < x < 1 + 2;

令 f '(x )< 0, 解得 x > 1 + 2 或 x < 1 - 2.

所以当 a = -

时,函数 f (x )的单调递增区间是 1 -

2,1 +

2 ,

2

2 和 1 + 2,+∞ .

(Ⅱ)法一: f '(x ) = 2ax 2 + 2 x + 1,

函数 f (x )在 1,3]上单调递增,等价于 f '(x ) = 2ax 2 + 2 x + 1 ≥ 0 在区间 x ∈ 1,3]恒成立,

等价于 a ≥ - 2 x - 1

? - 2 x - 1 ?

, x ∈ 1,3]

2 x 2

? ? max

令 g (x ) = - 2 x - 1 2 x 2 ? 1 ? 1 = - + ? x 2 x 2 ? ? x 2

+ 4 ?

?=

4 x 4 ? 1 x 2 + 1 x 3 > 0,

所以函数 g (x ) = - 2 x - 1 2 x 2

在区间 x ∈ 1,3

]上单调递增,故 g (x ) max

= g (3) = -

7

18

[

[

[

?h3=18a+7≥0,

解得-

2a<

0<1,

(

,解得a>0.综上所述a的取值范围是?-,+∞?.

?

22.(1)因为直线的极坐标方程为ρcos θ+

π?

?=-22,所以有

法二:f'

(x)=2ax2+2x+1,函数f(x)在1,3]上单调递增,

等价于f'

(x)=2ax2+2x+1≥0在区间x∈1,3]恒成立,令h(x)=2ax2+2x+1.则命题等价于h

(x)

min

≥0在区间x∈1,3

]恒成立.

?h1

)=2a+3≥0.7

(1)当a≤0时,由?()18≤a≤0;

(2)当a>0时因为函数图像的对称轴x=-

1

此时只有满足?

a

(

>

)

?h1=2a+3≥0

?7

?18

?

?

21解:(1)函数的定义域为R,求导函数可得f′(x)=,

当k<0时,令f′(x)>0,可得x<0或x>2;令f′(x)<0,可得0<x<2

∴函数f(x)的单调增区间为(﹣∞,0),(2,+∞),单调减区间为(0,2);

当k>0时,令f′(x)<0,可得x<0或x>2;令f′(x)>0,可得0<x<2

∴函数f(x)的单调增区间为(0,2),单调减区间为(﹣∞,0),(2,+∞);

(2)当k=1时,f(x)=,x>0,1nf(x)>ax成立,

等价于a<,设g(x)=(x>0)

存在x>0,使1nf(x)>ax成立,等价于a<g(x)

max

g′(x)=,当0<x<e时,g′(x)>0;当x>e时,g′(x)<0

∴g(x)在(0,e)上单调递增,在(e,+∞)上单调递减.

∴g(x)

max

=g(e)=﹣1,∴a<﹣1.

?

?4?

ρcosθ-ρsinθ+4=0,即直线l的直角坐标方程为:x-y+4=0

因为曲线 C 的的参数方程为 ? 3 ( α 为参数),经过变换后为 ? ? y = sin α

? 2 =

由此得,当 cos α + ? = 1 时, d 取得最大值,且最大值为 2 2 + 1 ? ? 1

3 x <- - ≤ x ≤

2 2

或 ? 2 ?- (2 x + 1)- (2 x - 3) ≤ 6 ?(2 x + 1)- (2 x - 3) ≤ 6

或 ? 2 ? -1 ≤ x < -

或 - ≤ x ≤

或 < x ≤ 2 .

?(2 x + 1)+ (2 x - 3) ≤ 6

? x = 2cos α 1 y = sin α

?

3

所以化为直角坐标方程为: x 2 + y 2 = 1

(2)因为点 Q 在曲线 C 上,故

可设点 Q 的坐标

为 (cos α ,sin α ) ,

2 ? ? x = cos α

( α 为参数)

从而点 Q 到直线 l 的距离为 d = cos α - sin α + 4

π

2 cos(α + ) + 4

4 2

?

π ? ?

6 ?

? ? 23(Ⅰ)原不等式等价于 ?

?

?

? 3 ? x > 1 1 3 3 2

2

2

2

?

故不等式的解集为{x - 1 ≤ x ≤ 2}

.

(Ⅱ)由三角不等式: f (x ) = 2x + 1 + 2x - 3 ≥ (2 x + 1)- (2 x - 3) = 4 ,所以函数 f (x )的

最小值为 4,

由恒成立关系,所以 a + 2 < 4 ? -4 < a + 2 < 4 ? -6 < a < 2 .

- 9 -

高三数学9月月考试题 理2

重庆市秀山高级中学2017届高三数学9月月考试题 理 一、选择题:本大题共12个小题,每小题5分,共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.已知命题p :12,=∈?x R x ,则p ?是.....................................................................( C ) A.12,≠∈?x R x B.12,≠??x R x C.12 ,0 0≠∈?x R x D. 12 ,0 0≠??x R x 2.若集合N M x y x N y y M x 则},1{},2{-====等于.............................( C ) A.),0(+∞ B.),0[+∞ C.),1[+∞ D.),1(+∞ 3.有下列四个命题: ①“若1=xy ,则y x ,互为倒数”的逆命题; ②“面积相等的三角形全等”的否命题; ③“若1≤m ,则有实根022 =+-m x x ”的逆否命题; ④“若B A B B A ?=则, ”的逆否命题,其中真命题是......................................( C ) A.①② B.②③ C.①②③ D.③④ 4. 已知函数???≤>=) 0(3)0(log )(2x x x x f x ,则)]41 ([f f 的值是.......................................( C ) A.9 1 - B.9- C.91 D.9 5.函数}3,2,1{}3,2,1{:→f 满足)())((x f x f f =,则这样的函数个数共有........( D ) A.1个 B.4个 C.8个 D.10个 6.设的定义域为,则)2 ()2(22lg )(x f x f x x x f +-+=..............................................( B ) A.)()(4,00,4- B.)()(4,11,4- - C.)()(2,11,2- - D.)()(4,22,4- - 7.若函数)(x f y =的值域是]3,21 [,则函数) (1 )()x f x f x F + =(的值域...............( B ) A.]3,21[ B.]310, 2[ C.]310,25[ D.]3 10,3[

福建省最新2021届高三数学10月月考试题

福建省罗源第一中学2021届高三数学10月月考试题 一、单选题(每小题5分) 1.复数 1 1i i -+(i 为虚数单位)的虚部是( ) A. -1 B. 1 C. i - D. i 2.αβ≠是cos cos αβ≠的( )条件. 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 3.已知sin(π+θ)=-3cos(2π-θ),|θ|<π 2 ,则θ等于( ) A .-π6 B .-π3 C.π6 D.π3 4.函数1ln sin 1x y x x +=?-的图象大致为( ) 5.已知a >0且a ≠1,函数f (x )=? ????a x ,x ≥1 ax +a -2,x <1在R 上单调递增,那么实数a 的取值范围是( ) A .(1,+∞) B .(0,1) C .(1,2) D .(1,2] 6.已知△ABC 中,AB =2,B =π4,C =π6 ,点P 是边BC 的中点,则AP →·BC → 等于( ) A .1 B .2 C .3 D .4 7.若函数f (x )=sin ? ????ωx -π6(ω>0)在[0,π]上的值域为???? ??-12,1,则ω的最小值为( ) A.23 B .34 C.43 D .3 2 8.在ABC ?中,已知点P 在线段BC 上,点Q 是AC 的中点, AQ y AB x AP +=,0,0>>y x ,则 y x 11+的最小值为( )

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