保密★用前
斌卷类型：4
2023届高三定时训练
数学试题
2022.1]
汴意事项：
1.茶卷前，乡非务将月心的姓、号宁等填红在答盟试按密定位晋上
2.山答逃桥题时，远东小题答案斤，川川笔把答题六对应题H的咨案冰马涂黑：如需
政动、州橡皮擦泸后，选泳其它袋茱标号，蓉选样越刑，将容笨与在然越
「在本试卷无效
3.考试站束，将不试卷和究卡一并父叫
·、中项逃择题：本题状8小题，每小题5分，共40分。在每小题给出的四个选项屮，
只有，项是符合题H婴求的。
1.议宋合A={1,2,3,4,7},B={x∈N2A.AUB=B
B.A∩B={2,3,4}
.C,B={2,7}
1).B A
2.下列函数中是俩数.儿在区创(0，+oo):是减对数时是
A.y日x|+1
B.y=x2
y=1
-x
I0.y=2
3.若ab>0,1la>b,
A.a2>b2
2
na+b>ab
a b
a b
2
2
4.心知函数f(x)=
x≤2.则fog,12)=
f(x-10,x>2.
1
1
1.
B.-6
C.
D.-3
3
6
5.已知足第叫象限加，化简
1+sina
-sin a
V1-sina
V1+sina
向.一数兴世第1页共斥页
A.-2tana
B.2tana
(tan a
1).-tan a
.已知函数f()=xsinx,x∈R.则f(兮：f0,f(-)的大小关系为
A.f-3>f⑩>f爱
3.f0>f(-爱>f爱
.f>f⑩>f(-
D.f(-2)>f>f0⑩
7.数列{a,}的通项a,=ncos”2-(nsin” ,共前n啄知为S。,则S0为
4.460
B.470
G.480
1).490
8.
已Ia=0.7e04,b=enl.4.c=0.98.则a,b.c人小关系是
A.c>a>b
B.b>a>c
(b>c>a
1).a>c>b
二、多项选择题：本题共4小题，每小题5分，共20分。在每小题给出的四个选项，
有多项符合题日要求。全部选对的得5分，部分选对的得2分，有选的得0分。
9.付丁w数f(x)sinx+cos2x,卜列论的母足
Af)的位瘦为[0，g
B.f)在0，]单洲递啪
C.f)的象关直线x=工对称D.f(的最小周期为π
4
10客名作“河内塔”邀、地面而立者三根柱子、布1号江上从上至下、从小列人套若
n个中：带孔的网盘.将一个市于最上方的一个网盘移动
创一个柱了，儿保何个柱了较大的贯性总正较小
的圆食卜训，视为一沃操.设将n个厨盆全部域1乃朴
于移动到3号千的以少操作数为an,则
向.一数邀第2页爪页2023届高三定时训练
log3 x 58
3 ( ,0) (2,+ )
x 3
q 0
x 2
(x 2)(x 3) 0 2 x 3
q x 2 x 3
x2 4ax +3a2 0 (x 3a)(x a) 0
a 0 a x 3a
p q q p p q
q x 2 x 3
0 a 2
3a 3
1 a 2
a 1 a 2
x2 mx 20 0 {x | 2 x n}
x2 mx 20 = 0 2 n
2+ n = m,
2 n = 20.
n =10 m =8
a b na+mb = 2 10a+8b = 2
5a+4b =1
1 1 1 1
+ = ( + )(5a + 4b)
5a b 5a b
4b 5a
=1+ + + 4 5+ 2 4 = 9
5a b
4b 5a 1 1
= a = b =
5a b 15 6
1 1
+ 9
5a b
f (x)
R x R x R
1
+1
2 x +1 x 1+ 2x
f ( x) = = 2 = = f (x)
2 x+1 + 2 2 2+ 2x+1
+ 2
2x
f (x)
2x = t t 0
t +1 (t +1)+ 2 1 1
y = = = + (t 0)
2t + 2 2t + 2 2 t +1
1 1 1 1 1
0 1 +
t +1 2 2 t +1 2
1 1
f (x) ( , )
2 2
1
f (x) log (2c 1) log9 (2c 1) 9
2
0 2c 1 3
1
c 2
2
1
c ( , 2)
2
an 3= 3Sn = (n+ 2)an
Sn n + 2
n 2 3Sn 1 = (n+ 2)an 1
3an = (n+ 2)an (n+1)an 1
an n +1=
an 1 n 1
an 1 n a n 1 a 4 a 3= n 2 = 3 = 2 =
an 2 n 2 an 3 n 3 a2 2 a1 1
an n +1 n n 1 4 3 n(n +1)= =
a1 n 1 n 2 n 3 2 1 2
n(n +1)
a1 =1 an =
2
n(n +1)
n =1 an =
2
n (n+1)
an =
2
an 3 n + 2 n(n +1)(n + 2)= Sn = an =
Sn n + 2 3 6
n 6 1 1
== = 6( )
Sn (n +1)(n + 2) n +1 n + 2
1 2 n 1 1 1 1 1 1
+ + + = 6( + + + )
S1 S2 Sn 2 3 3 4 n +1 n + 2
1 1
= 6( )
2 n + 2
1
0
n + 2
1 2 n 1
+ + + 6 = 3
S1 S2 Sn 2
ABD
1 BD
=
sin ADB sin 30
ACD
3 CD
=
sin ADC sin DAC
ADB+ ADC =180 sin ADB = sin ADC
D BC B 2BD=DC
1
sin CAD =
3
1
sin CAD =
3
1 2 2
cos CAD = 1 sin2 CAD = 1 ( )2 =
3 3
2 2
cos CAD =
3
sin BAC = sin (30 + CAD)
= sin30 cos CAD+cos30 sin CAD
1 2 2 3 1
= ( )+
2 3 2 3
2 2 + 3
= 0
6
2 2 2 2 + 3
cos CAD = sin BAC =
3 6
ABC
1
S = AB AC sin BAC
2
1 2 2 + 3 2 2 + 3
= 1 3 =
2 6 4
2 2 + 3
△ABC
4
f (x) = e 2x +2x+a ln(x+1)
a
f (x) = 2e 2x + 2+ (x 1)
x +1
f (0) =1 f (0) = a
f (x) x = 0 P ( 1,0)
f (0) 0
f (0) = a =1
0 ( 1)
f (x) (x +1)a e 2x + 2x (x+1)a a ln(x+1)
e 2x ln e 2x (x+1)a ln(x+1)a
x 0 2 a 0 0 e
2x 1 0 (x+1)a 1
g(x) = x ln x (0 x 1)
e 2x + 2x (x+1)a a ln(x+1)
g(e 2x ) g((x+1)a )
1
g (x) =1 0 g(x) (0,1)
x
2x a
x 0 2 a 0 e (x+1)
2x a ln(x +1)
x 0 2 a 0 2x + a ln(x +1) 0
h(x) = 2x + a ln(x +1) (x 0)
a
2 a 0 h (x) = 2+ (0,+ )
x +1
h (x) h (0) = 2+ a 0
h(x) (0,+ ) h(x) h(0) = 0
x 0 2 a 0 2x + a ln(x +1) 0
a
x 0 2 a 0 f (x) (x +1)
