北师大(新课标)高中数学必修5模块检测试卷（图片版含解析）
文档属性
| 名称 | 北师大(新课标)高中数学必修5模块检测试卷（图片版含解析） |
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| 格式 | |||
| 文件大小 | 136.5KB | ||
| 资源类型 | 教案 | ||
| 版本资源 | 北师大版 | ||
| 科目 | 数学 | ||
| 更新时间 | 2021-01-18 00:00:00 | ||
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文档简介
5
(3,1)口(-4,6)
A口「-7,+∞
∪(24,+∞
∪(24
B(3324a)(3-42a60(a+7)(a-24)<
an
a
an(n∈N
1
B
+
Ao
algx>big,
x>0
Co
a>
∈R
(√6+√2)
(6-√
20(√6+
(√6-3
05
√6
A
B口
9
5
A
6
<
45<80
>45
5
<
<
≠0
B
(a1+a20)=10
,
ta
Q=
a2+
a
a,
td=
a
+9
△
B口△AB
△AB
B
+2B
△
△
%
p
(1+n%)=(1+p%2口1+p%=√(1+m%(1+n%
+1
9
a
a+a0=a(a≠0),a
B
a
ag「a
△
C=2B
(120-C)
(120-C)=√3cosC+
90,A=30
C=90
0
(n≥2)uan=(q
≠
SnS也<
T
a
aB
ag
5
)2+(5k)2-2×(8k)×(5k)C
×16×10
4
4X+3
9X+a<0
X
4x+3<0
6X+8<0
4X+3<
+8<0
3)≤
(a+
(a+b)≥
va
a+-)+(b+-)≥√a+
(a+b)(a+b+1)≥√a(a+√6
(a+
a
5
S
40<
a;7<0
(a
fa=16(a
ta)T
BC
+tan
B
<7t
B
5
y≠
4(y-1)(y-n)≥
4n-1≤
4a,bn
=4
=-√4abn+
4
a,
+3d
=a,d
≠1口
+9d
=a,d
9d
(d-1)(d-1)(d+d+1)=3(d-1)
d°+d
d≠
d=-32
32)
2口-2
a1+(n-1)d=ad≌2+(n-1)(2)=2(-y2)
a
a=0
a<0
X
(x)+6a=a(X-1)(
+6a=0
(4a+2)Ⅹ+9a
△=[-4a+2)2-36
4a
<0
X
5
2囗f(x)=a(X-1)(X-3)-2X=aX
a
a
12a-(4a+2)y
-a-4
4
<
>0
√3)∪
A
B囗45
60
B=4
∠B
T
T
A
T
a=
2b
√3
8、x
T
B
>
+t
2(an+3)
a
a
0(n∈N
6x-2aβ+6β
+β
a-2aβ+6β=3
a
a
+
a
a
5
(a,
a
X
a
X
a
(an++1)=2g(an+1)
十
g(1+a
a
lg(1+a1)(1-
a
a
(3,1)口(-4,6)
A口「-7,+∞
∪(24,+∞
∪(24
B(3324a)(3-42a60(a+7)(a-24)<
a
an(n∈N
1
B
+
Ao
algx>big,
x>0
Co
a>
∈R
(√6+√2)
(6-√
20(√6+
(√6-3
05
√6
A
B口
9
5
A
6
<
45
>45
5
<
<
≠0
B
(a1+a20)=10
,
ta
Q=
a2+
a
a,
td=
a
+9
△
B口△AB
△AB
B
+2B
△
△
%
p
(1+n%)=(1+p%2口1+p%=√(1+m%(1+n%
+1
9
a
a+a0=a(a≠0),a
B
a
ag「a
△
(120-C)
(120-C)=√3cosC+
90,A=30
C=90
0
(n≥2)uan=(q
≠
T
a
aB
ag
5
)2+(5k)2-2×(8k)×(5k)C
×16×10
4
4X+3
9X+a<0
X
4x+3<0
+8<0
3)≤
(a+
(a+b)≥
va
a+-)+(b+-)≥√a+
(a+b)(a+b+1)≥√a(a+√6
(a+
a
5
S
40<
a;7<0
(a
fa=16(a
ta)T
+tan
B
<7t
B
5
y≠
4(y-1)(y-n)≥
4n-1≤
4a,bn
=4
=-√4abn+
4
a,
+3d
=a,d
≠1口
+9d
=a,d
9d
(d-1)(d-1)(d+d+1)=3(d-1)
d°+d
d≠
d=-32
32)
2口-2
a1+(n-1)d=ad≌2+(n-1)(2)=2(-y2)
a
a=0
a<0
X
(x)+6a=a(X-1)(
+6a=0
(4a+2)Ⅹ+9a
△=[-4a+2)2-36
4a
<0
X
5
2囗f(x)=a(X-1)(X-3)-2X=aX
a
a
12a-(4a+2)y
-a-4
4
<
>0
√3)∪
A
B囗45
60
B=4
∠B
T
T
A
T
a=
2b
√3
8、x
T
B
>
+t
2(an+3)
a
a
0(n∈N
6x-2aβ+6β
+β
a-2aβ+6β=3
a
a
+
a
a
5
(a,
a
X
a
X
a
(an++1)=2g(an+1)
十
g(1+a
a
lg(1+a1)(1-
a
a
同课章节目录