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100學年度國立中央大學數學系申請入學數學試題詳解

# 100學年度國立中央大學數學系申請入學數學試題詳解

Ratings: (0)|Views: 107|Likes:

### Availability:

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02/27/2013

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text

original

100/04/16

20111

(1)

x
16

x
28

(A)
x
(B)
x
2
(C)
x
3
(D)
x
4
(E)
x
6
[
]
(
x
16
,
x
28

)=
x
4
，∴
x
4

(D)
(2)

{1, 2, 3,
, 99, 100}

x

x
2
+
x
3

[
]
x
2
+
x
3

=
x
2
(1+
x
)
，我們只需要考慮
1+
x

2
2
, 3
2
, 4
2
,
, 10
2

∴有
9

(3)

5
x
+ 2
y
+ 2
z
=1172
x
+ 5
y
+ 2
z
=1172
x
+ 2
y
+ 5
z
=117 [
]
5
x
+ 2
y
+ 2
z
=117
…○
1
2
x
+ 5
y
+ 2
z
=117
…○
2
2
x
+ 2
y
+ 5
z
=117
…○
3
；○
+
+

9(
x
+
y
+
z
)=351
x
+
y
+
z
=39
…○

×
2
3
x
=117
78=39
x
=13
。同理可得

y
=13
z
=13
(4)

ABC

a
,
b
,
c
(
a
+
b
)
(
b
+
c
))
(
c
+
a
) = 4
6
5
，求
tan
A

[
]
Let

a
+
b
= 4

…○
1

b
+
c
= 5

…○
2

c
+
a
= 6

…○
3
>0
。將三式相加得
2(
a
+
b
+
c
)=15
a
+
b
+
c
=15
2
…○
4

c
=7
2
，○
2

a
=5
2
，○

b
=3
2
，由餘弦定理得：
cos
A
=
b
2
+ c
2

a
2
2
bc
=(3
2)
2
+ (7
2)
2
(5
2)
2
2(3
2)(7
2)=1314

tan
A
=3313
(5)
P
(
x
)

4

P
(2)=
P
(
2)=
P
(
3)=
1
P
(1)=
P
(
1)=1
，試求
P
(0)

[
]
P
(
x
)=(
ax
+
b
)(
x
2)(
x
+2)(
x
+ 3)
1
P
(1)=1=(
a
+
b
)( 1

2 )( 1 +2 )( 1

+ 3 )
1
12(
a
+
b
)=2

a
+
b
=
16
…○

100/04/16

20112
P
(
1)=1=(
a
+
b
)(
1

2 )(
1 +2 )(
1

+ 3 )
1
6(
a
+
b
)=2

a
+
b
=
13
…○

+

2
b
=
12
b
=
14
P
(0)=
b
×
(
2)(2)( 3)
1= (
14)
×
(
2)(2)( 3)
1=2
(6)

n

2011

416
，則
5
n

2011

[
]
Let
n
=2011
q
+416 then 5
n
=2011
×
(5
q
)+416
×
5=2011
×
(5
q
)+2080=2011
×
(5
q
+1)+ 69
5
n

2011

69
(7)

a

b

a

a
2
+
b
2
=18
，試求
a
+
b

[
]
a
= 4 +
b
a
2
+
b
2
= (4 +
b
)

2
+
b
2
=18
2
b
2
+8
b

2=0
，則
b
2
+4
b

1=0

b
=
2
±
5(

)
，∴
a
= 4 +
b
=2+5
a
+
b
=25(8)

log 3
x
+ 4 +12log( 5
x
+ 1 )=1 + log3

x
[
]
log 3
x
+ 4 +12log( 5
x
+ 1 )=1 + log3
12log(( 3
x
+ 4)( 5
x
+ 1)) =log30
log(15
x
2

+23
x
+ 4) =2log30
15
x
2

+23
x
+ 4=900
15
x
2

+23
x

896=0
(15
x
+128)(
x

7)= 0
x
=7,
12815(

x
>0 )

(1)

0.9999
=1

[
]
0.9999
=0.9+0.09+0.009+0.0009+
=9(0.1+0.01+0.001+0.0001+
)=9
×
0.11
0.1=1
，故命題為正確。
(2)(
a
,
b
,
c
)+[
a
,
b
,
c
]=854

a
b
c
=10
12
15
，求

a
+
b
+
c

[
]
a
=10
b
=12
c
=15
(
a
,
b
,
c
)=
[
a
,
b
,
c
]=60
61
=854
，∴
=14
a
+
b
+
c
=37
=37
×
14=518(3)

4

3

311
，求此袋中紅球的個數。
[
]

n

n
7
The text passage is a collection of mathematical problems and their solutions, primarily in the form of multiple-choice questions and word problems, with detailed solutions provided for each.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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Standard view
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Results for:
P. 1
100學年度國立中央大學數學系申請入學數學試題詳解

# 100學年度國立中央大學數學系申請入學數學試題詳解

Ratings: (0)|Views: 107|Likes:

### Availability:

See more
See less

02/27/2013

pdf

text

original

100/04/16

20111

(1)

x
16

x
28

(A)
x
(B)
x
2
(C)
x
3
(D)
x
4
(E)
x
6
[
]
(
x
16
,
x
28

)=
x
4
，∴
x
4

(D)
(2)

{1, 2, 3,
, 99, 100}

x

x
2
+
x
3

[
]
x
2
+
x
3

=
x
2
(1+
x
)
，我們只需要考慮
1+
x

2
2
, 3
2
, 4
2
,
, 10
2

∴有
9

(3)

5
x
+ 2
y
+ 2
z
=1172
x
+ 5
y
+ 2
z
=1172
x
+ 2
y
+ 5
z
=117 [
]
5
x
+ 2
y
+ 2
z
=117
…○
1
2
x
+ 5
y
+ 2
z
=117
…○
2
2
x
+ 2
y
+ 5
z
=117
…○
3
；○
+
+

9(
x
+
y
+
z
)=351
x
+
y
+
z
=39
…○

×
2
3
x
=117
78=39
x
=13
。同理可得

y
=13
z
=13
(4)

ABC

a
,
b
,
c
(
a
+
b
)
(
b
+
c
))
(
c
+
a
) = 4
6
5
，求
tan
A

[
]
Let

a
+
b
= 4

…○
1

b
+
c
= 5

…○
2

c
+
a
= 6

…○
3
>0
。將三式相加得
2(
a
+
b
+
c
)=15
a
+
b
+
c
=15
2
…○
4

c
=7
2
，○
2

a
=5
2
，○

b
=3
2
，由餘弦定理得：
cos
A
=
b
2
+ c
2

a
2
2
bc
=(3
2)
2
+ (7
2)
2
(5
2)
2
2(3
2)(7
2)=1314

tan
A
=3313
(5)
P
(
x
)

4

P
(2)=
P
(
2)=
P
(
3)=
1
P
(1)=
P
(
1)=1
，試求
P
(0)

[
]
P
(
x
)=(
ax
+
b
)(
x
2)(
x
+2)(
x
+ 3)
1
P
(1)=1=(
a
+
b
)( 1

2 )( 1 +2 )( 1

+ 3 )
1
12(
a
+
b
)=2

a
+
b
=
16
…○

100/04/16

20112
P
(
1)=1=(
a
+
b
)(
1

2 )(
1 +2 )(
1

+ 3 )
1
6(
a
+
b
)=2

a
+
b
=
13
…○

+

2
b
=
12
b
=
14
P
(0)=
b
×
(
2)(2)( 3)
1= (
14)
×
(
2)(2)( 3)
1=2
(6)

n

2011

416
，則
5
n

2011

[
]
Let
n
=2011
q
+416 then 5
n
=2011
×
(5
q
)+416
×
5=2011
×
(5
q
)+2080=2011
×
(5
q
+1)+ 69
5
n

2011

69
(7)

a

b

a

a
2
+
b
2
=18
，試求
a
+
b

[
]
a
= 4 +
b
a
2
+
b
2
= (4 +
b
)

2
+
b
2
=18
2
b
2
+8
b

2=0
，則
b
2
+4
b

1=0

b
=
2
±
5(

)
，∴
a
= 4 +
b
=2+5
a
+
b
=25(8)

log 3
x
+ 4 +12log( 5
x
+ 1 )=1 + log3

x
[
]
log 3
x
+ 4 +12log( 5
x
+ 1 )=1 + log3
12log(( 3
x
+ 4)( 5
x
+ 1)) =log30
log(15
x
2

+23
x
+ 4) =2log30
15
x
2

+23
x
+ 4=900
15
x
2

+23
x

896=0
(15
x
+128)(
x

7)= 0
x
=7,
12815(

x
>0 )

(1)

0.9999
=1

[
]
0.9999
=0.9+0.09+0.009+0.0009+
=9(0.1+0.01+0.001+0.0001+
)=9
×
0.11
0.1=1
，故命題為正確。
(2)(
a
,
b
,
c
)+[
a
,
b
,
c
]=854

a
b
c
=10
12
15
，求

a
+
b
+
c

[
]
a
=10
b
=12
c
=15
(
a
,
b
,
c
)=
[
a
,
b
,
c
]=60
61
=854
，∴
=14
a
+
b
+
c
=37
=37
×
14=518(3)

4

3

311
，求此袋中紅球的個數。
[
]

n

n
7
user
The text passage is a collection of mathematical problems and their solutions, primarily in the form of multiple-choice questions and word problems, with detailed solutions provided for each.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS