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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:
  
  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
explanation: The document appears to be a detailed solution set for a mathematics entrance examination for the Department of Mathematics at National Central University in Taiwan for the academic year 100 (2011). It contains multiple questions and their corresponding answers, each with a detailed step-by-step solution.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: Welcome to Scribd. Sign in or start your free trial to enjoy unlimited e-books, audiobooks & documents.Find out more
  Standard view
  Full 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
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document appears to be a detailed solution set for a mathematics entrance examination for the Department of Mathematics at National Central University in Taiwan for the academic year 100 (2011). It contains multiple questions and their corresponding answers, each with a detailed step-by-step solution.
  
  Label: multiple-question-and-answer