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Mathematics Paper 1 Memo - NATIONAL SENIOR CERTIFICATE GRADE 12

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Mathematics Paper 1 Memo - NATIONAL SENIOR CERTIFICATE GRADE 12 Powered By Docstoc
					                             NATIONAL
                        SENIOR CERTIFICATE



                                 GRADE 12



                             MATHEMATICS P1

                              EXEMPLAR 2008

                              MEMORANDUM




                     This memorandum consists of 12 pages.




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Mathematics Paper 1                                 2                    Grade 12 Exemplar 2008
                                            NSC : Memorandum

QUESTION 1

1.1.1   x 2 − 10 x = 24
                                                               standard form
        x 2 − 10 x − 24 = 0
                                                               factors
        ( x − 12)( x + 2) = 0
        x = 12 or x = −2                                       answers
        OR 2                                                                                 (3)
        ( x − 5) = 49 = 7 2
        ∴ x − 5 = 7or − 7
        ∴ x = 12 or - 2
1.1.2   x 2 − 6 x = 10(1 − 3x)
        x 2 − 6 x = 10 − 30 x
        x 2 + 24 x − 10 = 0                                    standard form
             −24 ± (24) 2 − 4(1)(−10)
        x=                                                     substitution
                      2(1)
            −24 ± 616                                          616
        x=
                 2
        x = 0, 41 or x = −24, 41                                 answers
                                                                                             (5)
1.1.3   ( x − 1)( x − 2) ≤ 6
        x 2 − 3x − 4 ≤ 0                                       standard form
        ( x − 4)( x + 1) ≤ 0                                   factors
         + 0 –             0     +
             –1             4
        −1 ≤ x ≤ 4                                               answer
                                                                                             (4)
1.2              x + 3y = 5
                      x = 5 − 3y                               solve for x
                                                               substitution
        (5 − 3 y ) y + y 2 = 3
         5 y − 3y2 + y2 = 3
                                                               standard form
                       0 = 3 − 5 y + 2 y2
                     0 = (3 − 2 y )(1 − y )                    factors
                    3
                 y=     or y = 1                                 y-answers
                    2
                    1
                 x=     or x = 2                               x-answers
                    2                                                                        (7)

                                                                                            [19]

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Mathematics Paper 1                           3                   Grade 12 Exemplar 2008
                                      NSC : Memorandum



QUESTION 2

2.1     FV = P (1 − i ) n
              V
                                                         formula
                                                         substitution
        0,25P = P(1 − 0,21) n
        n log 0,79 = log 0,25                            simplification
             log 0,25                                    n=
        n=                                               answer
             log 0,79
                                                                                       (5)
        n = 5,88 years
2.2.1   Fv = Pv(1 − i ) n
        491520 = 1200000(1 − i ) 4
                                                            substitution
        (1 − i ) 4 = 0,4096                               simplification
        i = 0,2
                                                         answer
        r = 20,00%                                                                     (4)
2.2.2   Fv(sinkingfund) = 1, 2000000(1,15) − 491520
                                       4                 substitution
        = R1607287,50                                     answer
                                                                                       (2)
2.2.3   (1,15) = 1,74900625
              4                                          substitution

        ∴ an increase of 74,90 %                         increase
                                                                                       (2)
2.2.4 Let x be the monthly repayment
            9                                            i
      i=        = 0,0075
          1200
                                                         1607287,50
                      ⎡ (1,0075) 48 − 1⎤
      1607287,50 = x ⎢                 ⎥                  1,0075
                      ⎣ 0,0075 ⎦                          substitution
        1607287,50 = x[57,5207111]                        57,5207111
                                                          answer
        x = R 27 942,76
                                                                                       (6)

                                                                                      [19]




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Mathematics Paper 1                                              4                     Grade 12 Exemplar 2008
                                                         NSC : Memorandum

QUESTION 3

3.1    2 ; 21                                                               2
                                                                            21
                                                                                                            (2)
3.2     (2 + 2 + ... + 2)+ (5 + 9 + 13 + ...)                               2 + 2 + ... + 2
                                                                                 for 50 terms
               for 50 terms              for 50 terms
             50         50                                                  5 + 9 + 13...
        = ∑ 2 + ∑ (4i + 1)                                                    for 50 terms

             i =1       i =1

                  ⎡ 50            ⎤                                         100
        = 2(50) + ⎢ (2(5) + 49(4))⎥                                         substitution
                  ⎣2              ⎦
        = 100 + 25(10 + 196)
        = 100 + 5150                                                        answer
        = 5250                                                                                              (5)

                                                                                                            [7]

QUESTION 4

4.1      130 ; 173
                                                                                                            (2)
4.2      5             18           37         62            93


                13             19        25             31                  a=3

                6      6        6
         The second difference is constant ∴ Tn is quadratic
                                                                            substitution
         ∴ an 2 + bn + c = Tn
         2a = 6
         a=3                                                                equation (ii)
                        Tn = 3n 2 + bn + c
                                                                            b=4
                          5 = 3(1) 2 + b(1) + c
                      b + c = 2 ... (i)                                     c=–2
                        18 = 3(2) 2 + b(2) + c
                     2b + c = 6 ... (ii)                                                                    (5)
         (ii) – (i):      b=4
                          c = −2
         Tn = 3n + 4n − 2
                   2




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Mathematics Paper 1                            5                   Grade 12 Exemplar 2008
                                       NSC : Memorandum
         OR
                  a(1) 2 + b(1) + c = 5
         (i)      a+b+c = 5
                  a(2) 2 + b(2) + c = 18
                                                          subst. x = 1, 2, 3
         (ii)     4a + 2b + c = 18
                  a(3) 2 + b(3) + c = 37
         (iii)    9a + 3b + c = 37
                                                          Equations (ii) – (i):
         (ii) – (i):     3a + b = 13
                         b = 13 − 3a
         Substitute b = 13 − 3a into (iii)                Substitution into (iii)
                 9a + 3(13 − 3a) + c = 37
                 9a + 39 − 9a + c = 37                    c=-2
                 c = −2
         Substitute b = 13 − 3a and c = – 2 into (ii)     b=4
                 4a + 2(13 − 3a ) + (− 2 ) = 18
                 − 2 a = −6
                    a=3
                ∴b=4

         OR
         Let Tn be the nth term of the sequence           Let Tn
         Then
         T2 − T1 = 13 ⎫
         T3 − T2 = 19 ⎪  ⎪
                                                          subtracting terms
                         ⎪
         T4 − T3 = 25 ⎬add both sides
         T5 − T4 = 31 ⎪                                   Tn − T
                         ⎪
         Tn − Tn −1 = ...⎪
                         ⎭
         Tn − T1 = 13 + 19 + 25 + ...(to n − 1 terms)     Tn − 5
                  ⎛ n −1⎞
         Tn − 5 = ⎜      ⎟[2(13) + (n − 2)6]              Tn = 3n 2 + 4n − 2
                  ⎝ 2 ⎠
         Tn = (n − 1)(3n + 7) + 5
         Tn = 3n 2 + 4n − 2

4.3      3n 2 + 4n − 2 = 1278                             substitution
         3n 2 + 4n − 1280 = 0
         (3n + 64)(n − 20) = 0                            factors
             − 64                                              − 64
         n=         or n = 20                             n≠
               3                                                3
             − 64                                         n = 20
         n=        is not valid                                                        (4)
               3
         ∴ n = 20                                                                     [11]

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Mathematics Paper 1                                    6              Grade 12 Exemplar 2008
                                               NSC : Memorandum

QUESTION 5

5.1    Pattern 3

        1            3          9                                 sum
             +            +                                        9
        4           16          64
                                                                   64
       Pattern 4
       1        3               9        27                       sum
           +              +          +                             27
       4       16               64       256
                                                                   256
                                                                                          (4)
5.2                                               n −1                nth term
        1            3          9                3
             +            +          + ... +
        4           16          64                4n
                                                                  sigma notation
         n
            3k −1
       ∑ 4k
       k =1                                                                               (3)
5.3          1                  3
       a=                 r=                                      r
             4                  4

                          1
              a           4
       S=         =             =1                                answer
             1− r           3
                         1−                                                               (2)
                            4
                                                                                          [9]




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Mathematics Paper 1                         7                         Grade 12 Exemplar 2008
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QUESTION 6

6.1                                                          answer
        b(1) =
               2   1
                   2
            1
        b=                                                   answer
             2                                                                               (2)
              1
        a1 =
              2
             1
        a=
             2
6.2     y = 2− x
                       x
          ⎛1⎞
        y=⎜ ⎟                                                substitution
          ⎝2⎠
                           y
          −1      ⎛1⎞
        f :x=⎜ ⎟                                               −1   ⎛1⎞
                                                                             y

                  ⎝2⎠                                         f :x=⎜ ⎟
        y = log 1 x                                                 ⎝2⎠
                   2
                                                             answer
                                                                                             (3)
6.3     The inverse is not a function because for example,     answer
              1                                                                              (2)
        g −1 ( ) = 1 or - 1
              2
6.4     x ∈ [0 ; ∞) or x ∈ (−∞ ; 0]                          x ∈ [0 ; ∞ )
                                                             x ∈ (−∞ ; 0]
                                                                                             (2)
6.5.1 0 < x < 1                                                answer

                                                                                             (2)
6.5.2   f ( x) − 1 = g ( x)                                  statement
        f ( x) − g ( x) = 1                                  answer
                                                                                             (2)
        x=0
                                                                                            [13]




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Mathematics Paper 1                           8                       Grade 12 Exemplar 2008
                                      NSC : Memorandum

QUESTION 7

7.1    x =3                                                      answer
       y=–1                                                      answer

                                                                                            (2)
7.2    x-intercept:
                   2
       0 = −1 +
                 x−3                                             substitution
       x−3 = 2
       x=5
       x-intercept (5 ; 0)                                       x-intercept
                    ⎛      5⎞
       y-intercept ⎜ 0 ; − ⎟
                    ⎝      3⎠                                    y-intercept
                                                                                            (3)
7.3                                   y                          shape
                                 6
                                                                 intercepts
                                                                 asymptotes
                                 4



                                 2
                                                                                            (3)
                                                             x
            -6        -4    -2              2      4     6


                                 -2



                                 -4



                                 -6
                                                                                            [8]




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Mathematics Paper 1                                9                      Grade 12 Exemplar 2008
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QUESTION 8

8.1       Period = 360º                                                 answer
                                                                                               (1)
8.2       The shift changes the range of g and will become now          –1
          [-1 ; 3]                                                      3
                                                                                               (2)
8.3        h( x) = cos( x + 30° − 30°)
           h( x) = cos x                                                answer
                                                                                               (1)

                                                                                               [4]

QUESTION 9

9.1                      1        1
                                −                                substitution
        f ′(x) = lim x + h x
                  h→0        h
                         x − ( x + h)
                                                                 common denominator
                           x ( x + h)
                = lim
                   h →0         h                                simplification
                         x − ( x + h)
                = lim                 ÷h
                   h → 0 x ( x + h)
                                                                 simplification
                             −h       1
                = lim               ×
                   h → 0 x ( x + h)   h                          answer
                             −1                                                               (5)
                = lim
                   h → 0 x ( x + h)

                   −1
                = 2
                    x
9.2.1      [
        Dx − 5 x + 2 x
                 2
                           ]
                                                                   answer
        = −10 x + 2
                                                                                              (2)
9.2.2                     1
        y=     x3 +
                         3x3
                                                                  3
               3
               1
        y = x + x −3
                                                                  2
               2                                                 x
               3                                                 1 −3
                     1                                             x
        dy 3 2                                                   3
          = x − x−4
        dx 2
                                                                   answer
                                                                                              (4)
                                                                                             [11]



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Mathematics Paper 1                            10                    Grade 12 Exemplar 2008
                                        NSC : Memorandum

QUESTION 10

10.1      f ′( x) = 3 x 2 − 8 x − 11                        derivative
                                                           derivative = 0
         0 = (3x − 11)( x + 1)
                                                           factors
                11
          x=         or x = −1                             x-values
                 3                                         points
                                ⎛ 11      ⎞
         A (− 1 ; 36 ) and B ⎜ ; − 14,81⎟
                                                                                         (5)
                                ⎝3        ⎠
         Or
         Some candidates may know Horners’ method
            ⎛ 11 ⎞ ⎛ ⎛ ⎛ 11        ⎞ 11 ⎞ ⎞ 11
          f ⎜ ⎟ = ⎜ ⎜ ⎜ − 4 ⎟ × ⎟ − 11⎟ × + 30
                        ⎜ 3             ⎟
            ⎝ 3 ⎠ ⎜⎝⎝ ⎝            ⎠ 3⎠   ⎟ 3
                                          ⎠
         (this makes calculator work much easier)
         i.e. f(x) = ((x – 4)x – 11)x + 30
10.2     (1 ; 36)                                          x=1
         ⎛ 17              ⎞                               y-values remain the same
         ⎜ ; − 14,81⎟                                                                    (2)
         ⎝ 3               ⎠
10.3     Average rate of change
             36 − (− 14,8)                                 formula
          =                                                substitution
                      11
                −1−
                       3
              50,8
          =
             − 4,6 &
          = −10,89                                         answer                        (3)
10.4      f ′(1) = 3x 2 − 8 x − 11
         = 3(1) 2 − 8(1) − 11                              m = – 16
         = −16
         f (1) = 13 − 4(1) 2 − 11(1) + 30                  point (1 ; 16)
         f (1) = 16
                                                           substitution
         y − 16 = −16( x − 1)
         y = −16 x + 32                                    answer                        (4)
10.5     − 16 x + 32 = x − 4 x − 11x + 30
                          3     2                          substitution
                                                           standard form
         0 = x3 − 4 x 2 + 6 x − 2
         0 = ( x − 1)( x 2 − 3x + 2)
         0 = ( x − 1)( x − 1)( x − 2)                      factors
         x = 1 or x = 2
                                                           answer                        (4)
         the tangent cuts the graph again at x = 2
10.6      k > 36 or k < −14,81                               answers
                                                                                         (2)


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10.7     f ′( x) = 3 x 2 − 8 x − 11
         f ′′( x) = 6 x − 8
         0 = 6x − 8
                4
         x=
                3
                                                                        f ′′( x) = 6 x − 8
                  ⎛4⎞                           ⎛4⎞                     f ′′( x) = 0
              f ′′⎜ ⎟ < 0                   f ′′⎜ ⎟ > 0
                  ⎝3⎠                           ⎝3⎠
                                 4
                                 3

                                      ⎛ 4 286 ⎞
         Point of inflection          ⎜ ;     ⎟                          argument
                                      ⎝ 3 27 ⎠

                                      or (1,33 ; 10,59)
                                                                       x-value
                                                                       y-value
                                                                                                         (6)
                                                                                                        [26]

QUESTION 11

11.1                 volume x 3 − 8 x 2 + 5 x + 50                     division by 5 - x
        area =              =                      = − x 2 + 3x + 10     answer
                     height         5− x
                                                                                                         (3)
11.2    f ( x) = x − 8 x + 5 x + 50
                  3       2

                                                                       f ′( x)
        f ′( x) = 3 x 2 − 16 x + 5
                                                                       =0
        0 = 3x 2 − 16 x + 5
        0 = (3x − 1)( x − 5)                                           factors
            1                                                          answers
        x=       or x ≠ 5                                              reject x = 5
            3                                                          dimensions
                  But x = 5 is not valid                                                                 (6)
                          1
                  ∴x =
                          3                                                                              [9]




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Mathematics Paper 1                              12                                   Grade 12 Exemplar 2008
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QUESTION 12

12.1     1     1
           x + y ≤ 480                                                        inequality
         2      5
         ∴ 5 x + 2 y ≤ 4800
         y 3
           ≥
         x 2
                 3x                                                           inequality
         ∴y≥                                                                                                (4)
                  2
12.2               y
and
12.3
         2000
            2000




         1500
            1500



         1000                                                                       constraints correctly
            1000
                                                                          plotted
                                                                                                            (3)
          500500



                                                                                      x
                       100    200   300     400   500   600   700   800      900


                                                                            feasible region
                                                                                                            (1)
12.4     P = 12 000x + 4 000y                                                 equation
                                                                                                            (2)
12.5     x = 600 and y = 900                                                  answer
                                                                                                            (2)
12.6     P = 12 000(600) + 4 000(900)                                       substitution
           = R 10 800 000                                                   answer
                                                                                                          (2)
                                                                                                         [14]

                                                                                                 Total: 150




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