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GRADE 12 NATIONAL SENIOR CERTIFICATE

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					                          NATIONAL
                     SENIOR CERTIFICATE




                            GRADE 12




                         MATHEMATICS P1

                     ADDITIONAL EXEMPLAR 2008

                           MEMORANDUM




MARKS: 150



                     This memorandum consists of 11 pages.




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Mathematics/P1                                               2             DoE/Additional Exemplar 2008
                                                     NSC - Memorandum

QUESTION 1

1.1.1    1      5
           +        =6
         x x −1
         x − 1 + 5 x = 6 x( x − 1)                                      simplification
                                                                        standard form
         x − 1 + 5x = 6 x 2 − 6 x
         0 = 6 x 2 − 12 x + 1                                           substitution
               − (−12) ±    (− 12)   2
                                         − 4(6)(1)                      simplification
         x=
                            2(6)
             12 ± 120
         x=
                 12
         x = 1,91 or x = 0,09                                            answers                      (6)

1.1.2    x 2 − 3x ≥ 28
                                                                        standard form
         x 2 − 3x − 28 ≥ 0                                              factors
         ( x − 7)( x + 4) ≥ 0                                           method



                    0       0
           +            –       +
                                              OR
                   –4       7




                   –4            7




         x ≤ −4 or x ≥ 7
                                              OR
         x ∈ (−∞;−4] ∪ [7; ∞)                                            answer
                                                                                                       (5)
1.2      2x − y = 3
         y = 2x − 3                                                     simplification
                                                                        substitution
         x 2 + 5 x(2 x − 3) + (2 x − 3) 2 = 15
         x 2 + 10 x 2 − 15 x + 4 x 2 − 12 x + 9 − 15 = 0                simplification
         15 x − 27 x − 6 = 0
               2


         5x 2 − 9 x − 2 = 0                                             standard form
         (5 x + 1)( x − 2) = 0                                          factors
                                                                        answers
                1
         x = − or x = 2
                5                                                       answers
                17
         y=−          or y = 1
                 5
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                                         OR

         2x − y = 3
             y+3
         x=
               2
         ⎛ y + 3⎞      ⎛ y + 3⎞
                     2

         ⎜       ⎟ + 5⎜         ⎟ y + y = 15
                                        2

         ⎝   2 ⎠       ⎝   2 ⎠
          y + 6 y + 9 5 y 2 + 15 y
           2
                       +               + y 2 = 15
               4               2
         y 2 + 6 y + 9 + 10 y 2 + 30 y + 4 y 2 − 60 = 0
         15 y 2 + 36 y − 51 = 0
         5 y 2 + 12 y − 17 = 0
         (5 y + 17)( y − 1) = 0
                17
         y=−          or y = 1
                 5
                1
         x = − or x = 2                                                                          (7)
                5                                                                              [18]



QUESTION 2

2.1      Tn = −5 + (n − 1)(4)                                     Tn = 4n − 9
         439 = −5 + 4(n − 1)                                      substitution of 439
         444 = 4(n − 1)
                                                                  answer
         n − 1 = 111                                                                             (3)
         n = 112

                                         OR

         Tn = 4n − 9
         439 = 4n − 9
         448 = 4n
         112 = n

2.2.1            27 p 2 p                                         ratio
           r=          =                                          simplification
                  81 p   3
                                                                                                 (2)

2.2.2          p                                                  setting up inequality
           −1 <  <1                                                 answer
               3
           −3< p < 3      p≠0                                                                    (3)



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                                          NSC - Memorandum

2.2.3    If p = 2 the sequence is 162 ; 108 ; 72 ; 48 ; ……..
                                           2
                  ∴ a = 162      ;     r=                           value of a and r
                                           3
                a
         S∞ =                                                       formula
               1− r
               162
         S∞ =                                                       substitution
                   2
               1−
                   3                                                answer
         S∞ = 486                                                                                 (3)
                                                                                                 [11]


QUESTION 3
3.1      Tebogo’s sequence will form a geometric sequence with        explanation (Tebogo)
         common ratio 3.
         Thembe’s sequence will form a quadratic sequence with a      explanation (Thembe)
         constant second difference 8.
                                                                                                   (4)
3.2          2                6               18               38
                          4          12                20
         2a = 8
         a=4
         Tn = 4n 2 + bn + c                                         a=4
         2 = 4+b+c
         −2=b+c                                                       setting up equations

         6 = 4(2) 2 + 2b + c                                        value of b
         − 10 = 2b + c
         b = −8
         c=6                                                        value of c
         Tn = 4n 2 − 8n + 6 (Thembe’s sequence)

                                     OR

         2a = 8
         a=4
         T0 = 6 = c
         T1 = 2 = 4 + b + 6
         b = −8
         Tn = 4n 2 − 8n + 6 (Thembe’s sequence)

         Tebogo’s sequence is
         Tn = 2.3n −1                                                 answer                       (7)
3.3      Tn = 4n 2 − 8n + 6                                         substitution
                                                                    answer
         T11 = 4(11) − 8(11) + 6
                      2


         T11 = 402                                                                                 (2)
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3.4
         Sn =
                 (
             a r n − 1)       )
                r −1

         531440 =
                          (
                   2 3n − 1       )                           substitution
                      3 −1
         531440 = 3 − 1
                    n
                                                              simplification
         531441 = 3 n
         312 = 3 n                                            answer
         n = 12                                                                                  (3)
                                                                                                [16]
QUESTION 4
4.1       f ( x) = −( x + 1) 2 + 4
                                                              f(x) = 0
         0 = −( x + 1) 2 + 4
         ( x + 1) 2 = 4                                       factors
         x + 1 = 2 or x + 1 = −2                                answer
                                                                                                  (4)
         x =1        or x = −3

                                      OR
          f ( x) = −( x + 1) 2 + 4
         0 = −x 2 − 2x − 1 + 4
         x 2 + 2x − 3 = 0
         ( x − 1)( x + 3) = 0
         x =1      or x = −3
         A(– 3 ; 0) and B(1 ; 0)
4.2      C(0 ; 3)
         y = a.3 x + 3                                        q=3
                                                              substitution
         4 = a.3 −1 + 3
           a
         1=
           3
         a=3
                                                              answer
         g ( x) = 3.3 x + 3                                                                       (3)
4.3       f ( x) = − x − 2 x + 3
                      2

                                                               f ′( x) = −2 x − 2
          f ′( x) = −2 x − 2
                                                              equating to 1
         1 = −2 x − 2
         3 = −2 x
                  3
         x=−                                                  answer for x
                  2
                                                              answer for y
                  ⎛ 3 15 ⎞
         Point ⎜ − ; ⎟                                                                            (4)
                  ⎝ 2 4⎠
4.4      k>4                                                    answer                           (2)
                                                                                                [13]

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QUESTION 5

5.1      p ( x) = a x
                                                                     substitution
         8 = a −3
              1
         8= 3
             a
               1                                                     simplification
         a3 =
               8
             1
         a=
             2                                                       answer
                                                                                                    (3)
5.2      y = log 1 x                                                 answer
                     2                                                                              (2)
                                          OR
         y = − log 2 x

                                          OR
                          1
         y = log 2
                          x
5.3      0< x<8                                                        answer
                                                                                                    (2)
5.4                           x −3                                     answer
                  ⎛1⎞
         q ( x) = ⎜ ⎟
                  ⎝2⎠
                                          OR
                         − x +3
         q ( x) = 2                                                                                 (2)
                                                                                                    [9]

QUESTION 6

6.1      y ∈ R − {5}                                                   answer
                                          OR                                                        (2)
         y ∈ (−∞;5) ∪ (5; ∞)
6.2       −3                                                         equating
               + 5 = −3 x + 2
         x +1
         − 3 + 5( x + 1) = −3 x( x + 1) + 2( x + 1)
                                                                     simplification
         − 3 + 5 x + 5 = −3 x 2 − 3 x + 2 x + 2
         3x 2 + 6 x = 0
         3x( x + 2) = 0
         x = 0 or x = – 2
         Points of intersection are
         (0 ; 2) and (– 2 ; 8)                                         answers                     (4)
6.3      Reflection about the asymptote x = – 1                        answer
                                      OR
         Reflection about the asymptote y = 5
                                      OR                                                            (2)
         Reflection about the x-axis and translated up by 10 units                                  [8]

Copyright reserved                                                                    Please turn over
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                                                  NSC - Memorandum

QUESTION 7

7.1      1800                                                                 answer                        (1)

7.2      x = – 450                                                              answer
                                                                                                             (2)
7.3      k ( x) = − sin 2 x                                                     answer
                                                                                                             (2)
                                                                                                             [5]

QUESTION 8

8.       let n = 2008
          2008 2 + 2009 × 2007 − 2006 × 2010 − 2016 × 2000                     let n = 2008
          = n 2 + (n + 1)(n − 1) − (n − 2)(n + 2) − (n + 8)(n − 8)            substitution
                                                                              simplification
         = n 2 + n 2 − 1 − n 2 + 4 − n 2 + 64                                 answer
         = 67

                                             OR


         Let n = 2000
         2008 2 + 2009 × 2007 − 2006 × 2010 − 2016 × 2000
         = (n + 8) 2 + (n + 9)(n + 7) − (n + 6)(n + 10) − n(n + 16)
         = n 2 + 16n + 64 + n 2 + 16n + 63 − (n 2 + 16n + 60) − (n 2 + 16n)
         = 64 + 63 − 60
         = 67                                                                                                [4]

QUESTION 9

9.1                  ⎛ 0,096 ⎞
                                        4n                                    4n
         35000 = 5000⎜1 +    ⎟                                                substitution
                     ⎝    4 ⎠
                              4n
             ⎛ 0,096 ⎞
         7 = ⎜1 +    ⎟
             ⎝    4 ⎠
                                   4n
                    ⎛ 0,096 ⎞                                                 log both sides
         log 7 = log⎜1 +      ⎟
                    ⎝      4 ⎠
                       ⎛ 0,096 ⎞                                              use of power law
         log 7 = 4n log⎜1 +     ⎟
                       ⎝     4 ⎠
         4n = 82,0486988...
         n = 20,51 years                                                      answer
                                                                                                             (5)




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9.2.1                 ⎡ ⎛ 0,12 ⎞ −60 ⎤                                  0,12
                    x ⎢1 − ⎜1 +     ⎟ ⎥                                  12
                      ⎢ ⎝       12 ⎠ ⎥
         192000 = ⎣                    ⎦                               substitution
                             0,12                                      60 months
                              12
                  ⎡ ⎛ 0,12 ⎞ −60 ⎤
         1920 = x ⎢1 − ⎜1 +       ⎟ ⎥
                  ⎢ ⎝
                  ⎣           12 ⎠ ⎥ ⎦                                 answer
         x = R 4 270,93                                                                               (4)
9.2.2    Balance Outstanding
                                      ⎡⎛ 0,12 ⎞ 45 ⎤
                               4270,93⎢⎜1 +     ⎟ − 1⎥                 0,12
                                      ⎢⎝     12 ⎠    ⎥
                            45
                  ⎛ 0,12 ⎞            ⎣              ⎦
         = 192000⎜1 +     ⎟ −                                           12
                  ⎝    12 ⎠              0,12                          45
                                          12                           4270,93
         = 300 443,66 – 241 226,7165…                                    substitution
         = R 59 216,95                                                   answer
                                                                                                     (7)
                                                                                                    [16]


QUESTION 10

10.1                    f ( x + h) − f ( x )
          f ′( x) = lim
                      h →0       h
                                                                       substitution
               ( x + h) − x
                        3      3
         = lim
           h→0         h                                               simplification
               ( x + h)( x 2 + 2 xh + h 2 ) − x 3
         = lim
           h→0                   h
               x + 3 x h + 3 xh 2 + h 3 − x 3
                 3       2                                             simplification
         = lim                                                         common factor
           h→0                   h
               h(3 x + 3 xh + h 2 )
                     2
         = lim
           h→0             h
         = lim(3 x + 3 xh + h 2 )
                   2
            h→0                                                        answer
         = 3x 2                                                                                       (5)

10.2.1            2
         y=            −3 x
             5 x
                  1    1
              2 −
          y = x 2 − x3                                                   simplification (one per term)
              5
                    3       2
         dy     1 −2 1 −3
            =− x − x                                                     answer
         dx     5       3
                  1        1
            =−        −
                    3
                5 x     3 x2
                          3
                                                                                                      (4)


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10.2.2       x 4 − 3x 2 + 7
         y=
                    x
                                                                                        simplification
         y = x − 3 x + 7 x −1
              3


         dy                                                                             answer
            = 3x 2 − 3 − 7 x −2
         dx
                          7
            = 3x 2 − 3 − 2                                                                                               (4)
                          x
                                                                                                                        [13]

QUESTION 11

11.1      f ( x) = x 3 + x 2 − 5 x + 3
                                                                                       f ( x) = 0
         0 = ( x − 1)( x 2 + 2 x − 3)
                                                                                      ( x − 1)
         0 = ( x − 1)( x − 1)( x + 3)
                                                                                      ( x 2 + 2 x − 3)
         x = 1 or x = – 3
         x-intercepts are (1 ; 0) and (– 3 ; 0)                                       x-intercepts
         y-intercept is (0 ; 3)                                                       y-intercept
                                                                                                                          (5)
11.2      f ′( x) = 3 x + 2 x − 5
                           2
                                                                                       f ′( x) = 3x + 2 x − 5
                                                                                                   2


         0 = 3x + 2 x − 5
                  2                                                                    f ′( x) = 0
         0 = (3 x + 5)( x − 1)                                                        factors
         Turning Points are
                       ⎛ 5 256 ⎞    ⎛ 5        ⎞
         (1 ; 0) and ⎜ − ;     ⎟ or ⎜ − ; 9,48 ⎟                                      answer
                       ⎝ 3 27 ⎠     ⎝ 3        ⎠                                      answer
                                                                                                                          (5)
11.3     f ′′( x) = 6 x + 2                                                            f ′′( x) = 6 x + 2
         0 = 6x + 2                                                                    f ′′( x) = 0
                 1
         x=−                                                                          answer
                 3
                                                                                                                          (3)
11.4                                                                                  intercepts
                               ⎛ 5 256 ⎞           y
                               ⎜− ;    ⎟
                                       10                                             turning points
                               ⎝ 3 27 ⎠ 9                                             shape
                                              8
                                                                                                                         (3)
                                                                                                                        [16]
                                              7

                                              6

                                              5

                                              4

                                              3

                                              2

                                              1

           -6    -5   -4       -3   -2   -1            1   2     3   4   5   6    7
                                              -1

                                              -2

                                              3


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QUESTION 12

12.1     Height of the cylinder = 2x
         By Pythagoras
                                ( )    2
         (radius of cylinder)2 = 5 3 − x 2                     substitution
                               = 75 − x 2
         V = πr 2 h                                            answer for r
         V = π (75 − x )2 x
                      2


         V = 150πx − 2πx 3                                     substitution
                                                                                              (3)
12.2     dV                                                     dV
             = 150π − 6πx 2                                        = 150π − 6πx 2
         dx                                                     dx
         0 = 150π − 6πx 2                                       dV
                                                                   =0
         x 2 = 25                                               dx
                                                               answer for x
         x = ±5
         x=5 x>0                                               answer
         Height of the cylinder = 2(5) cm                                                     (4)
                                = 10 cm                                                       [7]




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QUESTION 13

13.1   x + y ≤ 16                                                                                                  answer
       120 x + 60 y ≥ 1200                                                                                         answer
       2000 x + 3000 y ≥ 36000                                                                                     answer
                                                                                                                   answer
       x, y ∈ N 0
                                                                                                                                   (4)
             y
13.2    20                                                                                                             graphs of
        19                                                                                                       constraints
        18

        17                                                                                                         feasible region
        16                                                                                                                       (4)
        15

        14

        13

        12

        11

        10

         9

         8

         7

         6

         5

         4

         3

         2

         1
                                                                                                             x
                 1   2   3   4   5   6   7   8   9    10   11   12   13   14   15   16   17   18   19   20




13.3    C = 40000 x + 48000 y                                                                                      answer
                                                                                                                                   (1)
13.4        40000
        m=−
            48000
            5                                                                                                      gradient of
       m=−                                                                                                       search line
            6
       Minimum Cost at (6 ; 8) i.e. 6 Silver Jets and 8 Golden Flyers                                                answer
                                                                                                                                   (3)
13.5    C = 40000(6) + 48000(8)                                                                                    substitution
             = R 624 000                                                                                           answer
                                                                                                                                   (2)
                                                                                                                                  [14]

                                                                                                                 TOTAL:          150




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