# Mathematics Paper 1 Memo - NATIONAL SENIOR CERTIFICATE GRADE 12

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```					                             NATIONAL
SENIOR CERTIFICATE

MATHEMATICS P1

EXEMPLAR 2008

MEMORANDUM

This memorandum consists of 12 pages.

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]

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=
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
r = 20,00%                                                                     (4)
2.2.2   Fv(sinkingfund) = 1, 2000000(1,15) − 491520
4                 substitution
(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
x = R 27 942,76
(6)

[19]

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)
= 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

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]

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
1− r           3
1−                                                               (2)
4
[9]

Mathematics Paper 1                         7                         Grade 12 Exemplar 2008
NSC : Memorandum

QUESTION 6

b(1) =
2   1
2
1
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
(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]

Mathematics Paper 1                           8                       Grade 12 Exemplar 2008
NSC : Memorandum

QUESTION 7

(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]

Mathematics Paper 1                                9                      Grade 12 Exemplar 2008
NSC : Memorandum

QUESTION 8

(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
]
= −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
(4)
[11]

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 &
10.4      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
the tangent cuts the graph again at x = 2
10.6      k > 36 or k < −14,81                               answers
(2)

Mathematics Paper 1                               11                             Grade 12 Exemplar 2008
NSC : Memorandum

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
x=       or x ≠ 5                                              reject x = 5
3                                                          dimensions
But x = 5 is not valid                                                                 (6)
1
∴x =
3                                                                              [9]

Mathematics Paper 1                              12                                   Grade 12 Exemplar 2008
NSC : Memorandum

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