MATHEMATICS PAPER 1
GRADE 12
NOVEMBER 2017
MEMORANDUM
NATIONAL SENIOR CERTIFICATE

NOTE:

  • If a candidate answers a question TWICE, only mark the FIRST attempt.
  • Consistent accuracy applies in ALL aspects of the marking guidelines. Stop marking at the second calculation error.

QUESTION 1

1.1.1

x2 + 9x + 14 = 0
(x + 7)(x + 2) = 0
x = -7 or x = -2

 factors
✓ x = -7
✓ x = -2

(3)

1.1.2

4x2 + 9x - 3 = 0
1.12a

OR

1.12b

✓ substitution

 

✓ simplification

✓ x = 0,29

✓ x = - 2,54

 

OR

 

✓ for adding 81 on both sides
                    64

✓ simplification

✓ x = 0,29

✓ x = - 2,54

(4)

1.1.3

√x 2 - 5 = 2√x
x 2 - 5 = 4x
x 2 - 4x - 5 = 0
(x - 5)(x + 1) = 0
x = 5 or x = -1
x = 5

✓ x2 - 5 = 4x
✓ standard form
✓ both answers
✓ select x = 5
(4)

 

1.2

3x - y = 4
y = 3x - 4
x 2 + 2xy - y 2 = -2
x 2 + 2x(3x - 4) - (3x - 4)2 = -2
x 2 + 6x 2 - 8x - (9x 2 - 24x + 16)= -2
7x 2 - 8x - 9x 2 + 24x - 16 = -2
- 2x 2 + 16x - 14 = 0
x 2 - 8x + 7 = 0
(x - 7)(x - 1) = 0
x = 1   or   x = 7
y = 3(1) - 4      y = 3(7) - 4
y = -1   or    y = 17

OR

3x - y = 4
x y + 4
         3
x2 + 2xy - y2 = -2
x 2 + 2xy - y 2 = -2
1.2
y 2 + 8 y + 16 + 6 y 2 + 24 y - 9 y 2 = -18
- 2 y 2 + 32 y + 34 = 0
y 2 - 16 y - 17 = 0
(y - 17)(y + 1) = 0
y = -1        or        y = 17
x - 1 + 4               x 17 + 4
           3                             3
x = 1              or           x = 7

✓ y subject of formula

✓ substitution

✓ correct standard form

✓ factors

✓ x-values

✓ y-values

 

OR

✓ x subject of formula

✓ substitution

✓ correct standard form

✓ factors

✓ y-values

✓ x-values

(6)

1.3.1

x 2 + 8x + 16 > 0
(x + 4)(x + 4) > 0
x ∈ R, x ≠ - 4     or
x  (-∞; - 4) or x  (- 4;∞) or
x < - 4 or x > - 4

OR
1.31

✓ (x + 4)(x + 4)
✓✓ any one of the solutions

 

OR

✓ (x + 4)(x + 4)
✓✓ any one of the solutions

(3)

 

1.3.2

1.32

For two negative unequal roots:

0 < p < 16

OR

x2 + 8x + 16 = p
x
2 + 8x + 16 - p = 0
0 < 16 - p < 16
- 16 < - p < 0
0 < p < 16

OR
x 2 + 8x +16 - p = 0

x = - 8 ± √64 - 4(16 - p)
                    2
0 < 64 - 4(16 - p) < 64
0 < 4 p < 64
0 < p < 16

 

OR
x 2 + 8x +16 = p
x
2 + 8x +16 - p = 0
Roots are real and unequal :
82 - 4(16 - p) > 0
4 p > 0
p > 0
Roots are : - 8 ± √ 4 p
                         2

For both roots to be negative :
√4p < 8
4 p < 64
p < 16
0 < p < 16

✓0

✓ 16

 

✓✓ 0 < p < 16   (4)

 

OR

 

✓ 0

✓ 16

✓✓ 0 < p < 16               (4)

 

 

✓ 0

✓ 16

✓✓ 0 < p < 16               (4)

 

 

✓ 0

✓ 16

✓✓ 0 < p < 16              (4)

[24]


QUESTION 2

2.1.1

2.11

first differences: –9; –15; –21
second difference = – 6

✓ first differences

✓ – 6

(2)

2.1.2

Tn = an2+ bn + c 

a second difference = -3
                   2

3a + b = -9
3(- 3)+ b = -9
b = 0
a + b + c = 5
- 3 + 0 + c = 5
c = 8
Tn = -3n 2 + 8

OR

(n - 1)(n - 2)d

Tn = T1 + (n - 1)d1 + (n - 1)(n - 2)d2
                                            2
= 5 + (n - 1)(- 9) + (n - 1)(n - 2)(- 6)
                                          2
= 5 - 9n + 9 - 3n 2 + 9n - 6
T = -3n 2 + 8

✓ a = -3

✓ b = 0

✓ c = 8

✓ Tn = -3n2 + 8

 

OR

 

✓ a = -3

✓ b = 0

✓ c = 8

✓ Tn = -3n2 + 8

(4)

2.1.3

- 3n2 + 8 = -25 939
- 3n2 = -25947
n2 = 8649
n = -93 or n = 93
The 93rd term has a value of –25 939

✓ Tn= -25 939

✓ n2 = 8649

✓ answer

(3)

 

2.2.1

2k - 7 ; k + 8 and 2k -1
k + 8 - (2k - 7) = 2k -1- (k + 8)
- k +15 = k - 9
2k = 24
k = 12
2k - 7; k + 8 and 2k - 1
17 ; 20 ; 23.......
d = 3
T15 = 17 + 14(3)
= 59

✓ k + 8 - (2k - 7) = 2k -1- (k + 8)

✓ k = 12

✓ 17

✓ d = 3

✓ T15 = 59

(5)

2.2.2

Sequence is 17 ; 20 ; 23 ; 26 ; 29 ; 32 …….
Every alternate term of the sequence will be even
20 + 26 + 32 + .......
S30 = 30  [2(20) + (29)(6)]
           2
= 15[40 + 174]
= 3210

OR
T 30 = 20 + 29(6)
= 94
S30 = 30  (20 + 194)
           2
= 3210

✓ 20 + 26 + 32 + .......

✓ a = 20 d = 6

✓ subst into correct formula

✓ answer (4)

✓ a = 20 d = 6

✓ T30 = 94

✓ S30 = 30  (20 + 194)
           2
✓ answer (4)

[18]


QUESTION 3

3.1

a + ar = 2
a(1 + r ) = 2
a    2   
      1 + r

OR

  a    - 2 = 1 
1- r           4
4a - 8(1- r ) = 1- r
4a - 8 + 8r = 1- r
4a = 9 - 9r
9 - 9r
         4

OR
Sna(r n-1)
           r -1

2 = a(r 2 -1)
          r -1

2 = a(r -1)(r +1)
           r -1 2
= a(r +1)
a =  2   
      r +1

OR

  ar   2 = 1 
1 - r        4
a = 1 - r
       4r 2

✓ a + ar = 2

✓ a =   2   
         1+ r
(2)

✓    a   - 2 = 
    1- r          4

✓ a 9 - 9r
             4
(2)

 

OR

✓ 2 = a(r 2 -1)
             r -1

✓ a   2   
          1+ r
(2)

OR

ar  2  = 
   1- r        4

✓ a 1- r
          4r 2
(2)

3.2

3.2

OR
3.2b

OR

3.2c

OR

3.2d

✓ S= 2 +
                  4
✓   a    = 9 
    1- r     4

✓ substitution of a into the correct formula

✓ 9r 2 = 1
✓ r =
         3
✓ a =
         2

(6)

OR 

✓ S = 2 +
            4
   a    = 9 
    1- r     4
✓ r = 9 - 4a
        9

✓ substitution of a into the correct formula
✓ a =
          2
✓ r =
         3
(6)

OR

✓ r 2 - a
           a
   ar  2    = 1
    1- r          4
✓ substitution of a

✓ (2a - 3)(a - 3) = 0

✓ a = 3
          2
✓ r =
         3
(6)

OR

✓ S= 2 +
              4
  a    9
   1- r     4
✓ substitution of a
✓ 9r 2 = 1
✓ r =
         3
✓ a =
          2
(6)
[8]

 


QUESTION 4

4.1

f (x) = -ax2 + bx + 6
f / (x) = -2ax + b
- 2ax + b = 3
at x = -1
2a + b = 3               [1]

f (- 1) =
            2
- a - b + 6 =
                   2
- 2a - 2b + 12 = 7
2a + 2b = 5                [2]
[2]- [1]
b = 2
2a + 2 = 3
a = ½

 

OR
f /(x) = -2ax + b
3 = 2a + b
b
= 3 - 2a
 7 = -a(-1) + (3 - 2a)(-1)+ 6
 2
a + 3 =
            2
a = ½
b = 2

✓ – 2ax + b

✓✓ 2a + b = 3

✓ - a - b + 6 =
                   2

✓ solve simultaneously 

(5)

 

✓ – 2ax + b
✓✓ 2a + b = 3
  7 = -a(-1)2  + (3 - 2a)(-1)+ 6
     2
✓ solve simultaneously

(5)

4.2

f (x) = - ½ x 2 + 2x + 6
x - intercepts :
- ½ x 2 + 2x + 6 = 0
-x 2 + 4x +12 = 0
x 2 - 4x -12 = 0
(x - 6)(x + 2) = 0
(- 2 ; 0)      (6 ; 0)

✓ - ½ x2 + 2x + 6 = 0
✓ (- 2 ; 0)
✓ (6 ; 0)
(3)

 

4.3

4.3

OR
4.3b

 

 

4.3c 

OR

4.3d

 

 

4.4

4.6

4.4      

4.4:
f:
✓ shape
✓ x- intercepts
✓ y- intercept
✓ (2 ; 8)
(4)

4.6:
g:
✓ x- intercept
✓ y- intercept
(2)

4.5 0<x<4 or (0;4) ✓ 4
✓✓ 0<x<4 (3)

4.7

x ≤ -2 or - 1  x  6

OR/OF

(-∞ ; - 2] or [-1; 6]

✓ x ≤ -2

✓✓ - 1  x ≤ 6

(3)

[23]

 

5.1

y ∈ R ; y ≠ -1
OR
y < -1 or y > -1
OR
y ∈ (- ∞; - 1) or y ∈ (-1; ∞)
OR
R -{-1}

✓✓ answer (2)

5.2

D(2 ; - 1)
g (x) =  2   - 1
          x - 2

✓ D(2 ; - 1)
    2     - 1
    x - 2
(2)

5.3  

f(x) = log3x
log3t=1
 t=3

OR
g (x) =  2   - 1
          x - 2
1 =    2   - 1
       t - 2
2 =    2  
       t - 2
2t - 4 = 2
t = 3

✓ correct substitution of A
✓✓ t = 3
(3)

5.4

x = log3y
y = 3x

✓ interchange x and y
✓ y = 3x (2)

5.5

3x< 31
x < 1

OR
3x< 31
x ∈ (- ∞ ; 1)

✓ 3x< 31
✓ x < 1
✓ 3x< 31
✓ x ∈ (- ∞ ; 1)
(2)

5.6

Equation of the axis of symmetry: y = -x + 1
x-intercept of the axis of symmetry is at x = 1
f has an x-intercept at B(1 ; 0) which is the same as
the x-intercept of the axis of symmetry Point of intersection: B (1 ; 0)

OR
Since BE = ED = 1 and D lies on the axis of symmetry and the gradient of
the axis of symmetry is –1, B will also lie on the axis of symmetry. But B also
lies on f. Therefore B(1 ; 0) is the point of intersection between f and the axis
of symmetry with a negative gradient.

✓✓equation of axis of symmetry

✓ B or (1 ; 0)

OR

✓✓ BE = ED = 1
✓ B or (1 ; 0)


QUESTION 6

6.1

A = P(1 + i)n
6.1

✓  
   12
✓ n = 36

✓ correct substitution into formula

✓ 1 + = 36√36 1,214672
        12
✓ 6,5%
(5)

6.2.1

6.21 ✓ i = 0,11
          12
✓ n = 54
✓ correct substitution in P
✓ answer
(4)

6.2.2

Amount paid for the year : (5 536,95´12) = R 66 443,40
= 192 296,17
6.22

Interest = (5 536,95´12)- (235 000 -192 296,17)
= 66 443,40 - 42 703,83
= 23 739,57

OR
6.22b

OR
6.22c

✓ R 66 443,40
6.22d

OR

✓ R 66 443,40
✓ n = – 42
✓ substitution into correct formula
✓ R192 296,20
✓ R42 703,80
✓ R23 739,60
(6)

OR
✓ R62 648,18
✓ R172 351,82
✓ R192 296,17
✓ R66 443,40

✓ 235 000 – 192 296,17
✓ R23 739,57
(6)

[15]

 


QUESTION 7
 

7.1  

f (x + h) = 2(x + h)2 - (x + h)
= 2(x 2 + 2xh + h2 ) - x - h
= 2 x2 + 4 xh + 2h2 - x - h
f (x + h) - f (x) = 2x 2 + 4xh + 2h2 - x - h - 2x 2 + x
= 4 xh + 2h2 - h
f'(x) = lim f (x + h) - f (x)
h→0               h
= lim    4xh + 2h2 - h
h→0             h
= lim   h(4x + 2h -1)
 h→0         h
= lim (4x + 2h - 1)
 h→0
= 4x -1

OR

f / (x) = lim   f (x + h) - f (x)
           h→0         h
=  lim   2(x + h)2 - (x + h) - (2x2 - x
   h→0                h                                    

=  lim   2 x2 + 4xh + 2h2 - x - h - 2x2 + x
    h→0                     h
= lim     4xh + 2h2 - h
 h→0            h
= lim h(4x + 2h -1)
 h→0         h
= lim (4x + 2h - 1)
  h→0
= 4x -1

✓ 2x2 + 4xh + 2h 2 - x - h
✓ 4xh + 2h 2 - h
✓ f'(x) = lim f (x + h) - f (x)
   h→0               h
✓ subst. into formula
✓ lim (4x + 2h - 1)
   h→0
✓ 4x – 1

 

OR
✓ f'(x) = lim f (x + h) - f (x)
   h→0               h
✓ subst. into formula
✓ 2 x 2 + 4xh + 2h 2 - x - h
✓ 4xh + 2h 2 - h
✓ lim(4x + 2h -1)
   h→0
✓ 4x – 1
(6)

7.2.1

Dx[(x + 1)(3x - 7)]
= Dx(3x2 - 4x - 7)
= 6x - 4

✓ 3x2 - 4x - 7
✓ 6x - 4
(2)

7.2.2

7.22

7.22b

[12]


QUESTION 8

8.1

8.1
Point of inflection at x = 2

✓ x3 - 6x2 + 9x

✓ 3x 2 -12x + 9

✓ 6x -12

✓ 6x -12 = 0

✓ explanation

(5)

8.2

8.2

✓ shape

✓ (0 ; 0)

✓ (3 ; 0) as TP

✓ (1 ; 4)

(4)

8.3

f concave up for x > 2
y = - f (x) will be concave down for x > 2

✓✓ x > 2

(2)

8.4.1

(3;7)

✓ 3

✓ 7                             (2)

8.4.2

Do not agree with Claire as her statement is incorrect. Between x = 1
and x = 3 the graph of f is decreasing. Therefore at x = 2 the gradient
will have a negative value.

OR
f / (2) = 3(2)2 - 12(2) + 9
= -3
≠1

✓ no

✓ justification

(2)

[15]


QUESTION 9

y = x2 + 2
P(x ; x2 + 2 )
B(0 ; 3)
PB2 = (x - 0)2  + (x2 + 2 - 3)2
= x2 + x4 - 2x2 + 1
= x4 - x2 + 1
PB will be a minimum if PB2 is a minimum
d (PB2 ) = 4x32x 
    dx
4x3 - 2x = 0
x(2x2 -1)= 0
x = 0 or x2 = ½
x =
     √2
9
OR
Gradient of tangent to curve = 2x
Gradient of line joining B and the curve = x2+ 2 - 3
                                                                      x - 0
x2  -1
      2
Shortest distance will be where tangent to curve is perpendicular
to the line joining P and the curve.
x2 -1 = - 1 
  x         2x
2x(x2 -1)= -x
2x3 - 2x = 0
x(2x2 -1)= 0
x = 0   or  x2 = ½
x =
     √2
9b
OR
9c

 

✓ (x - 0)2  + (x2 + 2 - 3)2
✓ x 4 - x2 + 1
✓ 4x3 - 2x
d (PB2 ) = 0
        dx
✓ x =
        √2 
9d

answer

OR
✓ = 2x
✓ = x2 -1 
          x 
x2 -1 = - 1 
      x         2x
✓ 2x3 - 2x = 0
✓ x =
        √2
9e
OR
9f

[7]


QUESTION 10

10.1

10.1

8 values need to be placed in correct position:

2 or 3 correct: 1 mark
4 or 5 correct: 2 marks
6 or 7 correct: 3 marks
8 correct: 4 marks
(4)

10.2

(49 - x) + x + 8 + 4 + 5 + 2 + (60 - x) + 14 = 100
- x + 142 = 100
x = 42

✓ setting up equation
✓ answer

10.3

P (use only one application) = 7 + 2 + 18
                                                     100
=  2   or 27%
   100

7 + 2 + 18
         100
✓ answer (2)
[8]


QUESTION 11

11.1

5 x 5 x 10 x 9

= 2250

✓ 5 x 5
✓ 10 x 9
✓ 2250 (3)

 

11.2

 

No of digits used

Letters  Digits  Total 
 1  5 x 5  10  250
 2  5 x 5  10 x 9  2 250
 3  5 x 5  10 x 9 x 8  18 000
 4  5 x 5  10 x 9 x 8 x7  126 000
 5  5 x 5  10 x 9 x 8 x7 x 6  756 000

Codes of two letters and five digits will ensure unique numbers for 700 000 clients.

✓ 5 x 5 x 10 x 9 x 8 x 7 x 6
✓✓ five digits(3)
[6]

TOTAL : 150

Last modified on Friday, 30 July 2021 08:57