MATHEMATICS
PAPER 1
GRADE 12 
AMENDED SENIOR CERTIFICATE EXAMS
PAST PAPERS AND MEMOS
MAY/JUNE 2018

MEMORANDUM 

NOTE:

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

QUESTION 1

1.1.1

(3x −1 )(x + 4)= 0
x = 1  or x = −4
      3

 x = 1
          3 
✓ x=−4 (2)

1.1.2

2x2 +9x−14 = 0
1.1.2 iuhuihad

OR
1.1.2 b iuhiuah
x =1,22 or        x = −5,72

substitution into correct formula
simplification
✓  x =1,22
✓   x = −5,72 (4)

OR

✓ for adding 81 on both sides
                    16
✓  simplification
✓  x =1,22
✓  x = −5,72                     (4)

1.1.3

√3 − 26x = 3x
3 − 26= 9x2
9x2 + 26x −3 = 0
(9x−1)(x+3)= 0
= 1  or x = −3
      9
N/A

3 − 26= 9x2
 standard form
✓  factors
 answer with selection  (4)

1.1.4

(x −1 )(x − 4)> x +11
x2 −5x + 4 > x +11
x2 − 6x − 7 > 0
(x −7)(x +1)> 0
1.1.4 uihuhad
x < −1 or x > 7

✓ x2 −5+ 4
 standard form
 factors
✓  x < −1 or x > 7 (5)

1.2

4√x7 - 5√x7
     √x
= -√x7
    √x
= - x7/2
     x½
= −x3

OR

√x7(4−5)
      √x
= √x6 (−1)3

= −x3

OR

1.2 KGYGAD

 4√x7 - 5√x7
✓ -√x7
✓−x(3)

OR

✓ √x7(4−5)
✓ √x6 (−1)3
✓ −x3(3)

OR

✓ 4x7/2 - 5x7/2
✓ -x7/2
✓ −x(3)

1.3

x−2y−3=0
x = 2y+3 ................(1)
xy = 9  ................  (2)
Substitute (1) into (2)
(2y +3)y =9
2y2 +3y =9
2y2 +3y −9 = 0
(2y −3)(y +3)= 0
y = 3 or y = −3
      2
x = 6 or x = −3

OR

y = x - 3 ................  (1)
        2
xy = 9  ................  (2)
Substitute (1) into (2)
  x[ x−3] = 9
    [  2   ]
x2 −3x =18
x2 −3x −18 = 0
(x − 6)(x + 3) = 0
x=6 or x= −3
y= 3 or y = −3
     2

OR

x−2y−3=0
x = 2y+3  ......................... (1)
y= .........................(2)
     x
Substitute (2) into (1)

x = 2[9] +3
        [x]
x2 − 2(9)−3x = 0
x2 −3x −18 = 0
1.3 iugygad
x = 6  or x = - 3
y= 9 =1,5     or y = 9  = −3
     6                     −3

✓ x=2y+3
 substitution
 standard form
 y-values
✓  x-values                         (5)

OR

 y = x -3 
            2
 substitution
 standard form
✓  x-values
✓  y-values                         (5)

OR

✓ y = 9
          x
 substitution
 standard form
 x-values
✓  y-values                         (5)

1.4

x2 + 2xy + 2y2
=x2 + 2xy + y2 +y2
=  (x+ y)2 + y2
(x+y)2 ≥ 0 and y2 ≥ 0
Therefore (x+y)2 +y2 ≥ 0

✓  x2 + 2xy + y2 +y2
✓ (x+ y)2
✓  (x+y)2 ≥ 0 and y2 ≥ 0
 (x+y)2 +y2 ≥ 0 (4) [27]

QUESTION 2

2.1.1

2.1.1 hgygd
37;  50

✓  37
✓  50                                 (2)

2.1.2

a = second difference = 2  =1
                  2                  2
3a+b = 5
3+b = 5
b = 2
a +b + c = 5
1+ 2+c = 5
c = 2
Tn = an2 +bn+c
= n2 + 2n+ 2

✓  second difference of 2
✓  a =1
✓  b = 2
✓  c = 2 (4)

2.1.3

n2 + 2n + 2 =1765
n2 + 2n −1763 = 0
(n + 43)(n − 41)= 0
n = −43 or n = 41
N/A

OR

n2 +2n+2 =1765
n2 +2n−1763= 0
2.13 iughiuhad
n = −43 or n = 4
N/A

✓  equating Tn to 1765
✓  standard form
✓  factors
✓  answer with rejection (4)

OR

✓  equating Tn to 1765
✓  standard form
✓  subt in correct formula
✓  answer  with rejection (4)

2.2

Sum of multiples of 7 from 35 to 196:
a = 35;   d = 7
Sn = [a ]
        2
= 24 [ 35+196]
    2
= 12[231]
= 2772

Sum of all natural numbers from 35 to 196:
a = 35;  d = 1;  n = 162
Sn = n [a + ]
        2
162 [ 35+196]
    2
=81[231]
= 18 711
Sum of numbers not divisible by 7
= 18 711 - 2772
= 15 939

✓  correct a, d and substitution into correct formula
✓  answer
✓  162
✓  answer
✓  answer                          (5)  [15]

QUESTION 3

3.1

r = 0,94; a =100
T3 = ar2
=100(0,94)2
=88,36 km

✓ r = 0,94
 answer                          (2)

3.2

Sna(rn −1 )
           r −1
750 = 100(0,94n −1 )
                 0,94−1
750( −0,06)  = 0,94n −1
     100
0,94n =1− 9 or [47]n = 11
                20    [50]      20
0,94n    = 0,55
nlog 0,55
      log 0,94
= 9,66

He will pass the halfway point on the 10th day

✓ substitution into correct formula
 0,94n  = 0,55
 use of logarithms
 answer (4)

3.3

S∞ =    a    
         1− r
1500 <  100   
            1− r
1− r <  100   
          1500
r > 14 or 93,33%
     15

 use of S formula
 substitution
✓  answer (3) [9]

QUESTION 4

4.1

0< x ≤1 or (0 ;1]

✓ answer (2)

4.2

p = log4/3 16/9
[4]p = 16
[3]       9

[4]2 = [4]
[3]      [3]
p=2

✓  substitution
[4]2
       [3]
✓  answer (3)

4.3

f : y = log4/3 x
f -1 : x = log4/3 y
y = [4]x  
      [3]

✓ x = log 4/3 y

✓ y = [4]x   (2)
          [3]

 

4.4

y > 0   or     y∈(0;∞)

✓  answer (2)

4.5

[−2 ; 16/9]

✓ -2
 16   (2) [11]
     9

QUESTION 5

5.1

xR; x ≠ −1

✓ xR
  x ≠ −1 (2)

5.2

x-intercept of  f:
0 =  2   + 4
     x+1
    2    =  − 4
 x + 1
2 = -4x - 4
4x = −6
x = −3/2

 equating to 0
 answer  (2)

5.3

y     2    + 4
         x+1
14 =   2    + 4
3      k +1
  2    =  14 −4
k+1       3
    2   =  2  
 k +1     3
2k + 2= 6
k +1= 3
k = 2

✓ substitution
✓ simplification
✓  answer (3)

5.4

C(2; 4)

 2
✓ 4  (2)

5.5

y = a(x + p)2 + q
= a(x −2)2 + 4
Substitute (0 ; 0):
0 = a(0− 2)2 + 4
0 = 4a + 4
a =−1
y =−(x −2)2 + 4

✓ a(x−2)2 +4
 Substitute (0 ; 0)
 a = −1  (3)

5.6

x≤− 3 or −1<x < 0 or x > 4
       2

✓ x ≤ − 3/2
  −1< x < 0
✓  x > 4  (4)

5.7

2 − 5:  f  shifted 1 unit to the right and 9 units down.
x
( x -  3)2 -  5:  g  shifted 1 unit to the right and 9 units down.
Therefore the shift of both graphs took place relative to each other
They only intersect in the third quadrant.
Therefore there is only one point of intersection.

both shifted 1 unit to the right
  both shifted 9 units down
  relative shift
✓  one real root (4) [20]

QUESTION 6

6.1

A = P(1−i)n
0,5P = P( 1 - 0,15)n
(1 - 0,15)n =  0,5
(0,85)n =  0,5
n =   log0,5  or log0,855,5
       log0,85
= 4,27 years

A = 0,5P
 substitution into correct formula
 use of logs
 answer (4)

6.2

In account one month before his 55th birthday:
6.2 kujaygd
=3 478620,49
In account on his 55th birthday:
A = P(1+i)
3 478620,49 [1 + 0,092/12]1
= R3505289,91

OR

6.2 b jhfjuhad
= R3505289,91

✓ value of i
 value of n
 substitution into  correct formula
 adding last month's interest
 answer (5)

OR

 value of i
 value of n
 substitution into  correct formula
 adding last month's interest
 answer (5)

6.3

Invest  Rx  in account A paying 8,4% p.a. compounded quarterly.
A = P(1+i)n
= x [1 + 0,084/4]48
= 2,711662406x
Invest (R150 000 - x) in Account B paying 9,6% compounded monthly.
After 12 years, the amounts are equal:
= x [1 + 0,084/4]48 =  (150000 - x)(1 + 0,096/12)144
2,711662406x = 3,150044027(150000− x)
2,711662406x = 472506,6041−3,150044027x
5,861706433x = 472506,6041
x = R80 609,05
Invest R80 609 in Account A and R150 000 - R80 609,05 = R69 390,95  in Account B

OR

a = amount invested at 8,4% p.a. compounded quarterly
b = amount invested at 9,6% p.a. compounded monthly
a + b = 150 000
a = 150 000 - b
math sihjugad
b = R69 390,95
a = R80 609,05

✓ x [1 + 0,084/4]48
✓ (150000 - x)(1 + 0,096/12)144
equation
 R80609,05
 R69 390,95 (6)

OR

✓[1 + 0,096/12]144
✓ [150 000 - b][1 + 0,084/4]48
 equation
 b
 a      (6)  [15]

QUESTION 7
Penalize 1 mark for incorrect notation in the whole question.

7.1

7.1 a ohjiuhad
OR
7.1 b jhyuugujhgad

✓ 2−3x2 −6xh −3h2
✓ −6xh −3h2
✓ 
subst. into formula
✓ factorisation
answer                                     (5)

OR

✓ subst. into formula
✓ simplification
✓  −6xh −3h2
✓ 
common factor
✓  answer                                    (5)

7.2.1

Dx[4x +5)2]
= Dx(16x2 + 40x + 25 )
= 32x + 40

 16x2 + 40x + 25
 32x
 + 40                                        (3)

7.2.2

7.2.3 iuhiuhad

✓  x¼
✓  1 - 8x-2
✓  1/4x
✓  16x-3   
(4)  [12]

QUESTION 8

8.1

C(0;12)

 C(0;12)                                 (1)

8.2

3

x +13x +12 = 0

3

x    −13x −12 = 0

2

(x +1)(x x −12)= 0

(x +1 )(x − 4)(x +3)= 0

A( − ; )

B(4;0)

  f (x) = 0
 (x +1)
 (x2x−12)
  x = - 3 or 4
  clearly indicating A and B (5)

8.3

fI (x)= −3x2 +13
f//(x)= −6x
−6x = 0
x = 0
For f(x), point of inflection will be at (0 ; 12).
For g(x), point of inflection will be at (0 ; -12).

OR

g(x) = x3 - 13x -12
g/
(x)= 3x2 −13
g//(x)= 6x
6x = 0
x = 0
(0;−12)

OR

 f (x)= −3x2 +13
TP's where
−3x2 +13 = 0
x = 13/3
x = ± √ 13/3
= ±2,08
x-value of point of inflection: −2,08+2,08 = 0
                                                   2
For f(x), point of inflection will be at (0 ; 12).
For g(x), point of inflection will be at (0 ; -12).

✓  fI (x)= −3x2 +13
✓ f//(x)= −6x
  equating to zero
✓ (0;−12)                                     (4)

OF
✓ g/ (x)= 3x2 −13
 g//(x)= 6x
 equating to zero
✓ (0;−12)                                     (4)

OR
✓  f (x)= −3x2 +13
✓  −3x2 +13= 0
 x-values of TPs
✓ (0;−12)                                     (4)

8.4

FIx)= −3x2 +13
−3x2 +13 = −14
−3x2 = −27
x2 = 9
x=3 or x= −3

 equating derivative to - 14
 simplification
✓  answers      (4)  [14]

QUESTION 9

9.1.1

AC =  t - 30

✓ answer                                   (1)

9.1.2

302 =(t−30)2 + p2                  [Pythagoras]
p2 = 900 −(t −30 )2
p2 = 900−(t2 −60t +900)
p2 = 900−t2+ 60t −900
p2 = 60tt2

✓  p2 = 900 −(t −30 )2
✓ 
(t2 −60t +900)
  p2 = 60tt2 (3)

9.2

V (t)= 1/3 π r2t
= 1/3 π(60t t2 )t
= 20π t2 −  1/3πt 3

 substitution   (1)

9.3

V (t)= 20π t2 −  1/3πt3
V (t)= 40π t −π t2
40πt −π t2 = 0
t(40π −tπ)= 0
t = 0    OR   t = 40 cm
N/A

 40π t
✓ 
 −π t2
 answer with selection        (3)

9.4

Volume of cone
= 20(π)(40)2 −  1/3 π(40 )3
=10 666,67π     or   33510,33211
Volume of sphere
= 4/3 πr3
= 4/π(30 )3
= 36000π    or 113097,3355
10666,67π
     36000π
= 0,296296
≈ 29,63%

 volume of cone
 volume of sphere
✓  10666,67π
       36000π
 % cut out      (4)  [12]

QUESTION 10

10.1

10!
=3 628 800

✓ 10!
✓  answer  (2)

10.2

4!× 7!
= 120 960

OR

4!×6!×7
= 120 960

 4!
✓  7!
✓ 4!× 7! or 120 960 (3)

OR

 4!
 6!× 7
✓ 4!×6!×7 or 120 960 (3)

10.3

 6! 
10!
  1     or 0,000198
   5040

 6!
✓ 6! or     1     or  0,000198 (2) [7]
   10!       5040

 

QUESTION 11

11.1

P(tennis)× P(≤35years)= P(tennis and ≤ 35years )
  21 × 80  = α
 140  140   140
a =12

 statement
 substitution
 answer  (3)

11.2

P(gym or ≤35 years )
= P(gym)+ P(≤ 35years)− P(gym and ≤ 35years)
=  7080  − 40
   140  140 140
= 110
   140
 11  or   0,79
   14

✓  statement
70/140
80/140
40/140
110 or 11  or  0,79   (5) [8]
   140    14

 

TOTAL:   150

Last modified on Monday, 06 September 2021 07:27