MATHEMATICS PAPER 1
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
MAY/JUNE 2019

NOTE:

  • If a candidate answers a question TWICE, only mark the FIRST
  • Consistent Accuracy applies in all aspects of the marking

QUESTION 1

1.1.1

x 2 - 5x - 6 = 0
(x - 6)(x +1) = 0
x = 6 or x = –1

OR
x 2 - 5x - 6 = 0
x = 5 ± √(-5)2 - 4(1)(-6)
                2(1)
x 5 ± √49
           2
x = 6 or x = –1

?factors
?both answers
(2)
OR
?correct subst into correct formula
?both answers
(2)

1.1.2

(3x -1)(x - 4) = 16
3x 2 -13x -12 = 0
x = 13 ± √(-13)2 - 4(3)(-12)
                  2(3)
x 13 ± √313
           6
x = 5,12 or x = -0,78 

OR
x = 3x 2 -13x -12 = 0
 - 13 x = 4
    3
x13x + (13)2 = 4 + (13)2
         3        6                 6
(x - 13 )2313
       6          36 
x 13 ± √313
            6
x = 5,12 or x = -0,78

?standard form
?correct subst into correct formula
??answers
(4)

OR
?standard form
?adding (13)both sides 
                6
??answers
(4)

1.1.3

x(4 - x) ≥ 0
x(x - 4) ≥ 0    or     - x(x - 4) ≥ 0
x(x - 4) ≤ 0
1

0 ≤ x ≤ 4      or   x ∈  [0 ; 4]

factorisation 
0 ≤ x ≤ 4
(3)

1.1.4

52x - 1 =4
5x + 1 
(5x + 1)(5x - 1) = 4
       5x + 1
5x - 1 = 4
5x = 5
x = 1 

OR
52x - 1 = 4
5x + 1 
52x - 1 = 4.5x + 4
52x - 4.5x - 5 = 0
(5x - 5)(5x + 1) = 0
5x = 5 or   5x ≠ -1
x = 1

?factors in numerator
?5x - 1 = 4
?answer
OR

?standard form
?factors
?answer
(3)

1.2 x = 2 - 3y................................................ (1)
x 2 + 4xy - 5 = 0..................................... (2)
Substitute (1) in (2):
(2 - 3y)2 + 4 y(2 - 3y) - 5 = 0
4 -12 y + 9 y 2 + 8 y -12 y 2 - 5 = 0
- 3y 2 - 4 y -1 = 0
3y 2 + 4 y +1 = 0
(3y + 1)( y + 1) = 0
y = - 1 or   y = -1
        3
x = 3 or x = 5
OR

y = - x................................................. (1)
     3     3
x 2 + 4xy - 5 = 0..................................... (2)
Substitute (1) in (2):
x 2 + 4x ( 2  - ) - 5 = 0
               3      3 
3x 2 + 8x - 4x 2 - 15 = 0
- x 2 + 8x -15 = 0
x 2 - 8x + 15 = 0
(x - 5)(x - 3) = 0
x = 3 or x = 5
y = - 1 or   y = -1
        3

? y = 2 -
         3    3

?correct subst into correct formula
?either standard form 

?x – values
?y – values                         (5)

OR

? y = 2 -
         3    3

?correct subst into correct formula
?either standard form 

?x – values
?y – values                         (5)

1.3

ab = 2√10
bc = 3√2
ac = 6√5
ab.bc.ac = 2√10.6√5.3 √2
(abc)2 = 36√100
abc =360 = 6√10

OR
ac = 6√5   a = 6√5
                         c
bc = 3 2   b = 3√2 
                        c
ab = 2√10
(6√5)(3√2) = 2√10
   c      c
18√10 = 2√10.c2
c2 = 9
c = 3
Volume = abc = 2√10.3 = √360 = 6√10

? volume = abc 
?? ab.bc.ac = 2√10.6√5.3√2
? (abc)2 = 36√100
? answer
(5)

OR
? a = 6√5
           c
? b = 3√2
            c

? value of c
? Volume = abc
? answer (5)
[22]

 

QUESTION 2

2.1.1

59

? answer (1)

2.1.2

2

2a = -2
=-1
3(- 1) + b = 14
b = 17
(- 1) + (17) + c = 15
c = -1
Tn = -n2 + 17n - 1

?second difference of – 2
?a
?b
? c
(4)

2.1.3

T27 = -(27)2  + 17(27) - 1
= -271

? substitution
? answer
(2)

2.2.1

r - 18- 1
       36      2

answer (1)

2.2.2

Tn = 36 (-½)n-1
   9    = 36(-½)n-1 
4096
    1     = (-½)n-1
16384 
(-½)14 = (-½)n-1  

n = 15 

OR
36 ; - 18 ; 9 ; - 9 ;; - 9 ; ... ;    9   
                      2    4     8         4096
If you look only at the denominator: 2 ; 4 ; 8 ; … ; 4096

2k = 4096
2k = 212 
k = 12
n = 15 terms

? Tn = 36 (-½)n-1
?    1     = (-½)n-1
   16384 
? answer
(3)

OR

? 2k = 4096

? k = 12

? answer

(3)

2.2.3

=   a   
      1 - r
=
 36   
1 - (-½)
= 24

?correct subst into correct formula with – 1 < r < 1 
?answer if – 1 < r < 1
(2)

2.2.4 S250even   -18(( ¼) 250 -1)
                         ¼ - 1
= -24
S250sold   -36(( ¼) 250 -1)
                         ¼ - 1
= 48
Sodd =   48 
Seven     -24
= -2

T1 + T3 + T5 + T7 + ... + T499
T2 + T4 + T6 + T8 + ... + T500
a + ar 2 + ar 4 + ... + ar 498
    ar + ar 3 + ar 5 + ... + ar 499
=   a + ar 2 + ar 4 + ... + ar 498
     r(a + ar 2 + ar 4 + ... + ar 498 )
= 1 
   r
= -2

?r = ¼ and n = 250
? S 250 even = -24
?S 250 odd = 48
answer
(4)

?a + ar 2 + ar 4 + ... + ar 498
?ar + ar3 + ar5 + ... + ar 499
?r(a + ar 2 + ar 4 + ... + ar 498)
?answer  (4)
[17]


QUESTION 3

3.1.1

p + 6 – (2p + 3) = p – 2 – (p + 6)
p + 3 = – 8
p = 11

?equating i.t.o p
?simplifying
(2)

3.1.2

Tn = 25 + (n -1)(-8) = 33 - 8n
33 - 8n < -55
- 8n < -88
n > 11
Term 12 will be the first term smaller than  –55

?subst into Tn formula
?n > 11
?n = 12
(3)

3.2

S = n [a + l] = 6 [(x - 3) + (x -18)]6 
      2              2 
= 6x - 63
S = [a + l] = 9 [(x - 3) + (x - 27)]9 
      2              2 
9x -135
6x - 63 = 9x -135
3x = 72
x = 24
S15 = n [a + l] = 15 [(x - 3) + (x - 45)]
         2                2 
= 15 [2x - 48]
    2
= 15 [2(24) - 48]= 0 = RHS
    2

OR
3

= 21+18 +15 +.....+ -21.

Sn = n [a + l)]
        2
= 15 [21- 21]
    2
= 0 = RHS

OR
(x - 3) + (x - 6) + (x - 9) + (x -12) + (x -15) + (x -18)
= (x - 3) + (x - 6) + (x - 9) + (x -12) + (x -15) + (x -18)
+ (x - 21) + (x - 24) + (x - 27)
3x - 72 = 0
3x = 72
x = 24
4

= 21 + 18 + 15 +.....+ -21.
Sn = n [a + l]
        2
= 15 [21 - 21]
    2
= 0 = RHS

?6x - 63
?9x -135
?24
?15 [(x - 3) + (x - 45)]
    2
?substitution of x
(5)

OR
?expansion
?3x - 72 = 0
?24
?substitution of x
?sum of 15 terms
(5)

OR

?expansion
?3x - 72 = 0
?24
?substitution of x
?sum of 15 terms  (5)

[10]

 

 

QUESTION 4 

4.1

y > 0
OR
y ∈ (0 ; ∞)

?answer (1)
OR
?answer (1)

4.2

g;y = (½)x
g-1; x = (½)y 

y = log½x   or   y = - log2or y = log21/x

?x = (½)y 
?equation   (2)

4.3

Yes. The vertical line test cuts g–1 once
OR
Yes. For every x-value there is a unique y-value
OR
Yes. g is a one-to-one function
OR
Yes. The horizontal line cuts g only once

?yes
?valid reason  (2)
OR
?yes
?valid reason  (2)
OR
?yes
?valid reason  (2)
OR
?yes
?valid reason  (2)

4.4.1

y = -log2 x
2 = -log2 a
a = 2-2 = ¼ or a = (½)2 = ¼

?correct subst into correct formula (a ; 2)
?answer   (2)

4.4.2

M/(2;¼) or M(2;a)

?answer (1)

4.5

M//(-1; 9/4

?-1
9/??
(3)
[11]

 

QUESTION 5

5.1.1

x = -2
y = 3

?answer
?answer

(2)

5.1.2

(0;7/2)

?answer
(1)

5.1.3

  1    + 3 = 0
x + 2
1 + 3(x + 2) = 0
3x = -7
x = - 7
       3
x-intercept (-7/3 ; 0)

?y = 0
?answer
(2)

5.1.4

 5

?asymptotes at y = 3 and x = – 2
?intercepts at y = 3,5 and x = – 2,3
?shape (reasonable representation in correct quadrants)
(3)

5.2.1

- 2x + 4 = 0
2x = 4
x = 2
S(2 ; 0)

?y = 0

?x = 2
(2)

5.2.2

Equation of k:
y = a(x +1)2  +18
0 = a(2 +1)2  +18 or 0 = a(- 4 + 1)2 + 18
9a = -18
a = -2
y = -2(x +1)2  +18

?y = a(x +1)2  +18
?substitute (2 ; 0) or (– 4 ; 0)
?a
(3)

5.2.3 - 2x 2 - 4x +16 = -2x + 4
- 2x 2 - 2x +12 = 0
x 2 + x - 6 = 0
(x + 3)(x - 2) = 0
x = -3 or x = 2
y = -2(-3) + 4 = 10
T(–3 ; 10)

?equating
?standard form
?factors
?choosing x = –3
?answer
(5)

5.2.4 x < -3 or x > 2
OR
(-∞ ; - 3) ∪ (2 ; ∞)
?answer
OR
?answer
5.2.5(a) x < -1

OR
(-∞ ; - 1)

?answer
OR
?answer
5.2.5(b)  6 ? shape of cubic with local min tp moving to local max tp
? turning points at x = 2 and x = –4
? point of inflection at
x = –1
(3)
[25]

 

QUESTION 6 

6.1.1

A = P(1 - i)n
79866,96 = 180 000(1 - 0,15)n
(1 - 0,15)n = 79866,96
                    180 000
7

n = 4,999… years
n » 5 years

?substitution
?use of logs
?answer
(3)

6.1.2

A = P(1 + i)n
= 49 000(1 + 0.1 )20
                      4
= R80 292,21
The money will be enough to buy the car.

?values of and n
?substitution
?conclusion
(consistent with answer)
(3)

6.2.1

8
P
= R793 749,25

OR
Balance Outstanding 
9

= 841 885,56 - 48 136,62
= R793 748,94

?n = 234 ? i 0,1025
                        12
?substitution in present value formula
?answer
OR
?n = 6 in both
? i 0,1025
            12
? A – F
R793 748,94

6.2.2

A = P(1 + i)n
= 793749,25(1 + 0.1025)3
                              12
New instalment
10

 

?= 793749,25(1 + 0.1025)3
                              12
?n
= 231
?substitution of new P
?substitution of n and i into formula
?answer   (5)

[15]


QUESTION 7
 

7.1

f (x) = x2 + 2
f (x + h) = (x + h)2  + 2
= x2 + 2xh + h2 + 2
f (x + h) - f (x) = x2 + 2xh + h2 + 2 - (x2 + 2)
= 2xh + h2
f /( x) = lim f (x + h) - f (x)
h→0                h
=lim 2xh + h2
h→0      h

= lim h(2x + h)
 h→0      h

= lim (2x + h)
  h→0
= 2x

OR

f /( x) = lim f (x + h) - f (x)
h→0                h
= lim x + 2xh + h2 + 2 - (x2 + 2)
h→0            h
=lim 2xh + h2
h→0      h

= lim h(2x + h)
 h→0      h

= lim (2x + h)
  h→0
= 2x

?x 2 + 2xh + h2 + 2
2xh + h2

?=lim 2xh + h2
h→0      h

?= lim (2x + h)
  h→0
?answer
(4)

OR
?x2 + 2xh + h2 + 2
?=lim 2xh + h2
  h→0      h
? = lim h(2x + h)
   h→0      h
?answer
(4)

7.2.1 y = 4x 3 + 2x -1
dy = 12x 2 - 2x -2
dx
?+ 2x-1
?12x 2
?- 2x -2
(3)
7.2.2 z  
7.3 Point of contact:  (1 ; 5)
m = 2
y - y1 = m(x - x1 )       or     y = 2x + c
y - 5 = 2(x -1)                     5 = 2 + c
c = 3
y = 2x + 3                         y = 2x + 3
?m = 2
?substitution of (1 ; 5)
?answer (3)
[14]

 

QUESTION 8 

8.1

h(x) = -2(x +)(x -1)(x + 3)
                    2
h(x) = -(2x + 3)(x2 + 2x - 3)
h(x) = -2x3 - 7x2 + 9

OR
h(x) = -(2x + 3)(x -1)(x + 3)
h(x) = -(2x + 3)(x2 + 2x - 3)
h(x) = -2x3 - 7x2 + 9

?? – 2 (x +)(x -1)(x + 3)
                    2
? correct simplification
(3)

OR

?? - (2x + 3)(x -1)(x + 3)
? correct simplification
(3)

8.2

h /(x) = -6x2 -14x
- 6x 2 -14x = 0
- 2x(3x + 7) = 0
x = 0 or x = - 7
                    3

?first derivative
?= 0
?both answers (3)

8.3

x < - 7 or  x > 0
       3
OR

x ∈ (- ∞ ; - 7) ∪ ( 0 ; ∞)
                3

? ?answer
(2)

OR
??answer (2)

8.4

y = 4x + 7
- 6x 2 -14x = 4
0 = 6x2 +14x + 4
0 = 3x2 + 7x + 2
0 = (3x +1)(x + 2)
x = - 1 or x = -2 
       3

?y = 4x + 7
?h / (x) = 4
? standard form
? both answers

(4)

[12]

 

QUESTION 9 

9.1

Volume of Sphere
=π(8)3 or =2048π  or     = 2144,66
   3                    3 

? answer (1)

9.2

r 2 + x2 = 82      (Pythagoras)
r 2 = 64 - x2

? substitution or reason
Pythagoras (1)

9.3

Vcone 1/3 πr2h
=1/π(64 - x2 )(8 + x)
= π (512 + 64x - 8x2 - x3 )
   3
dV 64π - 16π x x2
dx      3        3          3
0 = 64 -16x - 3x2
0 = (8 - 3x)(x + 8)
x =          x ≠ 8
      3
11

?h = 8 + x
? 1/π(64 - x2 )(8 + x)
?expansion
dV 64π - 16π x x2
    dx      3        3          3
? x =
          3
? volume of the cone
 8 or 0,3
    27
(7)

[9]

 

QUESTION 10

10.1

12
P(One Red and One Blue)
= P(Red, Blue) + P(Blue, Red)

13

14
? addition of products
? answer
(4)

10.2.1

a = 0,48 x 250
a = 120

? answer                            (1)

10.2.2

b = 150 P(S) × P(F)
200 x 150
   250    250
= 0,48
= P(S and F)
These events are independent

? b
? P(S) × P(F)
? 200 and 150
    250        250
? conclusion
(with realistic probabilities)

(4)

   

[9]

 

QUESTION 11

11.1

10 × 9
= 90

??10 × 9
(2)

11.2.1

10!
=3 628 800

? 10!
(1)

11.2.2

2! × 2! × 2! × 2! × 2! ×4!
= 768

? 2! × 2! × 2! × 2! × 2!
? 4!
? 2! × 2! × 2! × 2! × 2!×4!
or 768
(3)

[6]

TOTAL: 150

Last modified on Friday, 24 September 2021 06:53