NATIONAL
SENIOR CERTIFICATE/
NASIONALE SENIOR
SERTIFIKAAT
GRADE 12/GRAAD 12
MATHEMATICS P1/WISKUNDE V1
FEBRUARY/MARCH/FEBRUARIE/MAART 2018
MARKING GUIDELINES/NASIENRIGLYNE
MARKS/PUNTE: 150
These marking guidelines consist of 17 pages./
Hierdie nasienriglyne bestaan uit 17 bladsye.
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2
NSC – Memorandum
DBE/November 2016
NOTE:
 If a candidate answers a question TWICE, only mark the FIRST attempt.
 Consistent accuracy applies in ALL aspects of the marking guidelines.
LET WEL:
 Indien 'n kandidaat 'n vraag TWEE KEER beantwoord, merk slegs die EERSTE poging.
 Volgehoue akkuraatheid is op ALLE aspekte van die nasienriglyne van toepassing.
QUESTION/VRAAG 1
1.1.1
1.1.2
x 2  6 x  16  0
x  8x  2  0
x  2 or x  8
 factors
 x  2
 x 8
(3)
2x  7x  1  0
2
x

 b  b 2  4ac
2a
 7  
subs into correct formula
7 2  42 1
22 
 7  57
4
 x  0,14
 x  3,64

 7  57
4
x  0,14 or x  3,64

OR/OF
7
49 1 49
x
 
2
16 2 16
x2 
NOTE: Penalise 1 mark if the
rounding to TWO decimal
places is incorrect.
for adding
2
7
57

x   
4
16

x
OR/OF
49
on
16
both sides
7
57

4
4
 7  57
x
4
x  0,14 or x  3,64
 7  57
4
 x  0,14
 x  3,64

(4)
1.2
x  25  0
x  5x  5  0
2
+
−
factors
+
+
−
+
inequality
5  x  5
x ={ –4 ; –3 ; –2 ; –1 ; 0 ; 1 ; 2 ; 3 ; 4}
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NOTE:
Final answer only
2/2
answer
(4)
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x  2y 1
1.3
DBE/November 2016
 x  2y 1
 substitution
2 y  1  2 7  y 2   y
4 y2  4 y  1  7  y2   y
3y2  3y  6  0
y2  y  2  0
 y  2( y  1)  0
y  2 or y  1
x  22   1 or x  2 1  1
x3
or x  3
OR/OF
 correct standard form
 factors
 y – values
 x – values
OR/OF
x 1
2
2
2
x  7  y  y
y
 y
 x 1
 x 1
x 7
  

 2 
 2 
 x2  2x  1   x  1
 
x 2  7  
4
2


x 1
2
2
2
 substitution
4 x 2  28  x 2  2 x  1  2 x  2
 correct standard form
3 x 2  27  0
x2  9  0
x  3x  3  0
x  3
 3 1
y
2
y  1
1.4
3 2018  3 2016
3 2017
3 2017 31  3 1

3 2017
1
 3
3
1
10
 3 or
3
3


OR/OF
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 factors
or
x3
or
y
or
3 1
2
y2
 x – values
 y – values
(6)
 common factor 32017
 answer
OR/OF
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4
NSC – Memorandum
3 2018  3 2016
3 2017
3 2016 3 2  1

3 2017
10

3


 common factor 32016
 answer
OR/OF
OR/OF
32018  32016
32017
32018 32016
 2017  2017
3
3
1
 3
3
1
10
 3 or
3
3
1.5.1
3x  5  0 and
5
x
and
3
DBE/November 2016
 dividing by 32017
 answer
(2)
x3
 3x  5  0
5
 x
3
 x3
x3
(3)
1.5.2
3x  5
1
x3
3x  5  x  3
 3x  5  x  3
3x  5  x  3
2
3x  5  x 2  6 x  9
x 2  9 x  14  0
x  7 x  2  0
x  2 or
x7
NOTE: If x = 2 is not
rejected, then
maximum 3 / 4 marks
 3x  5  x  3
2
 factors
x 7
(4)
[26]
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QUESTION/VRAAG 2
2.1.1
30 ; 10 ;
10
…….
3
a  30
r
1
3
r=
1
3
Tn  ar n 1
10
1
 30 
729
3
1
 31 n
2187
3 7  31 n
 7  1 n
n8
n 1
1
1
 
2187  3 
7
OR/OF
n 1
substitution into correct
formula
n 1
 37  31n or
1
1
   
3
3
7  n 1
n8
7
1 1
   
3 3
n1
or
use of logs
 n 8
(4)
2.1.2
2.2
a
1 r
30

1
1
3
 45
S 
substitution into correct
formula
answer
S n  a  (a  d )  ......  (a  n  2d )  (a  n  1d )
(2)
1
2
expanding S n
S n  (a  n  1d )  (a  n  2d )  .........  (a  d )  a
Adding both equations/Tel die twee vergelykings bymekaar:
2Sn  2a  n  1d  2a  n  1d  2a  n  1d  .............
 n2a  n  1d 
n
Sn  2a  n  1d 
2
reverse writing
 2Sn  n2a  n  1d 
n
 S n  2a  n  1d 
2
(4)
OR/OF
S n  a  (a  d )  ......  (a  n  2d )  Tn
1
2
S n  Tn  Tn  d   Tn  2d   .........  a
Adding both equations/Tel die twee vergelykings bymekaar:
2S n  a  Tn   a  Tn   a  Tn   .....  a  Tn 
n
a  Tn 
2
but Tn  a  (n  1)d
n
S n  2a  n  1d 
2
expanding S n
reverse writing
 2S n  na  Tn 
Sn 
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 Sn 
n
2a  n  1d 
2
(4)
[10]
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DBE/November 2016
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DBE/November 2016
QUESTION/VRAAG 3
3.1
–1; 2;5
Tn  1  (n  1)(3)
 3n
–4
 3n  4
3.2
T43  343  4
 125
3.3
(2)
OR/ OF
T43  1  43  13
 125
Tn  3n  4
NOTE:
Answer only 2 / 2
 subs of 43
 answer
(2)
n
n
S n   Tk  1  2  5  ...........  3n  4
 S n   Tk
k 1
k 1
n
 1  3n  4
2
n
 3n  5
2
3n 2  5n

2
Sn 
or
Sn 
n
 2  n  13
2
OR/OF
Tn  3n  4
n
 Tk  31  4  32  4  33  4  .........  3n  4
k 1
 31  2  3  .......  n   4n
3nn  1

 4n
2
3n 2  5n

2
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 substitution into correct
formula

2
n
3n  5 or 3n  5n
2
2
OR/OF
 1  4  32  4  33  4 
.........  3n  4
 31  2  3  .......  n  4n
3n 2  5n

2
(3)
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3.4
8
NSC – Memorandum
DBE/November 2016
T11  T11  T10   T10  T9   T9  T8     T3  T2   T2  T1   T1  generating sum
 29  26  23  2
125  29  26  23   2  T1
10

29  2
10
2
 29  2  T1
2
 155
NOTE:
 155  T1
 – 30
T1  30
Answer only 1 / 6
If they only use 3n – 4
breakdown 0 / 6
OR/OF
OR/OF
Tn  an 2  bn  c
 T11  121a  11b  c  125

Tn  Tn 1  an 2  bn  c  a n  1  bn  1  c
2
 an 2  bn  c  an 2  2an  a  bn  b  c
 2an  b  a

 121a  11b  c  125
Tn  Tn 1  3n  4
2a  3
3
a
2
and b  a  4
5
and b  
2
 calculating Tn  Tn1 in
terms of a, b and c
121a  11b  c  125
 a
3
 5
121   11    c  125
2
 2
c  29
 b
3 2 5
n  n  29
2
2
3 2 5
T1  1  1  29
2
2
 30
Tn 
3
2
5
2
 c  29
 –30
(6)
[13]
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QUESTION/VRAAG 4
4.1
4.2
x=4
 y = –9
E(4 ; –9)
(2)
f ( x)  x  4  9
2
 x  4 2  9  0
 x  4 2  9
 y=0
x  4  3
x  7 or x  1
A(1 ; 0)
 x  4  3
OR/OF
OR/OF
 A(1 ; 0)
f ( x)   x  4  9
2
0  x 2  8 x  16  9
 y=0
0  x2  8x  7
x  7 x  1  0
x  7 or x  1
A(1 ; 0)
4.3
4.4
C 0 ; 7 
M 8 ; 7 
70
or
04
7
m
4
7
y  0   x  4
4
7
y   x7
4
x  7x  1
 A(1 ; 0)
 C 0 ; 7 
x=8
y=7
NOTE:
Answer only 3 / 3
(3)
(3)
C0 ; 7 
D4 ; 0
m
4.5

 D4 ; 0
07
40
7
m
4
m
7
g : y   x7
4
7
g 1 : x   y  7
4
4 x  7 y  28
7 y  4 x  28
4
y   x4
7
OR/OF
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or 0  4m  7
m
7
4
 m
7
4
7
 y   x7
4
(3)
 interchange x and y
 simplification
4
 y  x4
7
OR/OF
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g 1 is the straight line through (0 ; 4) and ( 7 ; 0)
y  mx  4
0  7m  4
4
y  x4
7
4.6
DBE/November 2016
 straight line through (0 ; 4)
and ( 7 ; 0)
 substitution
4
 y  x4
7
(3)
x . f x   0
 x  0
 1  x  7
 x  0 or 1  x  7
(4)
[18]
QUESTION/VRAAG 5
5.1
5.2
a0  1
T(0 ; 1)
x=0
 y 1
g  x  ax
9 a
a3
5.3
5.4
1
y 
 3
(2)
 substitution
2
a=3
a0
x
1
 y   
 3
or y  3  x
30  3log3 x  31
1 x  3
1  x
x3
(2)
x
(2)
(2)
OR
y
x
0
1
1 x  3
3
1<x
x<3
(2)
[8]
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QUESTION/VRAAG 6
6.1
q 1
 q 1
(1)
6.2
Subs  0;0 
0
a
1
0 p
0 
a
 1
p
a  p
a
1
0 p
 a  p
Subs P :
a
1
2 2 p
a
2 2 p
2 1 
2
 substitution
NOTE:
Answer only 2 / 5
22 2  2p  a
22 2  a p 2  aa 2

 
2 1 2  a 1 2

 a=2
 p  2
a  2 ; p  2
(5)
y
6.3
f
y = 1
y=1
O
x
x = 2
x=2
shape
(0 ; 0)
(4)
[10]
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QUESTION/VRAAG 7
7.1
7.2.1


x 1  i   1
F
i
 0,06  60 
2500 1 
  1
12 



0,06
12
 R174 425,08
n
0,06
/ 0,005
12
 correct substitution
into correct formula
 answer
(3)
 n = 60 and i =
After eleven months, Genevieve will owe/
Na elf maande skuld Genevieve
11
 0,15 
A  82 0001 

12 

 R 94 006,79
 n = 11
 correct substitution
into correct formula
 answer
(3)
7.2.2
x 1  1  i  

P 
i
  0,15   n 
3 200 1  1 
 
12  
 

94 006, 79 
0,15
12
n
94 006, 79 0,15
 0,15 

 1  1 

3 200
12
12 

 94 006,79
 substitute into correct
formula
n
n
 0,15 
1 
  1  0,3672147...
12 

 0,15 
 n log 1 
  log 0, 6327852...
12 

 n  36,8382...
n  36,84
Genevieve will have to pay 36 installments of R3 200
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correct use of logs (logs to
be defined)
 n =36,84
 36 installments
(5)
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7.2.3
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NSC – Memorandum


x 1  1  i 
i
  0,15  0,83826912 

32001  1 

12 
 


0,15
12
P  2652
P
n
Outstanding balance after 36 installments is R2 652
Final payment will be:
1
 0,15 
A  2 652,001 

12 

 R 2685,00
OR/OF
 0,15  36 
3 200 1 
  1
36
12 


 0,15 
Balance : 94006,791 
 
0,15
12 

12
 R2 651,72
DBE/November 2016
n = 083826912
substitute into correct
formula
answer
 0,15 
 2 652,001 

12 

answer
1
OR/OF
36
 0,15 
 94006,791 

12 

 0,15 36 
3 200 1 
  1
12 



0,15
12
 2 651,72
Final payment will be:
 0,15 
A  2651,721 

12 

 R 2 685,00
1
Copyright reserved/Kopiereg voorbehou
 0,15 
 2651,721 

12 

answer
1
(5)
[16]
Please turn over/Blaai om asseblief
Mathematics/P1
14
NSC – Memorandum
DBE/November 2016
QUESTION/VRAAG 8
f x  h   4 x 2
8.1
 4x  h 
2
f  x  h   f  x   4 x  h   4 x 2
2


 4 x 2  2 xh  h 2  4 x 2
 4 x  8 xh  4h  4 x 2
2
2
 8 xh  4h 2
f x  h   f x 
f  x   lim
h 0
h
 8 xh  4h 2 
 lim 

h 0
h


 h8 x  4h  
 lim 

h 0
h

 8x
OR/OF
f x  h   f x 
f  x   lim
h0
h
 4 x  h 2  4 x 2 
 lim 

h0
h


2
2
 4 x 8 xh  4h  4 x 2 
 lim 

h0
h


 8 xh  4h 2 
 lim 

h0
h


 h8 x  4h  
 lim 

h0
h


 8x
8.2.1
 x2  2x  3
Dx 

 x 1 
 x  3x  1
 Dx 

x 1


 Dx x  3
1
8.2.2
f x   x  x
1
2
1
1 2
x
2
3
1 
f  x    x 2
4
f  x  
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 8xh  4h 2

f x  h   f x 
h
h8 x  4h 
h
 8x

OR/OF

f x  h   f x 
h
 4x  h 
2
 8xh  4h 2
h8 x  4h 
h
8x


(5)
x  3x  1
x 1
 x  3
1
(3)
1
 x2
1
1 2
 x
2
3
1 
 x 2
4
(3)
[11]
Please turn over/Blaai om asseblief
Mathematics/P1
15
NSC – Memorandum
DBE/November 2016
QUESTION/VRAAG 9
9.1
 f ( x)  x  2x  1x  4
f ( x)  x  2x  1 x  4 
 ( x 2  x  2) x  4 
 expansion
 x3  x 2  2 x  4 x 2  4 x  8
 x 3  3x 2  6 x  8
 x3  3x 2  6 x  8
b  3 ; c  6 ; d  8
(4)
f ( x)  x 3  3x 2  6 x  8
9.2
f / ( x)  0
 f / ( x)  0
3x  6 x  6  0
2
x2  2x  2  0
x

 3x 2  6 x  6
 b  b 2  4ac
2a
2
22  41 2
21
 substitution into correct
formula
2  12
2
x  0,73

 x  0,73
2
f  1   1  3 1  6 1  8 or f  1  1 2 5
 10
 10
3
2
f  1  3 1  6 1  6
3
y  10  3x  1
y  3 x  13
/
9.4
(4)
f ( x)  x  3x  6 x  8
9.3
3
 f  1  10
2
 f /  1  3
 substitution
 y  3x  13
f // x   6 x  6
(4)
 f // x   6 x  6
y
f //
x
(1; 0)
(0 ; -6)
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x- intercept
y- intercept
(3)
Please turn over/Blaai om asseblief
Mathematics/P1
9.5
16
NSC – Memorandum
f concave upwards
f // ( x)  0
6x  6  0
x 1
NOTE:
Answer only 2 / 2
DBE/November 2016
 f // ( x)  0
 x 1
(2)
[17]
QUESTION/VRAAG 10
.
f  x   3 x 3  x
 9x 2  1  0
1
x
3
1
M aximum of f will be at x 
3
1
or x  
3
  9x2 1  0
1
x
or
3
x
 Maximum at x 
1
3
1
3
3
1
1 1
f    3    
3
3 3
2

9
1
 3
 f 
2
9
M aximum of f  x   q will also be at x 
1
3
8
1
f q 
9
3
2
8
q 
9
9
6
q
9
2

3
8
For f x   q to have a maximum of the value of q
9
2
has to be .
3
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2
8
q 
9
9
2
q=
3

[6]
Please turn over/Blaai om asseblief
Mathematics/P1
17
NSC – Memorandum
DBE/November 2016
QUESTION/VRAAG 11
11.1.1 Let the event Veli arrive late for school be V.
Let the event Bongi arrive late for school be B. /
Laat V die gebeurtenis wees dat Veli Laat B die gebeurtenis
wees dat Bongi laatkom
P(V or B) = 1 - 0,7
= 0,3
11.1.2 P(V or B) = P(V) +P(B)  P(V and B)
0,3 = 0,25 + P(B)  0,15
P(B) = 0,2
11.1.3
P(V)  P(B) = 0,25  0,2
= 0,05
P(V)  P(B)  P(V and B)
V and B are NOT independent/
V en B is NIE onafhanklik nie.
 answer
(1)
 P(V or B) = P(V) +P(B)
–P(V and B)
substitution
 0,2
 P(V)  P(B) = 0,05
(3)
 P(V)  P(B)  P(V and B)
NOT independent
(3)
 6! or 720
11.2.1 6 ! =720
(2)
11.2.2 Number of arrangements
 3!  3!
2
 answer
 3!  3!  2
 72
(3)
11.2.3
3!  4!
6!
144

720
1
 or 0,2 or 20%
5
P(hearts next to each other) 
OR/OF
  3!  4!

1
or 0,2 or 20%
5
OR/OF
4  3!  3!
6!
144

720
1
 or 0,2 or 20%
5
P(hearts next to each other) 


1
or 0,2 or 20%
5
(3)
[15]
TOTAL/TOTAAL: 150
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