Downloaded f rom St anmorephysics. com
do
wn
l
oa
de
d
fr
om
st
an
m
or
ep
hy
si
c
s.
co
m
Downloaded f rom St anmorephysics. com
do
wn
l
oa
de
d
fr
om
st
an
m
or
ep
hy
si
c
s.
co
m
Downloaded f rom St anmorephysics. com
Downloaded f rom St anmorephysics. com
Downloaded f rom St anmorephysics. com
Downloaded f rom St anmorephysics. com
Downloaded f rom St anmorephysics. com
Downloaded f rom St anmorephysics. com
Downloaded f rom St anmorephysics. com
Downloaded f rom St anmorephysics. com
Downloaded f rom St anmorephysics. com
Downloaded f rom St anmorephysics. com
Downloaded f rom St anmorephysics. com
Downloaded f rom St anmorephysics. com
NATIONAL
SENIOR CERTIFICATE/
NASIONALE SENIOR
SERTIFIKAAT
SENIOR SERTIFIKAAT
GRADE/GRAAD 12
SEPTEMBER 2022
MATHEMATICS P1/ WISKUNDE V1
MARKING GUIDELINE/NASIENRIGLYN
MARKS/PUNTE:
150
This marking guideline consists of 16 pages.
Hierdie nasienriglyn bestaan uit 16 bladsye.
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
2
(EC/SEPTEMBER 2022)
NOTE/LET WEL:





If a candidate answers a question TWICE, mark the FIRST attempt ONLY.
Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord, merk SLEGS die EERSTE poging.
Consistent accuracy applies in ALL aspects of the marking guideline.
Volgehoue akkuraatheid geld deurgaans in ALLE aspekte van die nasienriglyn.
If a candidate crossed out an attempt of a question and did not redo the question, mark
the crossed-out attempt.
Indien ʼn kandidaat ʼn poging vir ʼn vraag deurgetrek het en nie die vraag weer
beantwoord het nie, merk die poging wat deurgetrek is.
The mark for substitution is awarded for substitution into the correct formula.
Die punt vir substitusie word toegeken vir substitusie in die korrekte formule.
QUESTION 1/VRAAG 1
1.1.1
x
2
 4 x  21  0
 factors / faktore
( x  3)( x  7 )  0
 x  3
x  7
or / of
 both x-values / beide x-waardes
(2)
OR/OF
x 
b 
b
2
OR/OF
 4ac
2a

2
4 
4  4 (1) (  2 1)
 substitution / vervanging
2 (1)

4 
100
2
 3
1.1.2
or / of
 both x-values / beide x-waardes
(2)
 7
x(2 x  7 )  3
2x
2
x 
 standard form / standaardvorm
 7x  3  0
b 
b
2
 4ac
2a

 (7) 
2
(  7 )  4 ( 2 )(  3)
2(2)

7 
 substitution / vervanging
73
4
 3, 8 9 o r / o f
 0, 39

x  3, 8 9
or/of  x
  0, 39
(4)
1.1.3
( 2 x  3 ) ( x  1)  6
2x
2x
2
 5x  3  6
 standard form / standaardvorm
2
 5x  3  0
 factors / faktore
( 2 x  1) ( x  3 )  0
 3  x 
1
2
  3  x  12
(Accuracy/Akkuraatheid)
(4)
Copyright reserved/Kopiereg voorbehou
Please turn over/Blaai om asseblief
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
3
(EC/SEPTEMBER 2022)
1.1.4
x  x  3
2
x  3 x
2
2
x

2
 (3  x )
4x  9  6x  x
2
 x
 squaring both sides
kwadreer beide kante
2
2
 standard form / standaardvorm
 factors / faktore
 both answers / beide antwoorde
 selection / keuse
(5)
OR/OF
 10 x  9  0
( x  1) ( x  9 )  0
 x  1 or / of x  9
OR/OF
x  x  3
2
x  2
x  3  0
L et k 
2
 k
 standard form / standaardvorm
In this question, CA marking applies
only if the k equation is quadratic.
In hierdie vraag, kan VA nasien slegs
toegepas word as die vergelyking van
k kwadraties is.
x,
 2k  3  0
( k  1) ( k  3 )  0
k  1 or / of k  3

x  1 or / of
k
2
 2k  3  0
(Accuracy/Akkuraatheid)
 factors / faktore
 both k-values / albei k-waardes
x   3 .....

x  0

 x 1
1.2

 selection of x-value /
keuse van x-waarde
(5)
2 y  x  3  0 ..................................... ..(1)
x
2
 y
2
 2 x y  1 ..................................( 2 )
x   2 y  3 ...................................... .....( 3 )

x  2 y  3
S u b s titu te (3 ) in to ( 2 ) / V e r v a n g (3 ) in ( 2 )
2
( 2 y  3)  y
4y
y
2
2
2
 2 y ( 2 y  3)  1
 12 y  9  y
2
 4y
2
 6y 1  0
 standard form / standaardvorm
 6y  8  0
 factors / faktore
( y  2 )( y  4 )  0
 y   4 or / of
 x  5
or / of
 substitution / vervanging
y  2
x 1
 y-values / y-waardes
 x-values / x-waardes
(6)
OR/OF
Copyright reserved/Kopiereg voorbehou
OR/OF
Please turn over/Blaai om asseblief
4
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
(EC/SEPTEMBER 2022)
2 y  x  3  0 .......................(1)
x
2
2
 y
 2 x y  1 ..................( 2 )
x  3
y  
...........................(3 )
2

y  
x  3
2
S u b s titu te (3 ) in to ( 2 ), V e r v a n g (3 ) in ( 2 ) :
2
x
x
2
2
 x  3 
 x  3 
 
  2x
 1
2 
2 


2
( x  3)

 substitution / vervanging
 x ( x  3)  1  0
4
4x
x
2
2
 x
2
 6x  9  4x
2
 12 x  4  0
 6x  5  0
( x  5 ) ( x  1)  0
 x  1 or / of
x  5
 y  2 or / of
y  4
 standard form / standaardvorm
 factors / faktore
 x-values / x-waardes
 y-values / y-waardes
(6)
1.3
1
x
K
1
x
 K
K

 K
1
x
1
 K

1
y

x
1
y
 1 2 .....  4 
1
y
 K
1

y
x  y
w 
1
w
...................  b o t h / b e i d e  1 2 
 equating / gelykstel
1
w
 exp. Law / eks. Wet
1
w

xy

 K
 multiplying /
vermenigvulduging
1
 simplification /
vereenvoudiging
w
xy
x  y
( r e c i p r o c a ls / o m g e k e e r d e s )
(4)
[25]
Copyright reserved/Kopiereg voorbehou
Please turn over/Blaai om asseblief
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
5
(EC/SEPTEMBER 2022)
QUESTION 2/VRAAG 2
2.1.1

3 d  ( 2 x  8 )  ( 3 x  1)
3 d  ( 2 x  8 )  ( 3 x  1)
3d   x  9
 substitution / vervanging
3(4)   x  9
 answer / antwoord
 x  3
2.1.2 (a)
T4  3 x  1
or / of
T7  2 x  8
 3( 3)  1
 2( 3)  8
 10
 2
 a  3d  10
or / of
a  3(4)   10
(3)
a  6d  2
a  6(4)  2
a  22
a  22
 substitution / vervanging

T4   1 0 o r / o f T7  2
 answer / antwoord
(3)
2.1.2 (b)
Sn 
n
2a
2
S 42 
42
2
 ( n  1) d
2(22) 

 formula / formule
( 4 2  1) ( 4 ) 
 substitution / vervanging
 2520
2.2.1
T2  3 9 a n d / e n T3  2 1
2.2.2
Tn  2 n
2
 answer / antwoord
(3)
 answers / antwoorde
(2)
 28n  87
T n  4 n  2 8
n 
A t/b y m in : 4 n  2 8  0
4n  28
 n  7
b
2a
OR/OF

 method/metode
 (28)
2(2)
 7

n  7
2
T7  2 (7 )  2 8 (7 )  8 7
 answer / antwoord
(3)
 11
OR/OF
Tn  2 n
2
 2(n
 28n  87
2
 14n  49  49)  87
OR/OF
 completing the square
voltooiing van vierkant
2
 2[( n  7 )  4 9 ]  8 7
2
 2(n  7)  98  87
2
 2(n  7)  11
 S m a lle s t v a lu e / k l e i n s t e w a a r d e   1 1
2.2.3
k  11
Copyright reserved/Kopiereg voorbehou
 simplification /
vereenvoudiging
 correct conclusion
korrekte gevolgtrekking
 answer / antwoord
(3)
(2)
[16]
Please turn over/Blaai om asseblief
6
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
(EC/SEPTEMBER 2022)
QUESTION 3/VRAAG 3
3.1.1
0 , 7  0 , 7 7 7 7 7 7 ...
 0 , 7  0 , 0 7  0 , 0 0 7  ...

0 , 7  0 , 0 7  0 , 0 0 7  ...
(1)
3.1.2
a  0, 7
Tn  a r
r  0 ,1
a  0, 7

T n   0 , 7   0 ,1 
r  0 ,1
and / en
n 1
  0 , 7   0 ,1 
n 1

  0 , 7   0 ,1 
 p 

n 1
n 1
 answer / antwoord
(3)
n 1
3.1.3
S 


a
1 r

0 ,7
1  0 ,1
7
0 ,7
1  0 ,1
 answer / antwoord
9
3.2
(2)
T 9  T1 0  6  T 8
8
ar  ar
7
9
 6  ar
ar
r

2
ar (r  r )
2
7
7
 6
8
ar  ar
9
 6  ar
7
 simplification / vereenvoudiging
 r  6  0
( r  3)( r  2 )  0
 r  3 or / of
r  2
Copyright reserved/Kopiereg voorbehou
 standard form / standaardvorm
 answers / antwoorde
(4)
[10]
Please turn over/Blaai om asseblief
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
7
(EC/SEPTEMBER 2022)
QUESTION 4/VRAAG 4
4.1
4.2
 answer / antwoord
p  3
0 
(1)
1 k
 substitution / vervanging
1 p
 0  1 k

k  1
k  1
(2)
4.3
f (x) 
y 
2
x  3
2
0  3
1
1
 substitution / vervanging
1
2
3
 y-value / y-waarde
 1
3
 y in te rc e p t is a t B
4.4
 0 ; 13 
/ y  a fs n it is b y B
 0 ; 13 
(2)

x  0 or / of 1  x  3
x  0
 1 
x  3
Accuracy / Akkuraatheid
OR/OF
x  (  ; 0] or / of
4.5
f (x) 



x  [1 ; 3 )
(3)
OR/OF
 x  (   ; 1] Accuracy/Akkuraatheid
 x  [1 ; 3 ) Accuracy/Akkuraatheid
(3)
x 1
x  3
( x  3)  2
x  3
2
x  3
2
x  3


( x  3)  2
x 3
x  3
x  3
1
 answer / antwoord
(2)
[10]
Copyright reserved/Kopiereg voorbehou
Please turn over/Blaai om asseblief
8
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
(EC/SEPTEMBER 2022)
QUESTION 5/VRAAG 5
5.1
 asymptote / asimptoot
 intercept / afsnit
 shape / vorm
(3)
5.2
y 1

,y
y  1 Accuracy / Akkuraatheid
(2)
OR/OF
y  (   ;1)
5.3
x
g ( x )   (  3  1)
x
 3 1
Answer only – Full Marks
Slegs antwoord – Volpunte
 A s y m p to te / A s im p to o t : y   1
OR/OF
 y  (   ;1) Accuracy/Akkuraatheid
(2)

x
3 1
 answer / antwoord
(2)
5.4
h(x)  3
x  3
x
y
 y  lo g 3 x

h(x)  3

x  3
x
y
 answer / antwoord
(3)
[10]
Copyright reserved/Kopiereg voorbehou
Please turn over/Blaai om asseblief
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
9
(EC/SEPTEMBER 2022)
QUESTION 6/VRAAG 6
6.1.1
x  
b
2x  4  0
2a
 
 subst. into correct formula
(4)
verv. in korrekte formule
(method mark / metodepunt)
2x  4
OR /OF
2 (1)
 2
x  2
 x-value / x-waarde
2
y  (2)  4(2)  11
 15
 y-value / y-waarde
D (2 ;  15)
(3)
6.1.2
/
g (x)  f (x)  2 x  4
c o o r d in a te s o f C / k o o r d in a te v a n C :
C (2 ; 0)
 coordinates of C
koördinate van C
OR/OF
Making connection between x-coordinate of T/P of the
function and the x-intercept of the derivative of the
function. Concluding that C ( 2 ; 0 ) .
Maak konneksie tussen x-koördinaat van draaipunt van
die funksie en die x-afsnit van die afgeleide van die
funksie. Gevolglik is C ( 2 ; 0 ) .
2
CN 
( 7  2 )  (1 0  0 )

125
 5
6.2.1
 substitution /
vervanging
2
 answer / antwoord
5
(4)
 answer / antwoord
1  x  7
(2)
6.2.2
g (x)  f (x)
 2x  4  (x
 x
2
2
 4 x  1 1)
 difference / verskil
 6x  7
F o r m a x im u m / V ir m a k s im u m :  2 x  6  0
 x  3
 derivative / afgeleide
 equating derivative to 0
stel afgeleide = 0
 answer / antwoord
(4)
[13]
Copyright reserved/Kopiereg voorbehou
Please turn over/Blaai om asseblief
10
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
(EC/SEPTEMBER 2022)
QUESTION 7/VRAAG 7
7.1
A  P (1  i )
n
R 2 7 7 6 3 , 1 2  P (1  0 , 1 7 )
P 
4
 substitution / vervanging
2 7 7 6 3,1 2
0,83
4
 R 58 500
 answer / antwoord
(2)
7.2
n
F 
x [ (1  i )  1]
i

0, 086 
x 1 

12 
 
R 300 000 
84

 1

0, 086
12
R 300 000 
x 

i 
0, 086
and / en n  84
12
 correct substitution into correct
formula / korrekte vervanging in
die korrekte formule
0, 086
12

0, 086 
1 

12 
 
84

 1

 answer / antwoord
 x  R 2 6 1 6 ,0 5
(3)
7.3.1
P 
x 1   1  i 

n


i


0 ,1 0 4 

R 8 9 0 1, 9 6  1   1 

12 


0 ,1 0 4
12
 R 950 000
300




0 ,1 0 4
and / en n  300
12
 correct substitution into correct
formula / korrekte vervanging in
die korrekte formule
 answer / antwoord
(3)
Copyright reserved/Kopiereg voorbehou
Please turn over/Blaai om asseblief
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
11
(EC/SEPTEMBER 2022)
7.3.2(a)
O u ts ta n d in g b a la n c e a f te r 2 0 4 p a y m e n ts :
U its ta a n d e b a la n s n a 2 0 4 b e ta lin g s
P 
x 1   1  i 

n


i


0 ,1 0 4 

R 8 9 0 1, 9 6  1   1 

12



96




n  96
 correct substitution into correct
formula / korrekte vervanging in
die korrekte formule
0 ,1 0 4
12
 R 5 7 8 5 5 1, 2 4
 answer / antwoord
(3)
OR / OF
OR / OF
O / B  A  Fv
x  1  i   1 


n
n
P (1  i ) 
i
0 ,1 0 4 

950 000 1 

12 

204


0 ,1 0 4 
8 9 0 1, 9 6   1 

12 
 
0 ,1 0 4
12
5 523 928, 831830547  4 945 376, 296008371
R 578 552, 54
204

 1
 n  204

 correct substitution into correct
formula / korrekte vervanging in
die korrekte formule
 answer / antwoord
(3)
7.3.2(b)
R 5 7 8 5 5 1, 2 4 

0 ,1 0 4 

R 7 5 0 0 1   1 

12 


n



0 ,1 0 4
0 ,1 0 4
12
1
R 7 500
 1513 
 

 1500 
P  R 5 7 8 5 5 1, 2 4
in P – formula / in P – f ormule
12
R 5 7 8 5 5 1, 2 4 

n
 lo g  1 5 1 3  0 , 1 3 3 1 5   n


 1500 
 simplification / vereenvoudiging
(isolating n / isoleer n)
 correct use of logs /
korrekte gebruik van logs
 n  127,97
 n  1 2 8 m o n th s /m a a n d e
 1 0 y e a rs 8 m o n th s
 answer / antwoord (months)
(4)
1 0 ja a r 8 m a a n d e
[15]
Copyright reserved/Kopiereg voorbehou
Please turn over/Blaai om asseblief
12
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
(EC/SEPTEMBER 2022)
QUESTION 8/VRAAG 8
8.1
f (x)  3x
2
 x
f (x  h)  f (x)
f  ( x )  li m
h  0
h
2
 3( x  h )  ( x  h )  ( 3 x
 li m
h  0
2
 x)
 substitution / vervanging
h
3x
 li m
2
 6 xh  3h
h  0
2
 x  h  3x
2
 x
 expansion / uitbreiding
h
 6 xh  3h
 li m
h  0
2
 h
 simplification /
vereenvoudiging
h
h (  6 x  3 h  1)
 li m
h  0
 factorisation / faktorisering
h
 li m (  6 x  3 h  1)
h  0
 answer / antwoord
OR/OF
  6x 1
OR/OF
f (x)  3x
2
 x
2
f ( x  h )  f ( x )   3( x  h )  ( x  h )  ( 3 x
  3( x
 3x
2
2
 2 xh  h )  x  h  ( 3 x
 6 xh  3h
2
 x
 x)
 expansion / uitbreiding
 h
h
6 xh  3h
h  0
 li m
 x  h  3x
2
f (x  h)  f (x)
h  0
 li m
2
2
 substitution / vervanging
 x)
2
 6 xh  3h
f  ( x )  li m
2
2
 h
h
h (  6 x  3 h  1)
h  0
 simplification /
vereenvoudiging
 factorisation / faktorisering
h
 li m (  6 x  3 h  1)
h  0
 answer / antwoord
  6x 1
8.2.1
Dx
(5)
4 

2
4
4
3 x  2  D x 3 x  4 x 




x 

3
 12 x  8 x
3

 4x
2
 12 x3 
8x
3
(3)
8.2.2
2
y  a x  6
2
y  a x  6x

dy
 a
2
x
1
2
 3x

1
 changing surd /
verandering van wortelvorm

2
dx
a
2

3x
1
2
(3)
[11]
Copyright reserved/Kopiereg voorbehou
Please turn over/Blaai om asseblief
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
13
(EC/SEPTEMBER 2022)
QUESTION 9/VRAAG 9
9.1
3
f (x)   x  3x  2
2
f '( x )   3 x
 3
A t tu rn in g p o in ts / B y d r a a ip u n te:
f '( x )  0
 3x
2


f '( x )  0

(1;  4)
2
f '( x )   3 x  3
 3  0
2
x 1
 x  1
3
y   (  1)  3 (  1)  2
3
y   (1)  3 (1)  2
or / of
  4
 0
 T u rn in g p o in ts / D r a a ip u n te:
(  1 ;  4 ) a n d / e n (1 ; 0 )
9.2
(1 ; 0 ) i s a n i n t e r c e p t / i s ' n a f s n i t
3
2
 x  2)
 ( x  1) ( x  1) ( x  2 )
x 1
or / of
x  2
1  x  1
OR/OF
x  (  1 ;1)
9.3.2
x p .o .i 
1  1
 0
f  ( x )  6 x  0
OR /OF
2


2
( x  1)( x  x  2 )
( x  1)( x  2 )
 values of x / waardes van x
(3)
  1  x  1
(2)
OR/OF
 x  (  1 ;1)
(2)
 x-coordinate / x-koördinaat
 x  0
 c o n c a v e d d o w n fo r / k o n k a a f a f v ir : x  0
9.4
(1 ; 0 )
(4)
f ( x )  x  3 x  2  ( x  1) ( x
9.3.1

 answer / antwoord
(3)
g ( x )  f ( x  3)
T u rn in g p o in t / D r a a ip u n t : (  1 ;  4 )  ( 2 ;  4 )
(1 ; 0 )  ( 4 ; 0 )
3
y  in te rc e p t / y - a fs n it : (  3 )  3 (  3 )  2   1 6
 x-intercepts / x-afsnitte
 y-intercept / y-afsnit
 turning points / draaipunte
 shape / vorm
(2;4)
(1;0)
(4;0)
(0;-16)
(4)
9.5
0  k  4
Copyright reserved/Kopiereg voorbehou
 answer / antwoord
(Accuracy / Akkuraatheid) (2)
[18]
Please turn over/Blaai om asseblief
14
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
(EC/SEPTEMBER 2022)
QUESTION 10/VRAAG 10
10.1
 OC
 OA
OC = x
OA = –3x + 9
(2)
10.2
C o o rd in a te s o f B a re / K o o r d in a te v a n B is :
(x ; y)  (x ;  3x  9)
A r e a , A o f r e c t a n g le O A B C / O p p e r v l a k t e , A v a n r e g h o e k O A B C
A  lb
 OC  OA
 x( 3 x  9)
 3x
2
 9x

3x  9 x

6 x  9
2
A r e a is m a x w h e n / O p p e r v la k te is m a k s . w a n n e e r :
dA
 0
dx
 6x  9  0
x 
 y  3

 B
3
2
3
2
9
9
2
 32 ; 92 
=0
x
y
(4)
[6]
Copyright reserved/Kopiereg voorbehou
Please turn over/Blaai om asseblief
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
15
(EC/SEPTEMBER 2022)
QUESTION 11/VRAAG 11
11.1.1
160  60  x  55  255  85  200  45  900
x  900  860
x  40
 addition and equating to 900
/ optel en gelyk stel aan 900
 answers / antwoorde
(2)
11.1.2
11.1.3
P ( o n ly / s le g s H ) 
200  2 


900  9 
P ( a t le a s t 2 / te n m in s te 2 ) 
 answer / antwoord
(2)
240
900
 answer / antwoord
P e r c e n ta g e / P e r s e n ta s ie  2 6 , 7 %
11.2.1
(a)
11.2.2
(b)
11.2.2
8 !  4 0 3 2 0 w ays

/ maniere
 2  7!
(2)
nd
(award 2 mark only if
multiplication is shown /
answer only – full marks)
(ken 2de punt toe slegs as
vermenigvuldiging getoon
word / slegs antwoord –
volpunte)
2  7 ! w a y s / m a n ie r e
 1 0 0 8 0 w a y s / m a n ie r e
P ( E v e n t / G e b e u r te n is ) 
6!  3!
8!

3
28
Copyright reserved/Kopiereg voorbehou
8 !  40 320
(2)
(1)
Answer Only – Full Marks
Slegs Antwoord – Volpunte


6!  3!
8!
(2)
Please turn over/Blaai om asseblief
16
Downloaded f romMATHEMATICS
St anmorephysics.
P1/WISKUNDEcom
V1
(EC/SEPTEMBER 2022)
11.3
P (R /G ) = P (R )× P (G )
1
5
1

5
4
5
x

3x
4x
4x 1
1
3x
4



4x 1

x
4x

3x
4x 1
 equating product to 0,2
stel produk gelyk aan 0,2
3x
4x 1
15 x  16 x  4
x  4
 N u m b e r o f b a lls / A a n ta l b a lle  1 6
 value of x / waarde van x
 answer / antwoord
(5)
[16]
TOTAL/TOTAAL:
Copyright reserved/Kopiereg voorbehou
150
Please turn over/Blaai om asseblief