# Mathematics Grade 12 P1 North West Mock 2023 Paper & Memo

> Grade 12 Mathematics Paper 1 question paper and memo from the North West Mock Exam 2023, 26 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2023-north-west-mock-exam-question-paper-and-memorandum-paper-1/5726debe-0faa-4735-b5bf-163a3ef18696
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: North West Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 26

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Mathematics · Grade 12 · North West Mock Exam · 2023. Question paper and memorandum, 26 pages. Read online or download the PDF.

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### Page 1

education

Depariment:

Education

North West Provincial Government
REPUBLIC OF SOUTH AFRICA

NATIONAL SENIOR CERTIFICATE

12611E

X10

This question paper consists of 9 pages and 1 information sheet.

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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1. Thisquestion paper consists of 11 questions.

2. Answer ALL the questions.

3. Number the answers correctly according to the numbering system used in
this question paper.

4. Clearly show ALL calculations, diagrams, graphs, etc. that you have used

in determining your answers.
5. Answers only will NOT necessarily be awarded full marks.

6. ‘You may use an approved scientific calculator (non-programmable and
non-graphical), unless stated otherwise.

7. If necessary, round off answers to TWO decimal places, unless
stated otherwise.

8. Diagrams are NOT necessarily drawn to scale.

9. -An.information sheet with formulae is included at the end of the .
question paper.

10. Write neatly and legibly.

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QUESTION 1

1

Solve for x:
1.192022 —
fa 5x
con,
1. Bipp x’ -2x-13=0 (correct to TWO decimal places.)
L13 (x-2)i-x) <0

LL4 2¥2x-l=x-ll

1.2 Solve for x and y simultaneously:
3x-y=4 and x +xy=24
1 1 i 1 1)
130 Let S=)14+—] 14=}}1+- ) 14+—}...)1+— | where meN and
TO 8 OJ 10 m)
7T<m<30.
Calculate all the value(s) of #2 for which S will be a natural number. (Show
ALL calculations)
QUESTION 2
2.1 Given a quadratic number pattern: -120 ; -99 ; -80; -63:...
2.1.1 Write down the next TWO terms of the pattern.
2.1.2 Determine the 7“"term.of the number pattern in the form
T,=an' +bn+c.
2.1.3. What valué must be added to T, for the sequence to have only one
value of w for whith TZ, = 0?
2.2 Given a finite arithmetic series: 9+ 14+ 19+... + 124
2.2.1 Determine the generabterm of this series in the form T, =dn+c.
2.2.2 Write the series in sigma notation. nny
4000
2.3 Given the arithmetic series: 2° +2°" +3.2° +2"? +... danny 3

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Calculate the value of x if the sum of the first 30 terms add up to 3 720.

NW/September 2023

(3)
(3)
Q)

(5)

(6)

(5)
[25]

2)

(4)

)

(4)
[19]

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QUESTION 3

3.1 Given: 5; 10; 20: ... a geometric sequence.
3.1.1Metermine the n” term.
3.1.2-—€alculate the sum of the first 18 terms.

3.2 The first and second term of a geometric series is given as (2x—4) and
(4x° -16) respectively. Determine the value(s) of x for which the
series will converge.

3.3. Aconvergent geometric series has a first term of 2 and r=

Calculate So :

2

QUESTION 4
Given: f(x)= 7 _»
x~1

4.1 Write down the equations of the asymptotes of f-

4.2. Draw the graph of /f. Clearly label ALL the intercepts with the axes and the
asymptotes on your graph.

4.3. Determine the equation of the line of symmetry of f for m<0.
44 — For which values of x will f(x) <—4?

QUESTION 5
The graph of g(x) = af 3) —5 passes through A(—2 ; —4).

5.1 Show that ae.
25

5.2. Determine the coordinates of the x-intercept of g.

5.3. Given: ms)-(4) :

5.3.1 Determine the equation of A(x) , the inverse of h, inthe form
yo...

5.3.2. Describe the translation from g to h.

qd)
(2)

(4)

(3)
[20]

@)

(A)
(2)

(2)
10}

2)
(3)

Q)

(2)
19

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QUESTION 6

The sketch below represents the following graphs: f(x) = ax’ 4 bx+¢ and

x)=-xdQT
gtx) 2 “in
* Tha of symmetry of f cuts the x-axis at B, which is also the
x ceptof zg.

® =27'0) is an x-intercept of f.
e PO; -12) is the »-intercept of /.

eh

6.1 Calculate the coordinates of B. @)

6.2 Determine the equation of f in the form f(x) =ax’ +bx+c. (4)

6.3 If f(x)=-x’ -8x—12, determine the coordinates of G if FH, the vertical length

between the points of intersection of f and g, is a maximum. 4)

64 For which values of x will x. f'(x)>0 ? (2)

Saar [12]
dena?
ura

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QUESTION 7

7.1 Convert an effective interest rate of 11,3% p.a. to its equivalent nominal rate per
annum;-compounded quarterly. G)

7.2 Lisa opened a savings account and deposited R10 000 immediately into the
account=The account paid interest at 5,3% per annum, compounded monthly.
She Started making additional monthly deposits of R500 into the account three
months after the account was opened. Her last monthly deposit of R500 was
made 5 years after the account was opened.

How much money was in the account 5 years after the account was opened? (6)
7.3 Sam wants to buy a house and takes out a loan of R860 000. He can only afford
to pay R7 200 per month starting one month after the loan is granted. The

interest rate is compounded monthly at 9,5% p.a.

7.3.1 Calculate the number of payments that Sam will make to repay the

loan. 4)
7.3.2. How much will Sam pay in the last month to settle the loan? (5)
17]
QUESTION 8
8.1 Determine f(x) from first principles if f(x) =1-2" (5)

8.2 Determine:

82.1. D, [se°-2] @)
x

8.2.2 2 if y=Vx(Vx-sx) “)

[12]

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QUESTION 9

The graph of _f(x) = ax® + bx* has stationary points at p(2 ; -4) and O(0; 0)

iN

NW/September 2023

9.1 Prove that qe and pei.
12 4

9.2 For which values of x will f be concave down?
9.3 Determine the equation of the tangent to f at x=—2 in the form J, =mx+n.
9.4 Use the graph to determine the values of k for which

Jeg 12 =0 will have three real roots.

12 4

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(A)

(3)
(4)

2)
[13]

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QUESTION 16

NW/September 2023

In the sketch below, AABC is an equilateral triangle with side AB = m units.

DEGF isa reptangle with BF = GC = x units.
Qo A

D E
B F G Cc
10.1 Prove that the area of the rectangle DEGF is /3x(m—2x) (3)
10.2 Determine, in terms of m , the maximum area of the rectangle. (4)
7)
QUESTION it
11.1 Given: P(A)=0,3, P(B)=0,43 and P(C)=0,18.
The events A and B are independent.
The events A and C are mutually exclusive.
11.1.1. What is the probability that A and C will happen simultancously? (1)
11.1.2 What is the probability that at least one of A or B will take place? GB)
11.2 A group of 250 Grade 12 learners participated in a survey. They were asked if
they used the social media app TikTok. The results are represented in the table
below:
Use TikTok Don’t use TikTok Fotal
Girls 105 25 130
Boys 68 52 120
Total 173 77 cei?
AUD]
11.2.1 A name is drawn randomly out of the group. What is thelQ]
probability that the name drawn will be a girl and using! Tok? 1)
11.2.2 Are the events being a girl and using TikTok independent? Motivate
your answer by doing appropriate calculations. (4)

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11.3 A group consisting of 5 boys, 6 girls and a teacher goes to the movies. They
buy 12 tickets in a row next to one another. One of the seats is next to an aisle.

11.3.1_In how many different ways can they be seated? qd)

To avoid disruptive behaviour, the teacher considers to place the
following restriction on how they may sit:

~The teacher at the end next to the aisle, and the children in any order as
long as Peter and John do not sit together.

Calculate in how many different ways they can be seated. GB)
11.3.3. Duncan wants to sit next to Suzie. If the whole group of twelve are
seated at random, what is the probability that the two of them will sit
together? (3)
[16]

TOTAL: 150

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ema NW/September 2023
INFORMATION SHEET: MATHEMATICS
_ -btvb* —4ac
A= P(i A= P(i—ni) A=P(t~iy” A= P(+i)"
T, =a Da 8, => Qa+(n-d)
T, =ar"™" Ss _ ar’ -1) 3 rel S,=-;-ler<l
. r-l l-r
pansy -1 poal-+iy")
F —_—
2 F(xth)— f(x)
yy = lim Lets)
Le) h>0 h
XX Yt+y.
dae —%) +0, -HY Mj 2 ; bo
2 2
y=mxte yoy, =m(x-x,) m=227*1 m=tan@
xX) —X,

(x-a¥ +(y—-bY =?

mnaspce; —2_=>_._©
sn A sinBo sinc

area AABC = 5 ab. sm C

sin(a + 8) = sina.cos f + cosa.sin 8
cos(a + h)= cosa.cos # —sina.sin
(cos? a-sin’ a

cos 2a =<1-2sin? @

[2c08* 1
_ x
a aie
n
n(A)
P(A) =
(A) lS)
poathx

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a? =b? +¢7 —2be.cos A

sin(@ - £) = sin a.cos 8 - cosaz.sin B

cos(a - f)= cosa@.cos § +sin a.sin §

sin 2a = 2sina.cosa

a 2

XG _ x)

o = 4}

W

P(A or B) = P(A) )— P(A and B)

pe ee FP)
dSi@-xY

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education

Department:

Education

North West Provincial Government
REPUBLIC OF SOUTH AFRICA

These marking guidelines consists of 16 pages.

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NOTE:

NSC-Marking guidelines

e Ifa candidate answers a question TWICE, only mark the FIRST attempt.

e Consistent Accuracy applies in ALL aspects of the marking memorandum.

QUESTION?
Ll | x Ls¢
x LSQk o ¥ standard form
x(¥=5) -0) Y factors
0 ~ v x-values (3)
x=0 or x=5
1.1.2 | x?-2x-13=0
2 Y substitution into the
~(-2)+ {2-413 su
x= (2) tVC2) OC!) correct formula

2(1)

Vx = 4,74
x=4,74 or x =—2,74 vx =-2,74 (3)
1.1.3. | (x-2)(1-x)<0 ae
Y critical values
xSlorx22 / xé(-;lJor[2;0) Yx<l
Vv x>2
(3)
OR
(x-2)(x-1)20
xSlorx22 / xe(—c;ljor[2;0)
114 | 2/2x-t=x-11
4(2x-1)=x° -22x+121 ¥ square on both sides
8x-4=x° —22x4121
x? 30x +125 =0 Y standard form
(x—25)(x-5)=0 ¥ factors/formula
x=25orx¥5 ¥ both answers
¥ reject x=5
OR (5)
2V2x-1l=x-11
x1 =
2
2x-1= x= 2244121 v square on both sides
4
8x—4=x° —22x +121
x’ -30x+125=0 Y standard form
(x-25)(x-5)=0 Y¥ factors/formula
-2 Y both answers
x= 25 orx#5 Y reject x=5 (5)

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1.2 3x-y=4
y=3x-4

x +xy =24
x cE 4)=24
—4x =24

“R x-24=0

x56 =0
(x-3)(x+2)=0
x=3 or x=-2
y =3(3)-4 or y=3(-2)-4

Y y=3x-4

¥ substitution
Y standard form

Y factors/formula

Y both x-values

y=5 y=-10 Y¥ both p-values
(6)
OR
3x=y+4
-2+4 VY xa t4
3 3
x +xy = 24
2
(4) +y(234)-24 ¥ substitution
2 2
y +8yt16 y +4Y 44
9 3
y +8y+1643y? +12y =216
Ay’ +20y—200 =0 Y standard form
y +5y-50=0
(v+10)(y-5)=0 Y factors/formula
y=-10 or y=5 Y both y-values
_(10)+4 +4
3 3
xy=2 x=3 Y both x-values (6)
* [se(ea ras} le)
a8 “O)ale}-
Gh (s}-C5) i
S= —_ —_ mt+1
7 9 \
7 ry
For S to be a natural number, m+1 must be a
multiple of 7 v fae
m+1=14 or m+1=21 or m+1=28 les 14,21,28
m=13 m=20 m=27 v answer (5)
[25]

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QUESTION 2
2.1.1 | -120 -—99 Ww Wy
~19 17
-2 -2 Y 48
t TWO terms: —48 ; —35 y -35 (2)
2.1.2 3a+b=21 a+b+c=-120 v 2”4 diff = —
3(-l)+b=21 (-1) +(24)+e=-120 Ya=-l
b=24 c=-143 vb =24
T, =-n? +24n-143 Y c= 143 (4)
2.1.3 | T'n=-2n+24=0 ¥ method
n=12 v¥ n=12
T, =—(12)? +24(12)-143
1,=1
A maximum of | v maximum |
Add -1 to Tn v-l
(4)
OR
_ 74) _
“21 v¥ method
T, =-(12)? +24(12)—143 ¥ n=12
[,=1
A maximum of 1 ¥ maximum |
Add -1 to Tn y-l (4)
OR
T, =-W +24n-143+k Y method
= (24) —4(-1)(k -143)
=576+4k —572
=4k+4 v A=4k+4
but A=0 YA=0
4k+4=0
k=-l ¥ _ (4)
2.2.1 | 94144+19+4...4124

Answer only:
full marks

T, =(9)+ (2-5)

T, =5n+4

v avon into the

rétt formula

7 asne4 (2)

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22.2 | 7 =5n+4=124 Y =124
5n=120
n=24 ¥ n=24
Ds 4) Y answer (3)
23 | 2 a
o Ea VY a=2*and d=2*
30 + x
Soo =F )+29(2 | Y substitution into the
3720 = 15(31.2") correct formula
248 = 31.2"
se24 Y 2° =8
2=2*
x=3 Y x=3 (4)
[19]
QUESTION 3
3.1.1 5;10;20;....
T,=ar"'
T, =()(2)"7 v answer (1)
3.1.2 _a(r"=1)
" r-l
5] (2)"*-1
f= s[e"=1] Y substitution into the
2-1 correct formula
Sig = 1310715 v answer (2)
3.2 22x44) (2x4) 4g Y r=2x+4
ed
Converge: -1< r< 1 Y -ler<l
-l< 2x+4<1 Y substitution
-5< 2x< -3
-2 <x< -5 Y answer (4)
anal
dona]
py

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3.3
v8,
v S$,
Y dividing
Y answer (3)
[10]
QUESTION 4
4.1 x=l Y x=l
y=-2 Vv ye? 0)
42
‘ | . v x-asymptote and
iV y-asymptote
v x-intercept
v y-intercept
v form-decreasing (4)
43 ya-xte OR y-y=-I(a-x)
—2=-(I)+e y=-x+1-2 v¥ method
re yeoxet v answer (2)
y=-x-l
Answer only:
full marks
4.4 Osx<l Y Oandl
v inequalities (2)
OR
¥) id 1
xeL0;1) ackets (2)
gy [10]
—

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QUESTION 5
5.1 1y"
=a|—| —5
y o( 2]
2
-4 4 -5 Y substitution
5) Y simplification (2)
5
5.2 x
lf
0=—|=] -5 =
#(5) ¥ y=0
s-3(3)
25\5 Y simplifying
125= (3
53 =5-*
x=-3 (-3:0) v answer G)
5.3.1 hive ly
Dye 5
LY Answer only:
«=-(4] full marks Y swop x and y
y=log, x or y=—logsx Y answer (2)
5
gy 3) (5
ny?
ec=(4] -5 Y rewriting g

2 units right and 5 units up

¥ 2 units right and 5 units up
(2)

19]

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QUESTION 6

6.1 I
= 2x42
vase

Y y=0

¥ answer (2)

6.2

x, )(x-x))
yaales 2)(x+6)
“D=c(0 +2)(0+6)

-12=12a
a=-l

ye-(x? +8x+12)
y=-x’ -8x-12

OR

(-2;0): 0 = a(-2)? + b(-2)-12

12=4a—2b
6=2a—b.......{1)
and ~_ 4
2a
—b=-8a
b =8a.....(2)
6=2a-8a
6=-6a
a=-
b=8(-1) =-8
y=-x’ -8x-12

Y coordinates of
A(-6;0)
Y substitution

¥Ya=-l

v answer (4)

¥ substitution

Y substitution

Ya=-l1

v answer (4)

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63 FH =x 8x-12-( 3702] Y FO)- 9%)
—-x* ~8x-12-1x-2
2
Taal Y FHitox
di -oy—17 _9 or yb v¥ method
2 2a
yy eT (85)
2 2(-l)
17 17
x=-— x=-—
4 4
17
o(-Z0) v answer (4)
6.4 4<x<0 v answer
Y inequility (2)
OR
v -4and0
xe(—4;0) Y inequility Q)
[12]
QUESTION 7
TA i” ”
1+i, =| 14+—
om m
i')’
1+0,113 [4] Y substitution
i
1,027...=1+7 Y 4" root
it
0,027...=—
4
0,1085..=i'
r=10,85% Y answer (3)

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7.2

Y substitution into the

0,054..= (14°95
12

—N = 104 097.) 9,054...
—n =—369, 212...
Sam will have 370 installments

0,053)"
A= 100001 + oe) correct formula
A=13026,71 ¥ n= 60 and (= 22>
Y substitution into the F
fanny formula
TT ¥ n=58
lon
Ee 970,51
Total =13026,71+32970,51
= R45997,22 Y answer (6)
OR
500] (14 0,053)" -] Y¥ substitution into the
0,053)” 12 correct formula
=10000{ } + 0.053 0,053
U9 v¥ n=60 and i=
12
Y substitution into the F
formula
v n=58
= R45997,22 ¥ answer (6)
=n Y substitution into the
7.3.1 ram) 1 | correct formula
-—__t_ . 0,095
860000 0,095 v i<~
12
0,945... -1-(14 228)
12
—0, 054... = (i + os |
12

Y correct use of log

v answer (4)

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dana?
foan!
fag

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7.3.2

309
A= 860.000{1 + oe)

A= R15782859,31

12
F = R15781334,69
Balance = R1524,62 (after 369 installments)

1
Last installment = 1524, 62 [ + o, 7)

= R1536,69

OR

> 0,2127679...
7200 i-{1+ "|
12

P= 0,095
12
P= R1524,62 (after 369 installments)

i
Last installment = 1524, 62 [ + o, 7)

= R1536,69

Y¥ loan and interest
¥ n= 369

Y payment and interest

v method

v answer (5)

v¥ method
v¥ n=-0,2127679...

¥ balance

¥ method
v answer (5)

[17]

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QUESTION 8
8.1 S@=l-x
=1-(x+h)
=1-(x° +2xh+h?)
=1-x?-2xh-h? ¥ =1—-x?-2xh—h?

f (x+h)-f (x)
h

1—x° —2xh-h’ -(1-x°)

= lim h Y substitution
—2xh—W*
_ -2xh-h?
ne) h v “h
- h(-2x-h)
— lim F Y -2x-h
= lim(—2x—h)
f= 2x v —2x (5)
8.2.1 bax =2]
x
Y —2x71
D. [3x7 -2x""] v 6x
=6x+2x? ¥ 2x7? @)

82.2 | yavx (ir —5r)

it -
yar? fi *4

> 3
yH=xo=Sx?
1 T
35 N53
dx 6 2

Y change from surd to
exponential form

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QUESTION 9

9.1

f (x)=ax’ +bx*
-5 pay +60)

SF (x) = 3ax’ + 2bx
f'(2)=3a(2)? +25(2) = 0

12a+4b=0
4b =-12a
b =-3a.....(2)

1
—==8a+4(-3
3 oe (—3a)

v substitution (2 3 -1)

Y f(x)=0
Y substitution x=2

Y solve simultancously

1 =8a-124
3
14g
3
1
a-—
12
1
b=—7 (4)
" 1 1
9.2 f (x)J=5%-5 Y fe)
1 1
3775 <0 Y f"(ay<0
x<l / xEe(-0 31) v answer (3)

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93 ~t. la
FO D* 4
Ion, 15 5 Y y=?
—?2) =—(-?)’ ——(-2)° =-= y=
S-) a! ) 4! ) 3 3
:-3)
> 3
(xjste—ty v gradient
4 2
1 | ¥ substitution of xy and
(2) =—(-2)? —=(-2 substitution of x,y and m
S'(-2)= 402-52)
m=2
y=2xt+e
32-2) +e
3
7
c==
3
y=2xt— v answer (4)
9.4 1 1
-<k<0 / ke -330 vv answer (2)
[13]
QUESTION 10
DF
10.1 tan 60° = — = v3
an x 8 Y tan60° = V3
DF = 3x Y DF =%3x
«. Area rectangle = DF x DE Y DE=(m-2x)
=V3x(m—2x) GB)

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10.2

Area =V3mx —2V3x2
“ = 3m -4V3x =0

x=—
4

M a= V3x(m —2x)

-5(F("--())

Be
m

8
OR

dA _
4 = Bm—aV3x=0
a Nee

m
x=—

4
V3mx —2V3(x)

2

m m
— Bm? 3m?
4. 8
_ 23m? —8V3m?
7 8
_ 3m’
8

Y fiy=0

v substitution

N32

vm (4)

Y f(x)=0
vy”
4

v substitution

m (4)

[7]

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QUESTION 11

11.1.1 | P(AandC)=0 v answer (1)
11.1.2 | P(AandB) = P(A)*P(B)
P(AgndB) = 0,3x 0,43
P(Adid B) = 0,129 Y P(AandB) =0,129
P(AGEB) = P(A) + P(B)—P(A and B)
P(AorB) = 0.3 +0,43-0,129 ¥ substitution
P(A or B) = 0,6 v answer (3)
11.2.1 P(GandT) = 105 = 21 < gv, Y answer (1)
250 50
11.2.2 | Independent: P(T and G) = P(T)x P(G)
P(T)xP(G) P(GandT) Y PT)= 173
173 _ 130 105 250
550 ~ 950 550. 130
250 250 250 Y P(G)=——
= 0.36 = 0.42 250
P(GandT) # P(G) x P(T) Y P(T)xP(G)=0,36
Events are not independent. ¥_answer (4)
11.3.1] 12! Y answer qd)
11.3.2 | Pieter and John next to one another =10!2! v 10!2!
11!-10!2! v 11!10!2!
= 32659200 ¥ answer (3)
11.3.3) 1112!
12! v 11!2!
1 v 12!
= 6 Y answer (3)
[16]
TOTAL: 150
son]

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