# Mathematics Grade 12 Paper 1 KwaZulu Natal Mock 2022 Memo

> Grade 12 Mathematics Paper 1 memorandum from the KwaZulu Natal Mock Exam 2022, 22 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/maths-grade-12-sept-2022-p1-and-memo/3b177b3f-c4bc-4629-a2fd-9e9f220c16a1
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2022
- Exam period: KwaZulu Natal Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 22

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### Page 1

othe RE SABER.

¢ MATHEMATICS Bi

PREPARATORY EXAMINATION

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SEPTEMBER 2022

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MARKS: 149
TIME: 3 hours

This question paper consists of % pages and 1 information sheet,

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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.
1. This question paper consists of 13 questions.

2, Read the questions carefully.

3. Answer ALL the questions.

4. Number your answers exactly as the questions are numbered.
5, Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining
your answers.
6. Answers only will NOT necessarily be awarded full marks. G

7, You may use an approved scientific calculator (non-programmable and wii,
unless stated otherwise.

8. If necessary, round off answers correct to TWO decimal places sins stated otherwise.
«

eS

5

cS

9. Diagrams are NOT necessarily drawn to scale.

10. Write neatly and legibly.

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Mathematips\Pin loaded from Stanmor ephysics. CORN Preparatory Examination 2022
NSC - GRADE 12

QUESTION i
1.1 Solve for x:
LLL @+5)@x-1)=0 (2)
112  -3°~7x=-8 (correct to TWO decimal places) (4)
1130 Vx+5+l=x 6)
1.14 (2x ~3)(x+5)<0 GB)
1.2 Solve for x and y simultaneously if: &
& ©)
x+3y=5 and xy+y?-3=0 . 2”
1.3 Simplify fully, without the use of a calculator: x
aT Ey)
eed Eee vea where @., : (4)

& [24]
QUESTION 2 Ke

Given the quadratic number pattern: 5 ; 9 cf 29 3.

21 Write down the 5" and 6” &. of the pattern. (2)
2.2 Show that the n'* tempafeshe quadratic pattern is given by 7 =2n*—2n+5 (4)
2.3 Is 2023 aterm OE patent Motivate your answer with relevant calculations, (4)

[10]

\
RY

QUESTION 3 )

Evaluate: 2 (30-4k) [4]

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mc mr BUEN H

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Mathematic# OWNloaded from Stanmorephysics. eayFreparatory Examination 2022
NSC — GRADE 12

QUESTION 4

4.1 Given the geometric series a + ar + a + ar +.., where a is the first term
and r is the common ratio. Prove that the sum to » terms of this series is given by

se a(r" -1)

na

srel
rol 4)

42 The first two terms of a geometric sequence with constant ratio r , and an arithmetic
sequence with constant first difference , d, is the same. The first term is 12.

4.2.1 Write down the second and third terms of EACH sequence in terms
of d and r. (2)

4.2.2 Ifitis further given that the sum of the first three terms of the geometric
sequence is three more than the sum of the first three terms of the
arithmetic sequence. Determine two possible values of the common ratio,
r, of the geometric sequence.

(5)
[7]
QUESTION 5
Sketched below is the graph of A(x) =1-5 . Ais y-intercept and B is the x-intercept of A.
x-

5.1 Write down the equations of the asymptotes of h. (2)
5.2 Calculate the coordinates of A and B. Q)
5.3 Write down the equation of the line of symmetry of h with positive gradient. (2)
5.4 Write down the range of A. Q)
18]

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QUESTION 6
SC 2; 0) and T(6 ; 0) are the x-intercepts of the graph of f(x) = ax’ +bx+c; a#0.
R is the y-intercept of f and g.

The straight line through R and T has the equation g(x)=-2x+d.

6.1 Calculate the value of d.
6.2 Show that I(x) =-x? +4x412.

6.3 Calculate the coordinates of the turing point of /
6:4 Determine for which values of x will:

6.4.1 F(x)- g(x) 20

6.4.2 xf (x) <0

6.5 Determine the coordinates of R' »the image of R on p if p(x)= —f(x-2),

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(2)
(4)

(3)

(2)
(3)

(2)
{16}

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Mathematied OWNIoaded from Stanmorephysics. saymreparatory Examination 2022
NSC — GRADE 12

QUESTION 7

Sketched below is the graph of p(x)=a";a>0; a% 1, The graph intersects the
y-axisat A. The point TS ; 32) lies on p.

a x
7A Write down the coordinates of the point A. (2)
7.2 Calculate the value of a. (2)
73 Write down the domain of p. (1)
7.4 Write down the equation of p', the inverse of p, in the formy=... (2)
75 Determine the values of x if p'(x) <5. (2)

9]

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“Mateyeipffloaded from Stanmore HY SICS. COM KZN Preparatory Examination 2022
NS'

— GRADE 12

QUESTION 8

8.1

8.2

Dipinda opened an account with an amount of R5000 on 1 June 2021. She then makes
monthly deposits of R600 at the end of every month. Her first deposit is made on the 30
June 2021 and her last deposit on 30 April 2023. The account earns interest of 14.25%
per annum compounded monthly. Calculate the amount that is in the account directly
after her last deposit is made into the account,

Molly wants to buy a house for her family for R800 000. She agreed to pay monthly
instalments of R10 000 on the loan which incurred interest at a rate of 13,35 % per
annum compounded monthly. The first payrnent was made at the end of the first month
after the loan was granted.

8.2.1\ Show that the loan will be paid back in full after in 200 months,

8.2.2 Suppose Molly encountered unexpected expenses and was unable to pay any
instalment at the end of the 120", 121", 122" and 123" months. At the end of the
124" month she increased her payment to still pay off the loan in 200 months by
making 77 equal monthly payments.

a) Calculate the balance on the loan after the 119** payment was made.

b) Calculate the new monthly instalment Molly must pay from the 124%
month to settle the loan in 200 months.

QUESTION 9

9.1

9.2

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Determine (x) from first principles if f(x) = = .
x

Determine:

92.1 g(x) if g(x)=(x+7)

ay, 4
92.2 — if pave

(6)

(4)

(3)

(4)

[17]

6)

(5)

(4)

[14]

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Mathematiesp1DOWNloaded From Stanmoerephysicszwerggnratory Examination 2022
NSC — GRADE 12

QUESTION 10

10.1 Given: f(x)=x° -12x-16

10.1.1 Calculate the x-intercepts of the graph of f- (5)
10.1.2 Determine the coordinates of the turning points of i (4)
10.1.3 Sketch the graph of f, indicating the intercepts with the axes and the

coordinates of the turning points. (4)
10.1.4 Determine the values of x for which the graph of / is concave up. (2)

10.2 Given: p(x)=—-x° -8x

Is it possible to draw a tangent with positive gradient to the graph of p?
Show all calculations to justify your answer. GB)
(18]

QUESTION 11

The lead, L, in metres, of a runner in the comrades’ marathon in the last ¢ minutes of the race,

where ft € [0:75] is given by equation:
2

L=1000+6¢-—

4

_ dL
11.1 Determine 7 2)
11.2 Calculate the time at which the runner has the greatest lead. (2)
11.3. At what sate is the runner’s lead decreasing when 1= 60 minutes? (2)
16]

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Mathomppticy Pi lad) from Stanmorephysies, COMEZN Preparatory Examination 2022

QUESTION 12

Given the word “BRACKET”. The letters of this word are randomly arranged to form new
arrangements of the letters,

12.1 How many unique arrangements of the letters can be made? (2)
12.2 Determine the number of unique arrangements of the'letters that aré randomly made
if the letters R and A must be together. (3)
{5}
QUESTION 13

13.1 Two different events A and B are mutually exclusive.
It is further given that:

e 3P(B)=P(A)
© P(A orB) = 0,63

Calculate P(B). (3)

13.2 Two cards from a regular pack of 52 playing cards (shown below) , are drawn at random
one after the other, without replacement.

a Ge ia Baa ar
* HI
: |
* f
Sgt Fi
e
&

ey ;

Hf oe ye aie 8}

Calculate the probability that:

13.2.1 both cards are picture cards. (2)
13.2.2 at least one of the cards is a picture card. (3)
3]

TOTAL: 150

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NSC - GRA 1

INFORMATION SHEET: MATHEMATICS

_-b tb? —4ac

2a
A=P(+ni) A=P(I-ni) A=P(-i)’” A=P(I+iy’
T, =a+(n-N)d =" 2a+(n-Nd}
my
T, =ar"* S, mau ) rel S, = 3-l<r<l
. r-l l-r
pil) -1| pa2lt+0"]
I L
\ _ | f(xt+h)- f(x)
= lim SE
F®) 50 h
d=y(x,-x) +(,-9,) M2427.) m= tan
y=mxte Y-y =M(x-H) ee
otal

(xa) +(9-)'=
b c

In AABC; —2- =—- =—
sinA sinB sinC

a? =b* +c7 -2be.cosA

area ABC=5absinC
sin(a+§) =sina.cosB + cosa.sinB sin(o.—B) =sina.cosB —cosa.sinB

cos(a +B) =cosa.cosB —sina.sinB cos(a—B) = cosa.cosB +sin a.sinB

cos’ o—sin’ a

_ 2 . .
cos 20, =41-2sin* o sin 2a = 2sina.cosa

2cos’ a-1
1" 2
X(%,-*
x= Le w= 2( )
nA n
P(A) = 8) P(A or B)=P(A)+P(B) —P(A and B)
- p= eH =P) ¥)y-F)
yp=atbx “Sa =a

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education

Department:
Education
PROVINCE OF KWAZULU-NATAL

NATIONAL
SENIOR CERTIFICATE

GRADE 12

MATHEMATICS P1

PREPARATORY EXAMINATION

SEPTEMBER 2022

MARKING GUIDELINE

MARKS: 150
TIME: 3 hours

NOTE:

If a candidate answered a QUESTION TWICE, mark only the FIRST attempt.

If a candidate crossed out an answer and did not redo it, mark the crossed-out answer.

Consistent accuracy applies to ALL aspects of the marking guidelines.
Assuming values/answers in order to solve a problem is unacceptable.

This marking guideline consists of 12 pages.

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NSC Marking Guideline

September 2022 Preparatory Examination

QUESTION 1
_ _i vo- vy i
Lit |x=-5 or x=5 A 5 A ; 0
1.1.2 | -3x?-7x+8=0 AY © standard form
—b +~vb? — 4ac (Penalize 1 mark
x= a if rounding off is | | CA¥ correct substitution into
—(-7) + JC)? — 4€3)(8) incorrect — once quadratic formula
x= 2(—3) here for entire
x= 3,17 or 0,84 paper) CAY CAY answers
(4)
13 |Veee41=x
Vvx+5=x-1 . .
(vexs)" = («—1) AVY _ isolating surd
x+5=x7-2x4+1
x? —3x-4=0 CAY standard form
(x + 1)(x-4) =0 CAY factors
x=-lorx=4 CAY x#-1
nla CAY x=4
()
114 |5<z7<3 AY critical value — 5
i. AY critical value ;
CA¥Y interval notation
(3)

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NSC Marking Guideline

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September 2022 Preparatory Examination

120 [x+3y=5 (1)
xy+y?-3=0 _+(2)
From (1): x =5=3y > (3) Av making x the subject
Substituting (3) into Q):
yS-3y)+y?-3=0 CAV’ substitution
—2y?+5y-3=0
2y? —5y+3=0 CAY standard form
(2y —3)(y—1) =0
3 CAY factors
yaa ory=1 CAY y-—values
x5 Or x= 2 CAY x-—values (6)
OR
x+3y=5 > (1) OR
xy+y?-3=0 -(2)
From (1): y= = > (3) V : .
Substituting (3) into (2): Av making y the subject
_ 2
x ( 5 *) + ( 3 *) —3=0 CAV’ . substitution
— — 2 _ =
3x(5—x)+25—-10x+x*-27=0 CAV’ standard form
—2x7+5x-2=0
2 =
2x°—Sx+2=0 CAV factors
(2x —1)(x -2) =0
x=5 orx=2 CAY x-—values
3 1 CAY ]
=s ory= -
y=5 y y— values 6)
13 | afton + ane2
52n + 4,57
_* 27,5" + 2", 22
~ | 524 4,57
_" 2(5" + 4) AY  factorising numerator
57(5" + 4) AY factorising denominator
n/2n
= lon CAY’ simplifying
2
== CAY answer
: (4)
[24]

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NSC Marking Guideline

QUESTION 2
2.1 45 ; 65 AAY Y answers (2)
2.2 5 9 17 29
1D \ / 8 12
2D 4 4
AY 2a=4
2a=4 .a=2 AY az
3atb=4 «b=-2 AY 3a+b=4
atb+c=5 “~c=5 AY a+b+c=5
Ty = 2n? —2n+5 (4)
2.3 Tn = 2n? — 2n +5 = 2023 AY — equating n" term to 2023
2n* — 2n — 2018 = 0
n? —n—1009 =0 CAV standard form
1+¥v1+ 4036
n= —_,= 32.27 or — 31.27 CAY n-—values
Since n is not a Natural Number, 2023 is not a term of the sequence. | CA Y conclusion (4)
[10]
QUESTION 3
26;22;18;... AV n~—value
_? _ AY — a~—value
Sn = ze + (a Dd] AY d-value
Sso = [2(26) + (50 ~ 1)(-4)] CAW answer my
Ss0 = —3600 OR
OR AY n—value
Sn = n [a+ Tn] AY a—value
2 AY Tso — value
50 v
Sso = > [26 + (-170)] CAV answer (4)
Sgo = 3600
[4]

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NSC Marking Guideline

September 2022 Preparatory Examination

QUESTION 4
41 Sy =atartar?+are+--+ar™ + (1) AY — equation (1)
rS, =ar+ar? +ar?+--+ar™1+ar" = (2) AY — equation (2)
(2)— (@):
TS, — Sy =ar™-—a AY © subtracting LHS and
Sn(r — 1) = a(r™ — 1) RHS terms
a(r™ — 1)
Sn=— aq ir el AY  factorising
7 (4)
4.2.1 | AS: 12+d;12+ 2d)... AY AS set up of terms
GS: 12r;12r?;... Av — GS set up of terms (2)
4.2.2 |12+d=12r > (1) AY equation (1) and (2)
36+ 3d4+3=124+ 1274 12r2 = (2)
From (1): d=12r—12 ~ (3) AY — making d the subject
Substituting (3) into (2), we have
36 + 3(12r — 12) + 3 = 12 42 + 12r2
12+ (12r -—12)+1=4+4r+4r?
4r2 —8r+3=0 CAV’ standard quadratic form
(2r —1)(2r -—3) =0 CAV factors
1 3
r= orr=5 CAY answers
(5)
[11]
QUESTION 5
5.1
yell
AY x=2
=2andy=1
x andy AY y=1 Q)
1 v i
5.2 y — intercept : ( 0:1 ) A y-intercept
x — intercept: x = 3 AY x=3
(3; 0) CAY coordinate form (3)
5.3 }y=x-1 AY Gradient value (2)
AY y-— intercept
5.4) yER y#1 AY answer qd)
OR OR
y € (—0; 1) U (1; 00) AY answer ()
[8]

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NSC Marking Guideline

September 2022 Preparatory Examination

QUESTION 6
6.1 0=-2(6)+d AY © substitution of point T(6 ; 0)
d-12 AY — answer (2)
Answer only full marks
6.2 c=12
y = a(x + 2)(x — 6) AY - subst. x — intercepts
12 = a(0+ 2)(0- 6) AV © substitution R(0 ; 12)
12 =—-12a
a=-1 AY a-—value
y= 1x + 2)(x— 6) AY — a—value and factors
f(x) = —-x? +4x +412 (4)
OR OR
x = 2...Axis of symmetry AY axis of symmetry value
y= a(x —2)?+q.....(A)
Substituting (0;12) into (A)
12 =a(0—2)?+q >12=4a+q...(1) | AY © substitution R(O ; 12)
Substituting (6;0) into (A)
0=a(6-2)?+q 70=16a+q.....(2)
(1) -(@):
12 =-12a y
-t=a A a—value
16=q
y = —1(x — 2)? + 16 AY — a—value and TP (4)
f(x) = —-x? + 4x4 12
= 77 - <—
6.3 2x 4+4=0 or x= or x= 2+6 A Using calculus/formula/midpoint
AD 2
x=2 . ae
CAY x=2 (provided it is +ve)
y = —(2)? + 4(2) +12 = 16 CAY y-value
(2 ; 16) (3)
6.4.1 |O<x<6 AAY “answer (2)
(penalty 1 mark for incorrect notation)
6.4.2 —2<x<0 or x>6 AAYY-2<x<0 AY x>6 (3)
6.5  |R/(2;-12 ) Avx— value (2)
AY y—value
[17]

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September 2022 Preparatory Examination

NSC Marking Guideline
QUESTION 7
7.1 A(0;1) AAY V answer (2)
7.2 y=a*
32=a° AY _ substitution of point T(5;32) 5
a=2 AVY answer (2)
7.3 xeER AY answer (1)
OR OR 1
x € (—00; 00) AVY answer i)
7.4 y = log, x CACAV V answer (2)
75 log,x=5 CAY end points
AY _ interval
x= 25 =32
O0<x< 32 Can be solved by log inequalities. (2)
Answer Only — Full marks
[9]
QUESTION 8
8.1
1 Jun 31 Jul 31Aug | oe 30 Apr
— 2021 2021 2023 A— formula
600 | 600 | 600 | sree 600 Av value of n
Dipinda’s final amount in the account: AY value of i
x[1+i"-1
=P(1+i)"+ IC D ] FV — formula
03 AY — value ofn
14.25%
23 600 1+ = 1|
= 5000 (1 + ae + ( 2 ) CAY correct substitution into A
12 = CA correct substitution into Fv
v
= R22 321,54 CAY answer (6)
821 |) x-a+o 7)
i
10000 [1 — (1+ 235%)" AY value of P, x and
800 000 = u value of i
13.35% x
= AY ‘substitution into
13.35%\" 11 formula
1+ =) =—=0,11
( 12 100
—n= log, ,2225%) 0.11 AY use of logs
n = 199,5083362 AY decimal value
Therefore the loan will be paid off in 200 months. (4)
N.B. Candidates can also substitute the value of 200 into the Pv
formula to show that the loan will be paid in 200 months.

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NSC Marking Guideline

8.2.2a pe x{1-(4+i)]
~ i
13.35%
12 AY — value ofn

a AY — value of i (3)
= R530 009,55 CAY answer

10 000 [.- (1+

ee]

If n = 81 is used and P = R532 010,58
Give a maximum of 2/3 marks
N.B. Candidates can also use the method of A -Fv

8.2.2b [A = P(1 +i)”

13.35 %)\*
) AY — value of n

A = R530 009,55 (1 +=
A = R553 991,4839 CAY answer

1-a4+i7
p20 +04

13.35 %\~77
x[1-(1+ 12 ) | AY — value ofn

553 991,4839 = 13.35%

12 CAY answer (4)
x = R10 748,55

[17]

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NSC Marking Guideline

September 2022 Preparatory Examination

QUESTION 9 (penalize 1 mark once for incorrect notation in this question)
h) — v
9.1 f/Ge) = imo +h) — f(x) A formula
h>0 h
202
f/@) = lim AY © substitution
2x-2(xth) 1 CAY LCD
/ = lim —_* x =
MOO =lim—Seah “<h
CAV simplification of
Ie) =i 2x —2x—2h x 1 numerator
P@)= jim—se@th) A
—2 CAY answer
N@= x5 ™
3x (5)
OR
OR
AVY — value of f(x + h)
+h)=
forth) 3(x +h) CAY value of
2 h) — f(x)
h)— =—* + fart
FOr +h) — FO) 3(x+h) 3x
fO+h-fG)_— -2
h 3x(x +h) CAY value of Fonte)
(x) = lim ———
£@) h-0 3x(x + h) AY — formula
—2
"y) - (5)
fo) 3x2 CAY _ answer
9.2.1 | g(x) =(x+7)3
g(x) = x? + 21x? + 147x + 343 AAV V (two terms correct 1
g! (x) = 3x? + 42x +147 mark, all terms correct 2
marks)
CACACAYY V each term (5)
922 | pe_ 4
ye 9x2
s 4 0, AAVY
y=xz-=—x an .
9 writing in exponential form
dy 93 8s (4)
dx 2 9 CACAY Y each term
[14]

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NSC Marking Guideline

QUESTION 10
10.1.1 | x? —12x -16 =0
(x + 2)(x? — 2x -8) =0 AY _ binomial factor
(x + 2)(x + 2)(x -4) =0 AAV ¥ factors
x=-2 orx=4 CA CA ¥ ¥ answers (5)
10.1.2 | f(x) =x? — 12x —16
f/ (x) = 3x? -12=0 AY derivative and
x?-4=0 equating to 0
(x +2)(x-2)=0
x=-2 or x=2 CAV factors
y=0 or y=~-32 CAY x—values
CAY_y—values (4)
10.1.3
(-2:0) , a of
o CAY Maximum and
ji Minimum points
\ CAY x-—intercepts
\ f AY — y—intercept
us AY — shape
\
\
of
(32)
(4)
10.1.4 | f//(x) = 6x > 0 AVY —_.2™ derivative
x>0 Answer only: full marks AY answer @)
10.2 | p/(x) = —3x4*7+8 Av derivative
2
—3x° <0 forall xeR AY reasoning
—3x?-8<0
The gradient of all tangents to the graph of p is always AV reasoning
negative.
(3)
[18]

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Mathemapownloaded from Stanmorephysics.com
NSC Marking Guideline

September 2022 Preparatory Examination

QUESTION 11

11.1 t?
L = 1000 + 6t — —
4
aL 1
—=6-- AAY ¥ for each term 2
dt 2 (2)
11.2 | For greatest lead: “=0 cAv 4=9
at dt
1 or
6—-=t=0 equating derivative
2
to0
t = 12 minutes (2)
CAY answer
11.3 | dL 1 CAV’ substitution of t = 60
— =6--5(60) ; a
dt t=60 2 into derivative and
dL _ os value of — 24
dt r=60
The runner’s lead is decreasing at 24 metres per minute CAY conclusion (provided
the derivative is — ve) Q)
[6]
QUESTION 12
12.1 7! or 5040 AY AY 7! or 5040
(2)
12.2 6! x 2! AAY ¥6! x 2!
= 1440 Av 1440 (3)
[5]

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Mathemaporwnloaded From Stanmorephysics.com _ September 2022 Preparatory Examination

NSC Marking Guideline
QUESTION 13
13.1 | P(A or B) = P(A) + P(B) AY — condition for mutually
0,63 = 3P(B)+P(B) exclusive events
4P(B) = 0,63 AY — correct substitution
P(B) = 0,16 Av P(B) value (3)
2. 12 #11 12 v1
13.2.1 P(Both Picture cards) = — x — Ay 52° 54
52 51
1 i =
= 557 = 000498 = 4,98 % AY 337 = 0,0498 = Q)
4,98 %
13.2.2 fo 280 | Outcomes Probabilities
12.11
5 parade ie
11/51 P PP 52 x 51
P
- ae Pie | :
ay 40517) NP | PNP 28
§2 51
40 12
? 12/51 P NPP —x—
war 52° 51
MS —
os
39/51 NP NPNP ga x 39
52°51
P (at least 1 picture card)
= 1-P (mo picture card) AY Method
40 39
=1- (5 x =) AY Correct Substitution
AY —=0,4118=
7 17
=> 0,4118 = 41,18 % 41,18 %
OR OR (3)
P (at least 1 picture)
12 11 12 40 40 12 AAY Y probabilities
= (6*a) + * a) * (a) ,
7 AY oid 0,4118 =
= 0,4118 = 41,18 % 41,18 %
(3)
[8]
TOTAL: 150

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