# Mathematics Grade 12 Paper 1 KZN June 2023 Paper & Memo

> Grade 12 Mathematics Paper 1 question paper and memo from the KZN June Exam 2023, 31 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2023-kzn-june-exam-question-paper-and-memorandum-paper-1/6d3b56be-6f0c-4ea1-9ade-c024a853644a
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: KZN June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 31

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## About this document

Mathematics · Grade 12 · KZN June Exam · 2023. Question paper and memorandum, 31 pages. Read online or download the PDF.

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### Page 1

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

ete Seba ees eee eeEs Be eeeeee eee eS
« “
Ls 1"
“ rh
“ "
" 1"
“" "
“ "
"
" "
“" 1"
" "
Lo ”
TCTTTITTTTTTTrTe te Ae 7 Pree 2?

MARKS: 150

TIME: 3 hours

### Page 2

June 2023 Common Test

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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1, This question-paper consists of 11 questions.
Mn
Ds Answer A’ ral estions.
1
& Clearly show ALL calculations, diagrams, graphs, et cetera that you have used in

determining your answers.
4. Answers only will not necessarily be awarded full marks.

5. An approved scientific calculator (non-programmable and non-graphical) may be
used, unless stated otherwise.

6. If necessary, answers should be rounded off to TWO decimal places, unless stated
otherwise.

7. Diagrams are NOT necessarily drawn to scale.

8. Number the answers correctly according to the numbering system used in this

question paper. Write neatly and legibly.

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Mathematics P1

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QUESTION 1

1.1

12

1.3

Solve the following equations.

1.1.1 x=0

1.1.2 =—l1x+3 (give your answer to 2 decimal places)

1.1.3 x7 -2x<8

1.1.4 [2023 5, 9200 y
apa pie —xX=x

The roots of a quadratic equation are:

ae 10+/-m-8
y

, for which values of m are the roots non real?

A rectangle has a length of (2x+ y) metres and a width of y metres as

shown in the diagram. The perimeter of the rectangle is 24 metres and the
area is 32m’.

2x+y
1.3.1 Show that the expression for the perimeter is: 24=4x+4y

1.3.2 Show that the expression for the area is : 32 =? +2xy

1.3.3 Determine the dimensions of the rectangle if x> 0.

QUESTION 2

T, =—2n? +40n+103 is the general term of a quadratic sequence.

2.1

2.2

23

2.4

Determine 7, the first term of the quadratic sequence.
Determine the second difference of this quadratic sequence.
Which term of the quadratic sequence has a value of 301?

Which term is the largest term in this quadratic sequence?

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June 2023 Common Test

(3)
(3)
(4)

(4)

(2)

ql)
()

(6)
[24]

(2)
(2)
(3)
(2)
19]

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Mathematics P1 4 June 2023 Common Test

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QUESTION 3

3.1

3.2

3.3

The following sequence is a combination of arithmetic and geometric

sequence: 3 Bis Os: 6:5, 15:3 12s swecessses
3.1.1 ite down the next two terms. (2)
3.1.2 termine Ty —T,, (5)

The given circle is completely divided into n sectors in such a way that the
angles are in arithmetic sequence. If the smallest angle is 8° and the largest
angle is 52°, calculate 7 ,the number of sectors.

JN

(4)
George is currently 115 kilograms and his doctor has advised him to take up
a healthy eating plan so as to lose some of the weight in order to maintain a
healthy body. His new diet will enable him to lose 5 kilograms in the first
week, thereafter he would lose : of the previous week’s weight loss each
week.
33:1 If he follows the new eating plan, how many kilograms will he
lose on the eighth day? Give your answer correct to 3 decimal
1 (2)
places.
3.3.2 If he follows the new eating plan indefinitely, write his weight loss
in sigma notation. (2)
33:3 If he follows the new eating plan indefinitely, what will his new
weight eventually be? @)
[18]

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QUESTION 4

June 2023 Common Test

The graph of h(x) = = + p Passes through points (2;-5) and (0;—1).

4.1 Write d € equation of the vertical asymptote of h. (1)
4.2 Write down the domain of h. (2)
4.3. - Determine the values of a and q. (6)

4.4 Draw the graph of h, showing the intercepts with the axes as well as the
asymptotes. GB)
[12]

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Mathematics P1 6 June 2023 Common Test

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QUESTION 5

The graphs of f(x)=-2(x-3)(x+1) and g (x) =mx +c are drawn below.

Q is the x - intercept_of both graphs and E is the y - intercept of both graphs.
D is the turning poi 7 . SP is parallel to the y - axis.

x

51 Show that the coordinates of E are (0;6) @
5.2 Determine the coordinates of D, the turning point of 7 . (4)
53 If h(x) =—f(x +3), determine the coordinates of the turning point of h. (2)
5.4 Write down the coordinates of Q ()
5.5 Determine the equation of g . (3)
5.6 Determine the value of x, for which the length of SP between E and Q is a
maximum.
(3)
5.7 Determine the value of k for which f(x)=k has one negative an
positive roots. (1)
[15]

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Mathematics P1 ‘ June 2023 Common Test

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QUESTION 6

The graph of A(x)=ax’ for x20 is drawn below. Point A (1;3) lies on the graph of h.

y
h
A(1 ; 3)
5 x

6.1 Determine the value of a. (2)
62 Determine the equation of h' the inverse of h in the form y = ....... (2)
6.3 Draw the graph of h”', showing the coordinates of another point. (2)
64 Is A” a function? Motivate your answer. (2)

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June 2023 Common Test

Mathematics P 1 8
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QUESTION 7
7A Convert a nominal interest rate of 7,12% p.a. compounded quarterly to a

yearly effective interest rate. (3)
72 A trans; mpany purchases a new lorry to the value of R 1 250 000.

7.2.1 ulate the trade-in value of the lorry in 6 years’ time if the

*s value decreases at 11% p.a. on a reducing balance. (2)

7.2.2 Calculate the value of a new lorry in 3 years’ time if the inflation

rate remains at 7,5% over the next 3 years. (2)

7.2.3. | How much money will the company need in order to replace the
lorry in 3 years’ time, if the old lorry will be used as a trade-in

towards buying the new lorry. (2)

7.2.4 The company decides to save money in a sinking fund. They
deposit equal amounts at the beginning of each year starting
immediately. The last deposit is made at the end of the third year.

Interest is calculated at 9,6% p.a. compounded monthly.

(a) | How many deposits will the company make? (1)
(b) How much will the company deposit each time? (5)
[15]

QUESTION 8

8.1 Given: f(x) =3x? +2x

8.1.1 Determine f'(x) from first principles. (5)

8.1.2 Determine the average gradient of f(x) between x =—1 and

x=2 GB)
8.2 Determine the following:
8.2.1 g(x) if g(x) =(7x-1)(3-x) (2)

ee
x

vx wx x 4)
[14]

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Mathematics P1 9

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QUESTION 9

June 2023 Common Test

Given: f(x)=(x-3)(x-3)(x+2)=2° —4x? -3x +18

9.1 Write do e Y—intercept of 7 . (ql)
9.2 Write he co-ordinates of the xintercepts of / . (3)
9.3 Determine the co-ordinates of the turning points of / . (4)

9.4 Draw the graph of 7 , clearly indicating the intercepts with the axes and the
co-ordinates of the turning points. (3)

9.5 Determine the equation of a tangent to / with a gradient of -3 at the point
where x is an integer. (4)
[15]

QUESTION 10
In 2020, the scientists stated that the number of people infected with the Covid 19 Virus in a

certain country was given by the equation N (¢) = -30 +3¢7,, where N(t) was the number of

infected people in thousands and ¢ was the number of months.

10.1 How many people were infected at the beginning? (at zero months). (1)
10.2 At what rate was the virus spreading in the second month? (2)
10.3 After how many months did the virus reach its peak? (4)
10.4 How many people were infected when the virus reached its peak? (2)

19]

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Mathematics P1 10
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QUESTION 11

Td A study on

June 2023 Common Test

favorite cell phone of Grade 11 and Grade 12 learners at Greendale

High Sc Ided the following results.
IPHONE SAMSUNG TOTAL
GRADE 11 a d 50
GRADE 12 b c 100
TOTAL 60 90 150

11.1.1 Are events Samsung and Grade 11 learners mutually exclusive?

Motivate your answer. (2)
11.1.2. Ifthe learners’ preference for cell phones is independent of their
grade, determine the value of d. (4)
11.2 During a TV Game Show, contestants have to answer randomly generated
questions within a fixed time:
° The probability of the first question being answered correctly is 0,8.
Whenever a question is answered correctly, the next question is more
difficult and the probability of a correct answer is reduced by 0,1.
. This means that the probability of getting the second answer correct is 0,7.
. Whenever the question is answered incorrectly, the next question is of the

same standard and the probability of it being answered correctly remains

unchanged.

11.2.1 After how many consecutive correct answers will the probability
of getting the next question correct, be zero?

11.2.2 Calculate the probability of getting the second answer corre

TOTAL

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(1)

(4)
(4)

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Mathematics P1

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June 2023 Common Test

INFORMATION SHEET: MATHEMATICS

INLIGTING BLADSY

_-bt Vb? -4ac

2a
A= P(1+ni) = P(l=ni) A=P(1-i)" A=P(1+i)"
T, =a+(n- S, =5Qa+(n-Dd)
T, =ar"" gy -W' ol) ; r#l Soe"; aler<

ee on l-r

pastel] pal-a+")

i i
f(x) = lim S(x+h)- f(x)

ho>0 h
d= (x, =x)? +0, -y,)? M By 5 FY me
2 2
y=met+c y-y, =m(x-x,) = 22771 m=tanO
X%2-%

i

snB snC

sin(a + £) = sin a.cos B + cosa.sin B

In AABC:

cos(a + 8) =cosa.cos # —sin a.sin B
cos* @—sin* a

cos2a =41—2sin? a
2cos? a@-1

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a? =b* +c? —2be.cos A area MABC = > ab.sin C

sin(a — £)=sin a.cos 8 —cosa.sin B

cos(a — 8) =cosa.cos B +sin a.sin B

sin 2a@ = 2sin a.cos @

a 2

X“& -x)

of =
n

P(A or B) = P(A) + P(B)

pa e-FY-3)
> @-xy"

and B)

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FINAL

KWAZULU-NATAL PROVINCE

E ATION
BLIC OF SOUTH AFRICA

NATION
NIOR CERTIFICATE

GRADE 12

MARKS: 150

This memorandum consists of 20 pages.

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June 2023 Common Test

1.1.2 on ee
-b+ IC

e-
2a
rel Van 46-3)
7 25)

x=0,25 or-2,45

QUESTION 1
LA x(x-2)=0 AY factors
x=0 or x=2 AY x=0
a Av x=2 (3)
AY standard form

CAY substitution

CAY answers ()

Lis x? -2x-8<0
(x-4)(x+2)<0

4

- 4

ire a

+ 4
xe[-2;4] OR -2<x<4

OR

x? -2x-8<0
(x-4)(x+2)<0

+ \ /

xe[-2;4] OR -2<x<4

AY’ standard form

CAY factors

CAV critical values

AY correct
notation

OR
AY standard form

CAY factors

CAV end values

AY correct
notation (4)

1.14 2 (241)
>

V3+x7-x=x

2
+X -xX=xX

(Se¥=a) <2

3+xX-x=x"

AY factors on the
numerator

CAY simplifying

CAY squaring

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Mathematics 3 r June 2023 Common Test
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3-x=0
x=3 CAY answer (4)
12 —m-8<0 AY
-8<m AY answer
R OR
-m-8<0 AY
—m<8
m>-8 AY answer (2)
1.3.1 Perimeter = 2/ + 2b
24=(2x+y)+y+(2x+y)+y AY qd)
24=4x+4y
1.3.2 Area = [b
32= y(2x+y) AY a)
32= y?+2xy
133 | 24=4x+4y...........(1)
6=x+y
x=6-y (3) AY equation 3
32= y? +2xy... 0.2... (2)
32=2y(6-y)+y" CAY substitution
32=12y-2y?+y?
32=12y-y’
y?-12y+32=0
(y-4)(y-8)=0 CAV factors
y=4 or 8 CAY y values
y=4only
x=6-4=2 CAY x value f
«length = 8m and width = 4m CAY answer o
[24]
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7 5 b
At turning point n =——
ig point n 2a

n= 40. =10
2(-2)

~. 4\9 has the maximum value.

OR

dT,
dn

0=-4n+40
4n= 40

=—4n+40

AY substitution

CAY answer

OR

AY derivative

QUESTION 2
2:1 T, = -2(1)° + 40(1) +103 AY substitution
T,=141 Av answer @)
pars 141 ; 17. IS 5 QS. 5; casease
345 303 26 3 nme AY first difference
re AY answer
The second difference is —4
OR OR
AY
2a = second difference
2-2) =-4
AY answer (2)
The second difference is —4
25 301 = —2n* + 40n+103 AY equating
2n? —40n+198=0 AY standard form
n? ~20n+99=0
(n-9)(n-11) =0
n=9 or ll CAY answers (3)
(conditional if
answers are
natural numbers)
2.4 The maximum value is at the turning point

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2023 Common Test

. T\) has the maximum value. CAY answer
T, =-2n? +40n +103 has a maximum value
AY T, =301 and
T, =301
T,, =301
T,, =301
af Ti, has imum value. CA answer
CAY answer
Answer only, full marks
(2)
(9]
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June

QUESTION 3 (DO NOT MARK 3.3.1)

CAY answer(on
condition, answer
is natural)

Sid 21; 24 AY AY (2)
S12 Ty of origi uence is 7, of GP AY Sub in formula
for GP
In the GP: 8 ; 12
Tar
Ty =3.2? =512 CAY 512
T,, of original sequence is 7,, of AP AY Sub in formula
for AP
339 $15
T, =a+(n-l)d
T,, =3+(11-6
T,, =63
‘AY 6.
Ty, -Ty, = 512-63 = 449 CANS ¥
CAY answer ©)
BZ a=
1=52°
5, = 360° AY Sn = 360"
7” AY values of a and
S,==(a+l) 1
2
360 = "(8 +52)
2 CAY substitution
360 = 30n
12=n (4)

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June 2023 Common Test

VYa=
ariel a=5 vm n=8 Ae aie
4
T, =ar" 1
2] 7
T= 5(3} 7g CAY answer (2)
332 « nt 2
23) “
a \4 ml
nt @)
a
avs(3)
4
33:3 2 nl S
us-s(3) =11s-, AL a
nal 1-2 i-=
a 4
CAY subtracting
His weight will eventually reach = 95kg CAY answer (3)
[18]
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th ‘ics ° 2023 Cor SI
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QUESTION 4 (DO NOT MARK QUESTION 4.3 AND 4.4)

4.1 x=3 AY a)
i—1
42 |xeR; **3 fan? AY AY (2)
vs
43 _5- a 4, AY sub (2;—5)
3-2
SH At Perevcsccesd 1
eeP ) CAY equation 1
-1=—+p AY sub (0;—1)
3-0
—3=a+3p .........(2) CAY equation 2
2H 2p cece (2)-()
p=l CAV value of p
—5=a+l
a=-6 CAV value of a (6)
os x-intercept 0 = J +1
3-x
6 4
3-x
6=3-x
x=-3 CAY x-intercept
y-intercept = Oy jo-1 CAY y-intercept
3-0
, CAY shape
G3)
h
f ee
rx 3 sel
[12]
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June 2023 Common Test

QUESTION 5

5.1

y=-2(0-3K0+1)
y =-2(-3\1.
y='S

E(0;6)

AY substitute x =0

1e9)

52

y= —2(x? -2x-3)
y= -2x° +4x+6

. . b
Turning point x =——
ig point x oF

x= =1

2(-2)
y=-2(l) +4(1I)+6=8
D(I;8)

OR

y=-A(x? -2x+1)-3-1]
y=-2(x-17 +8
DQ; 8)

OR
y=-2(x-3)(x+1)
x-intercepts = 3 or -1
midpoint is x = 1
y=-2(1-3)(1+1)
y=8
D1 ; 8)

AY equation

CAY substitution
CAY x value
CAY y value

OR

AY completing the
square

CAY simplifying
CAY x value
CAY y value
OR

AY x intercepts

CAY midpoint
value

CAY x value
CAY y value

(4)

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SP =-2x7 +6x
Maximum SP is at turning point

b

x=-—

2a

OR

SP

CAY substitution

CAY answer

OR

53 F(-2;-8) AY x coordinate
AY y coordinate (2)
54 0=-20- 3x41)
x=3 or -1l
AtQ x=3 re w
Q(3:0)
5.5 y=mx+e
y=mx+6 AVy=mx+6
0=m(3)+6 CAY substitute
-6=3m (3:0)
-2=m
y=-2x+6 CAY answer
OR OR
=n
oh
CAY substitute in
0-6 _ -6 gradient formula
m=5° 37 a)
CAY m=-2
c=6
y=-2x+6 GB)
CAY answer
5.6 SP =-2x7 +4x-6-(-2x+6) CAV expression for

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June 2023 Common Test

SP =-2x +4x-6-(-2x+6)

CAV expression for
SP

SP =-2x7 +6x
CAY derivative

ie —4x+6

dx

0=-4x+6

4x=6

ia 3 CAY answer

Z
OR
OR

SP =-2x7+4x-6-(-2x+6) CAV expression for
SP

SP =-2x7 +6x

SP =-2(x? -3x)
CAY completing

-3)) (-3yY the square
SP =-2|| x°-3x+| —] |-| —
a} 1)
3) 9
Pp =-2|| x-=] -=
2
sP --o{s-3) 42
z
. _. 3
Maximum SP is at x= 3 CAW answer )
7 k<6 AY (1)

[15]

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June 2023 Common Test

QUESTION 6
6.1] 3=a(l)’ AY substitution
a=3 ia CAY answer (2)
faYey
62) x=3y CAY swapping x
and y
Mia
3 Re
x
y= t fo
AS CAY answer(+ve) (2)
y=, [=
3
6.3 yp AY shape
rt
AY point (2)
6:
fo
6.4 | Yes, any vertical line cuts the graph at only one place. AY answer
OR CA explanation (2)
Yes, it passes the vertical line test
OR
Yes, it is a one to one function
[8]

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June 2023 Common Test

QUESTION 7

7A AY formula
(1+)
1+ om = AY substitution

CAY answer (3)

721) a= P(1-i)"

3
Aes 2500 000{1-1%)
100

A= R881 211,25

AY substitution into
correct formula

CAY answer (2)

72.2 | A=P(I+i)"

A=1 250 000( 14 7:5)
100

A=R1 552 871,10

AY substitution into
correct formula

CAY answer (2)

7.2.3 | R1 552 871,10 —R881 211,25
= R671 659,85

CAY subtraction
CAY answer (2)

7.2.4 | 4 deposits
a).

AY answer qd)

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June 2023 Common Test

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Teak Ty Tt % ty
b). f f hi |
* x x ¥
36
AY (1 +28)
1200
4
AY (1 5 eo
\ 200 200
9,6)"
ig a be av x(14 35 } +
671659,85 = x [1 °5 {1+ 25) {1+ | 1 1200
1200 1200 1200 6
X=R144 650,97 CAM setting up the
equation
CAY answer 6)
05]
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June 2023 Common Test

QUESTION 8

8.1.1

f@= 3x? +2x

f (x+h) =3( PRY 2x +h) = 3x7 + 6xh + 3h? +2x+2h

f= lim h

, 3x? +6xh+3h? +2x+2h-3x -2x
fe) = lim P

yy yy, OXN+ 3h? +2h
£@)=lim—~ ——

$1) = tim MOE)

S (x) = lim(6x + 3h +2)
hoo

Av calculating
fix +h)

CAY substitution
into the formula

CAY simplification

CAY factorising

(5)
FG) = 6x42 CAV answer
8.1.2 | Average gradient = 6x+3h+2 CAY formula for
average gradient
x=-l
AY value for h
h=2-(-]1)=3
CAY answer
Ave grad = 6(—1) +3(3)+2=5
OR
OR
Atx=-l y=X-1?+2(-I=1 Avy values
Atx=2  — y=3(2)?+2(2)=16
Ave grad = 2H
Hox,
Ave grad = 16-1 _15_5 CAY substituting in
2-C) 3 gradient la
CAV ans (3)

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June 2023 Common Test

8.2.1 | 9(x)=-7x° +22x-3 AY
g(x) =-14x4+22 CAV answer (2)
lS.
8.2.2 1
y=Sx+--
2 AY
1 ce
y=5x+—-Ox CAY 2x?
2
3 CAY CAY (two
dx (4)
[14]
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QUESTION 9
91 y=18 AY a)
9.2 O= (4-3) +2) AY equating to zero

x=3 or -2 Av AY (3)
93 Turning points: f (x) =3x° —8x-3 AY derivative

0=3x? —8x-3 CAY equating to

zero

0=(3x+1)(x-3)

x=3 or -3 CAWx values

y=(3-3)3-3)3+2)=0 (3:0)

y=(-4-3}(-1-3}/-142]=2-1852 (-1,502) | CAvy values

3 3 3 27 3 27
OR
OR

Turning points: f (x) =3x* -8x-3 AY derivative

0=3x? -8x-3 CAY equating to

0=(3x+I(x-3) ae

x=3 or -5 CAWx values

y= (3) -4(3)°-3(3)+18=0 — (3:0)

1y 1y 1 500 { 1 | CAVy values (4)
y=(-2) -4{-1) -3(-L)413-2 | -=;——
y=(-3} ~4{-3) (5) a (327

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fics

own

loaded from Stanmanaphysics.com

June 2023 Common Test

94

$00
27

Sy
(+33 )

18

(230) fe)

GB:

9)

AY shape

CAV turning points
labelled

CAV intercepts
labelled (3)

9.5 f'(20) = 3x7 -8x-3
—3 = 3x? -8x-3 CAY equating f (x)
to -3
O=3x" -8x
O= x(3x-8)
CAV factors
x=0Oor 8
2
. x=0
At x=0 y=18 and m=-3 CAY x=0
(4)
assets CAY answer
{15}
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June 2023 Common Test

QUESTION 10
10.1 | No people were infected AY (1)
—
102 | Nw - 3
AY substitution
NQ) =-3) 6(2) into the derivative
= 6000 people/month CAY answer (2)
10.3 Nw =-3P +6
=
Siz
O= “af * or CAY equating to
zero
O=f -12r
O=1(t-4
ee 2 CAY factors
as CAV both values
.. 4 months of t
CAV answer (4)
10.4 Virus reached the peak after 4 months
17,3 2
N(A)= (4) +3(4) CAY substitution
-. 16 000 people CAY answer (2)
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June 2023 Common Test

QUESTION 11 (DO NOT MARK 11.1.1)
11.1.1 AYNo
For mutually: sive events P(A and B) =0 Av explanation (2)
11.1.2 P(Grade 11 and Samsung) = <—Z AY
150
P(Grade 11) x P(Samsung) = 0. 0 AY
150 150
P(Grade 11 and Samsung) = P(Grade 11) P(Samsung)
4. 0,0 AY
“" 150 150° 150
da ex 9) 150 =30 CAY answer (4)
150 150
11.2.1] 8 AY (dl)
11.2.2
c ve
07,
a
0,3~NC cnc
08
og.-© NCC
0.2 Ne a
02\ Ne NCNC
P(CC or NCC) =0,8x0,7+0,2x0,8 AY 0,8x0,7
AY 0,2x0,8
AY addition
=0,72 CAY, (4)
[naar
(11)
TOTAL MARKS: 150

MARK THE PAPER OUT OF 137 THEN CONVERT TO 150

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