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

> Grade 12 Mathematics Paper 1 question paper and memo from the MP June Exam 2023, 21 pages. Read online or get the free PDF download.

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- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: MP June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 21

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## About this document

Mathematics · Grade 12 · MP June Exam · 2023. Question paper and memorandum, 21 pages. Read online or download the PDF.

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## Document text

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### Page 1

BF, education

MPUMALANGA PROVINCE
REPUBLIC OF SOUTH AFRICA

MARKS: 150
TIME: 3 hours

This question paper consists of 8 pages and a one-page information sheet.

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### Page 2

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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

l.

This question paper consists of 10 questions.
Answer ALDtt

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

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

Answers only will not necessarily be awarded full marks.

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

Diagrams and graphs are NOT necessarily drawn to scale.

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

Write neatly and legibly.

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QUESTION 1

1.1

1.2

1.3

QUESTION 2

Solve for x.

1.1.1

1.1.2

1.1.
1.1

(2+x)(-x+4)=0
? 2x44 (Correct to 2 decimal places)
-2V¥x-1=4

x-12 > -x?

Solve the following equations simultaneously:

x? —xy—Sy? =-5 and x+2y=1

Determine the values of ¢ for which the equation

5* =2-¢ will have real solutions,

Given the arithmetic series 3+ 8+ 13 +...... (to 253 terms)

Write down the 4" term of the series.

Calculate the 253 term of the series.

Express the series in sigma notation.

Calculate the sum of the series.

How many terms of the series are divisible by 4?

Given the quadratic sequence: 1; 3,7; p; ...

Calculate the value of p.
Determine the n'" term of the sequence.
The first difference between two consecutive terms is 62.

Calculate the values of these terms.

2.1

2.2

2.3

2.4

2.5

QUESTION 3 .

3.1
3.1.1
3.1.2
3.1.3

3.2

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Determine the largest integer value of m if 2.2") < 20.

kel

MDE/June 2023

(2)
4)
(6)
(4)

(6)

(3)
[25]

(1)
(3)
(2)
(2)
(4)
[12]

G3)
(4)

(3)

6)

[15]

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QUESTION 4

The following sketch shows the graphs of f(x) =x? +bx+c¢ and g(x) =ax+q.
The graph of fintersects the x-axis at A(-2 ; 0) and at B. C is the y-intercept of
both fand g, axis of symmetry of f is x=1.

ay 1
{
| g
f
{
i N .
A i B *
H
H
H
!
!
!
Cc
4.1 Determine
4.1.1 The coordinates of B (2)
4.1.2 The equation of fin the form y=x? +bx+c (2)
4.1.3. The values of a and q. (4)
4.1.4 The coordinates of the turning point of f. (2)

4.2 M is the reflection of C in the axis of symmetry off, Write down the

coordinates of M. (2)
43 For which values of x is f(x) > g(x)? (2)
4.4 Write down the coordinates of the turning point of y = —f(x = (2)

45 N is a point on the x-axis and T is a point on f such that TN
perpendicular to the x-axis. TN intersects g at P. Calculate ximum (4)
length of TP.
[20]

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QUESTION 5
Sketched below is the graph of f(x) =—log, x
y4
T >x
f
5.1 Write down the domain of f (i)
5.2 Write down the equation of f~ in the form y=.... (2)
5.3. Describe the transformation from f~' to hif A(x)=3* -5 (2)
5.4. Use the graph of f to solve for x if A(x) is the reflection of f about the
x-axis and k({x)23. (4)
[9]
QUESTION 6
The function f defined by f(x) = 4 +q has the following properties.
x+p
The range of fis ye R,y #2
¢ The axis of symmetry with a positive gradient is y = x +1
e The graph of fpasses through (0;—4)
6.1 Write down the value of q. (1)
6.2 Calculate the values of a and p . (4)
6.3 Sketch a neat graph of this function. Your graph must include tercepts
with the axes and asymptotes if any. (5)
[10]

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QUESTION 7

7.1 Thelma purchased a new car. The bank offered her a loan at an effective
interest rate of 10,4% p.a.

Deter: he nominal rate compounded monthly that she is required to pay. (4)

7.2 Ani r bought shares in a certain company, but found that his money
had become half of its original value after a period of 4 years. What was the
annual rate of decay, on a reducing balance method, for the shares he had
bought? (4)

7.3  Thini borrowed R80 000, exactly five years ago, at 21% interest p.a. She
paid back R25 000 exactly three years ago and R55 000 exactly a year ago.
She wants to settle her remaining debt today.

7.3.1 Calculate the value of the loan to date. (2)
7.3.2 How much must Thini pay to settle the debt today? (5)
[15]
QUESTION 8
8.1 Given f(x) =4x—2x’, determine /"(x) using FIRST PRINCIPLES. (5)

8.2 Determine:

1
8.2.1 pi» -4] 6)
dy. _ y2
8.2.2 x if xJy =x* —4x (4)
8.3 Find the equation of the tangent to the graph of f(x) = x? —6x? ich the
gradient of the tangent to fis equal to 36 and x < 0. ; (6)
[17]

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iC

QUESTION 9

The graph below represents the functions fand g with f(x) = ax’ +cex+2and

g(x) =—xt+ and (1; 0) are the x -intercepts of f The graph of fand g

intersect at Cc,
ay
B
A
y
ci; [9 a
9.1 Determine the coordinates of C. (1)
9.2 Show by calculation that a=-1 and c= 3, (4)
9.3. Determine the coordinates of B, the turning point of 7. (3)
9.4 Find the x-coordinate of the point of inflection of /. (2)
9.5 Write down the values of & for which f(x) = will have only one root. (2)
9.6 Write down the values of x for which f'(x) < 0. (2)

[14]

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QUESTION 10

10.1 Given that A and B are independent events, P(B only) = 0,4
P(A and B) = 0,1 ; P(A only) = x and P(not A or B) = y.

Calcul
10.1. e values of x and y. (4)
10.1 e probability that at least one of A or B occurs. @)

10.2 Inacity 10% of all crimes are violent. 95% of all violent crimes are reported,

buy only 45% of non- violent crimes are reported.

10.2.1 Draw a tree diagram showing all the possible outcomes. (3)
10.2.2 What is the probability that a random crime will be reported? (3)
[13]
TOTAL: 150

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INFORMATION SHEET

ea wet 6? -4ac
~ 2a
A=P(1+ in) A=P(\-in) A=P(1-i)' A=P(\+i)"
T,=a+(n S,= Sba+(r -1)d]
r, = ar"! 5-2 V ral S.= 74-3 -1<rcl
- - Tr
Pe xf(.+ iy" - 1] pate G+)" ee) = lim f(x +h)- f(x)
i i hoo h
d= (x) - 4) +Q2 -nF M222, 4 #2)
yamxte y-y= mx —x) m = 2M m=tan@
*

(x-a? +(y-b¥ =P?

a’ =b? +c? 2be.cosA

Area of AABC = Sabin c
sin(a + B)= sina.cos B + cosa.sin B sin(a - Bp) = sina.cos B — cosa.sin B
cosa - Bp) =cosa@.cosf + sina.sinB cosa + Bp) =cosa.cos# — sina.sinB

cos’ A — sin? A

cos2A =41—2sin? A sin2A = 2sin A.cosA
2cos? A-1
ae X<, -x/
=o ea
n
P(4)= en P(A or B)= P(A) )— P(A and B)

D=atbx _ of 9)

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### Page 10

Downloaded from Stanmorephysics.com

education

Ri s MPUMALANGA PROVINCE
RS) REPUBLICOFSOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

' GRADE 12

MATHEMATICS P1
JUNE 2023
MARKING GUIDELINES

This marking guideline consists of 12 pages.

Please turn over

### Page 11

‘powntsdded from Stanmecephysics.qam MDE/une 2023
QUESTION 1
T11] @-2Cx+4)=0
x-2=0 or-x+4=0 yea?
= =4 x=4

11.2] 3 ix+4 ¥ standard form
3. x-4=0
-(2)+, (27 -4(3)(-4) ¥ substitution in correct
“= 2(3)
ea 2452
6 vv values of x
x=-0,87 or x=1,54
(4)
1.1.3 | y-2/¢-1 34
—2yx-L=4~x Visolate the surd
. ovx—1} =(4—x)? ¥ square both sides
4x-4=16-8x+x"
x? —12x+20=0 Y standard form
¥ factors
(x—10Xx—2)=0 x-values
x=10 or x#2 ¥ x=10 only (6)
1.1.4] x?-x-12>0 y :
standard
(x-4)(x +3) >0 V factors
Y critical values
3 4
OR
—HO OT
3 4
x<-3 or x>4 ¥ solution (4)

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yobwpritegded from Stanmeveshysics.cam MDE/June 2023

1.2 |x+2y=1
x=1-2 ¥ subject of formulae
1- 1-2y)-Sy? =-5
( ¢ ¥) 2 y , Y substitution
1- y’ —yt+2y* —Sy* =-5S
¥ standard form
y? —Sy+6=0
(y-3)(y-2) =0 ¥ factors
y=3ory=2 ¥ y=3 or y=2

x =1-2(3) or x = 1-2(2)

x=-5S orx = -3
¥x=-Sor x=-3

13 | 5*>0 ¥ condition ©
2-t>0 roa
wt<2 (3)
[25]

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### Page 13

Yewprifsdded from StanmecedhysiGse cam MDE/June 2023
QUESTION 2
21 [18 V answer (1)
2.2 | T,=5n-2 ¥nth term
Toss 53)-2 ¥ substitution
Ths 3 ¥ answer (3)
23 |S Y limits
Dbn=2)
mi ¥5n-2 (2)
2 Fae
‘7 Sy, = 2531 +1263] substitution
2 v answer
Soo = 160149
OR
Sys3 = #120) +(253-1)5] ¥ substitution
¥ answer (2)

Sy5; = 160149

2.5 | 8428448 +,..+ 1248 ¥ terms divisible by 4
a=8,d=20
T, =20n-12 ¥ general term
20n—12 =1248
n=63 ¥ equating 4)
v value of 7.
[12]

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QUESTION 3

MDE/June 2023

3.1.1

¥ first differences

2 p-T
¥ equating
p-7-4 ¥p=13 @)
p-7-4=2
p=13
3.1.2 | 2a=2 3a+b=2 atb+c=l
a=l 3(l) +b =2 l-l+c=1 |“a=l
b=-1 c=! Yb=-1
v
_ te c=! (4)
T, =n? -n+1 VT, =n n+l
3.1312,4, 6...
a=2 d=2
T, =2n ¥ general term
on = 62 ¥ equating
n=31
Term number 31 and term number 32 Y both 31 and 32 @)
3.2
2.2"? < 2046
ka?
2.29 +2.2' 42.2? +2.2° +..+2.2"? < 2046 Y expansion
24+44+84+164...4
a=2, r=2,  n=m-(-2)41 at Od
=m+3
m+3
. 2(2"* -1) < 2046 ¥ substitution into
-1 correct formulae
2" 1 <1023
2" < 1024
ques <2!
m+3<10 ¥m<7
5
m<7 vm=6 ”
“m=6 faYaY
{15}

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. 4 MDE/June 2023
'‘Downtedded from Stanmereshysics.com
QUESTION 4
4.1.1 =2+x | ¥ method
2 v¥x=4
x=4
B(4: Answer only full marks 2)
4.1.21 yet Qer- ¥ method
y 2x-8 ¥ correct equation (2)
4.1.3 | C(0;-8) ¥ C(0;-8)
a= 8-0 ¥ substitution
0-4 Ya-2
a= : =2 ¥q=-8
(4)
g(x) =2x-8
a=2
q=-8
4.1.4} y=(1)? -2()-8 v value of y
= v
ye (;-9)
(13-9) 2)
4.2 | O+x_ 1 v¥x=2
>
=2
* YMQ2;-8) (2)
M(2;-8)
4.3 | x<0 or x>4 v¥x<0
Vx>4 (2)
44 |@;9) Vx=3
vy=9 (2)
45 | 7p =2x—-8-(x* -2x-8) ¥ top function —bottom
TP =-x? +4x function
TP' =-2x+4 v derivative
-2x+4=0 ¥ equating t
x=2 Y answer
TP =-(2)° +4(2)
TP =4

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i, MDE/June 2023
Downtedded from Stanmore sics.com une
Y factoring a negative

¥ completing the square

¥ “maximum 4 when

x=2 (4)
There is a maximum of 4 when x = 2
[20]
QUESTION 5
5.1 |x>0 OR xe (0;0) ¥ domain ()
5.2 | y=-log,x
y=log, x"
¥ interchanging x and y
x=log, y"
3" = y?
_{1y -*
Y\3) FYE 3 ¥ correct equation
Answer only full marks
(2)
5.3 | fis shifted 5 units vertically down v5 units
¥ down (2)
5.4 | log,x=3 %x=27
xa3 ¥ —log, x 2-3
x27 VV 0<xS27
“. -log, x 2-3 (4)
0<x<27 ia
i e

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ES

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Dowrtydded from Stanmeceshysicsscgn MDE/June 2023

QUESTION 6

61 | q=2 Y answer (4)
62 Tpt2
vp =-]

y=xtl
2=x+l1
x=l

OR ¥ sub (0;-4)

vane )

6.3

ya

------------- waa dq-------------- y=?

-2 vx intercept

vy intercept
¥ asymptote
‘ ¥ shape of graph (5)

[10]

[aYaYey

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Downisaded from Stanmerephysics.com MDE/June 2023

ig, Guidelines
QUESTION 7
TA bei -(1- 2304) ¥n=12
oe 12
iy, £49409103376 — 1 j= 2104
v 12
ig 24O191%
Y substitution into the
correct formula
¥10,91%
(4)
7.2 0,5P = P(i-i)* ¥ subs in correct formula
0,5=(1-1* ¥ divide by P
40,5 =1-i ‘ v4" root
i=1- 4 0,5
i=0,1591 ¥i=15,91%
i =15,91% (4)
7.3.1 | Loan with interest to date
= 80000(1 + 0,21)° ¥ subs in correct formula
= R207499,40 ¥ R20 7499,40
(2)
7.3.2 | Repayments with interest to date ¥ 25000(1 +.0,21)°
_25000(1+ 0,21)? +55000(1 + 0,21)’ ¥55000(1 40,21)!
= R101673,20
. a. ¥ R101673,20
Outstanding balance =Loan with interest to date- “method .
Repayments with interest to date ¥105826,20
=R207499,40 - R10 673,20
=R105826,20 (5)
15
— [iS]

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Downtsaded from Stanmor

Sics.cOomM MDE/June 2023
jarking Guidelines
QUESTION 8
8.1 P(x h) =4(x +h) 2x +h)?

=4x+4h-2x? -4xh—2h?
fG

mn LEH FH)

Y f(x+h)

4.0 h
r oy AetaN= 28? — ach 20? —(4x-2x?)
0 A ¥ subt. into correct
2
F'@)= tim Vivdxhn 2h formula
hoo h ¥ 4h 2
r'@) = lim A(4—4x—2h) 4h—4xh-2h
hoo h
S'(x) =4-4x ¥ factoring A
v answer (5)
8.2.1 | p jx! -x?] Vx
= 4x9 42x74 v4 v2xt  [G)
8.2.2 x? 4x ¥ dividing by x
vy = x x
vy =x-4 ¥ squaring
y=(x-4)?
y=x? -8x+16
@ 2-8 v¥ answer (4)
83 | m=36 Vv f'(x) = 36
f'(x) = 3x? -12x
3x* -12x =36
3x? -12x-36=0 Y factors
(x-6)(x +2) =0 Vx=-2
x #6 or x=-2
f(-2) = (-2)° - 6(-2)’ v¥ y=-32
f(-2)=-32
y—(-32) = 36(x -(-2))
5
y=36x+40 Y equation? ©)
cr

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EE

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### Page 20

Wobwritsaled from Stanmerephysics.com

MDE/June 2023
QUESTION 9
91 1CQ@:0) V answer @
9.2 ¥ substitute (—1 ; 0)
¥ subject of formula
¥ substitute (2 ; 0)
¥ sub in second
equation
(4)
9.3 | f(x) =-3x? +3 ¥ equating derivative
0=-3x? +3 to zero
0=-3(x-D(x+1 vx-value
x=l
£0) =-()’ +30) +2
=4
vy-value
BU: 4)
GB)
94 | -1+1_ 0 ¥ method
2
OR
f'() =-3x? +3
f"(x) = -6x
~6x=0 ¥ answer (2)
x=0
95 | k<Oork>4 v¥k<O0
Vk>4 e)
9.6 | x<-lor x>1 ¥ both critical values
Vx< Bing >1 (2)
|
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MDE/June 2023

QUESTION 10
10.1.1 | PB 1+ 0,4 ¥ P(B) =0,5
5
Pi (B) = 0,1 ¥ P(A) x P(B) = 0,1
(*¥0,1(0,5) = 0,1
x=0, ¥x=0,1
y=l-0,1
= 4
y=0.9 ¥y=0,9 @
10.1.2 | P(AorB) = P(A) + P(B) - P(A and B) ¥ formulae
= 0,2+0,5-0,1 v P(A) = 0,2
= 0,6 ¥ answer (3)
10.2.1
R VR
o% ¥ 0,1 0,95
Vv
0,1 0,05 ~ NR VNR
v 0,1 x 0,05
R
Py 0,45 NVR
NV ¥ non violent
0355S NR NVNR
(3)
10.2.2 | P(Violent crime rep) or P(Non -violent crime rep) | ¥ 0,1x0,95
= 0,1« 0,95+0,9 x 0,45 ¥ 0,9x 0,45
=0,5 ¥ answel
(3)
[aYaYay
[13]
TOTAL MARKS: 150

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