# Mathematics Grade 11 Paper 1 KZN November 2025 Paper & Memo

> Grade 11 Mathematics Paper 1 question paper and memo from the KZN November Exam 2025, 47 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/kzn-maths-grade-11-november-2025-p1-and-memo/0938d143-60ca-4200-aa87-7a717302934c
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2025
- Exam period: KZN November Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 47

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## About this document

Mathematics · Grade 11 · KZN November Exam · 2025. Question paper and memorandum, 47 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

GRADE 11

MARKS: 150

TIME: 3 hours

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

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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1.

10.

This question paper consists of 10 questions.
Answer ALL the questions in the SPECIAL ANSWER BOOK provided.

Clearly show numbering if additional space has been used to answer some question. If NOT,
cross out all rough work done in the additional space.

Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining
your answers.

Answers only will NOT necessarily be awarded full marks.

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

=)

If necessary, round off ans} O decimal places, unless stated otherwise.
Diagrams are NOT necessaril
An information sheet with fe d at the end of the question paper.

Write neatly and legibly.

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QUESTION 1
1.1 Solve for x:
11 4x°-9=0
1.2 2x? +6x-3=0 (answers correct to TWO decimal places)
2 _
113 pega lt te 6
x+2
1.14 x? -2x-1529
1.2 Two equations are given below:

1.3

1.4

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2”
1.2.1 Show that —=4" written as y=2x-2.
1722 Hence, or of e for both x and y.

The roots of a quadratic equation are given as: x=J/2—k , where ke (-9 : 0).

Determine the integral values of k for which the roots are rational.

where b <0 and

—bt
The roots of the equation ax” + bx +c =0 are cel

c=b+18. Calculate the values of a, band c.

KZN/November 2025

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(4)

(2)

(4)

(2)

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QUESTION 2

2.1

Simplify fully, WITHOUT using a calculator:

0
2 2x(27x° ) { = }

Die
es
-2
ai (6-288)
iB
m+4 4 5m+tl
2.2; It is given that a8
5.243
22.1 Simplify the above expression.
m+3
2.2.2 Hence, solve for x if a :
’ amass 4am! 125°
2.3 Solve for x:
QUESTION 3

KZN/November 2025

WhatsApp messages were sent using face emojis with two popping hearts per face (shown below).

(3)

(4)

(3)

(3)

(5)
[18]

MESSAGE 1

MESSAGE 2

MESSAGE 3

These faces and popping hearts in the different messages forms a linear number pattern.

3.1

3.2

3:3)

3.4

EB

How many popping hearts will appear in message 4?.

Determine the n” term (general term) of the popping hearts in the messages.

Calculate the number of popping hearts in the tenth (T,, ) message.

In which message will there be 710 popping hearts?

A new series of messages using the same emojis was created by:

e keeping the number of faces in message 1 and message 2 the same as the

original messages.
e adding 2 more faces to the original message 3.
e adding 6 more faces to the original message 4.

The new series of messages forms a quadratic number pattern.

(1)
(2)
(2)
(2)

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QUESTION 4

The first four terms of a quadratic number pattern are —7 ; m ; 5 ; n. The constant
second difference is 4.

4.1 Prove, by calculations, that m=-3 and n=17. (3)
4.2 Which two consecutive terms of the quadratic number pattern has a first
difference of 292? (3)
4.3 It is given that the general term of a quadratic number pattern is
T, =2n* —2n—7. Calculate the value of n if 7, =1 005. (4)
[10]
QUESTION 5

Consider the function f defined by f(x) - +5 . Graph f passes through (4 ; 0).
r xz

5.1 Write down the y-inter Se a (1)
52 Write down the equations ¢ ptotes of / (2)
5.3 Calculate the x-intercept (2)
5.4 Draw a neat sketch of f Clear ly ‘abel all the intercepts with the axes and
asymptotes. (3)
55 Determine the values of y for which f lies x<-—2 or 0<x<4. (2)
5.6 Graph / is reflected about the y-axis and shifted : units up to form p.
Write down the equation of p in the form of p(x) =.... (2)
[12]
QUESTION 6
6.1 The function h(x) =ax’ +bx +c has the following properties:
° h(2)=0
e the value of od is 5
2a
e therangeof his y< 10;yeR.
Draw a neat sketch of f. Clearly label the x-intercepts and turning point. (3)

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6.2 The graphs of g(x)=—-x°+9 and h(x) =ax? +bx-22 are sketched below.

6.2.1

6.2.2

6.2.3

6.2.4

6.2.5

6.2.6

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Ais an x-intercept of g and / and lies on the negative x-axis.

A and D are the points of intersection of graphs g and h.

C is the turning point of h.

B lies on the horizontal line that passes through the turning point of g such that BC is
parallel to the y-axis and passes through the other intercept of g.

BC = 18 units
¥
h
x

Calculate the coordinates of A. (3)

1 3
Show that a=— and b=-~. (4)

4 2
Write down the range of h. (2)

It is given that m(x)= x? —4. Explain, in words, the transformation
from the graphs of g to m. (2)

Calculate the coordinates of D. (5)

It is given that p(x)=k. Determine the integral value of k such

that the distance between the points of intersection of h and p is
8 units. (2)
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QUESTION 7
>

The graj (x)=2"+q is drawn below. . Graph f passes through the point (= : -13).

a

Inn a

f

(2)
72 If g=—8, calculate the coordinates of A. (3)
73 Determine the average gradient between A and B. (3)
x
7.4 Determine the values of x where <0 2
7) “
TS: Determine the values of k for which 8—2*—k =0 will have negative roots. (2)
[12]

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QUESTION 8

8.1

8.2

8.3

8.4

8.5

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interest at the rate of 10% per annum.

pine invests R15 000 for a period of 6 years in an account that pays
ulate the interest his investment earned at the end of 6 years.

Miss Coetzee bought a brand new car. The value of the car depreciates annually
at the rate of 8,25% p.a. on the reducing balance method. If the book value of the
car was R277 216-after 3 years, calculate the purchase price of the car.

An advertisement reads “XY Bank offers loans at an effective interest rate of 12%
per annum”. Calculate the nominal rate of interest if interest on the loan is
compounded quarterly.

Mr Dlezi invested R25 000 in an account that offers 9,25% per annum
compounded half-yearly. Calculate the accumulated amount in this account at the

end of nineteen half-years:

Mr Masango invested
13% p.a., compounded
rate changed to 10,5%
initial deposit, Mr M:
years, he had R928,30 moi
value of x, to the nearest rand

ings account that paid interest at the rate of
first two years. Thereafter, the interest
d quarterly. Three years after making the
from the account. At the end of 5
nt than his initial deposit. Calculate the

KZN/November 2025

(3)

(2)

(4)

(3)

(5)
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Mi i 5 KZN/N ber 2025
“SoWiloaded from Stanmerephysigs.com vember
QUESTION 9
In the Venn diagram below, events A, B and C are shown.
S
A B
iS

9.1 It is given that P(A)=0,47 and P(c)=3

9.1.1 (1)

9.1.2 (2)

"4 5

9.2 It is given that events A coon fil and P(A) = 5 PB).

9.2.1 Determine the P(A and B) in terms of P(B). (2)

9.2.2 It also given that P(A or B)=70%. Calculate P(B). (5)

[10]

QUESTION 10

A survey was conducted amongst 45 males and 85 females who were randomly selected.
This survey was about the brand of clothes they would love to buy between Gucci (G) and
Louis Vuitton (LV). The results are summarised below.

e 3y females said they would love to buy Louis Vuitton.
ey males said they would love to buy Gucci.

e The total number of males who said they would love to buy Louis Vuitton is double the
total number of females who said they would love to buy Gucci.

10.1 Calculate the total number of females who said they would love to buy Gucci.

10.2 Are the events of being a ‘male’ and having said that they would love to buy one
of the brands independent? Justify your answer with calculations.

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TOTAL MARKS:

(6)

(4)
[10]

150

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KZN/November 2025

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INFORMATION SHEET
Sab EAD’ -4a0
2a
A= P(1+ni) A=P(\-ni) A=P(1-i)” A= P(i+i)"
T.=a+(n-l)d 8, =F Pa+(n—1d]
T, =ar™"! 5 _ alr” 1) sr#l S,=-—4;-l<r<l
% r-1 l=?
pass" = paslal+i"
i i
"(x)= lim f(xt+h)- fo)
h>0 h
x, +x. +
d= (-—™)? +029) M22 Ae%)
y=mxt+e m=22 JL m= tan 0
X5—X,
(x-a) +(y-by =r?
In apc: 4 =
sn A sinB

a =b’ +c? —2be.cos A

area AABC = 5 ab.sin C

sin(a+f)=sina.cos B +cosa@.sin B
cos(a@ +f) =cosa@.cos B-sina.sin B
cos’ @—sin®> a
cos2a =41—2sin? a

2cos*a-1

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sin(a@— £)=sina.cos B —cosa.sin B

cos(a@—f) =cosa.cos # +sina.sin B

sin 2@ = 2sina@.cosa

Yo, —x/y

n

o

P(A or B) = P(A) + P(B) — P(A and B)

pa =D)
Le-H?

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL

SENIOR CERTIFICATE

GRADE 11

Full Name & Surname

For Educator / Moderator Use Only ( NOT for learners to complete)

QUESTION Ql Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 | Q10
Total Per
Question 27 18 13 10 12 21 12 17 10 10
Mark
Scored
Moderated
Mark
GRAND TOTAL
150

This answer book consists of 18 pages.

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CAPS-Grade 11 Answer Book

QUESTION
Marks
1 i 2 ”
1L3 ”
anmorephysics.com
1.1.4 =
(4)

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CAPS-Grade 11 Answer Book

(2)
(4)

L3
(2)

1.4
(5)
[27]

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CAPS-Grade 11 Answer Book
QUESTION 2
P =)
Marks
(3)
2.1.2:
anmorephysics.com
(4)
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CAPS-Grade 11 Answer Book

(3)
2.2.2
(3)
2.3
(5)
[18]

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CAPS-Grade 11 Answer Book

QUESTION 3
5 —_[ Solution Marks
— Q)
- - Q)
3.4 2)
a (2)
a mmoreph ysics.com
(6)
[13]

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le

CAPS-Grat Answer Book

QUESTION”
Marks
42 *
anmorephysics.com
43 *
(4)
[10]

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CAPS-Grade 11 Answer Book
QUESTION 5
Solution Marks
eal
Q)
5.2
(2)
5.3
ganmorephysics,com
(2)
5.4
ty
« 0 >
. @)
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CAPS-Grade 11 Answer Book

55
(2)
5.6
(2)
[12]
QUESTION 6
Solution Marks
6.1
4y
« 7 >
v
(3)

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CAPS-Grade 11 Answer Book
Py
(3)
6.2.2
(4)
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CAPS-Grade 11 Answer Book
(2)
(2)
6.2.5
Hahimorephysics.com
(5)
6.2.6
(2)
[21]
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CAPS-Grade 11 Answer Book

QUESTION 7

RL

Solution Marks

TA
(2)

72
(3)

a3
(6)

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le

CAPS-Grat Answer Book

74
_ (2)
(2)
[12]
QUESTION 8
Solution Marks
8.1
= Ranmorephysics.com G3)
- (2)
(4)

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CAPS-Grade 11 Answer Book
8.4 |
=—
a
Sar
Met TT
Er
GB)
8.5
7 ™?
Siafimorephysics.com
(5)
[17]
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le

CAPS-Gra Answer Book
QUESTION 9
| Solution Marks
()
9.2.1 .
92.2 .
Amaren &..
(5)
[10]

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CAPS-Grade 11 Answer Book

QUESTION 10

Solution Marks

10.1
(6)

10.2
(4)
[10]
TOTAL: 150

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CAPS-Grade 11 Answer Book
pAdlditionsl space Marks
aig
a
Nal
F fahimorephysics.com
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CAPS-Gra

ics.com

11 Answer Book

KZN/November 2025

wal

ditional
pAdd onal space

Marks

— |

=

IE

=)

:

Stafimorephysics.com

Copyright

Reserved

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‘ FINAL & CLOSED
education MARKING
Department: GUIDELINES
Education

PROVINCE OF KWAZULU-NATAL

NATIONAL
SENIOR CERTIFICATE

(“ann

MARKS: 150

These marking guidelines consist of 19 pages.

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QUESTION 1

CAPS — Grade 11
Marking Guidelines

KZN —November 2025

Ld

4x7 +920
(2x+3)(2x-3)=0

Accept subst. to quadratic formula

YA correct factors

pa laYe) lor x=— YCA answer Y¥CA answer
2 2 (3)
4x -9=0
Ve _./9 r ‘ 2
x =t 4 VA isolating x
i] 3
x=-> or x== YCA answer YCA answer
2 G3)
1.1.2 | -2x7+6x-3=0
ye —b +b? —4ac
2a Ha .
_ (6) + 4(6)° = 4(-2)(-3 _ + (6) -4(2)(3) | VA correct subst. into
x=. ORs x= :
2(-2) 2(2) quadratic formula
x=2,37 or x=0,63 Penalise 1 mark for incorrect rounding ¥CA answer ¥CA on)
1:1:3 _ _ 2x +x-6
x+2
(x-3)(x+2)=2x" +x-6
x? -x-6=2x? +x-6 YA multiplication
2 =
at + 2-0 YCA standard form
x(x+2)=0
x=0 or x¥-2 YCA answer VA rejection
(4)
1.1.4 | (x-5)(x+3)29

x? =2x-15-920
x’ —2x-2420

Final answer must be
given 2 or 0.

vA standard form

YCA critical value
(accept if shown on a sketch)

¥YCA both answers
(4)

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CAPS — Grade 11

KZN —November 2025

Marking Guidelines
12.1 | 1 gee
2Y j
2 LAA 1.2.1 is a proof, all SK 77 —98)
: marks are accuracy.
Var
-y=2-2x VA -y=2-2x
SPS 2x 2 (2)
1.2.2 | x+6=x7-y
x+6=x° -(2x-2) vA substitution
x’ -3x-4=0
(x-4)(x+1)=0 ¥CA factors
x=4 or x=—l ¥CA both values of x
y=2(4)-2 or y=2(-
y=6 or p=-4 YCA both values of y
(4)
OR
x+6=x-y
y=x-x-6
(2x-2)=x' -x-6 vA substitution
x’ -3x-4=0
(x-4)(x+1)=0 YCA factors
x=4 or x=-l ¥CA both values of x
y=2(4)-2 or y=2(-1)-2
y=6 or p=-4 YCA both values of y
(4)
1.3 A=2-(2)=4 & A=2-C7)=9
k=-2 or k=-7 YA k=-7 VA k=-2
(2)

YA value of a

YCA subst. a & c into A=72

Y¥CA standard form

YCA value of 5 with rejection

YCA value of c
(5)

[27]

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CAPS -Grade 11
Marking Guidelines

KZN — November 2025

QUESTION 2
2 )
2.1.1 3); (2%
2x.(27x? )3 | ——
m(2if)) (3)
2
=2x.(3°x')> x1 VAI
a2
=2x.(3x) 3
= 2x(3x)’
=2x(3°x") YA Px
=18x! ¥CA answer
3)
2.12 | (6-2v3)
iB
— 36- 243 +12 vA removing the brackets in
3 numerator
b= 243 3 vA rationalising denom. i.e. a3
BOS va
_ 483-243
3
3(16V3 = 24) YCA simplifying the numerator
—_——
= 16V3 24 ¥CA answer
(4)
m+4 m+
994 Dede
5.2+3
~ 2” edo") = Oi ee2”.2! vA expansion using exponent law
25 23 ans
4 , 93 (51
= 2" (2 ~4x2) a 2 (2 -1) vA factorising the numerator
2"5.8 225
ee or - or 0,2 ¥CA answer
40 eS)
Q)
OR OR
¥¥A factorising the numerator

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CAPS -Grade 11
Marking Guidelines

KZN — November 2025

qinr. = 42m
som ¥CA answer
4 3/51 @)
2" (24-4x2) 2m.25(2'-1) OR
saa OR =
275.8 oP 275
Ler or as or 0,2 vA simplifying terms in numerator
40 5)
OR
gmt4 4 pmtl vA adding like terms in numerator
5.2m+3
2” 16-8." ¥CA answer
5.28 (3)
_ 278 OR
5.2.8
=o. or Ee or 0,2
~ 40 5 > vA splitting to two fractions
gmt _ 4gmtl
5.2mt3
_ gms = ggmtl YA simplification
2758 ISB
gmt4-m gy gmtl-m YCA answer
40 40 (3)
16 8
40 40
=5 or iL or 0,2
40 5
1
222 “os YA equating ow to 5
1
5= 5
5:51
5 ise = 5° VA gas oy 5
1+3x=0
1
x= 3 ¥CA answer
GB)
OR OR

YA equating oe to 5

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CAPS — Grade 11
Marking Guidelines

KZN —November 2025

i YA 5=5*
125*
all
53 ¥CA answer
at @)
1=-3x.
1
x=——
3
vA isolating surds
vA squaring on both sides
¥CA standard form
x=11 or x#-3 YCA answers with selection.
(4)
[17]
QUESTION 3
Bl; 14 vA answer
Q)
3.2 2;6;10;14
T, =4n-2 YA 4n VA -2
(2)
3.3 Tj. =4(10)-2 or 7, =[2(10)-1]x2 YCA subst. of n=10
Ty =38 Answer only full marks vCA answer Q)
3.4 T, =4n-2
710 =4n-2 “CA, T= 710
4n=712 Answer only full marks OR
n=178 If n is a decimal, it must be YCA answer
rounded to whole number (2)
OR OR
¥GCA. T,=355

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CAPS — Grade 11
Marking Guidelines

KZN —November 2025

¥CA answer
Q)

2n~1-=355
2n=356 Answer only full marks OR
n=178 If 1 is a decimal, it must be
35 2p 614; 26 ccc hearts qi rounded to whole number
CS a) 0 first difference
MA, essscervssesveneoud second difference
2a=4 3(2)+b=4 2-2+c=2
a=2 b=-2 c=2
T, =2n? -2n+2

266 = 2n -2n+2
2n? —2n+2-266=0
n’—n-132=0
(n-12)(n+11)=0
n=12 or n#-I11

whole number

Only the 12" mes: pemnorephaemeom
OR
133 fF BEB eecccceee’ faces quad seq
. first difference

n must be a positive whole number

Only the 12" message

YA 14 and 26 in NEW pattern

VA ALL values of a, b&c
YCA general term
YCA T, =266

¥CA factors

¥CA both answers with rejection

(6)
OR

YA 7 and 13 in NEW pattern

YA ALL values of a, b&c
YCA general term

YCA T, =133

¥CA factors
¥CA both answers with rejection

(6)

[13]

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QUESTION 4

CAPS -Grade 11
Marking Guidelines

KZN — November 2025

4.1. Dlilfm; 5357
(m+7)-3 (5—m) ; (n—5)..
—2m-2 ; m+n-10..

first difference
..Second difference

vA sec diff. in terms of m and n

m2 = 4 “.m+n-10=4 2m = = -IO=
" un 4.1 is a proof, all VA -2m=6 VA -34+n-10=4
—2m=6 —3+n-l0=4 marks are accuracy. (3)
m=-3 n=17
4.2 FG RG LD: sssesoanas
4n=292 VA 4n=292
n=73 YCA value of n
-.T,, and T,, ina qu YCA Ty & Try
G3)

43 T, =2n?-2n-7
1005 = 2n? —2n—7
2n? —2n-1012=0
n’ —n-506=0
(n—23)(n+22)=0
n=23 or n#-22

VA T,=1005

vA standard form
¥CA factors
¥CA both answers with rejection

T,, =1005 (4)
[10]
QUESTION 5
Sel
. 3 1
pe: oe
(92
=a 1
0)= ie
f= 3*2
y=-l vA answer
Q)
5:2 yea? vA answer
armies vA answer
2 (2)

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### Page 37

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CAPS — Grade 11

KZN —November 2025

Marking Guidelines
53 ‘ 3 4641
ye=—_ + —
A ) Mi2 2
Ant
0=— 4+ — VA I =
see 2 et y=0
seg |
Alle 2
Px=2=26
= Answer only full marks,
ae and ALL marks are
x=4 accuracy since it is given vA answer
(2)
5.4
YA shape
¥CA both asymptotes
¥CA both intercepts
GB)
5.5 vv A answer
(2)
OR OR
1
ve(-»:5] or ye[-1; 0] ¥vA answer
2
(2)
5.6 | p(xy=—_42 VA VA
x-2 z-2
(2)
OR OR
3 -3
P(x)= +2 vA vA +2
ged —x+2
(2)
[12]

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### Page 38

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CAPS -Grade 11
Marking Guidelines

KZN —November 2025

VA TP (5510)

YA both x-intercepts.

vA shape
DO NOT penalise if the
y-intercept is shown
(3)
6.2.1 | -x2+49=0 YA let y=0
x -9=0
(x+3)(x-3)=0 YA correct factors
x=-3 or x=3 Answer only full marks
¥CA answer
A(-3;0) @)
OR OR
-x°+9=0 YA let y=0
x =9
cee YA isolating the square
x=-3 or x=3 Answer only full marks
A(-3; 0) ¥CA answer
@)

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### Page 39

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CAPS — Grade 11
Marking Guidelines

6.2.2 | Otherx—intercept inh is (9 0) YA identifying(9 ;0)
ysa(x—x,)(x-x,) oe os is a proof,
i -9 all marks are accuracy.
WAAL —9) VA substitution of »x—intercepts

= 2a G40+3,0-9) OR -9=a(3+3)(3-9) 27

VA subst.of (0 77) or

27
f= a(-27 OR —9=-36a =
4 al ) (3; 9)
1
a
4
i
a 7 DR 2)
yeni 9 vA simplification of A.
4
xp 2 ot 8
Pe 2
me 2 and b=->5 (4)
4 5
OR OR
C(3;-9) YA identifying C(3;—-9)

Since 6.2.2 is a proof,

_ 2
yaa(x+ py +q all marks are accuracy.

y=a(x-3)?-9 YA substitution of C(3;—9)
27 .
0=a(-3-3)-9 OR a = a(0—3)?—9 VA subst. (0 2) or (-3;0)
9
9 =a(36) OR 47%
1
a=—
4
=1(-37-9
74
1
y= at *—6x+9)-9 A simplification to h.
123,27
ar a ear
4
nase and ba 4)
4
623 | C(3;-9) VE od
yeR:y2-9 Answer only full marks VA answer
-- --9) om
, Answer only full marks YA Yo=W9
ye[-9; 0)
vA answer
(2)

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CAPS — Grade 11
Marking Guidelines

KZN — November 2025

Reflection about x-axis, and
shift-of 5 units up

6.2.4

OR

shift of 5 units down , and
reflection about the x —axis

Must be in this order

Must be in this order

YA reflection about the x—axis

YA Sunits up

(2)
OR

VA 5 units down
YA reflection about the x—axis

(2)

6.2.5 1, 3 27

=x? -x-—=-x7 49
4 2

Sepa: 63g

4 2; 4

5x° -6x-63=0

(x+3)(5x-21)=0

x=—3 or el - 4, 25
5

vA equating both equations

¥CA standard form

¥CA correct factors

ee
D(4,2 ; -8,64) YCA value of x YCA value of y
: (5)
6.2.6 | h(-l)=-5 & h(7)=-5 VA x-value of —1 or 7
“k=—-5 Answer only full marks “A answer
(2)
[21]
QUESTION 7
7A h (x) =2*+q
as 8 3
_ YA substitut 2; -7T—
“15227 +4 substitting { $ ;)
vA answer
=-8
2 Q)
72 2°-8=0 VA let y=0
2 =23 YA 2 =23
x=3
A(3;0) Answer only full marks || VA answer
GB)

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### Page 41

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CAPS —-Grade 11
Marking Guidelines

KZN — November 2025

73 fOy=2°-8
f@)=-7
B(0 >-7) YA coordinate of B
m=. Jn 7 VN
eX
0-(-7
m= 7) YCA substitution
3-0
—— Z ¥CA answer
3 G3)
4 O<x<3 vv AA answer
OR (2)
OR
xe[0;3) v¥VvAA answer
Q)
75 T<k<8 ¥¥CACA answer
(2)
OR OR
ke(7;8) ¥¥CACA answer
(2)
[12]
QUESTION 8
8.1. A=P(1+i.n)
A=R 15000(1+0,10x6) vA subst. into correct formula
A=R 24000 VA value of A
Interest = R24000 — R15000
=R 9000 ¥ CA answer
GB)
OR OR
Interest = pxixn vA correct formula
=15000x 0,16 YA subst. into correct formula
=R 9000 ¥CA answer
QB)
8.2 A=P(1-7)"
— ae 3
R29) vA subst. to correct formula
455625
(1—0,0825)°
= ¥CA answer
“. P= R358 922
(2)

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CAPS — Grade 11
Marking Guidelines

8.3 - \m
Lg fey fom YA correct formula
os m
‘ 4
140,12= ( + i) vA subst into correct formula
F =4(¥,12 = 1)x 100% YCA simplification
r=11,49% ¥CA answer
(4)
8.4 A= P(l +i)’
A=R250000{ 1+ vA D,0225) v n=19
A=R59022 ¥CA answer
GB)
8.5 12x2 Fi ) 4x2 0. 1B 24
6soo(1+- 933 [+ } =7428,30 | VA 6s00{1+%2)
12 ; 12
24 ; 0,105)”
v ,
6s00{1+%9 (i+° | —7428,30 “ [1.3 }
x= 8
( caer YA «(14 23]
1+ 4
YA all equated to 7428,30
x =R3300 ¥CA answer
(5)
OR OR
4 4 8 0,13)"
6500 —_ {1 ot) = (i) =7428,30 | “A 6s00{ +28)
12 4 4 ;
0,105
«os vA (re ) ox
-x= ae j -500{1+ 51> (6 } 4
Bs 8
Coe va (148385
—x=-R 3300 VA ALL equated to 7428,30
“x = R3300 ¥CA answer
(5)
OR OR

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### Page 43

MathemRawnloaded from Stanmorephysics.com

CAPS —-Grade 11
Marking Guidelines

KZN — November 2025

A,= e501 +

| 7428,30

of mos
[fp¥.t0>
4

pS wy

3) ( nies)
.| L+—— ] -x.
12 4

...reverse equation

forward equation

7428, 30

6500{1+

0,13)"
a} (It

1+

0,105 }

mes)
4 “(

24
vA es0o( 1+ 93?)
12

4
vA (122) —x

a108)
4

vA [+

VA ALL equated to 7428,30

4 4
oy = 148.30 659 )
( 0,10. }
1+ —
4
—x =—-R 3300
“x =R3300 Y CA answer
(5)
OR un
24 4
A, =6500( 1+ 923) (1-22°] soovenves forward equation 5
0,13
v ,
rane A 6s0o{ + 3
f= aT t+ Xeseesseed reverse equation ;
(1) En [1+ 2426) .
4 4
A,=P. 5
37 F3 . , VA (1-222)
6500 14.2:13 1 4.2:105 _ _7428,30 4
1D) js | 4) ¢ 0.05% YA ALL equated to 7428,30
1+ —
Gre
24 4
x=6500{1+ 3} (10808) ——7428,30_
12 4 ( a
ara
¥CA answer
[17]

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### Page 44

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CAPS — Grade 11
Marking Guidelines

KZN — November 2025

QUESTION 9
9.1.1 | P(AandC)=0 vA answer :
9.1.2

P(AorC)=P(A)+P(C)—P(A and C)

| P(A orc > Kore) =0,47+ 3-0
7

P(A or C)=0,90

P(A or C)=P(A) + P(

P(A or€)=0.47 +5

vA substitution to a correct rule
¥ A answer

(2)
OR

YA substitution to a correct rule

P(A orC)=0,90 vA answer -
9.2.1 | P(A and B)=P(A)xP(B vA P(A and B)=P(A)xP(B)
P(A and B)==P(B)xP(B)
P(A and B)==P(B)°
° vA answer
(2)
9.2.2, | P(A or B)= P(A) + P(B)— P(A and B)

0,7= =P) +P(B) -2P(By
= 1 xp) -2 P(By?
0.7= <P(B) ~=P(B)

206) ~ ; P(B) st =0 OR 4P(B)’-14P(B) +7 =0

T-Oe

P(B) =

‘)

5
2

op) = 2 eV) 40907)

2(4) If answer is CA,
P(B)=0,60 or P(B) # 2,90 0< P(B) <1

OR

YCA substitution to correct rule

¥CA equation in terms P(B)

YCA substitution into quadratic
formula

¥CA answer

(5)
OR

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CAPS — Grade 11
Marking Guidelines

KZN — November 2025

P(A or B) = P(A) + P(B) — P(A and B)
0,75 P(A) + SPA) = m(A).> P(A)

7 5
0,7= —P(A) -=P(A)?
pF) = MA)

5A)" ~7P(A) + =0 OR 25P(A)’ —35P(A) +7 =0

‘TO

P(A) =

¥CA substitution to correct rule

¥CA equation in terms of P(A)

YCA substitution into quadratic
formula

P(A) =

P(A) =0,24 or P(A) 4 YCA P(A)

P(A) = 2PCB)

0,24 = =P)

. _ 3 If answer is CA

-. P(B) = 0,60 or z 0<P(B) <1 ° ¥CA answer
[10]

QUESTION 10

10.1.1

Let the number of females who will love to buy Gucci be x

Gucci Louis
Vuitton
Males y 2x 45
Females a 3y 85
2x+3y 130
: x=85-3y
x+3(45-2x)=85 OR  y+2(85-3y)=45
x+135-6x =85 OR y+170-6y=45
—5x =-50 OR —S5y=-125
x=10 y=25
x=10

x must be a positive whole number

VA males : y+2x=45

VA females: x+3y=85

YA y=45-2x OR x=85-3y
¥CA substitution

YCA simplification

¥CA answer
(6)

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CAPS — Grade 11
Marking Guidelines

KZN — November 2025

OR
Gucci Louis
Vuitton
Males y 45-y 45
Females 85—3y 3y 85
85-2y 45+2y 130

45—y =2(85-3y)

OR

VA Mand LV = 45-y
VA Fand Gucci= 85-3

VAVA 45-y=2(85-3y)

45—y=170-6y
Sy =125
” 95 MCA values of y
_ , CA answer
x=10 females who would (6)
OR
OR
Allow n to represent total number
n(M)=85 & n(F)=45
n(M and G)=y
n(M and LV)=2(85-3y)=170-6y VA 2(85-3y) or 170-6y
n(F and G)=85-3y VA 85-3y
n(F and LV)=3y
-.2(85-3y)+y=45 YAVA 2(85—3y)+y=45
170-6y+y=45
Sy =-125
yeas CA value of y
x=10 females who would love Gucci items CA. answer.
(6)
10.2 25 25 5
P(M == VGA. — — 0,19
(M and G) 130 mag ge
=0,19
45 35 45. 35
og 76K 22 2
P(M)*P(G)= 739" 739 130 * 130
- ¥CA pica or — or 0,09
675 675 75
= 0,09
P(M and G) # P(M)xP(G)
Both events are NOT independent ¥CA NOT independent
(4)

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CAPS — Grade 11
Marking Guidelines

KZN —November 2025

OR OR
P(Mand Lv) = 72 2 2
130 ¥CA — or — or 0,19
bas B03
P(M)xP(LV) => x22 4s 95
130° 130 vca 4545
i71 130 130
676 ¥CA 1 oo 0,25
=0,25 676
P(M and LV) # P(M)xP(LV)
Both events are NOT indeper YCA NOT independent
(4)
[10]
TOTAL: 150
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