# Mathematics Grade 11 Paper 1 KZN November 2024 Paper & Memo

> Grade 11 Mathematics Paper 1 question paper and memo from the KZN November Exam 2024, 26 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-11-mathematics-2024-kzn-november-exam-question-paper-and-memorandum-paper-1/6f8dcc4b-3f3f-43e4-a6fd-4bac4da0ef94
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2024
- Exam period: KZN November Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 26

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## About this document

Mathematics · Grade 11 · KZN November Exam · 2024. Question paper and memorandum, 26 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

MLA’

Siahnorephysics.corn

MARKS:

TIME: 3 hours

This question paper consists of 8 pages and 1 information sheet.

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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

This question paper consists of 10 questions.
Answer ALL the questions.
Number the answers correctly according to the numbering system used in this question paper.

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 answers correct to TWO decimal places, unless stated otherwise.
Diagrams 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

11 Solve for x:

Orie) x?—7x=-1 (answers correct to TWO decimal places) (4)
1.12 =x? +2x+24<0 (4)
113 (2*-1)(2x-1)=0 (3)

(4)
(5)
xo +m? +4m
2 :
Calculate the smallest integral value of m for which the roots are non-real. (4)

1.4 Solve for x if mjrleadeclicdree (4)

[28]
QUESTION 2
2.1 Simplify fully, WITHOUT using a calculator.
(4)
(3)
2 4
2:2 Solve for x , WITHOUT using a calculator, x* —5x* +6=0 (4)
2.3 Solve for x and y, WITHOUT using a calculator, shy = ap (6)
[17]

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QUESTION 3

Consider the linear number pattern: 12 ; 9 ; 6;...

3.1

32

Determine the n“ term (general term) of the linear number pattern.

of the linear number pattern.

given linear number pattern? Justify your answer with an
e linear number pattern are the sequence of first differences

ce: 4x+y 5 —2x-1; y+3 3...
x and y.

QUESTION 4

Consider the quadratic number pattern: -13 ; —1 ; 73.0... 3; -2101.

4.1

4.2

43

44

4.5

Write down the FOURTH and FIFTH terms of the quadratic number pattern.

Show that the general term of the quadratic number pattern is
T, =-2n? +18n-29.

How many terms are in the above quadratic number pattern?
Determine the maximum value of Q, if it is given that Q, = 2.7,

Prove that all terms in the sequence of first differences are even numbers.

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(2)
(2)

(2)

(5)
[11]

(2)

(3)
(4)
(3)

(2)
[14]

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QUESTION 5

In the diagram, the graph of f(x) = 2 +q_ is drawn. D(4;3) isa point on f and Bis the y-
xX+p

intercept of f. The asymptotes of f intersect at (251).

D(4 ; 3)
f
7)
Sel Write down the equations of the asymptotes of /. (2)
52 Show that the equation of f is f(x) = oH (2)
x-
5.3 Calculate the coordinates of B. (2)
5.4 Determine the equation of the axis of symmetry which has a positive gradient. (2)
5:5 Determine the values of x for which f(x) <0 (4)
1 -t
5.6 The graph of / is transformed to obtain the graph of A(x) = (3 x) ‘
Describe the transformation, in words, from f to A. G3)
[15]

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QUESTION 6

The graphs of A(x) =ax° +bx+e and k(x) =—x+6 are drawn below. P and Q(6;0) are the

x-intercepts of A and S is the turning point of h. Graphs h and k have a common
y-intercept, RM lies on h and N lies on k such that MN is parallel to the y-axis.

x
6.1 Write down the coordinates of R. ()
6.2 It is further given that PQ is 8 units. Write down the x-coordinate of P. qd)
6.3 Prove that a= -5 , b=2 and c=6 (4)
6.4 Calculate the coordinates of S. (3)
6.5 Write down the range of h. qd)
6.6 Determine the values of x for which = >0 (3)
~*.

6.7 If g(x) =h(x)+d , determine the values of d for which g(x) will always

intersect the x-axis. (2)
6.8 Determine the maximum length of line MN when h(x) > k(x) (4)

[19]

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QUESTION 7

is defined by the following properties:

7A e “and g on same set of axes. Clearly show all intercepts
with the axes, and asymptotes, if any. (5)
7.2 The coordinates of the point of intersection of f and g are (-1; 4).
721 If f and gare reflected along the line x =1, write down the
coordinates of the point of intersection of the images of f and g
under this transformation. Q)
122 Determine the values of x for which 0< g(x)<4 (2)
72.3 Determine the values of k for which the roots of —2.f(x)=k are
negative. (2)
[11]
QUESTION 8
8.1 Miss Gwala invested R30 000 at an interest rate of 12% p.a., simple interest.
After n years, the accumulated value of her investment was R51 600.
Calculate the value of n. (2)
8.2 The total number of bees in an apiary are declining each year at the rate of 2,5%

of the total number of the previous year. There are currently 20 416 bees in the
apiary. How many bees in total were in the apiary 8 years ago?

(Round off your answer to the nearest whole number). (3)
8.3 Mrs Howard invested R10 050 on 1 January 2020 in an account that paid interest

at the rate of x % p.a., compounded quarterly. The accumulated amount in the

account on | January 2024 was R13 867. Calculate the value of x. (4)
8.4 Mr Mngayi invested R50 000 in a savings account for 7 years. The investment

earned interest at the rate of 14% p.a., compounded monthly for the first three
years. The rate of interest changed to 13% p.a., compounded half-yearly, for the
remaining four years. Mr Mngayi withdrew R6 032 from the account 5 years after
making the initial deposit.

8.4.1 Calculate the accumulated amount in the account three years after

making the initial deposit. (3)
8.4.2 Calculate the accumulated amount in the savings account at the end of

the seven-year period. (4)

[16]

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QUESTION 9

9.1 A and B are events such that P(A) = 0,38 , P(not B)=0,77 and
P(A or B) = 0,425. Calculate:
oat P(B) 2)
92) P(A and B) (2)

9:2) certain high school. The principal observed

sertain day. The results are summarised in the

rived late Absent Total
c 30
e m

h y

’ and ‘arrived late’ mutually exclusive? Give
(2)
9.2.2 The events ‘being a boy’ and ‘arrived on time’ are independent. It is
also given that P(being a boy and arrived on time) = =.

Calculate the value of y. (4)
[10]

QUESTION 10

10.1 A survey was conducted among 165 Grade 12 learners. These learners were asked
which car brand they preferred among BMW, Audi and VW.
The results of the survey are summarised below.
e 42 learners preferred a BMW.
85 learners preferred an Audi.
106 learners preferred a VW.
18 learners preferred a BMW and an Audi, but not a VW.
40 learners preferred a VW and an Audi, but not a BMW.
x learners preferred a BMW and a VW.
13 learners preferred all three car brands.
20 learners did not prefer any of these three car brands.

Draw a Venn diagram to represent the above information. (4)

10.2 x collected the mathematics workbook from each of the 25
al i Some of the learners in her class are boys and the remainder
are girl eC randomly selected the first book from the pile and marked it.

The te agai ndomly selected the second book from the pile and marked it.

The probability that the first two books selected belonged to boys is x

Calculate the number of girls in the class. (5)
19]
TOTALMARK [150]

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INFORMATION SHEET
_ -b+vb2 dae
2a
A= P(L+ni) A=P(\=ni) A= P(-i)" A=P(+i)"
T, =a+(n—1)d S, =F Ra+(—-pa]

T.=ar"! s, te) sr#l S,=—*-3 =ler<l
—_
pati -1 pals"

i i

rey ye F@th-S@)
i a

h
2 xX, +x +
d=y(%)-*%) +0.-»)° M Ata Me
2 2
y=mx+e y-y, =m(x-x,) m=22— m= tan 0
Fy Hy
(x-ay +(y—by =r?
invdBe =9§4_= 2b =e
sind sinB sinC
a =b? +c? —2be.cosA
area AABC => absin€
sin(a+ £)=sina.cos B +cosa.sin B sin(a@— £)=sina.cos B—cosa.sin B
cos(a+ 8) =cosa@.cos f —sina.sin B cos(a—f)=cosa.cos B +sina.sin B
cos’ a—sin’ a
cos2a =41-2sin’a@ sin2a@ = 2sina.cosa
2cos’a-l
x= i=l
n
P(Ay= 2)
n(S)
D=at+bx

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

150

These marking guidelines consist of 16 pages.

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G ll
Marking Guidelines
QUESTION 1
1.1.1 xl x=-1
x? hf i Fil= 0 VA standard form
is —b+ Vb? —4ac
2a
= =) + YET! ~ 400) YCA substitution
2(1)

x=0,15 or x=6,85

incorrect rounding

Penalise 1 mark for

¥CA answer ¥CA answer
(4)

x<—4 or x26
OR x€(—0;—4]U[6;0)

OR
—x? +2x4+24<0

x? -2x-2420
(x-6)(x+4)2>0
cevs:x=6 or x=-4

x<—4 or x26 —
OR xe(—-0;—4]U[6

VA factors

YCA critical values
( accept if shown on a sketch or
on a number line)

¥CA answer

(4)

YCA_ answer

OR

vA factors

YCA critical values
( accept if shown on a sketch or
on a number line)

YCA answer YCA answer

(4)

1.1.3 | (2*-1)(2x-1)=0
2*=2° or 2x=1
1

x=t
2

x=0 or

CA P=

VA answer YA answer

(3)

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GRADE 11

Marking Guidelines

KZN/November 2024

1.1.4 | J2x45=x43
(v2x+5 \ =(x+3)

QWxtS HN + 6x49

¥A_ squaring on both sides

x e4x44=0 YCA_ standard form
(¥#2)(¥+ 2) =0 ¥CA factors
x=-2 ¥CA answer
(4)
ie) Substitute equation | into equation 2:
y=2+5x and 2xy-7 =-x*
2x(24+5x)-7=-x° ¥A_ substitution
4x+10x? -7+x? =0
11x? +4x-7=0 ¥CA standard form
Y¥CA factors
YCA_ x-values
YCA_ y-values
(5)
OR OR
Substitute equation | into equation 2:
2xy-7=-x?
-2 —2y
(2?) y-7 - (=) YA. substitution
2 2
95] a7) ed HI os
5 25
5(2y° -4y)-175=-(y?-4y +4)
0=-y? -10y* +4y+20y—4+4175
O=-lly?+24y+171
O=Ily?-24y-171 VCA. standard form
0=(1ly—57)(y +3) YCA factors
y a or y=-3 ¥CA_ y-values
Gal
5— |-2
see td or valri}=2
soll or x=-l YCA_ x-values
1 (5)

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GRADE 11
Marking Guidelines
13 A=m> +4m
A<0
m+4m<0 VA m+4m<0
m(m+4)<0 ¥CA factors
cvs: m=0 or m=-4
<0 m must be || vCA_ interval solution for m
eeeeeenioreprveies com an integer YCA_ integer answer
: (4)
1.4
x= deirfert yx+ \x+ Vxtvkx....
x ae ae ee a YA. squaring on both sides
x =x4x VA x =x4+x
x = 2x
x’? -2x=0
x(x-2)=0 ¥CA factors
x=0 or x=2 ¥CA_ both answers
(4)
[28]

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GRADE 11

QUESTION 2

Marking Guidelines

KZN/November 2024

10**7.25*.8*
Ii (5.297? .5?*.2?*

i 2A 2 536+)
xt Ax+2 52x 73%
= |

PH2+3.x—4x-5

vA _ splitting to prime bases

YCA_ using the exponential laws
¥CA simplification

¥CA answer
(4)
2.1.2 3ie2 gio
_ 3303 VA 33
~ 37133 -6) YA correct factors in denominator
-
21
=? YCA_ answer
7 (3)
OR OR
glcargiol
324 6.3
3003
= Pg A 3203
3° (3'-6.3°) YA correct factors in denominator
1
3
3
=a ¥CA_ answer
GB)

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GRADE 11
Marking Guidelines
2 f
2.2 x3 =5x*+6=0
1 1
[er3}-3)-0 YA factors
i us
eater x? =3 ¥CA both equations

x =? or x3) =3° YCA raising both sides to power 3
x=8 or x27 ¥ CA both answers
(4)
OR OR
2 uf
x3 —5x3 +6=0
pit
let =k
k? -5k+6=0
(k-2)(k-3)=0 YA factors
k=2 or k=3
1 1
x=2 or 8 =3 ¥ CA both equations
1 1
yf ab? on tt ¥CA raising both sides to power 3
x=8 or x=27 ¥CA both answers
(4)
23 xalVy _f fp
2 2V2
3
_ p 3 vA =
33 72 ¥ A rationalising the denominator
2-2 °2 ie. multiplying by oH
_ fov2
4
_vova
V4
_3a2
2
ally _ 32
2 2
“x=3 and y=2 YCA x—value & YCA y—value
(6)
[17]
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GRADE 11

Marking Guidelines

KZN/November 2024

QUESTION 3
31. | L=—30tk
12=-3(l)+k vA substitution of constant
rae} difference and any given (” ; Tn)
Ty=A5=3n YCA_ answer
(2)
32 Ty =15=3(10) YA substitution of n=10
= ¥CA_ answer
Q)
33 T, =15—3n YCA T,=-316
—316=15—3n
—331=-3n
110,333 =n
.-316 is NOT a termina pattern ¥CA conclusion
(2)
3.4. 4x+y;—-2x-1; y+3 5.......
12 9
—2x-1—(4x+ y) =12
-2x-1-4x-y=12
-6x-y=13
FB — TS iirccssasncaanl (1) YA first equation
yt3—-(-2x-l=9
yt+34+2x4+1=9
YA_ second equation
YCA substituting (1) into (2)
YCA value of x
y=-6(-4,5)-13
y=14 YCA value of y
(5)
11]

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GRADE 11
Marking Guidelines
QUESTION 4
4.1. 118.7 vv AA answers (2)
4.2 HVA TIE Tf vsscsvccssccsz 3 —2101
T2854 Since Q4.2 is a proof,
mAl; —4 all marks are accuracy.
2a=-4 3(-2)+b=12 —24+18+c=-13 VA 2a=-4 VA 3(-2)+b=12
“.a=-2 b=18 c=-29 YA —24+18+c=-13
G3)
OR OR
2a=-4 VA 2a=—4
a=-2
T, =-2n? +bn+e
-1=-2(2) +b(2) +e YA -1=-2(2) +b(2)+e
7=2b+c
Fa Ee), (1)
T, =-2n? +bnt+c
7 =-2(3) +b(3)+e YA 7=-2(3) +b(3)+c
C= 25 —BDrecssssssssssssssssssssnssesseesesssseees (2)
“.7-2b =25-3b
b=18
ce=7-2(18)
c=-29 (3)
43 T, =-2n’ +18n-29
-2101=~-2n? +18 -29 VA equating T, to —2101
0 =-2n? +18n-29+ 2101
0 =-2n? +18n + 2072
0=n? —9n-1036 ¥CA standard form
0 = (n+ 28)(n-37) ¥CA factors
n#-28 or n=37
There are 37 terms in a pattern ¥CA answer with rejection
(4)
44 The value of a is negative. ». T, has a maximum value.
Ta= Ts'= 11. YCA T4=Ts=11
.. the maximum value of T, is 11. YCA maximum value of T, is 11.
The maximum value of Q, is 2(11) = 22 YCA answer
Answer only, full marks (3)
4.5 T, =—4n+16 YCA general term of first
=2(-2n+8) differences
2(-2n +8) is a multiple of 2. ¥CA common factor of 2
:. the sequence of first differences are even numbers. (2)

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Marking Guidelines

KZN/November 2024

[141 |
QUESTION 5
Si x=2 and y=1 YA answer & “A answer
Q)
5.2 : a
Sx) = +q
x+p
a
IQ)= at 1 “A substituting asymptotes(2;1)
3-2 43 “A substituting point (4;3)
4-2
2=2
2
4=a
4 Since the 5.2 is a proof,
“ fQ)= = i all marks are accuracy. 2)
x
53 4
0)= +1 YA let x=0
FO) 0-2
f0)=-1
B(O;—1) vA answer
(2)
5.4 m=1 m=1 vA value of m
y=lxt+e y-yY, =m(x-x)
1=12)+e OR y-1=1(x-2)
“jm yslx-2+1
yalx-l “yelx-1 YA answer
2)
55 | g2-4 41 VA let y=0
x-2
y= 40
x-2
—2=x YCA x—value
asa? ¥VCA answer
(4)
1 -I
5.6 A(x) =| —
(x) { i+]
h(x) = (4x)
h(x) =4).x1
h(x) = as VA_ simplified equation of h.
x

translation of 2 units to the left , and
1 unit down

vA translation of 2 units to the left

YA lL unit down

@)

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GRADE 11

Marking Guidelines

[15] |
QUESTION 6
6.1 R(O;6) VA coordinates of R
Q)
6.2 xp = 6-8
YA answer
Q)
Since the 6.3 is a proof, all WK Sabstof —2.ands
marks are accuracy.
vA substitution of (0;6)
n ya-Z (x4 2\0-6) v A substitution of 5
1
a aes (x -4x-12) YA simplification
ayaoge? +2x +6 @)
OR OR

YA substitution of (0;6)

YA. setting up 2 equations

YA solving equations
simultaneously

YA simplification

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GRADE 11

Marking Guidelines

paa(x+ p)+q
ySa(x-2)P +q
6=a(0—2)’+q

x= —2

0=a(-2-2/ +q
O0=1l6a+q
—1l6a = q... see(2)

-.6—4a=-l6a

12a =-6

l2a_-6

12) «12

1

a=-—

6.4

(4)

y=-3(2) +2(2)+6

=8

S(2;8) Answer only, full marks

¥A_ substitution

¥CA value of x “CA value of y

(3)

6.5

yell,y<8 OR ye(-~;8]

¥CA answer

()

6.6

x<-2 or 0<x<6 OR xe(-—«;-2) or xe[0;6)

¥vvAA answer

(3)

6.7

For roots of / to be real ,
h cannot be shifted downwards by more than 8 units.

2d2-8
OR

¥vCA answer
OR

(2)

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GRADE 11
Marking Guidelines
y= ris +2x+6+d
A=b?—4ac
AQ) 74{-F}(6+a)
A=16+2d ¥VCA answer (2)
16+2d =0 (A 20 for intersection with x-axis)
2d >-16
“d2-8
68 | MN(@)=A(x)—k)
= (-$# + 2x+6)-(6-x) vA subtracting two equations
= -5 x? 43x YCA_ expression of MN
b
s=-—
2a
__ 2
-3)
2
=3 YCA_ value of x
MN(x) = -50) +3(3)
) 1
=— or 4— or 4,5 ¥CA answer
2 2 (4)
[19]
QUESTION 7
dole
for g:
YA x-intercept
YA y-intercept
for f:
YA y-intercept
YA decreasing shape
YA asymptote
(5)
7.2.1 | (334) vA value of x “A value of y
(2)
7.2.2 | -3<xs<-l vv AA answer

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GRADE 11
Marking Guidelines
OR OR
x e[-3;-1] vv AA answer
(2)
723 | k<-6 ¥¥AA answer
Q)
OR OR
ke(-2 ;-6) vv AA answer
(2)
[11]
QUESTION 8
8.1. A=P(1+i.n)
51600 = 30000(1+0,12.7) YA _ subst. to correct formula
51600 _
30000
0,12 n must bea
cn =6 years positive integer ¥YCA answer
(2)
8.2 A=P(1-i)" vA reducing balance formula
20416 = P(1—-0,025)* YA. substitution
20416 _
(1-0,025)* Penalise a mark if NOT
Px 25000 bees rounded to whole number || /CA answer
GB)

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GRADE 11

Marking Guidelines

KZN/November 2024

8.3 A=P(1-+i)"

139867= 100s0{1 F “y YA subst. into correct formula

& YA : &16
13867 _\\4_
toos0 | YCA simplification
+. X¥=0,0813000046.....x 100%
x =8,13% ¥CA answer
(4)
os

8.4.1 aul _

Yo ¥ Ys Ys % OY % Y,

i Y A , J

14% p.a. monthly 13% p.a. half—yearly
A=P(1+i)”
0.14) vA _ subst into correct formula
A=s0o00{ 14214) du 1. Ga
12
A=R75913,30 YA answer
@)

gao | AqPati"

0,13

4
A, -75913,30{ 1+ 23)

A, = R97 659, 906—R6032

=R91627,906

4
A,= R91627,906{ 1+ “8)

answer

A, =R117876,22
(4)
OF OR
0,13)"
; ‘ YCA rsoiga0(1+%2
A,= 75913,30(1+%2) 6032 (1-28) :
: 0,13)'
YA-R6032 VA (1-23)
—_— Y CA answer (4)
[16]

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Marking Guidelines

11

KZN/November 2024

QUESTION 9

9.1.1. | P(B)=1—Pimot B)
“i-RiPl ¥A_ substitution to correct rule
0/23 YA answer

(2)

9.1.2

P(A or B) = P(A) + P(B)—P(A and B)
0,425 = 0,38 + 0,23 —P(A and B)
P(A and B) = 0,38 + 0,23 —0,425
P(A and B) = 0,185

P(A and B) = 0,19

Y A. substitution to correct rule

YCA_ answer
(2)

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GRADE 11
Marking Guidelines
9.2.1 NOT Mutually Exclusive, YA NO
P(girl and arriving late) # 0. vA reason
(2)
9.2.2. | P(boy and arriving on time) =P( boy )xP (arriving on time) vA tule of independent event
30 40
Bayne asl VR Phe =
16-—yl) y y y
3y° =19200
3y? _ 19200
5 3
y? = 6400
y= 6400
y=80 y must be a positive whole number ¥CA answer
(4)
[10]

QUESTION 10

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GRADE 11
Marking Guidelines
10.1.1 Ss
YA 20
YA 18 and 40
VA x-13
YA 24~-xand 66-x
(4)
10.2
Tx xol vA x va
20 25 24 25 24
7 xx vA picasa <
20 600 20
4200 = 20x — 20x
x?-x-210=0
(x-15)(x +14) =0
x=15 or x#-14 YA answer (boys)
total number of girls = 25-15
=10 vA answer (girls)
(5)
[9]
TOTAL: 150

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