# Mathematics Grade 11 Paper 2 KZN November 2023 Memo

> Grade 11 Mathematics Paper 2 memorandum from the KZN November Exam 2023, 37 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/maths-grade-11-november-2023-p2-and-memo/1340a28a-d72c-4dd4-b8c0-744d71b3998a
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2023
- Exam period: KZN November Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 37

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## About this document

Mathematics · Grade 11 · KZN November Exam · 2023. Memorandum, 37 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

      �----------------------- ------------------------�
     �:                                 MATHE       ATICS P2                                  ::
     II                                                                                       II
     II                                                                                       II
     II                                                                                       II
     II                                                                                       II
     II                                                                                       II
     �                                                                                        �
     �    �-----------------------------�----------------�                                    �

[ NAME:                                                                       CLASS:
                                                                                                         ]
    MARKS:           150                                                      TIME: 3 hours

                                         For educator use only.
  Question     Ql        Q2    Q3       Q4    Q5    Q6     Q7     QB     Q9     Ql0    Qll     Total
   Marks
  obtained
  Question
    total
                10       12    18       13    25     11     10     6     11      13     21         150

 This question paper consists of 24 pages, including pages with additional space and a formula sheet.

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Mathematics P2 . KZN/November 2023
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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1. This questio r consists of 11 questions.

2 Answer AL stions.

* 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.

S Diagrams are NOT necessarily drawn to scale.

8. Write your answers in the spaces provided.

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KZN/November 2023

QUESTION 1

Aphelele plays for his school’s cricket team. The number of runs scored by Aphelele in each of the
eight games that he batted in, is shown below. (Aphelele was given out in all of the games.)

aca} 8 19 7 15 32 14 12
La Determineithe \verage number of runs per game scored by Aphelele for these eight games. (2)
1:2 Determine the standard deviation for this data set. (1)

1.3 Aphelele wants to be selected for the District trials. The ds to have
at least one score above two standard devi S n y ‘eight games.
Will he be selected for the District trials? Mot culations. (2)

14 Aphelele hopes to achieve an average of 20 runs per game for his first 13 games.
What should his average number of runs per game be for the last five of these games in
order for him to reach this goal? (3)

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a3 Brian plays in the same team as Aphelele. After having played a few matches, he has an

average batting score of 37 runs per game, and the standard deviation of his scores is 13.

In the next two matches that he played, he scored respectively 20 runs and 0 runs.

Will the inclusion of these two scores cause the standard deviation of his scores to increase

or Gecraniey ty your answer. (2)

peal

[10]
QUESTION 2
The table below shows the height (in cm) of 250 Grade 11 learners.
4 Number of Cumulative
Height (cm) learners (f) frequency
145< x <150 26 6
150< x <155 A 29
60 89
74 B
52 215
32 247
3 250
250
yaa Calculate the values of A and B. (2)
2.2 Calculate the estimated mean height of the learners. (3)
23 Write down the modal class. (1)

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2.4 On the grid provided below, draw a cumulative frequency graph (ogive) to represent the
data on the height of the Gr. 11 learners. (3)
OGIVE
ey
S
s
S
=
2
8
s
£
=
1S)
140 145 150 155 160 165 170 175 180 185
Height of Gr. 11 learners in cm
2.5 Use the ogive to estimate the interquartile range of the data. Show all your calculations. (3)

[12]

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QUESTION 3
In the diagram below, the vertices of AABC are A, C(4;—4) and E(-6 ;-9).
The angle of inclination of AC is a. ACE = B. The equation of AC is 3x+ y—8=0.
y] A
a.
oO X
C(4;-4)
E(-6;-9)
3.1 Write down the gradient of AC. (2)
3.2 Calculate the size of a. (2)

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3.3 Calculate the size of /. (5)
3.4 Determine the equation of EF if F is a point on AC produced such that EFA = 90°. (4)
3.5 Calculate the length of EF. (5)
[18]

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QUESTION 4

Triangle PRS has vertices P(—1;3), R(3;—1) and S(a; b), as shown in the sketch below.

41 Calculate the coordinates of T, the midpoint of PR. (2)

4.2 If the perpendicular bisector of PR passes through S, show that a=b. (4)

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43 If a<0,b<0 and the area of APRS =12 square units, determine the (7)
coordinates of S.
[13]
QUESTION 5
5.1 If sin33° =m, determine the following, in terms of m, without the use of a calculator:
5.1.1 tan 33° (2)

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5.1.2 cos777° (3)
5.1.3 — sin(-237°) (3)

5.2 Given that 4tan 8+5=0 and £ <[0°;180°]. Determine, with the aid of a diagram,
and without the use of a calculator, the value of V41cos 7. (4)
5.3 5.3.1 Simplify the following expression to a single trigonometric ratio:

sin(180° — #). cos(90° — 2) -1
cos(-—A) a

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5.3.2 Hence, determine for which value(s) of #, where # <[0°;360°],
sin(L80° - ).cos(90°= 8)=1 Witt be undefined.
cos(—f) (3)
5.4 5.4.1 Prove the following identity:
tana.sina + cosa = —
cosa (2)
5.4.2 Hence determine the general solution of:
tana.sina + cosa = 3.
sina (4)
[25]

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QUESTION 6

KZN/November 2023

In the diagram, the graph of the function f (x) = cos( 3 is drawn for the interval x <[-180°;180°].

+

-180° -150° -1209 -909 = -60° = -30® 30° 60° go° 120° = 150° 180°
1
2:
6.1 Write down:
6.1.1 the amplitude of f. (1)
6.1.2 _ the period of f. (1)
6.2 Draw the graph of g(x) =sin(x—30°) for the interval xe [-180° : 180°] on the axes
provided above. Clearly indicate all intercepts with the axes and the turning point(s). (3)
6.3 Write down the values of x in the interval xe [-180° ; 180°] , for which f (x).g(x) 20. (2)
6.4 Write down the values of x in the interval x € [oe ‘ 180°], for which g(x) = f (x) +. (2)
6.5 The graph of h is obtained by reflecting the graph of g in the y-axis. Write down the
equation of h. (2)
[11]

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QUESTION 7
WA. AABC has AB = 11cm, BC = 6 cm and AC = 15 cm.
A
15cm
11cm Cc
6cm
B
Calculate the size of B. (4)

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Mathematics P2

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te In the diagram below, P, Q, R and S are points in the same vertical plane.
The angle of elevation of R from Q is a and the angle of elevation of Q from S is 0.

PS=xunits and QRS= 7.

--2

P x s
xsin(a@ +0)

Prove that: SR = -
cos @sin 2 (6)

[10]

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QUESTION 8
The diagram below shows an open rectangular box, made out of sheet metal.
Its measurements are as follows:
e length = 27 n_,
e breadth = 2 mio
© height = 10/¢r H
4200]
10 cm
27cm
24cm
8.1 Calculate the total surface area of sheet metal needed to manufacture this open box. (3)
8.2 4,5 liter water is poured into this box. Calculate the depth of the water in the box.
Take note: 1 liter = 1000 cm’. (3)
[6]

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Give reasons for ALL statements in QUESTIONS 9, 10 and 11.
QUESTION 9
9.1 O is the centre of the circle. A, B, C and D are points on the circumference. AOB is a

straight line@SAC and BC are drawn. OD is drawn parallel to BC, and intersects AC in E.
The radius Of-the circle is 10 cm, and AC = 12 cm.
Calculate-the!length of ED. (6)

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9.2 In the diagram below, JG is a tangent to circle FGH at G. FG, GH and FH are drawn.
JGF = x-5° and FGH =3x.
Calculate the value of x. (5)
[11]

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Mathematics P2

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QUESTION 10

In the diagram below, RS is a diameter of the circle centred at O. Chord ST is produced to W.
Chord SP produced meets the tangent RW at V. R, =50°.

10.1 Calculate the sizes of the following angles.

10.1.1 R, (3)
10.1.2 Ww (3)
10.13 =P (2)

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10.2 Prove that V, = PTS (4)
10.3 Hence, prove that WVPT is a cyclic quadrilateral. (1)
[13]

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QUESTION 11
11.1 Use the diagram to prove the theorem which states that P+ R =180° (5)
Pp
Ss
Q
R

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11.2 In the diagram two circles are intersecting at B and D. O is the centre of the smaller
circle. B, C, D and E are points on the circumference of circle O. A is a point on the
circumference of the bigger circle. DEA is a straight line. OB, OD, OA, BA, BD and

BE are drawn. OA and BE intersect at F. A, =x.

11.2.1 Name, with reasons, THREE other angles equal to x. (4)

11.2.2 Calculate C in terms of x. (4)

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11.2.3 Prove that AB = AE. (5)

11.2.4 Prove that AB is NOT a tangent to circle BCDE. (3)

[21]

TOTAL MARKS: 150

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Mathematics P2

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Mathematics P2 fy = way KZN/November 2023
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INFORMATION SHEET
x= —b+vb* —4ac
2a
A=P(1+ni) A=P(1-ni) A=P(1-i)" A=P(1+i)"
T, =a+(n—2)d s, =F [a+ (n-2d]
T, =ar"* Ss = M1). at S, =); -1<r<1
" re 1-r
i i
£"(x) = lim +h) = £00
hoo h
X+X% Wty
d= (x, -x,) +(Y2-y,)? Daa 3 +]
—' _ = Yo =
y=mx+c y-y, =m(x-x,) m=-2—+ m=tand
xy — X&
(xa)? +(y-by =r?
Inaape; —42_=—P___&_
sinA sinB~ sinc
a’ =b? +c? — 2bc.cosA
area AABC = sabsin Cc
sin(a + #) =sin acosf+cosasin 2 sin(@ — 8) =sinacos B—cosasin B
cos(a + #) = cosa cos f —sin asin 8 cos(a — 8) = cosacosf+sin asin B
cos’ a—sin’? a
cos2a@ =41-2sin’ a sin 2a = 2sinacosa
2cos* a-1
_ yx (x, — x)?
x a a=
n
P(A) = ae P(A or B) = P(A) +P(B) — P(A and B)
x-xX\y-V
ynanbx p-Dl-RKy-9)

Yi (x-xP

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} KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

6
~

MASELON

150

This marking guideline consists of 14 pages.
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Mathemawmaploaded from Stanmorephysics.com KZN/November 2023
GRADE 11
Marking Guideline
QUESTION 1
1d ee
_ 128 vy 128
= 8
-16 v answer
(2)
12 6=7.55 Y answer
(1)
13 Two standard deviation above the mean:
X-25 x -16 ¥ X4+15  -X+25
+ —_—_ | —}—+
0.9 8,45 16 23.55 31,1
X+25 =16+2x7.55=311 v311
Yes, he will make it to the trials, he has one score (32) that is ¥ yes and motivation
more than two standard deviations above the mean (2)
14 Total number of runs in 8 games = 128
Total number of runs in 13 games =13x 20 = 260 ¥ 260
Average in last 5 games = anes y 260-128
5
= 26,4
His average must be 26,4 or approximately 26 runs in his last5 | v 26,4 or 26
games. (3)
is Increase. y¥ increase

Both 0 and 20 are further than one standard deviation from the

mean, and will therefore cause an increase in standard deviation.

Y both 0 and 20 are further than
one standard deviation from the

mean

(2)

[10]

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GRADE 11
Marking Guideline
QUESTION 2
2.1 | A=29-6=23 ¥ A=23
B=89+74=163 v B=163
7 (2)
40630
= v 250
v answer
(3)
v answer
2.3
(1)
2.4
vA

all points correctly
plotted

¥ grounding
¥ shape
20
_ a 2 a, 2 i 3
3
140 145 150 155 Qhico 165 Q 170 175 180 185 (3)
25 Position Q1: 3 (250) = 63"score
Position Q3: = (250) =188"score
Q, =158 Note: YQ.
Q, =167 Quaccept 157 —- 159 vO
3
“IQR =167—158 Qs accept 166 — 168 ‘
IQR accept 7, 8, 9, 10 and 11 answer
=9 (3)
[12]

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KZN/November 2023

GRADE 11
Marking Guideline
QUESTION 3
*-
A Note: A learner may
introduce any point.
6,
Qe" -
a S
3.1 | y=-3x+8
: v answer
“Me =—3
3.2 | tana=m,, =-3 Y¥ tana =Mao
a =180° —71,57
=108, 43° v answer
(2)
_-4-(-9)
3.30) Mec = 4-(-6) Y substitution
1
3 v value of m.,
Let @be the angle of inclination of EC.
tan@ = 5
C7 ;
0 = 2657" 26,57
[vertically opp. 2 s]
[Zs ona straight line] ¥ 180° —108,43° = 71,57°
43° =71,57°
=71,57° +26,57° [ext. Z of ABCG] Y answer
= 98,14" (5)

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GRADE 11
Marking Guideline

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KZN/November 2023

34 | m,. =-3
Mig, [AC 1 EF] vm, =2
+) Mee = 3 EF 3
Equation of EF y= F X+C
Substitute E -9): -9= 38) c ¥ substitution
c=-7 v value of c
fn te-7 v answer
8 (4)
3.5 | To determine coordinates of F:
~3x+8= i X=7 Y equating the equations of EF
3 and AF
a x=15
3
x=4,5 V x-coordinate of F
y =-3x +8 =-3(4,5)+8=-5,5 ¥ y-coordinate of F
z bi Y substitution in distance
EF =,/| 4,5-(-6) | +] -5,5-(-9
il ( ) [ ( )] formula
=11,07em Y answer
(5)
[18]
QUESTION 4
ay
P(-1;3
4.1
Y substitution in midpt. formula
T(1;1) answer
(2)

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GRADE 11
Marking Guideline

KZN/November 2023

341
Men “3
ml
mM

+ TT bisector =

4.2

Subst. T (1;

d m=lin y=mx+c:
1=1+c

»<o

)
x
b

Y gradient of PR
Y gradient of | bisector

Y subst. S(a; b) and m=1

v y=x
(4)

43 | PR=\(-1-3)° +(8+1)°

= 42

Buta=b, .

Area APSR

ST = f(a-1)? +(a-2)?
ST = /2(a-1)?
=12 square units

a ; x base x height = 12

5xPRxST =12

5(4V2)( (2-2 2 ) =12
5x W2x V2 x ‘a—1)? =12

(a-1 =3
(a-1) =9
a-1=43

a=4(N/A) ora=-2

“ b=-2
“.S(-2;-2)

Y length of PR

¥ length of ST i.t.o. aand b

¥ length of ST i.t.o. a (or b) only

v ; x base x height =12

¥ substitution of PR and ST

v values of a and rejection

¥ answer

(2)

[13]

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GRADE 11
Marking Guideline
QUESTION 5
5.4
v vi-m
tan33° = ¥ answer
vi-m? (2)
5.1.2 | cos777° =cos{ 2(360') +57" | ¥ cos| 2(360') +57" |
=cos57° Y cos57°
=sin33°
=m v answer
(3)
5.1.3 | sin(—237°) =-sin 237° Y¥ —sin237°
=—-(-sin57°) v -sin57°
=y1-m? v answer (3)
OR OR
sin(—237°) = sin123° ¥ sin123°
=sin57° v sin57°
= V1_-m2 v answer (3)
5.2. 4tanB+5=0
5 P 5
tan B = “a ; Val Y tanB= “|
A,
-§ ©) x
Vv r=V4l
Y value of cos 8
v answer
(4)

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GRADE 11
Marking Guideline

531 sin(180°— 8). cos(90°— 8)-1
cos(—f)
_ Sin B.sinB-1 Y simplification (numerator)
¥ simplification (denominator)
Y¥ use of square identity
Y answer
(4)
5.3.2 | If cos(—f) =0 ¥ cos(—) =0
cos 8 =0
B=90° or f=270° v 90°; v 270°
(3)
5.4.1 tana.sina+Ccosa
= SIN® sing +cosa y sina
cosa cosa
sin’? a
= +COSa
cosa
ine 2
eel are Oe Y simplification
cosa
_o1
cosa (2)
5.4.2 1 8 yl _ 3
cosa sina cosa. sina
sina =3cosa@
tana =3 Y tane=3
ref Z = tan*(3)
= 71.57°
@ =71.57° + k.180, keZ ¥ a@=71.57°+k.180°
V¥keZ
(4)
[25]

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GRADE 11
Marking Guideline
QUESTION 6
6.1.1 | 1 v answer
(1)
6.1.2 | period = (20° oar v answer
(1)
" fay “]
(120°;1)
v¥ Shape
v x &y - intercepts
Y¥ turning points
-180
(—60°;1)
(3)
-15
6.3 —180°<x<-150° or 30°<x<180° v
—180° < x <-—150°
¥ 30°<x<180°
(2)
6.4 x=120° or x=180° ¥ 120°
y¥ 180°
(2)
6.5 h(x) = sin(—x — 30°) ¥ ¥ sin(—x —30°)
=—sin(x + 30°) OR sin(x+30°)
(2)
[11]
QUESTION 7
71 AC? = AB? + BC? — 2(AB)(BC)cosB Y applying the cosine rule
15? =11° +6? —2(11)(6)cosB Y substitution
= 11° +6? -157
cos B = ———___——
2(11)(6)
17 a
=-33 OR —0,5151..... Y value of cosB
B =180° —58,99°
=121,01 v answer
(4)

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GRADE 11
Marking Guideline

KZN/November 2023

72 In AQRS:
Q, =Q5P =0 [alt 2’s =] vs
2. QSR =(a+0)
SR Q
sin(a +0) = ¥ substitution in sine rule
rS, 3
gp — QS: Sina V sR = QS:Sin(a +4)
AY sin B
In AQPS:
PS
cos 0 = Os ¥ correct ratio
cos0 = x
Qs
Xx x
Sai, ¥ QS = ——
Q cos@ Q cos?
x _ sin(a+) a
~ cso * ~sinB ¥ substitution of QS
sR xeiilo:t)
cos @sinB (6)
[10]
QUESTION 8
8.1 Total surface area
= (length x breadth) + 2(length x height) + 2(breadth x height) | v formula for surface area
= (27x24) +2(27 x10) + 2(24 x10) Y substitution
= 1668 cm? v answer
(3)
8.2 Volume = length x breadth x height Y formula for volume
4500cm* = 27cm 24cm height Y substitution
height = 4900
24x27
=6,94cm Y answer
(3)
[6]

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GRADE 11
Marking Guideline
QUESTION 9
9.1 € =90° [zZ in a semi-circle] Y SIR
E, =C=90° [corr. 2’s, OEI|IBC] Y SIR
AE= AG [line from centre | to chord] vSIR
AE =2(1
2
AE =6cm
In AOAE:
OA? = AE? + OE? [Pythagoras] v SIR
10° = 6? + OE’
OE? = 64
OE=8 cm YS (length of OE)
ED = OD-OE
ED =10-8
ED=2cm v answer
—_ (6)
9.2 H=FGJ [tan-chord-theorem] ¥SYR
=x-5
H=F [2s opposite = sides] v SIR
=x-5
F+FGH +H =180 [sum of 2s of AFGH] vs
x-5 +3x+x-5 =180
5x =190°
x = 38" ¥ answer
(5)
[11]
QUESTION 10
10.1.1 | R, +R, =90° [radius | to tangent] VSYR
R, = 40° v answer
— (3)
10.1.2 S, +S, =50° [tan chord theorem] YSYR
W+WRS+§ =180° [sum of Zs of AWSR]
W +90° +50° = 180°
W =40° v answer (3)
OR OR
T, =90° [Z in semi-circle] ¥SVYR
w=7,-R, [exterior 2 of AWTR]
=90° -50° = 40 V answer (3)

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GRADE 11
Marking Guideline
10.1.3 | PR, =R, =40° [Zs in the same segment] VSYR
: (2)
10.2 T, =90° [zZ ina semi-circle] v¥ SIR
PTS =F, =90°+T,
Vv, +S, =90° +S, [ext 2 of A =sum of int opp zs] | v S/R
But: T, [4s in the same segment] YSVYR
V, =PTS (4)
10.3 | VY =PTS
.. WVPT is a cyclic quadrilateral [converse: exterior 2=intoppz] |“ R
(1)
[13]
QUESTION 11
11.1
Pp
Note:
No construction: ;
Q Ss

NN

Construction: Draw OQ and OS

Y construction

0, =2P [Z @ centre = 2xz @ circumf.] Y SIR
0, =2R [2 @ centre = 2x2 @ circumf.] vs
0, +0, =360° [Z’s around a point] Y SIR
J. 2P+ 2R = 360°
2(P +R) = 360° vs
2. P+R =180° (5)
11.2.1 | B,=A,=x [Zs in the same segment] “Ss
DB, =B, =x [2s opp. = radii] VS YR
A, =0, =x [Zs in the same segment] YSIR (4)
OR OR
B, =A, =x [2s in the same segment] vs
D, =B, =x [2s opp. = radii] YS YR
A, =A, =x [equal chords; equal 2s] VSIR (4)

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GRADE 11
Marking Guideline
11.2.2 | -. BOD =180° - 2x [sum of Zs of ABOD] YSVYR
é= i. BOD [2 at centre=2x Zatcircumference] | VR
v answer
: (4)
11.23 E,= [ext. 2 of cyclic quadrilateral] YSYR
- =E [ext 2 of AFEA] “S
= (90° -x)+x=90
B, =180 (ee A.) [sum of Zs of ABFA] vs
=90° -
AB = AE [sides opp. = Zs] yR (5)
OR OR
E, =C=90' -x [ext. Z of cyclic quadrilateral] YSYR
F, =180 -(é, + A,) [sum of 2s of AEFA] vs
=90
In AABF and A AEF:
1. BF=FE [line from centre 1 to chord] vSIR
2. AF=AF [common]
3. F=F [Zs on a straight line]
.. AABF = AAEF [s;2;s]
/AB = AE [=As] YR (5)
11.2.4 | B, +B, =(90° -x)+x=90 ¥ “showing that
.. OBA = B, +B, +B, = 90° +B, > 90 OBA > 90
For AB to be a tangent to circle BCDE, OBA should be equal to 90°. | “For AB tobea tangent,

[converse: tangent radius]
.. AB is not a tangent to circle BCDE.

OR
For AB to be a tangent to circle BCDE, DBA should be equal to

OBA should be = 90°
(3)

OR
Y AB will be a tangent if

€=90°-x  [converse: tan-chord-theorem]. DBA =C =90°- x

But: DBA =B,+B,
= 90°-x+B,
=€+B, ¥¥ showing that

.. DBA>C DBA>C
AB is not a tangent to circle BCDE. (3)
[21]
TOTAL: 150

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