# Mathematics Grade 11 Paper 2 KZN November 2015 Paper & Memo

> Grade 11 Mathematics Paper 2 question paper and memo from the KZN November Exam 2015, 18 pages. Read online or get the free PDF download.

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- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2015
- Exam period: KZN November Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 18

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## About this document

Mathematics · Grade 11 · KZN November Exam · 2015. Question paper and memorandum, 18 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

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### Page 1

MARKS:

TIME:

GR 11 MATHS — P2 NOVEMBER EXAMINATION Gss Page 1 of 9

basic education

Department:
Basic Education
REPUBLIC OF SOUTH AFRICA

EENBURY SECONDARY

SCHOOL

GRADE 11

MATHEMATICS PAPER TWO —— "
Ci
NOVEMBER 2015 "
S22 eee
150 EXAMINER: R. DEMRUGARAM
3 Hour MODERATOR: L. PILLAY

This question paper consists of 9 pages

### Page 2

GR 11 MATHS P2 - NOVEMBER EXAMINATION GSS Page 2 of 9

INSTRUCTIONS AND INFORMATION:

Read the following instructions carefully before answering the questions:

1.

2.

Nd

This question paper consists of 12 questions and 9 pages.

Answer all the questions.

. Clearly show all calculations, diagrams, graphs, ete which you have used in

determining the 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 your answers to TWO decimal places, unless stated
otherwise.

. Number your answers correctly according to the numbering system used in

this question paper.
Diagrams are not necessarily drawn to scale.

It is in your own interest to write neatly and legibly.

p.t.o. Question One...

### Page 3

GR 11 MATHS P2 ~- NOVEMBER EXAMINATION GSS Page 3 of 9

QUESTION ONE:

The following data represents a set of mathematics marks of 19 learners
in a Mathematics test. The maximum mark is 50,

32 9 49 11 731 22 14 25 38 36 40 42 10 19 42 8 6 16

1.1. Write down the 5 number summary. (5)
2. Draw a box and whisker plot of the data on the diagram sheet
provided. (3)
1.3. Comment on the distribution of the data. eo)
1.4. Calculate the mean. (2)
1.5. Determine the standard deviation. (2)
1.6. Determine the percentage of scores that lie within one standard
deviation of the mean. (2) [15]
QUESTION TWO:

The table below shows the performance of learners in a
grade 11 Mathematics test.

Class Interval Frequency Cumulative
Frequency

0<sx<10
10<x< 20
20 <x < 30
30.< x < 40
40 <x < 50
50s x < 60
60<x<70
70<x < 80
80<x< 90
90<x< 100

DHDiwie

a
MD tw

mR) Gb] co

2.1. Complete the table provided. — (3)
2.2. Draw an Ogive curve to represent this information. (4)
2.3, Estimate the median from the graph. (1) [8]

p.t.o. Question Three...

### Page 4

GR 11 MATHS P2 ~ NOVEMBER EXAMINATION GSS Page 4 of 9

QUESTION THREE:

In the diagram ABCD is a parallelogram with coordinates

A(*x;6); BC 93-2); C(8;y) andD(-452)

A(x;6) 93
D ¢ “4 > 2 Pe
; M_ (@ .
B(9;-2)
C(8sy)
3.1. Calculate the coordinates of M, the midpoint of BD. Q)
3.2. Hence, or otherwise show that x = - 3 units and y = - 6 units (3)
3,3, Calculate the gradient of BC. (2)
3.4. Calculate the gradient of BD. (2)
3.5. Hence show that ADBC is not right angled. (2)
3,6. Calculate the equation of AB in the form y = mx + c. (4)
3.7. Determine the value of DAB : (4) [19]
QUESTION FOUR: C
- 41, The distance between the points A(-6;-5)andB(2;x)
is 10 units. Determine the value(s) of x. " (4)

4.2. The equation of a straight line is y = px + 6. Determine the value
of p in each of the following cases if the given line:

4.2.1. passes through the point B(-1;-4) (2)
4.2.2. is perpendicular to the line 2y +x =4 (2)
4.2.3. is parallel to the x - axis. (4) [9]

p.t.o. Question Five...

### Page 5

GR 11 MATHS P2 - NOVEMBER EXAMINATION GSS

QUESTION FIVE:

In the figure point P has coordinates (1 ; 3)
Reflex XOP = @ and XOP =8

Page 5 of 9

y
a”
aa"
P(1;-v3)
Determine the following without the use of a calculator:
5.1. OP (3)
5.2. sin @ (2)
5.3. the value of B (3) [8]
QUESTION SIX:
6.1. Determine the value of the following expression without
using a calculator:

cos 10°, cos 120° (5)

sin 80°, sin 330°
6.2. Express as a single ratio of 6:

sin (90° — 9) tan ( 180°— 0) (5)

cos (—6).sin (180°+6)
6.3. Prove the following identity:
6.3.1. sinx + cost =— (4)

sin x sin x
6.3.2. For which value(s) of x is the identity undefined if
x€ [0°; 360°] 3) 117]

p.t.o. Question Seven...

### Page 6

GR 11 MATHS P2 -- NOVEMBER EXAMINATION GSS

Page 6 of 9
QUESTION SEVEN:
Given f(x) = sin (x - 30°)
a(x) = cos 2x
_7.1. Determine the general solution of (x) = g(x) (7)

7.2. Draw a neat sketch of fand g on the same set of axes

for x € [- 180°; 180°].

Clearly indicate all intercepts with the axes, turning points and

end points. (6)
7.3. Write down the period of g. qd)
7.4, Determine the values of x for which g(x) is strictly decreasing

as X increases. (2)
7.5. Iff is shifted 60° to the left to obtain the graph of h(x), write

down the resulting equation of A(x). ; (2) [18]
QUESTION EIGHT:
8.1. Complete the following rules for A PQR:
8.1.1, Area APQR= = pr (1)
8.1.2. cos Q= rare @ a)
8.2. In the figure D is the midpoint of AC.

AB=5em, BC =7 cm and AC = 6 cm.
A
Sem D 6cm
B Cc
Tom

8.2.1 Show by calculation that A =78.46° - B)
8.2.2. Calculate the area of AABD (3)
8.2.3. Calculate the size of C (3) [11]

p.t.o. Question Nine...

### Page 7

GR 11 MATHS P2 - NOVEMBER EXAMINATION GSS

QUESTION NINE:

A solid wooden hemisphere is carved from a cylindrical piece

Page 7 of 9

of wood. The radius of the hemisphere and the cylinder is 12 cm.

an

Calculate the volume of:

p.t.o.

9.1. the wooden hemisphere. G)
9.2. the volume of wood that has been carved away. (4) [7]
In Questions 10, J1and 12 please note:
1. Diagrams are not drawn to scale.
2. Give a reason for each new statement.
3. Show all necessary working in full.
QUESTION TEN:
10.1. Complete the statement so that it is valid:
The line drawn from the centre of a circle perpendicular to
the chord _ (1)
10.2. AB is the chord of circle centre O.
OE bisects chord AB.
AD = 12 cm, ED = 8 cm and OD = x cm
: E
—
Bom
A I2e Dp B
x
o
10.2.1. Determine the radius OB in terms of x. (1)
10.2.2. Hence calculate the diameter of the circle. (4) [6]

... Question II...

### Page 8

GR 11 MATHS P2 - NOVEMBER EXAMINATION Gss Page 8 of 9

QUESTION ELEVEN:

11.1. In the diagram, O is the centre of the circle.
A, B and D are points on the circle.

D

Using Euclidean geometry methods prove the theorem
which states that AOB = 2 ADB. Clearly indicate any
construction lines. (5)

11.2. In the diagram, O is the centre of the circle RMPS.
T is the midpoint of RM. R= 50°

11.2.1. Calculate, with reasons the value of the following:

a) O, (2)
b) 8 (2)
d ty (2) .

11.2.2. Show by calculations, that TOPM is NOT a cyclic quadrilateral.(2) [15]

p.t.o. Question Twelve...

### Page 9

GR 11 MATHS P2 ~ NOVEMBER EXAMINATION Gss

QUESTION TWELVE:

12.1. PT is a tangent to PORS.
TPR = 72°. Q=x ;

Calculate the value of x.

12.2. TP and TQ are tangents to the circle at P and Q respectively.
TY is drawn parallel to RQ with Yon RP. TPQ =x

12.2.1. Name with reasons 3 other angles equal to x.
12.2.2, Prove that TPYQ is a cyclic quadrilateral.
12.3.3. Prove that TY bisects PYQ.

12.2.4, Prove that ARQY is an isosceles triangle,

Page 9 of 9

(4)

(6)
(2)
(2)
(3) [17]

TOTAL : 150

### Page 11

Grape « Matuematics Paree 2

MaRiing MEMORAN DUID

Question ONE *
ne

67 8 9.) 11 ete 9) (@) 25 a1 82 36 (98) 40 42 42 49
ova

Ld.

13,
“A.
LS.
LG.

®.

13. S0% marks he belween 6 and 22

« ® ®

MIN =
& = 1044 TP [eames useal formalat, to obtained
Q. Fue antwert (full credit )
&3 7 38%"
2 4gvA (8)
MA “Tn Aca aca
| +
° 5 10 1G 290 28 80 385 40 45 50 (3)
Positively SkEwED
SKEWED To cuz! Aleut “4 / 50% marks lie bebwan io ¢ 3¢ ae (1)

4S
MEAN > = = 24,0504 Jam? (2?

O =(3,83 44 Careewibt with use of a table — 1 mark) (29
(% 40)

=(% -0 5, R40)

= (24,05 -13,835 24,05 +13,83)

= (10,22 3 37,88)“

22 x 100

19

= 41,37 To vA / TY, [ 4 4-7o

oe

els

Or.
So%, No an and 3F

actebe any one of the antweus «

### Page 12

Que.

STION TWo *

b

a7
CLASS
2a. | INTERVAL FRE farQueNcy|
; jos x £10 ; i 1
fosxe20 | 3 4 | gall covrecl 3 marks —
Zo £ 2430 6 lo @ | Evvoy 2 marks
feo 4 x 440 13 23 ’ C Evvos joowmark
40% X450 | 12 35 % Moye Van 2 Evry B marks
SO £ x sbo | 15 50
60 £ 470 8 | 58
|7o4x480 | 2 60
JPofxe% | 3 | 63 eo
904x400 | I bd. (3)
_ . 22 CUMULATIVE FREQUENC CURNE SHOWING
THE MARKS OF LEARNERS
- 6S ; .
A (upper bountl
+o a Mud)
+58. “
leg
4 Sante ic ~shape eo) ;
~ ~o eh BQ ee - i ~ - :
fo te :
a AS ~~ — pe to a —— a i
oe ~@--+20 . : oe ti. ee ~ ao 4
3 i I
oD mnb Gen - ; moophs ~ pene nr — - -
— tO” safe ~4!- a ~ - . - -
we | ;
aes , at il hy. ‘ A + ae
coil iAP to 2%, O40 0 © 710, 80 90 10@ (4)
“MARKS OF UBARNERS : oo
2% meta 246 accehh antyect [46 fur Jor [444 w fadow cole)
theveof «

### Page 13

Question Taree :

Bt. 8(,-2) 5 2 (-452)
M (4 i ad

(3,0)

32. A(%j6)) €(8)Y)

x+8 254;

2 2 w= | ov mene (a
eed 5 gt ~64" (4a ge = pal ee
33, mBC > S279) : (+74) 5 74 pln teet a
Kat p=iow4 (2)
_ 246. ° a 7
Cn ” "928 422. Dy sme at
) a4 Zh (2 ly 7-% 44
24, BD = os 7 a is nn
° andl (2) m, = 2"
a , D ME
aa 423. pzo-* On
fe Poec is not rt ome @.

MAD F so £2

i) .
- - 2 94S
= -2
A 7 2B)69°
PAB F 70,36° (os of .

ws
Z

= 2 Or ; —_

(2

Question Foua:

Al. ABS V (246)"+ (45)
V 644 x7 HOR +25 _
6A FI 410 F258
“(00 = oO

lo =
vA
100 =
x2 rox + es $29 -

a +lOx, ‘<tl

ors av — @
. age
_ siz. ses 3 _@)
_|503- Preflex x68 = 500% a

| Question Five

[Sur oF C1)", oy” ( (th of Pgtn)

### Page 14

Question Six;
G1. CoS 10° cos 120°

Sim 80° sin 330°

vA :
Sin Bo! ws ("tee 6c') oR Coste’. (Sin 30")
nel = Cos Or, (- 5a 30°)
7 Cos 60 vA ” ‘
- Sim acres

\

Tua

_

u

»

——?
- 4

2

pus

W

. (2)
62. sv (9o- 6) tom (180 -6)

ws (-e) Sv (180 + 5)

yl

cose , - ton ovr
OSS - smo om
si-y ef ; A
@s3o ne

+ va :

WSO

631 LHS =

cos", (5)
smo +

A Stn
: awl
Stn 70 + COS"

ye

sin xv
. |
. —— A
SMX (o
632. smx 26
2 : A aa Lh
m2 O ? ‘BO , 360

(3) fr).

### Page 15

Cos oc

Question Sevens}

| BLS Boe 289 (en 30%)

conensonenansemisncmesinaaswtn (ates tne a ho a ee AECL EH OOS rar

ook for shorte, Gechmasaty on Covatod

Leo rts omovies nyereammestesrontscerently nereewepcagcesnaeyf irl env ery ee Pee ON TERT ONAN ER

7 ws {90

~os-30)] % § Doe BO ohn 2 KB AD ZH Ee, a

_. | 8 Loe. 2 @S (l207x) “ wy 2-7-2 = 46" 200% oe 8,
ho 2 remap te
j 2 = I2o-2% +4360 Mor 2x = 360- (120-2) +4360, Kee
[Br F120 tK36o 2x = BEO-tao +t + k360
__ 2% = 40° +KI20% X= 240 rk3¢o// 2 2 ~I2p°- 3-4)
17.2. . _AY.
ae)
So remus
¥

Ch
oA —_ —_ |
_. [12. Igor” oth UA . 1) |

! wt
t4: 26 (ee 90°) 3 xe (0% 90°) (2)

(1S. hla) 2 sm (oc t3

wire

ay ; (2 Lie)

Ch

- (Question EicaT :

& 3ewy

8.22 Anca Dapp = 3 (AB)(AD) sin 78 40°

4 Ww 2.
| Z __ Cp nr 2 (8) (3) Sir 78 AL
; B Tew

(Bll. AREA OPaR=2

Bl2. cos @&

ees eee ee

rsivy Q(JB23. SMC 2 sm 78,46 VA

2 1135 em7% (3)

2PM

2
bt @-a

=: Ptt-ge — s 7 if

2be

smC = Ss ae
7

ae | cos & 2.6 45°-7" a - C 2 44,41° vot _ @ ~ |
eh a
a AL -1 ee 2 |

### Page 16

(Question Nine + . “

ce esen snare twine cif: omemeeneraeeenemtmers 5c th neg ene rere fet tndmcnaecteNe nn

| > 3619, I em? ch (3)

49.2 »_ Youume OF CYLINDER
> A(npP(2) “4

= 8428 67cm?“

VOLUME OF GARYED Woop
2 6428, 67- 3619 ,1)~%

1809 56 cm?“ (4)

[Question Ten | ap

sersnecrnmetomal sas mer scncnnnamrm ert fa Nte SEES

[lout bisecAs the chove ~* (1D

lo.24 OB 2 x48" (9
fe vam: r reer

lo22. (40)* = aha xt “A (th of Pyth) __.

iY

a x 416 +64 = 144 4907 _
ee
x 254 ”

— r5 = nnn susnnnneronmenenetnn |
__DiameteR = 26cm 4 (4) es} — oo
| — eommer wa ns a coves vowel
a (.
oiaeiees neeel aeineneneenianes wane — —
4 -
| — a
—_
|

tate nacienninaet nan
i
i

i

### Page 17

Question Eveven :

_AOB Ar CIRCUMFERENCE -

cent eenieae n e ULSD SHAR RELA RCT ian ets thn ent eitanseOmRENCRREASTEASEN |

AOB =

_2ADB

DRAW DC THROUGH DO”

CIRCLE OTH ACB A AT. CEN TRE “AND

teen cae nerhnmsteeriteateniainentnensnmninnntnasnciatiniel

(RADU)

DO= AD = O68

: 6.2 A+ By (ext 4 OF OD) . |

i = 2D) — _ |

| O27 0248 > __!

= oi in a a — = i
: AOB® = 2D, 42D,” a 7

f) 4 = 2(0) +P.) . _. _|
> Zapp |

BY og . 8) |

©

ifsle)

len 2) }n

(2)

vy

a
50°

0.8

(4a canrtre ) ov (48 in same seqm)

(29

40° vn (es of 1S OSC D5 radi)"

Chaat

L d)T = 90°A ( mielpt 1 contre ) (2) 1
= 40°+ 90° . _.
(>) t 10° VA i en
° - Tory is not egelicg wal (2) is).
anne |
in . - ssnsnscnttnangenceens nnn

### Page 18

UESTION TweLve +

wecwcananerrnccs none nadfenptaczuencr ea is a | an _ sea cenee _

at. $= 72°" (6 beto bm, chord)
Et Copp es of eyelte quad)

fazi. R=% 24 (tan chord theorem > v4
eo A, COWES ™ " NS
(ty u lee) we

a
Ya tv

BT Chen chord terre) 8

—
112.22. Ya 5 A SC mG Prowed) a
be woh ol _ vnnqenacmngeenee

“1 TeY@ is cycle (4S in Some Sey) (2)

yore wept Bes ootareaesisuinaniiecusinaton acetone

7 a) as
12.2.3. Py 2 Yo 270° (cs mr some 509m Qt)

TY bisects PI® __. _@

“ DRAY 18 wosceles (es) fo).
i

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