# Mathematical Literacy Grade 11 P2 2018 Memo

> Grade 11 Mathematical Literacy Paper 2 memorandum from the KwaZulu Natal June 2018, 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/maths-literacy-june-p2-qpmemo/1ea32eaa-a74a-4a77-afc5-7c020942647a
- Updated: 2026-09-27
- Subject: Mathematical Literacy
- Grade: Grade 11
- Document type: Memorandum
- Year: 2018
- Exam period: KwaZulu Natal June
- Paper: 2
- Curriculum: CAPS
- Pages: 14

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## About this document

Mathematical Literacy · Grade 11 · KwaZulu Natal June · 2018. Memorandum, 14 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

Dewnleadedfromhlayiso.com

education

Department:
Education
PROVINCE OF KWAZULU-NATAL

NATIONAL

SENIOR CERTIFICATE

GRADE 11

THE EMATICS P P2

COMMON TEST

JUNE 2018

MARKS: — 100

TIME: 2 hours

This question paper consists of 6 pages.

Please turn over

Copyright reserved
Sa ee ee =e

### Page 2

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Mathematics Paper 2 June 2018 Common Test

GRADE i
INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions:
1. This question paper consists of 5 questions.
2. Answer ALL the questions.

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

4. Clearly show ALL calculations, diagrams, graphs, et cetera, which you have used in
determining the answers.

5. Answers only will NOT necessarily be awarded full marks.

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

7. If necessary, round off answers to TWO decimal places, unless stated otherwise.
8. Diagrams are NOT necessarily drawn to scale.

9. Write neatly and legibly.

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Mathematics Paper 2 June 2018 Common Test

GRADE 11

QUESTION 1
A(O;w), BO; 3), C(4 ; 0) and D(10 : 8) are the vertices of a quadrilateral in the
Cartesian plane. AD is parallel to BC.
A(O;1v)
D(10;8)
B(0:3)
4
C(4:0) “
LL.
1.1.1 Calculate the gradient of CD. (2)
1.1.2. Hence, determine the angle of inclination of CD. (2)
1.2 Prove that BCD = 90°. (3)
1.3
1.3.1 Write down the gradient of AD. (1)
1.3.2 Hence, or otherwise, calculate the value of iv. (3)
1.4 Tfitis given that w= 5a, calculate the length of AD. (3)
1.5 Calculate the area of quadrilateral ABCD. (6)

(20]

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Mathematics Paper 2 June 2018 Common Test

GRADE 11

QUESTION 2

A, B(1;2), C(-2;p) and D(S5 ; 0) are the vertices ofa KITE.
M is the point of intersection of the diagonals of the kite.

ADB=45”.
.
D(5:0) &.
C(-2:p)
2.1 Determine the coordinates of M. (4)
2.2. Calculate the value of p. (6)
2.3. If p=~—9, determine the equation of AC. (5)
2.4 Determine the angle of inclination of AD. (5)
25 Determine the coordinates of A. (6)

[26]

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Mathematics Paper 2 5 June 2018 Common Test
GRADE 11

QUESTION 3

-

If tan@ = ; and OE [90":360"] , determine the value of 2sin6.cos@ without the

we

use ofa calculator. (4)

ve
io

Simplify without the use of a calculator:

sin (300° - x) + cos (90° + x]

3.2.1

sin (180° — 1) + tan 540° (5)
. 1

3.2.2 cos 330°.tan (-120") + sin 73°. | (6)
cos 197°,

[15]

QUESTION 4

(sin y=cos

4.1 The identity = =2tanx is given.
sin’ x-1
4.1.1 Prove the identity (5)
4.1.2. For which values of » in the interval 0° s x <360° will the identity in
4.1.1 not be defined? (3)
4.2 Solve for x if tan(3v+40°) =—] and x] -90°;90"] . (5)
43. Determine the general solution of 2sinx = V3+3cosy. (8)
[21]
Please turn over

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### Page 6

Mathematics Paper 2

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June 2018 Common Test

GRADE 11

QUESTION 5

wn

Copyright Reserved

wn

wa

aay

wn

The

€[-

i)
Nv

sketch represents the graphs of 7 (x) = acos(x +b) and g(x) =csinx for

180°; 1 80"| 5

bo

-180° 135° ~90° 45°

Write down the values of a, b and ce.

[fthe points of intersection of f and g are (14.64%) and (i:-0,51),
write down the values of k and m7.

For which values of x in the interval [-! 80°:0"] will

(a) f(x)-g(x) <0?

(b) f(x).g(x) =O?

Determine the minimum value of A(x) if A(x) =x) + 2.
Draw a sketch graph of m(x) = tan 2x for vE[0", 90") .

Describe (). the graph m has to be transformed to form the graph 1
where n( x)=tan(2 450").

TOTAL MARKS:

(4)

(2)
[18]

100

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education

5 Department:
Education
PROVINCE OF KWAZULU-NATAL

f MATHEMATICS | “
: COMMON TEST
" JUNE 2018

" MARKING GUIDELINE : |

- NATIONAL
SENIOR CERTIFICATE
[ GRADE 11 |
MARKS: 100 |
This marking guideline consists of 8 pages. |
Copyright reserved Please turn over

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J9AO UJN] BSE9|q

pasuasay JYBUAdOD

(¢)

JaMsue dD

uoneoyrdurs A

P[NULIOY
aneipenb ojur uoNnysgns ~

spun ¢Z1=

vl
(€) a
JOMSUR pa TL
4
—= KU
Ie
uonroryydurtts ce—“b = 08
(g—)b =0¢
Bynuoy ol-0 +
quoIpess UL WORN ySqns ~p au e | eet
7) :
JOMsue jp -= Mu
vel
(€)
Surpn[auos
pue [—Jjo Jomsue "06 = AQ aojo194]
[= =
poof
sjuoipess Surdpdyynu p a x= ex Pur
v
Od JO WoIpels p =
al = uy
Ore Ti
(2)
JoMsue a Sres=9
€ fo
gue), rs
w= guy} zy]
@ -
JaMsue pa =
BENULIO}
jualpels ul Sunnjnsqns ~ =
—.= Dy
A SA LUL

810c¢ dune Isay UOMO)

sUTPOpINy SUPE

T NOILLSANO

Z Jadeg sonemayye

### Page 9

Mathematics Paper 2

GRADE 11

Marking Guideline

Downloaded from hlayiso.com

Common Test June 2018

BC= (0-4) +(3-0) =5

CD = (10—4)’ +(8-0)° =10

Area of ABCD = area of trapezium
= (sum of parallel sides) x height

OR

>

BC= (0-4) +(3-0) =5

CD = (10-4) +(8-0)" = 10
Area of ABCD = area of rectangle + area of triangle
=(/xb)+(45h)

=(5x10)+|4x10x (124-5) |

= 874+ square units

OR

BC=,/(0—4) +(3-0)' =5

>

CD = y(10-4)' +(8-0) =10

Area of ABCD = area of ABCD + area of AABD
= (£x5«10)+[4x(154-3)x10]
=25 + 624

= 874 square units

Y length of BC
¥ length of CD

Y formula
Y substitution of(5+124)

¥ substitution of 10
v answer (6)

OR

Y length of BC

Y length of CD

Y substitution of (5x10)
¥¥ [4x 10x(124—5) |

v answer (6)

OR
Y length of BC
Y length of CD

v (4x5x10)
v v [4x(154-3)x10]

v answer (6)

[18]

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JONO UIN} ASBA|q

pavuasey 1ybuAdog

(s)
JOMsue jp cr g0l=
« Sb— Ch esl=
ab~ EMESIS Cy Jo uoneurpour jo o[suy
gq Jo uoyeurput fo afsue ep est =
LS‘9@— .081= Xd4
ada g
w= xaquer 1 = Xd ue}
&
dd Jo yworpess p T =
Irs = Ay
c-0 x4
(s) ,
uoyenbo g-xZ=a
6-=9
2 Jo onyea (-)
IA E-) 0 = 6-
BINULOF OUT] ZYSres}s UL ayNySqns ~» 2(6-5Z-) ammnsqns
24+x7= 0
JUOTPBIS JO aNTLA ~p c=
BINULIO} Juatpess ULaIMASAns ~
i
ee EG
(9) 6-=d
JOMSUB d
9E = 4 b-
sapis yjog Suuenbs ort .d =E1+ yd - a
Burdgyduns » 6pt+ dh =e1+dp-.d)
BynuULo}
souvysip UL UONNIASANSs SHY A a +or/ =ptdp— d+ of
BINULLOF _ fe ton _ = _—s \
aouvjsIp Ul UOMINSqNs SET {0 d)+ AS ¢ )h ~ Le d)+ nt ¢ H)
dO=08 “ qdo=04 TT
(y)
“A JO} Jomsue (LOW
Na Joy UONNYSgns ~ l= e=
Ny oj JaMsue G a
Ny Joy uoyNyNsqns a OFT orl
eet Me Ane Ne
cK + la bans Iy lz

SIOZ auny isa y, UOUTUTOD

ouTpopny Bupye PY
IL aavad
b

Z NOLLSANO

ZT dodeg sonemayey

### Page 11

Mathematics Paper 2

GRADE 11
Marking Guideline

Downloaded from hlayiso.com

Common Test June 2018

My, = tan 108, 43°

figrutomat Ade
y=-3xte
Substitute (5;0):
0=-3(5)+e
e=15
y=-3x+15

Solve simultaneous equations for AC and AD:
2xn-5=-3x415

| ¥ gradient of AD

Y substituting (5 ; 0)

Y equation of AD

Y solving simultaneously

5x =20
e4 Y value of x
y=3 v value of y
A(4;3) (6)
OR OR
DM = AM
= BM [sides opposite equal angles] ¥ DM=AM=BM= V5
AB=AD
= J10 [Theorem of Pythagoras]
AB = AD
W(e-1) +(v-2)' = (x5) +(y-0) ¥ yx) +(y-2y
xv -2xtlty?-4y44=x?-10x42545° = (x-5) +(y-0)
8x—4y = 20
ye QN—Soecccccee line 1 Y simplification to y=2x-5
AD=¥10
(e-3) 0-97 =Vi0 ¥ Io-5) +0) = si
Xx? -10K4 25497 =10 ee line 2
x -10x+25+(2x-5)' =10
x =10x +2544" —20x+25=10
5x° ~30x+40=0
x -6x+8=0
(x-4)(x-2)=0
x=2orx=4 vx=4
N/A
y=3 vy=3
(6)
[26]
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J@AO UN] BSBA|q

paruasay yybuAdog

[stl
(9) z
Jamsuv -
tL
l €.
san[ea afsur yeisads , ( I z
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|
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ors ur +(x— gt) us
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SIOZ uns Isoy, VOLUN 9 ZT Jodueg sonvemoayeyy

### Page 13

Mathematics Paper 2

QUESTION 4

GRADE ILL
Marking Guideline

Downloaded from hlayiso.com

Common Test June 2018

tan (3x-+40°)=-1

reference angle: 45°

3x+40° =180° -45° +7.360° or 3x+40° = 360° —45° + 1.360°

3x = 275° +1.360°
x =91,67 +7.120°
x =—28, 33°

3x = 95° +.360°
x = 31,67" +7.120°

x =—88,33° or 31,67°
where n <Z

OR

3x+40° = 180° —45° +.180°
3x = 95° +n.180°
x = 31,67 +7.60°

x = —88,33° or -28,33° or 31,67°
where n € Z

4.1.1 (sin. —cos
sin” x—-1
_ sin’ x~2sinxcos.x+ cos’ x1 Y multiplying out
sin’ x-1
_ —2sin xeosx+sin’ x+cos’ x1
: sit xl ¥ applying identity
—2sinxcosx+1-1 i 2
= es sin’ x+cos’ x =1
-(1-sin? x) v ~(1=sin’ x)
_ 72sin.xcos x Y applying identity
= cos" x 1-sin’ x=cos’ x
2sin x
~ cosx ¥ simplification
=2tanx - __ (5)
4.1.2 | sin?x—-1=0 Y sin’ x-1=0
sin? x=1
siny=-l or sinx=l
x=270° or aco? ¥ 270°
v 90°
(3)
4.2

¥ 3x+40° =180° —45° +1.360°
v

3x+40° = 360° —45° +.360°

v -88,33°
¥ 31,67
v¥ -28,33° (5)
OR
v 180° —45°
¥ +n.180°
v¥ -88,33°
v¥ 31,67
v¥ 28,33

(5) |

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### Page 14

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