PREPARATORY EXAMINATIONVOORBEREIDENDE EKSAMEN

2019MARKING GUIDELINES /

NASIENRIGLYNE

MATHEMATICS (PAPER 2) (10612)WISKUNDE (VRAESTEL 2) (10612)

22 pages / bladsye

10612 / 19

GAUTENG DEPARTMENT OF EDUCATION /

GAUTENGSE DEPARTEMENT VAN ONDERWYS

PREPARATORY EXAMINATION /

VOORBEREIDENDE EKSAMEN

MATHEMATICS / WISKUNDE(Paper 2 / Vraestel 2)

MARKING GUIDELINES / NASIENRIGLYNE

NOTE: If a candidate answers a question TWICE, only mark the FIRST attempt. If a candidate has crossed out an attempt of a question and has not redone the question, mark the

crossed out version. Consistent accuracy applies in ALL aspects of the marking guidelines. Stop marking at the

second calculation error. Assuming answers / values in order to solve a problem in NOT acceptable.

LET WEL: As ʼn kandidaat ʼn vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na. As ʼn kandidaat ʼn antwoord van ʼn vraag doodtrek en nie oordoen nie, sien die doodgetrekte

poging na. Volgehoue akkuraatheid word in ALLE aspekte van die nasienriglyne toegepas. Hou op nasien by

die tweede berekeningsfout. Aannames van antwoorde / waardes om ʼn probleem op te los, word NIE toegelaat nie.

GEOMETRY / MEETKUNDE

S

A mark for a correct statement(A statement mark is independent of a reason.)ʼn Punt vir ʼn korrekte bewering(ʼn Punt vir ʼn bewering is onafhanklik van die rede.)

R

A mark for a correct reason (A reason mark may only be awarded if the statement is correct.)ʼn Punt vir ʼn korrekte rede(ʼn Punt word slegs vir die rede toegeken as die bewering korrek is.)

S / R Award a mark if the statement AND reason are both correct.(Ken ʼn punt toe as beide die bewering EN rede korrek is.)

2

10612 / 19

QUESTION / VRAAG 1

Number of oranges in the bag Aantal lemoene in sak 18 16 20 15 14 13 14 10

Median mass of oranges in the same bag (to the nearest gram) / Mediaangewig van die lemoene in dieselfde sak (tot die naaste gram)

110 105 72 142 123 174 136 192

8 10 12 14 16 18 20 220

50

100

150

200

250

SCATTER PLOT / SPREIDINGSDIAGRAM

Number of oranges in a bag / Aantal lemoene in sak

Med

ian

mas

s of o

rang

es /

Med

iang

ewig

van

lem

oene

1.1 a = 307 , 20b = −11 ,70y = 307 ,20 − 11 ,7 x

a = 307 , 20b = −11 ,70 y = 307 ,20 − 11 ,7 x

(3)1.2 r =−0 , 93 r =−0 ,93

(1)1.3 See scatter plot above/sien spreidingsdiagram hierbo

(10 ; 190,2) (20 ; 73,2)

(10 ; 190,2) (20 ; 73,2)

(2)1.4 Negative strong association /

Negatiewe sterk assosiasieanswer / antwoord

(1)1.5 y = 307 , 20 − 11 ,7 (12)

= 166 ,8substitution / vervanganswer / antwoord

(2)[9]

3

10612 / 19

QUESTION / VRAAG 2

2.1.1 100 answer / antwoord (1)2.1.2 Median / Mediaan = ±62 answer / antwoord

(Accept / Aanvaar 61 / 62) (2)

2.1.3 Q1= 37/38Q3 = 72/73

Q2= 61/62 & min & max / min & maks (10 & 100)

(3)

2.1.4 Skewed to the left / Skeef na links left / links(1)

2.2 b = 20d − a2

= 8

2 a = d

sub 2a − a2

=8

a = 16d = 325+16+19+20+c+32+35=7×22∴ c = 27

b = 20

a = 16d = 32

c = 27 (4)[11]

4

10612 / 19

QUESTION / VRAAG 3

3.1

3.1.1 1=3+ x

22=3+ xx=−1

−2=4+ y2

−4=4 + yy=−8

B(−1 ; −8 )

1=3+x

2

−2= 4+ y

2

B(−1 ; −8 ) (3)

3.1.2 mCD=0−46−3

¿−43

substitution into gradient formula / vervang in gradient formule

mCD=−4

3 (2)3.1.3

y−2=−43

( x−11)

y=−43

x+503

OR / OF

substitute m /vervang m substitute / vervang Q(11;2) 

(2)

5

R

D(3 ; 4)

0 C(6 ; 0)

A

B

Q(11 ; 2)

P(1 ; )

y=−43

x+c

2=−43

(11 )+c

c=503

y=−43

x+503

10612 / 19

3.1.4

D is the

midpoint of PR / D is die middelpunt van PRC is the midpoint of PQ (line from midpoint of 1 side || to 2nd side)/C is die middelpunt van PQ (lyn van middelpunt van 1 sy || aan 2de sy)

RQ=2CD=10 (midpoint theorem / middelpunt stelling )

PK = RQ√ ( y + 2 )2 + ( 4 − 1 )2 = 10( y + 2 )2 + ( 4 − 1 )2 = 10 2

( y + 2 )2 = 91 or / ofy + 2 = ± √91y = ±√91 − 2y =−11 ,54or / ofy≠ 7 ,54

CD = 5 OR / OF R(5 ; 10)

statement / bewering

RQ = 10

correct substitution into distance formula / korrekte vervanging in die afstandsformule

simplification / vereenvoudiging

y =−11 ,54(6)

6

R (5 ; 10 ) midpoint / middelpuntRQ=√(2−10 )2 + (11 − 5 )2

RQ= 10

CD=√ ( y2− y1)2+(x2−x1 )2 OR / OF

¿√ (0−4 )2+(6−3 )2

¿√25CD=5

10612 / 19

3.2

3.2.1 mPQ=tan θtanθ=1θ=45 °

P¿

1=35 ° vertical opp ∠ s / regoorst ∠eQR || to the x-axis / aan die x-as

T¿

1=35 °+ 45° ext ∠ of Δ / buite ∠ v Δ

T¿

1=80°

α=T¿

1=80 ° corr∠ s ST||QR / ooreenkomstige ∠e ST||QR

OR / OF mPQ=tan θtanθ=1θ=45 °QR || to the x-axis / aan die x-as

S1

¿

=Q¿

=45 ° corr∠ ’s ST||QR / ooreenkomstige ∠ ’e ST||QR

P¿

1=35 ° vertical opp∠ s / regoorst ∠e

α=35 °+ 45°α=80 ° ext ∠ of Δ / buite ∠ v Δ

mPQ= tan θ

θ=45 °

P¿

1=35 °

T¿

1=80°

α=T¿

1=80 °

OR / OF

mPQ= tan θ

θ=45 °

S1

¿

=Q¿

=45 °

P¿

1=35 °

α=T¿

1=80 °

7

P

T

S

0

U

1 1

1

θ

10612 / 19

(5)3.2.2

x at/by U:

U2+(−8 )2

=1

∴U x=10 units / eenhedeQU =18 units / eenhede

x at/by W = x at / by U = 10

y at/by W:

y=10+23

=323

WU =

323

+5 =473

∴Area ΔQWU=12

(18 )(473 )

=141 square units /eenhede kwadraat

U x=10units / eenhedeQU = 18 units / eenhede

U = 10

WU =

473

correct substitution in area formula / korrekte vervanging in oppv. formule

141 square units /eenhede kwadraat

(6)[24]

8

10612 / 19

QUESTION / VRAAG 4

4.1 T U¿

S=180°−101, 31 °=78 ,69 ° adj supp∠ s / aangrensende suppl ∠e mTU=tan78 , 69 °=5c=6y=5 x+6

T U¿

S=78 ,69 °

mTU=5

y=5 x+6(3)

4.2 x−int /afsnit y=0−15

x + 45

= 0

−x+4=0x=4∴S( 4 ; 0 )

M=(−6+42

; 2+02 )

∴M (−1 ; 1)

OR / OF

S( 4 ; 0 )substitute correctly / korrekte vervangingM (−1 ; 1 )

9

OQ

y

L

M

Kx

U S

101,31°

10612 / 19

At M:

5 x+6=−13

x+45

265

x=−265

x=−1y=−1∴M (−1; 1)

CANDIDATE MUST SHOW CALCULATIONS TO GET MARKS IN THIS QUESTION / KANDIDAAT MOET BEREKENINGS TOON OM PUNTE IN HIERDIE VRAAG TE VERDIEN

5 x+6=−

13

x+45

x = –1

y = 1

(3)

4.3 ( x+1 )2+( y−1 )2=r 2

(−6+1 )2+(2−1 )2=r2

r2=26( x+1 )2+( y−1 )2=26

OR / OF( x+1 )2+( y−1 )2=r 2

(4+1 )2+ (0−1 )2=r2

r2=26( x+1 )2+( y−1 )2=26

substitute / vervang (−6 ; 2 )

r2=26

( x+1 )2+( y−1 )2=26OR / OF

substitute / vervang(4 ; 0 )

r2=26

( x+1 )2+( y−1 )2=26

(3) 4.4 mMP=−1

5 mMP×mKL=−1

mKL=5 radius⊥ tan /radius⊥raaklyn

mTU=5 proven / reeds bewys∴mTU=mKL=5KL||TU

mMP=−1

5

mKL=5

mTU=5

(3)4.5

VM = √(−1 + 12 )

2+ (1 − 7 )2

¿ 6,02

radius = √26 = 5,16,02 > 5,1

∴V (−12

; 7)does not lie within the circle . /l e nie binne die sirkel nie .

VM = 6,02

6,02 > 5,1

conclusion / gevolgtrekking

10

10612 / 19

(3)[15]

QUESTION / VRAAG 5

5.1.1 x2=(√1+k2)2−(1 )2 (Pythagoras )x2=k2

x=k

tan16°=1k

x=k

tan16 °= 1

k(2)

5.1.2 cos32 º=cos2 (16 ° )=2cos2 16 º−1

¿2(k√1+k2 )

2−1

OR / OF

cos32 º=cos2 (16 ° )=cos216 º− sin216º

¿(k√1+k2 )

2− (1

√1+k2 )2

OR / OF

cos32 º=cos2 (16 ° )=1 − 2sin216 º

¿1 − 2 (1√1+k2 )

2

cos2 (16 ° )2 cos216 º−1

correct substitution / korrekte vervanging

OR / OF

cos2 (16 ° )

cos2 16 º− sin2 16 º

correct substitution / korrekte vervanging

OR / OF

cos2 (16 ° )

1 − 2sin216 º

correct substitution / korrekte vervanging (3)

11

111

60º

10612 / 19

5.2 cos (90 ° + x ) sin ( x−180 º ) − cos2 (180 ° − x )cos (−2x )

¿(−sin x )(−sin x ) − cos2 xcos 2x

¿sin2 x − cos2 xcos2 x − sin2 x

OR / OF −cos 2 xcos 2 x

¿−( cos2 x − sin2 x )cos2 x − sin2 x

=−1

¿ −1

−sin x–sin x

–cos2 x

cos 2 x

–(cos2 x − sin2 x ) OR / OF–cos 2 x –1

(6)5.3 cos75 ° .cos 45 ° − cos15 ° . cos 45°

=cos75° . cos45 °−sin75 ° .sin 45 °=cos (75 ° + 45 ° )=cos120°=−cos60 °

=−12

OR / OF

cos75 ° .cos 45 ° − cos15 ° . cos 45°=sin15 ° .cos 45 °−cos 15° . sin 45 °=sin (15 ° − 45 °)=sin (−30° )=−sin 30°

=−12

cos75 ° .cos 45 °−sin 75 ° . sin 45 °

cos (75° + 45 °)

−cos60 °

− 1

2

OR / OF

sin 15 ° . cos 45°−cos15 ° . sin 45 °

sin (15 ° − 45 ° )

−sin 30 °

− 1

2(4)

5.4.1tanθ (sin 2θ + 3 cos2 θ

sin θ )¿ sin θ

cosθ (2 sin θ cosθ+3 cos2θsin θ )

¿2 sin2 θ+3 cosθ¿2 (1−cos2θ )+3 cosθ=−2 cos2θ + 3 cosθ + 2

2 sin θ cosθ and / en

sin θcosθ

simplification / vereenvoudiging

1−cos2 θ

(3)5.4.2 −2 cos2 θ + 3 cosθ + 2=0

2 cos2 θ−3cosθ−2=0(2 cosθ+1 ) (cosθ−2 )=0

factors / faktore (2 cosθ+1 ) (cosθ−2 )

both equations / beide

12

10612 / 19

cosθ=−12

or / of cos θ=2

no solution / geen oplossingref / verwy ∠ = 60 °

θ=±120 °+k 360° ; k∈Ζ OR / OF θ = 120 ° + k360 ° ;k∈Ζθ = 240 ° + k360 ° ; k∈Ζ

vergelykings

cosθ=−12

or / of cosθ=2

no solution / geen oplossing

θ=±120∘+k 360∘ k∈Ζ OR / OF

θ = 120 ° + k 360 °; k∈Ζθ = 240 ° + k 360 ° ; k∈Ζ

(4)5.5

cos (a + b) = −√22

ref ∠/ verw ∠= 45°

a + b = 180 ° − 45 °a + b = 135 ° .. .. . .. .. .. . (1)

cos ( a− 2b ) = 12

ref ∠ /verw ∠= 60 °

a − 2b = 60° .. .. . .. .. . .. .. . . (2)

3b = 75 ° (1)−(2)b = 25 °a = 110°

a + b = 135 °

a − 2b = 60 °

b = 25 °a = 110°

(4)[26]

13

10612 / 19

QUESTION / VRAAG 6

6.1

(45° ; −1

2 ) x-intercepts / x-afsnitte

shape / vorm

asymptotes / asimptote

(4)

6.2 y∈ [ 2 ;4 ] OR / OF 2≤ y≤4 y∈ [ 2 ;4 ] OR / OF 2≤ y≤4

(1)6.3 x∈ [ 135° ; 180 ° ] OR / OF 135 °≤x≤180°

x∈[ 225 ° ; 270 ° ) OR / OF 225 °≤x<270°

x∈ [ 135° ; 180 ° ] OR / OF 135 °≤x≤180 °x∈[ 225 °;270 ° ) OR / OF 225 °≤x<270 °

(2)[7]

14

0

1

2

-1

-2

225 ° 270 °

10612 / 19

QUESTION / VRAAG 7

7.1 D1=180°−60 °−(120 °−α ) (sum of ∠ s of Δ / som vd ∠e v Δ )D1=α

D1=180 °−60°−(120 °−α )

D1=α(2)

7.2 BDsin60 °

=4sinα

BD sinα=4 sin60 °

BD=4 (√3

2 )sinα

BD=2√3sinα

substitution into correct sin rule / vervang in korrekte sin reël

simplification / vereenvoudiging

answer / antwoord(3)

7.3

In Δ ADB:A { D B=θ (∠s of a Δ ) ¿ ABBD

=tan θ ¿AB=BD. tan θ ¿ =2√3sinα

. tan θ ¿AB=2√3 tan θsin α

¿¿A D B=θ

trig ratio / trig verhouding

substitution of BD / vervanging van BD

(3)[8]

15

D B1

C

A

4

10612 / 19

QUESTION / VRAAG 8

8.1 O2 = 50 °D1 = 25 °

∠ saround a point / ∠ e om ʼn punt∠centre = 2¿∠ at circumferencemidpts ∠ = 2 ¿ omtreks ∠

S

SR

(3)8.2 B3 = 25 ° tan chord theorem /

raaklyn koordstellingSR

(2)

8.3 BC D=120 ºB2=35ºO B C=O C B=65 º∴ B1=65 º−35 ºB1=30º

OR / OFBC D=120ºB2=35ºB1 + B2+ B3 = 90 °B1 = 30 °

opp∠ sof a cyclic quad /teenoorst ∠ e v kvh

sum of ∠ sof a triangle / som ∠ e v Δ∠ sopp. equal radii / ∠ e teenoor gelyke radiuse

OR / OF

opp∠ sof a cyclic quad/teenoorst ∠ e vkvh

sum of ∠ sof a triangle / som ∠ e v Δradius¿ tangent / radius ¿ raaklyn

S / R

SS

answer / antwoord

S / RS

Sanswer /

antwoord(4)

[9]

16

D1

221

CP

310º

3

2 1

2 1

O

BA

T

10612 / 19

QUESTION / VRAAG 9

9.1 Equal / gelyk. answer / antwoord (1)

9.2

9.2.1 D1 = F1=xF1 = F2=x

tan chord theorem / raaklyn koordstelling

= chords subtend = ∠ s = koorde onderspan = ∠ e

S / R

SR

(3)9.2.2 F2 = A=x

D1 = A=xABDC is a cyclic quad / ABDC is ʼn kvh

ext.∠of cyclic quad / buite ∠ v kvh

ext ∠= opp int ∠ OR converse of ext.∠of cyclic quad / buite ∠ = oorst binne ∠OF omgekeerde buite ∠ v kvh

SR

R

(3)9.2.3 B1 + B2 = A

A = D1

B1 + B2 = D1

BE||CD

tan chord theorem / raaklyn koordstelling

proved / reeds bewys

correspond ∠ s = / ooreenkomst ∠ e =

S

R(2)

17

A

E C

112

21

F

2 x

D1B

P

10612 / 19

9.2.4 C1 + C2 + F1 + F2 = 180 °

C1 = C2

F1 = F2

2 C1+ 2 F2 = 180 °C1 + F2 =90°E1 = 90°

FC is a diameter of circle FDCE.FC is ʼn middellyn van sirkel FDCE.

OR / OF

Let F1 = F2 = xC = 180 ° − 2x

C1 = C2 = 90° − x

In ΔFDC or / of ΔEFCD = 90 ° or / of E = 90 °

FC is a diameter of circle FDCE.FC is ʼn middellyn van sirkel FDCE.

opp∠ sof a cyclic quad / teenoorst ∠ e v kvh

diag rhombus bisect ∠ / diag ruit halveer ∠

proved / reeds bewys

sum of ∠ s of Δ / som van ∠ e v Δ

converse∠ in a semi circle /

omgekeerde∠ in half sirkel

proved / reeds bewys

opp∠ sof a cyclic quad / teenoorst ∠ e v kvhdiag rhombus bisect ∠ / diag ruit halveer ∠

sum of ∠ s of Δ / som van ∠ e v Δ

converse∠ in a semi circle /

omgekeerde∠ in half sirkel

SR

S

S

R

OR / OF

SR

S

S

R (5)

[14]

18

10612 / 19

QUESTION / VRAAG 10

10.1

NB: NO construction 0 / 5 / GEEN konstruksie 0 / 5

On sides KL and KM of Δ KLM mark points A and B respectively such that KA = GH and KB = GI. Draw AB

Op sye KL en KM van plaas A en B onderskeidelik sodat

KA = GH en KB = GI. Trek lyn AB.

Proof / Bewys

In ΔGHI and/ en ΔKABKA = GH construction / konstruksieK = G given / gegeeKB = GI construction / konstruksie∴ ΔGHI ≡ ΔKAB S∠S∴ A1 = Hbut { L

= H given / gegee ¿ ∴ A1=L ¿∴AB || LM corr .∠s =¿ooreenkomst . ∠e = ¿KLKA

= KMKB

line || one side Δ/ lyn || een sy v Δ ¿∴KLGH

= KMGI

¿¿

construction / konstruksie

S / R ΔGHI ≡ ΔKAB S∠ SS

R

S / R

(5)

19

K

L M

G

H I

o o

..

A B1

Δ KLM

10612 / 19

10.2

10.2.1 R1=60 º equilateral Δ / gelyksydige ΔW 1=P1+Q1 ext .∠of a Δ / buite ∠ v Δ¿60º + Q1

¿ R1+Q1

Q1=R2 ∠ s in the same segment/∠e in dieselfde segment

∴W 1=T R Q

S

S

S / R

(3)

10.2.2 In ΔTQR and / en ΔQRV1 . W 1=T R Q proved / reeds bewys2 . R1 = T Q R equilateral Δ/ gelyksydige Δ3 . Q2=T sum ∠s of Δ / som van ∠e v Δ ∴ΔWRQ ||| ΔRQT ∠∠∠

SS

R (3)

20

T

P

V

W

1

12

2

21

R

Q

1

1

10612 / 19

10.2.3

In ΔTPV and / en ΔWQR1 . P Q R= R1 both 60º / albei 60 ° P { Q

R=V 1 ext . ∠ of a cyclic quad . / buite ∠ v kvh ¿ V 1 = R1 ¿2 . P2=T R Q ext . ∠ of a cyclic quad . / buite ∠ v kvh ¿ but / maar W1=T R Q proved / reeds bewys ¿ { P ¿2=W1 ¿3 . T=Q2 sum of ∠s of Δ / som v ∠e v Δ ¿ΔVPT ||| ΔRWQ ∠∠∠ ¿VPRW

=PTWQ

=VTRQ

corresponding sides in prop/ ¿ ooreenkomstige sye in verhouding ¿∴PTWQ

=PVWR

¿¿

S

S / R

S

S

S R

(6)[17]

21

10612 / 19

QUESTION / VRAAG 11

11.1 PEEQ

= PDDO

x9

=

23

x

x + 3x2 + 3x = 6 xx2 − 3 x =0x ( x − 3 )= 0x = 0 or/ of x = 3N . A / n .v . tDO = 6DO = OROR = 6 units / eenhede

line || one side ΔPOQ OR prop theorem ED||OQ / lyn || een sy ΔPOQ OF eweredigheid stelling ED||OQ

radii / radiusse

SR

x = 3

OR = 6 (4)11.2 S is the midpoint of RE / S is die

middelpunt van REDE =2OSDE = 2,8 units / eenhede

midpoint theorem / middelpunt stelling

Ranswer (2)

11.3 Area ΔPEDArea ΔPER

= PDPR

= 214

= 17

same height (DE) / dieselfde hoogte (DE)

SR

17

Area ΔPER = 7 ¿ Area ΔPED = 18,9 units2 / eenhede2 18,9 (4)

[10]

TOTAL / TOTAAL [150]

22

P

D

+ 3E

O

S

QR

x +9