Mathematics P2 May-June 2016 Memo Afr & Eng

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SENIOR CERTIFICATE EXAMINATIONS/

SENIORSERTIFIKAAT-EKSAMEN
MATHEMATICS P2/WISKUNDE V2
JUNE 2016
MEMORANDUM
MARKS/PUNTE: 150
This memorandum consists of 21 pages./

Hierdie memorandum bestaan uit 21 bladsye.
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Mathematics P2/Wiskunde V2 2 DBE/2016

SCE/SSE – Memorandum
NOTE:
• If a candidate answered a question TWICE, mark only the FIRST attempt.
• If a candidate has crossed out an attempt to answer a question and did not redo it, mark the
crossed-out version.
• Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at the
second calculation error.
• Assuming answers/values in order to solve a problem is NOT acceptable.
LET WEL:
• Indien 'n kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na.
• As 'n kandidaat 'n poging om 'n vraag te beantwoord, doodgetrek en nie oorgedoen het nie,
sien die doodgetrekte poging na.
• Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing. Staak
nasien by die tweede berekeningsfout.
• Om antwoorde/waardes om 'n probleem op te los, te veronderstel, word NIE toegelaat NIE.
QUESTION/VRAAG 1
8 8 10 12 16 19 20 21 24 25 26
1.1 189 189

Mean/Gemiddelde =
11 Answer only: Full marks
= 17,18 Slegs antwoord: Volpunte  answer
(2)
1.2 Min = 8, max = 26  min, max
Median/Mediaan = 19  median
Q 1 = 10, Q 3 = 24  Q1 & Q3
∴ (8 ; 10 ; 19 ; 24 ; 26)
(3)
1.3
 box/boks/mond
8 19  whiskers/snor
24
6 10 14 18 22 26
(2)
1.4 The data is skewed to the left/Die data is skeef na links.  answer
OR/OF (1)
Negatively skewed/Negatief skeef  answer
(1)
1.5 SD/SA = 6,46  answer
(2)
1.6 17,18 + 6,46 = 23,64  interval
∴ 3 destinations/bestemmings  answer
(2)
[12]


QUESTION/VRAAG 2
Temperature at
midday (in °C)
18 21 19 26 32 35 36 40 38 30 25
Middaguur-
temperatuur (in °C)
Number of bottles of
water (500 mℓ)
12 15 13 31 46 51 57 70 63 53 23
Getal bottels water
(500 mℓ)
2.1 (30 ; 53) answer

(1)
2.2 a = – 38,51  value a
b = 2,68  value b
∴ ŷ = 2,68x – 38,51  equation
(3)
2.3 ∴ ŷ ≈ 36,53 bottles  answer
(2)
OR/OF
ŷ ≈ 2,68(28) – 38,51  substitution
 answer
≈ 36,53 bottles
(2)
2.4 Strong/Sterk  strong/sterk
The majority of the points lie close to the regression line./Die meerderheid  reason/rede
punte lê naby die regressielyn. (2)
OR/OF
Strong/Sterk  strong/sterk
r = 0,98  reason/rede
(2)
2.5 Temperature cannot rise beyond a certain point as this would be life  reason/rede
threatening OR there is only so much water one can consume before it (1)
becomes a risk to your health (hyponatremia)./Temperatuur kan nie hoёr
as 'n sekere punt styg nie, anders raak dit lewensgevaarlik. OF ‘n persoon
kan net ‘n sekere hoeveelheid water inneem, anders raak dit ‘n
gesondheidsrisiko [9]


QUESTION/VRAAG 3
B(8 ; 10)
A(–8 ; 6)
T C
10,54
D(–2 ; 0) O M x
10
3.1 y 2 − y1
mAD =
x 2 − x1
substitution
0−6
=
−2+8
−6 –1
= = −1
6 (2)
3.2 m BC = −1 [BC | | AD]  gradient
y = −x + c
substitute m and
10 = −8 + c (8 ; 10)
c = 18
y = − x + 18  equation
(3)
OR/OF
m BC = −1 [BC | | AD]  gradient
y − y1 = m( x − x1 )
y − 10 = −( x − 8) substitute m and
(8 ; 10)
y = − x + 18
 equation
(3)


3.3 y 2 − y1
m BD =
x 2 − x1
10 − 0 substitution
= =1
8+2  answer
m BD × m AD = 1 × −1 = −1  mBD × m AD = −1
∴ DB ⊥ AD (3)
OR
AD2 = 72 or AD = 72 or 6 2  calculating
all 3 sides
AB2 = 272 or AB = 272 or 4 17
BD 2 = 200 or BD = 200 or 10 2

2 2 2
∴ AB = AD + BD AB2 = AD 2 + BD 2
∴ AD̂B = 90° [converse Pyth th/ omgekeerde Pyth st] (3)
3.4 tan BD̂M = mBD = 1  tan BD̂M = mBD
∴ BD̂M = 45°  answer
(2)
OR
1
BM 10 1  sin BD̂M =
sin BD̂M = = = 2
BD 10 2 2
 answer
∴ BD̂M = 45°
(2)
3.5  x + x 2 y1 + y 2 
T( x ; y) =  1 ; 
 2 2 
 − 2 + 8 0 + 10 
= ; 
 2 2 
= (3 ; 5) T(3 ; 5)
T symmetrical about BM/T is simmetries om BM
∴ distance of T to BM = 5 units = distance from BM to C
value of x
∴ C(13 ; 5)
 value of y
(3)
OR/OF


3 13 − 0 1
m DF = =
8 − (−2) 3
Equation of DF: eq of DF
y − y1 = m( x − x1 )
1
y − 0 = ( x + 2)
3
1 2
y= x+
3 3
Equation of BC: y = –x + 18
1 2
x + = − x + 18
3 3
4 x = 52
x = 13
∴ y = –13 + 18 = 5 value of x
∴ C(13 ; 5)  value of y
(3)
3.6 area/opp ∆BDF = area/opp ∆BDM − area/opp ∆DFM  formula/method
 10 (DM)
=
1
(10)(10) − 1 (10) 10   10 (BM)
2 2 3 10 1
100 1  or 3 (⊥h)
= or 33 or 33,3 square units/vk eenh 3 3
3 3  answer
(5)
OR/OF
1  formula/method
area/opp ∆BDF = .BF.DM
2
1  20   BF
=  (10 )  DM
2 3 
100 1  answer
= or 33 or 33,3 square units/vk eenh
3 3 (5)
OR/OF


10
5−0 1 FM 1
tan FD̂M = mDC = = or tan FD̂M = = 3 =  gradient/ratio
13 + 2 3 DM 10 3
FD̂M = 18,43°
∴ BF̂D = 108,43° [ext ∠ ∆]  BF̂D
20 2
BF = or 6
3 3
2  DF
2  1
2
DF = (10) +  3  [Pyth ∆DFM]
 3
1000 10 10
DF = 10,54 or or
3 3
1
∴ area/opp ∆BDF = .BF.FD.sinBF̂D
2
correct
1  20  10 10 
=   (sin 108,43) substitution into
2  3  3  area rule
100 1  answer
= or 33 or 33,33 square units/vk eenh
3 3 (5)
OR/OF
20 2
BF = or 6  BF
3 3
BD = (10 − 0) 2 + (8 + 2) 2
= 200 or 10 2
 BD
1  formula/method
area/opp ∆BDF = .BF.BD.sinDB̂F
2
1  20  correct
=   200 (sin 45°) substitution into
2 3 
area rule
100 1
= or 33 or 33,33 square units/vk eenh  answer
3 3
(5)
OR/OF
area/opp ∆BDF  formula/method
= area/opp ∆BCD – area/opp ∆BCF  BD = 10 2
1
2
( )( ) 1  20 
= 10 2 5 2 −  (5)
2 3 
 BC = 5 2
20
 BF =
100 1 3
3 3
(5)
OR/OF

10
5− 0 1 FM 1
tan FD̂M = mDC = = or tan FD̂M = = 3 =  gradient/ratio
13 + 2 3 DM 10 3
FD̂M = 18,43°
BD̂F = 26,56°  BD̂F
area / opp ∆BDF
1
= .BD.DF. sin BD̂F  DF OR/OF
2 BD
1
(
= . 10 2 .
2
) 10 10 
3
. sin 26,56°

correct
substitution into
 
area rule
100 1
3 3
(5)
[18]


QUESTION/VRAAG 4
y R(k ; 21)
x
O C(3 ; –1)
P S
4.1 radius ⊥ tangent /raaklyn R

(1)
4.2 CR = TR + CT
2 2 2
(Pyth)
 substitution
CR = 20 + 10 = 500
2 2 2
CR = 500 or 10 5  answer
(2)
4.3 CR = ( x 2 − x1 ) + ( y 2 − y1 )
2 2 2
500 = (k − 3) 2 + (21 + 1) 2  substitution

k − 6k + 9 + 484 = 500
2
k 2 − 6k − 7 = 0  standard form
(k − 7)(k + 1) = 0
 factors
k = 7 or k ≠ −1 k = 7
(4)
OR/OF
CR 2 = ( x2 − x1 ) 2 + ( y2 − y1 ) 2
500 = (k − 3) 2 + (21 + 1) 2  substitution
(k − 3) 2 = 16  square form
k − 3 = 4 or k − 3 = −4  square root
k = 7
k = 7 or k ≠ −1 (4)


4.4 ( x − 3) 2 + ( y + 1) 2 = 100  answer

(2)
4.5 CS = 10 and CS ⊥ PS
∴ S(3 ; − 11)  S(3 ; − 11)
∴ y = – 11  answer
(2)
4.6.1 S(3;−11)
∴ 3(−11) − 4 x = 35  substituting
x = −17
∴ P(−17 ; − 11)  answer
(2)
OR/OF
4 35  equating
x+ = −11
3 3
4 − 68
x=
3 3
x = −17
 answer
P(−17 ; − 11) (2)
4.6.2 PT = PS [tangents from common point/rklyne vanaf dies pt]  S R
= 17 + 3 = 20 units  answer
(3)
OR
PC = (−17 − 3) 2 + (−11 + 1) 2
= 500 or 10 5  value of PC
PT 2 = PC 2 − TC 2 [Pyth th]
 using Pyth
= 500 – 100
= 400  answer
∴ PT = 20 (3)
OR
PC = (−17 − 3) 2 + (−11 + 1) 2
 value of PC
= 500 or 10 5  S/R or
∆ PTC ≡ ∆RTC [90°HS] proved
∴ PT = TR
∴ PT = 20  answer
(3)
4.7.1 M (3 ; − 16) answer
(1)


4.7.2 Radius = 4  answer

(1)
4.7.3 r1 + r2 = 10 + 4 = 14  r1 + r2
distance CM = (3 − 3) 2 + (−1 + 16) 2
= 225
= 15  15
CM > r1 + r2 explanation
Therefore the two circles do not intersect or touch./Daarom sny of
(3)
raak die twee sirkels nie.
[21]


QUESTION/VRAAG 5
5.1
P
T 2 R 3 S
1 5
5.1.1(a) sin T = = = 0,45 value
5 5
(1)
3 3 10
5.1.1(b) cos S = = = 0,95 value
10 10
(1)
5.1.2 cos(T + S) = cos T cos S − sin T sin S expansion
 2  3   1  1  2 1
=  
 −   5 10
 5  10   5  10 
6 1  simplification
= −
50 50
5 1 2
= or or  answer
50 2 2
(5)
5.2 1
− tan 2 (180° + θ )
cos(360° − θ ) sin(90° − θ )
 cos θ
1
= − tan 2 θ  cos θ
(cos θ )(cos θ ) tan2 θ
1  sin 2 θ  sin 2 θ
= −   
 2 
cos θ  cos θ 
2 cos 2 θ
1 − sin 2 θ
=
cos 2 θ
identity
cos 2 θ 1 − sin 2 θ
= OR
cos 2 θ 1 − sin 2 θ  answer
=1 (6)


5.3 2
3
(sin x − cos x) 2 =    squaring both
4 sides
9 expanding LHS
sin 2 x − 2 sin x cos x + cos 2 x =
16
9  using identity
1 − 2 sin x cos x =
16
7
2 sin x cos x =  simplifying
16
7
∴ sin 2 x =
16 answer
(5)
[18]


QUESTION/VRAAG 6
6.1 4 sin x + 2 cos 2 x = 2

2 sin x + cos 2 x − 1 = 0
2 sin x + (1 − 2 sin 2 x) − 1 = 0 using identity
2 sin x − 2 sin x = 0
2  standard form
2 sin x(sin x − 1) = 0
 factors
2 sin x = 0 or sin x − 1 = 0
sin x = 0 sin x = 1  sin x = 0 or
sin x = 1
x = k .180° or x = 90° + k .360 , k ∈ Z  k .180°

90° + k .360, k ∈ Z
(6)
6.2.1  turning point
(–90° ; –3)
1  turning point
g (90° ; 1)
 (–180° ; –1) &
– 180o – 90o O 90o 180o (0° ; –1)
–1
–2
f
–3
(3)
6.2.2 (−90° ; 0°)   answer
(2)
OR/OF
  answer
− 90° < x < 0° (2)
6.2.3 f ( x) = g ( x)
∴ −180°; 0° ; 90° ;180°
f ( x + 30°) = g ( x + 30°)  any ONE correct
∴ x = −30°; 60° ;150°  other 2 correct
(2)
[13]


QUESTION/VRAAG 7
81 87
D
64
B C
7.1 AB̂D = θ [alternate ∠ s; || lines]

BD 64 correct trig ratio
cos θ = =
AB 81 substitution into
θ = 38° correct ratio
 answer (to the
OR/OF nearest degree)
(3)
64 correct trig ratio
sin BÂD =
81 substitution into
BÂD = 52,18° correct ratio
 answer (to the
θ = 38° nearest degree)
(3)
7.2 BC = AB + AC − 2(AB)(AC) cos BÂC
2 2 2 use cosine rule
correct substitution
= 812 + 87 2 − 2(81)(87) cos 82,6° into cosine rule
= 12314,754...
BC = 110,97 m answer
(3)


7.3 sinDĈB sinBD̂C

=  use sine rule
BD BC
BD.sinBD̂C
sinDĈB =
BC
64. sin 110°  substitution
sin DĈB =
110,97
∴ DĈB = 32,82°  answer
(3)
[9]


QUESTION/VRAAG 8
8.1
18°
M O
1
148°
1 1
T 2
3 R
43o
8.1.1 P̂ = 32° [opp ∠s of cyclic quad/teenoorst∠e v koordevh] S

R
(2)
8.1.2 Ô1 = 2(32°) = 64° [∠ centre = 2 ∠ at circum/midpts∠ = 2 omtreks∠]  S R
(2)
OR/OF
reflex Ô = 296° [∠ centre = 2 ∠ at circum/midpts∠ = 2 omtreks∠]  S and R
S
Ô1 = 64° [∠s around a point/∠e om ' n punt ] (2)
8.1.3 OM̂S = 180° − (32° + 18° + 43°) [sum ∠s ∆ / som ∠e ∆] S
= 87°
S
(2)
8.1.4 R̂ 3 = TM̂P [ext ∠ cyclic quad/buite ∠ koordevh] R
= 87° + 18° − 6°
= 99° S
(2)


8.2
A 1
2 B
1 2
3
x
1 2 3 1
D C
E 2
8.2.1 corres ∠s/ooreenk ∠e; AB | | DC R

(1)
8.2.2 Ê 2 = x [tan - chord theorem/raakl − koordst ] S R
B̂2 = x [∠s opp = sides/∠e teenoor = sye] Any 3 ∠s

correct SR
Ê 3 = x [alt ∠s/verwiss ∠e; AB | | DC] SR
DÂB = x [opp ∠s ||m /teenoor ∠e ||m OR/OF alternate/verwiss ∠s/e; BC ||AD] (6)
8.2.3 D̂ = 180° − x [co - int ∠s suppl/ko − binne∠e suppl; AD | | BC] S  R
∴ B̂2 + D̂ = 180°
∴ ABED a cyc quad/kdvh [converse opp ∠s of cyclic quad/
omgek teenoorst ∠e koordevh] R
(3)
OR/OF
DÂB = x [opp ∠s/teenoor ∠e | |m ] OR/OF [alt∠s/verwiss ∠e; BC | | AD] S  R
Ê 3 = DÂB = x
∴ ABED a cyc quad/kdvh [converse ext ∠ of cyc quad/omgek buite∠ v koordevh] R
(3)
[18]


QUESTION/VRAAG 9
9.1 … in the alternate segment/…in die( teen)oorstaande segment  answer

(1)
9.2
A
1
2
D 2
1
1B
4 2
3
1 1 2
2 G F
C
9.2.1 Â 1 = D̂ 1 [tan chord theorem/raakl − koordst ] S  R
B̂ 4 = Â 1 + D̂ 2 [ext ∠ Δ/buite∠ ∆] SR

= D̂ 1 + D̂ 2 (4)
9.2.2 B̂4 = B̂2 [vert opp ∠s/regoorst ∠e] S
D̂1 + D̂ 2 = B̂2 [proven/bewys]

= Ĝ 2 [∠s in same segment/∠e in dies segment ] SR
R
∴ AGCD is cyc quad/kvh [converse ext∠ cyc quad/omgek buite∠ kvh] (4)
9.2.3 D̂1 = Â 2 [ ∠s in same segment/∠e in dies segment ] SR
Â 2 = F̂ [ ∠s in same segment/∠e in dies segment ]  S
∴ D̂1 = F̂
∴ DC = CF [sides opp = ∠s/sye teenoor = ∠e] R
(4)
[13]


QUESTION/VRAAG 10
M A
♦
B C K F
♦
V T
10.1 Constr/Konstr :
Draw line BC such that MB = AK and MC = AF  constr/konstr
Trek lyn BC sodat MB = AK en MC = AF
Proof/Bewys :
In ∆BMC and/en ∆KAF
MB = AK [constr/konstr ]
M̂ = Â [given/gegee]
MC = AF [constr/konstr ]
∆BMC ≡ ∆KAF [s ∠ s] S/R
∴ MB̂C = AK̂F or MĈB = AF̂K [ ≡ ∆] S

but /maar V̂ = K̂ or T̂ = F̂ [given/gegee]
∴ MB̂C = V̂ or MĈB = T̂ S
But these are corresponding ∠s/maar hulle is ooreenk ∠e
∴ BC | | VT [corr ∠s = /ooreenk∠e = ] S/R
MV MT S R
∴ = [prop theorem/eweredighst; BC | | VT]
MB MC
but /maar MB = AK and MC = AF [constr/konstr ]
MV MT
∴ = (7)
AK AF


10.2
E
M 2 1 H
1 2
F 1 2
3
G
K
10.2.1(a) In ∆KGH and ∆KEF

K̂ is common/gemeen S
Ĥ 2 = F̂ [ext ∠ cyclic quad/buite ∠ koordevh] S R
Ĝ 3 = Ê [sum∠s ∆ OR ext ∠ cyclic quad/som∠e∆ OR buite∠koordevh]  naming third
∴ ∆KGH ||| ∆KEF [∠∠∠ ] angle OR ∠∠∠
(4)
10.2.1(b) EF KE
= [||| ∆s] S
GH KG
EF KE
∴ = [KG = EF] S
GH EF
∴ EF 2 = KE.GH (2)
10.2.1(c) KG EM
= [prop theorem/eweredighst ; MG | | EK] S R
KF EF
but EF = KG [given/gegee]
KG EM S
=
KF KG
KG 2 = EM.KF (3)
10.2.2 KE.GH = EM.KF 
20 × 4 KE.GH = EM.KF
EM =
16
 substitution
= 5 units
 answer
(3)
[19]
TOTAL/TOTAAL: 150
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SENIOR CERTIFICATE EXAMINATIONS/

This memorandum consists of 21 pages./

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Mathematics P2/Wiskunde V2 2 DBE/2016

1.1 189 189

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Mathematics P2/Wiskunde V2 3 DBE/2016

2.1 (30 ; 53) answer

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Mathematics P2/Wiskunde V2 7 DBE/2016

Mathematics P2/Wiskunde V2 8 DBE/2016

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Mathematics P2/Wiskunde V2 9 DBE/2016

4.1 radius ⊥ tangent /raaklyn R

500 = (k − 3) 2 + (21 + 1) 2  substitution

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Mathematics P2/Wiskunde V2 10 DBE/2016

4.4 ( x − 3) 2 + ( y + 1) 2 = 100  answer

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Mathematics P2/Wiskunde V2 11 DBE/2016

4.7.2 Radius = 4  answer

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Mathematics P2/Wiskunde V2 12 DBE/2016

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6.1 4 sin x + 2 cos 2 x = 2

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7.1 AB̂D = θ [alternate ∠ s; || lines]

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7.3 sinDĈB sinBD̂C

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8.1.1 P̂ = 32° [opp ∠s of cyclic quad/teenoorst∠e v koordevh] S

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Mathematics P2/Wiskunde V2 18 DBE/2016

8.2.1 corres ∠s/ooreenk ∠e; AB | | DC R

B̂2 = x [∠s opp = sides/∠e teenoor = sye] Any 3 ∠s

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Mathematics P2/Wiskunde V2 19 DBE/2016

9.1 … in the alternate segment/…in die( teen)oorstaande segment  answer

9.2.1 Â 1 = D̂ 1 [tan chord theorem/raakl − koordst ] S  R

B̂ 4 = Â 1 + D̂ 2 [ext ∠ Δ/buite∠ ∆] SR

D̂1 + D̂ 2 = B̂2 [proven/bewys]

Â 2 = F̂ [ ∠s in same segment/∠e in dies segment ]  S

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Mathematics P2/Wiskunde V2 20 DBE/2016

∴ MB̂C = AK̂F or MĈB = AF̂K [ ≡ ∆] S

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Mathematics P2/Wiskunde V2 21 DBE/2016