grade 11 november 2015 mathematics p2 · 2016. 1. 5. · learner name national senior certificate...
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NATIONAL SENIOR CERTIFICATE
GRADE 11
NOVEMBER 2015
MATHEMATICS P2
MARKS: 150 TIME: 3 hours
This question paper consists of 14 pages.
*Imat2*
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2 MATHEMATICS P2 (EC/NOVEMBER 2015)
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INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 12 questions. 2. Answer ALL the questions in the SPECIAL ANSWER BOOK provided. 3. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in
determining the answers.
4. Answers only will NOT necessarily be awarded full marks. 5. You may use an approved scientific calculator (non-programmable and non-
graphical), unless stated otherwise.
6. If necessary, round off answers to TWO decimal places, unless stated otherwise. 7. Write neatly and legibly.
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3 MATHEMATICS P2 (EC/NOVEMBER 2015)
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QUESTION 1 The following table represents the heights, in centimetres, of 120 boys in a school.
HEIGHT (cm) FREQUENCY 150 55 4 155 160 22 160 165 56 165 170 32 6
1.1 Complete the cumulative frequency table in the SPECIAL ANSWER BOOK. (2) 1.2 Draw an ogive, using the diagram in the SPECIAL ANSWER BOOK, to represent
the information in the table. (4) 1.3 Determine, using the ogive, the five number summary. (5) 1.4 If the distribution of the data is represented by means of a box whisker diagram,
comment on the spread of the data. (1) [12]
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4 MATHEMATICS P2 (EC/NOVEMBER 2015)
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QUESTION 2 The following is a sample of weekly wages earned by ten people working for a small printing and design company.
R2 250 R2 250 R3 000 R3 300 R3 300
R3 600 R3 900 R4 350 R4 350 R5 250 2.1 Calculate the mean weekly wage. (2) 2.2 Calculate the standard deviation of the weekly wage. (1) 2.3 Determine the percentage of workers which lie within ONE standard deviation of
the mean. (4) [7]
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5 MATHEMATICS P2 (EC/NOVEMBER 2015)
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QUESTION 3 The points A(2a – 11; a +2), C(4; -1) and D(4p; p – 7) are the vertices of ∆ACD with B(-2; 3) on AC.
. .D
3.1 If points A, B and C are collinear, find the value of a. (4) 3.2 Determine the equation of the line AC. (3) 3.3 Hence, determine the co-ordinates of midpoint M of AB. (3) 3.4 Determine the value of p if CD is parallel to the x-axis. (3) [13]
.A(2a – 11; a + 2) ,...
x
y
(4p; p -7) C(4; -1)
B(-2; 3)
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6 MATHEMATICS P2 (EC/NOVEMBER 2015)
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QUESTION 4 In the diagram, M, N and P are vertices of ∆MNP, with N(-6; -12). M is a point on the y-axis. The equation of the line MN is . MR = NR and NQ ┴ MP. PR and NQ intersect at the origin O. 4.1 Calculate the gradient of NQ. (1) 4.2 Calculate the gradient of MP. (1) 4.3 Calculate the angle of inclination of MP. (3) 4.4 Hence, determine the equation of the line MP. (4) 4.5 Hence, determine the coordinates of P. (4) 4.6 Determine the co-ordinates of R. (3) [16]
M
Q
P
R
N(-6; -12)
x
y
O
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7 MATHEMATICS P2 (EC/NOVEMBER 2015)
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QUESTION 5 5.1 In the diagram below, P(1;3) is a point on the Cartesian plane,
OP = r and ̂ = θ. 5.1.1 Make use of the diagram to calculate the value of θ. (2) 5.1.2 Calculate the length of OP. Leave the answer in surd form. (2) 5.1.3 Determine the values of the following, without using a calculator: (a) sin θ (1) (b) cos (180° + θ) (2) 5.2 Determine the general solutions of:
(6)
5.3 Simplify: ( ) ( )
( ) (5)
5.4 Prove that:
(5)
[23]
P(1;3)
O θ
r
y
x
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8 MATHEMATICS P2 (EC/NOVEMBER 2015)
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QUESTION 6 6.1 COMPLETE: In ∆ABC … + … – …
(2)
6.2 In the diagram below, PQ is a straight line 1 500 m long. RS is a vertical tower
158 m high with P, Q and S points in the same horizontal plane. The angles of elevation of R from P and Q are and θ. ̂ .
6.2.1 Determine the length of PS. (3) 6.2.2 Determine the length of SQ. (3) 6.2.3 Hence, find the value of θ. (3) 6.2.4 Determine the area of ΔSPQ. (4) [15]
c b
R
Q P 1 500 m
158 m
25°
30°
S
θ
A
B C a
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9 MATHEMATICS P2 (EC/NOVEMBER 2015)
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QUESTION 7 Given: ( )
( )
7.1 Draw the graph of f and g for x in the SPECIAL ANSWER BOOK.
Show all the turning points and intercepts with the axes. Clearly show the asymptotes using dotted lines. (6)
7.2 Determine the values of x, for , for which f(x) > g(x). (6) 7.3 Write down the period of g(2x). (1) [13]
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10 MATHEMATICS P2 (EC/NOVEMBER 2015)
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GIVE REASONS FOR YOUR STATEMENTS AND CALCULATIONS IN QUESTIONS 8, 9, 10 AND 11. QUESTION 8 8.1 Complete: The line drawn from the centre of the circle perpendicular to the chord … (1) 8.2 In the figure below, AB and CD are chords of the circle with centre O. OE ┴ AB.
CF = FD. OE = 4 cm, OF = 3 cm and CD = 8 cm.
8.2.1 Calculate the length of OD. (3) 8.2.2 Hence calculate the length of AB. (4) [8]
. O D
C
B
A
F
E
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11 MATHEMATICS P2 (EC/NOVEMBER 2015)
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QUESTION 9 9.1 In the diagram O is the centre of the circle and ABC are points on the circle. Use the
diagram in your SPECIAL ANSWER BOOK to prove that: ̂ ̂ . (6)
9.2 In the figure below, ̂ and O is the centre of the circle. A, B, E C and D
are points on the circumference. Calculate, giving reasons, the sizes of:
9.2.1 ̂ (2) 9.2.2 ̂ (2) 9.2.3 ̂ (2) 9.2.4 ̂ (2) [14]
. A
C
B
O
A B
E
C
D
O
1
1
1
25°
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12 MATHEMATICS P2 (EC/NOVEMBER 2015)
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QUESTION 10 A, B, C and D are points on the circumference of the circle in the diagram below. ECF is a tangent at C, B1 = B2. 10.1 If B1 = x, find, with reasons, TWO other angles equal to x. (4) 10.2 Hence, show that DC bisects ̂ . (2) [6]
A
D
B
C
E
F
1 2 3
4 1 2
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13 MATHEMATICS P2 (EC/NOVEMBER 2015)
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QUESTION 11 11.1 Complete: Opposite angles of a cyclic quadrilateral … (1) 11.2 In the figure, ABCD is a cyclic quadrilateral. AB ║ DC in circle with centre O. BC
and AD produced meet at M. D3 = x
11.2.1 Show that MC = MD. (5) 11.2.2 If D3 = x, determine the value of ̂, in terms of x. (2) 11.2.3 Hence, show that BODM is a cyclic quadrilateral. (3) [11]
B
C
M
O
1 2
1 2 3
D A
1
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QUESTION 12 The solid in the diagram is made up of a right prism with square base, and a right pyramid on top of the prism. The length of the prism is 12 cm, the side of the base is 6 cm and the height of the pyramid is 8 cm.
12.1 Calculate the slant height of the triangular face of the pyramid. (3) 12.2 Calculate the area of one of the triangular faces. (3) 12.3 Calculate the total surface area of the solid.
TSA = area of slanted faces + area of right prism (5) [11] TOTAL: 150
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LEARNER NAME
NATIONAL SENIOR CERTIFICATE
MATHEMATICS P2
GRADE 11
NOVEMBER 2015
SPECIAL ANSWER BOOK
QUESTION MARK INITIAL MOD. 1 2 3 4 5 6 7 8 9
10 11 12
TOTAL
This answer book consists of 22 pages.
*Iwisab2*
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2 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015)
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QUESTION 1 Solution Marks 1.1
HEIGHT (cm) FREQUENCY CUMULATIVE FREQUENCY 150 55 4 155 160 22 160 165 56 165 170 32 6 (2)
1.2
(4)
1.3
(5) 1.4
(1) [12]
0
50
100
150
145 150 155 160 165 170 175 180
Cum
lativ
e Fre
quen
cy
Tabl
e
Heights (in centimetres)
Ogive
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(EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 3
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QUESTION 2 Solution Marks 2.1
(2) 2.2
(1) 2.3
(4) [7]
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QUESTION 3
. .D Solution Marks 3.1
(4) 3.2
(3) 3.3
(3)
.A(2a – 11; a + 2) ,...
x
y
(4p; p -7) C(4; -1)
B(-2; 3)
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(EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 5
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3.4
(3) [13]
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6 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015)
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QUESTION 4 Solution Marks 4.1
(1) 4.2
(1) 4.3
(3) 4.4
(4)
M
Q
P
R
N(-6; -12)
x
y
O
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(EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 7
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4.5
(4) 4.6
(3) [16]
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QUESTION 5 Solution Marks 5.1.1
(2) 5.1.2
(2) 5.1.3a
(1)
5.1.3b
(2)
5.2
(6)
P(1;3)
O θ
r
y
x
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(EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 9
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5.3
(5) 5.4
(5)
[23]
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10 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015)
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QUESTION 6 Solution Marks 6.1
(2) 6.2
6.2.1
(3) 6.2.2
(3)
A
B C
c b
a
Q P 1 500 m
158 m
25°
30°
S
θ
R
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(EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 11
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6.2.3
(3) 6.2.4
(4) [15]
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12 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015)
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QUESTION 7 Solution Marks 7.1
(6)
7.2
(6) 7.3
(1) [13]
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(EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 13
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QUESTION 8 Solution Marks 8.1
(1) 8.2.1
(3) 8.2.2
(4) [8]
. O D
C
B
A
F
E
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14 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015)
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QUESTION 9 Solution Marks 9.1
(6)
. A
C
B
O
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(EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 15
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9.2
9.2.1
(2) 9.2.2
(2) 9.2.3
(2)
9.2.4
(2) [14]
A B
E
C
D
O
1
1
1
25°
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16 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015)
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QUESTION 10
Solution Marks 10.1
(4) 10.2
(2) [6]
A
D
B
C
E
F
1 2 3
4 1 2
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(EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 17
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QUESTION 11 Solution Marks 11.1
(1) 11.2
11.2.1
(5)
B
C
M
O
1 2
1 2 3
D A
1
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18 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015)
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11.2.2
(2) 11.2.3
(3) [11]
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(EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 19
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QUESTION 12
Solution Marks 12.1
(3) 12.2
(3) 12.3
(5) [11]
6 cm
12 cm
8 cm
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20 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015)
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ADDITIONAL PAGES
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(EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 21
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22 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015)
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NATIONAL SENIOR CERTIFICATE
GRADE 11
NOVEMBER 2015
MATHEMATICS P2/WISKUNDE V2 MEMORANDUM
MARKS/PUNTE: 150
This memorandum consists of 8 pages. Hierdie memorandum bestaan uit 8 bladsye.
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2 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2015)
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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 not redone the question, mark the
crossed out version. • Consistent accuracy applies in ALL aspects of the marking memorandum. • Assuming answers/values in order to solve a problem is NOT acceptable. LET WEL: • Indien ʼn kandidaat ʼn vraag twee keer beantwoord, merk slegs die eerste poging. • Indien ʼn kandidaat ʼn antwoord doodgetrek het, maar nie oorgedoen het nie, merk die
doodgetrekte antwoord. • Volgehoue akkuraatheid geld in ALLE aspekte van die memorandum. • Aanname van antwoorde/waardes om ʼn probleem op te los, is Onaanvaarbaar. QUESTION/VRAAG 1: [12] 1.1
Height (cm) Hoogte (cm)
Frequency Frekwensie
Cumulative frequency
Kumulatiewe frekwensie
150 55 4 4 155 160 22 26 160 165 56 82 165 170 32 114 6 120
26
120 (2)
1.2
grounding at(150;0)/anker by (150;0)
plotting (155;4)/plot van (155;4)
plotting with upper limits/plot by boonste limiete
joining the points to form a smooth curve/verbind punte om glade kurwe te
vorm (4)
1.3 (150, 160,5, 163, 166, 175) OR/OF Min = 150 Q1 = 160,5 Q2 = 163 Q3 = 166 Max = 175
150 160,5 163 166 175
(5)
1.4 Skewed to the left / Skeefgetrek na links. answer/antwoord (1) [12]
0
50
100
150
140 150 160 170 180
Cum
lativ
e Fr
eque
ncy
/ K
umul
atie
we F
rekw
ensie
Heights/Hoogte (in cm)
Ogive/Ogief
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(EC/NOVEMBER 2015) MATHEMATICS P2/WISKUNDE V2 3
Kopiereg voorbehou Blaai om asseblief
QUESTION/VRAAG 2 [7] 2.1 ̅
̅
= R3 555
answer/antwoord (2) 2.2 answer/antwoord (1) 2.3 (3555 – 900,12 ; 3555 + 900,12) = (2654,88 ; 4455,12)
werkers lê binne een standaardafwyking of workers lie within one standard deviation.
van die werkers lê binne een standaardafwyking.
interval 7 answer/antwoord
(4)
[7] QUESTION/VRAAG 3 [13] 3.1 mAB = mBC = mAC
( ) ( )
mAB = mBC = mAC
substitute into equation/vervang in vgl
substituting into mAC/vervang in vgl mAC
answer/antw (4) 3.2 ( )
( )
equation/vgl subst of m and(-5;5) into
form/vervang van m en (4;-1) in formule
equation/vgl (3) 3.3
[
]
[
]
= [ ]
subst into correct
formula/vervang in korrekte formule
coordinates/coordinate (3)
3.4 mCD = 0
OR/OF p – 7 = -1 p = 6
correct m = 0 substitution into
eqn/vervang in vgl answer/antw
equating/ answer
(3) [13]
ANSWER ONLY : FULL MARKS SLEGS ANTWOORD : VOLPUNTE
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4 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2015)
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QUESTION/VRAAG 4: [13] 4.1
= 2
answer/antw
(1) 4.2 MNQ × MMP = -1 [NQ ┴ MP]
answer/antw
(1) 4.3
value of /waarde van (3)
4.4 ( )
( )
equation/vgl subst of m =
and
(0;6) into eqn/vervang m =
en (0;6) in vgl
answer/antw (4) 4.5
P (2;5)
subst into eqn/vervang in vgl
P (2;5) (4) 4.6 MR = NR
R = [
]
R = [
]
R = [-3;-3]
Substitution/vervang.
x value/waarde y value/waarde
(3) [16]
QUESTION/VRAAG 5: [23] 5.1.1
tan θ answer/antwoord (2)
5.1.2 ( ) ( ) √
OP2 answer/antwoord (2)
5.1.3a √
answer/antwoord (1) 5.1.3b ( )
√
ANSWER ONLY FULL MARKS/ SLEGS ANTWOORD:VOLPUNTE
-cosθ
answer/antwoord
(2)
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(EC/NOVEMBER 2015) MATHEMATICS P2/WISKUNDE V2 5
Kopiereg voorbehou Blaai om asseblief
5.2 ( ) ( )( )= 0
standard form/st vorm 1 – sin2 x
sin x = ½ or no soln (sin x = 2) 30° ; 150° k.360,
(6) 5.3
=
= -tan x
sin x -sin x cos2 x
tan x
(5) 5.4 ( )
( )
( )
( )
( ) ( )
numerator / teller denominator / noemer
expansion / uitbreiding
simplification / vereenvoudiging
factorisation / faktorisering (5)
[23] QUESTION/VRAAG 6: [11] 6.1 b2 = a2 + c2 – 2acCos B a2 + c2
2acCos B (2) 6.2.1
̂ PS = 338,83
Ratio for tan/verhou. vir
tan substitution 65°/verv van
65° answer/antwoord (3)
6.2.2 In ∆PQS = ( )( ) = 1484499,606 SQ = 1218,40 m2
use of cos
formula/gebruik van cos formule
Substitution/vervang SQ (3)
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6 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2015)
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6.2.3 In ∆RSQ:
value of /waarde van
(3) 6.2.4 Area of ΔSPQ = ½ SP.PQ.sin ̂
= ½ (338,83)(1 500)sin 30° =127061,25 m2
correct formula / korrekte formule
substitute / vervang (338,83), (1500)
Sin 300 answer / antwoord (4)
[15] QUESTION/VRAAG 7: [12] 7.1
f asymtotes /
asimptote min value / min
waarde max value /
maks waarde g
(-45°; -1) (45°; 1) (135° ; -1)
(6) 7.2 ( ) ( ) ( ) -45
45 0 90 135 180
(6) 7.3 90° answer /
antwoord (1)
[13]
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(EC/NOVEMBER 2015) MATHEMATICS P2/WISKUNDE V2 7
Kopiereg voorbehou Blaai om asseblief
QUESTION/VRAAG 8: [8] 8.1 bisects the chord. answer/antwoord
(1) 8.2.1 (Pythagoras)
(substitution/vervang)
method/metode
answer/antwoord
(3) 8.2.2 (Pythagoras)
(substitution/vervang) cm But AB = 2AE (OE ┴ AB) AB = 2(3) = 6 cm
method/metode
cm
S/R
answer/antwoord (4) [8]
QUESTION/VRAAG 9: [13] 9.1
CONSTR: Join CO, extend to D PROOF: In ∆AOC i) ̂ ̂ ̂ (ext of ∆/buitehoek van ∆) ii) ̂ ̂ ̂ (ext of ∆/buitehoek van ∆) iii) ̂ ̂ (AO = OC) iv) ̂ ̂ (BO = OC) ̂ ̂ ̂ ̂ ̂ = 2( ̂ ̂ ) ̂
construction/konstr
S/R S/R S/R S/R
conclusion /
gevolgtrekking (6)
9.2.1 ̂ ( radii equal/radiusse gelyk) S R (2)
9.2.2 ̂ (ext of ∆/buitehoek van ∆) S R (2)
9.2.3 ̂ (angles in same segment/ hoeke in dieselfde segment)
S R (2)
9.2.4 ̂ (opp angles of cyclic quad/ teenoorstaande hoeke van ʼn koordevierhoek)
S R (2)
[14]
. A
B
O
C
D
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8 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2015)
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QUESTION/VRAAG 10: [6] 10.1 ̂ ̂ = x (angles in the same segment/
hoeke in dieselfde segment)
̂ ̂ = x (tan chord/tan koord)
S R
S R (4)
10.2 ̂ ̂ (both equal to x/albei gelyk aan x) ̂
S/R conclusion /
gevolgtrekking (2) [6]
QUESTION/VRAAG 11 [11] 11.1 Are supplementary OR add to 180°. answer/antwoord
(1) 11.2.1 ̂ ̂ (Ext of cyclic quad/buitehoek
van koordevhk) ̂ = ̂ (corresponding angles, AB ║ DC/ Ooreenkomstige hoeke AB ║ DC) (base angles of ∆ equal/basis hoeke van ∆ gelyk)
S R
S R
R (5)
11.2.2 ̂ (angles of ∆/hoeke van ∆) S R
(2) 11.2.3 ̂ ( at centre = 2 at circumference/
by middle = 2 by omtreks) ̂ ̂
S R
S
(3) [11] QUESTION/VRAAG 12 [11] 12.1 Sh2 = (Pythagoras)
Sh2 = 64 + 9 (substitution/vervang) Sh = √ = 8,54 cm
S method/metode answer/antwoord
(3) 12.2 Area of ∆ face =
= ( )(√ )
= 25,63 cm2
formula/formule substitution/vervang answer/antwoord
(3) 12.3 TSA = area of slanted faces + area of right prism
TBO = oppervlakte van skuinsvlakke + oppervlakte van reghoekige prisma = 3(25,63) + 62 + 4(6 x 12) = 76,89 + 36 + 288 = 400,89 cm2
TSA substitute/vervang
answer/antwoord (5)
[11] TOTAL/TOTAAL: 150