GRADE 11 NOVEMBER 2015 MATHEMATICS P2 · 2016. 1. 5. · learner name national senior certificate mathematics p2 grade 11 november 2015 special answer book question mark initial mod

NATIONAL SENIOR CERTIFICATE

GRADE 11

NOVEMBER 2015

MATHEMATICS P2

MARKS: 150 TIME: 3 hours

This question paper consists of 14 pages.

*Imat2*

2 MATHEMATICS P2 (EC/NOVEMBER 2015)

Copyright reserved Please turn over

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.



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]



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]



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)



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



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



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



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]



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



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°



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



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



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

LEARNER NAME


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*

2 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015)


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

(EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 3



(2) 2.2

(1) 2.3

(4) [7]



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)



3.4

(3) [13]




(1) 4.2

(1) 4.3

(3) 4.4

(4)

M

Q

P

R

N(-6; -12)

x

y

O



4.5

(4) 4.6

(3) [16]



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



5.3

(5) 5.4

(5)

[23]




(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



6.2.3

(3) 6.2.4

(4) [15]




(6)

7.2

(6) 7.3

(1) [13]




(1) 8.2.1

(3) 8.2.2

(4) [8]

. O D

C

B

A

F

E




(6)

. A

C

B

O



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°



QUESTION 10

Solution Marks 10.1

(4) 10.2

(2) [6]

A

D

B

C

E

F

1 2 3

4 1 2




(1) 11.2

11.2.1

(5)

B

C

M

O

1 2

1 2 3

D A

1



11.2.2

(2) 11.2.3

(3) [11]



QUESTION 12

Solution Marks 12.1

(3) 12.2

(3) 12.3

(5) [11]

6 cm

12 cm

8 cm



ADDITIONAL PAGES


GRADE 11

NOVEMBER 2015

MATHEMATICS P2/WISKUNDE V2 MEMORANDUM

MARKS/PUNTE: 150

This memorandum consists of 8 pages. Hierdie memorandum bestaan uit 8 bladsye.

2 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2015)


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

(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



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)



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)



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]



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



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

Documents

GRADE 11 NOVEMBER 2015 MATHEMATICS P2 · 2016. 1. 5. · learner name national senior certificate mathematics p2 grade 11 november 2015 special answer book question mark initial mod