GRADE 12 SEPTEMBER 2015 MATHEMATICS P2 · 2016-09-03 · NATIONAL SENIOR CERTIFICATE GRADE 12 SEPTEMBER 2015 MATHEMATICS P2 MARKS: 150 TIME: 3 hours This question paper consists of

NATIONAL SENIOR CERTIFICATE

GRADE 12

SEPTEMBER 2015

MATHEMATICS P2

MARKS: 150 TIME: 3 hours

This question paper consists of 13 pages including 1 information sheet, and a SPECIAL ANSWERBOOK.

*MATHE2*

2 MATHEMATICS P2 (EC/SEPTEMBER 2015)

Copyright reserved Please turn over

INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 11 questions. 2. Answer ALL the questions in the SPECIAL ANSWER BOOK provided. 3. Clearly show ALL calculations, diagrams, graphs, et cetera that 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. Number the answers correctly according to the numbering system used in this

question paper. 8. Write neatly and legibly.

(EC/SEPTEMBER 2015) MATHEMATICS P2 3


QUESTION 1 The data in the table below represents the marks obtained by 10 Grade 12 learners for English Home Language (HL) and Afrikaans First Additional Language (FAL).

English HL 42 54 85 32 63 71 92 62 58 66

Afrikaans FAL 50 58 80 45 60 65 98 75 71 58

1.1 Draw a scatter plot of the data above by making use of the grid provided in the

SPECIAL ANSWER BOOK. (4) 1.2 Calculate the equation of the least squares regression line for this data. (3) 1.3 Calculate the correlation coefficient. (2) 1.4 Describe the correlation between English Home Language and Afrikaans First

Additional Language. (1) 1.5 Predict the final English Home Language mark for the learner who obtained 74

marks in Afrikaans First Additional Language.

(2) [12]



QUESTION 2 The weights (in kilogram) of the 20 boys in the hockey squad of School A are given below:

69 59 59 66 64 58 63 58 62 61

57 53 60 51 60 48 47 60 40 60

2.1 Determine the mean and variance for the weights of the School A squad. (3) 2.2 The following information was obtained from the School B boys’ hockey coach,

regarding the weights of the boys in his squad.

∑

∑

2.2.1 How many boys are in the School B squad? (1) 2.2.2 Determine the mean weight for the School B squad. (2) 2.2.3 Determine the standard deviation for the School B squad. (2) 2.3 If five boys of equal weight are added to the squad of School A so that the means

of both schools are the same, what must be the weight of each boy? (2) [10]



QUESTION 3

In the figure A(3 ; 5), B(x ; y), C(5 ; 3) and D( 1; 1) are the vertices of parallelogram ABCD. AC and BD, the diagonals of the parallelogram, intersect at E. 3.1 Determine: 3.1.1 The co-ordinates of E (2) 3.1.2 The co-ordinates of B (3) 3.1.3 The co-ordinates of the midpoint F, of CD and hence the equation of the line

passing through F, parallel to AD.

(5) 3.2 The points G(t +1 ; 2,5) ,D( 1; 1) and E(4 ; 4) and are collinear. Calculate the

value of t.

(4) 3.3 Determine, by calculations, whether ABCD is a rhombus or not. Give a reason for

your answer.

(5) [19]

B(x; y) y)Y) A(3; 5)

C(5; 3)

D ; )

O x

y

E



QUESTION 4 In the diagram below, M(3 ; 1), Q and N lie on the circumference of circle with centre P( 1 ; 4) and form ΔMQN. NPM is a straight line. 4.1 Determine the equation of the circle. (4) 4.2 Why is ? (1) 4.3 Show that the co-ordinates of Q are ( 4; 0). (3) 4.4 Calculate the gradient of MN. (2) 4.5 Hence, calculate the size of α. (5) 4.6 Determine the equation of a tangent to the circle at M. (5) [20]

M(3 ; 1)

P( 1 ;4)

y

x Q

N α

𝛽 𝜃



QUESTION 5 5.1 Prove, without the use of a calculator, that,

√

(4) 5.2 Determine the general solution of:

(7) 5.3 Prove the identity

(3) 5.4 Simplify

(6) [20] QUESTION 6 Given and for ; 6.1 Write down the period of g. (1) 6.2 Use the set of axes provided in the SPECIAL ANSWER BOOK, to draw sketch

graphs of f and g for x [ 90°;90°]. Clearly show all intercepts with the axes and the co-ordinates of all the turning points and end points of both curves.

(6) 6.3 Use the graphs to determine the value(s) of x for x ; ], where: 6.3.1 (2) 6.3.2 (2) 6.4 Determine the range of h(x) = 3f(x) – 1. (2) [13]



QUESTION 7 In the diagram below, C is a point on one side of the Buffalo River and is 3 m above the water. A is a point on the other side of the river directly opposite C on the higher bank. B is a boat on the river. A, B and C are in the same vertical plane. The angle of depression of B from A is 33,7°. The angle of depression C from A is 15, 60° and B from C is 16,7°. 7.1 Calculate the length of BC. (3) 7.2 Calculate the length of AB. (3) 7.3 Calculate the length of AD. (3) [9]

33,7o 16,7°

C

E B

A

D

3 m



Give reasons for ALL statements in QUESTIONS 8, 9, 10 and 11. QUESTION 8 In the diagram below, O is the centre of the circle which passes through P, T, R and S. PTRS is a cyclic quadrilateral and ST is drawn.

8.1 Express, giving reasons, each of the following angles in terms of x. 8.1.1 1 (2) 8.1.2 (2) 8.1.3 (2) 8.2 Hence, calculate the value of x if SOTR is a parallelogram. (3) [9]

R

S

T

O P

1 2 3

1 2

3

1

x



QUESTION 9 In the diagram below, M is the midpoint of chord PT of circle with centre O. OR is a radius passing through M. QR is produced to intersect tangent TA at A, such that TA A RA. T and R are joined.

Prove, stating reasons, that: 9.1 MTAR is a cyclic quadrilateral (4) 9.2 PR = TR (5) 9.3 1 = 2 (3) [12]

O M

T

R

P

1

2 3

Q

1 2



QUESTION 10 10.1 Complete the statement of the following theorem:

If two triangles are equiangular then their corresponding sides are … and the two triangles are similar. (1)

10.2 In the figure below, AB is a tangent to the circle with the centre O. AC = AO and

BA ║ CE. DC produced cuts tangent BA at B.

10.2.1 If A4 = x, determine with reasons three other angles equal to x. (3) 10.2.2 Prove that ∆ACF ||| ∆ADC. (3)

10.2.3 Prove that

(4)

[11]

A

B

C

E

D

O

F2 1 3

4

1 2

1 2 3

1

2 3 4

1 2

x



QUESTION 11 11.1 Make use of the diagram in the SPECIAL ANSWER BOOK, to prove the

theorem which states that if DE║BC then,

.

(6)

11.2 In the diagram below, DE ║ BC, AN A DE and BC.

Write down the values of: 11.2.1

(2) 11.2.2

(4) 11.2.3

(3) [15] TOTAL: 150

A

B C

D E

A

E D

N B C

M



INFORMATION SHEET: MATHEMATICS

aacbbx

242 �r�

)1( niPA � )1( niPA � niPA )1( � niPA )1( �

¦

n

i

n11

2)1(

1

� ¦

nnin

i

dnaTn )1( �� dnann )1(2

2S ��

1� nn arT � �

11

��

rraS

n

n ; 1zr

raS�

f 1; 11 �� r

� �> @iixF

n 11 �� [1 (1 ) ]nx iP

i

��

hxfhxfxf

h

)()(lim)('0

��

o

22 )()( 1212 yyxxd �� M ¸¹

·¨©

§ ��2

;2

2121 yyxx

cmxy � )( 11 xxmyy � � 12

12xxyy

m�

� Ttan m

� � � � 222 rbyax ��

In 'ABC: C

cB

bA

asinsinsin

Abccba cos.2222 �� CabABCarea sin.21

'

� � EDEDED sin.coscos.sinsin � � � � EDEDED sin.coscos.sinsin � �

� � EDEDED sin.sincos.coscos � � � � EDEDED sin.sincos.coscos � �

°¯

°®

�

�

�

1cos2sin21

sincos2cos

2

2

22

D

D

DD

D DDD cos.sin22sin

nfx

x ¦ � �

n

xxn

ii

2

2

1¦

� V

� �SnAnAP )()( P(A or B) = P(A) + P(B) – P(A and B)

bxay � ˆ � �¦

¦�

�� 2)(

)(xx

yyxxb

LEARNER NAME

GRADE


MATHEMATICS P2

GRADE 12

TRIAL EXAMINATIONS 2015

SPECIAL ANSWER BOOK

QUESTION MARK INITIAL MOD. 1 2 3 4 5 6 7 8 9

10 11

TOTAL This answerbook consists of 17 pages. *MATHE4*

2 MATHEMATICS SPECIAL ANSWERBOOK P2 (EC/SEPTEMBER 2015)


QUESTION 1 Solution Marks 1.1

English HL

42 54 85 32 63 71 92 62 58 66

Afrikaans FAL

50 58 80 45 60 65 98 75 71 58

(4) 1.2

(3) 1.3

(2) 1.4

(1) 1.5

(2) [12]

0

10

20

30

40

50

60

70

80

90

100

110

0 10 20 30 40 50 60 70 80 90 100

Afr

ikaa

ns F

irst

Add

ition

al L

angu

age

Mar

ks

English Home Language Marks

Scatter Plot for English HL and Afrikaans FAL Marks

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



(3) 2.2.1

(1) 2.2.2

(2) 2.2.3

(2) 2.3

(2) [10]



QUESTION 3 Solution Marks 3.1 3.1.1

(2) 3.1.2

(3) 3.1.3

(5)

B (x; y) A(3; 5)

C(5; 3)

D(-1;1)

O x

y

E



3.2

(4) 3.3

(5) [19]




(4) 4.2

(1) 4.3

(3) 4.4

(2)

𝛽

α

P(−1 ; 4)

M(3 ; 1)

Q

N

x

y

𝜃



4.5

(5) 4.6

(5) [20]




(4) 5.2

(7) 5.3

(3)



5.4 (6) [20]




(1) 6.2

(6) 6.3 6.3.1

(2) 6.3.2

(2) 6.4

(2) [13]



QUESTION 7

Solution Marks 7.1

(3)

33,7o 16,7o

C

E B

A

D

3 m



7.2

(3) 7.3

(3) [9]



QUESTION 8

Solution Marks 8.1 8.1.1

(2) 8.1.2

(2) 8.1.3

(2) 8.2

(3) [9]

R

S

Q

O P

1 2 3

1 2

3

1

x




(4) 9.2

(5) 9.3

(3) [12]

T

A

R

P Q

O M

1

2 3

1 2



QUESTION 10

Solution Marks 10.1

(1) 10.2

10.2.1

(3) 10.2.2

(3) 10.2.3

(4) [11]

E

1

x

A

B

C

D

F2

4

O

3

1 2

2 1

1 2 3

1

2 3 4




(6) 11.2

11.2.1

(2)

A

E D

N B C

M

A

B C

D E



11.2.2

(4) 11.2.3

(3) [15] TOTAL: 150


GRADE 12

SEPTEMBER 2015

MATHEMATICS P2 MEMORANDUM

MARKS: 150

This memorandum consists of 14 pages.



QUESTION 1 1.1

English HL

42 54 85 32 63 71 92 62 58 66

Afrikaans FAL

50 58 80 45 60 65 98 75 71 58

��999 plotting all ten points 9regression

line�

(4) 1.2 a = 18,03563…

b = 0,76 7429… 0,77x

9 value for A 9value for B 9equation (3)

1.3 r = 0,88 99Answer�

(2) 1.4 Strong positive correlation.

OR Linear trend OR If learner got high marks in English then they would achieve similar marks in Afrikaans

9Answer

(1) 1.5 74%

Accept answers from 73 to 75 99Answer

(2) [12]

0

10

20

30

40

50

60

70

80

90

100

110

0 10 20 30 40 50 60 70 80 90 100Afr

ikaa

ns F

irst

Add

ition

al L

angu

age

Mar

ks

English Home Language Marks

Scatter Plot for English HL and FAL Marks



QUESTION 2 2.1

∑

=

= 57,75 ( ) Answer only: Full Marks

9 Answer 9 square of

variance 9 Answer

(3) 2.2.1 22 students 9 Answer (1) 2.2.2

∑

=

= 60 Answer only: Full Marks

9 substitution 9 Answer

(2)

2.2.3

√

= 6,782 Answer only: Full Marks

9substitution 9 Answer

(2) 2.3

Each boy must be 69 kg

9

9Answer (2)

[10]



QUESTION 3

3.1.1 E = [

]

= [

]

E = (4; 4)

9substitution into correct formula 9 co-ordinates of

E.

(2) 3.1.2

mAB= mDC

=

=

&

B(9; 7) OR

x = 9 , y = 7 B(9; 7)

9equating two gradients. 9 simplification 9co-ordinates for B 9

9

9co-ordinates for B (3)

3.1.3

( ) ( )

F =

F = [2; 2]

y = x

9 co-ordinates of F 9 formula 9 correct value of m 9correct substitution into formula 9 Answer

(5) 3.2

mDE = mEG

3t – 9 = 7,5 3t = 1,5 t = 0,5

9 equating gradients 9 correct substitution 9 simplification 9 answer

(4)



3.3 AB √( ) ( ) √( ) ( ) √ √( ) ( ) √ ABCD is NOT a rhombus because AB AD OR mAC =

mDB =

mAC mDB =

=

ABCD is not a rhombus because: mAC mDB

9 substitution in formula 9 answer 9 answer 9 statement 9 reason 9 substitution in formula 9 answer 9 answer 9 statement 9 reason

(5) [19]



QUESTION 4 4.1 ( ) ( )

( ) ( )

= 16 + 9 r2 = 25 r = 5

( ) ( )

9 use of distance formula 9 r2 = 25 99 substituting values into

formula (4)

4.2 [angle subtended by diameter] 9answer (1) 4.3

( )( )

(mNQ). ( mQM) – 1

Q(-4; 0)

9substitution into formula for perpendicular gradients 9 simplification and solving trinomial 9 coordinates of Q

(3) 4.4 mMN =

=

=

9substitution in formula 9answer

(2) 4.5

mMN =

tan

mQM =

9 Gradient of MN 9value of 9 Gradient of QM 9 value of 9 value of

(5)



QUESTION 5 5.1 ( ) ( )

(√

√ )

√

9expression in terms of 30 and 450

9 use of compound identity 9substitution of

values 9 answer. (4)

5.2

( )( ) or (no solution)

9 cos 2A in terms of Sin A 9standard form 9 factors 9 for each value of sin A 9gen. sol in 1st quadrant 9gen. sol in 2nd quadrant 9gen. sol notation (7)

4.6

( )

( )

mMP =

=

mtan =

3

9 mMP 9mtan 9substitution in

formula 9simplification 9answer

(5) [20]



5.3

LHS=

LHS = RHS

9 double angle identity for sin2A 9 double angle identity for cos 2A 9 LHS = RHS

(3) 5.4

( )

9 9 9 9 9

9 (6)

[20]



QUESTION 6 6.1 120 9 answer. (1) 6.2

f 9 x-int 9 y-int 9 turning

point g 9 x-int 9y-int 9 turning

point (6)

6.3 6.3.1 ( 9 both

values correct

9 correct notation (2)

6.3.2 ( ) 9 both

values correct

9 correct notation (2)

6.4

OR [-4 ; 0,5]

9interval 9 values

(2) [13]



QUESTION 7 7.1

BC =

BC = 10,44

9 ratio for sin 9 simplification and

substitution 9 Answer

(3) 7.2

AB =

AB = 17,960

9 use of sine rule 9 simplification and

substitution 9 answer

(3) 7.3

= AD = 9,96m AD = 10m

9 ratio for sin 9 simplification &

substitution 9 answer

(3) [9]



QUESTION 8 8.1.1 [OS = OT, radii]

[ sum of � ’s of ∆] 9S/R 9 S/R (2)

8.1.2

[� at centre = 2 � at circumference]

=

=

9 statement and reason

9answer (2)

8.1.3 [SOQR is parallelogram]or

[ opp angles of parallelogram]

9 statement & reason 9answer (2)

8.2 [ opp angles of parallelogram]


9substitution of values

9 Answer (3) [9]

QUESTION 9 9.1 [line from centre to midpoint of chord]

[given QA┴TA] = MTAR is a cyclic quad [opp � ’s are supplementary]

9M2 = 90 9 reason 9M2 = A & reason 9conclusion and

reason

(4) 9.2 In

(i) M1 = M2 [ OR bisects PT] (ii) PM = MT [M is midpoint of PT] (iii) MR = MR [ common] ∆ PR = TR


9statement and reason


9 SAS 9conclusion (5)

9.3 = [ Tan chord theorem]

[Both equal to P]

9 = 9 Tan chord 9 conclusion (3)

[12]



QUESTION 10 10.1 Proportional 9 Answer (1) 10.2.1 A4 = D1 [tan chord theorem]

D1 = x D1 = E2 [angles in same segment] E2 = x A4 = C2 [alt int.]


9statement and

reason 9statement and

reason (3) 10.2.2 In ∆ACF &∆ADC

i) A3 = A3 [common] ii) C2 = D1 ∆ACF /// ∆ADC [equiangular]



9statement and reason (3)

10.2.3 ∆ACF /// ∆ADC

AF =

But AC = AO

AF =

9 sides in

proportion 9 choosing correct

proportion 9 simplification 9 answer

(4) [11]



QUESTION 11 11.1

RTP:

Constr: Draw height h from E to AD Draw height k from D to AE. Join BE and DC PROOF:

=

=

=

(same base and same height)

9 constr 9 ratio of area of ∆ADE: ∆BDE 9

9ratio of area of ∆ADE : ∆CED 9

9equating two areas 9conclusion

(6)

A

D E

B C

h k



11.2.1

[Prop. theorem; DE ║ BC] 9 answer

9 reason (2) 11.2.2 In ∆ADE & ∆ABC

i) [common] ii) [corresponding; DE ║ BC] ∆ADE/// ∆ ABC [equiangular]

9 Corresponding angles equal 9showing ∆ADE|||∆ABC 9 sides in proportion 9 answer

(4) 11.2.3

=

=

9Ratio of areas 9substitution of values in

denominator and numerator

9simplification and answer (3)

[15] TOTAL: 150

Corrections

Assessment and Examination Directorate Bundy Park, Private Bag 4571, King William’s Town, 5600 REPUBLIC OF SOUTH AFRICA, Website: www.ecdoe.gov.za E-mail:[email protected]

TO: CHIEF EDUCATION SPECIALISTS EDUCATION DEVELOPMENT OFFICERS DEPUTY CHIEF EDUCATION SPECIALISTS SENIOR EDUCATION SPECIALISTS PRINCIPALS OF SCHOOLS IN THE GET AND FET BAND TEACHER UNIONS/ORGANISATIONS SCHOOL GOVERNING BODIES

FROM: CES: INSTRUMENT DEVELOPMENT AND MODERATION

DIRECTORATE MS N. MBELEKI SUBJECT: ERRATA MATHEMATICS P2 GRADE 12 SEPTEMBER 2015 DATE: 28 SEPTEMBER 2015

Regretfully during the marking of Mathematics P2 it was discovered that certain changes had to be made in order to ensure that learners are not disadvantaged. All the above-mentioned are thus requested to ensure that the following errata is brought to the attention of those concerned. We regret any inconvenience caused. ERRATA: MATHEMATICS P2 2015 TRIAL QUES. CORRECTION MARK ALLOCATION 1.1 4 marks for scatter plot as the question paper

did not ask for regression line to be drawn 9first 4 points correct 9next 2 points correct 9next 2 points correct 9remaining 2 points correct (4)

1.5 Equation 𝒚 = 𝟏𝟖, 𝟎𝟒 +0,77x obtained in 1.2 may be used and the answer will be 72,68 Accept answers from 72 to 75

99answer (2)

Ref. No. 13/P Tel.: (043) 604 7708/082 391 1342 Enquiries: Ms N. Mbeleki Fax: 043 604 7789

6.4 Correct answer is: −𝟒 ≤ 𝒚 ≤ 𝟐 OR [- 4 ; 2]

9interval 9 values

(2)

8.1.3 Since it is stated in the next question that SOQP is a cyclic quad. Correct method is: �� + �� = 𝟏𝟖𝟎° (opp. angles of cyclic quad)

�� = 𝟏𝟖𝟎° − (𝟗𝟎° − 𝒙) �� = 𝟗𝟎° + 𝒙

9 statement and reason 9answer

(2)

10.2.3 𝐴𝐶𝐴𝐷

= 𝐴𝐹𝐴𝐶

= 𝐶𝐹𝐷𝐶

(∆ACF /// ∆ADC or similar Δs) 𝐴𝐶𝐴𝐷 =

𝐴𝐹𝐴𝐶

NOTE: Learners may continue: AF = 𝐴𝐶.𝐴𝐶

𝐴𝐷 But AC = AO

AF = AO2

AD

AF2 = AO4

𝐴𝐷2

Some may even say AF2 ≠ AO2

AD

Full 4 marks must be given if the learners accurately reached the step with AF (second line)!

9 statement 9 reason 99 choosing correct proportion (with

AF)

(4)

11.1 The last mark (for conclusion: ADDB

= AEEC

) must be removed since it is what they were asked to prove. That makes the question out of 6 marks as per the question paper!

(6) 11.2 Last line should be:

𝐀𝐃𝐀𝐁 =

𝐃𝐄𝐁𝐂 =

𝟑𝟓

Your co-operation in this matter is greatly appreciated.

MS N. MBELEKI CES: INSTRUMENT DEVELOPMENT AND MODERATION ASSESSMENT AND EXAMINATIONS DIRECTORATE

Documents

GRADE 12 SEPTEMBER 2015 MATHEMATICS P2 · 2016-09-03 · NATIONAL SENIOR CERTIFICATE GRADE 12 SEPTEMBER 2015 MATHEMATICS P2 MARKS: 150 TIME: 3 hours This question paper consists of