national senior certificate grade 12 · a particle moves along a straight line. the distance , s,...

Copyright reserved Please turn over

MARKS: 150 TIME: 3 hours

This question paper consists of 9 pages and 1 information sheet.

MATHEMATICS P1

NOVEMBER 2012

NATIONAL SENIOR CERTIFICATE

GRADE 12

Mathematics/P1 2 DBE/November 2012 NSC


INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

This question paper consists of 11 questions. Answer ALL the questions. Clearly show ALL calculations, diagrams, graphs, et cetera that you have used in determining your answers. Answers only will not necessarily be awarded full marks. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. If necessary, round off answers to TWO decimal places, unless stated otherwise. Diagrams are NOT necessarily drawn to scale. An information sheet with formulae is included at the end of the question paper. Number the answers correctly according to the numbering system used in this question paper. Write neatly and legibly.



QUESTION 1 1.1 Solve for x in each of the following: 1.1.1 ( )( ) 0412 =+− xx (2) 1.1.2 53 2 =− xx (Leave your answer correct to TWO decimal places.) (4) 1.1.3 0872 <−+ xx (4) 1.2 Given: 44 =− xy and 8=xy 1.2.1 Solve for x and y simultaneously. (6) 1.2.2 The graph of 44 =− xy is reflected across the line having equation

xy = . What is the equation of the reflected line?

(2)

1.3 The solutions of a quadratic equation are given by 7

522 +±−=

px

For which value(s) of p will this equation have:

1.3.1 Two equal solutions (2) 1.3.2 No real solutions (1)

[21] QUESTION 2 2.1 3x + 1 ; 2x ; 3x – 7 are the first three terms of an arithmetic sequence. Calculate the

value of x.

(2) 2.2 The first and second terms of an arithmetic sequence are 10 and 6 respectively. 2.2.1 Calculate the 11th term of the sequence. (2) 2.2.2 The sum of the first n terms of this sequence is –560. Calculate n. (6)

[10]



QUESTION 3 3.1 Given the geometric sequence: 27 ; 9 ; 3 … 3.1.1 Determine a formula for nT , the nth term of the sequence. (2) 3.1.2 Why does the sum to infinity for this sequence exist? (1) 3.1.3 Determine ∞S . (2) 3.2 Twenty water tanks are decreasing in size in such a way that the volume of each tank

is 21 the volume of the previous tank. The first tank is empty, but the other 19 tanks

are full of water.

Would it be possible for the first water tank to hold all the water from the other

19 tanks? Motivate your answer.

(4) 3.3 The nth term of a sequence is given by 18)5(2 2 +−−= nTn . 3.3.1 Write down the first THREE terms of the sequence. (3) 3.3.2 Which term of the sequence will have the greatest value? (1) 3.3.3 What is the second difference of this quadratic sequence? (2) 3.3.4 Determine ALL values of n for which the terms of the sequence will be

less than –110.

(6) [21]



QUESTION 4

4.1 Consider the function 62.3)( −= xxf . 4.1.1 Calculate the coordinates of the y-intercept of the graph of f. (1) 4.1.2 Calculate the coordinates of the x-intercept of the graph of f. (2) 4.1.3 Sketch the graph of f in your ANSWER BOOK.

Clearly show ALL asymptotes and intercepts with the axes.

(3)

4.1.4 Write down the range of f. (1)

x

y

S(−2 ; 0) T(6 ; 0)

Rf

g

0

4.2.1 Determine the value of d. (2) 4.2.2 Determine the equation of f in the form cbxaxxf ++= 2)( . (4) 4.2.3 If 124)( 2 ++−= xxxf , calculate the coordinates of the turning point

of f.

(2) 4.2.4 For which values of k will kxf =)( have two distinct roots? (2) 4.2.5 Determine the maximum value of 12)(3)( −= xfxh . (3)

[20]

4.2 S(–2 ; 0) and T(6 ; 0) are the x-intercepts of the graph of cbxaxxf ++= 2)( and R is the y-intercept. The straight line through R and T represents the graph of

dxxg +−= 2)( .



QUESTION 5 The graph of xxf 27)( −= for x ≥ 0 is sketched below. The point P(3 ; –9) lies on the graph of f.

x

y

P(3 ; − 9)

f

0

5.1 Use your graph to determine the values of x for which 9)( −≥xf . (2) 5.2 Write down the equation of 1−f in the form ...=y Include ALL restrictions. (3) 5.3 Sketch 1−f , the inverse of f, in your ANSWER BOOK.

Indicate the intercept(s) with the axes and the coordinates of ONE other point.

(3) 5.4 Describe the transformation from f to g if g(x) = x27 , where x ≥ 0. (1)

[9] QUESTION 6 The graph of a hyperbola with equation )(xfy = has the following properties:

• Domain: ∈x R, 5≠x • Range: ∈y R, 1≠y • Passes through the point (2 ; 0)

Determine ( )xf .

[4]

P(3 ; –9)



QUESTION 7 7.1 A business buys a machine that costs R120 000. The value of the machine depreciates

at 9% per annum according to the diminishing-balance method.

7.1.1 Determine the scrap value of the machine at the end of 5 years. (3) 7.1.2 After five years the machine needs to be replaced. During this time,

inflation remained constant at 7% per annum. Determine the cost of the new machine at the end of 5 years.

(3)

7.1.3 The business estimates that it will need R90 000 by the end of five years.

A sinking fund for R90 000, into which equal monthly instalments must be paid, is set up. Interest on this fund is 8,5% per annum, compounded monthly. The first payment will be made immediately and the last payment will be made at the end of the 5-year period. Calculate the value of the monthly payment into the sinking fund.

(5)

7.2 Lorraine receives an amount of R900 000 upon her retirement. She invests this

amount immediately at an interest rate of 10,5% per annum, compounded monthly. She needs an amount of R18 000 per month to maintain her current lifestyle. She plans to withdraw the first amount at the end of the first month. For how many months will she be able to live from her investment?

(6) [17]

QUESTION 8 8.1 Determine )(xf ′ from first principles if 52)( 2 −= xxf . (5)

8.2 Evaluate dxdy if

52 34 xxxy −+= − .

(3)

8.3

Given: 1

2)(2

−−+

=x

xxxg

8.3.1 Calculate )(xg ′ for 1≠x . (2) 8.3.2 Explain why it is not possible to determine )1(g ′ . (1)

[11]



QUESTION 9 9.1 The graph of the function 1616)( 23 ++−−= xxxxf is sketched below.

x

y

0

f

9.1.1 Calculate the x-coordinates of the turning points of f. (4) 9.1.2 Calculate the x-coordinate of the point at which )(xf ′ is a maximum. (3) 9.2 Consider the graph of 592)( 2 +−−= xxxg . 9.2.1 Determine the equation of the tangent to the graph of g at x = –1. (4) 9.2.2 For which values of q will the line y = –5x + q not intersect the parabola? (3) 9.3 Given: xxxh 54)( 3 += Explain if it is possible to draw a tangent to the graph of h that has a negative

gradient. Show ALL your calculations.

(3) [17]

QUESTION 10 A particle moves along a straight line. The distance, s, (in metres) of the particle from a fixed point on the line at time t seconds ( 0≥t ) is given by 45182)( 2 +−= ttts .

10.1 Calculate the particle's initial velocity. (Velocity is the rate of change of distance.) (3) 10.2 Determine the rate at which the velocity of the particle is changing at t seconds. (1) 10.3 After how many seconds will the particle be closest to the fixed point? (2)

[6]


Copyright reserved

QUESTION 11 A calculator company manufactures two kinds of calculators: scientific and basic. The company is able to sell all the calculators that it produces. A system of constraints has been developed for the production of the calculators. The feasible region is shaded below. Let x and y respectively be the number of scientific and basic calculators produced each day.

11.1 Is it possible for the company to manufacture 15 scientific calculators and 5 basic calculators in one day according to their system of constraints? Motivate your answer.

(1)

11.2 Write down all the algebraic inequalities which describe the constraints related to the

manufacturing of the calculators.

(6) 11.3 The profit Q (in hundreds of rands) is given by Q = yx 3+ . The dotted line on the

graph is a search line associated with the profit function.

11.3.1 Identify the point in the region where the profit is a maximum. Use only

A, B, C or D.

(1) 11.3.2 Write down the coordinates of a point on the dotted line (if the point

exists) at which the profit is greater than the profit at P.

(2) 11.3.3 Given that the profit, when given by Q byax += ( 0>a ; 0>b ), is a

maximum at B, determine the maximum value of ba .

(4) [14]

TOTAL: 150

5 10 15 20 25 30 35 40 45 50 55

5

10

15

20

25

30

35

40

x

y

A

B

CD

PFEASIBLEREGION

0

Scientific Calculators

Bas

ic C

alcu

lato

rs

Mathematics/P1 DBE/November 2012 NSC

Copyright reserved

INFORMATION SHEET: MATHEMATICS

aacbbx

242 −±−

=

)1( niPA += )1( niPA −= niPA )1( −= niPA )1( +=

∑=

=n

in

11

2)1(

1

+=∑

=

nnin

i

dnaTn )1( −+= ( )dnann )1(2

2S −+=

1−= nn arT ( )

11

−−

=rraS

n

n ; 1≠r

raS−

=∞ 1; 11 <<− r

( )[ ]iixF

n 11 −+= [1 (1 ) ]nx iP

i

−− +=

hxfhxfxf

h

)()(lim)('0

−+=

→

22 )()( 1212 yyxxd −+−= M

++2

;2

2121 yyxx

cmxy += )( 11 xxmyy −=− 12

12xxyy

m−

−= θtan=m

( ) ( ) 222 rbyax =−+−

In ∆ABC: C

cB

bA

asinsinsin

== Abccba cos.2222 −+=

CabABCarea sin.21

=∆

( ) βαβαβα sin.coscos.sinsin +=+ ( ) βαβαβα sin.coscos.sinsin −=−

( ) βαβαβα sin.sincos.coscos −=+ ( ) βαβαβα sin.sincos.coscos +=−

−

−

−

=

1cos2sin21

sincos2cos

2

2

22

α

α

αα

α ααα cos.sin22sin =

)sincos;sincos();( θθθθ xyyxyx +−→

nfx

x ∑= ( )

n

xxn

ii

2

2

1∑=

−=σ

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

bxay +=ˆ ( )∑

∑−

−−= 2)(

)(xx

yyxxb


MARKS: 150

This memorandum consists of 30 pages.

MATHEMATICS P1

NOVEMBER 2012

MEMORANDUM


GRADE 12

Mathematics/P1 2 DBE/November 2012 NSC – Memorandum

Copyright reserved

NOTE: • If a candidate answered a question TWICE, mark the FIRST attempt ONLY. • If a candidate crossed out an attempt of a question and did not redo the question, mark the

crossed out question. • Consistent accuracy applies in ALL aspects of the memorandum. QUESTION 1 1.1.1 ( )( ) 0412 =+− xx

4or21

−=x

answer answer

(2)

1.1.2

( ) ( ) ( )( )( )

14,1or47,16

611

3253411

24

05353

2

2

2

2

−=

±=

−−−±−−=

−±−=

=−−

=−

aacbbx

xxxx

OR

14,1or47,13661

61

3661

61

361

35

61

35

31

53

2

2

2

−=

±=

±=

−

+=

−

=−

=−

x

x

x

xx

xx

OR

standard form subs into correct formula answer

(4)

division by 3

3661

61

±=

−x

answer

(4)

Note: if a candidate has not rounded off correctly, penalise 1 mark

Note: if a candidate uses incorrect formula award max 1 mark (for standard form)



( ) ( ) ( )( )( )

14,1or47,12

1214

24

035

3

05353

961

31

352

31

31

2

2

2

2

−=

±=

−−−±−−=

−±−=

=−−

=−−

=−

aacbbx

xx

xxxx

standard form subs into correct formula answer

(4)

1.1.3

( )( ) 0180872

<−+<−+

xxxx

OR

x – 8 1 x + 8 – 0 + + + x – 1 – – – 0 + (x + 8)( x – 1) + 0 – 0 +

Therefore the solution is:

18 <<− x OR )1;8(−∈x OR OR

( )( ) 0180872

<−+<−+

xxxx

∴ 08 <+x and 01>−x or 08 >+x and 01<−x 8−<x and 1>x 8−>x and 1<x No solution Therefore the solution is:

18 <<− x OR )1;8(−∈x OR

factors – 8, 1 answer

(4)

factors – 8, 1 answer

(4)

–8 1 + 0 + 0 −

1 –8 OR

-8 1 x

-8 1 x



NOTE: In this alternative, award max 3/4 marks since there is no conclusion

( )( ) 0180872

<−+<−+

xxxx

factors – 8, 1 graph with bolded line

1.2.1

4or82or1

0)2)(1(022)1(8)44(

448and44

2

=−==−=

=−+=−−

=−=−

−===−

xxyy

yyyy

yyyy

yxxyxy

(x ; y) = (– 8 ; – 1) or (4 ; 2) OR

2)1(8)44(

448and44

=−=−

−===−

yyyy

yxxyxy

4or8 2or1 inspectionBy

=−==−=

xxyy

(x ; y) = (– 8 ; – 1) or (4 ; 2) OR

( ) ( ) ( )( )( )

4or82or112

21411

022)1(8)44(

448and44

2

2

=−==−=

−−−±−−=

=−−

=−=−

−===−

xxyy

y

yyyyyy

yxxyxy

(x ; y) = (– 8 ; – 1) or (4 ; 2)

44 −= yx substitution factors y-values x-values

(6)

44 −= yx substitution y-values x-values

(6) 44 −= yx substitution subs into correct formula y-values x-values

(6)

1 –8

Note: If candidate makes a mistake which leads to both equations being LINEAR award maximum 2/6 marks



OR

2or14or8

0)4)(8(0324

084

814

14

8and44

2

2

=−==−=

=−+=−+

=−+

=

+

+=

==−

yyxx

xxxx

xx

xx

xy

xyxy

(x ; y) = (– 8 ; – 1) or (4 ; 2) OR

( )( )( )

2or14or8

12321444

0324

084

814

14

8and44

2

2

2

=−==−=

−−±−=

=−+

=−+

=

+

+=

==−

yyxx

x

xx

xx

xx

xy

xyxy

(x ; y) = (– 8 ; – 1) or (4 ; 2) OR

( )( )

4or82or1

012020844

484

844and8

2

2

=−==−=

=+−=−−

=−−

=−

=

=−=

xxyy

yyyyyy

yy

yx

xyxy

(x ; y) = (– 8 ; – 1) or (4 ; 2)

14+=

xy

substitution factors x-values y-values

(6)

14+=

xy

substitution subs into correct formula x-values y-values

(6)

y

x 8=

substitution factors y-values x-values

(6)



OR

( ) ( ) ( )( )( )

4or82or112

21411

020844

484

844and8

2

2

2

=−==−=

−−−±−−=

=−−

=−−

=−

=

=−=

xxyy

y

yyyy

yy

yx

xyxy

(x ; y) = (– 8 ; – 1) or (4 ; 2) OR

( )( )

2or14or8

4803240

484

844and8

2

=−==−=

−+=−+=

=−

=

=−=

yyxx

xxxx

xx

xy

xyxy

(x ; y) = (– 8 ; – 1) or (4 ; 2) OR

( )( )( )

2or14or8

12321444

3240

484

844and8

2

2

=−==−=

−−±−=

−+=

=−

=

=−=

yyxx

x

xx

xx

xy

xyxy

(x ; y) = (– 8 ; – 1) or (4 ; 2)

y

x 8=

substitution subs into correct formula y-values x-values

(6)

x

y 8=

substitution factors x-values y-values

(6)

x

y 8=

substitution subs into correct formula x-values y-values

(6)



1.2.2 44 =− yx OR

44 −= xy OR

44+

=yx

OR

044 =−− yx OR

141

+= yx

interchanges x and y (2)

1.3.1

2552

052052

−=

−==+

=+

p

ppp

052 =+p or

052 =+p or 7

02 ±−

answer (2)

1.3.2

25

052

−<

<+

p

p

answer

(1) [21]



QUESTION 2 2.1

( ) ( )

362

71273132

2731322312

=−=−−=−−

−−=−−−−=+−

−=−

xx

xxxxxx

xxxxTTTT

OR

( ) ( )

326

6642

73132

231

2

==

−=

−++=

+=

xx

xx

xxx

TTT

OR

( )( ) ( ) ( )( )

362

22813221373

2 1213

==

−−=−+−=+−−

−=−

xx

xxxxx

TTTT

2312 TTTT −=− or

( ) ( ) xxxx 273132 −−=+− answer

(2)

2

312

TTT +=

or ( ) ( )2

73132 −++=

xxx

answer

(2)

( )1213 2 TTTT −=− or ( ) ( ) ( )( )13221373 +−=+−− xxxx answer

(2) 2.2.1 ( )

( )( )30

4111101

11

−=−−+=

−+=T

dnaTn

OR 10; 6; 2; –2; –6; –10; –14; –18; –22; –26; –30 … ∴ 3011 −=T

d = –4 answer

(2)

expands sequence answer

(2)



2.2.2 ( )[ ]

( ) ( )( )[ ]

( )( )14or20

014200280601120244

2441120

411022

560

122

2

2

2

−==+−=−−

=−−

+−=−

−−+=−

−+=

nnn

nnnn

nn

nn

dnanSn

only20=∴n OR

( )[ ]

( ) ( )( )[ ]

( ) ( ) ( )( )( )

14or2012

2801466

0280601120244

2441120

411022

560

122

2

2

2

2

−=

−−−±−−=

=−−

=−−

+−=−

−−+=−

−+=

n

n

nnnn

nn

nn

dnanSn

only20=∴n OR

( )[ ]

( ) ( )( )[ ]

( )( )14or20

01420

028060560122

24

24

220560

411022

560

122

2

2

2

−==+−

=−−

=−−

+−=−

−−+=−

−+=

nnn

nnnn

nnn

nn

dnanSn

only20=∴n

correct formula substitution of a and d subs 560−=nS 01120244 2 =++− nn or

01120244 2 =−− nn or 028062 =−− nn factors selects 20=n only

(6)

correct formula substitution of a and d subs 560−=nS 01120244 2 =−− nn or

01120244 2 =++− nn or 028062 =−− nn subs into correct formula selects 20=n only

(6)

correct formula substitution of a and d subs 560−=nS 0560122 2 =−− nn or

0560122 2 =++− nn or 028062 =−− nn factors selects 20=n only

(6)

Note: if candidate substitutes into incorrect formula, award 0/6

Note: if candidate writes answer only, award 1/6 marks



OR

11011 −=S n 12 13 14 15 16 17 18 19 20

Tn –34 –38 –42 –46 –50 –54 –58 –62 –66

Sn –144 –182 –224 –270 –320 –374 –432 –494 –560

20=∴n

11011 −=S sequence expanded series calculated answer

(6) [10]

QUESTION 3 3.1.1

1

1

3127

−

−

=

=n

nn arT

a = 27 and r =

31

substitute into correct formula

(2) 3.1.2 11 <<− r or 1<r

OR

The common ratio (r) is 31 which is between –1 and 1.

OR

1311 <<−

answer (1)

answer

(1)

answer (1)

3.1.3

41or5,40or281

311

271

=

−=

−=∞ r

aS

substitution answer

(2)

Note: The final answer can also be

written as n−43 or 4

31 −

n

Note: If candidate concludes series is not convergent, award 0 marks.

Note: If r >1 or r < – 1 is substituted then 0/2 marks.



3.2 Let V be the volume of the first tank.

.......8

;4

;2

VVV

VV

V

V

S

<=

=

−

−

=

9999980927,0524288524287

211

211

2

19

19

Yes, the water will fill the first tank without spilling over. OR Let V be the volume of the first tank.

.......8

;4

;2

VVV

VV

V

V

S

=⋅<

−=

−

−

=

1

211

211

211

2

19

19

19

Yes, the water will fill the first tank without spilling over. OR Let V be the volume of the first tank.

.......8

;4

;2

VVV

V

V

S

=

−=∞

211

2

Since the first tank will hold the water from infinitely many tanks without spilling over, certainly: Yes, the first tank will hold the water from the other 19 tanks without spilling over.

2V

substitute into correct formula answer conclusion

(4)

2V

substitute into correct formula observes that

1211

19

<

−

conclusion

(4)

2V

substitute into correct formula correct argument

(4)

Note: If candidate lets the volume of the first tank be a specific value (instead of a variable) and his/her argument follows correctly, award 4/4 marks Note: If candidate answers ‘Yes’ only with no justification: 1/4 marks



OR If the tanks are emptied one by one, starting from the second, each tank will fill only half the remaining space, so the first tank can hold all the water from the other 19 tanks.

Yes (explicit or understood from the argument.) argument

(4) 3.3.1 18)5(2 2 +−−= nTn

Term 1 = – 14 Term 2 = 0 Term 3 = 10

– 14 0 10

(3)

3.3.2 Term 5 OR n = 5 OR 5T answer (1)

3.3.3 Second difference = 2a

Second difference = 2(– 2) Second difference = – 4 OR

-14 0 10 14 10

– 4 Second difference = – 4

subs – 2 into 2a answer

(2)

first differences second difference

(2)

3.3.4

0)3)(13(039100782020128502020128)5(2

11018)5(2

2

2

2

2

2

>+−>−−

<++−

<+−+−

<+−−

−<+−−

nnnnnn

nnnn

n < – 3 or n > 13

Ν∈≥ nn ;14 OR Ν∈> nn ;13 OR

110−<nT standard form factors critical values inequalities 13>n (accept: 14≥n )

(6)

-3 13 + 0 + 0 −

13 –3

Note: Answer only award 2/6 marks



( )[ ] ( )[ ]0)3)(13(08585064)5(0128)5(2

11018)5(2

2

2

2

>+−>+−−−>−−

<+−−

−<+−−

nnnn

nnn

n < –3 or n > 13

Ν∈≥ nn ;14 OR Ν∈> nn ;13 OR

( )

13or385or85

645

128)5(211018)5(2

2

2

2

>−<>−−<−

>−

−<−−

−<+−−

nnnn

n

nn

Ν∈≥ nn ;14 OR Ν∈> nn ;13

OR

( )

0)3)(13(03910078202

11032202

32202

1852

2

2

2

2

2

>+−>−−

<−+−

−<−+−

−+−=

+−−=

nnnnnnnn

nnTnT

n

n

n < – 3 or n > 13

Ν∈≥ nn ;14 OR Ν∈> nn ;13 OR –14 ; 0 ; 10 ; 16 ; 18 ; 16 ; 10 ; 0 ; –14 ; –32 ; –54 ; –80 ; –110

Ν∈≥ nn ;14

110−<nT 064)5( 2 >−−n factors critical values inequalities 13>n (accept: 14≥n )

(6)

110−<nT 128)5(2 2 >−n 8 and – 8 85 >−n 85 −<−n 13>n (accept: 14≥n )

(6) 110−<nT standard form factors critical values inequalities 13>n (accept: 14≥n )

(6)

expansion conclusion of

14≥n (accept 13>n )

(6) [21]

-3 13 + 0 + 0 −

13 –3

-3 13 + 0 + 0 −

13 –3



QUESTION 4 4.1.1

)3;0(363

62.3 0

−−=−=

−=

yyy

answer

(1)

4.1.2

)0;1(12262.3

62.30

1

==

=

−=

x

x

x

x

y = 0 x-value (2)

4.1.3

-2 -1 1 2

-7

-6

-5

-4

-3

-2

-1

1

2

3

x

y

f

0

(0 ; -3)

(1 ; 0)

y = -6

intercepts asymptote shape

(3)

4.1.4 );6(OR6 ∞−−>y answer (1)

Note: If a candidate interchanges question 4.1.1 and 4.1.2: 0/3 marks Note: If a candidate says that xx 62.3 = (i.e. wrong mathematics ) s/he will arrive at correct answer BUT award max 1/2



4.2

4.2.1

12)6)(2(0

2

=+−=

+−=

dd

dxy

OR

( )( )

12262011

+−=−−=−−=−

xyxy

xxmyy

12=∴d OR

Since 2−=m and 6dm −

=

126

2

=

−=−

d

d

substitution answer (2)



4.2.2

124)124(

1)20)(60(12)2)(6(

2

2

++−=

−−−=

−=+−=+−=

xxxxy

aa

xxay

OR

122 ++= bxaxy ( ) ( ) 12220 2 +−+−= ba i.e. 12240 +−= ba ( ) ( ) 12660 2 ++= ba i.e. 126360 ++= ba

4

96240=

−=b

b

( )

1241

124240

2 ++−=

−=+−=

xxya

a

)2)(6( +−= xxay subs R(0 ; 12) a-value 1242 ++−= xxy

(4) 122 ++= bxaxy subs ( )0;2S − and ( )0;6T

b-value 1242 ++−= xxy

(4)

x

y

S(−2 ; 0) T(6 ; 0)

R f

g

0

Note: No marks for answer only.



OR ( ) qxay +−= 22 ( ) qa +−−= 2220 or ( ) qa +−= 2260 i.e. qa += 160 ( ) qa +−= 22012 i.e. qa += 412

16

11212

=−=−=

qa

a

( )( )

1241644

162

2

2

2

++−=

++−−=

+−−=

xxxx

xy

OR

( )( )( )( )

12412412426

2

2

2

++−=

−−−=

−−=

+−=

xxxxxxaxxay

( ) qxay +−= 22 subs R(0 ; 12) and S(– 2 ; 0) (or T(6 ; 0) ) a-value 1242 ++−= xxy

(4)

)2)(6( +−= xxay expand a-value 1242 ++−= xxy

(4) 4.2.3

1612)2(4)2(

2042

0

2

=++−=

==+−

=

yx

xdxdy

TP of f is (2 ; 16) OR

( )

1612)2(4)2(

212

42

2

=++−=

=−

−=

−=

y

abx

TP of f is (2 ; 16) OR

16)2()( 2 +−−= xxf TP of f is (2 ; 16)

x-value y-value

(2)

x-value y-value

(2)

x-value y-value

(2)



OR

1612)2(4)2(

22

62

2

=++−=

=

+−=

y

x

TP of f is (2 ; 16)

x-value y-value

(2)

4.2.4 16<k OR ( )16;∞−

answer (2)

4.2.5 Maximum value of 12)(3)( −= xfxh occurs at max value of f(x) Maximum value = 12163 − = 81 OR Maximum value of 12)(3)( −= xfxh occurs at max value of f(x)

81or333)2(

4

1216

12)2(

=

=

=−

−fh

OR ( )

( )xxxxxf

−=+−=−

4412 2

which has a maximum value of ( ) 42 =f ∴Maximum value of ( )xh is 43 or 81

subs 16 for f(x) 43 or 81

(3)

subs 16 for f(x) 43 or 81

(3)

subs f(2) = 4 43 or 81

(3) [20]



QUESTION 5 5.1 30 ≤≤ x OR [ ]3;0 x≤0

3≤x (2)

5.2 f -1 : yx 27−= yx 272 =

27

2xy = x ≤ 0 OR ( ;0]−∞

interchange x- and y- values

27

2xy =

x ≤ 0 or ( )0;∞−

(3) 5.3

x

y

P(−9 ; 3)

0

f − 1

shape end at origin any other point on the graph

(3) 5.4 Reflection about the x-axis

OR );();( yxyx −→ ; 0≥x

answer (1)

answer

(1) [9]

Note: if the candidate gives 30 << x , award 1/2 marks



QUESTION 6

15

3)(

33

1

15)2(

0

15

)(

+−

=

=−

=−

+−

=

+−

=

xxf

a

a

ax

axf

OR ( )( )( )( )

( )( )

15

33153

105215

+−

=

=−−==−−=−−

xy

yxk

kkyx

x – 5 + 1 substitution of (2 ; 0) a = 3

(4)

( )5−x ( )1−y substitution of (2 ; 0) 3=k

(4)

[4] QUESTION 7 7.1.1

86,88374R)09,01(000120

)i1(PA5

n

=−=

−=

i, n and P identified subs into correct formula answer

(3) 7.1.2

21,306168R)07,01(000120

)i1(PA5

n

=+=

+=

i, n and P identified subs into correct formula answer

(3) 7.1.3 Sinking fund needed: =vF R 90 000

68,1841R12085,0

112085,01

00090

]1)1[(

61

=

−

+

=

−+=

x

x

iixF

n

v

=vF R 90 000

i = 240017

12085,0

=

in annuity formula n = 61 subs into correct formula answer

(5)

NOTE:

( )52

−−

=xxxf as an alternative

simplified form.

NOTE: Incorrect formula (in 7.1.1 or 7.1.2) award max 1/3 marks

NOTE: Incorrect formula award max 2/5 marks



OR Consider the scenario as money deposited at the beginning of every month, but in the last month an additional payment was made at the end of the month:

( ) ( )[ ]

( ) ( )[ ]

68 R1184,

12085,01

12085,01

12085,01

12085,0000 90

1

12085,0

1 12085,01

12085,01

00090

1 1 11

1 11

60

60

=

+

−

+

+

=

+

−

+

+

=

+

−++=

+−++

=

x

x

iiix

xi

iixF

n

n

v

OR Present value of sinking fund needed:

03,51358R12085,0100090

61

=

+=

v

v

P

P

Using the present value formula:

68,1841R12085,0

12085,011

03,51358

])1(1[

61

=

+−

=

+−=

−

−

x

x

iixP

n

v

i = 240017

12085,0

=

in annuity formula n = 60 in annuity formula =vF R 90 000 subs into correct formula answer

(5)

i = 240017

12085,0

=

in annuity formula n = 61 03,51358R=vP subs into correct formula answer

(5)



7.2

months04,66

169log

12105,01

0001812105,0000900

1

12105,0

12105,01100018

000900

])1(1[

12105,01

=

=−

+=

−

+−

=

+−=

+

−

−

−

n

n

iixP

n

n

n

v

She will be able to maintain her current lifestyle for a little more than 66 months using her pension money. OR

months04,66169log

12105,01log

12105,01

0001812105,0000900

1

12105,0

12105,01100018

000900

])1(1[

=

=

+−

+=

−

+−

=

+−=

−

−

−

n

n

iixP

n

n

n

v

She will be able to maintain her current lifestyle for a little more than 66 months using her pension money.

x = 18 000

i = 0,10512

in

annuity formula subs into correct formula simplification use of logs answer in months

(6)

x = 18 000

i = 0,10512

in


(6)

Note: If vF formula used, possibly award one each for x, i, use of logs: max 3/6 marks If any other incorrect formula is used, award 0/6 marks

Note: If candidate rounds off early in Question 7.2 (and obtain 58 months), penalise 1 mark



OR

( ) ( )[ ]

months04,66

169log

12105,01

916

12105,01000900

12105,00001800018

00090012105,000018

12105,0100018

12105,01000900

12105,0

12105,010001800018

1800012105,0100018

12105,01000900

12105,0

12105,0

112105,0100018

12105,01000900

111

12105,01

=

=−

+=

+=

×−÷

×−

+=

+×−

+=

−

+=

+×

−

+

=

+

−+=+

=

+

n

n

iixiP

FA

n

n

n

nn

nn

n

n

nn

v

She will be able to maintain her current lifestyle for a little more than 66 months using her pension money.

x = 18 000

i = 12105,0 in


(6) [17]

QUESTION 8 8.1

x

hxh

hxhh

hxhxf

hxhxfhxfhxhx

hxhxfxxf

h

h

h

4

)24(lim

)24(lim

24lim)(

24)()(5242

5)(2)(52)(

0

0

2

0

2

22

2

2

=

+=

+=

+=′

+=−+

−++=

−+=+

−=

→

→

→

OR

substitution of of x + h simplification to 224 hxh + formula )24(lim

0hx

h+

→

answer

(5)

Note: If candidate makes a notation error Penalise 1 mark

Note: If candidate uses differentiation rules Award 0/5 marks



( ) ( ) ( )

( )[ ] ( )

( )[ ]

( )

( )x

hxh

hxhh

hxhh

xhxhxh

xhxhxh

xhxh

xfhxfxf

h

h

h

h

h

h

h

4

24lim

24lim

24lim

52]5242[lim

52522lim

5252lim

lim

0

0

2

0

222

0

222

0

22

0

0

=

+=

+=

+=

+−−++=

+−−++=

−−−+=

−+=′

→

→

→

→

→

→

→

formula substitution of x + h simplification

to h

hxh 224 +

)24(lim

0hx

h+

→

answer (5)

8.2

5164

5164

25

25

−+−

=

−+−= −

xx

xxdxdy

54 −− x 26x

51

−

(3) 8.3.1

( )( )11)(

121

)1)(2(1

2)(2

≠=′≠+=

−−+

=

−−+

=

xxgxx

xxx

xxxxg

simplification answer

(2) 8.3.2 The function is undefined at x = 1.

OR Division by zero is undefined. OR The denominator cannot be zero. OR

In the definition of the derivative , ( ) ( ) ( )h

ghggh

11lim10

−+=′

→, but ( )1g

does not exist.

answer

(1)

[11]

Note: candidates do NOT need to give their answer with positive exponents

Note: notation error penalise 1 mark



QUESTION 9 9.1.1 1616)( 23 ++−−= xxxxf

0)2)(83(0162316230

1623)(

2

2

2

=−+=−+

+−−=

+−−=′

xxxx

xxxxxf

38

−=x or 2=x

OR

1616)( 23 ++−−= xxxxf

( )( )( )32

163422

16230162301623)(

2

2

2

2

−−±−=

−+=

+−−=

+−−=′

x

xxxxxxxf

38

−=x or 2=x

1623)( 2 +−−=′ xxxf ( ) 0=′ xf or 16230 2 +−−= xx factors x values

(4)

1623)( 2 +−−=′ xxxf ( ) 0=′ xf or 16230 2 +−−= xx subs into formula x values

(4)

9.1.2

31

0260)(

−=

=−−=′′

x

xxf

OR

312

238

−=

+−=

x

x

OR

( )( )

31

322

1623)( 2

−=

−−−

=

+−−=′

x

xxxf

OR

26)( −−=′′ xxf 026 =−− x answer

(3)

2

238 +−

=x

answer

(3)

( )( )32

2−−−

=x

answer

(3)

Note: if neither ( ) 0=′ xf nor

16230 2 +−−= xx explicitly stated, award maximum 3/4 marks



( )( )

31

131

1616)( 23

−=

−−−

=

++−−=

x

xxxxf

( )( )13

1−−−

=x

answer (3)

9.2.1

( )

757

151255

9)1(494)(

125)1(9)1(2)1(

592)(

tan

2

2

+−==

+−−=+−=

−=−−−=

−−=′=

+−−−−=−

+−−=

xyc

ccxy

mxxg

gxxxg

OR

75)1(512

59)1(4

94)(12

5)1(9)1(2)1(592)(

tan

2

2

+−=+−=−

−=−−−=

−−=′=

+−−−−=−

+−−=

xyxy

mxxg

gxxxg

( )1−g = 12 94)( −−=′ xxg 5tan −=m answer

(4)

( )1−g = 12 94)( −−=′ xxg 5tan −=m answer

(4) 9.2.2

x

y

7

q > 7 OR

qxy +−= 5 and 592 2 +−−= xxy

( ) 712

59252

2

++−=

+−−=+−

xq

xxqx

∴ 7>q

sketch 7 correct inequality

(3)

method 7 correct inequality

(3)



OR

qxy +−= 5 and 592 2 +−−= xxy

48564

)2(2)5)(2(4164

05425925

2

2

qx

qx

qxxxxqx

−±−=

−−±−=

=−++

+−−=+−

70856

><−

qq

OR Since ( ) 121 =−g and at 1−=x , tangent equation is 75 +−= xy ,

qxy +−= 5 not intersecting g ⇒ ( )

qq

q

<<−

+−−<

7512

1512


(3)


(3) 9.3 512)( 2 +=′ xxh

For all values of x: 02 ≥x

05125512

012

2

2

2

>+

≥+

≥

xxx

For all values of x: 0)( >′ xh All tangents drawn to h will have a positive gradient. It will never be possible to draw a tangent with a negative gradient to the graph of h. OR

512)( 2 +=′ xxh Suppose ( ) 0<′ xh and try to solve for x :

125

0 5 12

2

2

−<

<+

x

x

but x2 is always positive ∴ no solution for x ∴ 0)( ≥′ xh for all x R∈ i.e. there are no tangents with negative slopes

512)( 2 +=′ xxh clearly argues that 0)( >′ xh conclusion

(3)

512)( 2 +=′ xxh clearly argues that ( ) 0<′ xh is

impossible conclusion

(3)



OR

512)( 2 +=′ xxh

x

y

5

h'

Since clearly 0)( >′ xh for all Rx∈ , it will never be possible to draw a tangent with a negative gradient to the graph of h.

512)( 2 +=′ xxh argues 0)( >′ xh by drawing a sketch conclusion

(3) [17]



QUESTION 10 10.1 45182)( 2 +−= ttts

184)( −=′ tts ( ) ( )

sms

/1818040

−=−=′

)(ts′ subs t = 0 into ( )ts′ formula answer

(3)

10.2 ( ) 4=′′ ts m/s2 answer (1)

10.3

seconds5,4orseconds291840184

=

==−

t

tt

OR

( )


29

292

2

=

+

−=

t

tts

OR

45182)( 2 +−= ttts

( )


2218

=

−−=

t

t

( ) 0' =ts answer

(2)

( )

29

292

2

+

−= tts

answer

(2)

( )2218−

−=t

answer

(2) [6]

Note: answer only award 0/3 marks



QUESTION 11

11.1 No, because (15 ; 5) does not lie within the feasible region.

OR No, because according to the constraints, the x-value (number of scientific calculators) must be at least 20.

answer (with motivation)

(1)

11.2

040502

20

≥≤+≤+

≥

yyx

yxx

OR

040

2521

20

≥+−≤

+−≤

≥

yxy

xy

x

OR

0

14040

15025

20

≥

≤+

≤+

≥

y

xy

xyx

OR

01250502516004040

20

≥≤+≤+

≥

yyxyx

x

20≥x 502 ≤+ yx 40≤+ yx 0≥y

(6)

11.3.1 A answer (1)

11.3.2 All points on the search line yield the same profit. Hence no such point exists. OR If such an (x ; y) exists, Q = x + 3y and

1531

+−= xy

so Qxy =+= 345 Q = 4 500 Hence, there is no such point.

No point exists

(2)

No point exists

(2)

5 10 15 20 25 30 35 40 45 50 55

5

10

15

20

25

30

35

40

x

y

A

B

CD

P

FEASIBLEREGION

0

Scientific Calculators

Bas

ic C

alcu

lato

rs


Copyright reserved

11.3.3

bQx

bay

byaxQ

+−=

+=

121

211

≤≤

−≤−≤−

ba

ba

The maximum value of ba is 1.

bQx

bay +−=

ba

−≤−1

1≤ba

answer (4)

[14]

TOTAL: 150



This question paper consists of 13 pages, 1 diagram sheet and 1 information sheet.

MATHEMATICS P2

NOVEMBER 2012


GRADE 12



INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions.

1. 2. 3. 4. 5.

This question paper consists of 13 questions. Answer ALL the questions. Clearly show ALL calculations, diagrams, graphs, et cetera which you have used in determining the answers. Answers only will not necessarily be awarded full marks. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise.

6. 7. 8.

If necessary, round off answers to TWO decimal places, unless stated otherwise. Diagrams are NOT necessarily drawn to scale. ONE diagram sheet for QUESTION 3.2 and QUESTION 7.3 is attached at the end of this question paper. Write your centre number and examination number on this sheet in the spaces provided and insert the sheet inside the back cover of your ANSWER BOOK.

9. 10. 11.

An information sheet with formulae is included at the end of this question paper. Number the answers correctly according to the numbering system used in this question paper. Write neatly and legibly.



QUESTION 1

The scatter plot below shows the age (in years) and the average height (in centimetres) of boys between 2 and 15 years.

1.1 Use the scatter plot to determine the average height of a 7-year-old boy. (1)

1.2 Describe the trend in the scatter plot. (1)

1.3 What is the approximate increase in the average height per annum between the ages of 2 and 15 years?

(3)

1.4 Explain why the observed trend CANNOT continue indefinitely. (1)

[6]

75 80 85 90 95

100 105 110 115 120 125 130 135 140 145 150 155 160 165 170 175 180

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Aver

age

heig

ht (i

n cm

)

Age of boys (in years)

Scatter Plot

[Source: www.fpnotebook.com/endo/exam/hgtmsrmnincharn.htm]



QUESTION 2

Abe plays for his school's cricket team. The number of runs scored by Abe in the eight games that he batted in, is shown below. (Abe was given out in all of the games.)

21 8 19 7 15 32 14 12

2.1 Determine the average runs scored by Abe in the eight games. (2)

2.2 Determine the standard deviation of the data set. (2)

2.3 Abe's scores for the first three of the next eight games were 22, 35 and 2 respectively. Describe the effect of his performance on the standard deviation of this larger set having 11 data points.

(2)

2.4 Abe hopes to score an average of 20 runs in the first 16 games. What should his average in the last five games be so that he may reach his goal?

(3) [9]

QUESTION 3

In a certain school 60 learners wrote examinations in Mathematics and Physical Sciences. The box-and-whisker diagram below shows the marks (out of 100) that these learners scored in the Physical Sciences examination.

3.1 Write down the range of the marks scored in the Physical Sciences examination. (1)

3.2 Use the information below to draw the box-and-whisker diagram for the Mathematics results on DIAGRAM SHEET 1. Minimum mark = 30 Range = 55 Upper quartile = 70 Interquartile range = 30 Median = 55

(4)

3.3 How many learners scored less than 70% in the Mathematics examination? (2)

3.4 Joe claims that the number of learners who scored between 30 and 45 in Physical Sciences is smaller than the number of learners who scored between 30 and 55 in Mathematics. Is Joe's claim valid? Justify your answer.

(2) [9]

Physical Sciences



QUESTION 4

As part of an environmental awareness initiative, learners of Greenside High School were requested to collect newspapers for recycling. The cumulative frequency graph (ogive) below shows the total weight of the newspapers (in kilograms) collected over a period of 6 months by 30 learners.

4.1 Determine the modal class of the weight of the newspapers collected. (1)

4.2 Determine the median weight of the newspapers collected by this group of learners. (1)

4.3 How many learners collected more than 60 kilograms of newspaper? (2) [4]

Weight of newspaper collected (in kilograms)



QUESTION 5

ABCD is a rhombus with A(– 3 ; 8) and C(5 ; – 4). The diagonals of ABCD bisect each other at M. The point E(6 ; 1) lies on BC.

5.1 Calculate the coordinates of M. (2)

5.2 Calculate the gradient of BC. (2)

5.3 Determine the equation of the line AD in the form y = mx + c. (3)

5.4 Determine the size of θ, that is CAB . Show ALL calculations.

(6) [13]

A(– 3 ; 8)

B

C(5 ; – 4)

D

M E(6 ; 1)

y

x

θ

P Q R S T



QUESTION 6

A circle centred at N(3 ; 2) touches the x-axis at point L. The line PQ, defined by the equation

34

34

+= xy , is a tangent to the same circle at point A.

x

y

6.1 Why is NL perpendicular to OL? (1) 6.2 Determine the coordinates of L. (1) 6.3 Determine the equation of the circle with centre N in the form

(x – a)2 + (y – b)2 = r2

(3) 6.4 Calculate the length of KL. (3) 6.5 Determine the equation of the diameter AB in the form y = mx + c. (4)

6.6 Show that the coordinates of A are

516;

57 .

(3)

6.7 Calculate the length of KA. (3) 6.8 Why is KLNA a kite? (2) 6.9 Show that .45KBA °= (3) 6.10 If the given circle is reflected about the x-axis, give the coordinates of the centre of

the new circle.

(1) [24]

Q

A

K

P

B

L

N(3 ; 2)

O



QUESTION 7

-12 -10 -8 -6 -4 -2 2 4 6 8 10 12

-4

-2

2

4

6

8

10

x

y

7.1 Describe the single transformation of ∆ABC to ∆A/B/C/. (2)

7.2 Write down the general rule of the transformation in QUESTION 7.1. (2)

7.3 ∆ A/B/C/ is enlarged by a scale factor of 2 to form ∆ A//B//C//. Draw the enlargement on DIAGRAM SHEET 1.

(2)

7.4 Write down the general rule of the transformation in QUESTION 7.3. (1)

7.5 ∆ABC is reflected about the x-axis and then it is reflected about the y-axis to form ∆DEF.

7.5.1 Write down the coordinates of D, where D is the image of A after the

transformation described above.

(2)

7.5.2 Write down the general rule of this transformation in the form: );();();( →→yx .

(2)

7.5.3 Describe a single transformation that ∆ABC undergoes to form ∆DEF. (2)

[13]

Consider the diagram below where A(– 5 ; 2), B(– 4 ; 1) and C(– 3 ; 3) are the vertices of ∆ABC.

A

0

B/

C

B

C/

A/



QUESTION 8

Answer this question WITHOUT using a calculator.

8.1 The point P(k ; 8) lies in the first quadrant such that OP = 17 units and α=POT as shown in the diagram alongside.

8.1.1 Determine the value of k. (2)

8.1.2 Write down the value of cosα . (1)

8.1.3 If it is further given that α + β = 180°, determine .cosβ (2)

8.1.4 Hence, determine the value of ).sin( αβ − (4)

8.2 Consider the expression: xx

xxcos2sin

sin2cos1−−−

8.2.1 Prove that: xxx

xx tancos2sin

sin2cos1=

−−−

(4)

8.2.2 The above expression is undefined if 0cos2sin =− xx . Solve this

equation in the interval °≤≤° 3600 x .

(4) [17]

QUESTION 9

9.1 Simplify as far as possible: °++°−° 45tan)90cos().180sin(

sin 2

θθθ

(5)

9.2 Simplify without the use of a calculator: °°−°°

412sin.38tan)115cos2(104sin

2

2

(8) [13]

α

P(k ; 8)

O

17

x

y

T



QUESTION 10 The graphs of )30sin()( °+= xxf and xxg cos2)( −= for °≤≤°− 18090 x are given below. The graphs intersect at point P and point Q.

-90 -60 -30 30 60 90 120 150 180

-2

-1

1

2

x

y

10.1 Calculate f (0) – g(0). (1)

10.2 Calculate the x-coordinates of point P and point Q. (7)

10.3 For which values of x will ?)()( xgxf ≥ (2)

10.4 Graph h is obtained by the following transformation of f: )60(2)( °+= xfxh . Describe the relationship between g and h.

(2) [12]

P

Q

0

g

f



QUESTION 11 ABCD is a parallelogram with AB = 3 units, BC = 2 units and θ=CBA for °≤<° 900 θ .

11.1 Prove that the area of parallelogram ABCD is .sin6 θ (3)

11.2 Calculate the value of θ for which the area of the parallelogram is 33 square units. (3)

11.3 Determine the value of θ for which the parallelogram has the maximum area. (2) [8]

3

D C

3

A B θ

2



QUESTION 12

A hot-air balloon H is directly above point B on the ground. Two ropes are used to keep the hot-air balloon in position. The ropes are held by two people on the ground at point C and point D. B, C and D are in the same horizontal plane. The angle of elevation from C to H is x.

x2BDC = and x−°= 90DBC . The distance between C and D is k metres.

12.1 Show that CB = 2k sin x. (5)

12.2 Hence, show that the length of rope HC is 2k tan x. (3)

12.3 If k = 40 m, x = 23° and HD = 31,8 m, calculate θ, the angle between the two ropes. (4) [12]

H

D

B

C k

90° – x

2x x

θ



QUESTION 13

The face of a standard clock is positioned such that the centre is at the origin. At a certain time, the end of the minute hand is at the point P(2 ; 4). 37 minutes later, the end of the minute hand is at the point P/ (a ; b).

13.1 Determine the value of a and b. (6)

13.2 OD is the position of the hour hand when the minute hand is at P and OD/ is the position of the hour hand when the minute hand is at P/. Calculate the angle between OD and OD/.

(4) [10]

TOTAL: 150

y

D/

D x

P/

P

O


Copyright reserved

CENTRE NUMBER:

EXAMINATION NUMBER:

DIAGRAM SHEET 1 QUESTION 3.2 QUESTION 7.3

-12 -10 -8 -6 -4 -2 2 4 6 8 10 12

-4

-2

2

4

6

8

10

x

y

Physical Science

Mathematics

A

0

B/

C

B

C/

A/


Copyright reserved

INFORMATION SHEET

aacbbx

242 −±−

=


∑=

=n

in

11

2)1(

1

+=∑

=

nnin

i

dnaTn )1( −+= ( )dnann )1(2

2S −+=

1−= nn arT ( )

11

−−

=rraS

n

n ; 1≠r

raS−

=∞ 1; 11 <<− r

( )[ ]iixF

n 11 −+=

hxfhxfxf

h

)()(lim)('0

−+=

→

22 )()( 1212 yyxxd −+−= M

++2

;2

2121 yyxx

cmxy += )( 11 xxmyy −=− 12

12xxyy

m−

−= θtan=m

( ) ( ) 222 rbyax =−+−

In ∆ABC: C

cB

bA

asinsinsin


CabABCarea sin.21

=∆



−

−

−

=

1cos2sin21

sincos2cos

2

2

22

α

α

αα



nfx

x ∑= ( )

n

xxn

ii

2

2

1∑=

−=σ


bxay +=ˆ ( )∑

∑−

−−= 2)(

)(xx

yyxxb

[1 (1 ) ]nx iPi

−− +=


MARKS: 150


MATHEMATICS P2

NOVEMBER 2012

MEMORANDUM


GRADE 12



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 unless indicated

otherwise QUESTION 1

1.1 Approximately 121cm (Accept 120 – 122) answer (1)

1.2 As the age increases, the height increases OR Every year the height increases by approximately 6,2 cm OR Straight line (linear) with a positive gradient OR Strong positive correlation OR Increase in height: increase in age is a constant

description (1)

1.3

23621588169height averagein increase eApproximat

,=−−

=

Range for numerator (87 – 89 ; 167 – 170) (Accept any answer between 6 and 6,4 cm)

reading off from graph numerator answer

(3)

1.4 Children stop growing when they reach adulthood. OR If the trend continues the boys would reach impossible heights OR The trend will start approaching a constant value. OR People cannot grow indefinitely

comment (1)

[6]



QUESTION 2

2.1 Average number of runs

168

128=== ∑

nx

x

128 16

(2)

2.2 Standard deviation = 7,55

7,55 (2)

2.3 Standard deviation = 9,71 Standard deviation increases. OR 2 and 35 are far from the mean, namely 16. Since the standard deviation depends on how far data points are from the mean, the standard deviation would be expected to increase.

9,71 increases

(2) 2 and 35 far from mean increase

(2) 2.4 Total number of runs required is 20 x 16 = 320

Total number of runs to be scored in last five games = 320 – 59 – 128 = 133 Average number of runs for last five games is

6,265

133=

OR

6,265

133133

320187

201659128

=∴

=∴=+

=++

xx

x

OR

6,261335

2016

559128

=∴=

=++

xx

x

320 133 26,6

(3)

320 133 26,6

(3) 320 133 26,6

(3) [9]

NOTE: Penalty of 1 mark for incorrect rounding off



QUESTION 3

3.1 Range = 85 – 30 = 55

55 (1)

3.2

max 85 Q3 = 70 Q1 = 40 Median = 55

(4)

3.3 From the information given for Mathematics, the value of the third quartile is 70%. Therefore 75% of learners got below 70%. Number of learners below 70% is expected to

be 45604360

10075

=×=× learners

75% of learners 45 learners

(2)

3.4 No, Joe's claim is invalid. 50% of the learners scored between 30% and 45% in Physical Sciences. 50% of the learners scored between 30% and 55% in Mathematics. Therefore the numbers will be equal. OR No, Joe's claim is invalid. Same number of learners (between min and median)

invalid/no median represents 50% of learners

(2)

[9]

QUESTION 4

4.1 Modal class is 50 ≤ x < 60 OR 50< x ≤ 60 OR 50 to 60

Correct class (1)

4.2 Median position is 15 learners (grouped data). Approximate weight is about 53 kg. (Accept from 52 kg to 54 kg )

53 kg (1)

4.3 30 – 23 = 7 learners collected more than 60 kg.

7 learners

(2) [4]

Maths

Phy Sc



QUESTION 5

5.1 Diagonals bisect each other at M:

22

)4(8;12

53=

−+==

+−= MM yx

M(1 ; 2)

1=Mx 2=My

(2)

5.2

55641

=−+

=

BC

BC

m

m

OR

56514

=−−−

=

BC

BC

m

m

substitution into gradient formula 5

(2)

6514

−−−

=BCm

5 (2)

5.3

235)3(58

5)3(8)( 11

+=+=−

==+=−

−=−

xyxy

mmxmy

xxmyy

BCAD

OR

substitute (–3 ; 8) gradients equal equation

(3)

Lines parallel

A(– 3 ; 8)

B

C(5 ; – 4)

D

M E(6 ; 1)

y

x

θ

P Q R S T



5==

AD

BCAD

mmm

23523

)3(585

+==

+−=+=

xyc

ccxy

gradients equal substitute (–3 ; 8) equation

(3) 5.4 ABCD is a rhombus, therefore

AB = BC

TSBSRA

CSRSRAACBˆˆ

ˆˆˆ

−=

−==θ

°=

−°=

−=

−−+

==

69,123ˆ...3099,56180ˆ

23ˆtan

5348ˆtan

SRA

SRA

SRA

mSRA AC

°=

==

69,78ˆ5ˆtan

TSB

mTSB BC

°=°−°==

4569,7869,123ˆ

θθ ACB

OR

°=

−=−−

+==

69,123ˆ23

5348ˆtan

SRA

mSRA AC

°=

==

69,78ˆ5ˆtan

RPA

mRPA AD

°==

°=°−°=

−=

45

ˆ45

69,7869,123

ˆˆˆ

RAP

RPASRARAP

θ

ACB ˆ=θ

23ˆtan −=SRA

123,69° 5ˆtan == BCmTSB 78,69° θ = 45°

(6)

23ˆtan −=SRA

123,69° 5ˆtan == ADmRPA 78,69° °= 45ˆP RA °= 45θ

(6)

Lines parallel

Exterior angle of a triangle

Diagonals of the rhombus bisect opposite angles



OR

°=

−=−−

+==

69,123ˆ23

5348ˆtan

SRA

mSRA AC

°=

=

69,78ˆ5ˆtan

RPA

RPA

°=°−°=

−=

=

4569,7869,123

ˆˆ

ˆ

θθθ

θ

RPASRA

RAP

OR

°=

−=−−

+==

69,123ˆ23

5348ˆtan

SRA

mSRA AC

°=

=

69,78ˆ5ˆtan

TSB

TSB

SCR ˆ=θ

SRA

CSRSCRTSBSCRˆ

ˆˆˆˆ

=

+=+

°=°−°=

−=

4569,7869,123

ˆˆ TSBSRAθ

OR ABCD is a rhombus, therefore AB = BC

CABBCA ˆˆ =∴

°==

+

−−

−

−=

+−

=

−=

=

451

15

8121

15

812

ˆtan.ˆtan1

ˆtanˆtan

)ˆˆtan(

ˆtantan

θ

θ

TSBSRATSBSRA

TSBSRA

BCA

23ˆtan −=SRA

123,69 5ˆtan == ADmRPA 78,69° RAP ˆ=θ °= 45θ

(6)

23ˆtan −=SRA

123,69 5ˆtan =TSB 78,69° SCR ˆ=θ °= 45θ

(6) CABBCA ˆˆ =

BCA ˆtantan =θ formula substitution 1tan =θ °= 45θ

(6)

Exterior angle of a triangle

Diagonals of the rhombus bisect opposite angles

BA=BC



OR From 5.1, M has coordinates (1 ; 2) Join ME

51

6112

−=−−

=MEm

From 5.2,

°=∴

−=×∴=

90ˆ1.

5

CEM

mmm

BCME

BC

( ) ( )( ) ( ) 261465

26126122

22

=−−+−=

=−+−=

EC

ME

∴MEC is a right-angled triangle. °= 45ˆMCE

ABCD is a rhombus, therefore AB = BC

°==∴ 45ˆMCBθ OR

( ) ( ) 1322813 22 =−+−−=AM Now to calculate the coordinates of B:

32

123

5348

=

−=×

−=−−

+=

BD

ACBD

AC

m

mm

m

34

32is ofEquation += xyBD

295 is ofEquation −= xyBC

BD and BC intersect at B. Solve equations simultaneously to get B(7 ; 6).

( ) ( )

°==∴

=

°=

=∴

==−+−=

451tan

tan

90ˆ Since

132522617 22

θθ

θAMBM

BMA

AMBMBM

gradient of ME gradient of BC °= 90ˆCEM 26=ME 26=EC °= 45ˆMCE

(6)

132=AM

34

32

+= xy

295 −= xy B(7 ; 6) 132=BM 45°

(6) [13]

diagonals bisect at right angles



QUESTION 6

x

y

6.1 The radius (NL) of a circle is perpendicular to the tangent (OL) at the point of contact.

radius ⊥ tangent

(1) 6.2 L(3 ; 0) (3 ; 0)

(1) 6.3 Centre N (3 ; 2) and r = NL = 2

Equation of the circle N:

4)2()3()()(

22

222

=−+−

=−+−

yxrbyax

r = 2 22 )2()3( −+− yx 4

(3)

6.4 Coordinates of K. K is the x-intercept of the tangent.

)0;1(14444034

340

34

34

−−=−=+=

+=

+=

Kxx

x

x

xy

KL = 3 – (–1) OR KL = 3 + 1 KL = 4

substitute y = 0 into

equation of tangent x = – 1 KL = 4

(3)

Q

A

K

P

B

L

N(3 ; 2)

O



OR

)0;1(14444034

340

34

34

−−=−=+=

+=

+=

Kxx

x

x

xy

416

)00()13(

)()(22

212

212

==

−++=

−+−=

KLKL

KL

yyxxKL

OR

For AK, m = 34 , c =

34

41

34ˆtan3

4

=∴=

==

KLOK

OKAOK

OR

)0;1(14444034

340

34

34

−−=−=+=

+=

+=

Kxx

x

x

xy

KN2 = NL2 + KL2 (– 1 – 3)2 + (0 – 2)2 = 4 + KL2 20 = 4 + KL2 16 = KL2 KL = 4

substitute y = 0 into

equation of tangent x = – 1 KL = 4

(3)

343

4

=OK

OK = 1 KL = 4

(3)

x = – 1

KN2 = NL2 + KL2 KL = 4

(3)

Theorem of Pythagoras



6.5

43

34

1

−=∴

=

−=×

AB

AK

AKAB

m

m

mm

417

43

48

49

43

)3(432

)( 11

+−=

++−=

−−=−

−=−

xy

xy

xy

xxmyy

OR

43

34

1

−=∴

=

−=×

AB

AK

AKAB

m

m

mm

( )

417

43

417

49

48

3432

43

+−=

=

+=

+

−=

+−=

xy

c

c

c

cxy

34

=AKm

43

−=ABm

substitution of point

(3;2) into equation equation

(4)

34

=AKm

43

−=ABm

substitution of point

(3;2) into equation equation

(4)

tangent ⊥ radius

tangent ⊥ radius



6.6 Point A lies on PQ and AB. Therefore

573525

51916164

1743

34

34

=

=+−=+

+−=+

x

xxx

xx

516

417

57

43

=

+

−=

y

y

516;

57A

OR Point A lies on PQ and the circle. Therefore

516

417

57

43

57

0)75(0497025

4)32

34()3(

4)234

34()3(

2

2

22

22

=

+

−=

=

=−

=+−

=−+−

=−++−

y

y

x

xxx

xx

xx

OR

equation 25x = 35 substitution of x (3) equation 0)75)(235( =−− xx

substitution of x

(3)

equation (5x – 7)2 = 0 substitution of x



Point A lies on the circle and line AB

57

0)75)(235(016115025

4)49

43(96

4)24

1743(96 :(1)in (2) Subs

(2)--------- 4

1743

(1)--------- 4)2()3(

2

22

22

22

=

=−−=+−

=+−++−

=−+−++−

+−=

=−+−

x

xxxx

xxx

xxx

xy

yx

516

417

57

43

=

+

−=

y

y

OR Using rotation: Let NKLNKA ˆˆ ==θ Move diagram 1 unit to the right. Then A/ is L/ rotated through 2θ .

53)

51()

52(sincos2cos

54)

52)(

51(2cossin22sin

21

42tan

2222 =−=−=

===∴

===

θθθ

θθθ

θKAAN

516)

53)(0()

54(42cos2sin

512)

54)(0()

53(42sin2cos

=−=+=

=−=−=∴

′′′′

′′′′

θθ

θθ

LLA

LLA

yxy

yxx

)5

16;5

12(A′

Now to get back to A, move back 1 unit to the left.

)5

16;57(A∴

OR

equation 0)75)(235( =−− xx substitution of x (3)

values of sin2θ and cos2θ substitution into rotation formulae

)5

16;5

12(A′

(3)



x

y

Let .ˆ θ=LKN So, 21

42tan ===

KNNLθ .

Hence 5

1sin =θ and 5

2cos =θ

Let axis-on M with axis xxAM −⊥

θ

θ

2ˆ

ˆˆ

=∴

==

∆≡∆

LKA

LKNNKA

NLKNAK

θθθ 2sin42sin2sin ==== KLAKAMy A

54

52

512cossin22sin =

== θθθ

516

544 =

=Ay

( )

57

583

2sin23

ˆ90sin23

ˆsin

=

−=

−=−°−=

−=

θKAM

NAMNAOLxA

tan 21

=θ

542sin =θ

solve for x and y

(3)

K L

N

A

θ θ

M



6.7

4

05

16157

)()(22

212

212

=

−+

+=

−+−= yyxxKA

OR

416

420

2024222

22

==

−=−=

=+=

KA

ANKNKAKN

OR KA = KL Tangents from a common point are equal KA = 4

distance formula substitution 4

(3)

20=KN 222 ANKNKA −=

4

(3) KA=KL reason 4

(3) 6.8 AN = NL Radii are equal

KA = KL ∴KLNA is a kite two pairs of adjacent sides are equal.

AN = NL KA = KL

(2) 6.9 AB = AN + NB = 2 + 2 = 4

AK = 4= AB

°=∴

°=

°=+

−∴°=

45ˆ90ˆ2

90ˆˆtriangleisoscelesangledrightaisΔAKB

90ˆ

KBA

KBA

KBABKA

BAK

OR

AB = 4 AK = AB °= 90ˆBAK

(3)

tangent ⊥ radius



N is midpoint of AB Let B be ( )BB yx ;

54

523

22

516

32

57

=∴=∴

=+

=+

BB

BB

yx

yx

∴

54;

523B

°=°−°=

−==

13,14387,36180

43tan

ββ

β ABm

°=

=+

−==

13,8

71

1523

054

tan

α

α KBm

°=°+°=

−°+=

4587,3613,8

)180(ˆ βαKBA

OR N is midpoint of AB Let B be ( )BB yx ;

54

523

22

516

32

57

=∴=∴

=+

=+

BB

BB

yx

yx

∴

54;

523B

24)54()1

523( 22 =++=KB

°=∴

=

−+=

4522cos

cos)32)(4(2)32(44 222

θ

θ

θ

143,13° 8,13° )180(ˆ βα −°+=KBA

(3) 4 2 substitution into cosine formula

22cos =θ

(3)

6.10 )2;3(/ −N )2;3(/ −N (1) [24]

α β K

A

B

4 2

4 4

A

B K



QUESTION 7 NOTE: CA not applicable in this question

7.1 Rotation about the origin through 90° in a clockwise direction. OR Rotation about the origin through 270° in an anti-clockwise direction. OR Rotation about the origin through -90°.

rotation of 90° clockwise direction

(2) rotation of 270° anti-clockwise

direction (2)

statement

(2)

7.2 );();( xyyx −→

(both)

);();( xyyx −→ (2)

7.3

-7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

1

2

3

4

5

6

7

8

9

10

x

y

one point correct all points correct and triangle drawn

(2)

7.4 )2;2();( yxyx → )2;2( yx (1) 7.5.1 )2;5()2;5()2;5( −→−−→− DA 5

– 2 (2) 7.5.2 );();();( yxyxyx −−→−→

(x ; – y) (– x ; – y)

(2) 7.5.3 Rotation of 180° through the origin in either direction.

OR Reflection about the origin.

rotation 180° (2) reflection origin (2)

[13]

0

A

B/ C

B

C/

A/ C//

A//

B//



QUESTION 8 No calculator allowed in this question

8.1.1 OT = k , PT = 8 and OP = 17

150

15225

64289178

2

2

222

=>

±==

−=

=+

kk

kkk

k

OR

150

15225

925)817)(817(

8172

222

=>

±==

×=+−=

−=

kkk

kk

substitution into Pythagoras k = 15

(2) substitution into Pythagoras k = 15

(2)

8.1.2 1715cos =α

1715

(1) 8.1.3

αββα

−°=°=+

180180

1715cos

)180cos(cos

−=

−=−°=∴

ααβ

OR

1715cos

)180cos(cos

−=

−=−°=∴

ααβ

)180cos( α−° or – cos α

1715

−

(2) )180cos( α−° or – cos α

1715

−

(2)

17 8 β α



8.1.4

289240

289120

289120

178

1715

1715

178

sincoscossin)sin(

=

+=

−−

=

−=−

αβαβαβ

OR

( )

289240

1715

1782

cos.2sin sin2

)2180sin()sin(2180

180

=

=

==

−°=−−°=

−−°=−

ααα

ααβα

αααβ

expansion

178sin =β

178sin =α

289240

(4)

substitute β 2sinαcosα

178sin =α

289240

(4)

8.2.1

RHSxxx

xxxx

xxxxx

xxxxx

xxxxLHS

==

=

−−

=

−−

=

−−−−

=

−−−

=

tancossin

)1sin2(cos)1sin2(sin

coscossin2sinsin2

coscossin2sin)sin21(1

cos2sinsin2cos1

2

2

OR

x2sin21− xx cossin2 either )1sin2(sin −xx or

)1sin2(cos −xx

xx

cossin

(4)



RHSxxx

xxxx

xxxxx

xxxxxxxx

xxxxx

xxxx

xxLHS

==

=

−−

=

−−

=

−−−

=

−−−

=

−−−−

=

−−−

=

tancossin


coscossin2sinsin2

coscossin2sin)cos1(2

coscossin2sincos2

coscossin2sin)1cos2(1

cos2sinsin2cos1

2

2

2

2

OR

RHSxxx

xxxx

xxxxx

xxxxxxxxx

xxxxxx

xxxxx

xxLHS

==

=

−−

=

−−

=

−−+

=

−−+−

=

−−−−

=

−−−

=

tancossin


coscossin2sinsin2

coscossin2sinsinsincoscossin2

sinsincos1coscossin2

sin)sin(cos1cos2sin

sin2cos1

2

22

22

22

1cos2 2 −x xx cossin2 either )1sin2(sin −xx or

)1sin2(cos −xx

xx

cossin

(4)

xx 22 sincos − xx cossin2 either )1sin2(sin −xx or

)1sin2(cos −xx

xx

cossin (4)



8.2.2

Zkkxkxx

xxxxx

xx

∈°+°=°+°==

=−=−

=−

360270or 360900cos

0)1sin2(cos0coscossin2

0cos2sin

or

k xkx

x

°+°=°+°=

=

360150or 360 30 21sin

x = 90° or x = 270° or x = 30° or x = 150° OR

( )

kxkxkxxkx

kxxZkkxxxx

xx

°+°=°+°=°++°=°+°=

°+−°−°=∈°+−°=−°=

=

3609012030360902360903

360901802or;360902)90sin(2sin

cos2sin

x = 30° or x = 150° or x = 270° or x = 90°

xx cossin2 0cos =x and

21sin =x

for two correct answers for four correct answers

(4)

)90sin( x−°

kx .12030 °+°= and

kx .36090 °+°= for two correct answers for four correct answers

(4) [17]



QUESTION 9

9.1

θθθ

θθ

θθθ

θθθ

2

2

2

2

2

2

2

tancossin

1sinsin

1)sin)((sinsin

45tan)90cos().180sin(sin

=

=

+−=

+−=

°++°−°

sinθ –sinθ 1 cos²θ θ2tan

(5) 9.2

( )

( )

338cos38sin

38cos38sin3

38cos38cos38sin

2338cos38sin2

52sin.38tan30cos.76sin412sin38tan

)115cos2(104sin

2

2

2

2

=°°

°°=

°

°°

°°

=

°°°°

=

°°−°°

OR

( )

( )

323.2

30cos2

52sin52sin52cos

30cos.52cos52sin2

52sin.38cos38sin

)115cos2).(52(2sin412sin38tan

)115cos2(104sin

2

2

2

2

2

=

=

°=

°

°°

°°°=

°°°

−°°=

°°−°°

sin 76° cos30°

°°

38cos38sin

sin52° 2sin38°cos38°

23

sin52°=cos38° 3

(8)

sin2(52°)

°°

38cos38sin

sin52° 2sin52°cos52° cos30° cos52°=sin38° and sin52°=cos38°

23

3 (8)

NOTE: • If cos 30° is missing: deduct

1 mark • Answer only: 0/8



OR

( )

( )

314cos

14cos3or76sin

76sin376sin104sin3

38cos38sin2104sin3

38cos38cos38sin

23)104(sin

52sin38cos38sin

30cos.104sin412sin38tan

)115cos2(104sin

2

2

2

2

=°

°°

°=

°°

=

°°°

=

°°

°

=

°

°°

°°=

°°−°°

OR

( )

( )

( )

3

104sin21

23.104sin

52sin52cos23.104sin

52sin52sin52cos

23.104sin

52sin.38cos38sin

30cos.104sin412sin38tan

)115cos2(104sin

2

2

2

2

=

°

°=

°°

°=

°

°°

°=

°°°

°°=

°°−°°

cos30°

°°

38cos38sin

sin52° cos2 38°

23

sin76° 3

(8)

cos30°

°°

38cos38sin

sin52°

23

cos52°=sin38° and sin52°=cos38°

cos520.sin520

°104sin21

3 (8)

[13]



QUESTION 10 10.1 5,2)2(5,0)0()0( =−−=− gf 2,5 (1) 10.2

35tan

cos5sin3

cos4cossin3

cos2cos21sin

23

cos230sin.cos30cos.sincos2)30sin(

−=

−=

−=+

−=

+

−=°+°−=°+

x

xx

xxx

xxx

xxxxx

kx °+°= 18011,109 ; k ∈Z °=°−= 11,10989,70 QP xandx

OR

35tan

cos5sin3

cos4sin3cos

cos2sin23cos

21

cos2sin60sincos60coscos2)60cos(

cos2)3090cos(cos2)30sin(

−=

−=

−=+

−=+

−=°+°−=−°

−=°−−°−=°+

x

xx

xxx

xxx

xxxxx

xxxx

kx .18011,109 °+°= ; k ∈Z °=°−= 11,10989,70 QP xandx

equation expansion of

sin(x+30°) substitution of

special angles simplification

3

5tan −=x

°−= 89,70Px °= 11,109Qx

(7) equation expansion of cos(60° – x) substitution of

special angles simplification

3

5tan −=x

°−= 89,70Px °= 11,109Qx

(7) 10.3 °≤≤°− 11,10989,70 x

OR ]11,109;89,70[ °°−

OR QP xxx ≤≤

angles correct interval

(2)

10.4 )(cos2)90sin(2)3060sin(2)( xgxxxxh −==°+=°+°+= h is the reflection of g about the x-axis. OR f is shifted to the left through 60° and then doubled. ∴ h is the reflection of g about the x-axis.

reflection about the x-axis or

line y = 0 (2) reflection about the x-axis or

line y = 0 (2) [12]



QUESTION 11 11.1

( )( )

θ

θ

sin6

sin23212

ΔABCArea2ABCDramparallelogArea

=

=

×=

OR

θθθ

θ

sin6sin2.33height base Area sin2

sin2

===×=∴=

=

hABCDh

h

OR Area of parallelogram ABCD = area of ΔABC + area of ΔADC

= ( )( ) θsin2321

+ ( )( ) θsin23

21

= 6 sin θ OR

Area = 21 (sum of // sides)× h

= 21 (3 + 3) × 2sin θ

= 6 sin θ

ABC area2 ∆ substitution into

area rule

(3)

θsin2

=h

θsin2=h b.h

(3)

sum of areas equal sides and equal angles

(3)

formula θsin2=h substitution

(3)

11.2 Area of parallelogram ABCD = 33

33sin6 =θ

°=

=

6023sin

θ

θ

OR 6 sin 60° = 33

∴ θ = 60°

33sin6 =θ

23sin =θ

60°

(3)

33sin6 =θ 60°

(3) 11.3 Maximum area of parallelogram occurs when sin θ = 1, that is

when °= 90θ

sin θ = 1 °= 90θ

(2) [8]

2

θ

h

3 NOTE: If no working is shown, then 0/3

NOTE: Deduct 1 mark if both 60° and 120° are given as answers



QUESTION 12

12.1

xkx

xxkx

xkCB

xk

x

sin2cos

cossin2.CB

)90sin(2sin.)90sin(2sin

CBDBCsin

CDCDBsin

CB

=

=

−°=

−°=

=

OR

kDBDCxxxBCD

==∴−°=+−°−°= 90)290(180ˆ

x−°90

xkCBCFCB

xkCF

xCDCF

sin22

sin

sin

BCDF Draw

===

=

⊥

OR

kDBDCxxxBCD

==∴−°=+−°−°= 90)290(180ˆ

CB2 = CD2 +BD2 – 2.CD.BD.cos2x

xkCBxk

xkxk

xkxk

xkkkCB

sin2)sin2(

sin4)sin2(2

))sin21(1(2)2cos1(2

2cos2

2

22

22

22

2

2222

==

=

=

−−=

−=

−+=

Using the sine rule

in triangle CBD

)90sin(2sin xk

xCB

−°=

)90sin(

2sin.x

xk−°

2sinx.cosx cos x

(5)

xCBDBCD −°== 90ˆˆ kDBDC ==

xFDC =ˆ xkCF sin= CB=2 CF

(5)

xCBDBCD −°== 90ˆˆ kDBDC == using cosine rule in triangle CDB factors simplification

(5)

D

B F C

k k

x x

x−°90



12.2

xkxxk

x

x

tan2cos

sin2cosBCHC

HCBCcos

=

=

=

=

OR

)90sin(

)90sin(90sin

xBCHC

xBCHC

−°=

−°=

°

xkx

xk

tan2cos

sin2

=

=

HCBCx =cos

x

BCHCcos

=

substitution of BC

(3)

)90sin( x

BCHC−°

=

substitution of BC sin(90° – x) = cos x

(3)

12.3 HC = xk tan2 = 2(40).tan(23°) = 33,9579... In ∆HCD:

°=∴=

−+=

−+=

−+=

85,74...2613,0cos

)8,31...)(9579,33(2408,31...)9579,33(

.2cos

cos..2

222

222

222

θθ

θ

θ

HDHCCDHDHC

HDHCHDHCCD

value of HC substitution into cos formula cos θ = 0,2613... 74,85°

(4) [12]



QUESTION 13

13.1 Angle that minute hand moves is:

°=

°×

222

3606037

OR

P is rotated by 360° - 222° = 138° in an anti-clockwise direction:

16,4138sin4138cos2

−=°−°=a

and

63,1138sin2138cos4

−=°+°=b

OR Angle that minute hand moves is:

°=

°×

222

3606037

P is rotated by 222° in a clockwise direction:

16,4222sin4222cos2

−=°+°=a

and 63,1

222sin2222cos4−=

°−°=b

OR

°==

43,632tan

αα

°=°−°+°=

°=−°+

43,2122218043,63

222180ββα

63,143,21sin20

16,443,21cos20

−=°−=

−=°−=∴

b

a

°× 3606037

222° substitution of 1380 into formula for x and y 16,4− 63,1−

(6)

°× 3606037

222° substitution of 2220 into formula for x and y 16,4− 63,1−

(6) 2tan =α °= 43,63α °=−°+ 222180 βα β = 21,43° 16,4−

63,1− (6)

4

β

2 α

180°- β

60 min : 360° 1 min : 6° 37 min : 37 × 6 = 222°

20

20

a

b


Copyright reserved

13.2 The minute hand moves through 360° in 60 minutes.

The hour hand moves through 30° in 60 minutes, that is, 121 that of

the minute hand. So when the minute hand moves through 222°,

the hour hand moves through °=° 5,18

12222

OR

The hour hand moves through °=° 30

12360 in 60 minutes

∴it moves through °=°× 5,18306037 in 37 minutes

360° 30°

121

18,5° (4)

360° 30°

°× 306037

18,5° (4)

[10]

TOTAL : 150



This question paper consists of 9 pages, 5 diagram sheets and 1 information sheet.

MATHEMATICS P3

NOVEMBER 2012


GRADE 12



INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

This question paper consists of 10 questions. Answer ALL the questions. Clearly show ALL calculations, diagrams, graphs, et cetera that you have used in determining your answers. Answers only will not necessarily be awarded full marks. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. If necessary, answers should be rounded off to TWO decimal places, unless stated otherwise. Diagrams are NOT necessarily drawn to scale. FIVE diagram sheets for answering QUESTION 1.2, QUESTION 3.1, QUESTION 7.1, QUESTION 7.2, QUESTION 8.2, QUESTION 9 and QUESTION 10 are attached at the end of this question paper. Write your centre number and examination number on these sheets in the spaces provided and insert them inside the back cover of your ANSWER BOOK. An information sheet with formulae is included at the end of the question paper. Number the answers correctly according to the numbering system used in this question paper. Write neatly and legibly.



QUESTION 1 A recording company investigates the relationship between the number of times a CD is played by a national radio station and the national sales of the same CD in the following week. The data below was collected for a random sample of 10 CDs. The sales figures are rounded to the nearest 50.

Number of times CD is played 47 34 40 34 33 50 28 53 25 46

Weekly sales of the CD 3 950 2 500 3 700 2 800 2 900 3 750 2 300 4 400 2 200 3 400 1.1 Identify the independent variable. (1) 1.2 Draw a scatter plot of this data on the grid provided on DIAGRAM SHEET 1. (3) 1.3 Determine the equation of the least squares regression line. (4) 1.4 Calculate the correlation coefficient. (2) 1.5 Predict, correct to the nearest 50, the weekly sales for a CD that was played 45 times

by the radio station in the previous week.

(2) 1.6 Comment on the strength of the relationship between the variables. (1)

[13] QUESTION 2 Each of the 200 employees of a company wrote a competency test. The results are indicated in the table below:

PASS FAIL TOTAL Males 46 32 78 Females 72 50 122 Total 118 82 200

2.1 Are the events PASS and FAIL mutually exclusive? Explain your answer. (2) 2.2 Is passing the competency test independent of gender? Substantiate your answer with

the necessary calculations.

(4) [6]



QUESTION 3 A company producing television sets decided to check the lifespan (in years) of their most popular model. They selected 50 sets of the most popular model at random for this test. The lifespan of each set was recorded. The information is represented in the table below.

LIFESPAN (IN YEARS)

FREQUENCY

4,95 ≤ x < 5,65 2

5,65 ≤ x < 6,35 6

6,35 ≤ x < 7,05 18

7,05 ≤ x < 7,75 17

7,75 ≤ x < 8,45 5

8,45 ≤ x < 9,15 2 3.1 Construct a histogram to represent the data. Use the grid provided on

DIAGRAM SHEET 2.

(3) 3.2 Calculate the estimated mean lifespan of the most popular model of television set. (3) 3.3 The data representing the lifespan of this batch of television sets is normally

distributed. This implies that approximately 68% of the data lies within one standard deviation of the mean, approximately 98% of the data lies within two standard deviations of the mean and approximately 100% of the data lies within three standard deviations of the mean. The standard deviation of this data set is 0,76 years. Calculate the lifespan of the most popular model of television set such that 98% of the lifespan of all the sets will exceed this value.

(3)

3.4 The company wants to issue a 5-year guarantee with this model of television set.

What would you recommend? Justify your recommendation.

(2) [11]



QUESTION 4

During summer in a certain city in South Africa the probability of a sunny day is 74 and the

probability of a rainy day is 73 .

• If it is a sunny day, then the probability that Vusi cycles to work is 107 , the probability

that Vusi drives to work is 51 and the probability that Vusi takes the train to work

is 101 .

• If it is a rainy day, then the probability that Vusi cycles to work is 91 , the probability that

Vusi drives to work is 95 and the probability that Vusi takes the train to work is

31 .

4.1 Draw a tree diagram to represent the above information. Indicate on your diagram the

probabilities associated with each branch as well as all the outcomes.

(5) 4.2 For a day selected at random, what is the probability that: 4.2.1 It is rainy and Vusi will cycle to work (2) 4.2.2 Vusi takes the train to work (3) 4.3 If Vusi works 245 days in a year, on approximately how many occasions does he

drive to work?

(4) [14]

QUESTION 5 Every client of CASHSAVE Bank has a personal identity number (PIN) which is made up of 5 digits chosen from the digits 0 to 9.

5.1 How many personal identity numbers (PINs) can be made if: 5.1.1 Digits can be repeated (2) 5.1.2 Digits cannot be repeated (2) 5.2 Suppose that a PIN can be made up by selecting digits at random and that the digits

can be repeated. What is the probability that such a PIN will contain at least one 9?

(4) [8]

QUESTION 6 6.1 Write down a recursive formula for the sequence: 1 ; 5 ; 13 ; 29 ; 61; ... (4) 6.2 Write down the next term of the given recursive sequence:

4 ; 7 ; 13 ; 24 ; 44; ...

(2) [6]



NOTE: Give reasons for all statements made in QUESTION 7, QUESTION 8,

QUESTION 9 and QUESTION 10. QUESTION 7

7.1 If in ∆ LMN and ∆ FGH it is given that FL = and GM = , prove the theorem that

states FHLN

FGLM

= .

(7) 7.2 In the diagram below, ∆VRK has P on VR and T on VK such that PT || RK.

VT = 4 units, PR = 9 units, TK = 6 units and VP = 2x – 10 units. Calculate the value of x.

(4)

[11]

9

2x – 10

6

9

K

T P

4 2x – 10

6

9

V

R

M

L

N

F

G H



M

P

W

T U

Q

R

c b

a

1

29°

75° 34°

1 1

2 3

d

QUESTION 8 8.1 Complete the following statement:

The angle between the tangent and the chord is equal ...

(1)

8.2 In the diagram points P, Q, R and T lie on the circumference of a circle. MW and

TW are tangents to the circle at P and T respectively. PT is produced to meet RU at U.

°= 75RPM

°= 29TQP

°= 34RTQ

Let a=WPT , b=TPR , c=QPM and d=UTR , calculate the values of a, b, c and d.

(9) [10]



QUESTION 9 O is the centre of the circle CAKB. AK produced intersects circle AOBT at T.

9.1 Prove that x2180T −°= . (3) 9.2 Prove AC || KB. (5) 9.3 Prove ∆BKT ||| ∆CAT (3)

9.4 If AK : KT = 5 : 2, determine the value of KBAC

(3) [14]

C x

A

O

B

K

T 1 2 3

4

1 2

1 2 3

x=BCA


Copyright reserved

QUESTION 10 In the diagram below, O is the centre of the circle. Chord AB is perpendicular to diameter DC. CM : MD = 4 : 9 and AB = 24 units.

10.1 Determine an expression for DC in terms of x if CM = 4x units. (1) 10.2 Determine an expression for OM in terms of x. (2) 10.3 Hence, or otherwise, calculate the length of the radius. (4)

[7]

TOTAL: 100

D

O

B

A

C M


Copyright reserved

CENTRE NUMBER:

EXAMINATION NUMBER:

DIAGRAM SHEET 1 QUESTION 1.2

0

500

1 000

1 500

2 000

2 500

3 000

3 500

4 000

4 500

5 000

0 5 10 15 20 25 30 35 40 45 50 55

Scatter plot showing the number of times a CD was playedvs the CD sales in the following week


Copyright reserved

CENTRE NUMBER:

EXAMINATION NUMBER:


0

5

10

15

20

4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5

Freq

uenc

y

Lifespan (years)

Histogram showing the frequency of the lifespan of the most populartelevision set (years)


Copyright reserved

CENTRE NUMBER:

EXAMINATION NUMBER:


QUESTION 7.2

M

L

N

F

G H

9

2x – 10

6

9

K

T P

4 2x – 10

6

9

V

R


Copyright reserved

M

P

W

T U

Q

R

c b

a

1

29°

75° 34°

1 1

2 3

d

CENTRE NUMBER:

EXAMINATION NUMBER:

DIAGRAM SHEET 4 QUESTION 8.2 QUESTION 9

C x

A

O

B

K

T 1 2 3

4

1 2

1 2 3


Copyright reserved

CENTRE NUMBER:

EXAMINATION NUMBER:

DIAGRAM SHEET 5 QUESTION 10

D

O

B

A

C M


Copyright reserved

INFORMATION SHEET: MATHEMATICS

aacbbx

242 −±−

=


∑=

=n

in

11

2)1(

1

+=∑

=

nnin

i dnaTn )1( −+= ( )dnan

n )1(22

S −+=

1−= nn arT ( )

11

−−

=rraS

n

n ; 1≠r

raS−

=∞ 1; 11 <<− r

( )[ ]iixF

n 11 −+= [1 (1 ) ]nx iP

i

−− +=

hxfhxfxf

h

)()(lim)('0

−+=

→

22 )()( 1212 yyxxd −+−= M

++2

;2

2121 yyxx

cmxy += )( 11 xxmyy −=− 12

12xxyy

m−

−= θtan=m

( ) ( ) 222 rbyax =−+−

In ∆ABC: C

cB

bA

asinsinsin


CabABCarea sin.21

=∆



−

−

−

=

1cos2sin21

sincos2cos

2

2

22

α

α

αα



nfx

x ∑= ( )

n

xxn

ii

2

2

1∑=

−=σ


bxay +=ˆ ( )∑

∑−

−−= 2)(

)(xx

yyxxb


MARKS: 100


MATHEMATICS P3

NOVEMBER 2012

MEMORANDUM


GRADE 12



NOTE: • If a candidate answered a question TWICE, mark the FIRST attempt ONLY. • If a candidate crossed out an attempt of a question and did not redo the question, mark the

crossed out question. • Consistent accuracy applies in ALL aspects of the memorandum. QUESTION 1 1.1 The number of times the CD was played.

Afrikaans: Getalkerewat die CD gespeel is.

answer (1)

1.2

0

500

1,000

1,500

2,000

2,500

3,000

3,500

4,000

4,500

5,000

0 5 10 15 20 25 30 35 40 45 50 55

Scatter plot showing the number of times a CD was playedvs the CD sales in the following week

all 10 points plotted correctly 2 marks if 5–9 points are plotted correctly 1 mark if 1–4 points are plotted correctly.

(3)

1.3 a = 293,06 (293,057554...) b = 74,28 (74,28057554...)

xy 28,7406,293ˆ +=

calculating a and b equation

(4) 1.4 r = 0,95 (0,9458185...)

answer

(2) 1.5

50)nearest the(to 36503635

66,3635)45(28,7406,293ˆ

≈≈≈

+≈y

substitution answer

(2)

1.6 There is a very strong positive relationship between the number of times that a CD was played and the sales of that CD in the following week.

strong (1)

[13]

Note: Penalise 1 mark for incorrect rounding off.



QUESTION 2 2.1 Yes. The events Pass and Fail are mutually exclusive.

It is not possible for pass and fail to take place at the same time. There is no intersection between the two sets. P(Pass and Fail) = 0 OR P(Pass) = 0,59 P(Fail) = 0,41 P(Pass) + P(Fail) = 0,59 + 0,41 =1 P(Pass and Fail) = 0 / No intersection of the sets The events Pass and Fail are mutually exclusive. Afrikaans Ja. Die gebeurtenisse Slaag en Druip is onderling uitsluitend. Dit is nie moontlik dat slaag en druip gelyktydig plaasvind nie. P(Slaag en Druip) = 0

Yes P(Pass and Fail) = 0 / no intersection between the sets.

(2)

Yes P(Pass and Fail) = 0 / No intersection between the sets

(2)

Ja P(Slaag en Druip) = 0 / geen snyding

(2) 2.2

PASS FAIL TOTAL Males 46 32 78 Females 72 50 122 Total 118 82 200

39,020078)Male( ==P

59,0200118)Pass( ==P

23,020046)PassandMale( ==P

(0,2301)23,059,039,0)Pass()Male(

=×=× PP

)PassandMale()Pass()Male( PPP =×∴

∴ Passing the competency test is independent of gender.

39,020078)Male( ==P or

59,0200118)Pass( ==P

23,0)PassandMale( =P 23,0)Pass()Male( =× PP conclusion

(4)

Note: If a candidate answers ‘No’ then award 0 marks

Note: If a candidate answers ‘No’ then award 0 marks



OR

61,0200122)Female( ==P

59,0200118)Pass( ==P

36,020072)PassandFemale( ==P

(0,3599)36,059,061,0)Pass()Female(

=×=× PP

)PassandFemale()Pass()Female( PPP =×∴ ∴ Passing the competency test is independent of gender.

61,0200120)Female( ==P or

59,0200118)Pass( ==P

36,0)PassandFemale( =P 36,0)Pass()Female( =× PP conclusion

(4) [6]



QUESTION 3 3.1

0

2

6

1817

5

2

00

5

10

15

20

Histogram showing the frequency of the lifespan of a television (years)

intervals 3 bars correct 6 bars correct

(3)

3.2 Lifespan (in years) Frequency Midpoint

4,95 ≤ x< 5,65 2 5,3

5,65 ≤ x< 6,35 6 6

6,35 ≤ x< 7,05 18 6,7

7,05 ≤ x< 7,75 17 7,4

7,75 ≤ x< 8,45 5 8,1

8,45 ≤ x< 9,15 2 8,8

years02,750

1,35150

8,821,854,7177,618663,52

=

=

×+×+×+×+×+×=x

( )022,7=x

frequencies ×

midpoints 50 answer

(3)

3.3 The required area is 98% to the right of some value. This value is at 2 standard deviations on the left of the mean.

)76,0(202,72

−=− σx

5,5= years

σ2−x )76,0(202,7 − answer

(3)

4,95 5,65 6,35 7,05 7,75 8,45 9,15

Note If the candidate draws a bar graph, award max 2 marks

Note: If candidate works out average ( )x of midpoints, answer is 7,05 then 0 marks



3.4 They can issue a 5-year guarantee.

The average lifespan of a set is 7,02 years - which is in excess of 5 years. 98% of the sets lasted for more than 5,5 years. Very few sets have lasted less than 5 years. The number of sets of this brand that will be returnedshould be minimal if a 5-year guarantee is issued. Afrikaans Hullekan ‘n 5 jaar-waarborguitreik. Die gemiddelde lewens duur van 'n televisiestel is 7,02 jaar –wat 5 jaar oorskry. 98% van die stelle het langer as 5,5jaargehou. 'n Klein aantal stelle het vir minder as 5 jaar gehou. Die aantal stele wat terug geneem sal moet word sal minimal wees indien 'n 5 jaar- waarborg uitgereik word.

Issue the 5-year guarantee reason

(2)

kan ‘n 5 jaar-waarborg uitreik rede

(2)

[11]



QUESTION 4 4.1

OR

Sunny branch Rainybranch cycle, drive, train branches on both weather types probabilities listed outcomes listed

(5)

Sunny

Rainy

Cycle

Drive

Train

0,7

0,2

0,1 0,571428...

0,42857...

Cycle

Drive

Train

0,111111...

0,5555...

0,33333...

Outcome(Sunny, cycle)

Outcome(Sunny, drive)

Outcome(Sunny, train)

Outcome(Rainy, drive)

Outcome(Rainy, train)

Outcome(Rainy, cycle)

Sunny

Rainy

Cycle

Drive

Train

7/10

1/5

1/10 4/7

3/7

Cycle

Drive

Train

1/9

5/9

1/3

Outcome(Sunny, cycle)

Outcome(Sunny, drive)

Outcome(Sunny, train)

Outcome(Rainy, drive)

Outcome(Rainy, train)

Outcome(Rainy, cycle)



4.2.1 P(Rainy, Cycle)

211

91

73

=

×=

OR P(Rainy, Cycle)

05,020476190476,0

...1111,0...428,0

≈=

×=

or 4,76%

91

73

×

answerin any form (must be from multiplication)

(2)

4.2.2 P(Train) P(Train)

%202,0

51

31

731,0

74

==

=

×+×=

OR %20

2,051

...1428,0...05714,031

731,0

74

==

=

+=

×+×=

1,074

× and 31

73

×

addition answer (in any form)

(3)

4.3 P(Drive)

95

732,0

74

×+×=

...35238,0

10537

=

=

Vusi drives for 24510537

× = 87 days (86,333...)

Accept: 86 days OR

P(Drive) 24595

732452,0

74

××+××=

333,5828 += = 87 days (86,333...)

Accept: 86 days

2,074

× and 95

73

×

addition

10537

answer (4)

2,074

× and 95

73

×

addition 333,5828 + answer

(4)

[14]

Note:

If 91

73

+ then 0 marks



QUESTION 5 5.1.1 Number of PIN codes

= 10 × 10 × 10 × 10 × 10 = 10

5 = 100 000

10 answer

(2)

5.1.2 Number of PIN codes = 10 × 9 × 8 × 7 × 6 = 30 240 OR Number of PIN codes

!5!10

=

= 30 240

multiplication answer

(2)

!5!10

answer (2)

5.2 Number of PINs that DO NOT contain 9s

0495999999

=××××=

P(at least one 9)

)9sno(P1 −=

41,0100000590491

=

−=

OR Number of PINs that DO NOT contain 9s

Number of PINs that contain AT LEAST one 9 = 100 000 – 59 049 = 40 951 P(at least one 9)

41,010000040951

=

=

9 59 049

100000590491−

answer (4)

9 59 049 40951 answer

(4) [8]

0495999999

=××××=



QUESTION 6 6.1 321 +=+ kk TT where 11 =T , 1≥k

OR

11 2 +

+ += kkk TT where 11 =T , 1≥k

OR

( ) 112 2 +++ +−= kkkk TTTT where 11 =T , 52 =T , 1≥k

321 +=+ kk TT

11 =T 1≥k

(4)

11 2 +

+ += kkk TT

11 =T 1≥k

(4)

( ) 112 2 +++ +−= kkkk TTTT

11 =T 52 =T 1≥k

(4) 6.2

4 7 13 24 44 2 + 1 222 + 323 + 424 +

The next term of the sequence is

815244 5

=++

OR 4 7 13 24 44 ? 4 7 13 24 44 79 3 6 11 20 35 3 5 9 15 2 4 6 The next term of the sequence is 79.

answer

answer

(2) [6]

Note: This sequence can be represented by the following recursive formula:

3 21

1 113 3n nT T n n n+ = + − + where 1 4 and 1T n= ≥



QUESTION 7

7.1 Draw a point P on FG such that FP = LM and a point Q on FH

such that FQ = LN. In ∆FPQ and ∆LMN

1. LF = (given) 2. FP = LM (construction) 3. FQ = LN (construction)

∴∆FPQ ≡∆LMN (SAS)

NMLQPF = (≡∆s) But NMLHGF = (given)

HGFQPF = PQ || GH (corresponding angles =)

FHFQ

FGFP

= (PQ || GH ; Prop Th)

FHLN

FGLM

=

construction All three statements must be given ∆FPQ ≡∆LMN (SAS) PQ || GH

(7)

NMLQPF =

HGFQPF =

FHFQ

FGFP

=

G H

F

P Q

L

M N

Note: No construction constitutes a breakdown, hence no marks



7.2

TKVT

PRVP

= (PT || RK;Prop Th)

8162610264

9102

===−

=−

xx

x

x

OR

VKVT

VRVP

= (PT || RK; Prop Th)

89612

4810020104

12102

==

−=−

=−

−

xx

xxx

x

TKVT

PRVP

=

(PT || RK; Prop Th) substitution answer

(4)

VKVT

VRVP

=

(PT || RK; Prop Th) substitution answer

(4) [11]

9

2x – 10

6

9

K

T P

4 2x – 10

6

9

V

R



76° 29°

63°

29°

34°

117°

76°

117°

41°

M

P

W

T U

Q

R

c b

a

1

29°

75° 34°

1 1

2 3

d

41°

105°

QUESTION 8 8.1 ... equal to the angle subtended by the chord in the alternate

segment. answer

(1) 8.2

°= 29a (tan ch.thm)

°= 34RPQ (∠s in same seg)

°= 41c °= 76b (adj∠s on str. line)

°= 76Q1 (∠s in same seg) °=105d (ext∠ cyclic quad)

OR

°= 29a (tan ch. thm)

c=1T (tan ch. thm) °=°+ 7534c (tan ch. thm)

°= 41c

°= 76b (adj∠s on str. line)

°=105d (adj∠s on str. line) OR An alternative solution for calculating d:

°== 76TPRQ1 (∠s in same seg) TPRRPQTQPQTR ++=+d (ext∠∆)

°+°+°=°+ 76342934d °= 105d

°= 29a tan ch. thm ∠s in same seg °= 41c °= 76b ext∠ cyclic quad

(9)

°= 29a tan ch. thm c=1T tan ch. thm °=°+ 7534c tan ch. thm °= 41c °= 76b

(9)

[10]

°= 34RPQ

°= 76Q1

°=105d

°=105d



C x

A

O

B

K

T 1 2 3

4

1 2

1 2 3

2x

360° – 2x

180° – x

x x

QUESTION 9

9.1 x2BOA = (∠circ centre = 2 ∠ circumference) x2180T −°= (opp∠ cyclic quad suppl)

x2BOA = ∠circ centre = 2 ∠ circumference opp∠ cyclic quad suppl

(3) 9.2 x=TAC (∠ sum ∆)

x=1K (ext∠ cyclic quad)

1KTAC = BK || AC (corresponding ∠s =) OR

x== CK1 (ext∠ cyclic quad) x=4B (∠ sum ∆)

x== CB4 BK || CA (corresponding ∠s =) OR

x=TAC (∠ sum ∆) x−°= 180AKB (opp∠ cyclic quad)

°=+ 180AKBTAC BK || AC (coint∠s supp)

x=TAC ∠ sum ∆ x=1K ext∠ cyclic quad corresponding ∠s =

(5) x== CK1 ext∠ cyclic quad x=4B ∠ sum ∆ corresponding ∠s =

(5) x=TAC ∠ sum ∆ x−°= 180AKB opp∠ cyclic quad co-int∠s supp

(5)



9.3 In ∆BKT and ∆CAT

1. 1KTAC = (= x) 2. T is common 3. 4BTCA = (∠ sum ∆)

∆BKT ||| ∆CAT (∠∠∠)

1KTAC = T is common ∠∠∠

(3)

9.4 KTAT

KBAC

= (||| ∆s)

27

KBAC

=

KTAT

KBAC

=

||| ∆s answer

(3) [14]


Copyright reserved

QUESTION 10

10.1 DC = 13x

CD = 13 x (1)

10.2 OD = x

213

OM = x25

OD = x2

13

answer

(2) 10.3 BO = OD (radii)

AM = MB = 12 units (line from circ cent ⊥ch) 22

2

213

2512

=

+ xx (Pythagoras)

)0(22

44

144144

4169

425144

2

2

22

>=±=

=

=

=+

xxx

x

x

xx

The radius = ( )22

13

= 13 units.

MB = 12

22

2

213

2512

=

+ xx

or 222 25,4225,612 xx =+

or 222

4169

42512 xx =+

answer answer

(4) [7]

D

O

B

A

C M 4x

6,5x

2,5x

national senior certificate grade 12 · a particle moves along a straight line. the distance , s,...

Documents