maths 2011 memo

7/29/2019 maths 2011 memo

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MARKS: 150

MATHEMATICS P1

NOVEMBER 2011

MEMORANDUM

NATIONAL

SENIOR CERTIFICATE

GRADE 12

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Mathematics/P1 2 DBE/November 2011NSC Memorandum

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.QUESTION 1

1.1.1

( )( ) 02306

6

6)1(

2

2

=+

=+

=+

=+

xx

xx

xx

xx

2or3=x

OR

062 =+xx

( )( )( )12

61411

2

4

2

2

=

=

a

acbbx

2or3=x

9 standard form

9 factors

9 answers

(3)

9 standard form

9 substitution

into correct

formula

9 answers

(3)

1.1.2

0843

843

2

2

=

=

xx

xx

( ) ( )( )( )

1124

6

96164

328344)4(

2

4

2

2

+=

=

=

a

acbbx

9 standard form

9 substitution intocorrect formula

9 112

Note:Answers by inspection:

award 3/3 marks

Note:

Answer only of 2=x :

award 1/3 marks

Note:

If candidate converts

equation to linear:

award 0/3 marks

Note:If candidate uses

incorrect formula:

maximum 1/4 marks

(for standard form)

Note:

If an error in subs and

gets:804

and

Note: Penalise

1 mark for

inaccurate

di ff

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Mathematics/PI 3 DBE/November 2011NSC Memorandum

OR

0843

843

2

2

=

=

xx

xx

( ) ( )( )( )

10,1or43,2

32

8344)4(

2

4

2

2

=

=

=

a

acbbx

9 standard form

9 substitution intocorrect formula

9 answer

9 answer

(4)

1.1.3

( )( ) 01140154

514

2

2

+

+

xx

xx

xx

4

1x or 1x OR [ )

;1

4

1;

OR

OR

NOTES:

If a candidate gives an answer of4

11 x then max 3/4 marks.

If a candidate gives an answer of 11

x then max 2/4 marks.

9 factors

9 both critical

values of41 and 1

9 orOR 9 answer

(4)

41 1

x

1

4

1

x

Note: If candidate giveseither of these correct

graphical solutions but

writes down the incorrect

intervals or uses AND:

max 3/4 marks

41 1

+ 0 +0

141

Note: Penalise 1 mark

for inaccurate

rounding off to ANYnumber of decimal

places if candidate

gives decimal answers

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1.2.1

( )( ) 023065 22

=++

=++

yxyx

yxyx

3

303

=

==+

y

x

yxyx

OR

2

202

=

==+

y

x

yxyx

9 factors

99 answers

(3)

OR

Let k=

y

x

0)2)(3(

065

065

065

2

2

22

=++

=++

=+

+

=++

kk

kk

y

x

y

x

yxyx

k= 3 or k= 2

3=y

xor 2=

y

x

OR

065 22 =++ yxyx

2

5

2

5

)1(2

)6)(1(4)5(5

2

22

yyx

yyx

yyyx

=

=

=

3

3

=

=

y

x

yx

or2

2

=

=

y

x

yx

OR

9 factors

99 answers

(3)

9 substitutes

correctly into

correct formula

99 answers

(3)

Note:

If a candidate gives

3=y

xor 2=

y

x

award 2/3 marks

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3

3

=

=

y

x

yx

or2

2

=

=

y

x

yx

OR

Let k=y

x

kyx =

( ) ( )

( ) 065065

065

065

22

2222

22

22

=++

=++

=++

=++

kky

ykyyk

ykyyky

yxyx

0)2)(3(

0652

=++

=++

kk

kk

k= 3 or k= 2

3=y

xor 2=

y

x

Note: (x;y) = (0;0) is also a solution, but in this casey

xis undefined

OR

Let 1=y ,

( )( ) 0320652

=++

=++

xx

xx

2=x or 3=x

2=

y

x

or 3=

y

x

99 answers

(3)

9 factors

99 answers

(3)

9 factors

99 answers

(3)

1.2.2

4

82

83

8

=

=+

=+

y

y

yy

yx

OR

8

8

82

8

=

=+

=+

y

y

yy

yx

9 substitution

x = 3y

9 subs yx 2=

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OR

xy

yx

=

=+

8

8

38

=x

xOR 2

8=

x

x

( )

4

12

242

324

83

=

=

=

+=

=

y

x

x

xx

xx

( )

8

16

16

216

82

=

=

=

+=

=

y

x

x

xx

xx

OR

( )( )

( )( ) 08288

8

032

=++=

=+

=++

yyyx

yx

yxyx

8=y or 4=y

16=x 12=x

OR

( ) ( )

( )( ) 04803212

064242

065401664

06858

8

2

2

222

22

=++=++

=++

=+++

=++

=

yyyy

yy

yyyyy

yyyy

yx

8=y or 4=y

16=x 12=x

9 xy = 8

9 substitution

99x values

correct

9 bothy values

correct

(5)

9 yx = 8

9 substitution

99y values

correct

9 bothx values

correct

(5)

9 yx = 8

9 substitution

9factors

9 bothy values

correct

9 bothx valuescorrect

(5)

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OR

( ) ( )

03212

064242

065401664

06858

8

2

2

222

22

=++

=++

=+++

=++

=

yy

yy

yyyyy

yyyy

yx

( )( )( )

2

1612

1232141212

2

=

=y

8=y or 4=y

16=x 12=x

OR

( ) ( )

( )

( )( ) 01216019228

0384562

016646540

08685

8

2

2

222

22

=

=+

=+

=+++

=++

=

xx

xx

xx

xxxxx

xxxx

xy

4

12

=

=

y

xor

8

16

=

=

y

x

OR

xy = 8

( ) ( )

( )0384562

016646540

08685

2

222

22

=+

=+++

=++

xx

xxxxx

xxxx

9 yx = 89 substitution

9 substitutes into

correct formula

9 bothy values

correct

9 bothx values

correct

(5)

9 xy = 8

9 substitution

9factors

9 bothx values

correct

9 both y values

correct

(5)

9 xy = 8

9 substitution

Note:

If a candidate uses the

formula and replacesx

fory and then answers

are swapped:

maximum 4/5 marks

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QUESTION 2

2.1.1

18

362

324

=

=

=

x

x

xx

OR

a = 4

a + 2d= 32

2d= 28

d= 14x = 14 + 4

x = 18

OR

18

2

324=

+=x

9 2312 TTTT =

9 answer

(2)

9 a + 2d= 32 and a = 4

9 answer

(2)

9 substitutes correctly

into arithmetic mean

formula i.e. 2

324 +

9 answers

(2)

2.1.2

128

128

32

42

=

=

=

x

x

x

x

28=x OR 31,11=x OR 27

2=x

ORa = 4

4432

44

4

2

22

=

=

=

x

xar

xr

92

3

1

2

T

T

T

T=

9 1282 =x

9 both answers(surd or decimal or

exponential form)(3)

9

2

4432

=

x

9 1282 =x

Note:

If only 128=x thenpenalty 1 mark

Note:

If answer only:

award 2/2 marks

Note:If candidate writes

4x x32 only(i.e. omits equality) :

0/2 marks

Note: If candidate

writes4

x

x

32only

(i.e. omits equality) :0/2 marks

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2.2

( )

81

797161

or81

40

9841or499841

13

133

3333

3333

3

134

8234

513535251

13

1

5

,

...

...

P

-

k

k

=

=

++++=

++++=

=

=

OR

81

797161or

81

409841or499841

6561...9

1

27

1

81

1

3333

3333

3

8234

513535251

13

1

5

,

...

...

P

-

k

k

=

++++=

++++=

++++=

=

=

9 43=a or81

1

9 3=r 9 subs into correct

formula

9 answer

(4)

99 expand the sum

9 13 terms in expansion

9 answer

(4)

2.3 [ ] [ ] ( )[ ] ( )[ ]

( )[ ] ( )[ ] [ ]( )[ ] ( )[ ] ( )[ ] ( )[ ]

( )[ ]

( )[ ]dnan

S

dnan

dnadnadnadnaS

adadnadnaS

dnadnadadaaS

n

n

n

n

122

12

1212...12122

...21

12...2

+=

+=

++++++++=+++++++=

+++++++++=

OR [ ] [ ][ ]

( )[ ]dnaanTaTaTaTaS

adadTTS

TdTdadaaS

nnnnn

nnn

nnn

1

...2

...)(

)(...2

++=

++++++++=

+++++=

+++++++=

9writing out Sn9

reversing Sn

9expressing 2Sn

9 grouping to get

( )[ ]dnanSn 122 += (4)

9writing out Sn9 reversing Sn

9expressing 2Sn9 grouping to get

( )[ ]

Note: If the candidate rounds

off and gets 9841,46 (i.e.

correct to one decimal place):

DO NOT penalise for the

rounding off.

Note:Correct answer only:

1/4 marks only

Note:

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QUESTION 3

3.1 21; 24 9 219 24

(2)

3.2 12 2.3

= kkT

and so 1006632962.3126

52 ==

T

36)1(6312 =+= kkT k

and so ( ) 153326651 ==T

100663143

1531006632965152

=

=TT

OR

Consider sequence P: 3 ; 6 ; 12 12.3 = nnP

1006632962.3 12626 ==

P

Consider sequence Q: 3 ; 9 ; 15

36 = nQn

( ) 153326626 ==Q

100663143

153100663296

26265152

=

=

= QPTT

912.3k

9 52T

9 36 k 9 51T

9 answer

(5)

912.3 = nnP

9 26P

9 36 = nQn

9 26Q

9 answer

(5)

Note:

If candidate writesout all 52 terms and

gets correct answer:award 5/5 marks

Note:

If candidate usedk= 52: max 2/5

Note: if candidate

interchanges order

i.e. does 5251 TT :

max 4/5 marks

Note: writes out all52 terms and

subtracts 5251 TT :

max 4/5 marks

Note:

If candidate writes 218 =T

247 =T : award 1/2 marks

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3.3 For all Nn , kn 2= or 12 = kn for some Nk

If kn 2= :1

2 2.3== kkn TT

If 12 = kn :

( )123

36

12

=

=

=

k

k

TT kn

In either case,n

T has a factor of 3,

so is divisible by 3.

OR

12.3 = nnP

Which is a multiple of 3

)12(3

36

=

=

n

nQn

Which is also a multiple of 3

Since 12 = kn QT or kn PT 2= for all Nn ,

nT is always divisible by 3

OR

The odd terms are odd multiples of 3 and the even terms are 3 times a

power of 2. This means that all the terms are multiples of 3 and are

therefore divisible by 3.

9 factors 12.3 k

9 factors ( )123 k

(2)

9 factors 12.3 n

9 factors ( )123 n (2)

9 odd multiples of 3

9 3 times a power of2

(2)

[9]

Note:

If a candidate only

illustrates divisibility

by 3 with a specific

finite part of the

sequence, not thegeneral term:

0/2 marks

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QUESTION 4

4.1 The second, third, fourth and fifth terms are 1 ; 6 ; T4 and 14

First differences are: 7 ; T4 + 6 ; 14 T4So T4 + 6 + 7= 14 2T4 6

T4 = 11

d = 11 + 6 + 7 = 2 or 14 + 22 6 = 2

OR

( ) ( ) ( )

( ) ( )

2

36

32115

772715

23344525

=

=

+=

++++=

++=

d

d

d

dd

TTTTTTTT

OR

1

66

14210

8216

14525

75

639

124

=

=

=+

=+

=++

=+

=++

=++

a

a

ba

ba

cba

ba

cba

cba

T2 T3 T4 T5

1 -6 -14

-7 -7+d -7+2d

d d

9 7

9T4 + 69 14 T4

9 setting up

equation( ) ( ) ( )

23344525 TTTTTTTT ++=

9 answer

(5)

9 7

9 7 + d

9 7 + 2d

9 setting upequation

( ) ( ) ( )23344525 TTTTTTTT ++=

9 answer

(5)

9 124 =++ cba 9 639 =++ cba

9 14525 =++ cba

9 solved

simultaneously

9 answer

(5)

Note: Answer only (i.e.

d = 2) with no working:

3 marks

Note: Candidate gives

114 =T and 2=d only:

award 5/5 marks

Note: Candidate uses trial

and errorand shows this:

award 5/5 marks

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OR

11

333

22013

=

=

=+

y

y

yy

Second difference = 2131113 =+=+y

T1 T2 T3 T4 T5x 1 -6 y -14

1 x -7 y + 6 -14 -y

- 8 +x y +13 - 20 2y

9 79 6+y

9 y14

9 setting up

equation

9answer

(5)

4.2

T1 = 10

OR

10

21)1(12)1(

2112

21

1)12(2)1(4

1

12

7)1(575

1

21

2

=

+=

+=

=

=++

=++

=

=+

=+

=

T

nnT

c

c

cba

b

bba

a

n

OR

T1 1 6

-9 -7

2

9 method

9T1 = 10

(2)

9 method

9T1 = 10

(2)

Note: Answer only:

award 2/2 marks

Note:

If incorrect din 4.1,

2/2 CA marks for

T1 = d+ 8 (since

dT = 71 1 )

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QUESTION 5

5.1.1

1

130

6

)0(

=

=

= fy

( )1;0 OR x = 0 andy = 1

9 1=y

9 0=x (2)

5.1.2

3

63

361

13

60

=

==

=

x

x

x

x

( )0;3

9y = 0

9 63 =x

9 answer

(3)

5.1.3

x

y

x = 3

y = 1

(0 ; 1)(3; 0)

0 3

1

9 shape

9 both intercepts

correct

9horizontal asymptote9 vertical asymptote

(4)

5 1 4 33

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5.1.5

5

2

)2(0

1

5

1

132

6

51

=

=

=

=

m

y

OR

5

2

20

1

)2(0

)2()0(

51

=

+

=

=

ffm

951

9 formula

9 substitution

9 answer

(4)

9 formula

9 = )2(f51

9 substitution

9 answer

(4)

5.20

2

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QUESTION 6

x

y

B

C(0 ; 4,5)

A

f

g

O

6.1

3

22

28

820

3

=

=

=

=

x

x

x

x

7

81

82)0( 0

=

=

=f

A(3 ; 0) B(0 ; 7)

9y = 0

9 answer for A

9x = 0

9 answer for B

(4)6.2 8=y OR 08 =+y 9 answer

(1)

6.3

( ) 8828)2()(

2 +=

+=

x

xfxh

x4= or x22

9 ( 822 x 9 answer of

( )=

xh

x

4 or

x2

2 (2)

6.4 yx 4= OR yx 22= xy 4log= xy 2log2 =

xy 2log2

1= OR xy 2log=

OR4log

logxy =

9 switch

x andy

9 answer in the

formy =

(2)

6 5 l)( O l)( 9

Note: no CA marks

Note: answer only:

award 2/2 marks

Note: answer only

award 2/2 marks

Note: candidate

works outf-1and gets

( )8log2 += xy award 1/2 marks

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6.6==

5

4

3

0

)()(kk

kgkg

)5()4()3()2()1()0( gggggg +++=

3=x is the axis of symmetry ofgby symmetry

)4()2( gg = and )5()1( gg =

Answer = )3()0( gg +

= 4,5 + 0

= 4,5

OR

==

5

4

3

0

)()(kk

kgkg

)3()2()1()0()(3

0

ggggkg

k

+++==

)5()4()(5

4

ggkgk

+==

3=x is the axis of symmetry ofgby symmetry

)1()5(

)2()4(

gg

gg

=

=

5,4

05,4

)3()0(

)()(5

4

3

0

=

+=

+=

==

gg

kgkgkk

OR

( )2

2

0)30(5,4

03)(

+=

+=

a

xaxg

9 )5()4()3()2()1()0( gggggg +++=

9 )4()2( gg = and )5()1( gg =

9 )3()0( gg +

9 answer

(4)

9 expansion

9 )4()2( gg = and )5()1( gg =

9 )3()0( gg +

9 answer

(4)

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5,225,0

)5()4()(5

4

=

+=

+==

ggkgk

5,4

5,27

)()(5

4

3

0

=

=

== kk

kgkg

OR

cbag

cbag

cbag

cg

cbkakkg

cbxaxxg

++=

++=

++=

=

++=

++=

39)3(

24)2(

)1(

)0(

)(

)(

2

2

cbakgk

4614)(3

0

++==

cbag

cbag

++=

++=

925)5(

416)4(

++==

5

4

2941)(k

cbakg

cbakgkgkk

2327)()(5

4

3

0

+===

( )

2

1

95,4

0)30(5,4

03)(

2

2

=

=

+=

+=

a

a

a

xaxg

9 5,27

9 answer

(4)

99 cba 2327 +

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QUESTION 7

7.1 ( )

( )

years55,9

93,0log

2

1log

93,0log21log

93,02

1

07,012

1

=

=

=

=

=

=

n

n

PP

iPA

n

n

n

OR

( )

( )

years55,9

2

1log

93,02

1

07,012

1

93,0

=

=

=

=

=

n

n

PP

iPA

n

n

n

92

PA =

9 subs into correct

formula

9log

9 answer

(4)

Note:If candidate interchangesA and P

i.e. uses2

AP = : max 2/4 marks

Note:

If candidate uses incorrect

formula: max 1/4 marks

for2

PA =

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7.2 Radesh:

( )( )5508

5085,0100061

=

+=

+=inPA

OR

5508

25506000

55106000

56000of%5,80006

=

+=+=

+=A

300

000605,0Bonus

=

=

8508R

3005508Received

=

+=

Thandi:

( )

68,9158R

4

08,010006

1

20

=

+=

+=n

iPA

Thandi's investment is bigger.

9 8 550

9 8508R

9n = 20

9i=4

08,0

9 answer

9 choice made

(6)

7.3 =vF initial deposit with interest + annuity

91,28215R

33,0324158,2501

12

15,0

112

15,0

1700

12

15,010001

18

18

=

+=

++

+=

OR

9 0125,0or80

1or

12

15,0=i

9n = 189n = 18

9

18

12

15,010001

+

9

+

1215,0

112

15,01

700

18

9 answer

(6)

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OR

91,28215R

74,9074117,375

12

15,0

1

12

15,01

7001215,01300

19

18

=

+=

+

+

+=vF

9 0125,0or80

1or

12

15,0=i

9n = 19 (correspondingto 700)

9n = 18 (corresponding

to 300)

9

18

12

15,01300

+

9

+

12

15,0

11215,01

700

19

9 answer

(6)

[16]

QUESTION 8

8.1( )

( ) ( )

( ) ( )

( )

( )

( )hxh

hxhh

hxh

h

xhxhx

h

xhxhx

h

xhx

h

xfhxfxf

h

h

h

h

h

h

h

48lim

48lim

48lim

4484lim

424lim

44lim

lim

0

0

2

0

222

0

222

0

22

0

0

=

=

=

+=

+++=

+=

+=

9 formula

9 substitution

9 expansion

9 hx 48

9 answer

(5)

Note:

Incorrect notation:

no lim written:

penalty 2 marks

lim written before

equals sign:

penalty 1 mark

Note:

A candidate who

gives 8x only:

0/5 marks

Note:

A candidate who omits

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x

hx

h

hxh

h

hxhxf

hxhxfhxf

hxhx

hxhxf

xxf

h

h

h

8

)48(lim

)48(lim

48lim)(

48)()(

484

)(4)(

4)(

0

0

2

0

2

22

2

2

=

=

=

=

=+

=

+=+

=

9 substitution

9 expansion

9 formula

9 hx 48

9 answer

(5)

8.2.1

21

2

2

1

2

3

22

3

xx

x

xy

=

=

xx

xxdx

dy

=

=

2

2

2

3

2

3

9 1

23 x

9 2

2

3 x

9 x

(3)8.2.2

112

14)1(98)1(

1498)(

11449

)17()(

2

2

=

+=

+=

++=

+=

f

xxf

xx

xxf

OR

2)17()( += xxf

( )( )7172)(f B h h i l

9multiplication

9 x98

914

9 answer

(4)

Note:

Incorrect notation in

8.2.1 and/or 8.2.2:Penalise 1 mark

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QUESTION 9

9.1 ( ) cbxaxxxf +++= 232

( )( )( )

( )60426

1076

256

26

2

2

2

+=

+=

=++=

xx

xx

xx

baxxxf

60

21

422

==

=

b

a

a

( ) ( ) ( )

43

2518

56052152)5(23

=

+=

++=

c

c

cf

OR

( ) ( ) ( )

43

529

26022122)2(23

=

+=

++=

c

c

cf

43;60;21 === cba

OR

9 ( ) baxxxf ++= 262

99 ( )( )256 xx

9b= 609 422 =a

9 subs (5 ; 18) or (2 ; -9)

9c = 43

(7)

( )

( )

)5(2)5(6)5(

424

4240

)2(2)2(62

26

2

2

2

++=

=

++=

++=

++=

baf

ab

ba

baf

baxxxf9

( ) baxxxf ++= 262

9 0)2( =f

9 0)5( =f

9

Note:

If derivative equal to

zero is not written:

penalize once only

Note:A candidate who substitutes

the values ofa, b and c and

then checks (by substitution)

that T ( )9;2 and ( )18;5S lieon the curve:

award max 2/7 marks

Note:A candidate who substitutes the values ofa, b and c into the

function i.e. gets 4360212)( 23 += xxxxf and then shows by

substitution that T ( )9;2 and ( )18;5S are on the curve and works

out the derivative i.e. gets ( ) 60426 2 = xxxf and shows (bysubstitution into the derivative) that the turning points are atx = 2

andx = 5 (assuming what s/he sets out to prove and proving what isgiven): award max 4/7 marks as follows:

9x = 2 from ( ) 0= xf OR subsx = 2 into the derivative and gets 09x = 5 from ( ) 0= xf OR subsx = 5 into the derivative and gets 09substitution ofx = 2 in fand gets 9

9substitution ofx = 5 in f and gets 18

7/29/2019 maths 2011 memo

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OR

9)2( =f i.e. 92416 =+++ cba

724 =++ cba 18)5( =f i.e. 18525250 =+++ cba

268525 =++ cba 261321 =+ ba

( ) baxxxf ++= 26 2 and ( ) 02 =f OR ( ) 05 =f 244 =+ ba 15010 =+ ba

21

9

189

1899

72312

=

=

=

=+

a

a

a

ba

OR

21

9

189

1899

450330

=

=

=

=+

a

a

a

ba

( )

60

1803

7232112

=

=

=+

b

b

b

( ) ( )43

7602214

724

=

=++

=++

c

c

cba

OR ( ) ( )43

2686052125

268525

=

=++

=++

c

c

cba

9 92416 =+++ cba

and 18525250 =+++ cba

9 ( ) baxxxf ++= 26 2 9 0)2( =f or ( ) 05 =f

9 1899 =a

9b = 60

9 subs (5 ; 18) or (2 ; -9)

9c = 43

(7)

9.2 60426)( 2 += xxxf

( ) ( )

24

60142162

tan

=

+=m

( ) ( ) ( ) ( )2

4316012112123

=

++=f

Point of contact is (1 ; 2)

( )1242 = xy )1(24224

+=

+=

c

cxy

9 60426)(2 += xxxf

9 subs ( )1f 9 24tan =m

9f(1) = 2

9 ( )1242 = xy O 2624

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2

7

2

52

=

+=

x

x

OR

( )

2

7

23

21

=

=x

92

7=x

(2)

9( )23

21

=x

92

7=x

(2)

[14]

QUESTION 10

x

y

4 0 1

y = f/(x)

10.1 x-value of turning point:

2

3

2

14

=

+=x

2

3>x OR

;

2

3x

92

3>x OR

;

2

3

(1)

10.2 f has a local minimum atx = 4 because:9 x = 4

99 graph

(3)

f' f

4 1

4

(1;y)

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OR

Gradient of f changes from negative to positive at 4=x

OR

0)4( =f

0)4( >f so graph is concave up atx = 4, sof has a local

minimum atx = 4.

9 x = 4

9gradient negative for

4

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QUESTION 12

Note: If the wrong inequality 50x + 25y 500 is used, candidate wrongly says that there are more

learners than available seats. Maximum of 10 marks.

12.1 Nyx,

8

5002550

15

+

+

y

yx

yx

OR

8

202

15

+

+

y

xy

xy

99 15+yx

99 8y

99 5002550 + yx

(6)

12.2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

1

2

3

4

5

6

7

8

9

0

1

2

3

4

56

7

8

9

0

1

2

x

y

9 15+yx

9 5002550 + yx

9 8y

9 feasible region(4)

12.3 yxC 300600 += 9 answer

(1)12.4.1 (6 ; 8) ; (7 ; 6) ; (8 ; 4) ; (9 ; 2) and (10 ; 0)

NOTE: The gradient of the search line is1

2=m

3 marks for all correct

solutions

2 marks if only 3 or 4

correct solutions

1 mark if only 1 or 2

Red buses

Bluebuses

Note: If candidate

gives 5002550 =+ yx :

max 5/6 marks

Note: for the inequalitys marks to be awarded,

the LHS and the RHS must be correct

M th ti /P1 28 DBE/N b 2011

7/29/2019 maths 2011 memo

28/28

Mathematics/P1 28 DBE/November 2011NSC Memorandum

Copyright reserved

QUESTION 12.2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

1

2

3

4

5

6

7

8

9

1011

12

13

14

15

16

17

18

19

20

2122

x

y

Red Buses

Blue Buses

Documents

maths 2011 memo