NATIONAL SENIOR CERTIFICATE GRADE … 322 2 1 2 2 2 2 2 = =− − ± − = + − = + − = + − = n n n n n n n n n The 25th term has a value of 322./Die 25ste term se waarde is

Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

MARKS: 150 PUNTE: 150

This memorandum consists of 18 pages. Hierdie memorandum bestaan uit 18 bladsye.

MATHEMATICS P1/WISKUNDE V1

FEBRUARY/MARCH/FEBRUARIE/MAART 2016

MEMORANDUM

GRADE 12/GRAAD 12

NATIONAL SENIOR CERTIFICATE

Mathematics/P1/Wiskunde V1 2 DBE/Feb.–Mar./Feb.–Mrt. 2016 NSC/NSS – Memorandum


NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent accuracy applies in ALL aspects of the marking memorandum.

LET WEL: • Indien 'n kandidaat 'n vraag TWEE keer beantwoord, sien slegs die EERSTE poging na. • Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing.

QUESTION/VRAAG 1

1.1.1

( )( )3or4

0340122

−===+−=−−

xxxxxx

OR/OF

3or4)1(2

)12)(1(4)1()1(2

4

012

2

2

2

−=

−−−±−−=

−±−=

=−−

aacbbx

xx

factors answers

(3) substitution into formula answers

(3)

1.1.2 ( )

2133

)1(2)1)(1(433

24

013013

2

2

2

±−=

−−±−=

−±−=

=−+

=−+

aacbbx

xxxx

standard form substitution into correct formula answer

(3) 1.1.3 ( )

4or004

><<−

xxxx

OR/OF

( )( )

4or00404

><>−<−

xxxx

xx

0<x 4>x or

(3) 0<x 4>x or

(3)

0 4

0 OR/OF 0 4

+ – +

4



1.1.4

( )( )

8908888888882

112

122

=+=

−−+

=

−−+

=

aa

aaa

aax

( )( )12 −+ aa answer (check ten eights written)/tien agtstes geskryf

(2) 1.2

( ) ( )( )

( )11

7)4(27324or3

0)4)(3(01270132771101514784127215492843721572372

:)2(in )1( substitute153

)1(..........7227

2

2

222

222

22

22

=−=−=−=

===−−=+−

=+−

=−+−++−

=+−++−

=−+−−

=+−

−==+

yyyyxx

xxxx

xxxxxxx

xxxxxxxxx

yxyxxyxy

OR/OF

432

712

7111

0)1)(1(0101111060121424914

1532

74

4914

1532

72

7

:)2(in )1( substitute)2(..........153

)1(..........2

727

2

2

222

222

22

22

==

+=

+−=

=−==+−=−

=−

=−+−−++

=++

−++

=+

+

−

+

=+−

+=

=+

xx

xx

yyyyy

yyyyyy

yyyyy

yyyy

yxyx

yx

xy

72 −= xy substitution standard form factorisation x-values y-values

(6)

2

7+=

yx

substitution standard form factorisation y-values x-values

(6)



1.3

011

1

2

2

=+−

+=

+=

xyxxxy

xxy

Since x is real, this equation has real roots./Omdat x reëel is, het die vergelyking reële wortels.

( )( ) 022040

2

≥+−≥−

≥∆

yyy

2−≤y or 2≥y

012 =+− xyx 0≥∆ 42 −y factors 2−≤y 2≥y

(6) [23]

QUESTION/VRAAG 2

2.1.1 –2 0 3 7 12 2 3 4 5 1 1 1 The next term of the sequence is 12./Die volgende term in die ry is 12.

answer

(1) 2.1.2

2112

=

=

a

a

21

2213

3 12

=

=+

−=+

b

b

TTba

3

221

21

1

−=

−=++

=++

c

c

Tcba

321

21 2 −+=∴ nnTn

OR/OF

value of a

2213 =+

b

value of b

221

21

−=++ c

value of c

(5)

–2 OR/ OF

2

+ – +

2 –2



2112

=

=

a

a

1line.........................25

...............212 1

2

−=+

++=−

++=

cb

Tcb

cbnanTn

2line.........................22................220 2

−=+++=

cbTcb

line 2 – line 1:

21

=b

substitute in line 1 or substitute in line 2

25

21

−=+ c

2212 −=+

c

3−=c

321

21 2 −+=∴ nnTn

OR/OF

321

21

123

21222

)21)(23(222

)1(2

)2)(1()2)(1(2

2)2)(1()1(TT

2

2

2

211

−+=

+−+−+−=

+−+−+−=

−−+−+−=

−−+−+=

nn

nnn

nnn

nnn

dnndnn

OR/OF

2112

=

=

a

a

21

2213

3 12

=

=+

−=+

b

b

TTba

3T0 −== c

321

21 2 −+=∴ nnTn

OR/OF

value of a

cb ++=−212

cb ++= 220 value of b value of c

(5) formula substitution value of a value of b value of c

(5) value of a

2213 =+

b

value of b c=0T value of c

(5)



Since T2 = 0, ( )2−n is a factor of Tn

( )( )knnacbnanTn

−−=++=

2

2

( )( )

( )

( )( )( )

( ) ( )

( )( )

321

21

3221213

33261332

33

12

3333

32331

212

1212

2

3

1

−+=

+−=

=

−=−=−−=−−

=−

−=

−=−−==

−=

−−=−−−=−=

nn

nnT

a

kkkkk

kk

ka

kakaT

ka

kakaT

n

( )( )knnaTn −−= 2 ( )( )ka −−=− 1212 ( )( )ka −−= 3233 value of k value of a

(5)

2.1.3

( )( )

26or252

6501411

06506446

322321

21

2

2

2

2

−==

−±−=

=−+

=−+

=−+

nn

n

nnnn

nn

The 25th term has a value of 322./Die 25ste term se waarde is 322. OR/OF

( )( )26or25

0262506506446

322321

21

2

2

2

−===+−=−+

=−+

=−+

nnnnnn

nn

nn

The 25th term has a value of 322./Die 25ste term se waarde is 322. OR/OF

322321

21 2 =−+ nn

standard form substitution into quadratic formula answer

(4) 3223

21

21 2 =−+ nn

standard form factors answer

(4)



( )( )

25232

2823236446

322321

21

2

2

==−

×=−+=−+

=−+

nn

nnnn

nn

3223

21

21 2 =−+ nn

( )( )23 −+ nn 2823× answer

(4) 2.2.1

32

23:104:8:

25

5

2

=

=−=+=+

d

dTTdaTdaT

8=+ da 104 =+ da answer

(3) 2.2.2

322

328

21

=

−=

−= dTT

( )

( )

( )∑=

−+=

+=

−+=

−+=

50

150 3

21322

3202

321

322

1

n

n

nS

n

n

dnaT

OR/OF

∑=

+

=50

150 3

202n

nS

322

1 =T

answer

(2)

(2)

2.2.3 ( )[ ]

( )

33550

32150

3222

250

122

50

=

−+

=

−+=

S

dnanSn

correct substitution into correct formula answer

(3) [18]



QUESTION/VRAAG 3

3.1

107

10070

=

=r

761

.100

76,11log1

10076,11

107

10710076,11

107

1

1

1

==−

=−

=

=

=

−

−

−

nn

n

arT

n

n

nn

During the 7th year/In die 7de jaar

OR/OF

107

10070

=

=r

( )

( )

761

7,0log1176,0log1

1176,0log7,0log11176,0

10076,117,0

7,010076,11

1

1

1

==−

=−

=−=

=

=

=

−

−

−

nn

n

n

arT

n

n

nn

During the 7th year/In die 7de jaar

value of r substitution in formula for nT use of logarithms answer

(4)

value of r substitution in formula for nT use of logarithms answer

(4) 3.2 ( ) ( )

( )( )

( )( )3,0

7,01100130

7,017,01100130

terms to...4970100130

n

n

nnh

−+=

−−

+=

++++=

130

terms to...4970100 n+++ answer

(3)



3.3 Eventual height of the tree/Uiteindelike hoogte van die boom

mm 3

1390 OR mm33,463

7,01100130

=

−+=

130 + 7,01

100−

answer

(3) [10]

QUESTION/VRAAG 4

4.1 ( )2;0 answer (1)

4.2

shape ( )2;0 asymptote

(3) 4.3 ( )

( )23121

52

1 =+=

=−

−f

f

Average gradient ( ) ( )( )21

21−−−−

=ff

673

523

−=

−=

( ) 52 =−f

( )231 =f

answer

(3) 4.4 Since the asymptote of f is 1=y ,

the asymptote of ( ) ( )xfxh 3= will be 3=y . Omdat die asimptoot van f 1=y is, sal die asimptoot van ( ) ( )xfxh 3= 3=y wees.

answer

(1) [8]

x

y

( )2;0 1=y



QUESTION/VRAAG 5 5.1 ( )

( ) ( )( ) 8104:)4;0(Substitute

81:8;1pointTurning2

2

2

−−=−−

−−=−

++=

a

xay

qpxay

( ) 814

81484

2 −−=

−=−==−=−

xy

qpaa

( ) 81 2 −−= xay substitute )4;0( − 4=a p and q values

(4)

5.2 Asymptote is 22 −=⇒−= dy Substitute ( ) :8;1 −

21

8 −+

=−r

k

( )rk +−= 16 rk 66 −−= ……………….line 1 Substitute ( ) :4;0 −

24 −=−rk

2−=

rk

rk 2−= …………..………line 2 Equating lines 1 and 2:

23

64266

−=

=−−=−−

r

rrr

Substituting into line 2 or line 1:

( ) 3232 =

−−=k 3

2366 =

−−−=k

2−=d rk 66 −−= rk 2−= rr 266 −=−− value of r value of k

(6) 5.3 ( ) ( )

10 ≤≤∴≥x

xfxg x≤0 1≤x

(2) 5.4 The line y = k must pass through f twice on the positive side of

the x-axis./Die lyn y = k moet twee keer deur f aan die positiewe kant van die x-as sny.

48 −<<− k

k<−8 4−<k

(2)



5.5 cxy +−=

Substitute the intersection point of the asymptotes, i.e.

− 2;

23 :

Vervang die snypunt van die asimptote, m.a.w.

− 2;

23 :

21

21232

−−=

−=

+−=−

xy

c

c

OR/OF

xy −= is translated 23 units right and 2 units down/

xy −= transleer 23 eenhede na regs en 2 eenhede na onder ⇒

21

223

−−=

−

−−=

xy

xy

cxy +−=

c+−=−232

answer (3) xy −=

223

−

−−= xy

answer (3)

5.6 By symmetry,

−=

−+−−+=

23;

215

1232;28

23Q

2

15=x

23

−=y

(2) [19]



QUESTION/VRAAG 6

6.1

21

2

41:

41:

yxf

xyf

=

=

−

xxfxxf

xy

xy

2)(OF4)(

4

4

11

2

−=−=

±=

=

−− OR/

interchanging x and y xy 42 =

answer

(3) 6.2

x

y

f

f -1

(-2; 1)

(1; -2)

both graphs pass through (0 ; 0) shape for both one additional point on both graphs

(3)

6.3 Yes. No value of x in the domain of f -1 maps onto more than one y-value. Ja. Geen waarde van x in die definisieversameling van f -1 assosieer met meer as een y-waarde nie. OR/OF

Yes. One to one function./Ja. Een-tot-een-funksie. OR/OF

Yes. Vertical line test holds./Ja. Die vertikale lyntoets werk.

yes reason

(2)

yes reason

(2) yes reason

(2) [8]



QUESTION/VRAAG 7

7.1.1 Quarterly interest rate/Kwartaallikse rentekoers

%5,24%10

=

=

answer

(1) 7.1.2 ( )

01,0926R100

5,215000

142

=

+=

+=×

niPA

n = 8

42

1005,215000

×

+

answer (3)

7.2.1 ( )[ ]

1214,0

000100008001

1214,01

1214,011

1214,0

00010000800

1214,0

1214,01100010

000800

11

×−=

+

+−=×

+−

=

+−=

−

−

−

−

n

n

n

n

v iixP

+

×−

=−

1214,01log

100001214,0800000

1log

n

4699962,233=n Motloi can make 233 withdrawals of R10 000./Motloi kan 233 onttrekkings van R10 000 maak.

1214,0

=i

substitute into present value formula

1214,0

000100008001

1214,01 ×−=

+

−n

use of logs 233

(5) 7.2.2

(a)

428757R36,57763854,0053961

1214,0

11214,0100010

1214,01000800F -A

48

48

v

=−=

−

+

−

+=

OR/OF

n = 48 in both formulae

1214,0

=i in both formulae

substitution into both formulae answer

(4)



( )[ ]

428757R

1214,01214,01110000

11

...4699962,185

=

+−

=

+=

−

−

ii-xP

n

v

n = –185,46996…

1214,0

=i

1214,01214,01110000

...4699962,185

+−

−

answer

(4) 7.2.2

(b) Let the purchase price of the house be y./Laat die koopprys van die huis y wees.

760524R23,0428757

3,0428757

%30428757

=

=

=

=

y

yy

OR/OF

Let the purchase price of the house be y./Laat die koopprys van die huis y wees.

760524R2

10030

428757

=

×=y

answer

(1) answer

(1) [14]



QUESTION/VRAAG 8

8.1

( )x

hxh

hxhh

hxhh

xfhxfxf

hxhxfhxfhxhx

hxhxhxhxf

h

h

h

h

2

2lim

)2(lim

2lim

)()(lim)(

2)()( 42

4)2(4)()(

0

0

2

0

0

2

22

222

−=

−−=

−−=

−−=

−+=′

−−=−+

+−−−=

+++−=++−=+

→

→

→

→

OR/OF

hxfhxfxf

h

)()(lim)(0

−+=′

→

( )

hxhxhx

hxhx

h

h

442lim

44)(lim

222

0

22

0

−++−−−=

+−−++−=

→

→

h

hxhh

2

0

2lim −−=

→

( )

hhxh

h

)2lim0

−−=

→ ( )hx

h−−=

→2lim

0 x2−=

finding )( hxf + 22 hxh −− formula factorisation answer

(5) formula finding )( hxf + 22 hxh −− factorisation answer

(5) 8.2.1 xxy 103 2 +=

106 += xdxdy

x6 10

(2) 8.2.2

3

22

22

2

182)(96

96

3)(

−

−

−=′

+−=

+−=

−=

xxxfxx

xx

xxxf

22 96

xx +−

29 −x x2 318 −− x

(3)



8.3.1

factor a is)2(0

84)2(80)2(23)2(2)2( 23

−∴=

−+−=

x

f

substitution of 2 into f value of 0

(2) 8.3.2

( )( )( )( )( )6722

4219228480232)(

2

23

−−−=+−−=

−+−=

xxxxxxxxxxf

42192 2 +− xx ( )( )( )6722 −−− xxx

(2) 8.3.3

( )( )

5 or 38

0583040233080466

80466)(

2

2

2

==

=−−=+−

=+−

+−=′

xx

xxxxxx

xxxf

80466)( 2 +−=′ xxxf ( ) 0=′ xf factors x-values

(4) 8.3.4

xy

f

-84

2 3,56

(2,67 ; y)

(5 ; y)

x-intercepts y-intercept shape

(3) 8.3.5

( )( )

1or320

012030202330404664080466

2

2

2

==

=−−=+−

=+−

=+−

xx

xxxxxxxx

But x must be an integer, so 1=x at the point where tangent touches f /x moet heelgetal wees so x = 1 by punt waar die raaklyn f raak:

2584)1(80)1(23)1(2)1( 23 −=−+−== fy

( )65;065

)1(4025

−=−

+=−+=

cc

cmxy

4080466 2 =+− xx factors x =1 y-value c+=− )1(4025 answer

(6)

[27]



QUESTION/VRAAG 9

9.1

2

2

340340

rh

hr

π

π

=∴

=

substitution into volume formula answer

(2) 9.2

12

22

2

6802

34022

22 A

−+=

+=

+=

rr

rrr

rhr

π

πππ

ππ

formula substitution of h

(2)

9.3 ( )( )

cm78,3orcm46804680

6804

06804

6804rA

6802A

3

3

2

2

2

12

π=

π=

=π

=−π

−π=′

+π=

−

−

−

r

r

rr

rr

rr

rrr

rπ4 2680 −− r

π4

6803 =r

answer

(4) [8]

QUESTION/VRAAG 10 10.1.1 160 answer

(1) 10.1.2

375,08316060)(

=

=

=MP

60 answer

(2) 10.1.3

304801616016

3160160

8083

)()()()()()(

==

=

=×

=×=×

bb

b

bKoffieenManlikPKoffiePManlikP

CoffeeandMalePCoffeePMaleP

formula

16080

160

b

answer

(4)


Copyright reserved/Kopiereg voorbehou

10.2.1

720123456

!6

=×××××=

6! answer

(2) 10.2.2

2402!5

Anees/ next to sits Xoliswa waysofnumber

=×=

sitAneesangs Xoliswa lere waaropgetal mani

OR/OF Regard Xoliswa and Anees as a single entity/Beskou Xoliswa en Anees as een Number of ways in which 5 passengers can be arranged = 5! Getal maniere waarop 5 passasiers gerangskik kan word = 5! So 5! different arrangements for XA and 5! different arrangements for AX So 5! verskillende rangskikkings vir XA en 5! verskillende rangskikkings vir AX

2402!5

Anees next to sits Xoliswa waysofnumber

=×=

sitAneesangs Xoliswa lere waaropgetal mani

2!5 × answer

(2)

!5!5 + answer

(2)

10.2.3

( )

31

!6!511!5row theof endan at isMary

!6tsarrangemen ofnumber total1!5right on the row theof endan at isMary waysofnumber !51left on the row theof endan at isMary waysofnumber

=

×+×=

=×=×=

P

( )

31

!6!511!5 isMary

!61!5

51

=

×+×=

=×=×=

van die ryaan einde P

ngsrangskikkigetaltotaleisregsrydievaneindey aan die waarop Mareregetal mani

!islinksrydievaneindey aan die waarop Mareregetal mani

both LHS and RHS ways 6 ! setting up probability answer

(4)

[15] TOTAL/TOTAAL: 150





NOVEMBER 2015

MEMORANDUM

GRADE 12/GRAAD 12


Mathematics/P1/Wiskunde V1 2 DBE/November 2015 NSC/NSC – Memorandum


NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent accuracy applies in ALL aspects of the marking memorandum.

LET WEL: • Indien 'n kandidaat 'n vraag TWEE keer beantwoord, merk slegs die EERSTE poging. • Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing.

QUESTION/VRAAG 1 1.1.1

( )( )

5or4

05402092

==

=−−=+−

xx

xxxx

factors 4=x 5=x (3)

1.1.2

59,0or 26,26

735

)3(2)4)(3(4)5(5

04532

2

=−=

±−=

−−±−=

=−+

xx

x

x

xx

OR/OF

59,0or26,26

735673

65

3673

65

3625

34

3625

35

2

2

=−=

±−=

±=+

=

+

+=++

xx

x

x

x

xx

standard form substitution into correct formula answers

(4)

for adding 3625 on

both sides

6

735 ±−=x

answers

(4) 1.1.3

( )

81or2

2

32

642

3

53

5

35

35

−

−

−

−

=

=

=

=

x

x

x

x

dividing both sides by 2 52 raising RHS to

53−

answer (4)



1.1.4

( )

( )( )

only212and12,1 if

2or1021023

44222

22

2

2

2

=−=−=−=

===−−=+−

+−=−

−=−

−=−

xxxx

xxxxxx

xxxxx

xx

OR/OF

( )

only212and12,1 if

2or102or12

22

222

=∴−=−=−=

===−=−

−=−

−=−

xxxx

xxxx

xx

xx

OR/OF

only2

2and2

02and02

22

=∴

≥≤

≥−≥−

−=−

x

xx

xx

xx

squaring both sides factors rejecting x = 1

2=x (4)

squaring both sides 02or12 =−=− xx rejecting x = 1

2=x (4)

02 ≥− x 02 ≥−x 2and2 ≥≤ xx 2=x

(4) 1.1.5

0)7(072

<+<+

xxxx

( )0;707 −∈<<− xx OF/OR

factors critical values inequality or interval

(3)

+ – +

-7 OR/ OF

-7 0 0



1.2 The square of any number is always positive or zero

So for the sum of two squares to be zero, both squares must be zero, i.e. Die kwadraat van enige getal is altyd positief of nul. Vir die som van twee kwadrate om nul te wees, moet beide die kwadrate nul wees, d.i. ( ) ( )

( )15

053

05and/03

05and/03 22

==−

==−=−

=−=−

yy

xxenyx

x enyx

5

( ) 03 2 =− yx ( ) 05 2 =−x and both values (4)

1.3

( )( )

41

04101410400

2

2

2

2

−<

<+<−−

<−

<∆=−+

=+

k

kkacb

kxxkxx

OR/OF Consider the functions xxy += 2

and ky = Beskou die funksies xxy += 2

en ky =

x

y

y = k

y = x2 + x

Turning point of/Draaipunt van xxy += 2 is

−−

41;

21

kxx =+2 does not have real roots when the line ky = does not intersect xxy += 2 .

kxx =+2 het geen reële wortels as die lyn ky = nie met xxy += 2 sny nie.

Therefore 41−

<k

standard form

0<∆

( )( )k−− 1412

41−

<k

(4) sketch

x = 21

−

y = 41

−

41−

<k

(4) [26]



QUESTION/VRAAG 2 2.1

21105

1

2

=

=

=TT

r

625,0or85

2125,15

=

=T

OR/OF

625,0or85

2110

4

5

=

=T

r = 21

answer

(2) 2.2 1

2110

−

=

n

nT geometric formula substitutes a and r values into formula

(2) 2.3 r =

21

11 <<− r

Therefore the sequence converges/Die ry konvergeer

r = 21

11 <<− r

(2) 2.4 ( )

n

n

n

n

n

n rra

ra

×=

×+−=

−−=

−

−

−−

=

−−

−−

=∞

2120

21202020

2112020

211

21110

211

10

11

1S - S

OR/OF

n

n

nnnnn TTT

×=

−

=

+++

=

+++= +++∞

2120

211

12110

41

211

2110

S - S 321

211

10

−

211

21110

−

−

n

n

×+−

21202020

answer

(4)

constructing the series

+++

41

211

2110

n

211

1

−

answer (4)

[10]



QUESTION/VRAAG 3 3.1 ( )

( )( )

118883

8131

−=−+−=−+−=−+=

kkk

dkaTk

d value answer

(2)

3.2 ( ) ( ) ∑∑∑−

=

−

==

−=−+−1

0

1

01)38(11)1(8118

n

k

n

k

n

kkkk OFOR/

for general term lower and upper values in sigma notation

(2)

3.3 ( )[ ]

( ) ( )[ ]

[ ]

[ ]

nnnn

nn

nn

nn

dnann

74)74(

1482

8862

)8(1322

122

S

2 −=

−=

−=

−+−=

−+−=

−+=

OR/OF

( )[ ]

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

nnnnn

nn

nn

dnann

74443

)4(13

)8(1322

122

S

2

2

−=

−+−=

−+−=

−+−=

−+=

OR/OF

[ ]

[ ][ ]

nnnn

nn

lann

7474

11832

2S

2 −=

−=

−+−=

+=

correct substitution into formula

[ ]8862

−+− nn

(4n – 7)

(3)

correct substitution into formula ( ) ( )[ ])4(13 −+− nn

nnn 443 2 −+−

(3)

correct substitution into formula

[ ]11832

−+− nn

(4n – 7) (3)



3.4.1 292113536Q6 ++++−−= 2113536 +++−−

29+ (2)

3.4.2

64634

)128(7)128(46

6Q2

128129

=

−+−=

+−= S

2)128(4)128(76 +−− answer

(3) [12]

QUESTION/VRAAG 4 Given: 82)( 1 −= +xxf 4.1 y = – 8 answer

(1) 4.2

x-intercept y-intercept shape asymptote

(4)

4.3 ( ) 82 1 −= +−xxg OR/OF

( ) 821 1

−

=

−x

xg

answer (1)

answer

(1) [6]

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

f

y = -8

0



QUESTION/VRAAG 5 Given ( ) 32 −= xxh for 42 ≤≤− x .

x

y

h

-3

OQ

-2 4P

5.1 For x-intercepts, y = 0

( )0;5,1Q5,1

032=

=−x

x

5,1=x y = 0

(2) 5.2

57: ofDomain 53)4(2:4

73)2(2:2:

1 ≤≤−

=−==−=−−=−=

− xhyxyx

h

–7

5

57 ≤≤− x (3)

5.3

x

y

h-1

1,5

0-3

(-7; -2)

(5 ; 4)

intercepts shape endpoints

(3)

5.4 ( ) 32 −= xxh For the inverse of h,

2332

+=

−=xy

yx

( ) ( )

393

3642

332

1

==

+=−

+=−

= −

xx

xx

xx

xhxh

OR/OF

2

3+=

xy

2

332 +=−

xx

3=x

(3)



( ) 32 −= xxh h and 1−h intersect when xy =

( )

332==−=

xxxxxh

( ) xxh = xx =−32

3=x (3)

5.5

( )

91259124

32

OP

2

22

22

222

+−=

+−+=

−+=

+=

xxxxx

xx

yx

minimum a be tohasOP minimum, itsat be toOPFor 2

Vir OP om minimum te wees, moet OP2

( )

( )

units1,34or 5

3or 599

5612

565 OP oflength Minimum

56

521201210

20

2

=+

−

=

=∴

−−==−

−==

x

x

abx

dxd orOP2

'n minimum wees

OR/OF

( )

( ) ( )units8,1or34,1

06,002,1OP

6,02,1212,1

6564

3221

21equation has OP

21(given)2

22

OP

=

−−+−=

−=−=

==

−=−

−=−

−=∴

−=

=

P

P

h

y

xx

xx

xx

xy

m

m

OR/OF

222OP yx +=

substitute 32 −= xy

9125 2 +− xx x-value answer

(5) \

21

OP−

=m

equation of OP

3221

−=− xx

substitution into distance formula answer

(5)



Minimum distance between the line ocbyax =++ and the point ( )OO yx ; Minimum afstand tussen die lyn ocbyax =++ en die punt ( )OO yx ;

( ) ( )( )( )

53

12

3010222

22

=

−+

−−+=

+

++=

ba

cbyax OO

formula substitution answer (5)

5.6.1

( )⇒−∈<′= 5,1;2for 0)()( xxfxh f is decreasing on the left of Q / f is dalend links van Q.

( )⇒∈>′= 4;5,1for 0)()( xxfxh f is increasing on the right of Q / f is stygend regs van Q. OR/OF

( )( )

023

022 Since

023

23

>

′′

>=′′=′

=

′=

f

xfxh

fh

f has a local minimum at 23

=x by the second derivative test.

f het 'n lokale minimum by 23

=x deur die tweede afgeleide toets.

OR/OF

cxxxf +−= 3)( 2 f has a minimum value since a > 0/f het 'n minimum waarde omdat a > 0

decreasing left of Q increasing right of Q

(2)

023

=

′f

( ) 02 >=′′ xf

(2)

cxxxf +−= 3)( 2 explanation

(2) 5.6.2

( ) 54)4( ==′= hfm answer (1)

[19]



QUESTION/VRAAG 6 6.1.1 )18;0(T answer

(1) 6.1.2

( )( ))0;3(Q

0330182 2

=+−=+−

xxx

OR/OF

)0;3(Q9

01822

2

=

=+−

xx

equating to 0

factors coordinates of Q

(3)

equating to 0

92 =x

coordinates of Q

(3)

6.1.3 x-coordinate of S is 4,5/x-koördinaat van S is 4,5 By symmetry about the line x = 4,5/Deur simmetrie om die lyn x = 4,5: R(6 ; 0)

0);R(6

(2)

6.1.4 For all R∈x

answer (2)

6.2 If C ( )yx; is the centre of the hyperbola/As C ( )yx; die middelpunt is van die hiperbool

6+= xy and 2−=x 462 =+−=∴ y

x

y

0

y = 4

x = -2

asymptote 4=y asymptote

2−=x shape

(4) [12]



QUESTION/VRAAG 7 7.1 R450 000 answer

(1) 7.2

p.a. % 14,21 ison depreciati of rate The

1421,0

45000090,736243

1

)1(45000090,736243

)1(

4

4

=

−=

−=

−=

i

i

i

iPA n

Die waardeverminderingskoers is 14,21% p.j.

substitution of 450 000 into correct formula

substitution of ( )90,736243;4

making i the subject answer

(4) 7.3

( )

( )

R614490,66a

0,0811450000y

i1PA

:TAt

4

n

=

+=

+=

i = 0,081 and n = 4 correct substitution into formula

answer (3)

7.4 753,76 R370

,90 736 R243 66 490, R614 Value Future=

−=

( )[ ]

R9397,1212062.0

112062.01

76,370753

11F

36

=

−

+

=

−+=

x

x

iix n

v

R370 753,76 substitution into correct formula

i = 12

0,062

n = 36 answer

(5) [13]



QUESTION/VRAAG 8

8.1

( )32

32lim

)32(lim

32lim

)()(lim)(

32 )3(332)()(

332)(3)()(

0

0

2

0

0

2

222

22

2

−=

−+=

−+=

−+=

−+=′

−+=

−−−−++=−+

−−++=

+−+=+

→

→

→

→

x

hxh

hxhh

hhxhh

xfhxfxf

hhxhxxhxhxhxxfhxf

hxhxhxhxhxhxf

h

h

h

h

OR/OF

32

)32(lim

)32(lim

32lim

3332lim

)3()(3)(lim

)()(lim)(

0

0

2

0

222

0

22

0

0

−=

−+=

−+=

−+=

+−−−++=

−−+−+=

−+=′

→

→

→

→

→

→

x

hxh

hxhh

hhxhh

xxhxhxhxh

xxhxhxh

xfhxfxf

h

h

h

h

h

h

finding f(x+h) hhxh 32 2 −+ formula factorisation

answer (5)

formula finding f(x+h) hhxh 32 2 −+ factorisation answer

(5) 8.2.1

53

44

44

2

22

44

2

12

1

−

−

−=

+−=

+−=

−=

xxdxdy

xxx

xy

xxy

44 12

xx +−

4−x 34x 54 −− x

(3)



8.2.2

[ ]12

1

1)1)(1(

2

2

+=++=

−

++−

xxxD

xxxxD

x

x

factorisation simplification 12 +x (3)

[11] QUESTION/VRAAG 9 9.1 Substitute Q(2; 10) into h(x):

1line..................921024810)2()2(2 23

=+=++−=++−

baba

ba

2line.............12404120)2(2)2(30)2(:QAt

23)(

2

2

=+=++−=++−

=′++−=′

bababa

hbaxxxh

6:1lineinSubstitute23

32:2line1line

=

=

=−

b

a

a

substitute Q into h finding derivative )2(h′ equating derivative to 0 solving simultaneously for a and b (5)

9.2 Average gradient/Gemiddelde gradiënt

PQ

PQ

xxxfxf

−

−=

)()(

( ) ( ) ( ) ( )

5,3

1612311 23

−=

−+−+−−=−f

5,4)1(2

)5,3(10 gradient/ Average

=−−−−

=gradiënt Gemiddelde

formula ( )1−f = –3,5

substitution answer (4)



9.3 ( ) 633 2 ++−=′ xxxh

)12(336)(−−=

+−=′′

xxxh

The concavity changes at 21

=x .

Die konkawiteit verander by 21

=x .

linking second derivative with concavity 36 +− x explanation

(3)

9.4 The graph of h has a point of inflection at 21

=x

Die grafiek van h het 'n buigpunt by 21

=x .

OR/OF The graph of h changes from concave up to concave down at

afkonkaafna

opkonkaafvanxbyveranderhvangrafiekDiex21/

21

==

answer

(1)

answer

(1) 9.5 Gradient of g is – 12/Gradiënt van g is –12

Gradient of tangent is/Gradiënt van die raaklyn is: ( ) 633 2 ++−=′ xxxh

only2 0)2)(3(0601833

1263312)(

2

2

2

−==+−=+−

=+−

−=++−

−=′

xxxxx

xxxx

xh

( ) 6323 ++−=′ xxxh 12)( −=′ xh factors

selection of x-value (4)

[17]

21

( ) 0<′′ xh ( ) 0>′′ xh

x




3

60tan

60tan

0

0

hr

hr

rh

=∴

=

=

060tan=rh

answer

(2) 10.2

3

2

2cone

91

331

31V

h

hh

hr

π

π

π

=

=

=

( )

/cmcm82,84or27

93131

3

2

9

2

π

π

π

=

=

=

=hdhd

hdhd

V

V

formula substitution of the value of r in terms of h simplified volume answer derivative answer with units

(5) [7]

QUESTION/VRAAG 11 11.1 P(A) ×P(B)

= 0,2×0,63 = 0,126 i.e. P(A) × P(B) = P(A and B) Therefore A and B are independent/Dus is A en B onafhanklik

0,2×0,63 P(A)×P(B)=P(A and B) conclusion

(3) 11.2.1 54382377 = 7

(2) 7

11.2.2 7!=5040 7!

(2)



11.2.3 There are 3 vowels ⇒3 options for first position There are 4 consonants ⇒4 options for last position The remaining 5 letters can be arranged in 12345 ×××× ways

( ) 14404123453 =×××××× Daar is 3 klinkers ⇒3 opsies vir die eerste posisie Daar is 4 konsonante ⇒4 opsies vir die laaste posisie Die oorblywende 5 letters kan as volg gerangskik word

12345 ×××× ways/maniere ( ) 14404123453 =××××××

×3 ×4 12345 ×××× answer

(4) 11.3 Orange OO

Orange 2+t

t

2+tt

22+t

Yellow OY

22+t

Orange YO

Yellow 2+t

t

22+t

Yellow YY

( ) ( )

( )( )30343012133

4413425

2513

444

44

10052

22

22

22

10052P(Yellow)P(Yellow)(Orange)P(Orange)P

2

22

22

2

==−−=+−

++=+

=++

+++

=

+

++

+

+

=+

ttttt

ttt

ttttt

tttt

tt

There are 3 orange balls in the bag/Daar is 3 oranje balle in die sak

10052

+

+ 22 t

tt

t

+

+ 2

22

2tt

( ) ( )4413425 22 ++=+ ttt 012133 2 =+− tt 3=t

(6)

[17] TOTAL/TOTAAL: 150 marks


SENIOR CERTIFICATE EXAMINATION/ SENIORSERTIFIKAAT-EKSAMEN

MARKS/PUNTE: 150

This memorandum consists of 19 pages./ Hierdie memorandum bestaan uit 19 bladsye.


2015

MEMORANDUM

Mathematics P1/Wiskunde V1 2 DBE/2015 NSC/NSS – Memorandum


NOTE: • If a candidate answers a QUESTION TWICE, only mark the FIRST attempt. • Consistent accuracy applies in all aspects of the marking memorandum.

LET WEL: • Indien ’n kandidaat ’n vraag TWEE keer beantwoord, sien slegs die EERSTE poging na. • Volgehoue akkuraatheid is DEURGAANS op ALLE aspekte van die memorandum van

toepassing.

QUESTION/VRAAG 1

1.1.1 ( )1or0

01==

=−xx

xx x = 0 1=x

(2) 1.1.2

( )

87,2or 87,04

564

)2(2)5)(2(4)4(4

05422

2

=−=

±=

−−−±−−=

=−−

xx

x

xx

OR/OF

( )

87,2or 87,0271

271

1251

0252

2

2

=−=

±=∴

±=−

+=−

=−−

xx

x

x

x

xx

correct substitution into correct formula answers

(3) completing the square/ voltooiing van die vierkant answers

(3) 1.1.3

355125

15

3

−==

=

−

x

x

x

OR/OF

35− answer

(2)



33

51

51

12515

3

−==−

=

=

−

xx

x

x

OR / OF

3125

1log

12515

5

−=

=

=

x

x

OR / OF

3035515511255125

15

03

3

−==+=

=⋅

=⋅

=

+

xx

x

x

x

x

3

51

51

=

−x

answer (2)

use of logs answer

(2)

35 answer

(2) 1.1.4 ( )( ) 023 >−− xx

32 << x

OR/OF ( )( )( )( ) 023

023<−−>−−

xxxx

32 << x

critical values solves an equality answer (3) critical values x<2 3<x (3)

2 OR/OF

3

– + –

2 3

2 2 3 OR/OF

3

– + –



1.2.1 3=x answer

(1) 1.2.2

( )( )( )( )

( )101

0120431

4313

41

2

2

==−

=+−

=+−+−=−+−−

=+

xx

xxxx

xxx

x

( )( ) 431 −=−+ xx standard form factors answer

(4)

1.2.3 Yes, the graph of f and the graph of g have equal roots at x = 1. Ja, die grafiek van f en die grafiek van g het gelyke wortels by x = 1. OR/OF Yes, the graphs of f and g intersect in only one point, which is at x = 1. /Ja die grafieke van f en g sny in slegs een punt wat by x = 1 is.

yes reason

(2)

yes reason

(2)



1.3 Speed/Spoed = y km/h Distance/Afstand = x km

Time/Tyd = yx

Speed/Spoed = 2

3y km/h

Distance/Afstand = x km Time/Tyd =

yx

yx

32

23

=

OR/OF

S D T

A to/na B y x yx

B to/na A 2

3y x yx

32

Average speed travelled/Gemiddelde spoed afgelê:

km/h5

656

323

232

2 taken/timeTotal

/ travelleddistanceTotal

yxxy

yxx

xyx

yx

xgeneem tyd Totale

gereis afstand Totale

=

=

+=

+=

time from A

to B is yx

time from B

to A is yx

32

x2

yx

yx

32

+

xxy

56

km/h5

6y

(6)

[23]

A

A

B

B



QUESTION/VRAAG 2

2.1 314 =T answer (1)

2.2

( )( )( )59

9141

59

−=−+=−+=

−=

nn

dnaT

nT

n

n

OFOR/

n9 – 5

(2)

4 ( )( )91−n

(2)

2.3

( ) ( )

( )

5500

440225

]18)24(42[225

])1(2[2

184

.......40;22;4

25

=

=

+=

−+=

==

dnanS

da

OR/OF

( )

[ ]5500

4364225436

7499

25

25

=

+=

=−=

S

T

OR/OF

5500436418400382364346328310292274256

23822020218416614813011294765840224

=++++++++++

++++++++++++++

d = 18 correct substitution into correct formula answer

(4)

43625 =T substitution into correct formula answer

(4)

43625 =T expands whole series answer

(4)



2.4 -6 -2 11 33 4 13 22 9 9

12

1929

12

19

62

19294

293

29

64392

2 −−=

−=−=

−=+−=+

=

−=++=+=

nnT

cb

cba

cbabaa

n

OR/OF

( ) ( )( )

( )( ) ( )( )( )

12

1929

218279446

2921416

221

1

2

2

211

−−=

+−+−+−=

−−+−+−=

−−+−+=

nn

nnn

nnn

dnndnTTn

OR/OF

2992

2

=

=++=

a

acbnanTn

( ) ( )

( ) ( )

12

19291219

1 line 2line...2

274

2line...2182

2229T

1line...296

1129

2

22

21

−−=

−=

−=

+=

++=−

++

=

++=−

++

=

nnT

c

b

- b

cb

cb

cb

cbT

n

sets up quadratic sequence

29

=a

2

19−=b

1−=c

(4) formula & substitution

29

=a

2

19−=b

1−=c (4)

sets up quadratic sequence

29

=a

2

19−=b

1−=c

(4) [11]



QUESTION/VRAAG 3

3.1 Given ( )∑

=−

21

43

p

p

3.1.1 ( )( )( ) 7293

2433

813

63

52

41

=−=

−=−=

=−=

T

T

T

81 -243 and 729 (2)

3.1.2 r = –3 answer (1)

3.1.3 ( )∑∞

=−4

3p

pwill NOT converge/sal NIE konvergeer.

To converge/om te konvergeer, 11 <<− r and r = – 3 OR/OF

( )∑∞

=−4

3p

pwill NOT converge/sal NIE konvergeer.

Because/Omdat 1−<r

NOT converge/ NIE konvergeer we do not have 11 <<− r

(2) NOT converge/ NIE konvergeer 1−<r

(2)

3.1.4 ( )( )( )

( )( )( )

x

xS

x

xS

882264845713

1381

8822648457313181

18

18

18

18

−=−−−−

=

−=−−−−

=

OFOR/

18=n xa 81= correct substitution into correct formula

(3) 18=n xa 81= correct substitution into correct formula

(3) 3.2.1

( )

( )

( )2

02

044124168

3 and 4 and1244

51241

512465

124;5;6

2

2

2

−==+

=++

+=++

−≥−≥+=+

−+=−

−+=−−

+−

xx

xxxxx

xxxx

xx

xx

xx

2312 TTTT −=−

1244 +=+ xx factorisation answer (5)

1241682 +=++ xxx



3.2.2

( )

( )19

3983

21224

58)2(6

10

3

2

1

−=−+=

−==

+−=

==−−=

Td

T

TT

3−=d correct substitution into correct formula answer

(3) [16]

QUESTION/VRAAG 4

4.1 2, −≠yR OR/OF ( ) ( )∞−∪−∞− ;22;

2−≠y (2)

( )2;−∞− ( )∞− ;2

(2)

4.2

325

21

5

21

)(

=+=

−−

=−

−−

=

aa

ax

axg

substitution of the point (0; –5) in to g(x) answer

(2) 4.3 For g, asymptotes intersect at/Vir g, asimptote sny by ( )2;1 −

∴ For/Vir ( ) 73 +−= xgy , asymptotes will intersect at/ asimptote sal sny by ( )72;31 +−+ i.e./d.i. at/by ( )5;4 OR/OF

( )

( )5;4

54

3

7213

373

21

)(

+−

=

+−−−

=

+−=

−−

=

x

x

xgyx

axg

( )2;1 − for g x = 4 y = 5

(3) subs x = 4 y = 5

(3) [ 7 ]



QUESTION/VRAAG 5

5.1

164

41

2

2

==

=

−

y

substitution answer

(2)

5.2

xxy

xf

y

y

x

441

1

logyorlog41:

41

−==

=

=

−

interchange x and y answer

(2) 5.3 f and f –1 are symmetrical about the line y = x, to obtain f –1, reflect f

in the line y = x. f en f –1 is simmetries om die lyn y = x, om dus f –1 te kry reflekteer f in die lyn y = x. OR/OF The x and y-coordinates of points on f may be swopped around to obtain the coordinates of the points on f –1. Two points that lie on the graph of f are (0 ; 1) and (–2 ; 16). The corresponding points that will lie on 1−f will therefore be (1 ; 0) and (16 ; –2). Die x- en y-koördinate van punte op f mag omgeruil word om die koördinate van punte op f –1 te kry. Twee punte op die grafiek van f is (0 ; 1) en (–2 ; 16). Die ooreenstemmende punte op 1−f sal dus (1 ; 0) and/en (16 ; –2) wees.

reflect in y = x

(1)

swop x and y

(1)

5.4

x

y

f -1

(1 ; 0)

shape of f –1

x-int of f –1at 1

(2)



5.5 0>x

OR/OF ( )∞;0

0>x (1)

( )∞;0

(1) 5.6 ( )

( )]16;0(or160

216 5.1, From2

1

1

∈≤<−=

−≥−

−

xxf

xf

0>x 16≤x

(2) 5.7.1

21

=q (using a calculator/gebruik 'n sakrekenaar)

OR/OF Without a calculator (not necessary)/Sonder sakrekenaar (nie nodig)

2112

2221

41

21log

12

41

=

==

=

=

−−

q

q

q

q

q

OFR/O

21

2log22log

41log

21log

21log

41

=

−−

=

=

=

q

q

q

q

21

=q

(1)

21

=q

(1)

5.7.2 At the intersection point of f and f -1, xy = (by symmetry). Thus need only solve ( ) xxf =−1 (instead of ( ) ( )xfxf 1−= ) By die snypunt van f en f -1 xy =, (deur simmetrie). Slegs nodig om ( ) xxf =−1 op te los (in plaas van ( ) ( )xfxf 1−=

=

=

=

=

21;

212121

5.7.1 from21

21log

log

41

41

y

x

xx

OR/OF

By/Van 5.7.1, 21log

21

41=

Which means that

21;

21 lies on the graph of 1−f ./

21log

21

41=

21

=x

21

=y

(3)

21log

21

41=



Wat beteken

21;

21 lê op die grafiek van 1−f .

But clearly,

21;

21 lies on xy = /Maar,

21;

21 lê op xy =

Hence

21;

21 is the intersection point of f and f -1 /Dus is

21;

21

die snypunt van f en f -1

21

=x

21

=y

(3) [14]

QUESTION/VRAAG 6

6.1

( )( )

units7AB2or5

0250103

030932

2

==−==−+=−+

=+−−

xxxx

xxxx

03093 2 =+−− xx factors answers AB = 7

(4) 6.2

( )( )3or2

0230601833

12123093

2

2

2

=−==+−=−−

=++−

+−=+−−

xxxxxx

xxxxx

At K, 0>x , hence ( ) 2412312 −=+−=y K ( )24;3 −

equating of equations 062 =−− xx factors x =3 y= –24

(5) 6.3 ( ) ( )

3or2 ≥−≤≤

xxxgxf

OR/OF ( ) ( )

);3[or]2;( ∞−−∞∈≤

xxgxf

OR/OF

( )

( )( ) 0230601833012123093

2

2

2

≥+−≥−−

≤++−

≤+−−+−−

xxxx

xxxxx

3or2 ≥−≤ xx

2−≤x 3≥x or

(3) ]2;( −−∞ );3[ ∞ or

(3) 2−≤x 3≥x or

(3)



6.4

( ) ( )

( )

4318

475

431818

213

213

CD/lengthMax CD/lengthMax

4318

213

1843

213

21

21

1821

213036

323

183CD02

1833)1212(3093CD

2

2

2

22

2

2

2

=

=

=+

+

−=

+

−−=

++

−−===

+

−

−−==+−

−−

=

+−==′−=

++−=

+−−+−−=

lengte Maks lengte Maks

x

xx

xx

xx-xfa

bx

xxxxx

OF

OFOF

OR/

OR/OR/

gf yy −= CD

1833 2 ++− xx method

21

=x max length

4318or

475CD =

(5) [17]

QUESTION/VRAAG 7

7.1 Anisha:

( )( )

00018R9005085,0100012

000 R12 of %5,71 value/investment Final

=+×+=

++= inPwaardebeleggings Finale

Lindiwe:

( )

54,27318R4085,0100012

1

/ valueinvestment Final

20

=

+=

+= niP

waardebeleggings Finale

Therefore Lindiwe will have a larger final amount./ Lindiwe sal 'n groter finale bedrag hê.

900 or 000 R12 of %5,7 ( )50,085100012 ×+ 00018R

20

40,085100012

+

54,18273R conclusion

(6)



7.2 ( )( )

( )

( )

years8000120

57,41611log

876,0000120

57,61141124,0100012057,61141

1

876,0

=

=

=

−=

−=

n

iPA

n

n

n

OR/OF ( )

( )

( )

( )

years8876,0log000120

57,41611log

876,0log000120

57,61141log

876,0000120

57,61141124,0100012057,61141

1

=

=

=

=

−=

−=

n

n

iPA

n

n

n

n

formula substitution

( ) 00012057,41611log 876,0=n

answer

(4)

formula substitution

876,0log000120

57,41611log=n

answer (4)

7.3

( ) ( )[ ]

22,96728R467,23022755,7366

1215,0

11215,01800

1215,015000

111

amount / final

24

24

=+=

−

+

+

+=

−+++=

iixiP

bedrag finale n

n

1215,0

=i

n = 24 (subs) (adding)

1215,0

11215,01800

1215,015000

24

24

−

+

+

+

answer

(5) [15]



QUESTION/VRAAG 8

8.1

( )( )

( )

( )

( )

( )

( )

( )

2

0

0

4

4lim

)()(lim)(

4

4

4)()(

4

444

44

44)()(

4)(

x

hxx

hxfhxfxf

hxx

hxxhh

hhxx

h

hxfhxf

hxxh

hxxhxx

hxxhxx-xhx

xfhxf

hxhxf

h

h

−=

+−

=

−+=′

+−

+−

=

+−

=−+

+−

=

+−−

=

++

=

−+

=−+

+=+

→

→

OR/OF

xhx44

−+

( )

( )hxxhxx

++− 44

( )hxx +− 4

formula answer

(5)



( )( )

( )

( )

( )

2

0

0

0

0

0

0

4

4lim

4lim

444lim

44

lim

44

lim

)()(lim)(

x

hxx

hxxhh

hxhxhxx

hhxx

hxxh

xhx

hxfhxfxf

h

h

h

h

h

h

−=

+−

=

+−

=

+−−

=

++−

=

−+=

−+=′

→

→

→

→

→

→

formula subst. into formula

( )

( )hxxhxx

++− 44

( )hxx +− 4

answer

(5)

8.2.1

510

255 2

+=

++=

xdxdy

xxy

10x 5

(2) 8.2.2

21

32

21

21

31

32

3 2

−=

−=

−

−

x

xxD

xxD

x

x

32

x

31

32 −

x

21

−

(3) 8.3 ( )

( ) 232

2

3

+=′

+=

xxpxxxp

0or / 03 22 ≥≥ xofx for all/vir alle R∈x 0223 2 >≥+∴ x for all/vir alle R∈x

i.e. ( ) 0>′ xp for all/vir alle R∈x i.e. all tangents to p have gradient greater than (or equal to) 2. Thus there is no tangent to p that has negative gradient. Alle raaklyne aan p sal dus ’n gradiënt groter (of gelyk aan) 2 hê. Daar sal dus geen raaklyn aan p wees met ’n negatiewe gradiënt nie.

( ) 23 2 +=′ xxp states & justifies ( ) 0>′ xp linking derivative to gradient of tangent/verband tussen gradiënt en afgeleide

(3) [13]



QUESTION/VRAAG 9

9.1 1=x or 3=x 3=x 1=x

(2) 9.2 31 << x answer

(2) 9.3 For a point x close to 3/Vir ’n punt naby aan 3:

If 3<x , ( ) 0<′ xf ⇒f decreasing/dalend If 3>x , ( ) 0>′ xf ⇒f increasing/stygend Therefore: f has a local minimum at/f het lokale minimum by 3=x OR/OF At x = 3, the gradient function changes from negative to positive therefore the function will have a local minimum point at x = 3/ By x = 3 verander die gradiëntfunksie van negatief na positief dus sal die funksie ’n lokale minimum punt hê by x = 3. OR/OF ( ) 03 =′f and ( ) 03 >′′f therefore the function will have a local

minimum point at x = 3 / ( ) 03 >′′f dus sal die funksie ’n lokale minimum punt hê by x = 3.

f dec for 3<x f dalend vir 3<x f incr for 3>x f stygend vir 3>x 3=x local min

(2)

at x = 3 gradient changes from neg to pos 3=x local min

(2)

( ) 03 >′′f 3=x local min

(2) 9.4 ( ) ( )xfvan draaipunt dieby xf ′=′′ of/point turningat the0

Using symmetry/Deur simmetrie 2

31+=x

= 2

answer

(1) 9.5 Concave up if/Konkaaf op as ( ) 0>′′ xf

x > 2 ( ) 0>′′ xf answer

(2) [9]

1 3



QUESTION/VRAAG 10

Given: 30009)( 23 +−= tttM ; 300 ≤≤ t 10.1 ( ) ( )

kggM

3or300030000900 23

=+−=

answer

(1) 10.2

( )9or0

0909300030009

2

23

23

===−

=−

=+−

tttt

tttt

Baby’s mass will return to the birth mass on the 9th day/ Baba se massa keer terug na massa by geboorte op die 9de dag.

( ) 3000=tM 093 =− tt factors 9=t

(4) 10.3 ( )

( )6or0

06301830

2

===−=−

=′

tttt

tttM

Baby’s mass will be a minimum on the 6th day/ Baba se massa sal ’n minimum wees op die 6de dag.

( ) 0=′ tM tt 183 2 − factors 6=t

(4)

10.4 ( )( )

31860186183 2

=−=−=′′−=′

ttttM

tttM

OR / OF

32

60 :/symmetry Using

=

+=t

simmitrieDeur

186 −t answer

(2)

2

60+

answer (2)

[11]




x−372 for Hockey only x−288 for Rugby only 56 outside of Hockey & Rugby

(3) 11.1.2 ( ) ( )

11660071660056288372

==−=+−++−

xx

xxx

OR/OF

( )

( ) ( ) ( ) ( )

116544288372

288372544RandHRHRor H

54456600Ror H

=−+=−+=−+=

=−=

xx

nnnn

n

setting up the equation answer

(2)

setting up the equation answer

(2) 11.1.3 No, they are not mutually exclusive.

There is an intersection between the two sets/ Nee, hul is nie onderling uitsluitend nie. Daar is ‘n snyding tussen die twee stelle

No justification

(2)

11.2.1 120!5 = answer (1)

11.2.2 12

!3!21=

×× !2

!3 answer

(3) 11.2.3

2311

!112!6!5

=

××

2!6!5 ×× division by !11

answer (3)


n(S)=600 Hockey Rugby

x 372 – x 288 – x

56





FEBRUARY/MARCH/FEBRUARIE/MAART 2015

MEMORANDUM


GRADE/GRAAD 12

Mathematics P1/Wiskunde V1 2 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum


NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent accuracy applies in all aspects of the marking memorandum.

LET WEL:

• Indien ’n kandidaat ’n vraag TWEE keer beantwoord, merk slegs die EERSTE poging. • Volgehoue akkuraatheid is DEURGAANS op ALLE aspekte van die memorandum van

toepassing. QUESTION/VRAAG 1 1.1.1

5or40)5)(4(=−=∴=−+

xxxx

factors/faktore answers/antwoorde

(2) 1.1.2 07112 2 =+− xx

( ) ( ) ( )( )( )

73,0or77,422

7241111

24

2

2

=

−−±−−=

−±−=

aacbb

x

OR/OF

73,077,4465

411or

465

411

1665

411

1656121

411

016121

27

411

02

11.21

27

211.

21

211

027

211

07112

2

2

222

2

2

==

−=+=∴

±=−

−=

−

=−+

−

=

−+

+−

=+−

=+−

xx

xx

x

x

x

xx

xx

xx

substitution into correct formula/substitusie in korrekte formule 4,77 0,73

(3) correct completion of the square/korrekte voltooiing van die vierkant 4,77 0,73

(3)



1.1.3

4or/51

0)4)(15(04215 2

><

>−−>+−

xofx

xxxx

or

standard form/ standaardvorm factors/faktore

51

<x

4>x of

(5) 1.1.4 162.622 =− xx

( )( ) 022820162.622

=+−

=−−xx

xx

22or/22 3 −== xx of 22orSolutionNoor/3 −≠= xofx

factors/faktore no solution to/ geen oplossing 22 −=x 322 =x answer/antw.

(4)

51

4 + – +

51

4



1.2

( ) ( )

( )( ) 01202

063371442

71212

12

2

2

222

22

=+−=−−

=−−

=+−++−

=−+−−

−=

xxxx

xxxxxxx

xxxx

xy

2=x or/of 1−=x 3=y or/of 3−=y

OR/OF

1221

23

21

23

3or30)3)(3(

090273

02842212:4

7224

124

721

221

2

21

2

2

2

222

222

22

−==

+−

=+=∴

−==∴=+−

=−

=−

=−+−−++×

=+−−++

=+

+−

+

+=

xx

xx

yyyy

yy

yyyyy

yyyyy

yyyy

yx

y the subject/ die onderwerp substitution/substitusie simplification/vereenv. factors/faktore x-values/waardes y-values/waardes

(6) x the subject/ die onderwerp substitution/substitusie simplification/vereenv. factors/faktore y-values/waardes x-values/waardes

(6)

1.3.1 k = – 2 or/of k = 2 answer/antw. (2)

1.3.2 k = – 3 – 3 (1)



1.4

( )

1006

2012

2012

22012

20122014

7.24.7

1248.7

12177

1277

=

=

=

−=

−

a = 2; b = 1006

( )12

177 22012 −

4.72012 10067.2 OR/OF a = 2 b = 1006

(4) [27]

QUESTION/VRAAG 2 2.1 ( ) ( ) ( )

( ) ( ) ( )( )( )

( )[ ]dnanS

dnanSadnadnadnaS

dnadadaaS

n

n

n

n

122

122...321

1...2

−+=

−+=++−++−++−+=

−+++++++=

first series/eerste reeks series reversed/reeks omgekeer sum/som division/deling

(4) 2.2

...919497)3100(50

1+++=−∑

=kk

1175

)]3(49)97(2[2

50

])1(2[2

5011503

971

=

−+=

−+=

=+−=−===

dnanS

nd

aT

n

OR/OF

( )

1175

]5097[2

50

][2

50115050503100

971

=

−=

+=

=+−=−=−=

==

lanS

nl

aT

n

a = 97 d = – 3 n = 50 answer/antwoord

(4)

a = 97 50−=l n = 50 answer/antwoord

(4)



2.3.1 (a) 45 TT − = 25 answer/antwoord (1)

2.3.1 (b) ( )( )415

616976970

=−+=− TT

n = 69 ( )( )61697 −+ answer/antw. (3)

2.3.2

( )[ ]9440

)6(1941522

20 terms20 to...421415

)(...)()( 8889707169706989

=

+=

++=−++−+−=− TTTTTTTT

−= 8969 TT (sum of the differences from/som van die verskille van 69T to 89T )

1415494402359469

=−=T

OR/OF

7 13 19 25

6 6 6

362

==∴

aa

273−=

=+bba

923594)89(2)89(3 2

89

=∴=+−=

ccT

141549)69(2)69(3

923

69

269

2

=∴+−=∴

+−=∴

TT

nnTn

expansion/uitbreiding n = 20 method/metode a = 415 answer/antwoord

(5)

a and/en b 89T (subst n = 89) Tn substitution/substitusie answer/antwoord

(5)



OR/OF

7 13 19 25

6 6 6

362

==∴

aa

213

11

167

01

−==+

=−++=−

=−

bb

ccbaTT

923594)89(2)89(3 2

89

=∴=+−=

ccT

141549)69(2)69(3

923

69

269

2

=∴+−=∴

+−=∴

TT

nnTn

a and/en b (subst n = 89) Tn substitution/substitusie answer/antwoord

(5) [17]

OR/OF

( )

[ ]

[ ]

( )

1415414144)66715(34

10

10235842359423584

687142

88

)1(22

)1(67

69

68

1169

1

88

11189

1

=∴=×+=

−=−∴

=−=∴=

×+=

−+=

−=−∴

−+=−

∑

∑

=+

=+

+

T

TTT

T

dnan

TTTT

nTT

nnn

nnn

nn

formula/formule value of/waarde van S88 first term value/ eerste term waarde substitution/substitusie answer/antwoord

(5) [17]

89T



QUESTION 3 3.1

9,045

5,40==r

( )

12,14...12147682,14

9,045 11212

=== −T

r = 0,9 substitution into correct formula/substitusie in korrekte formule answer/antwoord

(3) 3.2 9,0=r

19,01 <<−

answer/antwoord

(1)

3.3

4509,01

45

=−

=

∞

∞

S

S

substitution/substitusie 450

(2) 3.4 1<−∞ nSS

( )( )

( )( )( )

( )

( )

( )

( )...98,579,0log

4501log

4501log9,0log.

4501log9,0log

45019,0

19,0450

9,014504509,019,0145450

>

>

<

<

<

<

−−=−

−−

−=−

∞

∞

n

n

n

SS

SS

n

n

n

nn

n

n

Smallest value/Kleinste waarde: n = 58

( )( )

9,019,0145450

−−

−n

( )45019,0 =n

introducing/gebruik logs making n the subject/maak n die onderwerp n = 58 (5)

[11]




12−=−=

yx

2−=x 1−=y (2)

4.2.1

2

120

6)0(

=

−+

=g

y-intercept/afsnit (0 ; 2)

answer/antwoord (1)

4.2.2

462

261

12

60

==+

+=

−+

=

xx

x

x

x-intercept/afsnit (4 ; 0)

equating to/stel gelyk aan 0 answer/antwoord

(2) 4.3

asymptotes/asimptote intercepts/afsnitte shape/vorm

(3)

4.4 ( )21 +−=+ xy 3−−= xy

OR/OF Using general formula/Gebruik algemene formule:

( )

31)2(

−−=−+−=++−=

xyxy

qpxy

m = – 1 substitution of (–2 ; –1) answer

(3)

formula/formule substitution of p and q values/substitusie van p- en q-waardes answer/antwoord (3)

4.5 2−>x answer (2) [13]




39 2

==

aa

OR/OF

339

9log2log)(

22

1

=∴==

==−

aa

xxf

a

a

29 a= 3=a

(2)

29 a= 3=a

(2)

5.2 xxg −= 3)(

OR/OF x

xg

=

31)(

answer/antwoord (1)

answer/antwoord

(1) 5.3 9≥x

OR/OF

993

2loglog)(

23

31

≥∴==

==−

xx

xxxf

OR/OF

93

2log2

3

≥∴≥

≥

xx

x

answer/antwoord (2)

answer/antwoord

(2) answer/antwoord

(2) 5.4 Yes/Ja. For every y-value there is only one x such that/Vir

elke y-waarde is daar slegs een x sodanig dat ( )xfy = . OR/OF Yes/Ja. f is a one-to-one relation/is 'n een-tot-een-relasie.

Yes/Ja Reason/Rede (2) Yes/Ja Reason/Rede (2) [7]



QUESTION/VRAAG 6 6.1 23 ≤≤− x critical values/

kritiese waardes notation/notasie

(2) 6.2 ( )( )

( )( )1222232

248

)21)(31(8)2)(3(

:

2

21

−+=

−+==

−=−−+=−−+=

−−=

xxyxxy

aa

axxay

xxxxayf

b = 2 and/en c = – 12 OR/OF

12 and2

12222112

2122

2112

412

2112

212

5,12225)2(

4250

248:)2()1(

)2.....(498

2118

)1....(4250

2120

21

2

2

2

2

2

2

2

−==∴

−+=

−++=

−

++=

−

+=

−=−=−=

==−

+=−→+

+=−

+=→+

+=

+

+=

cb

xxy

xxy

xxy

xy

q

aa

qaqa

qaqa

qxay

OR/OF

)2)(3( −+= xxay substitute/vervang (1 ; – 8) a = 2 b = 2 and/en c = – 12 (5) equation/vergelyking 1 equation/vergelyking 2 a = 2 substitution/substitusie b = 2 and/en c = – 12 (5)



( )

( )

122

284:)2()1(

)2(828:8;1

)1(06039:0;3

0212

21

2)(

−=∴=⇒=∴

=−

−=+∴−=++−

=+∴=+−−

=∴

=+

−=

−′

+=′

cb

aa

cacba

cacba

ba

baf

baxxf

equation/vergelyking 1 equation/vergelyking 2 a = 2 b = 2 c = – 12 (5)

6.3

−−

−=

−−=

−=−=

−=

2112;

21TP

2112

12121

21

)2(22

2

y

y

x

abx

OR/OF

21

−=x

substitution/substitusie y-value/waarde

(3)



OR/OF

−−

−

+=

−

+=

−−

++=

−+=

5,12;21TP

5,12212

25,6212

1.2161.

212

]6[2

2

2

222

2

x

x

xxy

xxy

−−

−=

−

−+

−=

−=+−

=

5,12;21TP

2112

12212

212

21

223

y

y

x

OR/OF

−−

−=−

−+

−=∴

−=

−==+

+=

−+==

225;

21TP

22512

212

212

2124

02424)(

1222)(

2

/

2

y

x

xx

xxfxxyxf

method/metode x-value/waarde y-value/waarde

(3) method/metode x-value/waarde y-value/waarde

(3)

method/metode x-value/waarde y-value/waarde (3)

6.4 2

13=x

answer/i

(2) 6.5

62)1(4)1(

24)(/

=+=′=

+=

mfm

xxf

24/ += xy subst. x = 1 answer/antwoord

(3) [15]



QUESTION/VRAAG 7 7.1.1 )1244(400R ××

211200R=

)1244(400R ×× R211200

(2) 7.1.2 ( )[ ]

iixF

n 11 −+=

1208,0

11208,01400

528

−

+

=

42,5249431R=

400=x 528=n

1208,0

=i

substitution into correct formula/substitusie in korrekte formule

answer/antwoord (5)

7.1.3

iixP

n ])1(1[ −+−=

121,012

1,011

2000000

300

+−

=

−

x

01,17418R=x OR/OF

01,18174R12

1,0

112

1,01

121,012000000

300

300

=

−

+

=

+

x

x

P = 2000000

300=n and/en 12

1,0=i

substituting into correct formula/substitusie in korrekte formule answer/antwoord

(4) P = 2000000

300=n and/en 12

1,0=i

equating/stel gelyk answer/antwoord

(4) 7.2 Let PX and PY be the populations of the two towns at the

beginning of 2010./Laat PX en PY die bevolkings wees van die twee dorpe aan die begin van 2010.

YX AA = 33 )12,01()08,01( +=− YX PP

3

3

)08,01()12,01(

−+

=Y

X

PP

...778,0...404,1

=

1:8,1=

equating/stel gelyk 3)08,01( −= XX PA 3)12,01( += YY PA answer/antwoord (4)

[15]




( )

x

hxh

hxhh

hxhxf

hxhxhxhxxfhxf

hxhxhxhxf

h

h

h

4

24lim

)24(lim

24lim)(

24424242)()(

42424)(2)(

0

0

2

0

/

2

222

22

2

=

+=

+=

+=

+=

−−+++=−+

+++=

++=+

→

→

→

4242 22 +++ hxhx 224 hxh +

h

hxhh

)24(lim0

+→

x4 (4)

8.2.1 xxxf 53)( 2 +−= 21

2 53)( xxxf +−=

21

/

256)(

−+−= xxxf

21

5x x6−

21

25 −

x

(3) 8.2.2

xxxxpxxx

xxx

xx

xp

32166)(168

1681

41)(

37/

226

226

2

3

+−−=

++=

++=

+=

−−

−−

OR/OF

xxxxpxxxxp

xxxp

32166)()43)(4(2)(:rulechain theof use makingby

)4()(

37/

43/

23

+−−=

+−+=

+=

−−

−−

−

226 1681 x

xx++

226 168 xxx ++ −− answer/antwoord (4) )4(2 3 xx +− )43( 4 +− −x (4)

8.3.1 14143)( 2/ +−= xxxh finding/kry )(/ xh (1) 8.3.2 At/By B: 0)(/ =xh

014143 2 =+− xx

)3(2)14)(3(4)14(14 2 −−±

=x

an /22,3or45,1=

derivative equal to/ afgeleide gelyk aan 0 substitution into correct formula/substitusie in korrekte formule

x-value of/x-waarde van 1,45 (3)



8.3.3

( )( )( )421)86)(1(8147 223

−−−=+−−=−+−

xxxxxxxxx

( )0;4C OR/OF

cx > 3,22

h(4) = ( ) ( ) ( ) 08414474 23 =−+− ∴ cx = 4

)1( −x 862 +− xx ( )( )42 −− xx coordinates of/koördinate van C (4) 22,3>cx substitution of/ substitusie van 4 h(4) = 0 cx (4)

8.3.4

37

1460146

146)(14143)(

//

2/

<∴

<<−

−=

+−=

x

xx

xxhxxxh

37

=∴k

146)(// −= xxh 0146 <−x

37

=k

(3) [22]




2

2

66

rh

hr

π

π

=

=

2

6r

hπ

= (1)

9.2 ( )[ ]

rr

rr

r

rrhrrh

rrhrS

12060

60620

60206210

42210

2

22

2

2

22

+=

+

=

+=

+=

++=

π

ππ

π

ππ

ππ

πππ

OR/OF Area of/van 10 spheres/sfere = 22 40410 rr ππ =××× Area of/van 10 cylinders/silinders )22(10 2 hrr ππ +=

)622(10 22

rrrπ

ππ +=

rr 12020 2 += π

Total area/Totale area =r

rr 1202040 22 ++ ππ

rr 12060 2 += π

( )22 42210 rrhr πππ ++ 26020 rrh ππ + substitution/substitusie

(4) area of 10 spheres/ area van 10 sfere area of 10 cylinders/ area van 10 silinders substitution/substitusie simplification/vereenvoudiging

(4)

9.3

cm68,01120120

0120120

0120120

0120120

31

3

3

2

2/

=π

=∴

π=

=−π

=−π

=−π= −

r

r

rr

r

rrS

2120120 −−π rr 0=

π120

1203 =r

answer/antwoord

(4) [9]



QUESTION/VRAAG 10 10.1.1 d = 5

e = 4 f = 7 g = 5

d = 5 e = 4 f = 7 g = 5

(4) 10.1.2a

P(A and/en B and/en C) = 272

544=

272

544=

(1)

10.1.2b P(A or/of B or/of C) =

98

5448

= 98

5448

=

(1)

10.1.2c P(only/slegs C) =

547

547

(1)

10.1.2d P(that a country uses exactly two methods/dat 'n land

presies twee metodes gebruik) = 54

845 ++ =5417

5417

(1) 10.2.1

P(selects Midnight as drama/kies Midnight as drama) = 51

answer/antwoord (2)

10.2.2 Number of different selections of drama, romance and comedy/Aantal verskillende keuses van drama, liefdesverhale en komedie = 5 × 4 × 3= 60

product/produk answer/antwoord (2)

10.2.3 P(select Last Hero and Laughing Dragon/kies Last Hero en

Laughing Dragon) = 151

31

51

=×

OR/OF P(select Last Hero and Laughing Dragon/kies Last Hero en

Laughing Dragon) = 151

60141=

××

product/produk answer/antwoord

(2)

product/produk answer/antwoord

(2) [14]

TOTAL/TOTAAL: 150





NOVEMBER 2014

MEMORANDUM


GRADE/GRAAD 12

Mathematics P1/Wiskunde V1 2 DBE/November 2014 NSC/NSS – Memorandum


NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent accuracy applies in all aspects of the marking memorandum.

LET WEL:

• Indien ’n kandidaat ’n vraag TWEE keer beantwoord, merk slegs die EERSTE poging. • Volgehoue akkuraatheid is DEURGAANS op ALLE aspekte van die memorandum van


4or20)4)(2(−==

=+−xx

xx

x = 2 4−=x

(2) 1.1.2

85,1 or/52,261722

)3(2)14)(3(4)2(2

24

01423

2

2

2

−==

±=

−−−±=

−±−=

=−−

xofx

x

aacbbx

xx

OR/OF

85,1or/52,23

431343

31

943

31

91

314

91

32

2

2

−==

±=∴

±=−

=

−

+=+−

xofx

x

x

x

xx

standard form/standaardvorm substitution into correct formula/ substitusie in korrekte formule answers/ antwoorde

(4)

for adding 91 on

both sides/tel 91

by aan beide kante

3

431±=x

answers

(4)



1.1.3

2225202

20)12(22022

2

2

2

=∴=

=

=+

=++

x

x

x

x

xx

OR/OF

25.25.2

5.2)12(25.222.2

2

22

22

=∴=

=+

=+

x

x

x

xx

OR/OF

2242

202.52022.4

2

=∴==

=

=+

x

x

x

xx

common factor/gemeen. faktor simplification/ vereenvoudiging answer/antwoord

(3) common factor/gemeen. faktor simplification/ vereenvoudiging answer/antwoord

(3) 202.5 =x 42 =x answer/antwoord

(3) 1.2

)1;1()23;0(

10

3)1(23232

1or23

0)1)(32(0352

01624151027361216241510)9124(3

1624)32(5)32(3:)2(in )1(

)2(..........162453)1(..........32

2

22

22

2

2

−−

==

+−=+

−=∴

−=−=

=++=++

=−−−−++

+=−−++

+=+−+

+=−

+=

xorx

xorx

yy

yyyy

yyyyyyyyyy

yyyy

yxyxyx

OR/OF

substitution/substitusie

simplification/ vereenvoudiging standard form/ standaardvorm factorisation/faktorisering y-values/y-waardes x-values/x-waardes

(6)



123

100)1(

04816481556:2

2481624

21553

231624

2353

23

2

22

22

2

−=−=

===−=−

−+=+−×

−+=

−−

−

+=

−

−

−=

yory

xorxxx

xxxxxx

xxxx

xxxx

xy

substitution/substitusie simplification/ vereenvoudiging standard form / standard vorm factors/faktore x- values/x- waardes y-values/y-waardes (6)

1.3 ( )( )

0)4)(1(043623

621

2

2

<−+<−−

<+−

<−−

xxxxxxxx

41 <<− x or )4;1(−∈x

standard form/ standaardvorm factorisation/faktorisering critical values in the context of inequality / kritiese waardes in die konteks van die ongelykheid notation/notasie

(4)

1.4 4

04−≤

≥−−k

k

04 ≥−− k answer/antwoord

(2) [21]

OR/OF 1 4

+ – +

4



QUESTION/VRAAG 2 2.1 234 =T 23 (1) 2.2

1752)7)(1251(2

)1(251

=−+=

−+= dnaT

a = 2 and d = 7 subst. into correct formula /subt. in korrekte formule 1752 (3)

2.3 ∑=

−251

1)57(

nn

OR/OF

( )∑=

+250

027

pp

general term/ algemene term complete answer /volledige antwoord (2) general term/ algemene term complete answer / volledige antwoord (2)

2.4 [ ]

[ ]220127

175222

2512

=

+=

+=

n

n

S

lanS

OR/OF

[ ]

[ ]220127

)7)(1251()2(22

251

)1(22

=

−+=

−+= dnanSn

substitution/substitusie 220127 (2) substitution/substitusie 220127 (2)

2.5 The new series/Die nuwe reeks is 16 + 44 + 72 + …+1 752

63162

)1(2817361752)1(2816

=−=

−==−+

nn

nn

OR/OF 2 + 9 + 16 + 23 + 30 + 37 + 44 + 51 + ... + 1752

3T is divisible by /is deelbaar deur 4 Then 25115117 ,...,,, TTTT are divisible by 4, thus each 4th term is divisible by 4. Daarna is 25115117 ,...,,, TTTT deelbaar deur 4, d.w.s. elke 4de term is deelbaar deur 4.

∴number of terms divisible by 4 will be = 6314

3251=+

−

∴aantal terme deelbaar deur 4 sal wees = 6314

3251=+

−

OR/OF

generating new series divisible by 4/ vorming van nuwe reeks deelbaar deur 4 1752=nT 63 (4) 3T is divisible by 4/ is deelbaar deur 4 identifying terms divisible by 4/ identifiseer terme deelbaar deur 4 reasoning/redenering 63 (4)



Position of terms divisible by 4: 3 ; 7 ; 11 ; …; 247; 251

632524

25114

==

=−=

nn

nTn

generating sequence involving position of terms/vorming van reeks i.t.v. posisie van terme 251=nT 63 (4) [12]



QUESTION/VRAAG 3 3.1.1 – 1 ; – 7 ; – 11 ; p ; ...

– 6 – 4 p + 11 2 2

13215

2)4(11

−==+

=−−+

ppp

OR/OF – 1 ; – 7 ; – 11 ; p ; ... – 6 – 4 p + 11 2 2

13211

−=−=+

pp

p + 15 = 2

p = – 13 (2) first differences/ eerste verskille p = – 13 (2)

3.1.2

797

1911

96)1(3

63

122

2 +−=

=−=+−−=++

−=−=+−=+

==

nnTc

ccba

bb

ba

aa

n

OR/OF

( ) ( )( )

( )( ) ( )( )( )

792

462661

2221611

2211

2

2

211

+−=

+−++−−=

−−+−−+−=

−−+−+=

nn

nnn

nnn

dnndnTTn

1=a 9−=b 7=c answer/antwoord (4) formula/formule substitution of first and second differences/substitusie van eerste en tweede verskille simplification/vereenvoudiging answer/antwoord (4)



OR/OF 7; – 1 ; – 7 ; – 11 ; p ; ... – 8 – 6 – 4 p+ 11 2 2 2

79963

1227

2

0

+−=

−=∴−=+=∴=

==

nnTbba

aacT

n

OR/OF

1)2(21

==a

797:(1)in sub

963:)1()2(

)2.......(7247)1.......(111

2

2

1

2

+−=∴

=−=∴

−=+−−=++∴−=

−=++∴−=++=∴

nnTc

bb

cbTcbTcbnnT

n

n

c-value/c-waarde a-value/a-waarde b-value/b-waarde answer/antwoord (4) a-value/a-waarde b-value/b-waarde c-value/c-waarde answer/antwoord (4)



3.1.3 The sequence of first differences is/Die reeks van eerste verskille is: – 6 ; – 4 ; – 2 ; 0 ; ... – 6+(n – 1)(2) = 96 n = 52 ∴two terms are/twee terme is:

23397)53(953

22437)52(9522

53

252

=+−=

=+−=

TT

OR/OF The sequence of first differences is/Die reeks van eerste verskille is: – 6 ; – 4 ; – 2 ; 0 ; ... The formula for the sequence of first differences/Die formule vir die reeks van eerste verskille is 82 −= nTn 1st difference/1ste verskil: 2n – 8 = 96 2n = 104 n = 52 ∴two terms are/twee terme is:

23397)53(953

22437)52(9522

53

252

=+−=

=+−=

TT

OR/OF

( ) ( ) ( )[ ]

53106296799127996719179

96

22

22

1

===−−+−+−+−

=+−−−−+−

=− −

nn

nnnnnnnnn

TT nn

23397)53(953

22437)52(9522

53

252

=+−=

=+−=

TT

– 6+(n – 1)(2) = 96 52 2 243 2 339 (4) 2n – 8 = 96

52 2 243 2 339 (4) 961 =− −nn TT

53

2 243 2 339 (4)

OR/OF

( ) ( )[ ] [ ]

23397)53(953

22437)52(95252104296797991296797191

96

253

252

22

221

=+−=

=+−=

===−+−+−−++

=+−−++−+

=−+

TT

nn

nnnnnnnnn

TT nn

961 =−+ nn TT 52 2 243 2 339 (4)



3.2.1

1818

99

112

12

2or21or4or

41

4116

−−

−

=

=T

a = 16 and r =

41

subst. into correct formula/ subt in korrekte formule answer/antwoord (3)

3.2.2

33,21411

41116

10

10

=

−

−

=S

OR/OF

33,21

141

14116

10

10

=

−

−

=S

substitution into correct formula /substitusie in korrekte formule answer/antwoord (2) substitution into correct formula /substitusie in korrekte formule answer/antwoord (2)

3.3

502

10099

100...56

45

34

23

9911...

411

311

211

=

=

=

+

+

+

+

OR/OF

+

+

+

+

9911...

411

311

211

25

452

4112

234

23

311

23

23

211

3

2

1

=×=

+=

=×=

+=

=

+=

T

T

T

...25,2,

23 is an arithmetic sequence with

21 and

23

== da

502

10021)198(

23

98

==

−+=∴T

improper fractions/ onegte breuke

+

99100or

9911

answer/antwoord (4)

+

9911

giving the first three terms / gee die eerste drie terme answer /antwoord (4) [19]




11

==

qp

p value /waarde q value /waarde (2)

4.2

321

11

20

−==−−

++

=

xx

x

OR/OF Reflect (0 ; 3) across y = – x to get T( –3 ; 0)

3−=x Reflekteer (0 ; 3) om y = – 1 om T( –3 ; 0) te kry

3−=x

11

20 ++

=x

3−=x ( 2) reflect across/reflekteer om y = – x 3−=x (2)

4.3 Shifting g five units to the left shifts (– 1 ; 0) five units to the left. x = – 6

answer/antwoord (1)

4.4

eenhede

yxy

xx

xxxx

xx

units/45,2 6OS

633OS...73,13

0 S,at since 3

312

11

2

222

2

2

==∴

=+=+=

==

>=∴

=

+=++

=++

OR/OF

equating both graphs/stel grafieke gelyk

32 =x 3and3 == yx OS 2 = 6 answer/antwoord (5)



Translate g one unit down and one unit to the

right/Transleer g een eenheid af en een eenheid na regs

The new equation/Die nuwe vergelyking :x

xp 2)( =

Therefore the image of S is ( )2;2S/ / Daarom is die beeld van S nou ( )2;2S/ Now translate p back to g/Transleer p terug na g: ( )

( ) ( )eenhedeunits/45,26OS

122212221212OS

12;12S222

==∴

++++−=++−=

+−

x

xp 2)( =

coord. of/koörd. van /S coord. of/koörd. van S answer/antwoord (5)

4.5 k < 3 will give roots with opposite signs/ k < 3 sal wortels met teenoorgestelde tekens gee

k < 3 (1) [11]



QUESTION 5 5.1

33131

31log1

log

1

1

=∴

=

=

=−

=

−

−

a

a

a

xy

a

a

subt.

−1;

31

311 =−a or

1

31 −

=a

(2)

5.2 xy

yxh3

log: 3

=∴

=

swop x and y/ruil x en y answer/antwoord

(2) 5.3

xxg

xxg

xxg

31

3

3

log)(

1log)(

log)(

=

=

−=

OF

OF

OR/

OR/

OR/OF

yx −= 3 OR/OF

y

x

=

31

answer/antwoord (1)

answer/antwoord

(1) answer/antwoord

(1)

answer/antwoord (1) answer/antwoord (1)

5.4 0>x OR/OF ( )∞;0

answer/antwoord (1)

answer/antwoord

(1) 5.5

271271

3

3log3

3

≥

=

=

−=−

x

x

xx

exponential form/ eksponensiële vorm simplification/vereenvoudiging answer/antwoord (3) [9]




positive) is S of coordinate-( 22,123

064

2

2

xx

x

x

=

=

=−

y = 0 1,22 (2)

6.2 ( 0 ; – 6) 0 – 6 (2)

6.3.1

)64(2

)()(2−−=

−=

xx

xgxfQT

or 642 2+−= xx

correct formula/ korrekte formule substitution/substitusie (3)

6.3.2 QT 642 22

1

+−= xx

08QT of Deravitive 21

=−=−

xx x

x81

=

812

3

=x or 2641 xx=

32

81

=x

2

21

=x or

6413 =x

41

=x = 0,25

6414

412QTMax/

221

+

−

=Maks

eenhedeunits/75,6436 ==

derivative/afgeleide derivative equal to 0/ afgeleide gelyk aan 0

812

3

=x x-value/x-waarde substitution/substitusie answer/antwoord (6) [13]



QUESTION/VRAAG 7 7.1 niPA )1( −=

72 500 = 145 000 5)1( i−

5145000725001−=i

..1294,0= . ∴ Rate of interest/Rentekoers is 12,94 % p.a./p.j. OR/OF

( )

1294,0211

211

51

5

=

−=∴

=−

i

i

i

∴ Rate of interest/Rentekoers is 12,94 % p.a./p.j.

substitution/substitusie writing in terms of i herskryf in terme van i answer/antwoord (3) substitution/substitusie writing i.t.o i answer (3)

7.2.1 ( )[ ]

iixP

n−+−=

11

500 000 =

1212,01212,011

240

+−

−

x

+−

×=

−240

1212,011

1212,0500000

x

43,5505R=x

i = 1212,0

n =240 substitution into correct formula answer/antwoord (4)



7.2.2

( )[ ]i

ixPn−+−

=11

( )

( )65101,1

01,1101,06000

5000001212,0

1212,0116000

500000

−=

−=×

+−

=

−

−

−

n

n

n

01,1log61log

=− n

07,180=n ∴Melissa settles the loan in 181 months

6000 substitute into correct formula/substitusie in korrekte formule use of logs/gebruik van logs answer/antwoord (4)

7.2.3 Samuel He is paying off his loan over a longer period thus more interest will be paid./Hy betaal sy lening oor 'n langer tydperk af, dus sal hy meer rente betaal. OR/OF Samuel He will pay/Hy betaal R5505,43 × 240 – R500 000 = R821 303,20 She will pay between/Sy sal tussen R580 000 and/en R586 000,00 betaal.

Samuel reason/rede

(2) Samuel reason/rede

(2)

[13]




( )2

22

0

22

0

322

0

0

322

33223

3223

322223

223

3

33lim

)33(lim

33lim

)()(lim)(

33 33)()(

33 22

))(2()()(

x

hxhxh

hxhxhh

hxhhxh

xfhxfxf

hxhhxxhxhhxxxfhxf

hxhhxxhxhxhhxhxx

hxhxhxhxhxf

h

h

h

h

=

++=

++=

++=

−+=′

++=

−+++=−+

+++=

+++++=

+++=+=+

→

→

→

→

OR/OF

hxfhxfxf

h

)()(lim)(0

−+=′

→

h

xhxh

33

0

)(lim −+=

→

h

xhxhxh

32

0

))((lim −++=

→

h

xhxhxhxh

322

0

)2)((lim −+++=

→

h

xhxhhxxh

33223

0

33lim −+++=

→

h

hxhxhh

)33(lim 22

0

++=

→

( )22

033lim hxhx

h++=

→ 23x= OR

simplifying/vereenvouding formula/formule subst. into formula/subst. in formule factorization/faktorisering answer/antwoord

(5) formula/formule subst. into formula/subst. in formule simplifying/vereenvoudiging factorization/faktorisering answer/antwoord

(5)



hxfhxfxf

h

)()(lim)(0

−+=′

→

h

xhxh

33

0

)(lim −+=

→

hhxhxh

hxxhxhxhxxhx

h

h

)33(lim

)2)((lim

22

0

2222

0

++=

+++++−+=

→

→

( )22

033lim hxhx

h++=

→ 23x=

formula/formule subst. into formula/subst. in formule factorization/faktorisering simplifying/vereenvoudiging answer/antwoord

(5) 8.2 324)( xxxf +=′ 4x

32x (2)

8.3 12 612 +−= xxy 511 1212 xx

dxdy

−=

( )112 65−= xx

yx512=

simplification/vereenvoudiging derivative/afgeleide factors/faktore

(3)

8.4

314120412

0 whenup concave is 412

4461422

2

23

>

>>−∴

>

−=

+−=

−+−=

x

xx

(x)f as op konkaaf /isfx(x)f

xx(x)fxxxf(x)

//

//

/

first derivative/eerste afgeleide second derivative/tweede afgeleide 0)(// >xf

31

>x (4)

[14]



QUESTION/VRAAG 9

9.1 0383 2 =−−= xx)x(f / ( )( ) 0313 =−+ xx

31

−=x or 3=x

== 52,18or

271418or

27500 yy 0=y

Turning points are/Draaipunte is )0;3( and 27

500;31

−

derivative/afgeleide derivative/ afgeleide = 0 factors/faktore x-values/waardes each y-values/elke y-waarde (6)

9.2

x

y

− 52,18;

31

18

( ̶ 2 ; 0) (3 ; 0)O

x-intercepts/afsnitte y-intercept/afsnit turning points/ draaipunte shape/vorm

(4) 9.3

30or31

<<−

< xx

OR

)3;0()31;( ∪−−∞

31−

<x

both critical points/ beide kritieke-punte notation/notasie

(3)



QUESTION/VRAAG 10

10.1

hlhl

240402

−==+

answer (1) 10.2 10022 =+ hb

hb −= 50 lbhV =

)50)(240( hhhV −−=

10022 =+ hb hb −= 50 volume formula

(3) 10.3

cm80,8 possible, as large asbox afor 80,8or86,37)6(2

)2000)(6(4)280(280

02000280620001402

)240)(50(

2

2

23

2

=∴=≠

−−±=

=+−=′

+−=

−−=

hhh

h

hhVhhhV

hhhV

vir die grootste moontlike boks = 8,80 cm

simplifying/vereenvoudig derivative / afgeleide h-values in any form / h-waardes in enige vorm answer/antwoord

(5) [9]

QUESTION/VRAAG 11

11.1.1 P(male/manlik) = 18083 or 0,46 or 46,11%

answer/antwoord (1)

11.1.2 P(not game park/nie wildreservaat) = 1 – P(game park/wildreservaat)

= 180621−

=9059 or 0,66 or 65,56%

OR/OF P(not game park/nie wildreservaat)

65,56%or 0,66or 9059180118

18020

18098

=

=

+=

180621−

answer/antwoord (2)

18020

18098

+

answer/antwoord (2)



11.2 Events are independent if /Gebeure is onafhanklike indien P(male) × P(home) = P(male and home) P(manlik) × P(huis) = P(manlik en huis)

P(male/manlik) = 18083

and/en P(home/huis) = 18020 or 0,11 or 11,11%

P(male/manlik) × P(home/huis)

= 18083 ×

18020

=

162083

= 0,05123 or 5,12% P(male and home/manlik en huis)

= 18013

= 0,07222… or 7,22% Therefore P(male) × P(home) ≠ P(male and home) Dus P(manlik) × P(huis) ≠ P(manlik en huis) Thus the events are not independent./Dus is die gebeure nie onafhanklik nie OR/OF Home/Huis Not Home/

Nie huis

M 13 70 83 F 7 90 97 20 160 180

P(female/vroulik) × P(not home/nie huis)

= 18097 ×

180160

=

405194

= 0,479012345… or 47,90% P(female and not home/vroulik en nie-huis)

= 18090

= 0,5 or 50% Therefore P(female) × P(not home) ≠ P(female and not home) Thus the events are not independent. Dus P(vroulik) × P(nie-huis) ≠ P(vroulik en nie-huis) Dus is die gebeure nie onafhanklik nie.

P(m) × P(h) and their values/en hulle waardes answer of product P(m and/en h) value/waarde conclusion/afleiding (4) P(f) × P(not h) and their values/en hulle waardes answer of product P(f and/en not h) value/waarde conclusion/afleiding (4)

[7]



QUESTION/VRAAG 12

12.1.1 2223242526 ×××× = 7 893 600 OR/OF

!21!26

)!526(!26

526 =

−=P = 7 893 600

2223242526 ×××× 7 893 600 (2) formula/formule answer/antwoord (2)

12.1.2 222324 ×× = 12 144

222324 ×× 12 144

(2) 12.2.1 7×6×5×4×3×2×1

= 5 040 product/produk 5 040 (2)

12.2.2 (3×2×1)(5×4×3×2×1) = 720 OR/OF The five 'units' can be parked in 5×4×3×2×1 ways./Die vyf 'eenhede' kan op 5×4×3×2×1 maniere geparkeer word. The three silver cars can be parked in 3×2×1 ways./Die drie silwer motors kan op 3×2×1 maniere parkeer word. So there are (3×2×1)(5×4×3×2×1) = 720 ways to park the cars./Dus is daar (3×2×1)(5×4×3×2×1) = 720 maniere om die motors te parkeer. OR/OF Suppose for the moment the 3 silver cars are at one end./Veronderstel die drie silwer motors is op die punt. The 3 cars can be arranged in 3×2×1 = 6 ways./Die 3 motors kan op 3×2×1 = 6 maniere gerangskik word. For each of them the remaining four cars can be arranged in 4×3×2×1 = 24 ways./Die 4 oorblywende motors kan op 4×3×2×1 = 24 maniere rangskik word. So 6 × 24 = 144 ways if all 3 cars at one end./Dus is daar 6 × 24 = 144 maniere as die 3 motors op die punt is. Together, the silver cars can only occupy 5 different positions amongst the 7 positions. ./Saam kan die silwer motors slegs 5 verskillende posisies hê tussen die 7 moontlike posisies. ∴Total ways/Totale getal maniere = 5 × 144 = 720

3×2×1 5×4×3×2×1 720

(3) 5×4×3×2×1 3×2×1 720

(3) 6 × 24 = 144 5 × 144 720 (3)






EXEMPLAR 2014/MODEL 2014

MEMORANDUM


GRADE/GRAAD 12

Mathematics P1/Wiskunde V1 2 DBE/2014 NSC/NSS – Grade 12 Exemplar/Graad 12 Model – Memorandum


NOTE:

• If a candidate answers a question/vraag TWICE, only mark the FIRST attempt. • Consistent accuracy applies in all aspects of the marking memorandum.

LET WEL:

• Indien ’n kandidaat ’n vraag TWEE keer beantwoord, merk slegs die EERSTE poging. • Volgehoue akkuraatheid is DEURGAANS in ALLE aspekte van die memorandum van


( ) 043043 2

=−=−

xxxx

0or34

== xx

factors both answers

(2) 1.1.2

( ) ( ) ( )( )( )

65,5or35,02

286

1221466

026

026

2

2

==

±=

−−±−−=

=+−

=+−

xx

x

xxx

x

OR

( )( )

65,5or35,073

73

923

026

026

2

2

==±=

±=−

+−=−

=+−

=+−

xxx

x

x

xxx

x

0262 =+− xx

subs into

correct formula

35,0=x

65,5=x

(4)

0262 =+− xx

( ) 923 2 +−=−x

35,0=x

65,5=x

(4)

1.1.3

( )82

0;4

23

2

32

==

>=

xx

xx

OR

( )23

22=x 8=x

(2)



( )84

0;4

23

32

==

>=

xx

xx

OR

)0(88or8

2or)2(

022

04

4

33

31

31

32

32

>==−==−=

=

+

−

=−

=

xxxx

xx

xx

x

x

( )23

4=x 8=x

(2)

factors 8=x

(2)

1.1.4 ( )

505

positivealwaysis3053

<<−

<−

xx

xx

x

x3 > 0 x < 5

(2)

1.2

( )

( )( )

6441014

043263

26222and6

2

2

2

2

=−==−==+−

=−−

=++−

=−−−

=−−−=

yoryxorx

xxxx

xxxxx

yxxxy

OR

( )( )

6441014

043622

2222and6

2

2

2

=−==−==+−

=−−

−−=−

−==−−−=

yoryxorx

xxxx

xxxxy

yxxxy

subst 62 −−= xxy standard form factors x-values y-values

(6) 22 −= xy standard form factors x-values y-values

Answer only full marks



OR

( )( )

4164

0460242

2424

64

424

44

62

22

22

222and6

2

2

2

2

2

=−==−=

=+−=−−

−+=

−

+

−

++=

−

+

−

+

=

+=

=−−−=

xorxyory

yyyy

yyy

yyyy

yyy

yx

yxxxy

(6)

2

2+=

yx

standard form factors y - values x - values

(6) 1.3

8412

2234.3

2448.3

2

22

2

1

=−=

−=

−

+

+

x

x

x

x

OR

8412

22144

2448.3

2

22

2

1

=−=

−=

−

+

+

x

x

x

x

222 +x 4 answer

(3)

222 +x 4 answer

(3) 1.4.1 No, there will be no intersection between the graphs.

Min value of 5)1(3 2 +−x is 5 Nee, daar sal geen snyding tussen die grafieke wees nie. Min waarde van 5)1(3 2 +−x is 5 OR

32)1(

2)1(335)1(3

2

2

2

−≠−

−=−

=+−

x

xx

No, there will be no intersection between the graphs.

answer reason

(2)

reason answer



Nee, daar sal geen snyding tussen die grafieke wees nie. OR

( ) ( )( )

024

5346

056302)12(335)1(3

2

2

2

2

<−=

−−=∆

=+−

=++−

=+−

xxxx

x

No, there is no solution to the equation f(x) = g(x) Nee, daar is geen oplossing vir die vergelyking f(x) = g(x)

(2) reason answer

(2) 1.4.2

2/ of valuesreal allfor 02

2)1(335)1(3

2

2

>>−

−=−

+=+−

kxvanwaardeseleereallevirxk

kxkx

OR

( ) ( )( )

2241202412

/roots unequal realFor 2412

1260365346

056335363

2

2

2

>>>−

−=+−=

−−−=∆

=−+−

+=++−

kk

kwortelsongelykeeleereVir

kk

k

kxxkxx

answer (2) answer (2) [23]





482886

66183006)1(18300

)1(

==

−+=−+=−+=

nn

nn

dnaTn

a = 18 and d = 6 Tn = 300 answer

(3) 2.1.2

])1(2[2

dnanSn −+=

)]6(47)18(2[248

+=

7632=

substitution in formula answer

(2)

2.1.3 Sum of all numbers from 1 to 300 / Som van alle getalle van 1 tot 300

)]1(299)1(2[2

300+=

2

)301(300=

45150= Sum of numbers not divisible by 6 / Som van getalle wat nie deelbaar deur 6 is nie )1267632(45150 ++−= 37500=

substitution answer )1267632( ++ answer

(4) 2.2.1 16, 8; 4; ……

r =21

( )n

n

n

nn arT

−

+−

−

−

=

=

=

=

5

14

1

1

2222116

r =21

answer (in any format)

(2) 2.2.2

16 + 8 + 4 + 2 + 1 + 21 = 31

S5 = 31 n > 5 or n ≥ 6

16 + 8 + 4 + 2 + 1 +

21

S5 = 31 n > 5 / n ≥ 6

(3)



OR ( )

522

2321

213231

)21(3231211

21116

31

11

5

>>

>

−<−

−<

−

−

<

−−

=

−−

−

−

−

n

rraS

n

n

n

n

n

n

n

or n ≥ 6

Sn > 31 simplification n > 5 / n ≥ 6

(3)

2.2.3

32211

161

=

−=

−=∞ r

aS

OR

16 + 8 + 4 + 2 + 1 + 21 +

41 +

81 +

161 +

321 +

641 +

1281 …….

Answer gets closer and closer to 32 the more terms gets added together Antwoord beweeg nader en nader aan 32 hoe meer terme bymekaar getel word

substitution of a and r answer

(2)

expanding the series answer

(2) [16]



QUESTION/VRAAG 3 3.1.1 1; x ; y; z….. Tn = 4n + 6

10; 14; 18…… 1 x y 10 14

4 2a = 4 a = 2 OR Tn = 4n + 6 d = 4 2a = 4 a = 2

2nd difference = 4 2a = 4 a = 2

(2)

2a = 4 a = 2

(2) 3.1.2

1 x y 10 14

4

542

51421

4106103

2 −+=

−==++=++

==+=+

nnT

cccba

bbba

n

1st differences 10; 14; 18…… 103 =+ ba 1=++ cba 542 2 −+= nnTn

(4)



3.2 Consider the sequence made up by the first factors of each term: Beskou die ry wat deur die eerste faktore van elke term gevorm word: 1; 5; 9; 13; … 81 An arithmetic sequence / rekenkundige ry: dnaTn )1( −+= 4)1(1 −+= n 34 −= n To find the no. of terms: 3481 −= n Aantal terme: 844 =n 21=∴n The second factor is 1 more than the first factor / Tweede faktor is 1 meer as die eerste faktor:

24

134−=

+−=nnTn

OR Consider the sequence made up by the second factors of each term: Beskou die ry wat deur die tweede faktore van elke term gevorm word: 2; 6; 10; 14; …82 Also an arithmetic sequence / rekenkundige ry: dnaTn )1( −+= 4)1(2 −+= n 24 −= n In sigma notation:

∑=

−−21

1)24)(34(

nnn or ∑

=

−−21

1)12)(34(2

nnn or ∑

=

+−21

1

2 )62016(n

nn

nT 34 −= n no. of terms nT 24 −= n nT 24 −= n answer in sigma notation

(4) [10]





31

2)( −+

=x

xf

1

310

2)0(

−=

−+

=

= fy

( )1;0 −

subst 0=x ( )1;0 −

(2) 4.1.2

31

2331

23

31

20

−=

=++

=

−+

=

x

xx

x

− 0;

31

subs y = 0

− 0;

31

(2)

4.1.3

x

y

f

A(-1/3 ; 0)

B(0;-1)x = -1

y = -3

0

shape both intercepts correct horizontal and vertical asymptote

(3)

4.1.4

43)1(

−−=−+−=

xyxy

OR

( )

44

13

−−=−=

+−−=−+−=

xyk

kkxy

3)1( −+−= xy 4−−= xy

(2)

( ) k+−−=− 13 4−−= xy

(2)



4.2.1

( )

( )

( ) 322

31]1;1subs[3.1

1]2;0subs[3.2

3..

1

0

−=

=−=−

−−=

=−−=−

−=

+=

x

x

x

x

xfb

bby

ababay

qbay

subs q = – 3 1=a

2=b ( ) 32 −= xxf

(4) 4.2.2 A translation of 4 units up and 1 unit to the left.

'n Translasie van 4 eenhede na bo en 1 eenheid na links. OR Dilation by a factor of 2 and 7 units up. Verkleining deur faktor van 2 en 7 eenhede na bo.

4 units up 1 unit to the left

(2)

dilation by factor 2 7 units up

(2) [15]



QUESTION/VRAAG 5 5. 1

( )

125,6/849

3455

452

45

45

054)2(2

5

0 2

352)(

2

2

=

+

−−

−−=

−=−=

=−−

−−

−=

=′−=

+−−=

y

xx

xx

xfa

bx

xxxf

or

TP (45

− ;849 )

OR

849)

45(2

]1649)

45[(2

]23

1625)

45[(2

)23

252(-

2

2

2

2

++−=

−+−=

−−+−=

−+=

x

x

x

xxy

TP (45

− ;849 )

/2abx −= ( ) 0=′ xf

45

−=x

y = 125,6/849

(3)

]23

1625)

45[(2 2 −−+− x

45

−=x

y = 125,6/849

(3) 5. 2 mtangent = tan 135

= – 1 –4x –5 = –1 – 4x = 4 x = –1 y = – 2(–1) 2– 5(–1) + 3 = 6 Point of contact: P(–1; 6)

tan 135 = –1 –4x –5 = –1 x = –1 y = 6

(4) 5. 3 Eq of g: y – y1 = m(x – x1)

y – 6 = – 1(x + 1) y = – x +5

substitute in equation answer

(2) 5. 4 d > 5

answer

(1) [10]




2168

)8(4

)(

==

=

=

aa

a

axxg

subst (8 ; 4) 2=a

(2) 6.2 0≥x answer

(1) 6.3 0≥y answer

(1) 6.4

0;2

20;2

2

2

≥=

=

≥=

yxy

yxxxy

interchange x and y answer

(2) 6.5

( )( )2 or 8

28016100

168242

2

2

==−−=+−=

+−=

−=

xxxxxx

xxxxx

when x = 2, LHS = 2 but RHS = –2 Hence 8=x only

1682 2 +−= xxx

(squaring both sides)

factors

2 or 8 == xx

selects 8=x

(4)

6.6 80 << x 8<x x<0

(2) [12]




12,0000102/priceSelling =sVerkooppry

= 850 000

850 000

(1) 7.2

729,95 6 = 1209,01209,011

000748

])1(1[ =

240

x

x

iixP

n

v

+−

=

+−

−

−

OR

729,95 6 = 1209,0

11209,01

1209,01000748

]1)1[( =

240

240

x

x

iixF

n

v

−

+

=

+

−+

Pv =748 000

i = 1209,0

n = –240 x = R6 729,95

(4)

748000 240)1209,01( +

i = 1209,0

n = 240 x = R6 729,95

(4) 7.3 Total interest paid / Totale rente betaal

= (6 729,95 x 240) –748 000 = R 867 188

(6 729,95 x 240) 867 188

(2) 7.4

509,74 615 = 1209,0

1209,01195,6729

])1(1[ = Balance

155

x

iix n

+−

=

+−

−

−

OR

95,6729 n = –155 R615 509,74

(3)



77,509615962,153796732,6634111loan of Balance

962,153796 = 1209,0

11209,0195,6729

]1)1[( =

732,66341111209,01000748

85

85

=−=

−

+

=

−+

=

+=

iixF

A

n

v

OR

77,5096151209,0

11209,0195,6729

1209,01000748Balance

85

85

=

−

+

−

+=

732,6634111 n = 85 R615 509,77

(3) subs of 748 000 and

6729,95 n = 85 R615 509,77

(3) 7.5 New value of bond:

615 509,74

+

1209,01 4 or 615 509,77

+

1209,01 4

= 634 183,81 = 634 183,84

R615 509,74(1 +1209,0 )4

R634 183,81/ R634 183,84

(2)

7.6

634 183,81 =

1209,0

1209,0115008

+−

−n

log (0,44042605) = – n log

+

1209,01

n = 109,74 = 110 months OR

x = 8 500 subs into correct formula use of logs answer

(4)



)44042605,0(log1209,0

1209,0115008

1209,01

+

−

=−

+−

=

n

n

183,81 634

n = 109,74 = 110 months

x = 8 500 subs into correct formula use of logs answer

(4) [16]




x

hxh

hxhh

hxhxf

hxhxfhxfhxhx

hxhxfxxf

h

h

h

6

)36(lim

)36(lim

36lim)(

36)()(2363

2)(3)(23)(

0

0

2

0

2

22

2

2

=

+=

+=

+=′

+=−+

−++=

−+=+

−=

→

→

→

OR

( ) ( ) ( )

[ ] ( )

( )[ ]

( )

( )x

hxh

hxhh

hxhh

xhxhxh

xhxhxh

xhxh

xfhxfxf

h

h

h

h

h

h

h

6

36lim

36lim

36lim

23]2363[lim

23223lim

232)(3lim

lim

0

0

2

0

222

0

222

0

22

0

0

=

+=

+=

+=

+−−++=

+−−++=

−−−+=

−+=′

→

→

→

→

→

→

→

substitution of of x + h simplification to 236 hxh + formula taking out common factor answer

(5)

formula substitution of x + h simplification

to h

hxh 236 +

taking out common factor answer

(5) 8.2

52 4 xxy −= −

518 5 −−= −x

dxdy

58 −− x

51

−

(2) [7]



QUESTION/VRAAG 9 9.1 )2( −x is a factor of f / is ’n faktor van f.

answer

(1) 9.2

( )( )( )( )( )

( )( )( )5or2or3

05230)(

532152230114)(

2

23

==−==−−+=

−+−=−−−=

+−−=

xxxxxx

xfxxxxxxxxxxf

x-intercepts: (–3; 0); (2;0); (5;0)

( )1522 −− xx (–3; 0) (2; 0) (5; 0)

(4) 9.3

( )( )

( )( )

( )

−−

−−==

=−=

=+−=′

−−=′

+−−=

81,14;3

11and36;1aresTP'

)81,14(2740036

311or1

011130pointsturningAt

118330114)(

2

23

yy

xx

xxxf

xxxfxxxxf

( ) 1183 2 −−=′ xxxf ( ) 0=′ xf x - value x - value y - values

(5) 9.4

x

y

f

(-3 ; 0) (2 ; 0) (5 ; 0)

(-1 ; 36)

(0 ; 30)

(3,67 ; -14,81)

0

y and x - intercepts shape turning points

(3)



9.5 0)( <′ xf

if 67,31 <<− x OR (-1 ; 3,67)

extreme values notation

(2) extreme values notation

(2)

[15]




( ) ( )

1000080002500

1600800010000900

4010030

40100km40andkm30:hoursAfter

2

22

22

+−=

+−+=

−+=

−=∴==

tt

ttt

ttFC

tBCtCDtBFt

tBF 30= tBC 40100−= Pythagoras answer

(4)

10.2 FC is a minimum when FC2 is a minimum.

)minutes96(hrs6,150008000

080005000

10000800025002

22

==

=−=

+−=

t

tdt

dFCttFC

FC 2 =

1000080002500 2 +− tt

800050002

−= tdt

dFC

02

=dt

dFC

answer (4)

10.3

( ) ( )60

100006.180006.12500

10000800025002

2

=

+−=

+−= ttFC

They will be 60km apart.

subs into equation answer

(2)

[10]



QUESTION/VRAAG 11 11.1 P(A or B) = P(A) + P(B)

0,57 = P(A) + 2P(A) 0,57 = 3P(A) P(A) = 0,19 ∴P(B) = 2(0,19) = 0,38

P(A or B) = P(A) + P(B) P(A) = 0,19 answer

(3) 11.2.1

P A,P

A

5

3

2

1

5

2

Y A,Y

2

1

P B,P

B

9

5

9

4

Y B;Y

first tier second tier probabilities outcomes

(4)

11.2.2 P(AY) =

52

21

= 51

answer

(1)

11.2.3 P(P) =

+

95

21

53

21

4526

185

103

=

+=

53

21

95

21

answer (3)

[11]




5 4 3 2 1 Number of different letter arrangements: Aantal verskillende letter rangskikkings wat gevorm kan word: 5! = 5x4x3x2x1 = 120

5! 120

(2) 12.2 S and T can be arranged in 2! different ways.

The remaining three letters can be arranged in 3! different ways ∴Total number of different letter arrangements having S and T as the first two letters = 2!.3! S en T kan op 2! verskillende maniere rangskik word. Die 3 letters wat oorbly kan op 3! verskillende maniere rangskik word ∴ Totale aantal letterrangskikkings waarin S en T die eerste twee letters van die rangskikking sal wees = 2!.3!

101

1206.2

120!3.!2letters) first two as T and S P(having

=

=

=

!2 !3 answer

(3) [5]

TOTAL/TOTAAL: 150

NATIONAL SENIOR CERTIFICATE GRADE … 322 2 1 2 2 2 2 2 = =− − ± − = + − = + − = + − = n n n n n n n n n The 25th term has a value of 322./Die 25ste term se waarde is

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