n3 mathematics november 2016 memorandum...title: microsoft word - n3 mathematics november 2016...

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NATIONAL CERTIFICATE

NOVEMBER EXAMINATION

MATHEMATICS N3

21 NOVEMBER 2016

This marking guideline consists of 10 pages.

MARKING GUIDELINE

MARKING GUIDELINE -2- T870(E)(N21)H MATHEMATICS N3


✓ = 1 Mark QUESTION 1 1.1 1.1.1

(2) 1.1.2

(7)

1.1.3

(6)

( )

2

2

49 56 16

7 4

7 4

x x

x

x

+ +

= +

= +

1 21 13 3

133 2

13 2 23

3 2

12

3 2

2 23

133

( 2)

1 1( 2)

. 1 2 1

1 ( 1)

1( 1)

( 1) 1

a a a a

a aaa

a a a aaa

a aaa

a aa

a

a aora a

- - -

-

-

-

æ ö- + -ç ÷

è øæ ö

= - + -ç ÷è øæ öæ ö- - +

= ç ÷ç ÷ç ÷è øè ø

æ öæ ö- -= ç ÷ç ÷

è øè øæ ö-ç ÷=ç ÷ -è ø

=- -

( )

( )

1 21 13 3

113

2 13

12

23

2 23

13

( 2)

1 1( 2)

1 2 1

11

1

a a a a

a aa

a

a a aa

a

a aaa

aa

- - -

-

-

æ ö- + -ç ÷

è øæ öç ÷= - + -ç ÷è øæ öæ ö- - +ç ÷= ç ÷ç ÷è øè øæ ö-ç ÷=ç ÷ -è ø

=-

2 2

2 2

3 7 2 62 5 3 9(3 1)( 2) ( 3)( 3)(2 1)( 3) ( 2)( 3)3 12 1

x x x xx x xx x x xx x x xxx

- + + -÷

- - -- - - +

´+ - - +-

=+

✓

✓

✓ ✓ ✓

✓

ü

ü

ü

ü

ü ü

ü ü

ü ü

ü

ü

ü

ü

ü

ü

ü



1.2

(4) [19]

QUESTION 2 2.1

(3)

3 2

1If ( ) is divided by 2x+1, then the remainder is 2

1 1 1 1Hence 16 12 3 72 2 2 2

f x f

f p

æ ö-ç ÷è ø

-æ ö æ ö æ ö æ ö- = - - + - + = -ç ÷ ç ÷ ç ÷ ç ÷è ø è ø è ø è ø1 116 6 3 78 4

pæ ö- - - + = -ç ÷è ø

12 6 3 74p- - - + = -

1 7 54p- = - +

421

p = - ´-

8p =

12

12

3

3 32 4

0,125

18

12

2 4- -

æ ö= ç ÷è ø

æ ö= ç ÷è ø

= =

3 2

34

0,125 4

1 48

2 23 2234

0,125 4

x

x

x

x

x

-

-

=

\ =

\ =

\- =

\ = -

\ =

✓

✓

✓

✓

✓

✓

✓

✓

✓

✓



2.2 2.2.1

(6) 2.2.2

(5) [14]

2

10

2

0

2

2

2

3 32log + log log 2 04 4

34log 032

43410 32

43 1

8 33 8 33 8 3 0(3 1)( 3) 03 1 0 3 0

1(n/a) 33

x x

x

x

x

x

xxx xx xx xx x

x x

æ ö- + =ç ÷è ø

=+

=+

=+= +

- - =+ - =+ = - =-

= =

2

2

2

2

2

3 32log + log log 2 04 4

3 3log + log log 24 4

3 3log log 24 4

3 324 4

3 8 33 8 3 0(3 1)( 3) 03 1 0 3 0

1(n/a) 33

x x

x x

x x

x x

x xx xx xx x

x x

æ ö- + =ç ÷è øæ ö\ = +ç ÷è ø

æ ö æ ö\ = +ç ÷ ç ÷è øè ø

\ = +

\ = +

\ - - =\ + - =\ + = - =

-\ = =

( )

24 2 3 52 4 24 2 3 52 2 ( 2) 2

4( 2) 2 3 5( 2)4 8 2 3 5 103 217

xx x x

xx x x xx x xx x xx

x

-+ =

- - +-

- =- - + +

+ - + = -+ - + = -

- = -=

✓

✓

✓ ✓

✓

✓

✓

✓

✓

✓

✓

✓

✓

✓

✓

✓

✓



QUESTION 3 3.1

(5)

3.2 Let the number of women = ∴ the number of men =

and the number of children =

There are 92 women ✓

Let the number of men = let the number of women =

let the number of children =

There are 92 women ✓ (3) 3.3

(5)

[13]

2

2

2

2

18 99 18 0

81 819 184 4

9 2,2529 2,2523 6

x xx x

x x

x

x

x or x

- - =

+ + =

+ + = - +

æ ö\ + =ç ÷è ø

= - ±

= - = -

x34x -

60x +

34 60 3023 27692

x x xxx

\ + - + + =\ =\ =

x34+x

6034 ++x

581743

30212833029434

==

=+=++++\

xxx

xxx

CPV n =

PCV n =

PCV n loglog =

PCVn logloglog -=

VPCn

logloglog -

=

log18,3 log 2000log0,48

1,262 3,3010,319

6,395

n

n

n

-=

-=

-=

CPV n =2000 (0,48) 18,3

18,3(0,48)2000

18,3log(0,48) log200018,3log(0,48) log2000

log18,3 log 2000log0,48

1,262 3,3010,319

6,395

n

n

n

n

n

n

n

\ ´ =

\ =

æ ö\ = ç ÷è øæ ö\ = ç ÷è ø

-\ =

-\ =

-=

✓

✓

✓

✓

✓

✓

✓

✓

✓

✓

✓

✓ ✓

✓

✓

✓

✓

✓

✓



QUESTION 4 4.1

(2) 4. 2

(3) 4.3

(2) 4.4

(2) 4.5

(3)

2 1 1 2( ; ) ;2 2

1 1 ;2 2

M MM x y - + -æ ö= ç ÷è ø

- -æ ö= ç ÷è ø

..... diagonals of parm 1 1Midpoint of ( ; ) ;

2 22 2 1 1( ; ) ; ;

2 2 2 22 1 2 1 and

2 2 2 23 3

Coordinates of ( 3; 3

M M

M M

DM MB

DB M x y

x yM x y

x y

x yD

=

- -æ ö= = ç ÷è ø

+ + - -æ ö æ ö\ = =ç ÷ ç ÷è ø è ø

+ - + -\ = =

\ = - = -= - - )

2 22 1 2 1

2 2

( ) ( )

[1 2] [ 2 2]

1 16

17 units

BC x x y y= - + -

= - + - -

= +

=

1 1 o

2 ( 2) 42 1

tan tan (4) 75,964

BCm

ma - -

- -= =

-= = =

1The gradient of a line perpendicular to 4

The equation of the line is1 ( )4

11 ( 2)41 114 2

1 14 2

A A

BC

y y x x

y x

y x

y x

= -

- = - -

- = - +

- = - -

\ = - +

✓

✓

✓

✓

✓

✓ ✓

✓

✓ ✓

✓

✓



4.6 The gradient is not defined when a line is parallel to the y-axis(that is, vertical). ALL

x co-ordinates are equal is the equation of the line …………..equation 1

……………………….equation 2 Solve equation 1 and 2 simultaneously Substitute 1 into 2

Point of intersection

(3) [15]

QUESTION 5 5.1

1 mark for scale 2 marks for correct graph

(3) 5.2 5.2.1 Since the zeros of are 0, 6 and 6 (the x-axis is tangent to the curve):

(5)

2x\ = -2 5y x- =

2 ( 2) 532

y

y

\ - - =

=

3( 2; )2

-

236y x= - -

( )f x3

3 2

( ) ( 6)( 6) Coefficient of is (

-1 as given.) 12 3612, 36 and =0

f x x x xa

x

c

f x x x x

b

= - - -

= - + -= = -

✓

✓ ✓

✓

✓

✓ ✓

✓

✓

✓

✓



5.2.2

(4) 5.3

(4) [16] QUESTION 6 6.1

(7)

' 2

'

2

2

3 2

( ) 3 24 36 but at the turning point , ( ) 0

3 24 36 08 12 0

( 6)( 2) 06 or 2

If 2 then (2) (2) 12(2) 36(2) 32(2; 32)

f x x xf x

x xx xx xx xx fA

= - + -

=

\- + - =

\ - + =\ - - =\ = =

= = - + - = -\ -

12

2

122

3

3

3

31 621 62

y xx

y x xdy xdx x

xx

-

-

= -

= -

= +

= +

( )( )

2 o

o 2

2 o

o 2

2

2

2

cos (90 ) 1 cossin(90 ) 1 sin cos

cos (90 )sin(90 ) 1 sinsin

cos cos1 cos

cos (1 cos )1 cos 1 coscos (1 cos )

1 coscos

x xx x x

xLHSx x

xx x

xx x

x xx xxx

RHS

+ -=

- + -

+=

- + -

=+-

=+

- +=

+-

=

=

✓ ✓

✓ ✓

✓

✓

✓

✓ ✓

✓

✓

✓

✓

✓

✓



6.2

(3)

6.3

6.3.1

(5)

o

o

height of towertan 60150

height of tower 150 tan 60 259,808 m

=

= ´ =

o o o

o o o

o o o o

o o

o

o

ˆ 127 90 =37ˆ 270 255 15ˆ 180 15 37 128

sin sin0,67

sin 37 sin1280,67 sin 37

sin128 0,51 km

A

B

CBC ABA CBC

BC

= -

= - =

= - - =

=

=

´=

=

✓

✓

✓ ✓

✓

✓

✓

✓


Copyright reserved

6.4 6.4.1

Shape(1 each = 2) Min TP (1 each = 2) Max TP (1 each = 2) = 6

(6)

6.4.2 13° and 103° (2)

[23]

TOTAL: 100

✓ ✓

✓

✓

✓shape

✓

✓shape

✓

n3 mathematics november 2016 memorandum...title: microsoft word - n3 mathematics november 2016...

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