n2 mathematcis april 2016 memorandum - future …...title microsoft word - n2 mathematcis april 2016...

MARKING GUIDELINE -1- T...(E)(J)T MATHEMATICS N2

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NATIONAL CERTIFICATE

APRIL EXAMINATION

MATHEMATICS N2

APRIL 2016

This marking guideline consists of 11 pages.

MARKING GUIDELINE

MARKING GUIDELINE -2- T...(E)(A…)T MATHEMATICS N2


INSTRUCTIONS AND INFORMATION 1. 2. 3. 4. 5. 6. 7. 8.

ü equals 1 mark and . equals ½ mark. Half marks are not allocated, unless indicated otherwise. Where a formula is required, using the wrong formula is a principle error and NO marks are allocated. Students should show ALL formulae and intermediate steps and simplify where possible. ALL final answers must be rounded off to THREE decimal places (unless indicated otherwise). Questions may be answered in any order, but subsections of questions must be kept together. If subsections are separated, the student can be penalised by ONE mark. Where a student copied wrongly from the question paper, and the standard of the question is still the same, the student will be penalised by ONE mark. If the copying error simplifies the question and makes it easier, the student forfeits the marks. Questions must be answered in blue or black ink. Answers in PENCIL are not marked as it is regarded as rough work.

Ö



)2)(2)(4(2)4(8)4(2 22

+--=---xxa

xxa

QUESTION 1 1.1 1.1.1

(4) 1.1.2

(3) 1.2 1.2.1

(3)

1.2.2

(3)

1.2.3

(5) [18]

32882 22 +-- xaax

3 2

2

3 12 93 ( 4 3)3 ( 3)( 1)

x y x y xyxy x xxy x x

- +

= - += - -

2 2

2

2 2 33 2

12 4 96

12 56

x xx

x xxxx

+ -

+ -=

-=

313

3 33

4 33

aa

a aa

aa

--

- +=

-=

22 4 1 12 3 4 12 4 8

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

2

ax a xx x x ax a

a x x xx x a x

- +÷ ´

- - - -

- - += ´ ´

- + -

=

ü ü ü ü

ü

ü

ü

ü

ü

ü

ü

ü

ü ü ü

ü



QUESTION 2 2.1

(4)

2.2 2.2.1

(3)

2.2.2

(3) 2.3

(4) [14]

0,786 2,120x or x= = -

2

2

......................( )pqB xKxpqBKxpqxKB

=

\ =

\ =

.................( )

( )

nEI nR nr

IR Inr nEIR nE InrIR n E Ir

IRnE IRIRor n

IR E

=+

+ =\ = -\ = -

\ =--

=-

9 137,81.............(1) 14 198,26............(2)(1) (2) 5 60,45 = 12,09

From (1) 9(12,09) 137,81

cc

c

+ =+ =

- - = -\

+ =

!

!

!

!

137,81 108,81 29

cc

\ = -\ =

ü

ü

ü

ü ü

ü

ü

ü

ü

ü

ü

ü



322823

==

=

x

x

x

QUESTION 3 3.1

(4)

3.2

(3)

3.3 .

Show ALL the steps. Answers only are not acceptable. Without rounding off:

(5)

5 33 18 36

2 4

76 12

73 6

10 5

a b a ba ba a bba a bba b

- ´

= ´

= ´

=

242.3 =x

513,4 265 12,8 54,6

x ´=

´513,4 265log log

12,8 54,6x ´

\ =´

15 2log log13,4 log 265 log12,8 log54,6x\ = - - -

1log 5log13,4 log 265 log12,8 log54,62

x\ = - - -

4,003

log 4,0031010 069,317

xxx

\ =

\ =\ =

4,002744313

log 4,0027443131010 063,390

xxx

=

\ =\ =

ü

ü

ü

ü

ü

ü

ü

ü

ü

ü ü

ü

ü

ü



sradw

w

nw

/519,366

)603500(2

2

=

=

=

p

p

D

D

DnV

=

=

=

995,10

)603500(2015 p

p

3.4

(4)

[16] QUESTION 4 4.1

4.1.1

The angular velocity of the pulley in radians per second.

(3)

4.1.2

(3) 4.2 4.2.1

(3)

log 9 log 81 log 3 5log 2187 7

x x x

x

+ -=

2 4

7

log 9 log 81 log 3log 2187

log 3 log 3 log 3log 3

2log 3 4log 3 log 37 log 3

5log 37 log 357

x x x

x

x x x

x

x x x

x

x

x

LHS

RHS

+ -=

+ -=

+ -=

=

= =

10,955D cm=

o

22,5 where 22,5cm and 24 cm24

0,938 rad or 53,743

sr

s r

q

q

q

=

= = =

=

o

Final round off: 0,9375 rad or 53,715q =

ü

ü

ü

ü

ü

ü

ü

ü

ü



oq

q

q

824,17952,0cos5,134

128cos

=

=

=

4.2.2

(2) 4.3

4.3.1

(2) 4.3.2

Follow up with height of 4.3

(3) 4.4

(4)

[20] QUESTION 5 5.1 5.1.1

*without rounding off :

(3)

2

2

2

12

1 (24) (0,938)2270,144 cm

A r q=

=

=

2

2

or1 (24) (0,9375)2270 cm

=

=

2 2 2

2 2(13,2) (8)110,24

10,5 cm

AD AB BD

AD

= -

= -=

=

2

2

3

13

1 (8) (10,5)3703,717 cm 703,717 m

V r hp

p

=

=

== !

3 or 703,7167544 cm703,7167544 m= !

2

0 0[ 12 13 21 24 23 18] 0,82

88,8cm

A += + + + + + + ´

=2or A 8880mm=

oq 885,17=

ü

ü

ü

ü

ü

ü

ü

ü

ü ü

ü

ü

ü

ü

ü

ü

ü

ü



307,4125,17061285,134

2

222

==

-=

ACACAC

17,41824,17sin.5.1345,134

824.17sin

=

=

o

opp

)812,0549,46(

812,0

552,46726,0sin9,56307,41sin

1

radiansor

radians

==

=

=

=

=

-

bb

b

b

b

b

o

o

565,296435,63360

565,116435,631802tan

]360;0[ ; 073,5tanA

=-=

=-=-=

Î=+

A

AA

A oo

5.1.2

or

or

Or

or

(3)

5.1.3

Or

(3)

5.2

(3)

o

tan128128 tan128 tan17,88541,306 cm

In ABCAC

ACACAC

q

q

D

=

\ =

\ =\ =

o

tan128128 tan128 tan17,82441,155 cm

In ABCAC

ACACAC

q

q

D

=

\ =

\ =\ =

o

sin134,5134,5sin134,5sin17,88541,306 cm

In ABCAC

ACACAC

q

q

D

=

\ =

\ =\ =

o

o

46,3260,808 rad

46,3480,809 rad

or

bb

bb

=\ =

=\ =

ü

ü

ü ü

ü

ü

ü

ü

ü

ü

ü

ü

ü

ü

ü

ü

ü

ü

ü

ü

ü



5.3

(4) [16] QUESTION 6 6.1

shape x-intercepts y-intercept turning points ½ mark(Ö) 60°,300°

½ mark& ½ mark Ö Ö 0,5 ½ mark Ö

(180°,-1,5) ½ markÖ (360°, 0,5) ½ markÖ

6 x ½ marks=3 marks

(3) 6.2 6.2.1

(2)

o

o

tan 5725

25 tan5738,497m

AC

ACAC

=

\ =\ =

1 os for 0 3602

y c x xo o= - £ £

( )

2 4 5 01 ( 5) 01 or 5

x xx xx x

- - =

+ - =

= - =

ü

ü

Ö

Ö

Ö

Ö

Ö=shape

ü ü ü

ü

Ö

ü



6.2.2

(1)

6.2.3

(1)

6.2.4

(2)

TP- ½ mark

x-intercepts(½mark & ½ mark) Y-intercept(½ mark) Shape (1 mark)

(3)

let 05

xy

== -

2( 4)2

2

bxa-

=

- -=

=

2

If 2 then(2) 4(2) 59

therefore, TP is (2; 9)

xy

=

= - -= -

-

ü

ü

ü ü

Ö

Ö Ö

Ö Ö

Ö

Ö

ü shape of parabola

Ö

üshape of straight line

ü

Ö

Ö

Ö ü

Ö

MARKING GUIDELINE -11- T...(E)(J)T MATHEMATICS N2

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6.3

x-intercept ½ mark y-intercept ½ mark shape 1 mark

(2)

6.4

(0;–5) and (5;0) (2) [16]

TOTAL: 100

Ö

Ö ü

n2 mathematcis april 2016 memorandum - future …...title microsoft word - n2 mathematcis april 2016...

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