N3 Mathematics November 2016 Memorandum...Title: Microsoft Word - N3 Mathematics November 2016...
Transcript of 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
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✓ = 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
- + + -÷
- - -- - - +
´+ - - +-
=+
✓
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✓ ✓ ✓
✓
ü
ü
ü
ü
ü ü
ü ü
ü ü
ü
ü
ü
ü
ü
ü
ü
MARKING GUIDELINE -3- T870(E)(N21)H MATHEMATICS N3
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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
-
-
=
\ =
\ =
\- =
\ = -
\ =
✓
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MARKING GUIDELINE -4- T870(E)(N21)H MATHEMATICS N3
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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
-+ =
- - +-
- =- - + +
+ - + = -+ - + = -
- = -=
✓
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✓ ✓
✓
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✓
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✓
✓
✓
✓
✓
✓
✓
✓
✓
MARKING GUIDELINE -5- T870(E)(N21)H MATHEMATICS N3
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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
-\ =
-\ =
-=
✓
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✓
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✓
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✓
✓ ✓
✓
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✓
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✓
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MARKING GUIDELINE -6- T870(E)(N21)H MATHEMATICS N3
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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
= -
- = - -
- = - +
- = - -
\ = - +
✓
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MARKING GUIDELINE -7- T870(E)(N21)H MATHEMATICS N3
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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
= - - -
= - + -= = -
✓
✓ ✓
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MARKING GUIDELINE -8- T870(E)(N21)H MATHEMATICS N3
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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
+ -=
- + -
+=
- + -
=+-
=+
- +=
+-
=
=
✓ ✓
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✓
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✓
MARKING GUIDELINE -9- T870(E)(N21)H MATHEMATICS N3
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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
= -
= - =
= - - =
=
=
´=
=
✓
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✓ ✓
✓
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✓
✓
MARKING GUIDELINE -10- T870(E)(N21)H MATHEMATICS N3
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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
✓