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National Senior Certificate
Grade 12
Mathematics
Paper 2
MEMORANDUM
Other products for Mathematics available from Learning Channel:
2
1.1
1.2
1.3
1.4
1.5
2010 Mathematics Grade 12 Paper 2: Memorandum
)
(
3 + (–1)
M = ​_
​ –12+  3 
​; _
​  2   
​ 
​✓
M = (1; 1) ✓
4 + (–2) _
4 + (–2)
Midpoint FH = ​_
​     
​; ​     
​ 
​✓
(
2
2
)
= (1; 1) ✓
\ midpoint FH = midpoint EG
\ lines bisect each other.
3 – (–1) _
4   ​= –1 ✓
mEG = _
​ –1 – 3  ​ 
 
= ​ –4
4 – (–2) _
​ 
 ​ 
 
= ​ 6 ​ = 1 ✓
mFH = _
4 – (–2) 6
\ mEG × mFH = –1
Þ EG^FH ✓
\ using 1.2, diagonals of EFGH bisect at 90° ✓
\ EFGH is a rhombus.
OR
____________________
lengthEH = √
​ (–1 – (–2))2 + (3 –  
(–2))2 ​
______
= ​√1 + 25 ​ 
___
= ​√26 ​ ✓
________________
lengthEF = ​√(–1 – 4)2 + (3  
– 4)2 ​
______
=√
​ 25 + 1 ​ 
___
= ​√26 ​ ✓
\ using 1.2, EFGH is a parallelogram (diagonals bisect) ✓
But EH = EF ✓
\ EFGH is a rhombus.
mEG = –1 from above ✓
\ y = –x + c Substitute (–1; 3): ✓
3 = –(–1) + c
2=c
\ y = –x + 2 ✓
y = –x + 2 Let x = ​ _52 ​. 
y = – ​ _52 ​ + 2 ✓
y = – ​ _1 ​ ✓
(
2
)
\ ​ _​ 52 ​;  _
​ –3
  ​ ​does not lie on the line. ✓
4
1.6
4 – (–1) _
mFH = _
​  4 – 3   
​ 
= ​ 51 ​ = 5 ✓
\ tan a = 5
A = tan–1(5) ✓
= 78,69° ✓
(2)
(2)
(4)
(3)
(3)
(3)
2010 Mathematics Grade 12 Paper 2: Memorandum
1.7
1.8
2.1
2.2
2.3
2.4
2.5
3
__________________
lengthEG = √
​ (3 – (–1))2 + (–1  
– 3)2 ​
___
= ​√32 ​ ✓
___________________
lengthHM = √
​ (1 – (–2))2 + (1 –  
(–2))2 ​
___
= ​√18 ​ ✓
___
___
\ area ▵EGH = ​ _12 √
​​  18 ​ × ​√ 32 ​ ✓
= 5,05 units2 ✓
G ® H back 5 down 1
\ E ® P is the same ✓
\ P(–6; 2) ✓ (or equivalent)
▵EDC is right angled at C (tangent, radius)
ED2 = EC2 + DC2 ✓
132 = 122 + DC2
25 = DC2
5 = DC ✓
DC2 = (a – 1)2 + (2 – (–1)2 ✓
25 = a2 – 2a + 1 + 9 ✓
0 = a2 – 2a – 15 ✓
a = –3; a = 5 ✓
By inspection, for this sketch a = 5. ✓
2 – (–1) _
m =_
​ 
  
​ 
= ​ 3 ​ ✓
4
5–1
–4
_
mtangent = ​  3  ​ ✓
_  ​x  + c Substitute (1; –1):
y = ​ –4
3
_  ​x  + c ✓
–1 = ​ –4
3
​ _13 ​ = c ✓
_  ​x  + _​ 1 ​ 
y = ​ –4
3
3
(4)
(2)
(2)
(5)
DC
y-axis is tangent to circle at A.
\ AD is horizontal \ A(0; 2) ✓✓ (inspection)
(x – a)2 + (y – b)2 = c2 ✓
Substitute (1; –1) and (0; 2):
(0 – 1)2 + (2 – (–1))2 = c2 ✓✓
10 = c2 ✓
(x – 1)2 + (y + 1)2 = 10 ✓
(4)
(2)
(5)
4
2010 Mathematics Grade 12 Paper 2: Memorandum
3.1
and
3.2
y
B′
C′(5; 3)
A′
D′
x
A & D′′
D & A′′
B
C′′
B′′(3; –4)
C
3.3
3.4
4.1
P(x; y)
® P′(–x; –y)
® final image (–45x; –45y) ✓✓
Linear factor ​ _54 ​ 
Þ area factor _
​ 16 ​ ✓
25
\ area = _
​ 16 ​p  units2 ✓
25
(x cos q – y sin q; x sin q + y cos q) ✓
= (6 cos 60°
– 3 sin
60°; 6 sin 60° + 3 cos 60°) ✓✓
__
__
√
√
= ​ __
​ 6 – 3​  3 ​  
​; __
​ 3 + 6​  3 ​  
​ 
​ ✓✓
)
( 2
2
OR
= (​ 3 – _
​ 3​2 3 ​ 
​; 3​√3 ​+ _​ 32  ​)​
__
√  
4.2
5.1
y = –x
__
 
(6; 3) ® (3; 6) ® (3; –6) ✓✓
y
5.1.1 tan q = ​ _ ​ = _​ a ​  ✓
x b
5.1.2
cos (–q) = cos q = ​ _xr  ​ ✓✓
(10)
(2)
(2)
(5)
(2)
(1)
______
r=√
​ a2 + b2 ​ 
(Pythagoras) ✓
a
\ cos (–q) = __
​  ______
  2 
 ​ ✓
2
5.2
5.2.1
5.2.2
​√ a + b  ​ 
cos 53°
= sin (90° – 53°)
= sin 37° ✓
=k ✓
sin (–74°)
= –sin 74°
= –2sin 37°cos 37° ✓✓
_____
= –2k √
​ 1 – k2 ​ ✓
(4)
(2)
(4)
2010 Mathematics Grade 12 Paper 2: Memorandum
5.3
5.3.1
5.3.2
LHS
asin 2a
__
= ​ sin
​ 
+ cos 2a ✓
cos a   
a2sin acos a
__
= ​ sin
+ 1 – 2sin 2a ✓✓
cos a    ​ 
=1 ✓
= RHS
LHS
sin (x – 90°)cos (90° – 2x)
234° ____
=_
​ sin
 ​ 
 
– ​ 
  
    ​
cos 36°
sin (x – 360°)
5
(4)
54° __
 ​ 
 
– ​  sin x   
 
​ ✓✓ ✓✓
=_
​ –sin
cos 36°
(–cos x)sin 2x
54° ___
sin xcos x
  
​ 
– ​ cos x · 2sin
     
​ ✓✓
=_
​ –sin
x
sin 54°
5.4
6.1
6.2
= –1 + 2cos2 x ✓
= cos 2x
= RHS
3cos2 x + 5sin x = 3
3(1 – sin2 x) + 5sin x = 3 ✓
3 – 3sin2 x + 5sin x = 3
0 = 3sin2 x – 5sin x
0 = sin x (3sin x – 5) ✓
sin x = 0 or sin x = ​ _35 ​ ✓✓
\ x = 0° + n180° or x is undefined ✓✓
(n Î )
^
A​C​B
  = 110° – 50°
= 60° ✓
2
\ AB = 1502 + 2602 – (2.150.260)cos 60° ✓
AB2 = 51 100 ✓
\ AB = 226,05 ✓
AB = 226 km ✓
6.2.1 C​^
D​ B = 180° – (q + 30°) ✓
p
6.2.2 In ▵ABC:
    
​
tan q = ​ _
CB
CB tan q = p ………….(i) ✓
CB
___
In ▵CBD:​ sin [180° – (q 
   ​ = _
​ sin8 q  
​✓
+ 30°)]
(8)
(6)
(5)
(1)
CB
   ​ 
=_
​  8    
​✓
​ __
(sin q + 30°) sin q
8(sin q + 30°)
 
​………….(ii) ✓
CB = __
​  sin q   
Combining (i) and (ii):
8sin (q + 30°)
p = __
​ 
 
 
 
​ · tan q
sin q
8
sin
(q + 30°) _
sin q
 
​ · ​ cos
 
 ​ 
✓
p = __
​  sin q   
q
8(sin q + 30°)
​ 
✓
p = __
​  cos q   
(6)
6
7.1
7.2
2010 Mathematics Grade 12 Paper 2: Memorandum
sin 2x = cos (x – 45°)
sin 2x = sin [90° – (x – 45°)]
sin 2x = sin (135° – x) ✓
\ 2x = 135° – x + n · 360° (n Î ) ✓
3x = 135° + n · 360°
x = 45° + n · 120° ✓
or
2x = 180° – (135° – x) + n · 360° (n Î ) ✓
2x = 45° + x + n · 360°
x = 45° + n · 360° ✓
\ for x Î [–180°; 180°]
x = 45°; 165°; –75° ✓✓ ✓
(8)
y
1
g
–180° –135° –90° –45°
45
90
135
180
x
f
–1
7.3
✓✓ ✓for g ✓✓ ✓for ƒ
7.3.1 g(x) ≤ ƒ(x) for [–180°; 90°]
Þ –180° ≤ x ≤ –75° ✓✓ ✓
ƒ(x)
7.3.2 ​ _  ​ 
undefined Þ g(x) = 0 ✓
g(x)
8.1
8.1.1
8.1.2
8.1.3
\ x = –45° only for [–180°; 90°] ✓
_
x​ ​  = 65,27 ✓✓ ✓
Using stats mode on calculator or manually
SD = 8,71 ✓✓
Upper boundary
= 65,27 + 8,71
= 73,98 ✓
Lower boundary
= 65,27 – 8,71
= 56,56 ✓
\ reject 50; 45; 80
i.e. 3 bags would be rejected ✓
(6)
(3)
(2)
(3)
(2)
(3)
2010 Mathematics Grade 12 Paper 2: Memorandum
8.2
Ordered list
11 000
12 600
14 200 Q1 = 14 200
14 500
15 300
Median = 15 350
15 400
16 500
16 800 Q3 = 16 800
18 600
19 600
\ Minimum = 11 000 ✓
Lower quartile = 14 200 ✓
Median = 15 350 ✓
Upper quartile = 16 800 ✓
Maximum = 19 600 ✓
8.2.2 Scale in 100s
7
8.2.1
(5)
110 120 130 140 150 160 170 180 190 200
✓✓ ✓✓
(3)
600
8.2.3 Maximum per day = _
   
​ 
= 700 patients per day ✓
​ 1928
9.1
000
   
​ 
= 392 patients per day ✓
Minimum per day = _
​ 1128
Marks
20 ≤ x ≤ 29
30 ≤ x ≤ 39
40 ≤ x ≤ 49
50 ≤ x ≤ 59
60 ≤ x ≤ 69
70 ≤ x ≤ 79
80 ≤ x ≤ 89
90 ≤ x ≤ 100
F
4
12
30
82
55
35
24
3
CF
4
16
46
128
183
218
242
245
(2)
(2)
8
2010 Mathematics Grade 12 Paper 2: Memorandum
9.2
✓✓ ✓✓
9.3
9.4
​ 
= 123
Median at _
​ 2452+  1 
Median approximately 59 ✓✓
Data is grouped, so original raw data is lost.
\ mean or median will be approximate. ✓✓
(4)
(2)
(2)