Grade 11 Mathematics: Memorandum Paper 2
[Pages:3]Mathematics(NSC)/Grade 11/ P2
86
MEMORANDUM
Grade 11 Mathematics: Memorandum Paper 2
Exemplar
1.1.1 AB (5 2)2 (4 0)2 D
626 D
2.3
25 20 5
25
=5D
2
1.1.2 Both pointshavethesamex-valuetherefore
x = 4D
1
mAB 11 0 11 DD
m AD
20 0 0 1 0
2 DD
4
1.1.3
m =
5 2 40
=
3 4D
2.4 No AC andBD arenotequaldiagonals. D
?y
3 4
x
2
D
1.1.4
tan T
3 4D
mAB mBC z1u? AB andBC arenot
2 3.1
perpendicular toeach other. D A'(5;3) DD
2
B'(4;8) DD
1.2 1.3.1 1.3.2 1.4
1.5
1.6.1
?T 36,87D D
2 m = 3 DD
0,81 D -1,92 DD
sinA cosA DD
= -tanA D
tan2x
=
1 3D
?Referenceangle:18,43?D
?2x = 18,43?+ 180?nD
?x = 9,22?+ 90?nD n ZD
?x = 9,22?or 99,22?or 189,22?
KT
5
DD
sin40D sin60D
2 3.2
2 3.3 1 2
3
3.5
4
C'(2;2) D
5
(-y ;x) DD
2
M
idpointofBB?is(82-4
4+8 ;2 )
=
(2;6) DD
mBB'
1 3
D
Equationofperpendicular:
y = 3x + cD
?6= 6+ cD
?0= c
?y = 3x
5
Aypointofintersection-4x = 3xD
? 7x = 0
? x = 0D
? y = 0D
? (0;0) isthepointofintersectionofAA?
and BB?.
3
? KT = 3,71 cm D 1.6.2 PT2 = 72 + 52 -2(7)(5)cos30?DD
3 3.6 A''(3;5) D B''( 8 ;4 ) D
? PT= 3,66cm D 1.7 BasicshapeD
M inimum = 10D
3 4.1
C''( 2;2) D
3
M edianandlower quartileD
Upper quartileandmaximum D
ScaleshownD
10 11
17
20
30
5
1.8 h = 12 D (Pythagoras) V = 1 r2h 3
1 = 3 (5)?(12) D
= 314.16mm3 D
3
2.1 DiagonalsareequalD
Adjacentsidesareperpendicular D
2
2.2
AC (21 0)2 (205)2 D
P'(3;6) D Q'(12;12) D R'(18;3) D S '(9;3) D
666 D
LinesofenlargementDD
BD (11 10)2 (25 0)2 D
P?Q?R?S?ongraph D 4
7
Copyright reserved
Mathematics(NSC)/Grade 11/ P2
87
MEMORANDUM
Exemplar
4.2 PQ 4 1 2 4 2 2 13 D P' Q ' 3 12 2 6 12 2 117
D
cos2 sin 2 TTD
1 sin 2 sin 2 TT 1 2sin 2 T DD
or cos2 T 1 cos2 2cos2 TT1DD
7.1 39,69 cmD
1
9 13 u3 13 Area PQRS 13 u 13 = 13D Area P?Q?R?S? 3 13 u 3 13 D
7.2
sin18x
3 5
D
Reference angle is 36,87?D
18x 216,87 3?60DDn
= 9?13 = 117 The length of the sides of PQRS increase by a factor of 3 to give the length of the sides of
x 12 2?0DDn D OR
P?Q?R?S?. D
18x 323,13 3?60DDn
The area of PQRS increased by a factor of 9
x 18 2?0DDn D
to give the area of P?Q?R?S? . This is 3? i.e
? x 12 , 18, 32 or 38DD
6
the square of the increase of the length of the sides. D
8.1 y - xDD 8.2 In ?PAB:
2
6
PB
5
5.1.1
tan x.cos x tan x sin x sin x
DDDD
DD
sin 90D x sin( y x)
sin x . cos x sin x . 1 D cos x sin x cos x sin x
PB ? 5cos x D
sin( y x)
3
5.1.2
1 1 cos x
or
cos x 1 cos x D
cos 60D DD
tan 45D
8.3 6
In ?PBT:
sin y
PT PB D
PT ?5cos x sin y D sin( y x)
2
1
2 1
DD
1 2
9.1
1 2
bc
sin
x
D
1
9.2
DA^ K 360 90 90DDDx
4
180D x D
5.2.1 cos x (2 cos x ?1) D 5.2.2 cos x = 0 D
1
DAK '1?bcsin(180D x) D
2
? x = 90? + 360?n or 270? + 360?n D n Z (add on the period of the cos graph i.e.360?n
1 2
bc
sin
x
D
to get general solution)
' ABC
3
OR 1
cos x = 2 D
10.1 Sum of lengths is 42,4D
Mean length is 4,24D
2
10.2
? x = 60? + 360?n or 300? + 360?nD,
n ZD
5
Length (cm)
xi x
xi x 2
5.3.1 sin (180?+58?) = - sin 58?D = - kD
2
5.3.2 sin2 58? + cos2 58? = 1 D
? cos2 58? = 1 ?k2
3,2
-1,04
1,0816
3,6
-0,64
0,4096
5
0,76
0,5776
cos58D 1 k?2 DD
3
4,1
-0,14
0,0196
6.1
0,5
or
1 2
D
1
6.2 Sipho, Ray and Vishnu get - 0,17DD
Lorraine gets 0,23DD
4
4,3
0,06
0,0036
4,7
0,46
0,2116
3,4
-0,84
0,7056
6.3
1
sin 2 cos 2
T T
1
sin 2 T cos2 T
D
5,2
0,96
0,9216
4,6
0,36
0,1296
4,3
0,06DD 0,0036DD
4,064D
cos2 sin 2 TT
cos 2
sin
2
D TT
Standard deviation = 4,064 0,67 D
7
9
6
Copyright reserved
Mathematics(NSC)/Grade 11/ P2 MEMORANDUM
10.3
11.1 11.2
Length to width comparison of 10 shells
4
3
W idth (mm)
2
1
0
0
1
2
3
4
5
6
Length (mm)
y
1 2
x
1 2
DD
Line on graph D
90, 330, 740, 940, 1000 DD
Length of pebble/cumulative frequency graph
88
3 2
Cumulative frequency
Length of pebble
D Values plotted at ends of intervals
D D Accurate points
D Accurate curve
D Labels (Length of shell, cumulative
frequency, title)
5
11.3 Median: 49 (47 ? 51) D
Upper quartile: 61 (59 ? 63) D
Lower quartile: 35 (33 ? 37) D
3
Exemplar
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