mathematics p2 memo march 2010 eng
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This memorandum consists of 14 pages.
MATHEMATICS P2
FEBRUARY/MARCH 2010
MEMORANDUM
NATIONALSENIOR CERTIFICATE
GRADE 12
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QUESTION 1
1.1 Range = 26 4 =22 maximum and minimumvalues
answer ANSWER ONLY: Full Marks(2)
1.2 Mean
=12
71417232625221913854 +++++++++++
=12
183
method
183answer
= 15,25 (3)
1.3 Standard deviation = 7,6 (7,59522..) answer (2)
1.4.1Increase in mean =
( ) ( )12
1953 +
C2= per month. answer (2)
1.4.2 The maximum value increases by 1C and theminimum value increases by 5C. This implies thatthe range of the range of the data will now decrease.This will result in the standard deviation gettingsmaller. (new SD = 6,27..)
decrease in range
decrease in standard deviation(2)
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QUESTION 2
2.1.1Survey of training and goals scored
0
2
4
6
8
10
12
1416
0 5 10 15 20 25
Hours of training
N u m
b e r o
f g o a
l s s c o r e
d
plotting the points
All 9 pointcorrect 3 marks5 or 7 pointscorrect 2 marks1 or 2 pointscorrect 1 mark 0 points correct 0 marks
(3)
2.1.2 A(indicated on the graph) answer (1)
2.1.3 8 Goals answer (2)
2.2 Let the mean time for all 560 learners be x.Then the mean time for the learners living in neighbourhood C is also
x.
560
)200()32225()24135( x x
++=
29
10440360
20072003240560
==
++=
x
x
x x
equal meantimes
mean number
simplification
answer (4)
[10]
A
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4 DoE/Feb.
QUESTION 3
0 10 20 30 40 50 60
1511
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QUESTION 4
4.121
4002 =
=PQm
substitution(1)
4.2A:
++2
02;
2
40
A (2 ; 1) x-coordinate y-coordinate
(2)4.3 1. =PQ AB mm
1)2/1.( = ABm , 2= ABmEquation of AB is y = 2 x + c 1 = 2(2) + cc = 3
Equation of AB is y = 2 x 3 .
OR1. =PQ AB mm
1)2/1.( = ABm , 2= ABm
32
421
)2(21
===
x y
x y
x y
1. =PQ AB mm 2= ABmequation of AB
y = 2 x 3 c = 3
(5)
1. =PQ AB mm 2= ABm
gradient of ABsubstitution into
formulaequation of AB
(5)4.4 B is the point (0 ; 3)
5
)03(4)-(0BQ 22
=
+=
coordinates of Bsubstitutionanswer
(3)4.5
5
)23(0)-(0BP 22
=+=
BP = BQBPQ is isosceles.ORBP = 2 + 3
= 5BP = BQBPQ is isosceles
BP = 5BP = BQ
(2)
BP = 5BP = BQ
(2)4.6 If PBQR is a rhombus then A is the midpoint of BR.
Let the coordinates of R be ( x ; y)
4
22
0
=
=+
x
xand
5
12
3
=
=
y
y
R(4 ; 5)
OR RQ || PB so 4= R x
RQ = PB = 5, so 5= R y R(4 ; 5)
A is the midpoint of BR
x coordinate y coordinate
(3)
RQ || PB x coordinate
y coordinate(3)
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QUESTION 5
x
y
P(5 ; 4)
45
BC
D
A
O
5.1AB is defined as 0535 = x y which can be written as 1
53 += x y
53= ABm
Let be the inclination of AB.
53tan =
= 96.30 .
Let be the inclination of CD = 45 + 30,96
= 75,96
Gradient of CD = tan 75,96 = 4.
OR
4
tan
4
153
1
153
45tan.tan145tantan
)45tan(tan
===
+=
+=
+=
CD
CD
m
m
53= ABm
53
tan =
= 96.30
= 75,96
gradient of CD(5)
expansiontan 45 = 1tan =
53
substitutionanswer
(5)
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5.2 Equation of CD is y = 4 x + c 4 = 4(5) + cc = 16
Equation of CD is y = 4 x 16 .
OR
164
2044
)5(44
===
x y
x y
x y
y- interceptequation of CD
(2)
substitutionequation of CD
(2)[7]
QUESTION 6
6.1
58)2()4(3841644168
03848
22
22
22
=+++++=+++++
=+++
y x y y x x
y x y x
Centre is (4 ; 2) and the radius is 58
completing thesquare (both or one)
factor formcentreradius
(4)6.2 Centre of second circle is (4 ; 6)
Distance between centres is 31,11128)26()44( 2 ==+++ centredistance
(2)6.3 Sum of radii = 71,122658 =+
Distance between centres is 11,31.
sum of the radii > distance between the centres
the circles must overlap and hence the circles must intersect.
sum of radii
conclusion(3)
6.4 Equation of second circle:( )
026128
263612168
26)6(4
22
22
22
=++=+++
=+
y y x x
y y x x
y x
Let ( x ; y) be either of the two points on intersection.Then
026128
0384822
22
=++=+++
y x y x
y x y x
Subtract 0641616 =+ x y
4+= x y Both points of intersection lie on this line. y = x + 4 is the equation of the common chord.
OR
equation of circle inform = 0
statement two points of intersection
subtracting
simplification(4)
and
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Check that the line y = x + 4 cuts the two circles at the same points:
( )
0)1)(3(
032
0642
2644168
26)2(4
2
2
22
22
=+==
=++++=+
x x
x x
x x
x x x x
x x
x = 3 or x = 1
1or 3032
0642
0384168816
038)4(48)4(
03848
2
2
22
22
22
===
==++++
=+++=+++
x x x x
x x
x x x x x
x x x x
y x y x
substitution
answer
substitution
answer (4)
[13] QUESTION 7
7.1.1 )2;5(/ P answer (1)
7.1.2 )2;5(/P x coordinate y coordinate
(2)7.2.1
)7;6()12;14(:
)9;5()10;18(:
)8;4()8;16(:
)9;3()6;18(:
)2;2()4;14(:
E E
L L
H H
U U
K K
So halve and interchange or interchange and halve.
Reflection across y = x followed by contraction by
2
1
OR
Contraction by21
followed by reflection across y = x.
reflectedthe line y = x enlarged
scale factor of 21
(4)
7.2.2)4;8()8;16(
21/ == H
OR )16;8(/ H
(8 ; 4)
(8 ; 16)(2)
7.2.3Area KUHLE : Area =////////// ELHUK
2
12
= 4 : 1 answer
(2)[11]
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QUESTION 8
8.1 For anti-clockwise rotation:
2323
23
221
3
60sin2)60cos(3
120sin2120cos3
sincos/
=
=
=
== y x x
2233
212
233
)60cos(260sin3
120cos2120sin3
cossin/
=
+
=
+=+=
+= y x y
2
233;
2323/P
formula
simplificationsubstitution
answer
simplification
answer
(6)8.2
=
23
21
2 y x
y x 34 = equation 1
+
=
21
23
0 y x
y x += 30 x y 3= equation 2
Substitute equation 2 into equation 1
32
2
24
34
334
===
+==
y
x
x
x x
x x
Q 32;2
y x 34 =
x y 3=
x-coordinate y-coordinate
(4)[10]
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QUESTION 9
9.1.153
sin = and54
cos =
57
cossin =+
correct quadrantand values.
53
sin =
54
cos =
answer (4)
9.1.2
724
2592516
54
53
2
sincoscossin2
2cos2sin
2tan 22
=
=
==
OR
724
43
1
432
tan1tan2
2tan
2
2
=
=
=
4
53
2cos2sin
sin 2 = 2sin .cos cos 2 = cos 2
sin2 substitution
answer (5)
expansion
substitution
answer (5)
9.2.1
x
x x x
x
x x x x
x x x x
cos1
sin1
cossin
)(cos
)sin)(sin())(tan(cos
)90cos().180sin(tan).360cos(
22
2
2
2
=
=
=
+
cos x sin x
sin x
x x
2
2
cossin
answer (5)
9.2.2 x = 30
3
2
23
130cos
1 ==
x = 30
answer (2)
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QUESTION 10
10.1.1
q p +=
+=+=
12sin36cos12cos36sin
)1236sin(48sin
writing 48 in terms of
36 and 12 expansionanswer
(3)
10.1.2
q p ==
=12sin36cos12cos36sin
)1236sin(24sin
OR
q p ==
=12sin36cos12cos36sin
)1236sin(24sin
writing 24 in terms of 36 and 12
expansionq p =24sin
(3)
writing 24 in terms of 36 and 12 expansion
q p =24sin (3)
10.1.3
)(224cos
24cos)(2
24cos24sin248sin
q pq p
q pq p
+=
=+=
OR
( )( )
++=
+=+=
=
2
2
2
1121
48sin1121
248cos1
24cos
124cos248cos
q p
OR
= 24sin124cos 22 22 )(124cos q p =
2)(124cos q p =
124cos248cos 2 =q p +=48sin
answer (3)
124cos248cos 2 =q p =24sin
answer (3)
= 24sin124cos 22q p =24sin
answer (3)
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10.2
( )
23
)20cos20(sin23
20sin21
20cos23
20sin
20sin41
20sin20cos23
20cos43
20sin41
20sin20cos23
20cos43
20sin
20sin21
20cos23
20sin21
20cos23
20sin
)20sin60cos20cos60(sin20sin60cos20cos60sin20sin
))2060(sin())2060(sin(20sin
80sin40sin20sin
22
222
2
2222
22
2
222
222
222
=
+=
++=
++
+++=
++
+=
+++=+++=
++
OR
Use2
2cos1sin 2
=
( ){ }
( ){ }
( ){ }
23
023
20cos20cos21
221
23
20cos20cos60cos221
23
160cos40sin60sin40cos.60cos40sin60sin40cos.60cos21
23
160cos80cos40cos21
23
=
=
=
=
+++=
++=
LHS
40= 60 -20
80= 60 +20
expansions
substitution
simplification
factorisation
(7)
40= 60 20
80= 60 +20
expansionof cos 40
expansionof cos 60
simplification
simplification
answer for bracket
(7)
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10.3.1
x
x x x
x x
x x
x x x x
x x x x
cos1
)cos1()cos1)(cos1(
cos1cos1cos1
sin
cos1)cos(sinsin
cos1cossinsin
2
2
222
224
=+
+=
+=
+=
++=
++
factorisation
1cossin 22 =+ x x
identity
factorisation
(4)
10.3.2
Z k k x
x
x
+===+
;360.180
1cos
0cos1
Undefined for Z k k x += ;360.180 .
0cos1 =+ x
180 + k.360 (2)[22]
QUESTION 11
11.1
0)sin21(sin
0sin2sin
sin21sin1
2cossin1
2
2
=+=+
=+=+
x x
x x
x x
x x
0sin = x or `21
sin = x
180.k x = or 360.30 k x += Z k 360.210 k x +=
{ } 036;330;210;180 x
OR
0)sin21(sin
0sin2sin
sinsin1sin1
sincossin1
2cossin1
2
22
22
=+=+
=+=+
=+
x x
x x
x x x
x x x
x x
0sin = x or `21
sin = x
180.k x = or 360.30 k x += Z k 360.210 k x +=
{ } 036;330;210;180 x
expansion
factorisation
equations
180.k x =solution for
`21
sin = x
answers(7)
expansion
factorisation
equations
180.k x =solution for
`2
1sin = x
answers(7)
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11.2
45 90 135 180 225 270 315 360
-3
-2
-1
1
2
3
x
y
f
g
1+sin x max and min
valuesshape
cos2 xamplitudeintercepts
(4)
11.3 210180 x or 360330 x answer (3)
[14]QUESTION 12
12.1
cos)sin(sin
DE
cosDF
DEDE
DFcos
)sin(sin
DF
DFBC but
)sin(sin
BC
sin)sin(BC
sinBC
)(180sin[
+=
=
=
+=
=
+=
=+
=+
b
b
b
b
b
sine rule
)(180 +=C B A
BC =
BC = DF
manipulation
DE =
(6)
12.2 substitutionnumerator
m1559,50 27cos.79sin
43sin2000DE
=
=
substitutiondenominator
answer (3)
[9]
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