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2018 MATHS P2

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N
12/26/2018
15° N
A
40°
70° E
E
35 °S
B
C
S
๐‘ฆ
1
๐‘ฆ = cos ๐œƒ
90°
270°
180°
360°
-1
MATHEMATICS
PAPER 2
SUGGESTED SOLUTIONS
b (i)
Z
b (ii)
5.5 cm
T
7 cm
P (c)
38 °
X
9 cm
Y
FULL SOLUTIONS | 2018
QUESTION ONE
Question
Number
Solution
suggest
|83 ๐‘ฆ5| = |41
1 (a) (i)
|
โ€’5
2
(40) - (3๐‘ฆ) = 8 โ€’ ( โ€’ 5)
40 โ€’ 3๐‘ฆ = 13
โ€’ 3๐‘ฆ = 13 โ€’ 40
โ€’ 3๐‘ฆ = โ€’ 27
๐‘ฆ=9
(a)(ii)
๐‘ฉ=
(๐Ÿ–๐Ÿ‘ ๐Ÿ—๐Ÿ“)
|๐‘ฉ| = ๐Ÿ๐Ÿ‘
๐‘ด
1 (b) (i)
โ€’๐Ÿ
(
๐Ÿ
๐Ÿ“
= ๐Ÿ๐Ÿ‘ โ€’ ๐Ÿ‘
โ€’๐Ÿ—
๐Ÿ–
)
P(both black) = P(ww )
๐Ÿ”
๐Ÿ“
= ๐Ÿ๐Ÿ“ × ๐Ÿ๐Ÿ’
๐Ÿ
=๐Ÿ•
P(different colours) = P(WG or GW )
๐Ÿ”
๐Ÿ—
๐Ÿ—
๐Ÿ”
= ๐Ÿ๐Ÿ“ × ๐Ÿ๐Ÿ’ + ๐Ÿ๐Ÿ“ × ๐Ÿ๐Ÿ’
=
๐Ÿ“๐Ÿ’ + ๐Ÿ“๐Ÿ’
๐Ÿ๐Ÿ๐ŸŽ
๐Ÿ๐Ÿ–
= ๐Ÿ‘๐Ÿ“
2
By T.H Musondela
QUESTION TWO
Question
Number
Solution
2 (i)
๐Ÿ
(๐š)
๐’™=
๐’™=
๐Ÿ‘๐’™ โ€’ ๐’™ โ€’ ๐Ÿ“ = ๐ŸŽ
โ€’ ( โ€’ ๐Ÿ) ± ( โ€’ ๐Ÿ)๐Ÿ โ€’ ๐Ÿ’(๐Ÿ‘)( โ€’ ๐Ÿ“)
๐Ÿ(๐Ÿ‘)
๐Ÿ ± ๐Ÿ”๐Ÿ
๐Ÿ”
๐’™ =1.47 or -1.14
๐Œ๐š๐ญ๐ก๐ž๐ฆ๐š๐ญ๐ข๐œ๐ฌ
๐‚๐ก๐ž๐ฆ๐ข๐ฌ๐ญ๐ซ๐ฒ
๐Ÿ’
๐ŸŽ
๐Ÿ“
3
๐Ÿ
0
๐Ÿ”
๐๐ก๐ฒ๐ฌ๐ข๐œ๐ฌ
or
M
P
C
4
5
3
2
6
(b) (ii)
(a) 6
(๐’ƒ) ๐Ÿ“ + ๐Ÿ + ๐ŸŽ = ๐Ÿ•
(c) ๐Ÿ
3
By T.H Musondela
QUESTION THREE
Question
Number
3 (a)
Solution
Your Comments
โˆซ๐Ÿ (๐Ÿ + ๐’™ โ€’ ๐’™๐Ÿ)๐’…๐’™
โ€’๐Ÿ
=
[
๐Ÿ ๐Ÿ ๐Ÿ ๐Ÿ‘
๐Ÿ๐’™ + ๐Ÿ๐’™ โ€’ ๐Ÿ‘๐’™
[
๐Ÿ
๐Ÿ
๐Ÿ
]
โ€’๐Ÿ
๐Ÿ
= ๐Ÿ(๐Ÿ) + ๐Ÿ(๐Ÿ) โ€’ ๐Ÿ‘(๐Ÿ)
= (๐Ÿ’ + ๐Ÿ โ€’
๐Ÿ‘
] โ€’ [๐Ÿ( โ€’ ๐Ÿ) + ๐Ÿ๐Ÿ( โ€’ ๐Ÿ)๐Ÿ โ€’ ๐Ÿ๐Ÿ‘( โ€’ ๐Ÿ)๐Ÿ‘]
๐Ÿ–
๐Ÿ
๐Ÿ
)
โ€’
(
โ€’
๐Ÿ
+
+
)
๐Ÿ‘
๐Ÿ
๐Ÿ‘
=
๐Ÿ๐ŸŽ
๐Ÿ•
โ€’
(
โ€’
)
๐Ÿ‘
๐Ÿ”
=
๐Ÿ๐ŸŽ
๐Ÿ•
+
๐Ÿ‘
๐Ÿ”
๐Ÿ—
=๐Ÿ
๐Ÿ
= 4.5 or ๐Ÿ’๐Ÿ
(a) ((ii)
(b) ๐’š = ๐’™ + ๐Ÿ’๐’™
๐’…๐’š
=
๐’…๐’™
1-
โ€’๐Ÿ
๐Ÿ’
๐Ÿ
๐’™
๐Ÿ’
๐Ÿ‘
๐’Ž = ๐Ÿ โ€’ ๐Ÿ๐Ÿ” = ๐Ÿ’
๐’Ž๐’Ž๐Ÿ =โ€’ ๐Ÿ
๐Ÿ’
โˆด ๐’Ž๐Ÿ =โ€’ ๐Ÿ‘
๐’Ž๐’™ + ๐’„ = ๐’š at (๐Ÿ’,๐Ÿ“)๐’™ = ๐Ÿ’ and y= 5
๐Ÿ’
โ€’ ๐Ÿ‘(๐Ÿ’) + ๐’„ = ๐Ÿ“
โ€’ ๐Ÿ๐Ÿ” + ๐Ÿ‘๐’„ = ๐Ÿ๐Ÿ“
๐’„=
๐Ÿ‘๐Ÿ
๐Ÿ‘
๐Ÿ’
โˆด ๐’š =โ€’ ๐Ÿ‘ +
๐Ÿ‘๐Ÿ
๐Ÿ‘
or 3y + ๐Ÿ’๐’š = ๐Ÿ‘๐Ÿ
4
By T.H Musondela
QUESTION FOUR
Question
Number
Solution
Your Comments
b (i)
Z
b (ii)
5.5 cm
T
7 cm
P (c)
38 °
9 cm
X
Y
QUESTION FIVE
Question
Number
Solution
5 (a)
You Comments
๐’ƒโ€’๐’‚
๐Ÿ
๐Ÿ(๐’ƒ โ€’ ๐’‚)
๐Ÿ
๐’‚ โ€’๐’ƒ
= (๐’‚ โ€’ ๐’ƒ)(๐’‚ + ๐’ƒ)
โ€’ ๐Ÿ(๐’‚ โ€’ ๐’ƒ)
= (๐’‚ โ€’ ๐’ƒ)(๐’‚ + ๐’ƒ)
โ€’๐Ÿ
= ๐’‚+๐’ƒ
5(b)(i)
๐Ÿ
(๐Ÿ๐’Œ โ€’ ๐Ÿ๐Ÿ“)(๐’Œ + ๐Ÿ’) = ๐ค
๐Ÿ
๐Ÿ
๐Ÿ๐’Œ + ๐Ÿ–๐’Œ โ€’ ๐Ÿ๐Ÿ“๐’Œ โ€’ ๐Ÿ”๐ŸŽ = ๐’Œ
๐Ÿ
๐’Œ โ€’ ๐Ÿ•๐’Œ + ๐Ÿ”๐ŸŽ = ๐ŸŽ
๐Ÿ
๐’Œ โ€’ ๐Ÿ๐Ÿ๐’Œ + ๐Ÿ“๐’Œ โ€’ ๐Ÿ”๐ŸŽ = ๐ŸŽ
๐’Œ(๐’Œ โ€’ ๐Ÿ๐Ÿ) + ๐Ÿ“(๐’Œ โ€’ ๐Ÿ๐Ÿ) = ๐ŸŽ
(๐’Œ โ€’ ๐Ÿ๐Ÿ)(๐’Œ + ๐Ÿ“) = ๐ŸŽ
๐’Œ = ๐Ÿ๐Ÿ
5
By T.H Musondela
5(b) (ii)
๐‘ป๐Ÿ = ๐’Œ + ๐Ÿ’ = ๐Ÿ๐Ÿ + ๐Ÿ’ = ๐Ÿ๐Ÿ”
๐‘ป๐Ÿ = ๐’Œ = ๐Ÿ๐Ÿ
๐‘ป๐Ÿ‘ = (๐Ÿ๐’Œ โ€’ ๐Ÿ๐Ÿ“) = ๐Ÿ(๐Ÿ๐Ÿ) โ€’ ๐Ÿ๐Ÿ“ = ๐Ÿ—
โˆด 16, 12, 9
5(b) (ii)
๐’‚
๐‘บโˆž = ๐Ÿ โ€’ ๐’“
๐‘บโˆž =
๐Ÿ๐Ÿ”
๐Ÿ‘
๐Ÿโ€’๐Ÿ’
= 64
QUESTION SIX
Q no
Solution
6 (i) (a)
(a)
๐Ÿ
๐‘จ๐‘ฌ = ๐Ÿ‘ ๐‘จ๐‘ช
You Comments
๐Ÿ
= ๐Ÿ‘(๐’‚ + ๐Ÿ๐’ƒ)
(b) ๐‘ฉ๐‘ฌ = ๐‘ฉ๐‘จ + ๐‘จ๐‘ฌ
๐Ÿ
๐Ÿ‘
= โ€’ ๐’‚ + (๐’‚ + ๐Ÿ๐’ƒ)
๐Ÿ
= ๐Ÿ‘ ( โ€’ ๐Ÿ‘๐’‚ + ๐’‚ + ๐Ÿ๐’ƒ)
๐Ÿ
= ๐Ÿ‘ ( โ€’ ๐Ÿ๐’‚ + ๐Ÿ๐’ƒ)
๐Ÿ
๐Ÿ
= ๐Ÿ‘ ( โ€’ ๐’‚ + ๐’ƒ) or = ๐Ÿ‘ (๐’ƒ โ€’ ๐’‚)
(c) ๐‘ฉ๐‘ซ = ๐‘ฉ๐‘จ + ๐‘จ๐‘ซ
= โ€’๐’‚+๐’ƒ
๐Ÿ
(b) ๐‘ฉ๐‘ฌ= ๐Ÿ‘ ( โ€’ ๐’‚ + ๐’ƒ) and ๐‘ฉ๐‘ซ =
โ€’๐’‚+๐’ƒ
๐Ÿ
๐‘ฉ๐‘ฌ= ๐Ÿ‘ ๐‘ฉ๐‘ซ โˆด the points B, E and D are colinear
6
By T.H Musondela
6 (b)
Start
Enter x, y
M= sqrt (x^2+y^2)
Is M < ๐ŸŽ?
Yes
Error โ€˜Mโ€™ must be positive
no
Display M
Stop
7
By T.H Musondela
QUESTION SEVEN
Question
Number
7 (a) (i)
Solution
Comments
3
2
(a) ๐‘ฆ = 2๐‘ฅ โ€’ 3๐‘ฅ + 5
When ๐‘ฅ =โ€’ 2,
3
2
๐‘ = 2( โ€’ 2) โ€’ 3( โ€’ 2) + 5
๐‘ =โ€’ 23
7 (a) (ii)
7 (a) (iii)
๐‘ฆ=๐‘ฅ
X= -1.1
8
By T.H Musondela
(a) (1.5,5) and (2,7.3)
7 (a) (iv)
๐‘ฆ2 โ€’ ๐‘ฆ1
M =๐‘ฅ
7.3 โ€’ 5
= 2 โ€’ 1.5
2 โ€’ ๐‘ฅ1
=
2.25
0.5
= 4.5
7 (b)
3
2
โ€’
x+1 xโ€’1
=
3(x โ€’ 1) โ€’ 2(x + 1)
(x + 1)(x โ€’ 1)
=
3x โ€’ 3 โ€’ 2๐‘ฅ โ€’ 2
(x + 1)(x โ€’ 1)
xโ€’5
= (x + 1)(x โ€’ 1)
QUESTION EIGHT
Question
Number
Solution
Comments
๐พ๐‘…
8 (i)
80
(a) sin 60 = sin 52
๐พ๐‘… =
80sin 60
sin 52
= 87.92015
= 87.9°
(b) A=
1
(80)(50)๐‘ ๐‘–๐‘› 60
2
= 1732.0508
2
= 1700 ๐‘š (3sf)
8 (ii)
๐Ÿ
(iii) ๐Ÿ (sd)(KN) = Area of ฮ” KNR
๐ฌ๐ =
๐Ÿ × 3260
๐Ÿ–๐ŸŽ
= 81.5 (m)
9
By T.H Musondela
8 (B)
๐‘ฆ
1
๐‘ฆ = cos ๐œƒ
90°
180°
270°
360°
๐‘ฅ
-1
8 (c)
3
12๐‘‘๐‘›
3
15๐‘๐‘‘
3
÷
9๐‘ ๐‘›
2 2
10๐‘ ๐‘‘
12๐‘‘๐‘›๐‘›๐‘›
10๐‘๐‘๐‘‘๐‘‘
×
15๐‘๐‘‘๐‘‘๐‘‘
9๐‘๐‘๐‘๐‘›
๐Ÿ
=
๐Ÿ–๐’
๐Ÿ
๐Ÿ—๐’„
QUESTION NINE
Question
Number
8 (a)
Solution
Comments
๐’‡
2
10
15
23
30
10
๐’™
15
25
35
45
55
65
โˆ‘๐’‡ = 90
๐Œ๐ž๐š๐ง (๐—) =
โˆ‘๐‘“๐‘ฅ
๐Ÿ
๐’‡๐’™
30
250
525
1035
1650
650
๐’‡๐’™
450
6250
18375
46575
90750
42250
โˆ‘๐‘“๐‘ฅ = 4140
โˆ‘๐‘“๐‘ฅ = 204650
2
4140
โˆ‘๐‘“ = 90
= 46
SD =
=
๐œฎ๐’‡๐’™
โˆ‘๐’‡
๐Ÿ
- (๐—)
๐Ÿ๐ŸŽ๐Ÿ’๐Ÿ”๐Ÿ“๐ŸŽ
๐Ÿ—๐ŸŽ
๐Ÿ
- (46)
๐Ÿ
= ๐Ÿ๐Ÿ“๐Ÿ•.๐Ÿ–๐Ÿ–๐Ÿ–๐Ÿ–๐Ÿ–๐Ÿ–๐Ÿ–
= ๐Ÿ๐Ÿ.๐Ÿ“๐Ÿ”๐Ÿ“๐Ÿ‘๐Ÿ–
= 12.6
(3sf)
10
By T.H Musondela
(b)(i)
Marks(x )
โ‰ค10
โ‰ค 20
โ‰ค 30
โ‰ค 40
โ‰ค 50
โ‰ค 60
โ‰ค 70
CF
RCF
0
0
2
0.02
12
0.13
27
0.3
50
0.56
80
0.89
90
1
๐Ÿ
0.๐Ÿ—
0.๐Ÿ–
0.๐Ÿ•
0.๐Ÿ”
0.๐Ÿ“
0.๐Ÿ’
0.๐Ÿ‘
0. ๐Ÿ
0. ๐Ÿ
๐ŸŽ
๐Ÿ๐ŸŽ
(iii)
65
× 1 = 0.65
100
๐Ÿ๐ŸŽ
๐Ÿ‘๐ŸŽ
๐Ÿ’๐ŸŽ
,
โˆด 65 percentile 53
11
By T.H Musondela
๐Ÿ“๐ŸŽ
๐Ÿ”๐ŸŽ
๐Ÿ•๐ŸŽ
QUESTION 10
Question
Number
Comments
Solution
(ii) Un anticlockwise Rotation of 270 about (0,0)
10 (a)
Or
clockwise Rotation of 90 about (0,0)
10 (b)
it is enlargment, centre (0,0) and scale factor 2
10 (c)
This is a stretch in which y axis is invariant and k = -2
โˆด the required matrix is
( โ€’02 01)
Or
(
)( 21
๐‘Ž ๐‘
๐‘ ๐‘‘
4
1
)
=
( โ€’14
โ€’8
1
)
Solving the simultaneous equation
๐‘Ž =โ€’ 2 , b = 0 , c = 0 and d = 1
โˆด the required matrix is
10 (d)
(i)
(
)( 21
1 0
โ€’2 1
( โ€’02 01)
2 4
4 1
)
(
2 2
= โ€’3 0
โˆด the coordinates are
(2,-3) ,
(2,0) ,
( 4,-7)
12
By T.H Musondela
4
โ€’7
)
QUESTION ELEVEN
Question
Number
11 (a)
Solution
Comments
}
๐‘ฅ + ๐‘ฆ โ‰ค 60
๐‘ฅ + ๐‘ฆ โ‰ฅ 35
๐‘–๐‘› ๐‘Ž๐‘›๐‘ฆ ๐‘œ๐‘Ÿ๐‘‘๐‘’๐‘Ÿ
๐‘ฆโ‰ฅ6
๐‘ฆ โ‰ค 14
11 (b)
70
๐’š โ€’ ๐’‚๐’™๐’Š๐’”
60
50
๐’™ + ๐’š = ๐Ÿ”๐ŸŽ
40
๐’™ + ๐’š = ๐Ÿ‘๐Ÿ“
30
20
๐’š = ๐Ÿ๐Ÿ’
Solution
set
10
๐’š=๐Ÿ”
0
11 (c)
10
20
30
40
(i) 10 teachers
(ii) 52 learners
d
K 30.00
13
By T.H Musondela
50
60
70
QUESTION 12
Question
Number
Solution
Comments
1
3 × 11.4 × [14 × 10 + 8 × 4 +
14 × 10 × 8 × 4 ]
1
(3 possible answers do exist here)
= 3 × 11.4 × [140 + 32 + 4480 ]
1
= 3 × 11.4 × [2389328 ]
= 907.94484
N
12 (b)
15° N
A
40°
70° E
E
C
B
35 °S
S
(i)
Differ in latitude = 15+35 = 50°
XZ =
=
50
× 2 × 3.142 × 6370
360
2,001,454
360
= 5,559.5944444 km
= 5,560 km (3sf)
(ii) (a)
๐œƒ
× 2 × 3.142 × 6370cos 35
360
= 900
900 × 360
๐œƒ= 2 × 3.142 × 6370cos 35
= 9.881088 (3sf)
= 9.9° (1dp)
(ii) (b)
Longitude of Q is (70-9.881088) E = 60.1
โˆด Q (35 °S, 60.1 °E)
14
By T.H Musondela
This not a frustum. Me I donโ€™t know
teacher
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