GRADE 12 EXAMINATION
NOVEMBER 2008
ADVANCED PROGRAMME MATHEMATICS
MARKING GUIDELINES
Time: 3 hours
300 marks
These marking guidelines are prepared for use by examiners and sub-examiners,
all of whom are required to attend a standardisation meeting to ensure that the
guidelines are consistently interpreted and applied in the marking of candidates'
scripts.
The IEB will not enter into any discussions or correspondence about any marking
guidelines. It is acknowledged that there may be different views about some
matters of emphasis or detail in the guidelines. It is also recognised that,
without the benefit of attendance at a standardisation meeting, there may be
different interpretations of the application of the marking guidelines.
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MODULE 1
Page 2 of 18
CALCULUS AND ALGEBRA
QUESTION 1
1.1
(a)
(b)
or
(x – 1) (x 2 + x + 1) 9
x = 19
x2 + x + 1 = 0 9
– 1 ± 1– 4
2
– 1 + 3i
∴ x=
or
2
(1)
(5)
∴ x=
1.2
– 1 – 3i
2
999
(x − 1 − 2 )(x − 1 + 2 )99
= x 2 − x + 2 x − x + 1 − 2 − 2 x + 2 − 2 999 or
( x − 1) 2 − ( 2 ) 2
= x 2 − 2 x − 1 is a factor. 99
By inspection:
x 4 − 2 x 3 + 4 x 2 − 10 x − 5 = (x 2 − 2 x − 1)(x 2 + 5) =
( x 2 – 2 x – 1 )( x –
5 i )( x + 5i ) 999
(10)
16 marks
QUESTION 2
2.1
(a)
(b)
(c)
log 2 x − log( x − 20 ) = 1 9
2x
∴ log
=1
x − 20
99
2x
∴
= 10
x − 20
∴ 2 x = 10 x − 200
∴ 8 x = 200
∴ x = 25 9
(4)
e x = 5e x − 5 9
∴ 4e x = 5 9
∴ e x = 1,25 9
∴ x = ln 1,25 = 0,223... 9
(4)
Let k = x 9
(k − 6)(k + 2) = 0 99
x =6
∴ x = ±6
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99
x ≠ −2 9
(6)
GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS – MARKING GUIDELINES
2.2
Page 3 of 18
⎛ L ⎞
107 = 10 log⎜ −16 ⎟ 99
⎝ 10 ⎠
10,7 = log L − log 10 −16 99
∴ − 5,3 = log L
99
∴ L = 10 −5,3
= 5,01 × 10-6
(6)
20 marks
QUESTION 3
Prove: x k +1 − y k +1 9 9is divisible by x − y .
x k .x − y k . y 99
= p ( x − y ) + y k .x − y k . y from Step 299
= px( x − y ) + xy k . − y k . y 99
= px( x − y ) + y k ( x − y )
= (x − y )( px + y k ) 99
(
(
)
)
So we have shown that provided that x k − y k is divisible by x − y then so is x k +1 − y k +1 .
But since the statement is true for n = 1, then by the argument above it is true for n = 2, and hence
n = 3 and so on for all natural values of n. 99
12 marks
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QUESTION 4
4.1
f ( x) =
(a)
(
)
ex ex −1
3e x
1
1
∴ f ( x) = e x − 99
3
3
(b)
y
(2)
99
9
O
x
− 13 9
4.2
Page 4 of 18
(4)
At Q, log(2 x + 3) = 1 9
∴ 2 x + 3 = 10
99
7
∴x =
2
At P, log(2 x + 3) = −1
∴ 2x + 3 = 1
10
∴ x = − 29
20
999
(6)
12 marks
QUESTION 5
5.1
5.2
3 − a(1) 2 = −4(1) + 5 999
∴ a = 2 99
⎧ 3 − 2 x 2 if
f ( x) = ⎨
⎩− 4 x + 5 if
x ≥1
x <1
lim f ' ( x) = −4 999
h→0 −
(5)
lim f ' ( x) = −4(1) = −4 999
h →0 +
Therefore differentiable at x = 1. 9
99method
(9)
14 marks
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Page 5 of 18
GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS – MARKING GUIDELINES
QUESTION 6
2 x 2 − 5x + 2
2x 2 − x − 1
(2 x − 1)(x − 2) 9
=
(2 x + 1)(x − 1)
2 x 2 − 5x + 2 = 0
(2 x − 1)(x − 2) = 0 9
x = 1 or 2 9
2
g ( x) =
6.1
x=
6.2
1
2
or x = 2 9
x →∞
Horizontal:
2 − 5x + x22
2 − 1x − x12
x = 1 99
999 = 1
y = 19
(7)
2 x 2 − x − 1) ( 4 x − 5 ) − ( 2 x 2 − 5 x + 2 ) ( 4 x − 1)
(
999999
g '( x) =
2
2
2
x
−
x
−
1
(
)
=
6.4
(2)
2x2 – x – 1
(2 x + 1) (x − 1) 9
x = − 12 and
Vertical:
lim
6.3
or
8 x 2 − 12 x + 7
(2 x 2 − x − 1) 2
99
(8)
Δ = 144 − 4(8)(7) = −80 9
8 x 2 − 12 x + 7 ≠ 0 9
Therefore no local max or min turning point. 9
(3)
y
6.5
9
9
9
–2
9
x= − 12
1
2
29
9
x=1 9
y=1
x
(7)
27 marks
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Page 6 of 18
QUESTION 7
7.1
7.2
(
)
⎛ 1 − 12 ⎞
⎜1 + x ⎟ 999999
⎜ 2
⎟
⎝
⎠
cos(a − x ). cos x + sin x(− sin (a − x )(− 1)) 9999
= cos(a − x ) cos x + sin( a − x) sin x 9
= cos(a − x − x) 9
= cos(a − 2 x) 9
1
f ' ( x) = x + x
2
−
1
2
(6)
(7)
13 marks
QUESTION 8
dy
8.1
= 2t 9
dt
dy
1
= 2t × 99
dx
3
8.2
dx
= 39
dt
= 2t
3
(4)
1
2 = 2t × 99
3
t = 39
x = 3(3) + 1 = 10 9
y = 32 = 9
(10;9)
9
(5)
9 marks
QUESTION 9
9.1
3 solutions99
(2)
9.2
x = 0,59
(1)
9.3
− x + 1 = x 3 − 5x 2 + 6 x
∴ 0 = x 3 − 5x 2 + 7 x − 1
x r +1
999
x 3 − 5x 2 + 7 x − 1
9999
= xr −
3 x 2 − 10 x + 7
x1 = 0.59
x2 = 0
x3 = …
………
x = 0,160713 (6 dp) 99
(10)
13 marks
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Page 7 of 18
QUESTION 10
10.1
y ' = 3ax 2 + 2bx + c
999
y ' ' = 6ax + 2b
0 = 6a (2) + 2b
b = −6a
10.2
99
(5)
− 3 = 3a (0) + 2b(0) + c
999
∴ c = −3
Passes through (0;0) ∴ d = 0 9
Passes through (2; -22)
∴ −22 = 8a + 4b − 6 99
∴ b = −2a − 4
∴ −6a = −2a − 4
∴a =1 9
∴ b = −6 9
y = x 3 − 6 x 2 − 3x 9
(9)
14 marks
QUESTION 11
11.1
11.2
π⎞
⎛
P = 3 × ⎜ 2 × ⎟ 99
3⎠
⎝
= 6,28 cm 9
(3)
9999999
1
π
π
π⎤
1
⎡1
A = Δ ABC + 3 segments = × 2 × 2 × sin + 3⎢ × 2 2 × − × 2 × 2 × sin ⎥
3 2
3⎦
2
3
⎣2
2
= 2,82 cm 9
or
A = 1 Sector + 2 segments
= 1 × 2 2 × π + 2 ⎡ 1 × 2 2 × π − 1 × 2 × 2 × sin π ⎤
⎢⎣ 2
2
3
3 2
3 ⎥⎦
or
A = 3 sectors – 2 triangles
= 3 ⎡ 1 × 2 2 × π ⎤ − 2 ⎡ 1 × 2 × 2 × sin π ⎤
⎢⎣ 2
⎢⎣ 2
3 ⎥⎦
3 ⎥⎦
(8)
11 marks
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GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS – MARKING GUIDELINES
QUESTION 12
3
(
)
2
∫ x x −1
12.1
−
1
2
dx 9
or
u = x2 − 1
Let
2
3
=
(
)
1
−
1
2
2 dx 99
2
x
x
−
1
2 ∫2
du = 2 x
dx
3
1
= ⎡⎢(x 2 − 1)2 ⎤⎥ 9999
⎣
⎦2
dx = du
2x
3
=
∫
8 − 3 = 1,10 units999
2
8
x
x −1
2
dx = ∫
3
1 du
2 u
8
⎡ 1⎤
= ⎢u 2 ⎥
⎣ ⎦3
= 8− 3
= 1,10
12.2
5−3
= 0,4 99
5
(a)
x1 = 3, 4
h1 = ln 3, 4
Area1 = 0, 4895
x2 = 3,8
h2 = ln 3,8
Area2 = 0,5340
x3 = 4, 2
h3 = ln 4, 2
Area3 = 0,5740 9999999
x4 = 4, 6
h4 = ln 4, 6
Area4 = 0, 6104
x5 = 5
h5 = ln 5
Area5 = 0, 6438
2
TOTAL = 2,852 units 9
(b)
(10)
(10)
[x ln x − x]53 9
= 5.ln 5 − 5 − 3ln 3 + 3 9999
= 2, 751 units 2 9
(6)
26 marks
QUESTION 13
1
1
1
V = ∫ π (4 − 4 x )dx − ∫ π (1 − x )dx 9999
2
−1
2
−1
1
V =
∫ π (3
3x 2 ) dx
1
1
⎡
⎡
x3 ⎤
4x3 ⎤
= π ⎢4 x −
⎥ − π ⎢ x − ⎥ 9999
3 ⎦ −1
3 ⎦ −1
⎣
⎣
= π [3x x 3 ] 1
4⎤
1⎤
⎡ 4
⎡ 1
= π ⎢ 4 − + 4 − ⎥ − π ⎢1 − + 1 − ⎥ 9999
3⎦
3⎦
⎣ 3
⎣ 3
= π [(3 1) ( 3 + 1)]
= 4π
units 3 9
1
= 4π units 3
13 marks
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GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS – MARKING GUIDELINES
MODULE 2
Page 9 of 18
STATISTICS
QUESTION 1
1.1
1.2
H 0 = 3118
H1 ≠ 3118
(4)
3333
Testing for change therefore involves two rejection regions. 5% level of
significance means each region is 2,5%. 3 The critical value that corresponds
is 1,96. 3
Z=
x−μ
σ
3
n
Z=
3345 − 3118
850,5
40
0.025
333
0.025
1,96
Z = 1,688
This value of Z falls within the acceptance region, therefore not enough evidence to
reject H 0 . Accept H 0 3
1.3
1,96 =
(7)
x − 3118
33
850,5
40
x = 3381,6 g 33
(4)
15 marks
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Page 10 of 18
QUESTION 2
2.1
As x increases y decreases. 33
2.2
(a)
(b)
2.3
b=
r = −0,98 3333
very strong correlation 3
n∑ xy − ∑ x ∑ y
n ∑ x 2 − (∑ x ) 2
(2)
(4)
(1)
33
n = 12 3
12(3291,88) − 511(78,52)
12(289490) − (511) 2 333
b = −0,0072
y − y = b( x − x )
b=
Now for a: y − 6,54 = −0,0072( x − 42,58) 3333
y = −0,0072 x + 6,85
2.4
(a)
(b)
y = 6,85 − 0,0072(95)
333
y = 6,17
unreliable, outside of range, extrapolation33
(10)
(3)
(2)
22 marks
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Page 11 of 18
QUESTION 3
3.1
Proportion not in favour =
Proportion in favour =
202
3
700
498
700
90% confidence interval z= 1,6453
p (1 − p )
3
n
202 498
×
202
700
700 333
± 1,645
700
700
[0,260399; 0,316742] 3333
p ± 1,645
3.2
(a)
105,89 + 110,11
= 108 33
2
(b)
105,89 = 108 − 1,96
σ = 28,48 33
3
3.3
3
σ
700
(10)
(2)
3
⎛10 ⎞
⎛10 ⎞
1 − ⎜⎜ ⎟⎟(0,55)1 (0,45) 9 − ⎜⎜ ⎟⎟(0,45)10 33
⎝1 ⎠
⎝0 ⎠
= 0,99533
(6)
3333
33
(9)
27 marks
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Page 12 of 18
QUESTION 4
4.1
4.2
4.3
8! – 7! x 2! 99
= 30240 9
(3)
0,1 1
9
=
0,5 5
0,4 4
(b)
= 9
0,5 5
(c)
P (A) . P (B) = 0,4 x 0,5
= 0,2 9
and P (A∩B)
= 0,1 9
∴ not equal
∴ not independent 9
⎧ ⎛ 4 ⎞⎛ 8 ⎞
⎟⎟
⎪ ⎜⎜ ⎟⎟⎜⎜
⎪⎪ ⎝ x ⎠⎝ 3 − x ⎠
(a)
P ( X = x) = ⎨
for
⎛12 ⎞
⎜⎜ ⎟⎟
⎪
⎝3 ⎠
⎪
⎪⎩0 elsewhere
⎛ 4 ⎞⎛ 8 ⎞
⎜⎜ ⎟⎟⎜⎜ ⎟⎟
⎝ 2 ⎠⎝1 ⎠ = 12 33333
(b)
55
⎛12 ⎞
⎜⎜ ⎟⎟
⎝3 ⎠
(a)
(1)
(1)
(3)
x = 0,1,2,3
formula 33333
(6)
3
(5)
6
4.4
(a)
∫ (0,2 − 0,02 x)dx 33
2
x − x2
5 100
6
99
2
= 6 − 36 − ( 2 − 4 ) 33
5 100 5 100
= 0,48 33
Probability delay would be between 2 and 6 hours is 0,48
(b)
x − x2
5 100
(8)
m
33 = 0,533
0
m m2
−
= 0,5 3
5 100
20m 2 − 400m + 1000 = 0
3333
m = 2,93 or m = 17,07( N / A)
(9)
36 marks
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GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS – MARKING GUIDELINES
MODULE 3
Page 13 of 18
FINANCE AND MODELLING
QUESTION 1
1.1
9
9
5
2
7
P(1 − 0,2) (1 − 0,4) = P(1 − i ) 9
0,1179648 = (1 − i )
0,7368... = 1 − i
i = 26,31%
9
7
9
9
(6)
1.2
A
B
C
D
Paying more than the required amount in order to pay back more quickly. 99
Cash flow problem - Paying only sufficient to cover interest. 99
Interest rates rise. Paying less than the required amount. 99
Gets promotion, bigger salary. Able to pay more per month and finish the loan in
the right time. 99
(8)
14 marks
QUESTION 2
2.1
9
99
−n
⎡
0
,095 ⎞ ⎤
⎛
99
3
9 ⎢1 − ⎜1 + 12 ⎟ ⎥
⎛ 0,095 ⎞
⎝
⎠ ⎥
6000000⎜1 +
⎟ = 55000 ⎢
⎥
⎢
0
,
095
12 ⎠
⎝
⎥
⎢
12
⎥⎦
⎢⎣
n = 273,5 months 999
So 276 months in total = 23 years 9
2.2
(10)
Amount remaining =
9
⎡ ⎛ 0,095 ⎞ 21 ⎤
99 24 99 ⎢ ⎜1 +
⎟ − 1⎥
12 ⎠
⎛ 0,095 ⎞
⎝
⎢
⎥ 99
6000000⎜1 +
⎟ − 55000
⎢
⎥
0
,
095
12 ⎠
⎝
⎢
⎥
12
⎣⎢
⎦⎥
= R7 250 071,51 99 – R1 251 189,89 99
= 5 998 881,62 9
(12)
22 marks
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Page 14 of 18
QUESTION 3
3.1
3.2
99 9
9
9
999
60
59
58
⎛ 0,16 ⎞
⎛ 0,16 ⎞
⎛ 0,16 ⎞
25000⎜1 +
⎟ + 25000⎜1 +
⎟ + 25000⎜1 +
⎟ = R 163860,42
12 ⎠
12 ⎠
12 ⎠
⎝
⎝
⎝
9 9
⎡ ⎛ 0,16 ⎞ 58 ⎤
⎟ − 1⎥
9
⎢ ⎜1 +
12
⎝
⎠
⎥ 999
163860,42 = x ⎢
0
,
16
⎢
⎥
⎢
⎥
12
⎣
⎦
x = R1890,08 999
(8)
(9)
17 marks
QUESTION 4
4.1
132 = 30q − 3(6 ) 999
q=5 9
30 = 5(6 ) − 2 p 999
p=0 9
99 9
4.2
Tn +1
9 9
0
,
14
⎛
⎞
m
= 100 × 1,1n + ⎜1 +
⎟Tn 9
12 ⎠
⎝
9
(8)
(7)
15 marks
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Page 15 of 18
GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS – MARKING GUIDELINES
QUESTION 5
5.1
5.2
(a)
Rabbits = 30, 9 Foxes = 5 9
(2)
(b)
No. 9 The number of rabbits and foxes are tending to a limit. 999
(4)
(c)
Rabbits = 55. 9 Foxes = 8 9
(2)
(d)
9
9
30 < r < 55
(4)
(a)
(b)
9
9
4<f<8
In the pred-prey model one must include the influence of the contact 9
between rabbits and foxes. The term represents the number of rabbit
deaths 9 and is a function of the product of the number of rabbits and
foxes. 9
c
R=
99
fb
0,048
9
=
0,12 × 0,008
= 50 9
a⎛
c ⎞
99
⎟
F = ⎜⎜1 −
b⎝
fbK ⎟⎠
(3)
0,64 ⎛
0,048
⎞
⎟ 9
⎜1 −
0,008 ⎝ 0,12 × 0,008 × 400 ⎠
9
= 70
=
(8)
23 marks
QUESTION 6
6.1
A Logistic 9 model since there is a limit 9 to the number of ants over time.
6.2
(a)
(b)
2 500 99
250 ⎞
⎛
375 = 250 + 250r ⎜1 −
⎟ 999
⎝ 2500 ⎠
0,5 = 9r 9
10
r=5 9
9
(2)
(2)
(5)
9 marks
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MODULE 4
Page 16 of 18
MATRICES AND GRAPHS
QUESTION 1
1.1
(a)
(b)
1.2
(a)
(b)
1.3
9
9
Enlargement scale factor 2
99
⎛ 2 0⎞
⎜
⎟
⎝ 0 2⎠
99
⎛1 0 ⎞
⎜
⎟
⎝1 1 ⎠
9
9
9
9
9
Rotation –900 and stretch factor 2, parallel to x-axis.
9
9
999
2
0
0
1
⎛
⎞⎛
⎞ ⎛ 0 2⎞
9⎜
⎟⎜
⎟=⎜
⎟
⎝ 0 1 ⎠ ⎝ −1 0 ⎠ ⎝ −1 0 ⎠
9
9
9
Shear factor 1, invariant y-axis:
99
99
⎛ cos 60 − sin 60 ⎞ ⎛10 ⎞ ⎛⎜ 5 − 3 3 ⎞⎟
⎜⎜
⎟⎟ ⎜⎜ ⎟⎟ =
99
⎝ sin 60 cos 60 ⎠ ⎝ 6 ⎠ ⎜⎝ 3 + 5 3 ⎟⎠
(4)
(5)
(5)
(6)
(6)
26 marks
QUESTION 2
2.1
6
4 ⎞
⎛9
⎜
⎟
cofactors : ⎜ 4 −12 −8 ⎟ 9999
⎜ −3 −2 −16 ⎟
⎝
⎠
−3 ⎞
⎛9 4
⎜
⎟
adj = ⎜ 6 −12 −2 ⎟ 9
⎜ 4 −8 −16 ⎟
⎝
⎠
−3 ⎞ ⎛ 4 2 −1 ⎞ ⎛ 44 0 0 ⎞
⎛9 4
⎜
⎟⎜
⎟ ⎜
⎟
⎜ 6 −12 −2 ⎟ ⎜ 2 −3 0 ⎟ = ⎜ 0 44 0 ⎟ 9999
⎜ 4 −8 −16 ⎟ ⎜ 0 2 −3 ⎟ ⎜ 0 0 44 ⎟
⎝
⎠⎝
⎠ ⎝
⎠
−3 ⎞
⎛9 4
1 ⎜
⎟
inverse = ⎜ 6 −12 −2 ⎟ 9
44 ⎜
⎟
⎝ 4 −8 −16 ⎠
(10)
2.2
9
99
9
99
−3 ⎞ ⎛ 3 ⎞ ⎛ 0,5 ⎞
⎛9 4
1 ⎜
⎟⎜ ⎟ ⎜
⎟
6 −12 −2 ⎟ ⎜ 4 ⎟ = ⎜ −1 ⎟
⎜
44 ⎜
⎟⎜ ⎟ ⎜
⎟
⎝ 4 −8 −16 ⎠ ⎝ 7 ⎠ ⎝ −3 ⎠
(6)
16 marks
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Page 17 of 18
GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS – MARKING GUIDELINES
QUESTION 3
(1)
(2)
(3)
Two planes parallel. 999
det = –3, so unique solution 999
det = 0, planes not parallel, so this is the one. 9999
10 marks
QUESTION 4
4.1
18, since 9 edges 99
(2)
4.2
Both the vertices of order five must go to all other vertices 99hence the minimum
order of any other vertex must be 2 9
(3)
4.3
One with order 5 99 four with an order of 3 99 one with an order of 1 9
(5)
10 marks
QUESTION 5
5.1
B, D, H, I 99
5.2
(Chinese Postman Problem)
– Recognises the need to travel a route twice 9
– (Routes B E D and I H repeated is not a minimum route)
– Routes B E I 99 and D G H 99 are repeated
– A solution that starts and finishes at A 99
– Every edge has been travelled at least once 99
– Solution for the shortest inspection route:
e.g. A B C D E B E G D G F H G H I E I G C A 99 9
5.3
5.4
(2)
(14)
Summing all the edges 9
+ B E I repeat + D G H repeat = 51,4 + 5,6 9+ 4,3 9 = 61,3 km 9
(4)
Yes 99 BD would remove two of the vertices with odd order 9 and HI = 3,4 99
Total travelled would be shorter than before at 51,4 + 6,4 + 3,4 = 61,2 km 9
(6)
26 marks
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GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS – MARKING GUIDELINES
Page 18 of 18
QUESTION 6
6.1
Students must demonstrate that they have used Prim's algorithm, rather than just an
intuitive approach or Kruskal's algorithm – students lose 50% of the mark if they do
not demonstrate use of the correct algorithm 99
1.
2.
3.
4.
6.2
E2 → Office (488 m) 99
E2 → E4 (587 m) 99
E4 → E3 (523 m) 99
Office → E1 (1514 m) 99
488 + 587 + 523 + 1514 = 3112 m 99
(10)
(2)
12 marks
Total: 300 marks
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