2003 Mathematics
Advanced Higher
Finalised Marking Instructions
2003 Mathematics
Advanced Higher – Section A
Finalised Marking Instructions
Advanced Higher 2003: Section A Solutions and marks
A1.
(a) Given f (x) = x (1 + x)10, then
f ′ (x) = (1 + x)10 + x.10 (1 + x)9
1,1
= (1 + 11x) (1 + x)9 *
1
ln y = x ln 3
1
(b) Given y = 3x, then
1 dy
= ln 3
y dx
1
dy
= ln 3 y = ln 3 . 3x.
dx
A2.
n
Sn =
∑
1
n
uk =
k=1
∑ (11
− 2k)
k=1
n
=
∑ 11
n
− 2
k=1
∑k
1
k=1
= 11n − 2 × 12 n (n + 1)
1,1
= −n2 + 10n.
−n2 + 10n = 21
(n − 3) (n − 7) = 0
The sum is 21 when there are 3 terms and when there are 7 terms.†
1
1
Alternative for first 2/3 marks
Using results for Arithmetic Series.
Sn =
A3.
*
†
a = 9, d = −2
1
n
[18 + (n − 1) (−2)]
2
1
y3 + 3xy =
dy
dy
+ 3x
+ 3y =
3y2
dx
dx
dy
=
dx
Thus at (2, 1), the gradient is 1
and an equation is (y − 1) = 1 (x − 2).
i.e. x = y + 1 or y = x − 1.
3x2 − 5
6x
6x − 3y
2x − y
= 2
2
3y + 3x
y + x
The (1 + x)9 must be pulled out.
If trial and error is used, both values are needed for the last mark.
Page 2
1,1
1
1
A4.
Let z = x + iy.
z + i = (x + iy) + i = x + (1 + y) i
∴
|z + i| =
1
x2 + (1 + y)2
1
∴
x2 + (1 + y)2 = 4
which is a circle, centre (0, −1) radius 2.
(The centre could be given as −i .)
1
x = 1 + sin θ
A5.
dx = cos θ dθ
1
θ = 0 ⇒ x = 1; θ = π / 2 ⇒ x = 2
π/2
2 1
cos θ
∫0 (1 + sin θ )3 dθ = ∫1 x3 dx
=
2
∫1 x
−3
1
1
dx
2
 x−2 
=  
 −2  1
−1
−1 
3
−
= 
 =
8
2
8
( )
1
*
1
x + y + 3z = 1
3x + ay + z = 1
x + y + z = −1.
A6.
Hence
x + y + 3z = 1
(a − 3) y − 8z = −2
−2z = −2
When a ≠ 3, we can solve to give a unique solution.
6
6
;
x = −2 +
.
z = 1;
y =
a − 3
3 − a
When a = 3, we get z = 14 from the second equation but z = 1‡ from the
third, i.e. inconsistent§ .
*
†
‡
§
optional
1 off for lower triangular form
1 for identifying the two values for z
1 for conclusion
Page 3
2†
2E1
2
A7.
f (x) =
f ′ (x) =
x
1 + x2
(1 + x2) − 2x2
1,1
(1 + x2)2
1 − x2
=
(1 + x2)2
f ′ (x) = 0 ⇒ 1 − x2 = 0 ⇒ x = ±1.
The graph of f (x) has two stationary values:
(1, 12 ) and (−1, − 12 )
and passes through (0, 0).
1
1
1
1
1
Thus g has two turning points (1, and (−1,
and its third critical value is (0, 0) {as its gradient is discontinuous}.
1
2)
A8.
A9.
1
2)
Statement A is true: p(n) = n(n + 1) and one of n and (n + 1) must be even.
Or: n2 and n are either both odd or both even. In either case n2 + n is even.*
Statement B is false: when n = 1, n2 + n = 2.
3
1
1
1
1
cosθ − i sin θ
cosθ − i sin θ
=
=
×
=
w cosθ + i sin θ cosθ + i sin θ cosθ − i sin θ cos2 θ − i 2 sin 2 θ
=
1
cos θ − i sin θ
= cos θ − i sin θ †
1
−1
wk + w−k = wk + (wk)
1
(cos θ + i sin θ )k
1
= cos kθ + i sin kθ +
cos kθ + i sin kθ
= cos kθ + i sin kθ + cos kθ − i sin kθ
= 2 cos kθ
= (cos θ + i sin θ )k +
(w
+ w−1) = w4 + 4w2 + 6 + 4w−2 + w−4
4
(2 cos θ ) = (w4 + w−4) + 4 (w2 + w−2) + 6
16 cos4 θ = 2 cos 4θ + 8 cos 2θ + 6
1
1
3
cos 4θ + cos 2θ + .
so
cos4 θ =
8
2
8
*
†
4
1
1
1
2E1
1,1
1
First mark needs attempt at justification; alternative proofs (e.g. induction) are acceptable
needs justifying.
Page 4
A10.
(a)
I1 =
1
∫0
xe−xdx = x ∫ e−xdx −
0
= [ −xe−x − e−x] 0
= −e−1 − e−1 − (0 − 1)
2
= 0.264 .*
= 1 −
e
(
(b)
1
−x
∫ 1. ∫ e dx. dx
1
)
(
)
1
1
n −x
n
−x
n−1
−x
∫0 x e dx = x ∫ e dx − ∫ nx ∫ e dx dx0
1
2E1
3E1
1
1
= [ −xne−x] 0 + n ∫ xn − 1e−xdx

0
1
= −e−1 − (−0) + n ∫ xn − 1e−xdx
0
= nIn − 1 − e−1
I3 = 3I2 − e−1
= 3 (2I1 − e−1) − e−1
= 3 (2 − 4e−1 − e−1) − e−1
= 6 − 16e−1 ≈ 0.1139.
(c)
dV
dt
A11.
dV
∫ V (10 −
V)
1
10
ln 5 −
1
10
=
∫ 1 dt
ln V − ln (10 − V)
V
ln
10 − V
V
10 − V
V
10t
V (1 + e )
2
1
1
= 10t
) = 10t
= e10t
= 10e10t − Ve10t
= 10e10t
10e10t
V =
1 + e10t
10t
10e
10
V =
= −10t
10t
e
1 + e
+ 1
→ 10 as t → ∞.
[END OF MARKING INSTRUCTIONS]
*
1
ln 5 = 0 + C
C = 0
(
1
1
1
= V (10 − V)
1 1
1
+
dV = ∫ 1 dt
∫
10 V
10 − V
1
(ln V − ln (10 − V)) = t + C
10
1
1
ln V −
ln (10 − V) = t + C
10
10
V (0) = 5, so
1
optional
Page 5
2E1
1
1
2003 Mathematics
Advanced Higher – Section B
Finalised Marking Instructions
Advanced Higher 2003: Section B Solutions and marks
B1.
Let
x − 3
y − 2
z + 1
=
=
= t
4
−1
2
x = 3 + 4t
then
y = 2 − t
2E1
z = −1 + 2t.
Thus
2 (3 + 4t) + (2 − t) − (−1 + 2t) = 4
1
9 + 5t = 4
t = −1
so the point is (−1, 3, −3).
1
A2 = 4A − 3I
B2.
A3 = 4A2 − 3A
1
= 16A − 12I − 3A
= 13A − 12I
1
A4 = 13A2 − 12A
1
= 52A − 39I − 12A
= 40A − 39I
1
i.e. p = 40, q = −39.
Alternative
A4 = (A2)
2
B3.
1
= 16A2 − 24A + 9I
1
= 64A − 48I − 24A + 9I
1
= 40A − 39I*
1
Let λ represent a fixed point, then
{
1
7
λ +
2
λ
7
2λ = λ +
λ
7
λ =
λ
λ =
}
2
λ = 7.
The fixed points are 7 and − 7.†
*
†
Using I = 1 at any stage costs a mark.
For full marks, exact values are needed. Iterative method giving ±2.645 gets 2 marks.
Page 2
1
1
1
B4.
f (x)
f ′ (x)
f ″ (x)
f ′′′ (x)
f ′′′′ (x)
either
= sin 2 x
= 2 sin x cos x
= 2 cos2 x − 2 sin 2 x
= −4 cos x sin x − 4 sin x cos x
= −8 cos2 x + 8 sin 2 x
or
=
=
=
=
f (0) = 0
f ′ (0) = 0
f ″ (0) = 2
f ′′′ (0) = 0
f ′′′′ (0) = −8
sin 2x
2 cos 2x
−4 sin 2x
−8 cos 2x
2E1
f (x) = 0 + 0.x + 2.
= x2 −
x2
x3
x4
+ 0. − 8.
2
6
24
1
1 4
x
3
1
Since cos2 x + sin 2 x = 1,
cos2 x = 1 − x2 + 13 x4
1
OR
1 3
x + …
3!
1
1
∴ (sin x)2 = x − x3 + … x − x3 + …
6
6
1
= x2 − 2 × x4 + … = …
6
sin x = x −
(
B5.
(a)
)(
1
)
1
1,1
When n = 1, LHS = 0, RHS = 0 × 1 × 2 = 0. Thus true when n = 1.
1
k
Assume
∑ 3 (r2 − r) = (k − 1)k (k + 1) and consider the sum to k
+ 1.
1
r=1
k+1
∑
r=1
k
3 (r2 − r) = 3 ((k + 1)2 − (k + 1)) +
∑ 3 (r 2
r=1
− r)
= 3 (k + 1)2 − 3 (k + 1) + (k − 1) k (k + 1)
1
= (k + 1) [ 3k + 3 − 3 + k − k ]
2
= (k + 1) (k + 2k ) = k (k + 1) (k + 2)
2
((k
+ 1) − 1) (k + 1) ((k + 1) + 1) .
Thus true for k + 1. Since true for 1, true for all n ≥ 1.
1
(b)
40
∑ 3 (r 2
r = 11
−
r)
40
=
∑ 3 (r 2
r=1
−
r)
10
−
∑ 3 (r 2
r=1
− r)
1
= 39 × 40 × 41 − 9 × 10 × 11
= 63960 − 990
= 62970.
Page 3
1
B6.
Consider first
d 2y
dy
+ 4y = 0.
− 4
2
dx
dx
Auxiliary equations is
m2 − 4m + 4 = 0
1
(m − 2)2 = 0
1
y = (A + Bx) e2x.
1
Thus the complementary function is
To find the particular integral, let f (x) = aex. Then
1
f ′ (x) = ae and f ″ (x) = ae .
x
x
1
aex − 4aex + 4aex = aex
a = 1.
1
y = (A + Bx) e2x + ex
1
so
Therefore the general solution is
dy
= Be2x + 2 (A + Bx) e2x + ex
dx
1
Initial conditions give
2 = A+ 1
1 = B + 2A + 1
i.e. A = 1 and B = −2.
The required solution is
y = (1 − 2x) e2x + ex.
[END OF MARKING INSTRUCTIONS]
Page 4
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1
2003 Mathematics
Advanced Higher – Section C
Finalised Marking Instructions
Advanced Higher 2003: Section C Solutions and marks
C1.
C2.
C3.
P(Breast cancer | Mammogram positive)
=
P(Breast cancer and Mammogram positive)
P(Mammogram positive)
1
=
P(Mammogram positive | Breast cancer)P(Breast cancer)


P(M+ | BC).P(BC) + P(M+| BC) . P( BC)
1
=
0·9 × 0·01
0·9 × 0·01 + 0·1 × 0·99
1
1
=
0·009
1
(= 0.083)*
=
0·108
12
1
(a)
X ∼ Bin (20, 0·25)
(b)
P (X ≤ 3) = 0·2252
(c)
The hypothesis p = 0.25 cannot be rejected at the 5% significance level
since the probability calculated in (b) exceeds 0.05.
Thus there is no evidence from the data to support the manager's belief.
(a)
1,1
1
5
(X − 32)
9
5
⇒ E (Y) = (104 − 32) = 40
9
Y =
()
5
V (Y) =
9
2
⇒ σ =
3
(b)
*
†
‡
1
1
1†
2
× 1·22
For central 95% probability, z = 1·96.
⇒ limits are 40 ± 1·96 × 0·667‡
i.e. (38·7, 41·3)
optional
for conclusion
not sufficient, limits have to be evaluated
Page 2
M1,1
1
1
1
1
C4.
(a)
ˆp ± 1·96
ˆp (1 − ˆp)
n
1,1
(b)(i) We require
(
0·3 × 0·7
1.8
=
n
n
2 × 1·96
(ii)
)
*
1.8
≤ 0.1
n
( )
1.8 2
⇒ n ≥ 324
0.1
Thus a sample size of 324 or greater is required.
n ≥
C5.
1,1
1
1
(a)
¯x − µ
σ/ n
530 − 502
=
63 / 25
z =
= 2·22
H0 : µ = 502
H1 : µ > 502
The critical region is z > 2·33
Since 2·22 lies outside in the critical region, the null hypothesis is
accepted
i.e. there is no evidence of an increase
1
1
1
1
(b)
p-value = P (Z > 2·22) = 0·0132
which is greater that 0·01 confirming the conclusion
1
1
(c)
The central limit theorem guarantees that sample means will be
approximately normally distributed so the test will still be valid
1
1
[END OF MARKING INSTRUCTIONS]
*
1
optional
Page 3
2003 Mathematics
Advanced Higher – Section D
Finalised Marking Instructions
Advanced Higher 2003: Section D Solutions and marks
D1.
L(2·5)
=
(2·5 − 4)(2·5 − 6)
(2·5 − 1)(2·5 − 6)
(2·5 − 1)(2·5 − 4)
3·2182 +
4·0631 +
3·1278
(−3)(−5)
(3)(−2)
(5)(2)
= 1·1264 + 3·5552 − 0·7038 = 3·9778
D2.
f (x) = ln (3 + 2x)
f ′ (x) =
3
2
(3 + 2x)
f ″ (x) =
−4
(3 + 2x)2
Taylor polynomial is
2h
2h2
−
.
p (x) = p (1 + h) = ln 5 −
5
25
For ln 5·4, take f (1·2), h = 0·2; p(1·2) = 1·6094 + 0·08 − 0·0032 = 1·6862.
Coefficient of h in Taylor polynomial is substantially smaller than 1.
Hence f (x) is likely to be very insensitive to small changes in x near x = 1.
D3.
2
1
2f 0 = f 1 − f 0 = (f 2 − f 1) − (f 1 − f 0) = f 2 − 2f 1 + f 0
3f 0 = (f 3 − 2f 2 + f 1) − (f 2 − 2f 1 + f 0) = f 3 − 3f 2 + 3f 1 − f 0
Maximum error is ε + 3ε + 3ε + ε = 8ε.
This occurs when f 1 and f 3 have been rounded up and f 0 and f 2 rounded down
by the maximum amount, or vice versa.
D4.
3
2
1
1
(a) Difference table is:
i
0
1
2
3
4
5
x
0·3
0·6
0·9
1·2
1·5
1·8
f (x)
1·298
1·195
1·323
1·700
2·346
3·280
diff1
−103
128
377
646
934
diff2
231
249
269
288
diff3
18
20
19
3
(b) 2f 3 = 0·288
1
(c) Third degree polynomial would be suitable.
(Differences are approximately constant (well within rounding error).)
1
(d) p = 0·1
(0·1)(−0·9)
(0·1)(−0·9)(−1·9)
f (0·63) = 1·195 + 0·1(0·128) +
(0·249) +
(0·020)
2
6
= 1·195 + 0·013 − 0·011 − 0·001 = 1·196
(from f (0·3) with p = 1·1,
f (0·63) = 1·298 − 0·113 + 0·013 − 0·000 = 1·198).
Page 2
3
D5.
(a)
1
1
∫x0 f (x)dx = ∫0 f (x0 + ph) h dp = h ∫0 [ f (x0) + f ′(x0)ph + 12 f ″ (x0)p2h2] dp
x1
1
f ′(x0)hp2 f ″ (x0)h2p3 

= h f (x0)p +
+


0
2
6
f ′(x0)h f ″ (x0)h2 

+
= h f (x0) +


2
6 
1
1
f ″ (x0)h2 f ″ (x0)h2 

+
= h f (x0) + f (x1) − f (x0) −


2
2
4
6 
=
h(f 0 + f 1) h3f ″ (x0)
−
(using f (x1) = f (x0) + hf ′(x0) + 12 h2f ″ (x0) +…)
2
12
5
First term is trapezium rule; second term is principal truncation error.
(b) Trapezium rule calculation is:
x
π/4
5π / 16
3π / 8
7π / 16
π/2
f (x)
0·5554
0·8163
1·0884
1·3480
1·5708
m
1
2
2
2
1
m f (x)
0·5554
1·6326
2·1768
2·6960
1·5708
8·6316
Hence I = 8·6316 × π / 32 ≈ 0·8474.
(c) f ″ (x) = 2 cosx − x sin x whose magnitude has maximum on [π / 4, π / 2] at
x = π / 2 since f ″ (π / 2) = −π / 2 = −1.571 and f ″ (π / 4) = 0·859 and
f ′′′ (x) ≠ 0 on the interval.
|maximum truncation error| = (π / 16)2 × π / 4 × 1·571 / 12 ≈ 0·0040.
Hence estimate for I is I = 0·85.
[END OF MARKING INSTRUCTIONS]
Page 3
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2
1
2003 Mathematics
Advanced Higher – Section E
Finalised Marking Instructions
Advanced Higher 2003: Section E Solutions and marks
E1.
(a)
Given that
Since
(b)
d 2s
ds
= at + c.
= a,
2
dt
dt
ds
ds
= U when t = 0, c = U . Thus
= U + at .
dt
dt
⇒ s = ∫ (U + at ) dt = U t + 12 at 2 + c′.
When t = 0, s = 0 so c′ = 0, hence s = U t + 12 at 2.
Use s = 12 gt 2. When s = H, t = 6 so
H = 18g.
Hence, when s =
1
1
1
1
2H
1 2
2 gt
=
1
2H
t 2 = 18
t = 3 2 ≈ 4·2 seconds
E2.
1
B
q
1
210
050°
A
f
40°
30
By the sine rule
sin θ°
sin 40°
=
30
210
1
⇒ sin θ° = 0·092
⇒ θ° ≈ 5·3°
Then φ = 180 − (40 + 5·3) = 134·7
and the required bearing is 134·7° − 90° = 044·7°.
1
1
braking
force
E3.
(a)
By Newton II
m
so
(
d 2s
t
=
−2m
1
+
dt 2
4
(
d 2s
t
= −2 1 +
2
dt
4
Page 2
)
)
1
ds
t2
= −2t −
+ c.
dt
4
= 12 when t = 0, c = 12. So
Integrating gives
Since
(b)
ds
dt
ds
t2
= 12 − 2t − .
(∗)
dt
4
The car is stationary when ds
dt = 0, i.e. when
t2
− 2t − 12 = 0
4
⇒
t 2 + 8t − 48 = 0
⇒
(t + 12) (t − 4) = 0
⇒ t = 4
(As t ≥ 0, the root −12 is ignored.)
Integrating (*) gives
t3
(as s (0) = 0)
s = 12t − t 2 −
12
The stopping distance is
s (4) = (48 − 16 −
R
E4.
16
3)
1
1
1
1
F
5
a
mg
a
3
4
Resolving perpendicular to the plane
R = mg cos α = 45 mg
and hence F = µR = 45 µmg.
Resolving parallel to the plane
mg sin α = P + F = P +
1
4
5
µmg
P = mg ( 35 − 45 µ)
⇒
=
(b)
1
= 26 23 m .
P
(a)
1
1
1
1
mg (3 − 4µ) .
5
As in (a) F = 45 µmg
Resolving parallel to the plane
3
4
mg + µmg
5
5
mg
P =
(3 + 4µ) .
10
To find µ, equate these expressions for P.
2P =
1
10
⇒
(3 + 4µ) =
1
5
(3 − 4µ)
1,1
1
3 + 4µ = 6 − 8µ
⇒
Page 3
µ =
1
.
4
1
E5.
(a)
V = V (cos 45°, sin 45°) =
V
(1, 1) .
2
From the equations of motion
¨x = 0
⇒
¨ y = −g
⇒
Substituting t =
V
t
2
V
1
y =
t − gt 2.
2
2
x =
2x
, gives
V
1
1
1
V 2x
1
y =
− g
2
2 V
( )
2x
V
2
gx2
.
V2
To hit A, we require y = h when x = 10h.
1
= x −
(b)
⇒
g (10h)
= h
V2
g
⇒
9h = 2 (10h)2
V
10h −
⇒
V =
10
gh.
3
At B: y < h when x = 11h
⇒
1
V2
102
=
gh
9
⇒
(b)
1
2
1
1
g (11h)2
< h
V2
11h −
⇒
gh 112
10 <
V2
⇒
V2
112
<
gh
10
V
11
<
.
gh
10
⇒
So
10
<
3
V
11
<
.
gh
10
[END OF MARKING INSTRUCTIONS]
Page 4
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