2005 Mathematics
Advanced Higher
Finalised Marking Instructions
These Marking Instructions have been prepared by Examination Teams
for use by SQA Appointed Markers when marking External Course
Assessments.
General Marking Principles
These principles describe the approach taken when marking Advanced Higher Mathematics
papers. For more detailed guidance please refer to the detailed Marking Instructions.
1
The main principle is to give credit for the skills demonstrated and the criteria met. Failure
to have a correct method may not preclude a candidate gaining credit for their solution.
2
The answer to one part of a question, even if incorrect, is accepted as a basis for subsequent
dependent parts of the question.
3
The following are not penalised:
• working subsequent to a correct answer (unless it provides firm evidence that the
requirements of the question have not been met)
• legitimate variation in numerical values / algebraic expressions.
4
Full credit will only be given where the solution contains appropriate working. Where the
correct answer might be obtained by inspection or mentally, credit may be given.
5
Sometimes the method to be used in a particular question is explicitly stated; no credit will
be given where a candidate obtains the correct answer by an alternative method.
6
Where the method to be used in a particular question is not explicitly stated, full credit will
be given for an alternative valid method.
In the detailed Marking Instructions which follow, marks are shown alongside the line for
which they are awarded. There are two codes used, M and E. M indicates a method mark, so in
question 1, 1M, 1, 1 means a method mark for the product rule (and then a mark for each of the
terms). E is shorthand for error. In question 3, 2E1, means that a correct answer is awarded 2
marks but that 1 mark is deducted for each error.
page 2
Advanced Higher Mathematics 2005
Marking Instructions
1.
(a)
(b)
f (x) = x3 tan 2x
f ′ (x) = 3x2 tan 2x + x3 (2sec2 2x)
1 + x2
y =
1 + x
dy
2x (1 + x) − (1 + x2) .1
=
dx
(1 + x)2
x2 + 2x − 1
=
(1 + x)2
Alternative 1
1 + x2
2
= x − 1 +
y =
1 + x
1 + x
dy
2
= 1 −
or 1 − 2 (1 + x)−2
dx
(1 + x)2
Alternative 2
1 + x2
= (1 + x2) (1 + x)−1
y =
1 + x
dy
= 2x (1 + x)−1 + (1 + x2) (−1) (1 + x)2
dx
dy
2x
1 + x2
=
−
dx
(1 + x)
(1 + x)2
1M, 1, 1
1M,1
1
M1,1
1
M1,1
1
2y2 − 2xy − 4y + x2 = 0
2.
dy
dy
dy
− 2x
− 2y − 4
+ 2x = 0
1M, 1
dx
dx
dx
dy
= 0, so −2y + 2x = 0, i.e. y = x.
For a horizontal tangent
1
dx
This gives 2x2 − 2x2 − 4x + x2 = 0 ⇒ x (x − 4) = 0, i.e. x = 0 or 4.
1
[Both x values are needed.]
4y
3.
x2
x3
x4
+
+
+…
2!
3!
4!
x4
x2
2
e = 1 + x +
+…
2!
2
2
e x + x = e xe x
x2
x3
x4
x4
+
+
+… 1 + x2 +
+…
= 1 + x +
2
6
24
2
x2
x3
x4
x4
x4
+
+ x3 +
+
+
+…
= 1 + x + x2 +
2
6
2
2
24
4
= 1 + x + 23 x2 + 67 x3 + 25
24 x +…
[1 mark for first 3 terms, 1 mark for final two terms.]
2
Alternatives for f (x) = ex + x :
1: Applying Maclaurin directly: 1 method mark and then 2E1.
2: Using (x + x2) in the expansion of ex: 1 method mark and then 2E1.
ex = 1 + x +
(
)(
page 3
)
2E1
1
M1
1,1
u1 = S1 = 8 − 1 = 7
4.
5.
u2 = S2 − S1 = 12 − 7 = 5
1
u3 = S3 − S2 = 15 − 12 = 3
It is an arithmetic series.
Method 1: a = 7; d = −2.
un = 7 + (n − 1) (−2) = 9 − 2n.
1
1
1
1
Method 2: un = S(n) − S(n − 1) = 8n − n2 − (8(n − 1) − (n − 1)2)
= 8n − n2 − 8n + 8 + n2 − 2n + 1 = 9 − 2n
1
1
u = 1 + x ⇒ dx = du
When x = 0, u = 1 and when x = 3, u = 4
3
4u − 1
x
dx = ∫
du
∴∫
0
1 u1/2
1 + x
1
1
=
4
∫1 [ u
1
− u−1/2] du
1/2
4
u1/2 
 u3/2
−
= 

3 / 2
1 / 2 1
2 × 8
2
− 2 × 2 −  − 2
= 
 3

 3
1
= 143 − 2 = 2 23 (2.67 is acceptable).
Alternative for the final 2 marks:
4u − 1
4
−1/2
du
=
∫1 u1/2
∫1 (u − 1) u du

= (u − 1) ∫ u−1/2du −

6.
(
1
2
3
1
λ
3
2
1
9
∫ 1.
u1/2
1
2
1
4

du
1
= [ (u − 1) 2u1/2 − 43 u3/2] 1
4
1
= [ 12 −
= 2 23
1
1
0
5
32
3
) (
⇒
]
1
0
0
− [0 −
4
3
1
λ − 2
0
z = 23 ;
(λ − 2) y − 2 = −2 ⇒ y = 0; x = 1 − 0 − 43 = − 13 .
]
2
−3
3
1
−2
2
)
When λ = 2, the second and third rows of the second matrix are the same,
so there is an infinite number of solutions.
(†) Use of ‘redundant’ is worth a mark.
Interpretation in geometrical terms can be given both the marks.
page 4
2E1
1
1
1(†)
1
7.
A =
(
) (
0 4 2
1 0 1 ⇒ A2 =
−1 −2 −3
0 4 2
1 0 1
−1 −2 −3
(
( )
2 −4 −2
−1 2 −1
1 2 5
2
A + A =
2 0 0
0 2 0
0 0 2
=
−1
A
(A2
−1
2A
A−1
8.
)( ) (
)( )
0 4 2
1 0 1
−1 −2 −3
=
2 −4 −2
−1 2 −1
1 2 5
0 4 2
1 0 1
−1 −2 −3
+
2E1
1
= 2I
1
+ A) = 2A−1
= A+ I
= 12 A + 12 I
1
1
Using z = t , then
1
1M
x − 4y = 1 − 2t
x − y = t − 5
Subtracting:
)
3y = 3t − 6 ⇒ y = t − 2
x = y + t − 5 ⇒ x = 2t − 7
The line of intersection is given by: x = 2t − 7, y = t − 2, z = t .
1
1
x + 2y − 4z = 2t − 7 + 2 (t − 2) − 4t
= 2t − 7 + 2t − 4 − 4t = −11
1
Alternative for first 4 marks:
The normals to the planes are i − 4j + 2k and i − j − k and the vector
product of these will give the direction of the line of intersection.
(i − 4j + 2k) ∧ (i − j − k) = 6i + 3j + 3k
Now obtain a point (e.g. let z = 0 and solve the equations x − 4y = 1,
x − y = −5 gives (−7, −2, 0)).
Then write down the symmetric equation:
x + 7
y + 2
z
=
=
= λ
6
3
3
which leads to x = 6λ − 7, y = 3λ − 2, z = 3λ.
9.
Let z = a + ib so ¯ z = a − ib
1M,1
1
1
1
z + 2iz¯
a + ib + 2ia + 2b
a + 2b
2a + b
3a
a
z
page 5
=
=
=
=
=
=
=
8 +
8 +
8
7
6
2; b
2 +
7i
7i
1
M1
= 3
3i.
1
10.
When n = 1, LHS =
1
1
1
1
1 1
1
= , RHS = −
= −
= .
1 × 2 × 3
6
4 2 × 2 × 3 4 12 6
Thus true for n = 1.
Assume true for n = k and consider n = k + 1.
k +1
k
1
1
1
∑ r (r + 1)(r + 2) = ∑ r (r + 1)(r + 2) + (k + 1)(k + 2)(k + 3)
r=1
r=1
=
1
1
1
−
+
4 2(k + 1)(k + 2) (k + 1)(k + 2)(k + 3)
=
1
−(k + 3) + 2
+
4 2(k + 1)(k + 2)(k + 3)
=
1
k+1
−
4 2(k + 1)(k + 2)(k + 3)
1
1
−
4 2 ((k + 1) + 1)((k + 1) + 2)
Hence, since true for n = 1, true for all positive n.
1
1
1
1
=
n
lim
1
1
= .
+ 1) (r + 2)
4
∑ r (r
n→∞ r=1
11.
(a) x = 2. [2 alone is not enough.]
1
1
1
(b)
x3
y =
x − 2
dy
3x2 (x − 2) − x3
x2 (2x − 6)
=
=
= 0
dx
(x − 2)2
(x − 2)2
when x = 0 and when x = 3.
Stationary points are (0, 0) and (3, 27).
2E1
1
1
(c)
|
|
x3
+ 1
x − 2
Stationary points are (0, 1) and (3, 28).
y =
page 6
1, 1
12.
(a) z4 = (cos θ + i sin θ )4
= cos4 θ + 4 cos3 θ (i sin θ) + 6 cos2 θ (i 2 sin 2 θ) + 4 cosθ (i 3 sin 3 θ) + i 4 sin 4 θ M1
(b)
(c)
= cos4 θ + 4i cos3 θ sin θ − 6 cos2 θ sin 2 θ − 4i cosθ sin 3 θ + sin 4 θ
1
= (cos4 θ − 6 cos2 θ sin 2 θ + sin 4 θ ) + i (4 cos3 θ sin θ − 4 cosθ sin 3 θ )
1
(cos θ + i sin θ )4 = cos 4θ + i sin 4θ
1
Equating the real parts from (a) and (b):
4
2
2
M1
4
cos 4θ = cos θ − 6 cos θ sin θ + sin θ
1
2
cos 4θ
2
2
2 sin θ
=
cos
θ
−
6
sin
θ
+
sin
θ
cos2 θ
cos2 θ
1
= cos2 θ − 6 (1 − cos2 θ ) + (1 − cos2 θ )
1 − cos2 θ
cos2 θ
2E1
= 7 cos2 θ − 6 + (sec 2 θ − 2 + cos2 θ )
= 8 cos2 θ + sec 2 θ − 8
p = 8, q = 1, r = −8.
13.
1
A
Bx + C
+ 2
x
x + 1
2
= A (x + 1) + (Bx + C) x
⇒ 1 = A ⇒ A = 1
⇒ 1 = 2 + B + C
⇒ 1 = 2 + B − C
⇒ C = 0, B = −1
1
1
x
−
=
x3 + x
x
x2 + 1
k
k
1
1
x
− 2
dx = ∫
I (k) = ∫ 3
dx
1 x + x
1 x
x + 1
k 1
1 k 2x
dx
= ∫ dx − ∫ 2
1 x
2 1x + 1
k
1
= [ ln x] k1 − [ ln (x2 + 1)] 1
2
1
2
= [ ln k − 0] − [ ln (k + 1) − ln 2]
2
= ln k − ln k 2 + 1 + 12 ln 2
k 2
= ln 2
k + 1
k
2
eI(k) =
k2 + 1
2
=
→ 2
1 + k −2
1
+ x
1
x = 0
x = 1
x = −1
x3
=
(
page 7
1
1
1,1
)
2E1
1
1
1
1
14.
Let y = emx, then the auxiliary equations is
m2 − 3m + 2 = 0
(m − 1) (m − 2) = 0
m = 1 or m = 2
The Complementary Function is y = Aex + Be2x.
For the Particular Integral, try y = a sin x + b cos x.
dy
= a cos x − b sin x
dx
d 2y
= −a sin x − b cos x
dx2
Substituting:
(−a sin x − b cosx) − 3(a cosx − b sin x) + 2(a sin x + b cosx) = 20 sin x
1
1
1
1
1
(−a + 3b + 2a) sin x + (−b − 3a + 2b) cosx = 20sin x
a + 3b = 20;
− 3a + b = 0
a = 2; b = 6.
1
y = Aex + Be2x + 2 sin x + 6 cos x
1
The general solution is
dy
= Aex + 2Be2x + 2 cos x − 6 sin x
dx
y = 0 when x = 0 so A + B + 6 = 0.
dy
dx = 0 when x = 0 so A + 2B + 2 = 0.
B = 4;
A = −10
The particular solution is
y = −10ex + 4e2x + 2 sin x + 6 cos x.
page 8
1
1
1
15.
(a)
(b)
f (x) = (sin x)1/2 ⇒ f ′ (x) =
f (x) =
1 cos x
2 (sin x)1/2
g (x) = [ g (x)] 1/2. Thus
1
g′ (x)
i.e. k = 2
f ′ (x) = g′ (x) [ g (x)] −1/2 =
2
2 g (x)
x
−2x
∫ 1 − x2 dx = − ∫ 2 1 − x2 dx
= − 1 − x2 + c
Alternative methods for working out ∫ 1 x− x2 dx include use of substitutions.
For example u = 1 − x2; u2 = 1 − x2; x = sin θ . Marks should be
awarded:
For setting up an integral.
For carrying out the integration.
For returning to an answer in x.
(c)
1/2
∫0
sin −1 x dx = sin −1 x

∫
1.dx −
= x sin −1 x −

∫
1, 1
1M,1
1
1
1
1
1/2
d
(sin −1 x) dx
0
dx
1/2
x
dx
1 − x2 0
∫x
= [ x sin −1 x + 1 − x2] 0
1
1
3
sin −1
− (0 + 1)
+
=
2
2
4
1π
3
π
3
+
− 1 =
+
− 1
=
26
2
12
2
1/2
()
[END OF MARKING INSTRUCTIONS]
page 9
1
1
1
1
1