2008 Mathematics Advanced Higher Finalised Marking Instructions
© Scottish Qualifications Authority 2008 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is to be used for any other purposes written permission must be obtained from the Assessment Materials Team, Dalkeith. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre’s responsibility to obtain the necessary copyright clearance. SQA’s Assessment Materials Team at Dalkeith may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.
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, can be 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 in the question paper, full credit is available for an alternative valid method. (Some likely alternatives are included but these should not be assumed to be the only acceptable ones.)
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 4, M1 means a method mark for correct form of partial fraction. E is shorthand for error. For example, 2E1, means that a correct answer is awarded 2 marks but that 1 mark is deducted for each error.
page 2
Advanced Higher Mathematics 2008 1.
Let the common difference be d . General term is a + (n − 1) d . So 2 + 19d = 97 ⇒ d = 5. Sum of an arithmetic series is n2 [ 2a + (n − 1) d ] . Required sum is 502 {4 + 49 × 5} = 6225.
2.
(a)
1 1 1 1
f (x) = cos−1 (3x) −1 × 3 1 − (3x)2
f ′ (x) =
1,1
−3 1 − 9x2
= (b)
dx = 2 sec θ tan θ, dθ dy = dx
dy dθ dx dθ
=
dy = 3 cos θ dθ
3 cos θ 2 sec θ tan θ
1,1 1
3 cos3 θ = 2 sin θ
2
3.
1 for inverting 1
1 for translation -2
-1
0
1
2
3
4
1 for showing asymptotes
-1
-2
Asymptotes are y = −1 and x = 1.
page 3
1
12x2 + 20 A Bx + C + 2 = 2 x (x + 5) x x + 5 2 2 12x + 20 = A (x + 5) + x (Bx + C) = (A + B) x2 + Cx + 5A ∴ 5A = 20 ⇒ A = 4 ⇒ B = 8 C = 0 2 12x + 20 4 8x + 2 = 2 x (x + 5) x x + 5
4.
Hence 2
∫1
12x2 + 20 dx = x (x2 + 5)
∫1 x
+
x2
= [ 4 ln x + 4 ln (x2 + 5)] 1 2
= 4 [ ln x (x2 + 5)] = 4 [ ln 18 − ln 6] = 4 ln 3 (≈ 4·39) 2 1
5.
2E1
8x dx + 5 24 2 2x = ∫ dx + 4 ∫ 2 dx 1 x 1 x + 5 2
4
1M
xy2 + 3x2y dy dy + 6xy + 3x2 y2 + 2xy dx dx (2xy + 3x2) dy dx dy dx
1
= 4 = 0
1M,1
= −y2 − 6xy =
−y2 − 6xy 2xy + 3x2
When x = 1, y2 + 3y = 4 ⇒ y2 + 3y − 4 = 0 ⇒ (y + 4) (y − 1) = 0 ⇒ y = 1 since y > 0 Hence at (1, 1) dy −7 = dx 5 Tangent is (y − 1) = − 75 (x − 1) 5y + 7x = 12. Alternative for the first 3 marks. xy2 + 3x2y = 4 4 y2 + 3xy = x dy dy 4 + 3y + 3x = − 2 2y dx dy x (2y + 3x) dy = − 42 − 3y dx x 4 − dy x2 − 3y = dx 2y + 3x page 4
1,1
1
1 1 1
1 1 1
6.
(a)
( )
1 x = 4 − x2 x 4 A matrix is singular when its determinant is 0, hence x = ±2. 1 2 (b) When x = 2, A = 2 4 det
( ) ( )( ) ( 1 2 2 4
A2 =
1 2 2 4
=
5 10 10 20
)
1 1
= 5A.
1
2 4 2 2 A = (A ) = (5A) = 25A = 125A. 125 250 = 125A was accepted. 250 500 2
Evaluating A4 =
7.
∫ 8x
2
(
)
sin 4x dx = 8x2 = 8x2
∫ sin 4x dx − ∫ 16x (∫ sin 4x dx) dx
1M,1
( −14 cos 4x) − ∫ 16x × −14 cos 4x dx
= −2x2 cos 4x + 4 x
1
1
∫ cos 4x dx − ∫ 4 sin 4x dx
= −2x2 cos 4x + x sin 4x +
8.
2E1
1 4
cos 4x + c
1 1
The rth term is
( )
()
10 ( 2)10 − r 1 x r x
( )
r
10 20 − 3r x r 20 − 3r = 14 ⇒ r = 2 =
term is 45x or
( ) () 10 ( 2)r 1 x r x
10 − r
=
1 for form 1 for powers 1 for simplifying
14
( )
10 3r − 10 x r
1 1 1,1,1
3r − 10 = 14 ⇒ r = 8 term is 45x14
1 1
d (tan x) = sec2 x. dx
1
9. 1+ tan2 x = 1 +
∫ tan
2
sin 2 x cos2 x + sin 2 x = = sec 2 x. cos2 x cos2 x x dx =
∫ (sec
2
x − 1) dx
= tan x − x + c
page 5
1 1 1
10.
x′ (t) = t 3 − 12t 2 + 32t x″ (t) = 3t 2 − 24t + 32 which is 32 when t = 0. (b) x′ (t) = t 3 − 12t 2 + 32t ⇒ x (t) = 14 t 4 − 4t 3 + 16t 2 + c Since x (0) = 0, c = 0, so at time t (a)
x (t) =
1 4 4 t
− 4t 3 + 16t 2.
1 1 1
At O, x = 0
− 4t 3 + 16t 2 = 0 t (t − 16t + 64) = 0 t 2 (t − 8)2 = 0 The body returns to O when t = 8. 1 4 4t 2 2
11.
(a) (b)
12.
Counter example m = 2. So statement is false. Let the numbers be 2n + 1 and 2m then (2n + 1)3 + (2m)2 = 8n3 + 12n2 + 6n + 1 + 4m2 = 2 (4n3 + 6n2 + 3n + 2m2) + 1 which is odd. OR Proving algebraically that either the cube of an odd number is odd or the square of an even number is even. Odd cubed is odd and even squared is even. So the sum of them is odd.
1 1 1,1 1M 1 1
1 1 1
f (x) = x ln (2 + x) so x 2 1 + ln(2 + x), f ″ (x) = , + 2 2+x (2 + x) 2+x 4 1 f ′′′(x) = − − 3 (2 + x) (2 + x)2 and so f ′ (0) = ln 2, f ″ (0) = 1, f ′′′ (0) = − 34 . x2 x3 Thus f (x) = (ln 2) x + − +… 2 8 x ln (2 − x) = −f (−x) x2 x3 = (ln 2) x − − +… 2 8 x ln (4 − x2) = x ln (2 + x) + x ln (2 − x) x3 +… = (2 ln 2) x − 4 Alternative for first three marks: f (x) = x ln (2 + x) = x (ln 2 + ln (1 + 2x )) 2 = x (ln 2 + 2x − x8 +… ) 2 3 = x ln 2 + x2 − x8 + … f ′(x) =
page 6
1 1 1 1 1 1 1 1 1 1
13.
d 2y dy + 2y = 2x2 − 3 2 dx dx m2 − 3m + 2 = 0 (m − 1) (m − 2) = 0 m = 1 or m = 2 Complementary function: y = Aex + Be2x For particular integral try y = ax2 + bx + c ⇒
1 1 1 1M
dy d 2y = 2ax + b; 2 = 2a dx dx
Hence require 2a − 3 (2ax + b) + 2 (ax2 + bx + c) = 2x2
1
2ax2 + (−6a + 2b) x + (2a − 3b + 2c) = 2x2 ⇒ a = 1; b = 3; c = General solution is: y = Aex + Be2x + x2 + 3x + When x = 0, y =
1 2
7 2
1 1
dy and dx = 1. 1 2
dy dx
7 2
= A+ B+
7 2
⇒ A + B = −3
= Aex + 2Be2x + 2x + 3 ⇒ 1 = A + 2B + 3 ⇒ A + 2B = −2 B = 1
A = −4
Particular solution is y = −4ex + e2x + x2 + 3x + 27 .
page 7
1 1 1
14.
(a) →
→
AB = i − 2j
→
→
AB × AC =
|
i j k 1 −2 0 −1 2 2
|
AC = −i + 2j + 2k
1
= (−4 − 0) i − (2 − 0) j + (2 − 2) k
1
= −4i − 2j
Equation is −4x − 2y = k = −4 (1) − 2 (1) = −6 i.e. −2x − y = −3
1
2x + y = 3 (b)
In π1: 2 × 0 + a = 3 ⇒ a = 3. In π2: 0 + 3a − b = 2 ⇒ b = 3a − 2 = 7. Hence the point of intersection is (0, 3, 7). Line of intersection: direction from
|
i j k −4 −2 0 1 3 −1
|
1 1
= 2i − 4j − 10k
1
x = 0 + 2t; y = 3 − 4t; z = 7 − 10t
1
There are many valid variations on this (including symmetric form) and these were marked on their merits. (c)
Let the angle be θ, then −4 − 6 (−4i − 2j) . (i + 3j − k) cos θ = = = 2 2 2 2 2 20 × 11 4 + 2 1 + 3 + 1 or (−4i − 2j) × (i + 3j − k) sin θ = 42 + 22 12 + 32 + 12
|
| |
|
=
|
|
22 + 42 + 102 = 20 11
Hence θ ≈ 47.6°.
120 = 20 × 11
6 11
5 55
1M, 1
1M 1 1
page 8
15.
(a)
(b)
(c)
16.
( ) ( )
1 × ln x − x × 1x d x ln x − 1 = = . 2 dx ln x (ln x) (ln x)2 2 ln x 2 1 d2 x x × (ln x) − (ln x − 1) × x = dx2 ln x (ln x)4 ln x − 2 ln x + 2 2 − ln x = = x (ln x)3 x (ln x)3 Stationary points when ln x = 1, giving x = e and y = e. At (e, e), the second derivative is 2 − 1 > 0 e × 13 so (e, e) is a minimum. 2
d y 2 When dx 2 = 0, ln x = 2 ⇒ x = e . x = e2 ⇒ y = 12 e2.
Adding the expressions for zk and z1k gives zk + z1k = 2 cos kθ so cos kθ = 12 (zk + z−k). Subtracting the expressions for zk and z1k gives zk − z1k = 2i sin kθ so sin kθ = 2i1 (zk − z−k). cos2 θ sin 2 θ = (cos θ sin θ )2
((
z + 1z ) (z − 1z ) 4i 1 2 1 2 = − z − 2 . 16 z
=
(
(
)
)
)
i.e. a = OR A correct trigonometric proof that cos2 θ sin 2 θ = 1 8
1 8
−
1 8
[END OF MARKING INSTRUCTIONS]
page 9
1
1 1 1
1 2E1
1 1
1
2
1 2 1 z − 2 = z4 + 4 − 2 = 2 cos 4θ − 2 z z ⇒ cos2 θ sin 2 θ = 18 − 18 cos 4θ, and b = 18 . 2
1
1 1
zk = cos kθ + i sin kθ , 1 1 cos kθ − i sin kθ = = cos kθ − i sin kθ . so k = z cos kθ + i sin kθ cos2 kθ + sin 2 kθ
For k = 1
1M,1
cos 4θ .
2E1
1 1
1