©
2012 Mathematics Advanced Higher Finalised Marking Instructions
Scottish Qualifications Authority 2012 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 SQA’s NQ Delivery: Exam Operations. 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 NQ Delivery: Exam Operations 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.
Advanced Higher Mathematics 2012 Marks awarded for
1. (3,4)
f (x) =
(a)
3x + 1 x2 + 1 1M for quotient rule (or product) 1 for two correct terms 1 for third correct term
3 (x2 + 1) − (3x + 1) 2x f ′ (x) = (x2 + 1)2
(b)
3x2 + 3 − 6x2 − 2x = (x2 + 1)2 −3x2 − 2x + 3 = (x2 + 1)2 g (x) = cos2 x etan x
g′(x) = 2 cosx(− sin x)e
tan x
+ (cos x)(sec x) e 2
2
tan x
1M 1 1 1
product rule first correct term second correct term
= − sin 2x etan x + etan x simplification tan x = (1 − sin 2x) e (b) alternative g (x) = cos2 x exp (tan x) ln (g (x)) = ln (cos2 x) + tan x 1M = 2 ln (cos x) + tan x Differentiating g′ (x) (− sin x) = 2 + sec 2 x 1 g (x) cos x 1 − 2 sin x cos x g′ (x) = cos2 x exp (tan x) 1 2 cos x = (1 − sin 2x) tan x 1
(
2. (5)
)
a = 2048 and ar3 = 256 ⇒ r3 = 18 ⇒ r = 12 . a (1 − rn) Sn = 1 − r 1 n 1 − (2) 4088 ⇒ = 1 1 − 2 2048 511 = 256 n 1 511 1 511 ⇒ 1 − = × = 2 256 2 512 1 511 1 = 1 − = 2n 512 512 n ⇒ 2 = 512 ⇒ n = 9
1M valid approach 1
correct answer only, 2 marks
1M for sum formula
()
1 1
any valid method
Marks awarded for
3. (6)
4. (5)
5. (5)
Since w is a root, w = −1 − 2i is also a root. The corresponding factors are (z + 1 − 2i) and (z + 1 + 2i) from which ((z + 1) − 2i)((z + 1) + 2i) = (z + 1)2 + 4 = z2 + 2z + 5 z3 + 5z2 + 11z + 15 = (z2 + 2z + 5) (z + 3)
1
for conjugate
1 1
evidence needed
The roots are (−1 + 2i) , (−1 − 2i) and −3.
1
for stating roots together
1 1
for two correct points for third correct point
The general term is given by: 1 r 9 (2x)9 − r − 2 r x 9−r 9−r 2 x (−1)r 9 = × r x2r 9 = × (−1)r 29 − rx9 − 3r r The term independent of x occurs when 9 − 3r = 0, i.e. when r = 3. 9! (−1)3 26 The term is: 6! 3! = −5376.
() () ()
( )
1 1 1 1
1
Method 1 → → PQ = 3i + j + 4k and QR = 2i − 2j − 2k A normal to the plane: i j k → → PQ × QR = 3 1 4 2 −2 −2 3 4 3 1 1 4 = i − j + k −2 −2 2 −2 2 −2 = 6i + 14j − 8k Hence the equation has the form: 6x + 14y − 8z = d. The plane passes through P (−2, 1, −1) so d = −12 + 14 + 8 = 10 which gives an equation 6x + 14y − 8z = 10 i.e. 3x + 7y − 4z = 5.
|
| |
|
| |
|
|
1
1M
1 1
1
→
PR could be used
Marks awarded for Method 2 A plane has an equation of the form ax + by + cz = d . Using the points P, Q, R we get −2a + b − c = d 1M a + 2b + 3c = d 3a + c = d Using Gaussian elimination to solve these we have or other valid method −2 1 −1 d −2 1 −1 d ⇒ 1 1 2 3 d 0 5 5 3d 3 0 1 d 0 6 8 2d
|
|
|
|
−2 1 −1 d 0 5 5 3d 0 0 2 − 85 d 4 7 3 ⇒ c = − d, b = d, a = d 5 5 5 These give the equation ( 35 d) x + ( 75d ) y + (− 45 d ) z = d i.e. 3x + 7y − 4z = 5 ⇒
|
|
1 1
1
Marks awarded for
6. (5)
Method 1 ex = 1 + x +
+
x2 2
+ …
x3 6
1
(1 + e ) = 1 + 2ex + e2x x 2
= 1 + 2 (1 + x +
1M
+ … ) 3 2 + (1 + 2x + (2x) + (2x) 2 6 + … ) 2 2 1 3 4 3 = 1 + 2 + 2x + x + 3 x + 1 + 2x + 2x + 3 x +… +
x2 2
x3 6
= 4 + 4x + 3x2 + 53 x3 + …
1 1 1
Method 2 ex = 1 + x +
x2 2
+
+ …
x3 6
1
( 1 + e x) = 2 + x + 2 + 6 + … (1 + ex)2 = (2 + x + x2 + x6 + … )(2 + x + x2 + x6 + … ) x2
2
x3
3
2
3
1 1M
= 4 + 4x + 3x2 + 13 x3 + 12 x3 + 12 x3 + 13 x3 + …
1
= 4 + 4x + 3x2 + 53 x3 + …
1
Method 3 ex = 1 + x +
x2 2
f (x) = (1 + ex)2 f ′ (x) = 2ex (1 + ex) = 2ex + 2e2x f ″ (x) = 2ex + 4e2x f ′′′ (x) = 2ex + 8e2x
+
+ …
1
f (0) = 4 f ′ (0) = 4
1
x3 6
f ″ (0) = 6 f ′′′ (0) = 10
1 1
can award marks for correct columns.
f (x) = f (0) + f ′(0)x + f ″ (0) x2 + f ′′′(0) x6 +… 2
3
(1 + ex)2 = 4 + 4x + 3x2 + 53 x3 + …
1
y = |x + 2|
7. (4)
y
(a)
(0,2)
(−2,0)
1 1
for shape for coordinates
1 1
for both horizontal lines for values: 1, −1, −2
x
y
(b)
1 −2 −1
x
Marks awarded for x = 4 sin θ ⇒ dx = 4 cos θ dθ
8. (6)
x = 0 ⇒ θ = 0 x = 2 ⇒ θ = π6
1 1
2
∫0 16 − x2 dx π/6
16 − (4 sin θ )2 . 4 cosθ dθ
= ∫0
π/6
1
16 (1 − sin 2 θ ) . 4 cosθ dθ
= ∫0
π/6
= ∫0
16 cos2 θ . 4 cos θ dθ
π/6
= ∫0 16 cos2 θ dθ = 8
π/6 ∫0
(1
= 8 [θ +
9. (4)
1
+ cos 2θ ) dθ π/6
sin 2θ ] 0
1 2
1
for applying trig. identity and integrating
1
numerical approx. allowed
1
for multiplying by A
1
for rearranging A + A−1 = I
A2 = −A−1
1
for subtracting I
A3 = −I, i.e. k = −1
1
for multiplying by A
1
for rearranging
1
for squaring
1
for multiplying by A
=
8π 6
+ 4 sin π3
=
4π 3
+ 2 3 (≈ 7·65)
Method 1 A + A−1 = I A2 + I = A −1
A + I = I − A 2
Hence
Method 2 A + A−1 = I A = I − A−1 A2 = I − 2A−1 + (A
)
−1 2
A3 = A − 2I + A−1 A = (A + A
−1
3
Hence
)
− 2I = I − 2I
A = −I, i.e. k = −1 3
1
Method 3 A + A−1 = I A = I − A−1
1
for rearranging
A = A − A
1
for multiplying by A2
A3 = (A − I) − A
1
using A2 = A − I
3
2
= −I, i.e. k = −1 Plus other valid methods.
1
Marks awarded for
10. (3)
Method 1 1234 = 7 × 176 + 2 176 = 7 × 25 + 1 25 = 7 × 3 + 4
1
123410 = 34127
1
1
Hence
Method 2 1234 = 7 × 176 + 2
answer only, 1 of 3
1
= 7 × (7 × 25 + 1) + 2
11. (1,4)
= 7 × (7 × (7 × 3 + 4) + 1) + 2
1
= 3 × 73 + 4 × 72 + 1 × 7 + 2 Hence 123410 = 34127
1
d (sin −1 x) = dx
(a) (b) ∫ sin −1 x.
12. (5)
x 1 − x2
dx =
sin −1 x ∫
x 1 − x2
dx − ∫
= sin −1 x ∫
x 1 − x2
dx − ∫
(
(
d dx
1 1 − x2
(sin −1 x) ∫
1 1 − x2
∫
x 1 − x2
= sin −1 x (− 1 − x2) − ∫
(
1 1 − x2
x 1 − x2
1
dx dx
)
1
)
1
)
dx dx
(−
1 − x2) dx
= sin −1 x (− 1 − x2) − ∫ (−1) dx
1
= x − sin −1 x. 1 − x2 + c
1
dr = −0.02; dt
answer only, 1 of 3
dh = 0.01 dt
for ∫
x dx = − 1 − x2 1 − x2
1
V = πr2h
( )
dV dr dh = π 2r h + πr2 dt dt dt = π (2 × 0.6 × (−0.02) × 2 + 0.36 × 0.01)
1M for implicit differentiation 1 for accuracy 1
= π (−0.048 + 0.0036) = −0.0444π (≈ −0.14) The rate of change in the volume is −0.0444π m3 s−1.
1
units required
Marks awarded for
13. (10)
x = 2t + 12 t 2
⇒
y =
⇒
1 3 3t
− 3t
dx dt dy dt
= 2 + t
1
= t − 3
1
2
dy t − 3 = dx 2 + t d dy 2t (2 + t) − (t 2 − 3) t 2 + 4t + 3 = = dt dx (2 + t)2 (2 + t)2 2
1
( )
1
d 2y t 2 + 4t + 3 1 t 2 + 4t + 3 = × = dx2 (2 + t)2 2+t (2 + t)3
1
Stationary points when
dy dx
= 0, i.e.
t2 − 3 = 0 ⇒ t = ± 3 d 2y 3 + 4 3 + 3 = >0 dx2 (2 + 3)3 which gives a minimum. When t =
1
3,
d 2y 3 − 4 3 + 3 = <0 dx2 (2 − 3)3 which gives a maximum.
1
no marks for using a nature table
1
no marks for using a nature table
When t = − 3,
2
d y At a point of inflexion, dx 2 = 0.
1
In this case, that means t 2 + 4t + 3 = (t + 1)(t + 3) = 0 and this has exactly two roots. Note that this is a slimmed-down version of the complete story of points of inflexion.
1
need to show 2 values exist
Marks awarded for
14. (5, 1,3)
(a)
| | |
4 0 6 2 −2 4 −1 1 λ 4 0 6 0 4 −2 0 4 6 + 4λ
1 −1 2 1 3 9
|
1
| |
4 0 6 1 0 4 −2 3 0 0 8 + 4λ 6 6 3 z = = 8 + 4λ 2 (2 + λ) 18 + 6λ 4y = 3 + 2z ⇒ 4y = 4 + 2λ 3λ + 9 ⇒ y = 4 (2 + λ) 2λ − 14 4x = 1 − 6z ⇒ 4x = 4 + 2λ λ − 7 ⇒ x = 4 (2 + λ) (b) When λ = −2, the final row gives 0 = 6 which is inconsistent. There are no solutions.
for augmented matrix
1
1
triangular form needed
1
first root
1
other two roots
1
(c) λ = −2.1; x = 22.75; y = −6.75; z = −15 1,1 1 for first 2 values; 1 for third Although the values of λ are close, the values of x, y and z are quite different. The system is ill-conditioned near λ = −2. 1
Marks awarded for
15. (4,7)
(a)
1 A B C = + + (x − 1)(x + 2)2 x − 1 x + 2 (x + 2)2
1M
1 = A(x + 2)2 + B(x − 1)(x + 2) + C (x − 1) x = 1 ⇒ A =
1 9
1
x = −2 ⇒ C = − 13 x = 0 ⇒ 1 = ∴
(b)
4 9
− 2B +
(
1 3
1
⇒ B = − 19
1 1 1 1 3 = − − 2 (x − 1)(x + 2) 9 x − 1 x + 2 (x + 2)2
(x − 1)
)
1
dy x − 1 − y = dx (x + 2)2
dy 1 1 − y = dx x − 1 (x + 2)2
1M for rearranging
Integrating factor: exp (∫ − x −1 1 dx)
1
= exp (− ln (x − 1) = (x − 1)−1
1
1 dy 1 1 − y= 2 (x − 1) dx (x − 1) (x − 1) (x + 2)2
( )
d y 1 = dx x − 1 (x − 1)(x + 2)2 =
(
1 1 1 3 − − 9 x − 1 x + 2 (x + 2)2
(
1
)
1
)
y 1 3 = ln|x − 1 | − ln|x + 2 | + +c 1 x−1 9 x+2 x−1 3 y= ln|x − 1 | − ln|x + 2 | + + c(x − 1) 1 9 x+2 x − 1 |x − 1 | 3 = ln + + c(x − 1) |x + 2 | x + 2 9
(
(
)
)
constant of integration needed.
Marks awarded for
16. (6,4)
(a) For n = 1, the LHS = cos θ + i sin θ and the RHS = cos θ + i sin θ . Hence the result is true for n = 1.
1
Assume the result is true for n = k , i.e. (cos θ + i sin θ )k = cos kθ + i sin kθ .
1
Now consider the case when n = k + 1: (cosθ + i sinθ )k + 1 =(cosθ + i sinθ )k (cosθ + i sinθ ) = (cos kθ + i sin kθ) (cos θ + i sin θ )
1 1
= (coskθ cosθ − sinkθ sinθ) + i (sinkθ cosθ + coskθ sinθ) 1 = cos(k + 1)θ + i sin (k + 1)θ Thus, if the result is true for n = k the result is true for n = k + 1. Since it is true for n = 1, the result is true for all n ≥ 1. 1
11π cos 11π (cos 18π + i sin 18π )11 18 + i sin 18 (b) = cos π9 + i sin π9 (cos 36π + i sin 36π )4
=
11π cos 11π cos π9 − i sin π9 18 + i sin 18 × cos π9 + i sin π9 cos π9 − i sin π9
π 11π π cos 11π 18 cos 9 + sin 18 sin 9 = + imaginary term cos2 π9 + sin 2 π9 11π π = cos − + imaginary term 18 9 π = cos + imaginary term 2 Thus the real part is zero as required.
(
)
END OF SOLUTIONS
1
working with n is penalised.
for applying the inductive hypothesis multiplying and collecting
using result from above
1
1
1
or equivalent