©
2012 Mathematics
Advanced Higher
Finalised Marking Instructions
 Scottish Qualifications Authority 2012
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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