Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Paper 2 2005
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
1.
Given
Put in a form that we can integrate
⌠
⎮
⎮
⎮
⌡
4x
−1
3
x
2
dx
⌠
⎮
⎮
⎮
⌡
4x
3
x
2
−
1
x
2
dx
⌠
⎮
−2
⎮ 4x − x dx
⌡
( n + 1)
⌠
x
n
⎮
Now we can integrate using the rule
⎮ x dx → ( n + 1)
⌡
We get
⌠
⎮
−2
⎮ 4x − x dx
⌡
4x
−1
x
−
+C
2
−1
2
2x
2
+
1
x
Page 1 of 13
+C
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Paper 2 2005
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
2.
Given diagram
(a)
sin( p + q)
sin( p) cos( q) + cos( p) sin( q)
15 8
⋅
17 10
120
170
(b)(i) cos( p + q)
+
6
48
170
84
170
85
cos( p) cos( q) − sin( p) sin( q)
⋅
8
64
−
17 10
170
tan( p + q)
⋅
17 10
168
8
(b)(ii)
8
+
−
15 6
⋅
17 10
90
170
−26
−13
170
85
sin( p + q)
84
85
−84
cos( p + q)
− 13
13
85
Page 2 of 13
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Paper 2 2005
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
3.
Given diagram
(a)
The perpendicular bisector
passes through the midpoint
of AB.
⎛ 1 + 5 , 4 + 0⎞
2 ⎠
⎝ 2
Midpt
( 3 , 2)
The gradient of AB is given by
m
4−0
4
5−1
4
1
The perpendicular gradient is given by
mab ⋅ mpg
−1
1⋅ m
pg
−1
mpg
−1
So equation of perpendicular gradient is
y−b
mpg ( x − a)
y−2
−1 ( x − 3)
y−2
−x + 3
x+ y
5
Page 3 of 13
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Paper 2 2005
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
(b)
Give the tangent equation
x + 3y
1
Gradient is
y
−1
3
x+
1
mT
3
−1
3
Gradient of radius line CA is
mT ⋅ mCA
−1
−1
3
⋅ mCA
−1
mCA
3
Equation of radius line CA is
y−b
y−0
3x −
m ( x − a)
3( x − 1)
y
3
(c)(i) By solving simultaneous equations we can find
co-ordinates of centre C.
x+ y
3x −
y
5
4x
8
x
2
3
Page 4 of 13
y
5−2
3
C ( 2 , 3)
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Paper 2 2005
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
(ii)
Equation of circle is given by
( x − a) + ( y − b )
2
( a , b)
r
2
( 2 , 3)
( x − 2) + ( y − 3)
2
4.
2
r
2
( 3 − 0) + ( 2 − 1)
2
2
10
10
Given diagram
(a)
⎯
→
TA
In component form
⎛23 ⎞ ⎛ 28 ⎞
⎜ 0 − ⎜−15
⎜
⎜
⎝8 ⎠ ⎝ 7 ⎠
⎛−5 ⎞
⎜ 15
⎜
⎝1 ⎠
Page 5 of 13
⎯
→
TB
⎛−12 ⎞ ⎛ 28 ⎞
⎜ 0 − ⎜−15
⎜
⎜
⎝9 ⎠ ⎝7 ⎠
⎛−40 ⎞
⎜ 15
⎜
⎝2 ⎠
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Paper 2 2005
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
(b)
Angle between both beams is
cos( θ )
cos( θ )
cos( θ )
θ
a⋅b
a⋅b
a ⋅ b
427
251⋅ 1829
[ ( −5) ⋅ ( −40) + 15⋅ 15 + 1 ⋅ 2]
a
( −5) + ( 15) + 12
b
( −40) + ( 15) + 22
2
2
427
0.63
251⋅ 1829
50.9o
Page 6 of 13
2
= 0.63
2
251
1829
427
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Paper 2 2005
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
y
5.
2x
2
−9
y
Given diagram
Points of intersection
are given by
2x
−9
2x
−x
x
9
2
2
2
x
2
2
9
x = 3 and -3
Area between curves is given by
3
⌠
⎮ x2 − 2x2 − 9 dx
⌡− 3
(
)
3
⌠
⎮ − x2 + 9 dx
⌡− 3
( )
3
⎡−⎛ 33 ⎞ + 9 × 3⎤ − ⎡⎢ −( −3) + 9⋅ ( −3) ⎥⎤
⎣ ⎝3⎠
⎦ ⎣ 3
⎦
[ −( 9) + 27] − ( 9 − 27)
54 − 18
= 36
Page 7 of 13
⎤
⎡ ⎛ x3 ⎞
⎢−⎜
+ 9x⎥
⎣ ⎝ 3⎠
⎦
x
2
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Paper 2 2005
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
6.
Given diagram
The equation of the tangent
at P is found by
24
y
x> 0
x
−1
y
d
dx
24x
2
−3
y
−12x
2
When x = 4 the gradient is
d
dx
−3
y
−12 ( 4)
2
−12
−3
8
2
When x = 4 the y co-ordinate is
y
24
P ( 4 , 12)
12
4
Equation of tangent is
y−b
3x + 2y
m ( x − a)
y − 12
−3
2
( x − 4)
36
Page 8 of 13
2y
− 24
−3x + 12
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Paper 2 2005
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
7.
Solving using the rules for logs we get
log4 ( 5 − x) − log4 ( 3 − x)
⎛ 5 − x⎞
log4 ⎜
3−x
⎝
⎠
2
x< 3
2
take antilog
5−
x
3−
x
5−
x
15x
x
16
48 − 16x
43
43
15
Page 9 of 13
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Paper 2 2005
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
8.
Question 8
Given Sketch
f ( x)
ksin ( 2x)
g( x)
sin( x)
A
k> 1
B
0
For points of intersection we have
k ⋅ sin( 2x)
k ⋅ 2 sin( x) ⋅ cos( x)
sin( x)
sin( x)
Since x is not 0 at A or C (from the diagram) we have
k ⋅ 2⋅ cos⋅ ( x)
sin( x)
sin( x)
sin( x)
cos⋅ ( x)
k ⋅ 2⋅ cos⋅ ( x)
1
2k
Page 10 of 13
1
180
D
C
360
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Paper 2 2005
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
9.
Given information and equation
V
(a)
The initial value ( at launch ) is when t = 0.
V
(b)
− 0.06335 t
252⋅ e
− 0.06335× 0
252⋅ e
252⋅ 1
£252million
When V = 20 million we have
20
20
252
− 0.06335 t
252⋅ e
− 0.06335 ⋅ t
e
⎞
loge⎛
252
⎝ ⎠
20
−0.06335⋅ t
⎞
loge⎛
⎝ 252 ⎠
−0.06335
20
t
t
40years
Page 11 of 13
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Paper 2 2005
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
10.
Given diagram and information
Keywords equilateral triangle
b perpendicular to a and c
a
c
3units
2units
Evaluating
a ⋅ ( a + b + c)
we get
a⋅a
a ⋅ ( a + b + c)
11.
b
(a)
3⋅ 3
a⋅b
9
9 + 0 + 4.5
0
13.5
Given x = -1 then we have
x + px + px + 1
3
2
0
( −1) + p ( −1) + p ( −1) + 1
3
2
−1 + p − p + 1
0
0
0
0
OR
By synthetic division
1
-1
1
p
p
1
-1
1-p
-1
p-1
1
0
Page 12 of 13
a⋅c
3 ⋅ 3 ⋅ cos( 60)
9⋅
1
2
4.5
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Paper 2 2005
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
11.
(b)
From synthetic division above we can write
( x + 1) ⎡⎣x + ( p − 1)x + 1⎤⎦
x + px + px + 1
3
2
2
Using the discriminant we have
a
b
1
( p − 1)
For real roots we must have
b − 4ac ≥ 0
2
( p − 1) − 4 ⋅ 1 ⋅ 1 ≥ 0
2
p − 2p + 1 − 4 ≥ 0
2
p − 2p − 3 ≥ 0
2
( p − 3) ( p + 1) ≥ 0
Hence p must be in the range
p ≤ 1 and
p≥ 3
Page 13 of 13
c
1