Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Exam 2002 Paper 2
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
y
1.
(a)
Given the diagram opposite we have
A(-1,6)
Mid-point from AB is
C(5,2)
O
⎡ [ −1 + ( −3) ] , [ 6 + ( −2) ] ⎤ ( −2 , 2)
⎢
⎥
2
2
⎣
⎦
x
B(-3,-2)
Equation of the median line p from C is y = 2
(b)
Gradient of BC
mbc
[ 2 − ( −2) ]
4
1
[ 5 − ( −3) ]
8
2
Gradient of bisector is -2 since
mbc ⋅ mbisector
−1
Midpoint of BC is
⎡ [ 5 + ( −3) ] , [ 2 + ( −2) ] ⎤ ( −1, 0)
⎢
⎥
2
2
⎣
⎦
Equation of bisector q is
y−0
2x + y
(c)
−2( x − 1)
2
Intersection co-ordinates of p and q are
−2x + 2
2
x=0
Co-ordinates are (0, 2)
Page 1 of 9
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Exam 2002 Paper 2
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
2.
Given A (6, 0, 0) and D (3, 3, 8)
(a)
C is (0, 6, and 0)
(b)
→
⎯
DA
→ ⎯
→ ⎯
→
⎯
DC + CO + OA
→
⎯
DB
( 6 , 6 , 0) − ( 3 , 3 , 8)
(c)
( −3 , 3 , −8) + ( 0 , −6 , 0) + ( 6 , 0 , 0)
( 3 , 3 , −8)
Angle ADB
Using the dot product rule we have
cos ( ADB)
cos ( ADB)
angle ( ADB)
⎯
→
→⎯
DA⋅ DB
DA ⋅ DB
64
32
82
41
cos
( 3 , −3 , −8) ⋅ ( 3 , 3 , −8)
( 9 + 9 + 64) ⋅ ( 9 + 9 + 64)
32 ⎞
o
⎜ 41 38.7
⎝ ⎠
− 1⎛
Page 2 of 9
( 3 , −3 , −8)
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Exam 2002 Paper 2
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
3.
Given y 2x − 7x + 4x + 4
(a)
Max / Mini is given by
3
y
y = f(x)
2
A
o
d
y
dx
0
d
y
dx
6x − 14x + 4
2
(2,0)
0
This factorises to give
( 3x − 1) ( 2x − 4)
0
x
x
2
and
1
3
From the sketch it is clear that
x
(b)
2
is a minimum
x
and
From the sketch clearly (x-2) is a factor.
Using synthetic division we get
3
2
2x − 7x + 4x + 4
2
2x − 3x − 2
(
2
)
( x − 2) 2x − 3x − 2
( 2x + 1) ( x − 2)
Hence we have
( 2x + 1) ( x − 2)
2
Page 3 of 9
1
3
is a maximum
x
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Exam 2002 Paper 2
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
(c)
The point A is
⎛ −1 , 0⎞
⎜ 2
⎝
⎠
: x from part (b) 2x-1 = 0
Hence
3
2
2x − 7x + 4x + 4 < 0
4.
x<
for
−1
2
From Information given we can write:Using the recurrence formula
U
n +1
= a ⋅U
n
+ C
We have
U n +1 = U n + 0.5 − 0.2 ⋅ U n
U n +1 = 0.8 ⋅ U n + 0.5
(a)
Limiting value is found by
L
0.8⋅ L + 0.5
rearranging we get
(b)
L
0.5
( 1 − 0.8)
2.5metres
To make sure they do not grow more than 2 metres we need:L
0.5
( 1 − a)
2
rearranging we get
a
0.75
Hence we need to trim the hedges by at least 25 % each year.
Page 4 of 9
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Exam 2002 Paper 2
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
5.
Given the sketch and the equations:y
(a)
2
1 + 10x − 2x
y
To get the points of intersection we have
2
1 + 10⋅ x − 2⋅ x
2
x − 5x
2
1 + 5⋅ x − x
0
x⋅ ( x − 5)
x
(b)
2
1 + 5x − x
0
0
and
x
5
To find shaded area we have:5
(
5
(
) (
)
A
⌠
2
2
⎮ 1 + 10⋅ x − 2⋅ x − 1 + 5⋅ x − x dx
⌡0
A
⌠
2
⎮ −x − 5⋅ x dx
⌡0
A
)
2
⎛ x3
x ⎞
⎜ − − 5⋅
2⎠
⎝ 3
2
3
2
⎛ 53
5 ⎞ ⎛ 0
0 ⎞
⎜ − − 5⋅
− ⎜ − − 5⋅
2 ⎠ ⎝ 3
2 ⎠
⎝ 3
125
6
Page 5 of 9
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Exam 2002 Paper 2
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
6.
Given:For
y
π⎞
π
x
6⎠
3
π
x
3
⎛ π − π⎞
⎝3 6⎠
2⋅ sin ⎜
y
⎛
⎝
2⋅ sin ⎜ x −
⎛ π⎞
⎝6⎠
2⋅ sin ⎜
1
⎛π ⎞
⎜ ,1
⎝3 ⎠
Hence co-ordinates P is
Value of gradient at x
π
is given by d y ⎜⎛ ⎞
3
dx ⎝ 3 ⎠
π
Using the chain rule we get
d
y
dx
⎛
⎝
2⋅ cos ⎜ x −
d ⎛ π⎞
y⎜
dx ⎝ 3 ⎠
π⎞
6⎠
⎛ π − π⎞
⎝3 6⎠
2⋅ cos ⎜
We have gradient
⎛ π⎞
⎝6⎠
2⋅ cos ⎜
3
3
⎛ π , 1⎞
⎝3 ⎠
P⎜
Hence equation of the tangent is
y−1
⎛
⎝
3⋅ ⎜ x −
π⎞
3⎠
y
⎛
⎝
3⋅ x + ⎜ 1 −
Page 6 of 9
π
⎞
3⎠
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Exam 2002 Paper 2
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
7.
Given: -
y
log 3( x − 2) + 1
y
0
(the x axis)
They intersect when
log 3( x − 2) + 1
0
Using the rules of logs we take antilog of each side.
log 3( x − 2) + 1
log3( x − 2)
0
−1
( x − 2)
−1
3
x
2+
1
7
3
3
1
8.
Given: - a 2⋅ ( 4 − t) 2 and the point starts from REST.
To get the velocity equation we integrate the acceleration
equation above.
v
v
⌠
⎮ a dt
⌡
−4
3
⌠
1
⎮
2
⎮ 2⋅ ( 4 − t) dt
⌡
⋅ ( 4 − t)
3
2
3
2
2
2⋅ ⋅ ( 4 − t) ⋅ ( −1) + C
3
+C
when t = 0 then v = 0 (starting from rest) then we have
0
−4
3
3
⋅ ( 4 − 0)
2
+C
C
Hence equation is v
32
3
−4
3
⋅ ( 4 − t)
3
2
Page 7 of 9
+
32
3
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Exam 2002 Paper 2
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
9.
Given:-
2
( 1 − 2k) ⋅ x − 5⋅ kx − 2⋅ k
0
To have real roots for all integer values of k we need
2
b − 4ac ≥ 0
Hence we have
2
( −5k) − 4⋅ ( 1 − 2k) ( −2k) ≥ 0
2
2
25k + 8k − 16k ≥ 0
2
9k + 8k ≥ 0
Since 9k2 is always positive for all values of k and 9k2 > 8k for all
integers values of k, then the equation has real roots for all
integer values of k.
Page 8 of 9
Scottish Higher Still Course
www.mathsrevision.com
Higher Still Level Exam 2002 Paper 2
Created by
Graduate Bsc (Hons) MathsSci (Open) GIMA
10.
y
Given the sketch:-
(0,6)
cos ( θ )
(a)(i)
8
a
10
L
10
Rearranging we get L
5
4
L
⋅a
O
(b)(ii) sin ( θ )
b
6
( 8 − a)
10
Rearranging we get b
Area rectangle
L⋅ b
3
5
(a,0)
(8,0) x
⋅ ( 8 − a)
3
⋅ a⋅ ⎢⎡ ⋅ ( 8 − a)⎥⎤
4 ⎣5
⎦
5
θ
θ
3
4
⋅ a⋅ ( 8 − a)
Maximum / Minimum values are when A'(a) = 0.
3
A
4
⋅ a ⋅ ( 8 − a)
d
A
da
6−
6⋅ a
4
6a −
3
4
⋅a
2
0
rearranging we get
a
24
6
4
This is a maximum value because the constant term for a2 in the
equation A (a) is negative (-3/4).
Page 9 of 9