Prelim Examination 2009 / 2010
(Assessing Units 1 & 2)
MATHEMATICS
Advanced Higher Grade
Time allowed - 2 hours
Read Carefully
1.
2.
3.
Calculators may be used in this paper.
Candidates should answer all questions.
Full credit will only be given where the solution contains appropriate working.
All questions should be attempted
4 x
1.
2.
3.
Given that f ( x)
cos
1
x e 4 x , 0 < x < 1, obtain and simplify f x .
Find the term independent of u in the expansion of
(a)
6x 4
Express the function f x
Ax
B
C
x
D
E
x 1
x 1
2
u3
4
8
4
3u .
x 3 5x 4
in the form
x3 x
, where A, B, C, D and E are integers.
4
3
(b)
Hence show that
f ( x ) dx
4
16 ln 6
2
4.
4 2i
3 i
(a)
Express z
1 2i 3 i in the form x iy where x and y are real numbers.
(b)
(i)
Show that
(ii)
Write down a second root of the equation z z 3
(iii)
Find the other two roots of the equation z z 3
2i is a root of the equation z z 3
8z 2
8z 2
8z 2
36 z 32
36 z 32
36 z 32
5.
Prove by induction that 6 n 1 is divisible by 5 for all natural numbers n.
6.
A curve is defined by the parametric equations
x 10t , y
1 12t
t3
128
128
2
0.
0.
0.
1
3
5
for all t.
(a)
Find the coordinates of the stationary points of this curve.
(b)
Obtain an expression for
points found in (a).
128
3
4
d2y
and use this to determine the nature of the stationary
dx 2
3
[Turn over for Questions 7, 8, 9 and 10
7.
ax 3
The function f is defined by f x
bx 2
cx 1 where a, b and c are constants.
It is known that the graph of f passes through the point (1, 0) and has a stationary point at (-2, 9).
(a)
Deduce that a, b and c must satisfy the system of equations
a + b + c = –1
4a – 2b + c = –4
12a – 4b + c = 0.
(b)
8.
9.
Use Gaussian elimination to find the values of a, b and c.
The function g is defined by g ( x)
(a)
4
x 2
9
x 2
,x
2, x
R.
(i)
Write down the equations of the asymptotes of g.
(ii)
Find the coordinates of the point where the graph of y
the y- axis.
2
g (x) crosses
1
(b)
Find the coordinates and nature of the stationary points of g.
5
(c)
Sketch the graph of y = g(x), indicating the features found in (a) and (b).
3
(a)
The sum of the first n terms of a series is given by S n
th
Find an expression for the n term, u n .
(b)
10.
5
Obtain the sum of the series 64 32 16 ...
Use the substitution 5x
2
5
4 cos
6 n.
3
1
.
4
3
to show that
16 25 x 2 dx
0
n2
4
6 3
15
(Note that cos 2 A 1 2 sin 2 A. )
[ END OF QUESTION PAPER ]
7
Marking Scheme – Advanced Higher Grade 2009 / 2010
Prelim (Assessing Units 1 & 2)
Illustrations for awarding each mark
Give one mark for each
1
ans:
e
4 x
4 x
8 cos
1
4
2 x 1 x
(or equivalent)
xe
x
4 x
4 x
g f " (stated or implied)
"fg
2
4 x
x
e4
4 marks
2 x 1 x
knows to use product rule for differentiation
differentiates correctly
differentiates correctly
Simplifies correctly
8 cos
1
xe
4
x
e
4 x
4 x
2
4 x
4 x
8 cos
1
xe
4 x
2 x 1 x
(or a simplified equivalent)
2
ans: 81,648
finds correct general term
simplifies to find correct expression for
power of u
solves for r correctly
finds correct term
ans:
f ( x)
6x 1
4
x
3
1
x 1
x 1
4 marks
starts division correctly
completes division correctly
starts to find partial fractions correctly
expresses f(x) in correct form
r
ans: Proof
integrates correctly
substitutes correctly
continues correctly
completes proof correctly
4 marks
8 r
2
u3
3u
r
u 4r 24
r 6
81,648
x3
x 6x 4
f ( x)
6x 2
x3
6x 1
4x
5x 4
6x 2 4x 4
x3 x
4
C x 1 x 1
f ( x)
3(b)
2
4 marks
8
3(a)
4 x
4 x
3x 2
Dx x 1 Ex x 1
4
3
1
6x 1
x x 1 x 1
x
4 ln x
3 ln x 1
ln x 1
3
2
27 3 4 ln 3 3 ln 4 ln 2
12 2 4 ln 2 3 ln 3 ln 1
16 ln 3 5 ln 2 3 ln 4 (or equivalent)
... 16 ln 6
4(a)
ans:
4
4i
3 marks
4 2i 3 i
3 i 3 i
5 5i
4 4i
knows how to deal with quotient
finds product correctly
correct answer
4(b)
(i)
ans: Proof
2 marks
4
2i
8 2i
(or equivalent)
... 0
starts to show that -2i is a root correctly
completes proof
4(b)
(ii)
ans: 2i
5
ans:
36
2i
2
32
2i
128
1 mark
correct answer
4(b)
(iii)
3
4
4i
2i
3 marks
starts to find ‘second quadratic factor’
correctly
finds correct ‘second quadratic factor
finds correct roots
z 2 4 z 2 ...
(or equivalent)
z 2 8z 32 0
4 4i
ans: Proof
61 1 5
6 k 1 5m
verifies result for n = 1 (e.g.)
states correct assumption for n k
states correct result for n k 1
continues proof correctly
concludes proof correctly
5 marks
6k
6k
1
1
1 5p
1 ... 6 6 k
1
5
... 5 6m 1
5p
Since T for n = 1 &
(T for n = k
T for n = k+1), the result is T
n
N
6(a)
Give one mark for each
ans: 20, 15 & 20,17
Illustrations for awarding each mark
4 marks
substitutes correctly into correct formula
solves for t correctly
correct coordinates
correct coordinates
6(b)
MaximumT .P.
3 marks
finds second derivative correctly
correct nature
correct nature
ans: Proof
0
t
2
20, 15
20,17
ans: a 1, b
d2y
dx 2
60t
1000
3t
50
d2y
>0 MinimumT .P.
20, 15
dx 2
d2y
<0 MaximumT .P.
20,17
dx 2
4 marks
obtains first equation
obtains second equation
differentiates correctly
obtains third equation
7(b)
12 3t 2
10
d2y
60t
3t
2
1000
50
dx
MinimumT .P. &
ans: 20, 15
20,17
7(a)
dy
dx
dy
dx
2, c
4
correct augmented matrix
correct first modified system
correct second modified system
correct third modified system
correct solution
a + b + c = -1
4a – 2b + c = -4
f ( x) 3ax 2 2bx c
12a – 4b + c = 0
5 marks
1
1
1
1
4
2 1
4
12
4 1
1
1
0
6
12
4
1
1
0
6
0
16
1
1
0
1
1
3 0
1
0
1
3
1
0
11 12
1
1
0 96 48
0
0 96 66
72
a 1, b
2, c
4
8(a)
(i)
Give one mark for each
ans: x
2, y x 2
Illustrations for awarding each mark
2 marks
x
y
correct equation
correct equation
8(a)
(ii)
ans: 0,2 5
1 mark
0,2 5
correct coordinates
8(b)
ans:
5, 10
1,2
MaximumT .P. &
g ( x) 1
MinimumT .P.
5 marks
x
differentiates correctly
solves g (x) 0 correctly
correct y-coordinates
correct formula for second derivative
correct nature of both points
8(c)
ans: Correct graph
ans: u n
2n 7
ans: 42 75
identifies correct common ratio
identifies correct number of terms
correct answer
x 2
2
5,1
5, 10 & 1,2
18
g ( x)
3
x 2
MaximumT .P. &
g ( 5) < 0
MinimumT .P.
g (1) > 0
See graph at end of Marking Scheme
3 mark
S1
starts correctly
identifies correct sequence
correct answer
9(b)
9
3 marks
turning points shown
y-intercept shown
correct behaviour at asymptotes
9(a)
2
x 2
5, S 2
5, 3, 1,...
u n 2n 7
3 marks
r
n
1
2
9
42 75
8, S 3
9,...
10
Give one mark for each
ans: Proof
starts substitution correctly
changes limits correctly
manipulates expression correctly
manipulates expression correctly
integrates correctly
substitutes limits correctly
completes proof correctly
Illustrations for awarding each mark
7 marks
2
4
16 25 cos
5
x
0
2
2
5
,x
43
4 1 cos 2
5
4
sin
5
d
3
sin d
2
16
5
3
1
2
1
cos 2
2
d
2
16 1
5 2
1
sin 2
4
3
2
16
5
1
23
1
sin 2
4
3
16
5
1
22
1
sin 2
4
2
...
4
6 3
.
15
TOTAL MARKS = 70
Graph for Question 8(c)
y
y=x–2
2.5
(1, 2)
0
(-5, -10)
x = -2
x