General Certificate of Education
Advanced Level Examination
January 2011
Mathematics
MPC4
Unit Pure Core 4
Monday 24 January 2011
9.00 am to 10.30 am
d
For this paper you must have:
*
the blue AQA booklet of formulae and statistical tables.
You may use a graphics calculator.
e
s
Time allowed
*
1 hour 30 minutes
n
Instructions
*
Use black ink or black ball-point pen. Pencil should only be used for
drawing.
*
Fill in the boxes at the top of this page.
*
Answer all questions.
*
Write the question part reference (eg (a), (b)(i) etc) in the left-hand
margin.
*
You must answer the questions in the spaces provided. Do not write
outside the box around each page.
*
Show all necessary working; otherwise marks for method may be
lost.
*
Do all rough work in this book. Cross through any work that you do
not want to be marked.
e
d
n
o
C
Information
*
The marks for questions are shown in brackets.
*
The maximum mark for this paper is 75.
Advice
*
Unless stated otherwise, you may quote formulae, without proof,
from the booklet.
P38267/Jan11/MPC4 6/6/6/
MPC4
2
Express 2 sin x þ 5 cos x in the form R sinðx þ aÞ , where R > 0 and 0 < a < 90 .
Give your value of a to the nearest 0.1 .
(3 marks)
1 (a)
(b) (i)
Write down the maximum value of 2 sin x þ 5 cos x .
(1 mark)
(ii) Find the value of x in the interval 0 4 x 4 360 at which this maximum occurs,
giving the value of x to the nearest 0.1 .
(2 marks)
The polynomial f ðxÞ is defined by f ðxÞ ¼ 9x 3 þ 18x 2 x 2 .
2 (a)
(i)
Use the Factor Theorem to show that 3x þ 1 is a factor of f ðxÞ .
(ii) Express f ðxÞ as a product of three linear factors.
(iii) Simplify
(b)
(3 marks)
9x 3 þ 21x 2 þ 6x
.
f ðxÞ
(3 marks)
When the polynomial 9x 3 þ px 2 x 2 is divided by 3x 2 , the remainder is 4 .
Find the value of the constant p.
3 (a)
(b)
Express
(2 marks)
3 þ 9x
A
B
in the form
þ
, where A and B are integers.
ð1 þ xÞð3 þ 5xÞ
1 þ x 3 þ 5x
(3 marks)
Hence, or otherwise, find the binomial expansion of
including the term in x 2 .
(c)
(2 marks)
3 þ 9x
up to and
ð1 þ xÞð3 þ 5xÞ
(7 marks)
Find the range of values of x for which the binomial expansion of
valid.
3 þ 9x
is
ð1 þ xÞð3 þ 5xÞ
(2 marks)
P38267/Jan11/MPC4
3
A curve is defined by the parametric equations
4
x ¼ 3et ,
(a) (i)
y ¼ e2t e2t
Find the gradient of the curve at the point where t ¼ 0 .
(3 marks)
(ii) Find an equation of the tangent to the curve at the point where t ¼ 0 .
(b)
(1 mark)
Show that the cartesian equation of the curve can be written in the form
y¼
x2
k
2
k x
where k is an integer.
(2 marks)
A model for the radioactive decay of a form of iodine is given by
5
m ¼ m0 2
18 t
The mass of the iodine after t days is m grams. Its initial mass is m0 grams.
(a)
(b)
Use the given model to find the mass that remains after 10 grams of this form of
iodine have decayed for 14 days, giving your answer to the nearest gram. (2 marks)
m
A mass of m0 grams of this form of iodine decays to 0 grams in d days.
16
Find the value of d.
(c)
(2 marks)
After n days, a mass of this form of iodine has decayed to less than 1% of its initial
mass.
Find the minimum integer value of n.
(3 marks)
s
Turn over
P38267/Jan11/MPC4
4
6 (a) (i)
Given that tan 2x þ tan x ¼ 0 , show that tan x ¼ 0 or tan2 x ¼ 3 .
(3 marks)
(ii) Hence find all solutions of tan 2x þ tan x ¼ 0 in the interval 0 < x < 180 .
(1 mark)
(b) (i)
Given that cos x 6¼ 0 , show that the equation
sin 2x ¼ cos x cos 2x
can be written in the form
2 sin2 x þ 2 sin x 1 ¼ 0
(3 marks)
(ii) Show that all solutions of the equation 2 sin2 x þ 2 sin x 1 ¼ 0 are given by
pffiffiffi
31
, where p is an integer.
sin x ¼
p
7 (a) (i)
Solve the differential equation
dx pffiffiffi t ¼ x sin
to find x in terms of t.
2
dt
(3 marks)
(3 marks)
(ii) Given that x ¼ 1 when t ¼ 0 , show that the solution can be written as
x ¼ ða cos btÞ2
where a and b are constants to be found.
(3 marks)
The height, x metres, above the ground of a car in a fairground ride at time t seconds
dx pffiffiffi t ¼ x sin
.
is modelled by the differential equation
2
dt
(b)
The car is 1 metre above the ground when t ¼ 0 .
(i)
Find the greatest height above the ground reached by the car during the ride.
(2 marks)
(ii) Find the value of t when the car is first 5 metres above the ground, giving your
answer to one decimal place.
(2 marks)
P38267/Jan11/MPC4
5
The coordinates of the points A and B are ð3, 2, 4Þ and ð6, 0, 3Þ respectively.
2
3
2
3
2
3
The line l1 has equation r ¼ 4 2 5 þ l 4 1 5 .
3
4
8
(a) (i)
!
Find the vector AB .
(2 marks)
!
(ii) Calculate the acute angle between AB and the line l1 , giving your answer to the
nearest 0.1.
(4 marks)
The point D lies on l1 where l ¼ 2 . The line l2 passes through D and is parallel
to AB .
(b)
(i)
Find a vector equation of line l2 with parameter m .
(2 marks)
(ii) The diagram shows a symmetrical trapezium ABCD , with angle DAB equal to
angle ABC .
line l1
D
A
C
line l2
B
The point C lies on line l2 . The length of AD is equal to the length of BC .
Find the coordinates of C.
(6 marks)
Copyright ª 2011 AQA and its licensors. All rights reserved.
P38267/Jan11/MPC4
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
Key to mark scheme abbreviations
M
m or dM
A
B
E
or ft or F
CAO
CSO
AWFW
AWRT
ACF
AG
SC
OE
A2,1
–x EE
NMS
PI
SCA
c
sf
dp
mark is for method
mark is dependent on one or more M marks and is for method
mark is dependent on M or m marks and is for accuracy
mark is independent of M or m marks and is for method and accuracy
mark is for explanation
follow through from previous incorrect result
correct answer only
correct solution only
anything which falls within
anything which rounds to
any correct form
answer given
special case
or equivalent
2 or 1 (or 0) accuracy marks
deduct x marks for each error
no method shown
possibly implied
substantially correct approach
candidate
significant figure(s)
decimal place(s)
No Method Shown
Where the question specifically requires a particular method to be used, we must usually see evidence
of use of this method for any marks to be awarded.
Where the answer can be reasonably obtained without showing working and it is very unlikely that the
correct answer can be obtained by using an incorrect method, we must award full marks. However,
the obvious penalty to candidates showing no working is that incorrect answers, however close, earn
no marks.
Where a question asks the candidate to state or write down a result, no method need be shown for full
marks.
Where the permitted calculator has functions which reasonably allow the solution of the question
directly, the correct answer without working earns full marks, unless it is given to less than the degree
of accuracy accepted in the mark scheme, when it gains no marks.
Otherwise we require evidence of a correct method for any marks to be awarded.
3
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
MPC4
Q
Solution
1(a)
R = 29
Rsinα = 5 or Rcosα = 2 or tan α =
α = 68.2 º
(b)(i) (maximum value =)
(ii)
Marks
B1
5
2
29
sin ( x + α ) = 1
Total
M1
A1
3
Condone α = 68.20 º
B1ft
1
ft on R
Or x + α = 90, x + α =
M1
x = 21.8 º only
Comments
Accept 5.4 or 5.38, 5.39, 5.385….
A1
Total
4
2
6
No ISW
π
2
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
MPC4 (cont)
Q
Solution
2 (a)(i) f − 1 = 9 − 1 3 + 18 − 1 2 − − 1 − 2
( 3) ( 3)
( 3) ( 3)
Marks
Total
Comments
f (−
M1
= 9 ( − 271 ) + 18 ( 19 ) − ( − 13 ) − 2
1
3
) attempted
NOT long division
= − 13 + 2 + 13 − 2 = 0
⇒ ( 3 x + 1) is a factor
(ii)
(f ( x) =) (3x + 1) ( 3x 2 + kx − 2 )
Shown = 0 plus statement
3 and − 2
A1
3
(f ( x) =) (3x + 1)(3x − 1)( x + 2)
A1
9 x3 + 21x 2 + 6 x = x ( 9 x 2 + 21x + 6 )
M1
x and attempt to factorise quadratic
equation.
A1
Correct factors
= 3 x ( 3x + 1)( x + 2 )
(b)
2
M1
k=5
(iii)
A1
9 x3 + 21x 2 + 6 x
3x
=
f ( x)
3x − 1
A1
9 ( 23 ) + p ( 23 ) − 23 − 2 = −4
M1
p = −9
A1
3
2
3
cso
no ISW
Condone missing brackets, but must have
= −4
2
10
2(a)(ii) Alternative
Using long division
3x 2 + 5 x − 2
(M1)
3x2 + ax + b
(A1)
3x 2 + 5 x − 2
3 x + 1) 9 x3 + 18 x 2 − x − 2
9 x3 + 3x 2
15 x 2 − x
15 x 2 + 5 x
−6 x − 2
− 6x − 2
(f ( x ) =) ( 3x + 1)( 3x − 1)( x + 2 )
(A1)
5
(3)
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
MPC4 (cont)
Q
2(a)(iii) Alternative
Solution
f ( x) + q( x)
, where q is a quadratic
f ( x)
expression
( 3x + 1)( x + 2 )
= 1+
( 3x + 1)( 3x − 1)( x + 2 )
=1+
1
3x − 1
Marks
Total
(M1)
(A1)
(A1)
6
(3)
Comments
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
MPC4 (cont)
Q
Solution
3(a) 3 + 9 x = A ( 3 + 5 x ) + B (1 + x )
Marks
M1
x = −1
x = − 53
m1
A=3
B = −6
A1
Total
Comments
PI by correct A and B
Substitute two values of x and solve for A
and B.
3
Alternative
Equating coefficients
3 + 9 x = A ( 3 + 5 x ) + B (1 + x )
3 = 3A + B
9 = 5A + B
A=3
B = −6
(M1)
(m1)
(A1)
Set up simultaneous equations and solve.
Condone 1 error.
(3)
Alternative
Cover up rule
3
x=−
5
3−9
3−5
3 − 275
B=
1 − 53
A=3
B = −6
x = −1
(1 + x )
(b)
A=
−1
(M1)
x = −1 and x = − 53
and attempt to find A and B.
(A1
A1)
(3)
= 1 − x + kx 2
= 1 − x + x2
( 3 + 5 x ) = 3−1 (1 + 53 x )
−1
2
(1 + 53 x ) = 1 − 53 x + ( 53 x )
−1
−1
= 1 − 53 x + 259 x 2
M1
A1
B1
Condone missing brackets; allow one sign
error
M1
3 + 9x
(1 + x )( 3 + 5 x )
SC NMS
A and B both correct; 3/3
One of A and B correct 1/3
A1
25 2 ⎞
⎛ 5
= 3 (1 − x + x 2 ) − 6 × 3−1 ⎜1 − x +
x ⎟
9
⎝ 3
⎠
Use PFs and simplify to a + bx + cx
or expand product of ( 3 + 9x ) and
M1
binomial expansions and simplify to
1
23
= 1 + x − x2
3
9
a + bx + cx 2
A1
7
7
2
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
MPC4 (cont)
Q
(c)
Solution
5 x < 1 oe or 5 x > −1 oe
3
3
x<
3
3
3
or − < x <
5
5
5
Marks
Total
M1
A1
8
Comments
Condone ≤ instead of <
CAO
2
12
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
MPC4 (cont)
Q
4(a)(i) dx
dt
Solution
= 3et
t=0
(ii)
(b)
gradient =
y=
e 2t =
4
( x − 3)
3
Marks
M1
dy
= 2e 2t + 2e −2t
dt
A1
4
3
oe
x2
x
⎛ x⎞
or 9e 2t = x 2 or et = or e 2t = ⎜ ⎟
9
3
⎝3⎠
⎛ x2 ⎞
⎛ x⎞
or t = ln ⎜ ⎟ or 2t = ln ⎜ ⎟
⎝3⎠
⎝ 9 ⎠
2
9
x
− 2
y=
9 x
Total
Comments
Both derivatives attempted and one
correct
Both correct
Condone
dy 4
=
dx 3
A1
3
cso
B1ft
1
ft on non-zero gradient
2
Equation required
2
M1
A1
6
9
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
MPC4 (cont)
Q
5(a)
Solution
m = 10 × 2
14
−
8
≈ 3 (gm)
(b)
(c)
d
8
Marks
Total
M1
A1
2
Condone 2.97 or better
NOT 2.9 as final answer
2
cso
1
16
M1
d
= 4 ⇒ d = 32
8
A1
2
−
=
0.01m0 = m0 × 2
−
t
8
t
ln ( 0.01) = − ln ( 2 )
8
t = 53.15
n = 54
Comments
M1
m0 can be numerical
M1
Take logs correctly from their equation
leading to a linear equation in t.
A1
10
3
7
cso
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
Q
6(a)(i)
Solution
Marks
2 tan x
tan 2 x =
1 − tan 2 x
2 tan x + tan x (1 − tan 2 x ) = 0
Total
Comments
B1
Condone numerator as tan x + tan x
M1
Multiplying throughout by their
denominator
tan x = 0
or ( 2 + 1 − tan 2 x ) = 0 ⇒ tan 2 x = 3
A1
3
AG Must show tan x = 0 and tan2x = 3
Alternative
tan 2 x =
sin 2 x
2sin x cos x
=
cos 2 x cos 2 x − sin 2 x
2sin x cos x
sin x
+
=0
2
2
cos x − sin x cos x
(B1)
2sin x cos 2 x + sin x ( cos 2 x − sin 2 x ) = 0
sin x(2 cos 2 x + cos 2 x − sin 2 x) = 0
⇒ sin x = 0 ⎫ and
⎬
⇒ tan x = 0 ⎭ and
(ii)
(b)(i)
(M1)
3cos 2 x = sin 2 x ⎫
⎬
tan 2 x = 3 ⎭
(A1)
(3)
x = 60 AND x = 120
B1
1
2sin x cos x = cos x.f ( x)
M1
2sin x cos x = cos x (1 − 2sin 2 x )
( cos x ≠ 0 )
Condone extra answers outside interval
eg 0 and 180
Where f(x) = cos2x – sin2x
or 2cos2x – 1 or 1 – 2sin2x
A1
2
2sin x = 1 − 2sin x
2sin 2 x + 2sin x − 1 = 0
A1
11
3
AG
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
(ii)
sin x =
−2 ± 4 − 4 × 2 × ( −1)
2× 2
−2 ± 2 3
sin x =
4
⎫
−1 − 3
has no solution ⎪
⎪
2
⎬
3 −1
⎪
sin x =
⎪⎭
2
M1
Correct use of quadratic formula or
completing the square or correct factors
A1
12 must be simplified and must
have ±
sin x =
E1
3
10
12
Reject one solution and state correct
solution.
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
MPC4
Q
Solution
7
(a)(i)
∫
Marks
dx
⎛t⎞
= ∫ sin ⎜ ⎟ dt
x
⎝2⎠
B1
⎛t⎞
2 x = −2 cos ⎜ ⎟ (+ k )
⎝2⎠
⎛
⎞
⎛t⎞
x = ⎜ − cos ⎜ ⎟ + C ⎟
⎝2⎠
⎝
⎠
(ii)
(1, 0 )
Condone missing + k
2
A1
2 = −2 + k or 1 = ( −1 + C )
⎛
⎛ t ⎞⎞
x = ⎜ 2 − cos ⎜ ⎟ ⎟
⎝ 2 ⎠⎠
⎝
2
Must have previous line correct
M1
Use (1, 0 ) to find a constant
A1ft
ft on C = p – q from (a)(i)
A1
A1ft
⎛t⎞
cos ⎜ ⎟ = 2 − 5
⎝2⎠
M1
t = 2 cos −1 2 − 5 = 3.6 (seconds 1dp)
3
cso applies to (a)(ii)
2
ft is (their a + 1)2
M1
Greatest height = 9 (m)
)
3
2
(b)(i) Greatest height when cos ( bt ) = −1
(
Comments
Correct separation;
condone missing integral signs.
⎛t⎞
p x = q cos⎜ ⎟
⎝2⎠
M1
k = 4 or C = 2
(ii)
Total
A1
cos bt = a − 5
2
10
13
condone 3.6 or better (3.618…..)
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
MPC4 (cont)
Q
8(a)(i)
Solution
⎡ 6 ⎤ ⎡ 3 ⎤ ⎡ 3⎤
JJJG ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
AB = ⎢ 0 ⎥ − ⎢ −2 ⎥ = ⎢ 2 ⎥
⎢⎣ 3⎥⎦ ⎢⎣ 4 ⎥⎦ ⎢⎣ −1⎥⎦
⎡ 3⎤ ⎡ 2 ⎤
⎢ 2 ⎥ • ⎢ −1⎥ = 6 − 2 − 3 = 1
⎢ ⎥ ⎢ ⎥
⎢⎣ −1⎥⎦ ⎢⎣ 3⎥⎦
(ii)
sp
14 14
1
cosθ =
θ = 85.9º
14
cosθ =
(b)(i)
⎡ 3⎤
⎡ 2⎤ ⎡ 7 ⎤
JJJG ⎢ ⎥
OD = ⎢ −2 ⎥ + 2 ⎢⎢ −1⎥⎥ = ⎢⎢ −4 ⎥⎥
⎢⎣ 4 ⎥⎦
⎢⎣ 3⎥⎦ ⎢⎣ 10 ⎥⎦
line l2
(ii)
⎡ 7⎤
⎡ 3⎤
⎢
⎥
r = ⎢ −4 ⎥ + μ ⎢⎢ 2 ⎥⎥
⎢⎣ 10 ⎥⎦
⎢⎣ −1⎥⎦
⎡ 1+ 3 p⎤
JJJG JJJG JJJG ⎢
BC = OC − OB = ⎢ −4 + 2 p ⎥⎥
⎢⎣ 7 − p ⎥⎦
⎡ 4⎤
JJJG ⎢ ⎥
AD = ⎢ −2 ⎥
⎢⎣ 6 ⎥⎦
JJJG
BC = 56
(1 + 3 p )2 + (− 4 + 2 p )2 + (7 − p )2 = 56
Marks
Total
Comments
JJJG JJJG
± OB − OA implied by 2 correct
(
M1
components
A1
2
M1
Scalar product with correct vectors; allow
one component error.
A1ft
ft on AB
m1
Correct form for cos θ with one correct
modulus
A1
JJJG
4
cso
Implied by 2 correct components
A1ft
⎡x⎤
r = or ⎢⎢ y ⎥⎥ required
⎣⎢ z ⎦⎥
2
M1
μ = p at C
JJJG
Find BC in terms of p
B1ft
PI
5
and p = 1
7
4 2⎞
⎛ 1
C is at ⎜ 9 , −2 ,9 ⎟
7 7⎠
⎝ 7
JJJG
ft on AB
JJJG
B1 is for BC = 56
m1
JJJG
m1
ft on BC
Simplification to quadratic equation with
all terms on one side
A1
Exact fraction required
( 7 p − 5)( p − 1) = 0
p=
85.9 or better
M1
14 p 2 − 24 p + 66 = 56
7 p 2 − 12 p + 5 = 0
)
A1
6
14
14
cso
Accept as column vector
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
MPC4 (cont)
Q
Solution
8(b)(ii) Alternative : Using equal angles
Marks
(M1)
⎡ 1+ 3 p⎤
JJJG JJJG JJJG ⎢
BC = OC − OB = ⎢ −4 + 2 p ⎥⎥
⎢⎣ 7 − p ⎥⎦
⎡ 4⎤
JJJG ⎢ ⎥
AD = ⎢ −2 ⎥
⎢⎣ 6 ⎥⎦
Total
JJJG
BC = 56
Comments
μ = p at C
JJJG
Find BC in terms of p
(B1ft)
⎡ −3⎤ ⎡ 1 + 3 p ⎤
⎢ −2 ⎥ • ⎢ −4 + 2 p ⎥
⎥
JJJG JJJG ⎢ ⎥ ⎢
BA • BC ⎢⎣ 1 ⎥⎦ ⎢⎣ 7 − p ⎥⎦ 1
(cos θ ) =
=
=
14
14 56
14 56
JJJG
(m1)
Condone AB used.
JJJG
Allow BC in terms of p, in which
case previous B1 is implied
−3 − 9 p + 8 − 4 p + 7 − p = 2
5
p=
7
4 2⎞
⎛ 1
C is at ⎜ 9 , −2 ,9 ⎟
7 7⎠
⎝ 7
Reduce to linear or quadratic equation
in p.
(m1)
(A1)
(A1)
15
(6)
Mark Scheme – General Certificate of Education (A-level) Mathematics – Pure Core 4 – January 2011
MPC4 (cont)
Q
Solution
8(b)(ii) Alternative : using symmetry (i)
Marks
JJJG JJJG
AD = BC = 56
(B1ft)
JJJG JJJG JJJG
JJJG
DC = AB − AD cos θ − BC cos θ
(M1)
JJJG
10
DC =
14
JJJG
JJJG
10
= p 14
DC = p AB ⇒
14
5
p=
7
4 2⎞
⎛ 1
C is at ⎜ 9 , −2 ,9 ⎟
7 7⎠
⎝ 7
Total
Comments
⎡ 4⎤
JJJG ⎢ ⎥
AD = ⎢ −2 ⎥
⎢⎣ 6 ⎥⎦
Substitute values and evaluate
JJJG JJJG
JJJG
AB − AD cos θ − BC cos θ
JJJG
(A1ft)
F on AB and cos θ
(m1)
Set up equation in p
(A1)
(A1)
D
C
(6)
Alternative using symmetry (ii)
JJJG
AD = 56
(B1ft)
A
JJJG JJJG
1
2
AE = AD cos θ = 56 × =
14
14
JJJG
JJJG
2
= q 14
AE = q AB ⇒
14
JJJG JJJG
and AE = FB ⇒ p = 1 − 2q
(M1)
(A1ft)
2
5
p=
14
7
4 2⎞
⎛ 1
C is at ⎜ 9 , −2 ,9 ⎟
7 7⎠
⎝ 7
JJJG
for AD cos θ . F on cos θ
Set up equation to find p
(A1)
(A1)
TOTAL
(6)
75
16
F
B
Substitute values and evaluate
(m1)
q=
E
AQA – Core 4 – Jan 2011 – Answers Question 1: Exam report a) 2Sinx  5Cosx
R  22  52  29
5
 2

 29 
Sinx 
Cosx   29  Cos Sinx  Sin Cosx 
29
 29

2
5
with Cos 
and Sin 
so   68.2o
29
29
2Sinx  5Cosx  29Sin( x  68.2o )
b) 2Sinx  5Cosx  29 Sin( x  68.2o )
For all x,  1  Sin( x  68.2o )  1
 29  29Sin( x  68.2)  29
so
The maximum value of 2Sinx  5Cosx is 29
ii ) The maximum occurs when Sin( x  68.2)  1
which means x  68.2  90 so x  21.8o
This question provided a good start for the candidates with over half the entry gaining full marks, although some candidates omitted it altogether. Part (a) The most common error was to equate tanα to 2/5 instead of 5/2. Some candidates didn’t make explicit their value for R so didn’t answer the question, and so lost a mark even if they gave the correct answer to part (b)(i). Part (b) Despite having quoted R as √29 in part (a), some candidates went on to state the maximum of 2sinx + 5cosx as 5, and said the maximum occurred at x = 0, or other similar errors. Some candidates had the correct answer for where the maximum occurred in their answer, but then went on to give more than one value, and so lost the mark. A few candidates correctly found the minimum value and where it occurred, but gained no credit. Almost all candidates heeded the instruction to give values of angles to the nearest 0.1°. Question 2: Exam report a) f ( x)  9 x  18 x  x  2
3
2
1
i ) 3 x  1 is a factor of f ( x)  f ( )  0
3
3
2
1
 1
 1  1
f ( )  9     18         2 
3
 3
 3  3
1
1
  2 2  0
3
3
3 x  1 is a factor of f ( x)
ii ) f ( x)  (3 x  1)(3x 2  5 x  2) (long division)
f ( x) = (3x  1)(3 x  1)( x  2)
9 x 3  21x 2  6 x
3 x(3x 2  7 x  2)
3 x(3 x  1)( x  2)
iii )


(3 x  1)(3 x  1)( x  2) (3 x  1)(3 x  1)( x  2)
f ( x)
9 x3  21x 2  6 x
3x
Simplifying :

3x  1
f ( x)
b) g ( x)  9 x 3  px 2  x  2
2
the division of g ( x) by 3 x  2 has remainder  4  g ( )  4
3
3
2
8 4p 8 4p
2
2
2 2
 
g( )  9   p     2  
3 9 3 9
3
3
3 3
4p
And
  4  p  9
9
Overall, this question was answered well and was a good source of marks for most candidates. Part (a)(i) Many candidates lost a mark here through not giving a full proof of the result. Although most recognised the need to find f(‐
1/3) many candidates merely wrote this in an unsimplified form followed by = 0 . An extra line of working was required as well as a concluding statement. A few candidates used the method of long division which gained no marks as the question asked for use of the Factor Theorem. Part (a)(ii) This was answered very well and most candidates obtained the correct three factors and presented them as a product. Methods used, though, varied considerably from those who just wrote the factors down by inspection, to those who used long division to find a quadratic factor and factorised that, to those who used a complete method of undetermined coefficients. All methods are valid, although some are much more efficient than others. Part (a)(iii) Although there were many correct solutions, some candidates lost either a factor of 3 or x or sometimes both so could not complete correctly. Also, some stopped after correct cancelling of one factor, for which they gained some credit. Part (b) The majority of candidates found f(2/3) and equated it to −4, although a few equated it to 4. However, arithmetic errors often occurred so some did not reach the correct answer. A few candidates attempted this part by long division which, although valid, is a most inefficient method. Question 3: a)
Exam report A
B
3  9x


(multiplying by (1  x)(3  5 x))
(1  x)(3  5 x) 1  x 3  5 x
3  9 x  A(3  5 x)  B (1  x)
The majority of candidates did well on this question, with nearly half the candidates scoring 10 of the 12 marks or more. Part (a) Candidates were generally confident and competent in finding partial fractions and A and B were usually found correctly. The most common method was to substitute x = −1 and x = −3 / 5. The la er occasionally caused arithmetic problems. However, those who chose to use simultaneous equations or a mix of the two usual methods were just as successful. Part (b) It was pleasing to see that candidates could generally use the binomial expansions correctly and accurately and there were many complete correct solutions. The most common errors occurred in the expansion of (3+5x)‐1 . for x  1 , 3  9  A(3  5) so A  3
3
27
3
for x   , 3 
 B(1  ) so B  6
5
5
5
3  9x
3
6


(1  x)(3  5 x) 1  x 3  5 x
3  9x
b)
 3(1  x) 1  6(3  5 x) 1 ( factorise by 3)
(1  x)(3  5 x)
 5 
 3(1  x) 1  6  31  1  x 
 3 
1 2 2
x  ...  1  x  x 2  ...
(1  x) 1  1  x 
2
1
1
Instead of 31 1  5 x  some used 


3 
1
 5 
3 1  x  . Some omitted the brackets  3 
2
around 
3  9x
25 2
 5

x  ... 
 3 1  x  x 2  ...  2  1  x 
(1  x)(3  5 x)
9
 3

10
50
 3  3 x  3 x 2  2  x  x 2  ...
3
9
1
23 2
 1  x  x  ...
3
9
1
c) (1  x) is valid for  1  x  1
2
5x 
  too, although some continued  3 
1 2  5 
5
5
5
25 2
(1  x) 1  1  x 
 x   ...  1  x  x  ...
3
3
2 3 
3
9
5
5
3
3
(1  x) 1 is valid for  1  x  1    x 
3
3
5
5
3
3
For both expansion to be valid, we need to have   x 
5
5
correctly from here. Almost all knew they were to combine the two series to get to the final result, although many sign and arithmetic errors were seen in attempting to do this. Very few misused the constants A and B found earlier, with just a few raising their result to these powers. Few candidates attempted this question by multiplying their expansions together. Part (c) Many candidates showed they were not familiar with finding the range of valid values of x. Some just stated −1< x <1 while others assumed the answer to be −5 / 3 < x < 5 / 3 . Some didn’t even use an inequality but gave specific values for which they believed the expansions were not valid. Those who did find the correct limits often included the equality, so lost a mark. Question 4: Exam report dy dy dx 2e  2e
4e
4e

 
 t 
t
dx dt dt
3e
3e
3
dy 4

When t  0,
dx 3
ii ) At t  0, x  3e0  3 and y  e0  e0  0
The equation of the tangent to the curve at t  0 is 2t
2t
2t
t
a)i )
4
4
y  0  ( x  3)  y  x  4
3
3
x
x2
t
t
2t
t 2
b) x  3e so e 
and e  (e ) 
3
9
2
x
9
This gives: y   2
9 x
This question was also done well, with again nearly half the entry gaining full marks. Part (a)(i) Most candidates found the derivatives dx/dt and dy/dt successfully and went on to divide to find dy/dx correctly. The most common, although rare, error was the omission of the minus sign when differentiating e−2t. Occasionally a candidate used dx dy
/ . dt dt
Part (a)(ii) Almost all candidates gained this mark for a correct tangent equation using their numerical gradient from part (a)(i). Only a few took the negative reciprocal to find the equation of the normal. Part (b) Most candidates were successful here too. Whilst some used the quickest method of substituting et 
x
x 2 , or e 2t 
3
9
others used logarithms. Candidates were required to conclude with the correct equation, which some failed to do; a verification was given no credit. Question 5: Exam report a) m  m0  2
1
 t
8
m0  10 and t  14
m  10  2
b)

14
8
 2.9730  3g
d
d


m0
1
 m0  2 8 
2 8
16
16
  d8 
1
Composing by "ln": ln    ln  2 
 16 


d
8ln16
ln 2  d 
 32
8
ln 2
c) m  1% of m0  0.01m0
 ln16  
0.01m0  m0  2

n
8
 0.01  2

n
8
n
8ln 0.01
 53.15
ln(0.01)   ln 2  n 
ln 2
8
n  54 days
Most candidates scored well on this question although some misinterpreted the information given. Part (a) Most candidates substituted and evaluated correctly. A few, having written the unsimplified form, could not cope with the index and combined the 10 and 2 incorrectly. Some misinterpreted the information and found a mass that decayed to 10 grams in 14 days. Part (b) Most candidates substituted the 16 correctly and then, usually, took logs in order to solve for d. Only a minority recognised that 1/16 = 2‐4 and so d/8=4. Some candidates began 
d
incorrectly by writing 2 8  16 which led to a negative answer. Most of these then just dropped the minus sign rather than looking for an error in their working. Part (c) Most candidates set up the correct equation and used logarithms to various bases to solve it. The most common answer here was 53.15, as candidates failed to heed the request to give the answer as an integer. Many also rounded down to 53, misinterpreting what the question was asking. Not all understood the question and assumed the mass had decayed by 1%, so used 99%. Question 6: a )i ) Tan 2 x  Tanx  0
2Tanx
 Tanx  0
1  Tan 2 x
2Tanx  Tanx(1  Tan 2 x)  0
Exam report 2Tanx  Tanx  Tan3 x  0
3Tanx  Tan3 x  0
Tanx  3  Tan 2 x   0
so Tanx  0 or Tan 2 x  3
ii ) Tanx  0 for x  0o or 180o (outside the range)
Tan 2 x  3  Tanx  3 or Tanx   3
Tan 1 ( 3)  60o and Tan 1 ( 3)  60
so x  60  k 180 or x  60  k 180
Because 0o  x  180o
we select x  60o or x  120o
b)i ) Sin2 x  CosxCos 2 x
2 SinxCosx  Cosx(1  2 Sin 2 x)
Because Cosx  0, we can divide both sides by Cosx :
2 Sinx  1  2 Sin 2 x  2 Sin 2  2 Sinx  1  0
ii ) This is a quadratic equation in "Sinx", using the quadratic formula
2  12 1  3
1  3
( 0.37) or Sinx 
(  1.37)

4
2
2
3 1
1  3
 1.
The solutions are given by Sinx 
only because
2
2
Sinx 
There was a wide range of responses to this question and relatively few candidates gained full marks. Part (a)(i) Most candidates wrote down a correct form for tan 2x with only a few writing tan2 x instead of 2tan x . Most multiplied by the denominator to obtain a cubic equation in tan x. However, many divided throughout by tan x to get tan2 x = 3 and failed to demonstrate that tan x = 0 is also a solution. Some used verification which did not warrant any marks. Part (a)(ii) Very few candidates gained this mark. The solution of 120° was almost always omitted. Most wrote 0° and 180° too. This in itself was not penalised this time as these values were outside the range of x requested. Part (b)(i) This was answered very well, as candidates showed familiarity with double angle formulae and handled the algebra involved with confidence. Part (b)(ii) The quadratic formula was used correctly by most candidates and most demonstrated where the 3 came from. The majority of candidates, though, just seemed to ignore the negative root and gave no explanation as to why. If they found only one root then they lost 2 marks. Some gave unsatisfactory explanations for the exclusion of the negative root, such as sin x cannot be negative or that it was >1. Those who used completion of the square to solve the equation almost invariably just took the positive square root. Question 7: dx
t
a)i )  xSin separating the varaibles
dt
2
t
t
1 dx
1
 Sin , integrating 
dx   Sin dt
2
2
x dt
x
t
c
t
2 x  2Cos  c  x   Cos
2
2
2
2
t
c

Squaring both sides : x   a  Cos  with  a
2
2

ii ) When t  0, x  1 so 2 1  2Cos 0  c
 c  4 and a  2
2
t

x   2  Cos 
2

t
t
b)i ) For all t ,  1  Cos  1  1  2  Cos  3
2
2
The maximum height is therefore xMAX = 32  9
ii ) We need to solve
5  2  Cos
Cos 1 (2  5)  1.81
so
t
t
 Cos  2  5
2
2
t
 1.81  t  3.6 seconds
2
Exam report A full range of marks were seen in this question. Some candidates had little or no idea as to how to solve a differential equation, whereas some gave concise and correct solutions and could use their solution to solve the problem in the given context. Part (a)(i) Most candidates separated the variables correctly, although some poor notation was seen and some mishandled the x . Many also made errors in the integration, often getting the coefficient wrong in one or both integrals. Although many did add an arbitrary constant, many also failed to complete the question as requested and by writing x in terms of t. Of those that did, many failed to square their expression in t correctly; (f(t) +C)2 = (f(t))2 +C2was a common error. Many apparently didn’t review their solution when given the form of answer in part (a)(ii). When manipulating an expression containing a constant some, for example, renamed as C/2 as C and sometimes then confused themselves; whereas others correctly introduced a new letter for the revised constant. Part (a)(ii) Many went on to find a correct value for the constant and give the required form, as they used a previous line of working from part (a)(i) rather than their final answer. Any attempt to find a constant using the information given was credited, but many tried to force their solution into the form of the answer given, rather than review their work for an error. Part (b)(i) This question confused a lot of candidates and many assumed the greatest height occurred at cosbt = 0 and not −1. Many candidates didn’t attempt this part. Part (b)(ii) For those who set up a correct equation, the most common error here was to use degrees instead of radians; so the more common answer was 207.3, it apparently not occurring to candidates that a time of over 3 minutes was rather long for a fairground car to reach a height of 5 m. Many did not set up a correct form of the equation and so could gain no marks Question 8: 6 3  3 
 
   
a)i ) AB   0  2    2  . AB  32  22  (1) 2  14
 3  4   1

  
Exam report 2 
 
Let ' s call u   1 the direction vector of the line. u  14
3 
 
3  2 

   
ii ) The scalar product AB  u   2    1  6  2  3  1
 1  3 
   

AB  u
1
1
1
Cos  

 .   Cos 1 ( )  85.9o
14
14  14 14
AB  u
b) For   2, r  3i  2 j  4k  2  2i  j  3k   7i  4 j  10k
The point D has coordinates (7,  4,10)
7 
3 
 
 
i ) A vector equation of the line l2 is : r   4     2 
10 
 
 
 1
ii ) AD 2  (7  3) 2  (4  2) 2  (10  4) 2  16  4  36  56
BC 2   7  3  6    4  2  0   10    3
2
2
2
 (1  3 )2  (2  4)2  (7   ) 2
expanding the brackets:
BC2  1  6  9  2  4  2  16  16  49  14   2
BC 2  14  2  24   66  56
14  2  24   10  0  7  2  12  5  0
5
(7   5)(   1)  0 so   1 or  
7
  1 will produce a parallelogram, not a trapezium
5
so we select  = and C has coordinates  9 17 ,  2 74 ,9 72 
7
The first part of this question was found to be fairly straight forward by most candidates, but then the second part defeated most; although some fully, or almost fully correct solutions were seen. Part (a)(i) This was answered well, although some candidates made arithmetic errors. A few lost a mark for giving their answer in coordinate form and a few found OA−OB. Part (a)(ii) Most candidates worked with the correct vectors, although some used the position vectors of A and/or B. The scalar product was usually found correctly and the formula for finding the cosine of the angle was well known. However, having set up the correct equation, some candidates equated the product of √14√14 to 2√14, so found an incorrect angle. Some made an error in finding a modulus. Part (b)(i) Almost all found the vector OD correctly. However very few candidates gained the mark for the equation of the line as most wrote l2 = instead of r = so they didn’t actually have a vector equation of the line. Part b)(ii) This problem required the candidates to both devise a strategy for its solution and carry it out. The key to the solution was realising that the moduli of AD and BC were equal, which could then be used directly or in setting up an equation involving equal angles. However, many candidates either took the vectors AD and BC to be equal, or that AD = −BC. They got a wrong result for the coordinates of C very quickly from this assumption. Those few candidates who got as far as finding p = 5/7 or 1 usually opted for the latter root (some stating μ had to be an integer) and apparently didn’t think about the geometry of the trapezium. Those who did have an appropriate strategy often made an error in setting up an equation and/or made an algebraic or numerical error in solving it. Component title Core 4 – Unit PC4 Max mark 75 A* 68 GRADE BOUNDARIES A B 61 54 C 47 D 41 E 35