Version 1.0
General Certificate of Education (A-level)
January 2011
Mathematics
MPC4
(Specification 6360)
Pure Core 4
Mark Scheme
Mark schemes are prepared by the Principal Examiner and considered, together with the relevant
questions, by a panel of subject teachers. This mark scheme includes any amendments made at the
standardisation events which all examiners participate in and is the scheme which was used by them
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scripts: alternative answers not already covered by the mark scheme are discussed and legislated for.
If, after the standardisation process, examiners encounter unusual answers which have not been
raised they are required to refer these to the Principal Examiner.
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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