Version 1.
General Certificate of Education (A-level)
June 2011
Mathematics MPC4
(Specification 6360)
Pure Core 4
Final
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
in this examination. The standardisation process ensures that the mark scheme covers the
candidates’ responses to questions and that every examiner understands and applies it in the same
correct way. As preparation for standardisation each examiner analyses a number of candidates’
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.
It must be stressed that a mark scheme is a working document, in many cases further developed and
expanded on the basis of candidates’ reactions to a particular paper. Assumptions about future mark
schemes on the basis of one year’s document should be avoided; whilst the guiding principles of
assessment remain constant, details will change, depending on the content of a particular examination
paper.
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Key to mark scheme abbreviations
M mark is for method
m or dM mark is dependent on one or more M marks and is for method
A mark is dependent on M or m marks and is for accuracy
B mark is independent of M or m marks and is for method and accuracy
E mark is for explanation
or ft or F follow through from previous incorrect result
CAO correct answer only
CSO correct solution only
AWFW anything which falls within
AWRT anything which rounds to
ACF any correct form
AG answer given
SC special case
OE or equivalent
A2,1 2 or 1 (or 0) accuracy marks
–x EE deduct x marks for each error
NMS no method shown
PI possibly implied
SCA substantially correct approach
c candidate
sf significant figure(s)
dp 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.
Q Solution Marks Total Comments
1(a) ( f ( −2 ) = ) 0 B1 1 ISW ( 0 seen is B1)
3
(b) 3 3 3 M1 Clear attempt at f ( 32 ) with 3
f = 4 − 13 + 6
2 2 2 terms
Factor theorem required;
NOT long division
27 3 Must see this, or equivalent
4× − 13 × + 6 or 13.5 − 19.5 + 6
8 2
= 0 ( 2 x - 3) is a factor A1 2 Shown = 0 and statement.
(c) Any appropriate method to find third factor M1 Full long division
Compare coefficients
Factor Theorem f ( 12 )
( x + 2 )( 2 x − 3)( 2 x − 1) A1 Or (2x 2
+ x − 6 ) ( 2 x − 1)
NMS M1A1
SC1 (2x + 1) or (1 – 2x) or
( x − 12 ) or ( 12 − x ) for third factor
2 x 2 + x − 6 = ( x + 2 )( 2 x − 3) M1 Factorise numerator correctly
or cancel 2 x 2 + x − 6
2x2 + x − 6 1
=
f ( x) 2x −1 A1 4 No ISW
7
Q Solution Marks Total Comments
2(a)(i) ( A =)80 B1 1 Ignore units
(ii) 2000 = A × k 25 M1 A or their value from (a)(i)
1
k = 25 25 or 25 25
or k = 100.04log 25 or e0.04ln 25 Correct expression for k,
k = 1.137411 AG A1 2 or 1.13741146….seen,
and correct answer to 6 d.p.
(b) 100000 Take logs correctly.
ln = t ln k M1
their A Condone miscopied k
ln1250 = t ln k or t = logk1250
t = 55.38 A1 Condone 55.3 or 55.4 PI
2016 A1 3
6
2(b) Alternative
By trial and improvement
1250 = k t M1 Attempt to calculate
k 55 and k 56 .
t = 56 or 55 < t < 56 A1
2016 A1 3
Q Solution Marks Total Comments
1
3 1
(1 − x )
3
(a)(i) = 1 − 13 x M1 Condone 13 + − 13 x for M1
= 1 − 13 x − 19 x 2 2 Must simplify coefficients
A1
including signs
1 1
1
1
27 3 May have 5 instead of 1253
(ii)
(125 − 27 x )
3
3 = 125 1 − x B1
125
1
27 3 1 27 1 27
2
M1 Attempt to replace x by ± 125
27
x
1 − x =
1 − × x − x condone missing brackets,
125 3 125 9 125
or start binomial again.
9 81 2 −9 −81 2
= 5− x− x A1 3 Condone 5 + x+ x
25 3125 25 3125
(b) 2
x= used in answer to (a)(ii) M1 Condone x = 276
9
or x = 0.222 or better
2
9 2 81 2
3
119 ≈ 5 − × −
25 9 3125 9
= 4.91872 A1 2 This answer only and must
follow from correct expansion
7
Alternative using ( a + bx )
3(a) n
(ii) Allow one error; condone
1
1 −
2
1
M1 missing brackets
(125 − 27 x ) = 125 + ×125 3 × ( −27 x )
3
3
3
5
1 21 −
+ − ×125 3 ( −27 x )
2
3 3 2
9 81 2
= 5− x− x A2 3
25 3125
Q Solution Marks Total Comments
4
dx dy
(a)(i)
= − 6sin 2θ , = − 2sin θ M1 ( ddθx = ) p sin 2θ or r sin θ cos θ
dθ dθ
( dy
dθ )
= q sin θ
A1 Both correct.
dy
dy −2sin θ M1 Use chain rule dθ
;
= dx
dx −6sin 2θ dθ
condone one slip
2sin θ 1 k = 6 must come from correct
= = A1 4
6 × 2sin θ cos θ 6 cos θ working seen AG
1
(ii) π 1 B1ft ft on k 1
θ= mT = k×2
3 3
k need not be numerical
mN = −3 B1ft ft on mT
( x, y ) = ( − 32 ,1) B1
3
B1 4 CAO; any correct form, ISW.
Normal y − 1 = −3 x +
2 2 y + 6x + 7 = 0
(b) sin 2 x = 1
2 (1 − cos 2 x ) M1 p + q cos 2 x ; Allow different
A1 letters for x or mixture eg θ even
for A1and the following A1ft
p dx = px q cos 2 x = 1
2 q sin 2 x A1ft Both integrals correct;
π ft on p and q
4
x 1
π sin
2
x dx = − sin 2 x
−
2 4
4
π 1 π 1 m1 Correct use of limits;
= − −− −−
8 4 8 4 F ( π4 ) − F ( − π4 ) or 2 F ( π4 )
π
F ( x ) = px + r sin 2 x and sin ,
2
π
sin − must be evaluated
2
correctly for m1
π 1
= − A1 CSO OE ISW
4 2 5
13
4 (b) Alternative
sin x dx = − sin x cos x − − cos x cos x dx
2
M1 Use parts; condone sign slips
= − sin x cos x + 1 − sin 2 x dx m1 Use cos 2 x = 1 − sin 2 x
2 sin 2 x dx = − sin x cos x + x A1
π π
4
2 sin 2 x dx = G − G −
π 4 4 m1 Correct use of limits
−
4
π
4
π 1
π sin
2
x dx = −
4 2 A1 5
−
4
Q Solution Marks Total Comments
5 (a) 4 5 −1
B1 (
± OA − OB )
AB = −1 − 1 = −2 Co-ordinate form only is B0
3 −2 5 Condone one component
incorrect
Line through A and B M1 OA + λ d or OB + λ d where
d = AB or BA all in
components and identified.
5 −1 4 −1 x
r = 1 + λ −2 or
r = −1 + λ −2
A1 3 OE r or y required
−2 5 3 5 z
Condone missing brackets on
OA or OB
(b)(i) 5 − λ = −8 + 5 μ Clear attempt to set up and solve
1 − 2λ = 5 M1 at least two simultaneous
equations in μ and a different
−2 + 5λ = −6 − 2 μ
parameter. Allow in column
vector form.
λ = −2 μ =3 A1 One of λ or μ correct OE
−2 + 5 × −2 = −12 − 6 − 2 × 3 = −12 Verify intersect, λ and μ correct
E1
Both equal −12 so intersect or verify ( 7,5, −12 ) is on both
lines; statement required
7
P is ( 7,5, −12 ) B1 4 CAO condone P = 5 OE
−12
and missing brackets
(ii) −8 + 5 μ 4
BC = 5 − −1 B1 BC = OC − OB
or
−6 − 2 μ 3 CB = OB − OC
3
6 •
BC =0 M1 Clear attempt at
−15 ± BP or ± AB or ± AP in
components sp with BC = 0
−36 + 15μ + 36 + 135 + 30μ = 0 m1 Linear equation in μ using their
BC and solved for μ .
Condone one arithmetical or
sign slip
μ = −3 A1
C is ( −23,5, 0 ) A1 5 CSO Condone column vector.
12
Q Solution Marks Total Comments
6
2 1
(a) (C =) or 2e −1 or 2 or 2 ( e −1 ) B1 1 One of these answers only.
e e Not 0.736 but allow ISW.
(b) d dy B1
(2y) = 2
dx dx
d 2x 2 dy M1 Product; 2 terms added, one
dx
( e y ) = 2e 2 x y 2 + e 2 x 2 y
dx dy
with ;
dx
A1 A1 for each term
A1
d 2 B1
dx
( x + C ) = 2x
Solve their equation correctly
dy M1
= dy
dx for
dx
x − e2 x y 2 Condone factor of 2 in both
A1 7
e2 x y + 1 numerator and denominator.
ISW
1
(c) dy 1 M1 Substitute x = 1 and y = into
Evaluate at 1, e
dx e dy
numerator of ; allow one slip
dx
numerator = 1 − e2 e −2 = 0 stationary point A1 2 Conclusion required; must score
full marks in part (b)
Allow 1 − 1 = 0 or 2 – 2 = 0
10
Q Solution Marks Total Comments
Q7 dA
(a) dt B1
= −k B1 2
(b)(i) A = − kt (+ C ) M1 Integrate
A1 C correct from A = ± kt + C
C = 4π × 602
4π × 302 = −9k + 4π × 602 m1 Use r = 30 t = 9 and attempt to
find k, as far as k = …
k = 1200π
A = −1200π t + 14400π
= 1200π (12 − t ) A1 4 AG CSO
(ii) t = 12 ( days ) B1 1
7
Q Solution Marks Total Comments
Q8 1 = A (1 − x ) + B (1 − x )( 3 − 2 x ) + C ( 3 − 2 x )
2
M1 Attempt to clear fractions
(a)
3
x =1 x= x=0
2
m1 Use any two (or three) values of
x to set up two (or three)
2
1 equations
C =1 1 = A − 1 = A + 3B + 3C
2
A=4 B = −2 C =1 A1 Two values correct
A1 4 All values correct
(b) 1 4 2 1
2 y
dy = − +
3 − 2 x 1 − x (1 − x )2
dx B1ft Separate using partial fractions;
correct notation; condone
missing integral signs
but dy and dx must be in correct
place.
ft on their A, B, C and on each
integral.
1 k
2 y
dy = y= B1 OE y
dy = 2k y is B1
− 2 ln ( 3 − 2 x ) B1ft
Condone missing brackets on
+ 2 ln (1 − x ) B1ft one ln integral.
1 B1ft Condone omission of +C
+ ( +C )
1− x
x = 0 y = 0 0 = −2 ln 3 + 0 + 1 + C M1 Use ( 0, 0 ) to find C. Must get to
C = …..
C = 2 ln 3 − 1 A1 Correct C found from correct
equation. C must be exact, in
any form but not decimal.
3 − 3x 1 m1 Correct use of rules of logs to
y = 2 ln + −1 progress towards requested form
3 − 2x 1 − x
of answer . C must be of the
form r lns + t
1
3 − 3x x
y 2 = 2 ln + A1 9 OE
3 − 2x 1− x
CSO condone B0 for separation
13
TOTAL 75
Q8 Alternative
1 = A (1 − x ) + B (1 − x )( 3 − 2 x ) + C ( 3 − 2 x )
2
(a) M1
1 = A + 3B + 3C Set up three simultaneous
m1
0 = −2 A − 5 B − 2C equations
0 = A + 2B
A1 Two values correct
A=4 B = −2 C =1 A1 4 All values correct
