Aqa mpc1-w-ms-jan13

Version
General Certificate of Education (A-level)
January 2013
Mathematics
(Specification 6360)
MPC1
Pure Core 1
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 students’ 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 students’ 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 students’ 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.
Further copies of this Mark Scheme are available from: aqa.org.uk
Copyright © 2013 AQA and its licensors. All rights reserved.
Copyright
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Registered address: AQA, Devas Street, Manchester M15 6EX.

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.

MPC1 - AQA GCE Mark Scheme 2013 January
MPC1
Q Solution Marks Total Comments
1(a) (i) 21 5 1k  condone 3 7 5 1k  
4k   B1 1 AG condone 4y  
(ii) (x =) 2 B1
(y =) –1 B1 2 midpoint coords are (2, –1)
(b) 1 3
–
5 5
y x M1 obtaining
3
5
y a x 
or
y
x


=
4 2
7 3
 
 
or
1 2
2 3
 
 
or
4 1
7 2
  

condone one sign error in expression
(Gradient AB =)
3
5

A1 2
allow – 0.6 ,
6
10
etc for A1 & condone
error in rearranging if gradient is correct .
(c) Perp grad =
5
3
M1 –1/ “their” grad AB
 
5
– 2 3
3
y x 
or
5
, 7
3
y x c c   etc
A1
correct equation in any form
(must simplify x – –3 to x+3
or c to a single term equivalent to 7)
5x –3y + 21 = 0 A1 3
or any multiple of this with integer
coefficients –terms in any order but all
terms on one side of equation
(d) 3x + 5y = 1 and 5x + 8y = 4 must use correct pair of equations and
P x Q  or R y = S M1 attempt to eliminate y (or x) (generous)
12x  A1
7y   A1 3 (12, –7)
Total 11

MPC1 (cont)
2(a)
3
d 4
– 2
d 8
y t
t
t
 
 
 
M1
A1 2
one of these terms correct
all correct (no + c etc)
(b)(i)
d 4
1 – 2
d 8
y
t
t
   M1 Correctly sub t = 1 into their
d
d
y
t
1
2
= 1 A1cso 2 must have
d
d
y
t
correct ( watch for t3
etc)
(ii)
d
0
d
y
t
 must have used
d
d
y
t
in part (b)(i)
 (height is) decreasing (when t = 1) E1 1
must state that “
d
0
d
y
t
 ” or “–1.5 < 0”
or the equivalent in words
FT their value of
d
d
y
t
with appropriate
explanation and conclusion
(c)(i)
2
2
2
d 4
3 2
d 8
y
t
t
 
   
 
M1
Correctly differentiating their
d
d
y
t
even if wrongly simplified
2
2
d
2, 4
d
y
t
t
 
  
 
A1cso 2
Both derivatives correct and simplified to
4
(ii)  minimum E1 1
FT their numerical value of
2
2
d
d
y
t
from
part (c) (i)
Total 8

MPC1 (cont)
3(a)(i) 18 3 2 B1 1 Condone k = 3
(ii) 2 2
3 2 4 2
M1
attempt to write each term in form 2n
with at least 2 terms correct
A1 correct unsimplified
=
2
7
A1 3
or
2
2
 M1
integer terms
4
6 8


A1
=
2
7
A1
(b)
7 2 – 3 2 2 3
2 2 3 2 2 3


 
M1
(numerator =)
14 2 2 6 7 6 3   
m1
correct unsimplified but must simplify
 
2
2 ,  
2
3 and 2 3 correctly
(denominator = 8 – 3 =) 5 B1 must be seen or identified as denominator
giving
25 5 6
5

(Answer =) 5 6 A1cso 4 m = 5, n = 6
Total 8

MPC1 (cont)
4(a)(i) 2
( 3)x  M1 or 3p  seen
2
( 3)x  + 2 A1 2
(ii) 2
( 3)x  = –2 M1 FT their positive value of q
not use of discriminant
No (real) square root of –2 therefore
equation has no real solutions
A1cso 2
for graphical approach see below to see if
SC1 can be awarded
(b)(i) x = ‘their’ p or y = ‘their’ q M1 or 3x  found using calculus
Vertex is at (3, 2) A1cao 2
(ii) B1
y intercept = 11 stated or marked on y-
axis (as y intercept of any graph)
M1  shape (generous)
A1 3 above x-axis , vertex in first quadrant
crossing y-axis into second quadrant
(iii) Translation E1 and no other transformation
through
–3
2
 
  
M1 FT negative of BOTH ‘their’ vertex
coords
A1 3
both components correct for A1; may
describe in words or use a column vector
Total 12
y
x
11

MPC1 (cont)
5(a)   3 2
p 1 ( 1) 4 ( 1) 3( 1) 18         M1 p(–1) attempted not long division
( 1 4 3 18)     = 16 A1 2
(b)(i)   3 2
p 3 3 – 4 3 3 3 18     M1 p(3) attempted not long division
 p 3 27 – 36 9 18 0 3 is a factorx      A1 2 shown = 0 plus statement
(ii) Quadratic factor 2
( )x x c  or 2
( 6)x b x  M1 –x or –6 term by inspection
or full long division by 3x 
or comparing coefficients
or p(–2) attempted
Quadratic factor  2
– 6x x  A1
correct quadratic factor (or x+2 shown
to be factor by Factor Theorem)
    p 3 3 ( 2)x x x x      A1 3 or    
2
p 3 ( 2)x x x    
must see product of factors
(c)
M1
cubic curve with one maximum and
one minimum
A1
meeting x-axis at –2 and touching
x-axis at 3
Final A1 is dependent on previous A1 and can
be withheld if curve has very poor curvature
beyond x = 3, V shape at x = 3 etc
A1 3
graph as shown , going beyond
2x   but condone max on or to
right of y-axis
Total 10
3–2
y
x

MPC1 (cont)
6(a) (Gradient = 10 – 6 + 5) = 9 B1 correct gradient from sub x =1 into
d
d
y
x
4 " 9" ( 1)y their x  
or " 9"y their x c  and attempt
to find c using x =1 and y = 4 M1
must attempt to use given expression for
d
d
y
x
and must be attempting tangent
and not normal and coordinates must be
correct
9 5y x  A1 3 condone 9 ,... 5y x c c   
(b)   5 310 6
5
5 3
y x x x C    M1 one term correct
A1
A1
another term correct
all integration correct including + C
4 = 2 – 2 + 5 + C
1C   m1
substituting both x =1 and y = 4
and attempting to find C
5 3
2 2 5 1y x x x    A1cso 5
must have y = ... and coefficients
simplified
Total 8

MPC1 (cont)
7(a) 2
0 4 12 ( 0)x y y     M1 sub x = 0 & correct quadratic in y
or 2
( 2) 16y   or 2
( 2) 16 0y   
( 6)( 2) ( 0)y y   A1 correct factors
or formula as far as
4 64
2

or 2 16y   
– 2, 6y  A1 3 condone (0, –2) & (0, 6)
(b)
2 2
( 3) 9 ( 2) 4 ( 12)x y      M1
correct sum of square terms and attempt
to complete squares ( or multiply out)
PI by next line
 2
9 4 12r    A1  2
25r  seen on RHS
  5r  A1 3 25r  or 5r   scores A0
(c)(i)  2 2 2
(2 3) (5 2)CP      M1 condone one sign slip within one bracket
  34CP  A1 2 34n 
(ii) 2 2 2
34 25PQ CP r    M1
Pythagoras used correctly with values
FT “their” r and CP
  3PQ  A1 2
Total 10

MPC1 (cont)
8(a) 2
2 1 2 3x x kx k    equated and multiplied out
2
2 1 2 3 0 OEx x kx k     and all 5 terms on one side and = 0
2
2 (2 1) 3 1 0x k x k      B1 1
AG
(correct with no trailing = signs etc)
(b)(i) 2
(2 1) 4 2(3 1)k k    M1 clear attempt at 2
4b ac
2
(2 1) 4 2(3 1)k k    > 0 B1 discriminant > 0 which must
appear before the printed answer
2
4 4 1 24 8k k k    > 0
2
4 20 9 0k k    A1cso 3 AG (all working correct with no
missing brackets etc)
(ii) 2
4 20 9 (2 9)(2 1)k k k k     M1 correct factors or correct use of
formula as far as
20 256
8

critical values are 1 9and
2 2
A1
condone
4 36
,
8 8
etc here but must
combine sums of fractions
M1 sketch or sign diagram including
values
1 9,
2 2
k k 
Take their final line as their answer
A1 4 fractions must be simplified
condone use of OR but not AND
Total 8
TOTAL 75
+ – +
0.5
k

Aqa mpc1-w-ms-jan13

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