A-LEVEL

Mathematics
Pure Core 4 – MPC4
Mark scheme

6360
June 2015

Version 1.1: Final

Mark schemes are prepared by the Lead Assessment Writer 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 associates 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 associate understands and applies it in the same
correct way. As preparation for standardisation each associate 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, associates encounter unusual answers which have not been
raised they are required to refer these to the Lead Assessment Writer.

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

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 MARK SCHEME – A-LEVEL MATHEMATICS – MPC4 – JUNE 2015

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.

Page 3 of 12
 MARK SCHEME – A-LEVEL MATHEMATICS – MPC4 – JUNE 2015

Q1 Solution Mark Total Comment

(a) 19 x  2  A 1  6 x   B  5  x  M1 Correct equation and attempt to find a

value for A or B.

A3 A1
B  1 A1
NMS or cover up rule; A or B correct SC2
3 A and B correct SC3.

(b) 3 1
 5  x  1 6x dx
Condone missing brackets
 p ln  5  x   q ln 1  6 x  M1 OE Either term in a correct form
 3ln  5  x  A1ft ft on their A
 16 ln 1  6x  A1ft ft on their B
4 Substitute limits correctly in their integral;
  3ln1 
0
1
6 ln 25  3ln 5  ln1
1
6
m1 F(4)  F(0)

  16 ln 25  3ln 5 ACF. ln1  0 PI

A1
8 CSO
 ln 5 Condone equivalent fractions or recurring
3
A1 6 decimal

Total 9

Q2 Solution Mark Total Comment

(a) R  29 B1 Allow 5.4 or better
Their 29
29 cos   2, 29 sin   5 or tan   52 M1
Note cos   2 or sin   5 is M0
  1.19 A1 3 Must be exactly this

(b)(i) R cos  x     R or cos  x     1

Candidate’s R and 
or x    2 or x    0 or x   M1
 x  5.09 A1 Must be exactly this
2

(ii) 1
cos  x      M1 Candidate’s R and  ; PI
R
Rounded or truncated to at least 2 dp;
 x     1.75757... and 4.52560... A1
Ignore ‘extra’ solutions

Condone x  0.568 ;
x  0.567 and x  3.34 A1 3 x  3.34 must be correct
NMS is 0/3
A0 if extra values in interval 0  x  2
Total 8

Page 4 of 12
 MARK SCHEME – A-LEVEL MATHEMATICS – MPC4 – JUNE 2015

Q3 Solution Mark Total Comment

(a) f   12   1  3  1  d  2 Attempt to evaluate f   12  and equated

M1
to 2
d 1 A1 2 NMS is 0/2

(b)(i)  2 x  1 is a factor B1 OE  x  12 

g( x)   2 x  1  4 x 2  bx  3
Attempt to find quadratic factor or a
M1
second linear factor using Factor Theorem

g( x)   2 x  1  4 x 2  8 x  3 OE if  x  12  is used
g( x)   2 x  1 2 x  1 2 x  3 OE; must be a product

NMS : SC3 if product is correct

A1 3 SC1 if one or two factors are correct

(ii) 4x2 1 1
 B1
g( x) 2x  3
d  1  k
  M1 Attempt to differentiate simplified h
dx  2 x  3   2 x  3 2
2
 A1 Correct derivative
 2 x  3
2

(Derivative is) negative, or  0 Explanation and conclusion required

E1
hence decreasing 4 Derivative must be correct

Total 9

(b)(ii) Special case

h  x   2 x13 B1

2 x  3 is an increasing function, so 1

Award only if h  x  
2 x3
1
is a decreasing function E1 2 2 x 3 is correct

Page 5 of 12
 MARK SCHEME – A-LEVEL MATHEMATICS – MPC4 – JUNE 2015

Q4 Solution Mark Total Comment

(a) 1
1   5 x  kx 2 M1 k any non-zero numerical expression
5
1  x  2x2 A1 2 Simplified to this

(b) (i) 
2
2 
2 2 1
8  3x  1  x  
 
8
3
3 3 3 3 B1 ACF for 8
8
4
2

 3  3
1  x 
 8 
2 Expand correctly using their 83 x
 2   3  1  2   5  3 
 1      x         x  M1 Condone poor use of or missing brackets
 3   8  2  3   3  8 
1 1 5 2 1 1 5 2
 x x A1 3 Accept  1  x  x 
4 16 256 4 4 64 

(ii) 1
x M1 x  13 used in their expansion from (b)(i)
3

0.2313 (4dp) A1 2 Note 3 in 4th decimal place

Total 7

Page 6 of 12
 MARK SCHEME – A-LEVEL MATHEMATICS – MPC4 – JUNE 2015

Q5 Solution Mark Total Comment

(a)  dx   dy 
    2sin 2t    cos t B1 Both correct
 dt   dt 
 dy  cos t Correct use of chain rule with their
  M1
 dx  2sin 2t derivatives of form a sin 2t , b cos t
 1
At t  gradient mT   A1 3
6 2

(b) Gradient of normal mN  2 B1ft ft gradient of tangent; mN  1

mT

For mN , allow their mT with a change of

  2      
 y  cos  6    mN  x  sin  6   M1 sign or the reciprocal at
       sin 6 ,cos 26  or  12 , 12 
1
y  2x  A1 3 Must be in this y  mx  c form
2
Alternative for M1
   2  Use y  mx  c to find c with their
sin    2 cos  c
6  6  gradient mN at  sin 6 ,cos 26  or  12 , 12 
(c) cos 2q  1  2sin 2 q B1 Seen or used in this form
Use parametric equations and candidate’s
sin q  2 1  2sin 2 q  
1
M1 cos 2q in the form 1  k sin 2 q
2
Collect like terms; must be a quadratic
8sin 2 q  2sin q  3  0 OE A1
equation
 1 3 Must come from a correct quadratic
 sin q   sin q   A1 equation with the previous 3 marks
 2 4 awarded
1
 x  
8 A1 5 Previous 4 marks must have been awarded

Total 11

Page 7 of 12
 MARK SCHEME – A-LEVEL MATHEMATICS – MPC4 – JUNE 2015

Mark scheme Alternative

Q5 Solution Mark Total Comment

(a) dy dx Find a correct Cartesian equation and

x  1  2 y2 1  4 y or  4 y B1
dx dy differentiate implicitly correctly
dy 1
 M1 Use y  sin 6 or y  12 in their
dy
; PI
dx 4sin 6 dx

 1
At t  gradient mT   A1 3 CSO
6 2

(b) Gradient of normal  2 B1ft ft gradient of tangent, mN  1

mT

For mN , allow their mT with a change of

  2     
 y  cos  6    mN  x  sin    M1 sign or the reciprocal at
     6   sin 6 ,cos 26  or  12 , 12 
1
y  2x  A1 3 CSO
2
Alternative for M1
   2  Use y  mx  c to find c with candidate’s
sin    2 cos  c
6  6  gradient mN at  sin 6 ,cos 26  or  12 , 12 
x  1 2 y2 PI by x  1  2  2 x  12 
2
(c) B1
y  12 Use their Cartesian equation and normal
1  2 y2  M1
2 to eliminate x
4y  y  2  0 
2 3

Collect like terms; must be a quadratic

8sin 2 q  2sin q  3  0 A1 equation
 1 3 Must come from a correct quadratic
 sin q   sin q   A1 equation with the previous 3 marks
 2 4
awarded
1
 x   A1
8 5 Previous 4 marks must have been awarded

Total 11

Page 8 of 12
 MARK SCHEME – A-LEVEL MATHEMATICS – MPC4 – JUNE 2015

Q6 Solution Mark Total Comment

(a)  2  2 
 
 AB    4 
  B1
Or  BA    
 4
 6   6 
3
  
AB  1  (2  3)  ( 4  1)  (6  2) 
  M1 Correctly ft on “their” AB
 2 
56 14 cos BAC  14 m1
Correct use of formula with consistent
vectors; ACF
angle BAC  60 A1 4 or  / 3 ; NMS 60 scores 0/4

(b)  3  3  5 
  
 
BC   2     1   2  B1

 BC ACF
10   2   4 
  
AB  BC  Correct scalar product with their AB ,
M1 
2  3  2   4    4   6  2  6   0 their BC , equate to 0 and solve for 
14  56  0    4 A1
C is at 15, 6, 2  A1 4
Accept as a column vector
NMS (15,6,2) scores 0/4

(c) E1 is at 11,0,0  B1 Accept as a column vector

15  2   17 
        
OD  OC  AB   6    4    2  B1
 2   6   4 
 17   3
   Correct vector expression with their  and
OE 2   2   2  4  1
1
M1 
their OD
 4   2 
E2 is at  23, 4, 8 A1 4 Accept as a column vector
Total 12

(b) Alternative by Pythagoras

 3  3  5 
  
 
BC   2     1   2  B1

 BC ACF
10   2   4 
 3        2  AC 2  AB 2  BC 2 Correct Pythagoras
2 2 2

M1 expression, attempt to expand and solve

 56   2  3    4      6  2 
2 2 2
for 

112  28  0   4 A1
C is at 15, 6, 2  A1 4 Accept as a column vector

Page 9 of 12
 MARK SCHEME – A-LEVEL MATHEMATICS – MPC4 – JUNE 2015

(b) Alternative by cos 60  1

2

1 AB 56
   B1
2 AC  3        2 
2 2 2

1 56
 M1 Square and simplify
4 14 2
 2  16    4  or   4  A1
C is at 15, 6, 2  A1 4 Accept as a column vector

(c) Alternatives
Alt (i)
 5  12 
     1  
OE1  OB  12 AC   2    4 
2
 4   8
E1 is at 11, 0, 0  B1
 5  6 
     M1 Correct vector expression with their BE1
OE2  OB  3BE1   2   3  2 
 4   4  B1` All correct

E2 is at  23, 4, 8 A1 4

Alt (ii)
 5  12 
      
OD  OB  AC   2    4 
 4   8
D is at 17, 2, 4  B1
 17   12 
  1    1   Correct vector expression with their
OE2  OD  AC   2    4  M1  
2 2 OD and their AC
 4   8
E2 is at  23, 4, 8 A1
 5  12 
     1  
OE1  OB  12 AC   2    4 
2
 4   8
E1 is at 11, 0, 0  B1 4

Page 10 of 12
 MARK SCHEME – A-LEVEL MATHEMATICS – MPC4 – JUNE 2015

Q7 Solution Mark Total Comment

(a) 3 3
1 1 1 1 1
k     2e 3ln 2   ln 2 Clear use of    and e 3ln 2 
2 2
B1 1 2 8 8
1 1 1 2
   ln 2   ln 2 Accept  ln 2
8 8 4 8

(b) dy
3 y2 B1
dx
dy
pye3 x  qe3 x M1
dx

dy A1
6 ye3 x  2e 3 x
dx

1  0 B1 Both required
1 and no other terms

3 dy 1 1 1 dy Substitute
 6   2 1   0 m1 x  ln 2 or e3 x  18 and y  1
into their
4 dx 8 2 8 dx 2
expression
dy 11
 or 1.375 A1 6
dx 8

Total 7

Page 11 of 12
 MARK SCHEME – A-LEVEL MATHEMATICS – MPC4 – JUNE 2015

Q8 Solution Mark Total Comment

(a)(i) 1 1
 4  5x
dx  
5 1  t 
2
dt B1
Correct separation and notation seen on a
single line somewhere in their solution
1  1 
a  4  5 x  2 or b 1  t  OE a 4  5 x or b 
1
M1 
 1 t 
1
2 2
 4  5x  2 A1 OE 4  5x
5 5
1 1
 (1  t ) 1  C  OE 
5 1  t 
5 A1

x0,t0  C 1 m1 Use  0, 0  to find a constant

1
2 1
 4  5 x  2  1  1  t 
1
A1 OE
5 5
2
1  4  5t  4
2
5  1  t   4
1

x  1    A1 ACF eg x    
4  5  5 20  1  t  5
 7

(b)(i) dr
B1 Seen; allow R for r
dt
1 1
M1 seen ; allow R for r
r2 r2
dr k c
 Any constant k including but not
dt r 2 A1 3 
including variable t
Must use R or r consistently

(ii)  dr  k c dr
 4.5 with r  1 to find a value
   4.5  2 or 4.5  Use
 dt  1  12 M1 dt
for the constant
4.5
0.5  2  r  3 (metres) A1 2
r

Total 12

Page 12 of 12