Pure Mathematics P3 Mark scheme

Question Scheme Marks

1 9x2 – 4 = (3x − 2) (3x + 2 ) at any stage B1
Eliminating the common factor of (3x + 2) at any stage
2 (3x  2) 2 M1

(3x  2) (3x  2) 3x  2
Use of a common denominator
2(3x  2)(3x  1) 2(9 x 2  4) 2(3x  1) 2(3x  2) M1
2
 2
or 
(9 x  4)(3x  1) (9 x  4)(3x  1) (3x  2)(3x  1) (3x  1)(3x  2)
6 6
or 2 A1
(3x  2)(3x  1) 9 x  3x  2
(4)
(4 marks)
Notes:
B1: For factorising 9 x2  4 (3x  2)(3x  2) using difference of two squares. It can be
awarded at any stage of the answer but it must be scored on E pen as the first mark.
B1: For eliminating/cancelling out a factor of (3x+2) at any stage of the answer.
M1: For combining two fractions to form a single fraction with a common denominator. Allow
slips on the numerator but at least one must have been adapted. Condone invisible brackets.
Accept two separate fractions with the same denominator as shown in the mark scheme.
Amongst possible (incorrect) options scoring method marks are
2(3x  2) 2(9 x 2  4)
 . Only one numerator adapted, separate fractions
(9 x 2  4)(3x  1) (9 x 2  4)(3x  1)
2  3x  1  2  3x  2
Invisible brackets, single fraction.
(3x  2)(3x  1)
6
A1:
(3x  2)(3x  1)
This is not a given answer so you can allow recovery from ‘invisible’ brackets.

Alternative
2(3x  2) 2 2(3x  2)(3x  1)  2(9 x 2  4) 18x  12 has scored 0,0,1,0 so far
 2
 
2 2
(9 x  4) (3x  1) (9 x  4)(3x  1) (9 x  4)(3x  1)

6(3x  2)
 is now 1,1,1,0
(3x  2)(3x  2)(3x  1)
6
 and now 1,1,1,1
(3x  2)(3x  1)

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 Question Scheme Marks
2(a) x3 + 3x2 + 4x – 12 = 0  x3 + 3x2 = 12 – 4x
 x2 (x + 3) = 12 – 4x M1
12  4 x dM1
 x2 
( x  3)
4(3  x)
x A1*
( x  3)
(3)
(b)  4(3  1)  M1 A1
x1 
  1.41
 (3  1) 
awrt x2 1.20
  x3 1.31 A1
(3)
(c) Attempts f (1.2725) 
()0.00827... f (1.2715) 
0.00821.... M1
Values correct with reason (change of sign with f(x) continuous) and A1
conclusion ( α = 1.272)
(2)
(8 marks)
Notes:
(a)
M1: Moves from f(x) = 0, which may be implied by subsequent working, to x2 ( x  3)  12  4 x
by separating terms and factorising in either order. No need to factorise rhs for this mark.
dM1: Divides by ‘(x+3)’ term to make x2 the subject, then takes square root. No need for rhs to
be factorised at this stage.
A1*: CSO. This is a given solution. Do not allow sloppy algebra or notation with root on just
numerator for instance. The 12−4x needs to have been factorised.
(b)
M1: An attempt to substitute x0  1 into the iterative formula to calculate x1 . This can be awarded for
4(3  1) 8
the sight of , , 2 and even 1.4
(3  1) 4
A1: x1  1.41 . The subscript is not important. Mark as the first value found, 2 is A0

A1: x2 awrt
 1.20 x3 awrt 1.31 . Mark as the second and third values found. Condone 1.2
for x2
(c)
M1: Calculates f(1.2715) and f(1.2725), or the tighter interval with at least 1 correct to 1 sig fig
rounded or truncated. Accept f(1.2715) = -0.008 1sf rounded or truncated. Also accept
f(1.2715) = -0.01 2dp. Accept f(1.2725) = (+) 0.008 1sf rounded or truncated. Also accept
f(1.2725) = (+)0.01 2dp
A1: Both values correct (see above),
A valid reason; Accept change of sign, or >0 <0, or f(1.2715) ×f(1.2725)<0
And a (minimal) conclusion; Accept hence root or α=1.272 or QED or

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 Question Scheme Marks
3(a) 1 M1
Uses −2(3 − x) + 5 = x + 30
2
Attempts to solve by multiplying out bracket, collect terms etc. M1
3
x = 31
2
62 A1
x only
3
(3)
(b) Makes the connection that there must be two intersections. M1
Implied by either end point k  5 or k 11

5 < k  11 A1

(2)
(5 marks)
Notes:
(a)
1
M1: Deduces that the solution to f (
x) x  30 can be found by solving
2
1
2(3  x)  5 x  30
2
M1: Correct method used to solve their equation. Multiplies out bracket/ collects like terms.
62
A1: x only. Do not allow 20.6
3
(b)
M1: Deduces that two distinct roots occurs when y  k intersects y  f ( x) in two places. This
may be implied by the sight of either end point. Score for sight of either k  5 or k 11
A1: Correct solution only k : k  , 5  k 11

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 Question Scheme Marks
4(i) 1 1 M1 A1
 
 2 x  1
dx
2
ln(2 x  1)

13
1 1 1 1  25  dM1
 5  2 x  1
dx  ln 25  ln 9  ln  
2 2 2  9 
5 A1
 ln  
3
(4)
(ii) 1 M1
Integrates to give  cos2 x   sec x c where   0,   0
3
 1 1 
  2 cos 2 x  3sec 3 x c
 
 1  π  1 π   1  Substitutes limits of 0 dM1
  cos  2    3sec        cos  0   3sec  0   and 𝜋𝜋 and subtracts the
 2  2  3 2   2  2
correct way around
 2 32 A1
(3)
(7 marks)
Notes:
(i)
1
M1: For
 
 2 x  1
dx k ln(2 x  1) where k is a constant.

1 1
A1: Correct integration
 
 2 x  1
dx
2
ln(2 x  1)

dM1: Scored for substituting in the limits, subtracting and using correctly at least one log law.
 25 
You may see the subtraction law k ln 25  k ln 9  k ln   or the index law
 9 
1 1
ln 25  ln 9 ln 5  ln 3
2 2
5
A1: cao ln  
3
(ii)
1
M1: Integrates to a form  cos2 x   sec x c where   0,   0
3
𝜋𝜋
dM1: Dependent upon the previous M1. It is scored for substituting limits of 0 and 2 and
subtracting the correct way around.
A1: cao 2 3  2

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 Question Scheme Marks
5 5 x 2  10 x  9
y
( x  1)2
Differentiates numerator to 10x −10 and denominator to 2(x − 1) o.e. B1

Uses the quotient rule M1 A1


2 2

dy ( x  1) 10 x  10   5 x  10 x  9 2  x  1 
dx ( x  1)4

Takes out a common factor from the numerator and cancels M1

 
dy  x  1 ( x  1) 10 x  10   5 x  10 x  9 2

2

dx ( x  1) 4 3

Simplifies the numerator by multiplying and collecting terms M1

dy


10 x 2  20 x  10  10 x 2  20 x  18 
dx ( x  1)3

dy 8 A1

dx ( x  1)3

(6)
(6 marks)
Notes:
B1: See scheme.

dy ( x  1)
2
 Ax  B    5 x2  10 x  9   Cx  D 
M1: Uses the quotient rule to reach a form  o.e.
dx ( x  1)4
Alternatively uses the product rule to reach a for
dy
dx  
 ( x  1)2  Ax  B   5 x 2  10 x  9 C  x  1
3

dy
A1: Fully correct If the product rule is used
dx
dy
dx  
( x  1)2 10 x  10   5 x 2  10 x  9 2  x  1
3

dy g( x)
M1: This is for using a correct method to reach a form  . See scheme when using the
dx ( x  1)3
quotient rule. If the product rule is used it is for combining the terms using a common
denominator.
M1: Scored for simplifying the numerator (By multiplying out and collecting terms).
dy 8
A1: 
dx ( x  1)3

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 Question Scheme Marks
6(a) f(x) > 2 B1
(1)
(b) fg( x) eln x  2 , x  2 M1 A1
(2)
(c) e 2 x3
26e 2 x3
4 M1 A1
 2x  3 ln 4
ln 4  3 3
x or ln 2  M1 A1
2 2
(4)
(d) Let y  e x  2  y  2  e x  ln( y  2)  x M1
A1
f 1 (
x) ln( x  2), x > 2
B1ft
(3)
(e) Shape for f(x) B1

(0, 3) B1

Shape for f -1(x) B1

(3, 0) B1
(4)
(14 marks)
Notes:
(a)
B1: Range of f(x)>2. Accept y>2, (2,∞), f>2, as well as ‘range is the set of numbers bigger
than 2’ but don’t accept x >2
(b)
M1: For applying the correct order of operations. Look for eln x  2 . Note that ln e x  2 is M0
A1: Simplifies eln x  2 to x  2 . Just the answer is acceptable for both marks.
(c)
M1: Starts with e2 x3  2 6 and proceeds to e2 x3  ...
A1: e2 x3  4
M1: Takes ln’s both sides, 2 x  3 ln.. and proceeds to x=….

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 Question 6 notes continued
ln 4  3 3
A1: x oe. eg ln2 - Remember to isw any incorrect working after a correct
2 2
answer.
(d)
M1: Starts with y e x  2 or 
x e y  2 and attempts to change the subject. All ln work must be
correct. The 2 must be dealt with first. Eg. y  e x  2  ln y  x  ln 2  x  ln y  ln 2 is
M0.
A1: f 1 ( x) 
ln( x  2) or y= ln( x  2) or y= ln x  2 There must be some form of bracket.
B1ft: Either x >2, or follow through on their answer to part (a), provided that it wasn’t y 
Do not accept y>2 or f-1(x)>2.
(e)
B1: Shape for y=ex. The graph should only lie in quadrants 1 and 2. It should start out with a
gradient that is approx. 0 above the x axis in quadrant 2 and increase in gradient as it moves
into quadrant 1. You should not see a minimum point on the graph.
B1: (0, 3) lies on the curve. Accept 3 written on the y axis as long as the point lies on the curve.
B1: Shape for y= ln x. The graph should only lie in quadrants 4 and 1. It should start out with
gradient that is approx. infinite to the right of the y axis in quadrant 4 and decrease in
gradient as it moves into quadrant 1. You should not see a maximum point. Also with
hold this mark if it intersects y=ex.
B1: (3, 0) lies on the curve. Accept 3 written on the x axis as long as the point lies on the curve.

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 Question Scheme Marks
7(a) p = 4π2 or (2π)2 B1
(1)
(b) dx M1 A1
x = (4y – sin 2y)2  = 2(4y – sin 2y) (4 – 2cos 2y)
dy
π dx dy 1 M1
Sub y 24π (= 75.4) OR (= 0.013)
2 dy dx 24π
π 1 M1
Equation of tangent y x 4π 2
2 24π
π 1 π M1 A1
Using y x 4π 2 with x 0 y cso
2 24π 3
(6)
Alternative I for first two marks
2
x  4 y  sin 2 y   x0.5 4 y  sin 2 y
dx M1A1
 0.5 x 0.5 4  2cos 2 y

dy
Alternative II for first two marks
x 
16 y 2  8 y sin 2 y  sin 2 2 y 
dy dy dy dy M1A1
 1 32 y
  8sin 2 y 16 y cos 2 y  4sin 2 y cos 2 y
dx dx dx dx
Or 1dx 
32 y dy  8sin 2 y dy 16 y cos 2 y dy  4sin 2 y cos 2 y dy

(7 marks)
Notes:
(a)
2
B1: B1: p 4π 2 or exact equivalent 2π . Also allow x 4π 2
(b)
M1: Uses the chain rule of differentiation to get a form
A(4 y  sin 2 y)  B  C cos 2 y  , A, B, C  0 on the right hand side.
Alternatively attempts to expand and then differentiate using product rule and chain rule to
  dx
a form x  16 y 2  8 y sin 2 y  sin 2 2 y   Py  Q sin 2 y  Ry cos 2 y  S sin 2 y cos 2 y P, Q, R, S  0
dy
A second method is to take the square root first. To score the method look for a
differentiated expression of the form Px0.5... 4  Q cos 2 y
A third method is to multiply out and use implicit differentiation. Look for the correct
terms, condoning errors on just the constants.

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 Question 7 notes continued

dx dy 1
A1: 2  4 y  sin 2 y  4  2cos 2 y  or
  with both sides
dy dx 2  4 y  sin 2 y  4  2cos 2 y 
correct. The lhs may be seen elsewhere if clearly linked to the rhs. In the alternative
dx
32 y  8sin 2 y  16 y cos 2 y  4sin 2 y cos 2 y

dy

M1: M1: Sub y  into their dx or inverted dx . Evidence could be minimal, eg y   dx ...
2 dy dy 2 dy
2
x  4 y  sin 2 y 
It is not dependent upon the previous M1 but it must be a changed
M1:
  
M1: Score for a correct method for finding the equation of the tangent at  ' 4 ',  .
2

 2
π 1
Allow for y x their 4π 2
2 their numerical dx
dy
π
Allow for y their numerical dx x their 4π 2
2 dy
π 1
Even allow for y x p
2 their numerical dx
dy

1  
c as long as  ' 4 ',
2
It is possible to score this by stating the equation y is
2 
x
24π 
used in a subsequent line.
M1: y mx  c and stating the value of 'c'
Score for writing their equation in the form 
π 1 y.
or setting x  0 in their y x 4π 2 and solving for
2 24π
Alternatively using the gradient of the line segment AP = gradient of tangent.

y 1
Look for 2 2   y .. Such a method scores the previous M mark as well.
4 24
At this stage all of the constants must be numerical. It is not dependent and it is possible to
score this using the ''incorrect'' gradient.
π  
A1: cso y . You do not have to see  0, 
3  3

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 Question Scheme Marks
8(a) aT b  log10 N 
N log10 a  log10 T b M1
 log10 N log10 a  b log10 T so m = b and c  log10 a
A1

(2)
(b) Uses the graph to find either a or b a  10intercept or b = gradient M1
Uses the graph to find both a and b a  10intercept and b = gradient M1
Uses T  3 in N  aT with their a and b
b
M1
Number of microbes  800 A1
(4)
(c) States that 'a' is the number of microbes 1 day after the start of the experiment. B1
(1)
(7 marks)
Notes:
(a)
M1: Takes log10' s of both sides and attempts to use the addition law. Condone log  log10 for
this mark.
A1: Proceeds correctly to log
10 N log10 a  b log10 T and states m = b and c  log10 a
(b) Way One: Main scheme
M1: For attempting to use the graph to find either a or b using a  10intercept or b = gradient. This
may be implied by a  101.75 to 1.85 or b = 2.27 to 2.33
M1: For attempting to use the graph to find BOTH a and b (See previous M1)
M1: Uses T  3 in N  aT b with their a and b
A1: Number of microbes  800
Way Two: Alternative using line of best fit techniques.
M1: For log10 3  0.48 and using the graph to find log10 N
M1: For using the graph to find log10 N (FYI log10 N  2.9 )
M1: For log10 N k  N 10k
A1: Number of microbes  800
(c)
B1: See scheme.

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 Question Scheme Marks

9(a) 1 sin 2 A B1
sec 2 A tan 2 A
cos 2 A cos 2 A
1 sin 2 A M1
cos 2 A
1 2 sin A cos A M1
cos2 A sin 2 A
cos2 A sin 2 A 2 sin A cos A
cos2 A sin 2 A
(cos A sin A)(cos A sin A) M1
(cos A sin A)(cos A sin A)
cos A sin A A1*
cos A sin A
(5)
(b) 1 cos θ sin θ 1
sec 2θ tan 2θ
2 cos θ sin θ 2
 2cos θ + 2sin θ = cos θ – sin θ
1 M1 A1
 tan θ
3
dM1
 θ = awrt 2.820, 5.961
A1
(4)
(9 marks)
Notes:
(a)
1 sin 2 A
B1: A correct identity for sec 2 A or tan 2 A .
cos 2 A cos 2 A
1
It need not be in the proof and it could be implied by the sight of sec 2 A
cos A
2
sin 2 A
M1: For setting their expression as a single fraction. The denominator must be correct for their
fractions and at least two terms on the numerator.
This is usually scored for 1 cos 2 A tan 2 A or 1 sin 2 A
cos 2 A cos 2 A
M1: For getting an expression in just sin A and cos A by using the double angle identities
sin 2 A 2sin A cos A and cos 2 A cos2 A sin 2 A , 2 cos2 A 1 or 1 2sin 2 A .
Alternatively for getting an expression in just sin A and cos A by using the double angle
2 tan A sin A
identities sin 2 A 2sin A cos A and tan 2 A with tan A .
1 tan 2 A cos A
2 sin A
For example 1 cos A is B1M0M1 so far
cos2 A sin 2 A 1 sin 2
A
cos2 A

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 Question 9 notes continued

M1: In the main scheme it is for replacing 1 by cos2 A sin2 A and factorising both numerator and
denominator.
A1*: Cancelling to produce given answer with no errors. Allow a consistent use of another
variable such as θ , but mixing up variables will lose the A1*.
(b)
M1: For using part (a), cross multiplying, dividing by cosθ to reach tan θ k
Condone tan 2θ k for this mark only.
1
A1: tan θ
3
dM1: Scored for tan θ k leading to at least one value (with 1 dp accuracy) for  between 0
and 2π . You may have to use a calculator to check. Allow answers in degrees for this mark.
A1: θ awrt 2.820, 5.961with no extra solutions within the range. Condone 2.82 for 2.820.
You may condone different/ mixed variables in part (b)

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 Question Scheme Marks
10(a) Subs D  15 and t = 4 x = 15e−0.2×4 = 6.740 (mg) M1 A1

(2)
(b) M1
15e−0.2×7 + 15e−0.2×2 = 13.754 (mg)
A1*

(2)
−0.2×T −0.2×(T+5)
(c) 15e + 15e = 7.5 M1

15e−0.2×T + 15e−0.2×Te−1 = 7.5

7.5
15e0.2T (1+e1 ) 
7.5  e0.2T  dM1
15 1+e1 

 7.5  2  2 
T
5ln   5ln
  A1 A1
e 
 15 1  e  
 1 


(4)
(8 marks)
Notes:
(a)
0.2 t
M1: Attempts to substitute both D  15 and t =4 in x  De . It can be implied by sight of
0.8 0.24
15e , 15e or awrt 6.7. Condone slips on the power. Eg you may see -0.02
A1: Cao. 6.740 (mg) Note that 6.74 (mg) is A0
(b)
M1: Attempt to find the sum of two expressions with D =15 in both terms with t values of 2
and 7. Evidence would be 15e0.27  15e0.22 or similar expressions such as
15e1  15 e0.22 . Award for the sight of the two numbers awrt 3.70 and awrt 10.05,
followed by their total awrt 13.75. Alternatively finds the amount after 5 hours,
15e−1 = awrt 5.52 adds the second dose = 15 to get a total of awrt 20.52 then multiplies this
by e0.4 to get awrt 13.75. Sight of 5.52+15=20.52  13.75 is fine.
0.27
A1*: Cso so both the expression 15e  15e0.22 and 13.754( mg ) are required
Alternatively both the expression 15e0.25  15  e0.22 and 13.754 (mg) are required.
Sight of just the numbers is not enough for the A1*
(c)
M1: Attempts to write down a correct equation involving T or t. Accept with or without correct
bracketing Eg. accept 15e0.2T 15e0.2(T 5)  7.5 or similar equations
15e  15 e
1 0.2T
7.5
dM1: Attempts to solve their equation, dependent upon the previous mark, by proceeding to
e0.2T  .. An attempt should involve an attempt at the index law x m
n
x m  x n and
taking out a factor of e 0.2T
Also score for candidates who make e 0.2T
the subject using
the same criteria.

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and 135
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
 Question 10 notes continued
 7.5 
A1: Any correct form of the answer, for example, 5ln  
 15 1  e1  
 
2
A1: Cso.
 T 5ln  2   Condone t appearing for T throughout this question.
 e 
(c)
Alternative 1
1st Mark (Method): 15e0.2T  awrt 5.52e0.2T  7.5  e0.2T  awrt 0.37
7.5
2nd Mark (Accuracy): T=-5ln  awrt 0.37  or awrt 5.03 or T=-5ln  

 awrt 20.52 
Alternative 2
7.5 
1st Mark (Method ): 13.754e0.2T  7.5  T  5ln   or equivalent such as 3.03
 13.754 
7.5 
2nd Mark (Accuracy): 3.03 + 2 = 5.03 Allow 5ln  2
 13.754 
Alternative 3 (by trial and improvement)
0.25
1st Mark (Method): 15e  15e0.210 
7.55 or 15e0.25.1  15e0.210.1 
7.40 or any value
between.
2nd Mark (Accuracy): Answer T =5.03.

136 Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018