abc
General Certificate of Education
Mathematics – Pure Core
SPECIMEN UNITS AND
MARK SCHEMES
ADVANCED SUBSIDIARY MATHEMATICS (5361)
ADVANCED SUBSIDIARY PURE MATHEMATICS (5366)
ADVANCED SUBSIDIARY FURTHER MATHEMATICS (5371)
ADVANCED MATHEMATICS (6361)
ADVANCED PURE MATHEMATICS (6366)
ADVANCED FURTHER MATHEMATICS (6371)
General Certificate of Education
Specimen Unit
Advanced Subsidiary Examination
MATHEMATICS
Unit Pure Core 1
MPC1
abc
In addition to this paper you will require:
• an 8-page answer book;
• the AQA booklet of formulae and statistical tables.
You must not use a calculator.
Time allowed: 1 hour 30 minutes
Instructions
• Use blue or black ink or ball-point pen. Pencil should only be used for drawing.
• Write the information required on the front of your answer book. The Examining Body for this
paper is AQA. The Paper Reference is MPC1.
• Answer all questions.
• All necessary working should be shown; otherwise marks for method may be lost.
• Calculators (scientific and graphical) are not permitted in this paper.
Information
• The maximum mark for this paper is 75.
• Mark allocations are shown in brackets.
Advice
• Unless stated otherwise, formulae may be quoted, without proof, from the booklet.
[2]
2
Answer all questions.
1
The line L has equation
x+ y =9
and the curve C has equation
y = x2 + 3
(a) Sketch on one pair of axes the line L and the curve C. Indicate the coordinates of their
points of intersection with the axes.
(3 marks)
(b) Show that the x-coordinates of the points of intersection of L and C satisfy the equation
x2 + x − 6 = 0
(2 marks)
(c) Hence calculate the coordinates of the points of intersection of L and C.
2
(4 marks)
The line AB has equation 5 x − 2 y = 7.
The point A has coordinates (1, − 1) and the point B has coordinates (3, k) .
(a)
(i) Find the value of k.
(1 mark)
(ii) Find the gradient of AB.
(2 marks)
(b) The point C has coordinates (– 6, –2) . Show that AC has length p 2 , where p is an
integer.
(3 marks)
3
(a) Given that (x – 1) is a factor of f (x ), where
f ( x ) = x 3 − 4 x 2 − kx + 10
show that k = 7.
(2 marks)
(b) Divide f (x ) by (x – 1) to find a quadratic factor of f (x ) .
(2 marks)
(c) Write f (x ) as a product of three linear factors.
(2 marks)
(d) Calculate the remainder when f (x ) is divided by (x – 2) .
(2 marks)
[3]
3
4 The number x satisifies the equation
x 2 + mx + 16 = 0
where m is a constant.
Find the values of m for which this equation has:
(a) equal roots;
(2 marks)
(b) two distinct real roots;
(2 marks)
(c) no real roots.
(2 marks)
5 The diagram shows a part of the graph of
y = x − 2x 4
(a)
(i) Find
dy
.
dx
(2 marks)
(ii) Show that the x-coordinate of the stationary point P is 12 .
(iii) Find the y-coordinate of P.
(3 marks)
(1 mark)
(b) Find the area of the shaded region.
(5 marks)
Turn over ►
[4]
4
6 A circle C has equation
x 2 + y 2 − 10 x = 0
(a) By completing the square, express this equation in the form
( x − a )2 + y 2 = r 2
(3 marks)
(b) Write down the radius and the coordinates of the centre of the circle C.
(2 marks)
(c) Describe a geometrical transformation by which C can be obtained from the circle with
equation
x2 + y2 = r 2
(2 marks)
(d) The point P, which has coordinates (9, 3), lies on the circle C.
(i) Show that the line which passes through P and the centre of C has gradient
3
4
(2 marks)
(ii) Find the equation of the tangent to the circle C at the point P.
Give your answer in the form y = mx + c .
7
(a) Express
2 +1
2 −1
in the form a 2 + b , where a and b are integers.
(4 marks)
(4 marks)
(b) Solve the inequality
(
)
2 x– 2 < x + 2 2
(4 marks)
[5]
5
8 A curve has equation
y = x 4 − 8 x 3 + 16 x 2 + 8
(a) Find
d2 y
dy
and 2 .
dx
dx
(5 marks)
(b) Find the three values of x for which
dy
= 0.
dx
(c) Determine the coordinates of the point at which y has a maximum value.
END OF QUESTIONS
[6]
(4 marks)
(5 marks)
abc
MPC1 Specimen
Question
Solution
1(a) Sketches of L and C
Coordinates (0, 9), (9, 0) indicated
Coordinates (0, 3) indicated
Marks
M1
A1
A1
(b) Equating expressions for y
M1
A1
2
x + x−6 = 0
(c) Solving quadratic
x = −3 or x = 2
Points are (−3, 12) and (2, 7)
M1
A1
A1A1
Total
2(a)(i) k = 4
(ii) Gradient =
=
5
2
B1
4 − (−1)
3 −1
A1
M1
A1
A1
Total
3(a) Use of factor theorem
1 − 4 − k + 10 = 0, so k = 7
4(a)
B2
B2
B1
B1
Total
2
Two solutions needed
4
9
1
2
ft wrong value of k
stated or used
3
2
or complete division
convincingly shown (ag)
2
2
B1 if −3x or −10 correct
B1 if signs wrong
2
8
ft wrong value for f(2)
2
m 2 − 64 > 0
M1
A1
2
m 2 − 64 < 0
M1
A1
2
−8 < m < 8
A1 if not clearly paired
or use of equation of AB
M1
A1
m < −8 or m > 8
(c)
oe
convincingly shown (ag)
m − 64 = 0
m = ±8
(b)
2
6
M1
A1
(b) Quotient is x 2 − 3 x − 10
(c) f(x) = (x − 1)(x + 2)(x − 5)
(d) f(2) = −12
so remainder is −12
3
Comments
General shape
Accept labels on sketch
ditto
M1
(b) Distance formula
AC = 50
=5 2
Total
Total
6
[7]
MPC1 (cont)
Question
5(a)(i)
Solution
y′ = 1 − 8x
Marks
M1A1
3
(ii) SP ⇒ y ′ = 0
⇒ x3 =
(iii)
(b)
yP =
∫
convincingly shown
Substitution of x =
=
1
2
A1
3
B1
1
− 801
A1
A1
Total
6(a) Use of ( x − 5) 2 = x 2 − 10 x + 25
a = 5, r = 5
(b) Radius 5
Centre (5, 0)
(c) Translation
5 units in positive x direction
(d)(i) Use of formula for gradient
Grad 34 convincingly shown
(ii) Grad of tangent is −
ag; 2/3 for verification
M1 if at least one term correct
m1
9
80
B1
B1
M1
A1
M1
A1
4
3
First A1 awarded if at least one term
correct
3
Condone RHS = 25
2
ft wrong value for a
ditto
Condone ‘transformation’ if clarified
ft wrong value for a
2
ag
4
ft wrong gradient
ft wrong gradient
2
B1
M1
A1
A1
4
i.e. y = – x + 15
3
5
11
M1
A1A1
Tangent is y − 3 = − 43 ( x − 9)
Total
12
M1
m1A1
A1
B1
7(a) Rationalising denominator
Numerator becomes 2 2 + 3
Denom = 1, so ans is 2 2 + 3
2x − 2
2x − x < 2 2 + 2
(b) LHS =
4
m1
A1
2 2+2
4
2 −1
8
Total
[8]
ft one small error in numerator
Allow 2 x − 4
for isolating x terms
M1
Reasonable attempt at division
x<
PI
M1A1
y dx = 12 x 2 − 52 x 5 (+ c)
1
8
Comments
M1 if at least one term correct
A1
3
8
Area =
2
M1
1
8
1
2
⇒ x=
Total
ft error in expanding LHS
MPC1 (cont)
Question
8(a)
Solution
3
Marks
M1
A2
m1
A1
B1
M1
A2
2
y ′ = 4 x − 24 x + 32 x
y ′′ = 12 x 2 − 48 x + 32
(b)
y ′ = 0 if x = 0
(c)
or if x 2 − 6 x + 8 = 0
ie x = 2 or x = 4
y ′′ = 32, −16, 32
Total
5
Comments
at least one term correct
A1 with one error
at least one term correct
ft numerical error in y′
4
A1 if only one small error
M1
A1
Relationship between sign of y" and
maximum/minimum
Max at x = 2
y = 24
M1
A1
A1
Total
TOTAL
[9]
M1 if at least one correct
ft numerical error in y′′
5
15
75
abc
General Certificate of Education
Specimen Unit
Advanced Subsidiary Examination
MATHEMATICS
Unit Pure Core 2
MPC2
In addition to this paper you will require:
• an 8-page answer book.
• the AQA booklet of formulae and statistical tables.
You may use a graphics calculator.
Time allowed: 1 hour 30 minutes
Instructions
• Use blue or black ink or ball-point pen. Pencil should only be used for drawing.
• Write the information required on the front of your answer book. The Examining Body for this
paper is AQA. The Paper Reference is MPC2.
• Answer all questions.
• All necessary working should be shown; otherwise marks for method may be lost.
• The final answer to questions requiring the use of tables or calculators should normally be given to
three significant figures.
Information
• The maximum mark for this paper is 75.
• Mark allocations are shown in brackets.
Advice
• Unless stated otherwise, formulae may be quoted, without proof, from the booklet.
[10]
2
Answer all questions.
1
The diagram shows triangle ABC.
The lengths of AB and AC are 7cm and 8cm respectively. The size of angle BAC is 60°.
Calculate the length of BC, giving your answer to 3 significant figures.
(3 marks)
2
The diagrams show a square of side 6 cm and a sector of a circle of radius 6 cm and angle θ
radians.
The area of the square is three times the area of the sector.
(a) Show that θ = 23 .
(2 marks)
(b) Show that the perimeter of the square is 1 12 times the perimeter of the sector.
(3 marks)
3 The nth term of an arithmetic sequence is un , where
un = 10 + 0.5n
(a) Find the value of u1 and the value of u2 .
(b) Write down the common difference of the arithmetic sequence.
(c) Find the value of n for which un = 25.
(2 marks)
(1 mark)
(2 marks)
30
(d)
Evaluate
∑u
n =1
n
.
(3 marks)
[11]
3
4
(a) Given that
logax = loga5 + 2 loga3
where a is a positive constant, show that x = 45.
(b)
(i) Write down the value of log2 2.
(3 marks)
(1 mark)
(ii) Given that
log2 y = log42
find the value of y.
5
(2 marks)
The curve C is defined by the equation
y = 2x 2 x +
1
x4
for x > 0
(a) Write x 2 x in the form xk, where k is a fraction.
(b) Find
dy
.
dx
(1 mark)
(3 marks)
(c)
Find an equation of the tangent to the curve C at the point on the curve where x = 1.
(4 marks)
2
d y
.
(2 marks)
(d) (i) Find
dx 2
(ii) Hence deduce that the curve C has no maximum points.
6
(2 marks)
The amount of money which Pauline pays into an insurance scheme is recorded each year. The
amount which Pauline pays in during the nth year is £An. The first three values of An are given
by:
A2 = 650
A3 = 530
A1 = 800
The recorded amounts may be modelled by a law of the form
An +1 = pAn + q
where p and q are constants.
(a) Find the value of p and the value of q.
(5 marks)
(b) Given that the amounts converge to a limiting value, £V, find an equation for V and hence
find the value of V.
(3 marks)
Turn over ►
[12]
4
7
(a) Express
x5 + 1
in the form x p + x q , where p and q are integers.
x2
(2 marks)
3
2
 x5 + 1
(b) Hence find the exact value of ∫  2  dx .

1 x
(5 marks)
8 The angle θ radians, where 0 ≤ θ ≤ 2π , satisfies the equation
3 tan θ = 2 cos θ
(a) Show that
3 sin θ = 2 cos2θ
(1 mark)
(b) Hence use an appropriate identity to show that
2 sin2θ + 3 sin θ − 2 = 0
(c)
(i) Solve the quadratic equation in part (b). Hence explain why the only possible value
of sin θ which will satisfy it is 12 .
(3 marks)
(ii) Find the values of θ for which sinθ = 12 and 0
(d)
(3 marks)
θ
2π .
(2 marks)
Hence write down the solutions of the equation
3 tan 2x = 2 cos 2x
that lie in the interval 0°
x
180°.
(3 marks)
[13]
5
9 The diagram shows a sketch of the curve with equation y = 4 x .
(a) (i) Use the trapezium rule with five ordinates (four strips) to find an approximation
2
for ∫ 4 x dx .
(4 marks)
0
(ii) By considering the graph of y = 4 x , explain with the aid of a diagram whether your
approximation will be an overestimate or an underestimate of the true value
2
for ∫ 4 x dx .
(2 marks)
0
(b) Describe the single transformation by which the curve with equation y = 5 × 4x can be
(2 marks)
obtained from the curve with equation y = 4 x .
(c) Sketch the curve with equation y = 4 − x .
(1 mark)
(d) The two curves y = 5 × 4x and y = 4 − x intersect at the point P.
(i) Show that the x-coordinate of the point P is a root of the equation 4 2 x = 0.2
(2 marks)
(ii) Solve this equation to find the x-coordinate of the point P. Give your answer to
5 significant figures.
(3 marks)
END OF QUESTIONS
[14]
abc
MPC2 Specimen
Question
1
2(a)
Solution
BC = 7 + 8 − 2 × 7 × 8 × cos 60o
= 49 + 64 − 56
= 57 ⇒ BC = 57 = 7.55 to 3sf
Total
2
2
2
Marks
M1
m1
Total
A1
3
3
1
{Area of square} = 3 × 62θ
62 =
3
6 2θ ⇒ θ =
Use of
M1
2
Comments
2
1
2
r 2θ PI
2
ag
A1
3
ag
B1B1
5
2
sc B1 for 10, 10.5
(b) Common difference is 0.5
B1
1
(c) 10 + 0.5n = 25 ⇒ 0.5n = 25 – 10
⇒ n = 30
M1
A1
2
2
(b) Arc length = 6θ
A1
3
B1
Perimeter of sector = (12 + 6θ )
M1
1
2
= 1 12 + 6 ×  = 24
2
3
= perimeter of square
Total
3(a) u1 = 10.5;
(d)
30
∑u
n =1
n
u2 = 11
= sum of AP with n = 30
M1
30
(10.5 + 25)
2
= 532.5
=
4(a) logax = loga5 + loga32
logax = loga[5 × 32]
⇒ x = 45
m1
A1
Total
⇒ y=
= 2
3
8
PI
M1
m1
A1
(b)(i) log22 = 1
(ii) {log2 y =} log42 = 0.5
1
22
oe
B1
3
1
B1
B1
Total
[15]
2
6
ag convincingly found
MPC2 (cont)
Question
5(a)
(b)
Solution
x2 x = x
5
2
Total
B1
1
Accept k = 2.5
M1
A1A1
3
One correct index ft
A1 for each correct term
B1
M1
m1
A1
4
Accept any valid form
2
ft each term provided equivalent demands
ie indices one fractional one ‘negative’
y = 2 x + x −4
3
(c) When x = 1, y = 2+1 = 3
When x = 1, y ′ = 5−4 = 1
Eqn. of tangent: y − 3 = 1(x − 1)
2
d y 15
=
dx 2 2
1
x2
+
20
x6
B2,1
(ii) Since x>0, y ′′(x) is >0
so any turning point must be a minimum
ie C has no maximum points
E2,1,0
Total
M1
6(a) 650 = 800 p + q
A1
530 = 650 p + q
m1
4
p = ; q = 10
A1A1
5
(b)
Comments
5
2
dy
4
= 5x 2 − 5
dx
x
(d)(i)
Marks
V = pV + q
q
V =
1− p
V = 50
7(a)
x 3 + x −2
(b)
x4
− x −1
4
2
12
For either equation
Need both
Full valid method to solve simultaneous
equations
5
M1
m1
A1
Total
M1
A1
M1
3
8
2
Index raised by 1 (either term)
One term correct ft p, q.
All correct
Use of limits
m1
67
192
A1
Total
5
7
[16]
ft on 1 numerical slip in (a)
One power correct
A1
A1
  3 4

 

  2  − 2  −  1 − 1


 4
3 4





=1
E1 for attempt to find the sign of y ′′(x)
Must be an exact value
MPC2 (cont)
Question
8(a)
(b)
(c)(i)
Solution
sin θ
3
= 2 cosθ ⇒ 3 sin θ = 2 cos 2 θ
cosθ
sin 2 θ + cos2 θ = 1
3 sin θ = 2 (1 − sin 2 θ )
Marks
Total
B1
M1
1
ag convincingly found
oe seen
3
ag convincingly found
M0 for verification
oe eg use of the formula.
A1
A1
M1
A1
2 sin 2 θ + 3 sin θ − 2 = 0
Attempt to solve for sin θ
(2 sin θ − 1)(sin θ + 2) = 0
Comments
Since −1≤sin θ ≤ 1, the only possible
value for sin θ is
(ii)
(d)
1
2
θ
= 0.5235…
θ
= 2.6179…
3 tan 2 x = 2 cos 2 x ⇒ sin 2 x =
A1
ag convincingly found and explained
B1
In (ii) accept 3sf and in terms of π;
ft on their 0.5235…
B1
1
M1
2
2x = 30° or 150°
⇒ x = 15°
or x = 75°
A1
A1
Total
9(a)(i) h = 0.5
Integral = h/2 {……}
{….}=
1
 1
f (0) + 2[f   + f (1) + f 1 ] + f (2)
 2
 2
{….}=1+2[2+4+8]+16
Integral =11.25
(ii) Relevant trapezia drawn on a copy of
given graph.
Overestimate
(b) Stretch in y-direction
scale factor 5
(c) Sketch showing the reflection of the graph
of the given curve in the y-axis
(d)(i) 5 4 x = 4− x ⇒ 5 4 x × 4 x = 1
( )
2x
( )
⇒ 5 × 4 = 1 ⇒ 42 x = 0.2
(ii) ln 42x = ln 0.2
ln 0.2
2x =
ln 4
x = − 0.58048(20…)
2
Links with previous parts
3
12
B1
M1
A1
A1
M1
A1`
M1
A1
4
2
2
B1
M1
1
A1
M1
2
ag convincingly found
oe using base 10
3
14
75
Need 5sf or better
A1
A1
Total
TOTAL
[17]
abc
General Certificate of Education
Specimen Unit
Advanced Level Examination
MATHEMATICS
Unit Pure Core 3
MPC3
In addition to this paper you will require:
• an 8-page answer book;
• the AQA booklet of formulae and statistical tables;
• an insert for use in Question 5 (enclosed).
You may use a graphics calculator.
Time allowed: 1 hour 30 minutes
Instructions
• Use blue or black ink or ball-point pen. Pencil should only be used for drawing.
• Write the information required on the front of your answer book. The Examining Body for
this paper is AQA. The Paper Reference is MPC3.
• Answer all questions.
• All necessary working should be shown; otherwise marks for method may be lost.
• The final answer to questions requiring the use of tables or calculators should normally be
given to three significant figures.
Information
• The maximum mark for this paper is 75.
• Mark allocations are shown in brackets.
Advice
• Unless stated otherwise, formulae may be quoted, without proof, from the booklet.
[18]
2
Answer all questions.
1 Find
dy
when:
dx
(a) y = x tan 3 x ;
(b) y =
(3 marks)
sin x
.
x
(3 marks)
2 A curve has equation y =
x
. The region R is bounded by the curve, the x-axis
+2
from the origin to the point (1,0 ) and the line x = 1 .
(x
3
)
(a)
Explain why R lies entirely above the x-axis.
(1 mark)
(b)
Use Simpson’s Rule with five ordinates (four strips) to find an approximation for the
area of R, giving your answer to 3 significant figures.
(4 marks)
(c)
Find the exact value of the volume of the solid formed when R is rotated through
2π radians about the x-axis.
(5 marks)
3 A curve has equation y = e 2 x − 4 x .
1
ln 2 .
2
Find the
(a)
Show that the x-coordinate of the stationary point on the curve is
(b)
corresponding y-coordinate in the form a + b ln 2 , where a and b are integers to be
determined.
(6 marks)
2
d y
and hence determine the nature of the stationary
Find an expression for
d x2
point.
(3 marks)
(c)
Show that the area of the region enclosed by the curve, the x-axis and the lines x = 0
and x = 1 is
1
2
(e
2
)
−5 .
(5 marks)
[19]
3
4 (a)
Describe a sequence of geometrical transformations that maps the graph of y = sin x
onto the graph of y = 3 + sin 2 x .
(4 marks)
π
.
6
(3 marks)
(b)
Find the gradient of the curve with equation y = 3 + sin 2 x at the point where x =
(c)
(i) Find
∫ x sin 2 x d x .
(ii) Hence show that
(4 marks)
π (3π + 2)
∫ x(3 + sin 2 x )d x = 8 .
π
2
0
(2 marks)
5 [An insert is provided for use in answering this question.]
The curve with equation y = x3 − 4 x 2 − 4 intersects the x -axis at the point A where x = α .
(a)
Show that α lies between 4 and 5.
(b)
Show that the equation x 3 − 4 x 2 − 4 = 0 can be rearranged in the form x = 4 +
(c)
(i) Use the iterative formula xn +1 = 4 +
(2 marks)
to three significant figures.
4
.
x2
(2 marks)
4
with x1 = 5 to find x3 , giving your answer
xn2
(3 marks)
4
and y = x and the position of x1 .
x2
On the insert provided, draw a cobweb or staircase diagram to show how
convergence takes place, indicating the positions of x 2 and x3 .
(3 marks)
(ii) The sketch shows the graphs of y = 4 +
Turn over ►
[20]
4
6
Solve the equation
4 cot 2 x + 12 cosec x + 1 = 0
giving all values of x to the nearest degree in the interval 0
x 360 o .
(7 marks)
7 The functions f and g are defined with their respective domains by
f (x ) =
4
, x>0
3+ x
g( x ) = 9 − 2 x 2 , x ∈ 3
(a)
Find fg( x ) , giving your answer in its simplest form.
(2 marks)
(b)
(i) Solve the equation g ( x ) = 1 .
(2 marks)
(ii) Explain why the function g does not have an inverse.
(1 mark)
(c)
Solve the equation g( x ) = 1 .
(d)
The inverse of f is f −1 .
(3 marks)
(i) Find f −1 (x ) .
(3 marks)
(ii) Solve the equation f −1 ( x ) = f ( x ) .
(4 marks)
END OF QUESTIONS
[21]
Surname
abc
Other Names
Centre Number
Candidate Number
Candidate Signature
General Certificate of Education
Specimen Unit
Advanced Level Examination
MATHEMATICS
MPC3
Unit Pure Core 3
Insert for use in answering Question 5.
Fill in the boxes at the top of this page.
Fasten this insert securely to your answer book.
[22]
abc
MPC3 Specimen
Question
1(a)
(b)
Solution
3x sec 3x + tan 3 x
Marks
M1
M1
A1
2
x cos x − sin x
x2
2(a)
b)
(c)
1
× 0.25
3
“Outside multiplier”
Product Rule
3
sec 2
Correct
Quotient Rule
B1
d(sin x )
= cos x
dx
Correct
E1
3
6
1
B1
1
× 0.25{y (0) + y (1) +
3
4 [ y (0.25) + y (0.75)] + 2 y (0.5)}
M1
= 0.3246193….
= 0.325 to 3 sf
A1
A1
1
x2
dx
x3 + 2
0
k ln(x3 +2)
Comments
M1
A1
Total
y ≥ 0 when x ≥ 0 so R is above x-axis
Total
y(0) = 0; y (0.25) = 0.17609;
y(0.5) = 0.34300; y (0.75) = 0.48193;
y(1) = 0.57735
Correct to at least 2 sf
4
B1
V =π∫
1
3
π
3
ln( x 3 + 2)
M1
A1
(ln 3 − ln 2)
m1
A1
Total
Integration correct
Correct use of limits
5
10
[23]
MPC3 (cont)
Question
3(a)
Solution
dy
= 2e 2 x − 4
dx
dy
= 0 ⇒ e 2x = 2
dx
2 x = ln 2
Marks
M1
A1
1
ln 2
2
A1
B1
B1
y = 2 − 2 ln 2 a = 2
b = −2
(b)
d y
dx
2
= 4e 2 x
2
d y
2
dx
1
2
=8
2
ag
6
M1
⇒ Minimum Point
1
ke
B1
x = ln 2
(c)
A1
M1
A1
e2 x − 2x 2
3
1 2
 1

 e − 2 −  − 0
2
 2

1
1
2
2
(e
2
1
2
−5
2
A1
Total
Use of limits 0 and 1
5
14
[24]
e2 x
− 2x 2
m1
)
ft if negative for maximum
ke2 x
1
A1
= e2 − 2 =
Comments
2x
M1
⇒x=
2
Total
ag
MPC3 (cont)
Question
Solution
4(a) One way stretch in x − direction
scaling factor
(ii)
∫
Total
Comments
1
2
translation (in y-direction)
0 
of  
3 
(b) dy
= 2 cos 2 x
dx
π
1
dy
= 2x = 1
When x = ,
dx
2
6
 x

(c)(i) − cos 2 x 
2


+
Marks
M1
1
cos 2 xdx
2
x
1
= − cos 2 x + sin 2 x
2
4
2
3x
+ previous result and attempt at limits
2
3π 2  −π  3π 2 π
+−
+
=
8  4 
8
4
π (3π + 2)
=
8
Total
A1
M1
A1
4
M1
A1
A1
M1
A1
cos
Correct
3
Integration by parts attempt
1
− cos 2 x
2
A1
A1
4
Ignore − [0]
M1
A1
2
13
[25]
(ignore +c)
ag all integration must be correct for A1
MPC3 (cont)
Question
5(a)
Solution
f (4) = − 4
Marks
f (5) = 21
;
Total
f (x ) = x3 − 4 x 2 − 4
M1
change of sign ⇒ root between 4 and 5
4
(b)
x−4− 2 = 0
x
4
⇒ x = 4+ 2
x
4
(c)(i) x = 4 +
2
52
= 4.16
⇒ x 3 = 4.23 (to 3 sf)
A1
2
Divide by x 2
M1
A1
Comments
2
ag
M1
A1
A1
3
(ii)
M1
A1
A1
6
(
2
Total
)
3
10
Cobweb ‘to curve first’
Thin line ⇒ x2 marked
Next iteration ⇒ x3 marked
4 cosec x − 1 + 12 cosec x + 1 = 0
M1
Attempt at cot 2 x = cosec2 x − 1
4 cosec 2 x + 12 cosec x − 3 = 0
A1
or may use cos 2 x = 1 − sin 2 x
2
after 4 cos2 x + 12 + 1 = 0 etc
sin x sin x
− 12 ± 192
8
= 0.23205 − 3.23205
1
cosec x =
sin x
sin x = − 0.3094
cosecx =
x = 198°
342°
M1
A1
B1
A1
A1
Total
[26]
7
7
MPC3 (cont)
Question
7(a)
Solution
=
(b)(i)
4
fg (x ) =
Total
A1
2
Comments
3 + 9 − 2x 2
2
6− x
Marks
M1
and no further wrong ‘simplification’
2
9 − 2x 2 = 1 ⇒ x 2 = 4
x = ±2
(ii) Two values of x map onto one ⇒ not
one-one
(c) Two values from g(x) =1 x = ±2
9 − 2 x 2 = −1 ⇒ x 2 = 5
x=± 5
4
(d)(i)
y=
⇒ 3 y + yx = 4
3+ x
x = 4 −3
y
4
f −1 ( x ) = − 3
x
(ii) their f −1 ( x ) = f ( x )
and multiplying up
x 2 + 3x − 4 = 0
⇒ x = − 4, 1
x>0
⇒ only solution is
x =1
M1
A1
E1
B1
M1
A1
M1
x = 2 scores M1 only
2
1
or many-one
3
multiplying out and attempt to make x
the subject
A1
A1
3
or f (x ) = x
⇒ 4 = x(3 + x )
(x + 4)(x − 1) = 0
M1
A1
A1
A1
Total
TOTAL
[27]
4
15
75
General Certificate of Education
Specimen Unit
Advanced Level Examination
MATHEMATICS
Unit Pure Core 4
MPC4
abc
In addition to this paper you will require:
• an 8-page answer book;
• the AQA booklet of formulae and statistical tables.
You may use a graphics calculator.
Time allowed: 1 hour 30 minutes
Instructions
• Use blue or black ink or ball-point pen. Pencil should only be used for drawing.
• Write the information required on the front of your answer book. The Examining Body for this
paper is AQA. The Paper Reference is MPC4.
• Answer all questions.
• All necessary working should be shown; otherwise marks for method may be lost.
• The final answer to questions requiring the use of tables or calculators should normally be given
to three significant figures.
Information
• The maximum mark for this paper is 75.
• Mark allocations are shown in brackets.
Advice
• Unless stated otherwise, formulae may be quoted, without proof, from the booklet.
[28]
2
Answer all questions.
1 The polynomials f(x) and g(x) are defined by
f (x) = 4x + 4x − 3
2
g (x) = 4x − x
3
(a) By considering f ( 12 ) and g ( 12 ), or otherwise, show that f (x ) and g(x )
have a common linear factor.
(b) Hence write
2
f (x )
as a simplified algebraic fraction.
g(x )
(3 marks)
1
(a) Obtain the binomial expansion of (1 + x ) 2 as far as the term in x 2 .
(b)
(2 marks)
1
(i) Hence, or otherwise, find the series expansion of (4 + 2 x ) 2
as far as the term in x 2 .
(3 marks)
(ii) Find the range of values of x for which this expansion is valid.
3
(3 marks)
(1 mark)
(a) Express 4 sin θ − 3 cos θ in the form R sin (θ – α ), where R is a positive constant and
0o < α < 90o.
Give the value of α to the nearest 0.1o.
(3 marks)
(b) Hence find the solutions in the interval 0 o < θ < 360 o of the equation
4 sin θ – 3 cos θ = 2
Give each solution to the nearest degree.
[29]
(4 marks)
3
4
(a) Express 4 sin 2 x in the form a + b cos 2 x, where a and b are constants.
(2 marks)
π
12
(b) Find the value of ∫ 4 sin 2 x dx.
(4 marks)
0
(c) Hence find the volume generated when the part of the graph of
y = 2 sin x
between x = 0 and x = π is rotated through one revolution about the x-axis.
12
(2 marks)
5 The population P of a particular species is modelled by the formula
P = Ae -kt
where t is the time in years measured from a date when P = 5000.
(a) Write down the value of A.
6
(1 mark)
(b) Given that P = 3500 when t = 10, show that k ≈ 0.03567.
(4 marks)
(c) Find the value of the population 20 years after the initial date.
(3 marks)
(a) Solve the differential equation
dx 10 − x
=
dt
5
given that x = 1 when t = 0.
(4 marks)
(b) Find the value of t for which x = 2, giving your answer to three decimal places. (2 marks)
Turn over ►
[30]
4
7
(a) Express
25 x + 1
(2 x − 1)(x + 1)2
in the form
A
B
C
+
+
2 x − 1 x + 1 ( x + 1)2
(4 marks)
(b) Hence find the value of
2
25 x + 1
∫ (2 x – 1)(x + 1)
2
dx
1
giving your answer in the form p + q 1n 2.
(6 marks)
8 A curve is defined by the parametric equations
2
2
x = t2 + , y = t2 − , t ≠ 0
t
t
(a)
(i) Express x + y and x – y in terms of t.
(2 marks)
(ii) Hence verify that the cartesian equation of the curve is
(x + y )(x – y )2 = 32
(b)
(i) By finding
(2 marks)
dx
dy
dy
and
at the point for which t = 2.
, calculate the value of
dt
dt
dx
(5 marks)
(ii) Hence find the equation of the tangent to the curve at this point.
Give your answer in the form ax + by = c, where a, b and c are integers.
[31]
(3 marks)
5
 x   3  4
9 The line l1 has equation  y  = − 2 + t 4 .
 z   1  3
 x   9   – 1
The line l 2 has equation  y  = − 4 + s  3  .
 z   0   2 
(a) Show that the lines l1 and l 2 intersect and find the coordinates of their point of
intersection.
(5 marks)
(b) The point P on the line l1 is where t = p, and the point Q has coordinates (5, 9, 11).
(i) Show that
 4
QP . 4 = 41 p − 82
3
(4 marks)
(ii) Hence find the coordinates of the foot of the perpendicular from the point Q to the
line l1 .
(3 marks)
END OF QUESTIONS
[32]
abc
MPC4 Specimen
Question
Solution
1(a) f( 1 ) = 0, g( 1 ) = 0
2
2
Marks
M1A1
A1
So 2x − 1 is a common factor
(b)
f ( x) = (2 x − 1)(2 x + 3)
2(a)
(b)(i)
(1 + x)
B1
1
( 4 + 2 x)
2
= 2(1 + 12 x)
1
2
1
16
... = 2 + x −
1
2
2
x + ...
(ii) Valid if −2 < x < 2
3(a) R = 5
cos α or sin α =
sin(θ − α ) =
3
ft numerical error
M1A1
6
2
M1 if two terms correct
B1
M1
A1
3
B1
Total
or
3
5
M1
A1
M1
2
5
One solution is α + sin −1
2
5
m1
A1A1
Solutions 60° and 193°
Total
4(a) Use of cos 2 A ≡ 1 − 2 sin 2 A
4 sin 2 x ≡ 2 − 2 cos 2 x
(b) ∫ ... dx = 2 x − sin 2 x (+ c)
Use of sin π6 =
π
12
π
3
PI
4
1
2
Accept awrt 60 or 61, and awrt 193
PI
2
M1 if at least one term correct
M1A1
m1
1
0
2
A1
B1
2
(2 sin x) = 4 sin x
So volume is π
(π6 − 12 )
B1
Total
5(a) A = 5000
(b) 3500 = 5000e −10 k
ln 0.7 = −10k
k = − 101 ln 0.7
(c)
1
6
7
M1
A1
∫ ... dx = 6 − 2
(c)
Reasonable attempt
B1
4
5
α ≈ 36.9°
(b)
B1
Total
= 1 + 12 x − 18 x 2 + ...
2
3
Comments
or other complete method
or x − 12
B1
g( x) = (2 x − 1)(2 x 2 + x)
f ( x)
2x + 3
= 2
So
g( x) 2 x + x
1
Total
B1
B1
M1
A1
... ≈ 0.03567
P = 5000 (0.7) 2 or 5000e −0.7134
... = 2450
A1
M1A1
A1
Total
4
2
8
1
4
3
8
[33]
Accept 0.0236
Accept 0.0741
ft one error
oe; ft wrong value for A
oe
ft one numerical error
convincingly shown (ag)
M1 if only one small error
Accept awrt 2450
MPC4 (cont)
Question
Solution
6(a) Attempt to separate variables
Marks
M1
m1
dx
∫ 10 − x dx = ±k ln(10 −x) (+ c)
t = −5 ln(10 − x) + c
c = 5 ln 9
(b) x = 2 ⇒ t = 5 ln 9 − 5 ln 8
... ≈ 0.589
7(a)
A1
A1
M1
A1
Total
25x + 1 ≡ A( x + 1) 2 + B(2 x −1)( x + 1) + C (2 x −1)
A = 6, B = −3, C = 8
(b)
∫
2
∫
8
(+ c)
x +1
... = 3 ln 3 − 3 ln 3 + 3 ln 2 − ( 83 − 4)
1
... =
4
3
Total
8(a)(i)
(ii)
4
A1 if only one error
ft wrong values for
coeffs throughout this question
6
x + y = 2t , x − y = 4 / t
B2
10
2
( x + y )( x − y ) 2 = (2t 2 )(16 / t 2 )
M1
A1
2
Convincingly shown (ag)
5
ft numerical or sign error
3
12
ft one numerical error
2
... = 32
(b)(i)
2
6
m1
A1
+ 3 ln 2
Comments
4
M1
A1
B1
B1
... = 3 ln(2 x − 1)...
... − 3 ln( x + 1) −
B1
M1A2
Total
2 dy
2
dx
= 2t − 2 ,
= 2t + 2
dt
dt
t
t
M1A1
m1
Use of chain rule
When t = 2,
dy 9
=
dx 7
A2,1
B1
M1
(ii) The point is (5, 3)
The tangent is y − 3 = 97 ( x − 5)
ie 9x − 7y = 24
A1
Total
[34]
MPC4 (cont)
Question
Solution
9(a) 3 + 4 t = 9 − s
−2 + 4t= −4 + 3s
1 + 3t = 2s
Solve two to obtain s = 2, t = 1
Check in 3rd equation
Point of intersection is (7, 2, 4)
(b)(i)
4 p−2
Marks
M1
A1
m1
A1
A1F
QP =  4 p − 11 
3 p – 10
Total
5
M1A1
4
QP.4 = 4(4 p − 2) + 4(4 − 11) + 3(3 − 10)
3
m1
A1
4
... = 41p − 82
(ii)
Comments
oe
M1
QP ⊥ l1 ⇒ 41 p − 82 = 0
... ⇒ p = 2
Foot of perpendicular is (11, 6, 7)
A1
A1
Total
TOTAL
[35]
3
12
75
Convincingly shown (ag)