Further Maths Summer Transition Work

ΑΒ
AQA Qualifications
AQA Level 2 Certificate

FURTHER MATHEMATICS
Level 2 (8365)
Miscellaneous Worksheet
Version 2.0
Our specification is published on our website (www.aqa.org.uk). We will let centres know in writing
about any changes to the specification. We will also publish changes on our website. The definitive
version of our specification will always be the one on our website, this may differ from printed
versions.
You can download a copy of this resource from our All About Maths website
(http://allaboutmaths.aqa.org.uk/).
Copyright © 2018 AQA and its licensors. All rights reserved.
AQA retains the copyright on all its publications, including the specifications. However, registered
centres for AQA are permitted to copy material from this specification booklet for their own internal use.
AQA Education (AQA) is a registered charity (number 1073334) and a company limited by guarantee registered in England and Wales
(number 3644723). Our registered address is AQA, Devas Street, Manchester M15 6EX.
M Miscellaneous
Question 1 (Spec ref 2.13)
A (0, 8) and B (–3, 1) are points on y = abx as shown.
A
B
O
x
By working out the values of a and b, show that the equation of the curve can be written in
the form y = 2x + 3
(4 marks)
Here are five cards.
2 3 4 5 6
Using four or five of the cards, how many numbers greater than 4000 can be made?
(4 marks)
Question 3 (Spec ref 2.9/2.20)

Two sequences S and T have nth terms
2n + 3 30
Sn = and Tn =
n 3n + 4
Use an algebraic method to work out the value of n when Sn + Tn = 3
(5 marks)
3
Question 4 (Spec Ref 2.18)
By expanding and simplifying, solve
2
 5 1
 4
 2 x 2 − x 2  =x(1 + 4 x ) + 108
 
 
(5 marks)

In the expansion of ( a + 5 x )4 where a > 0
The coefficient of x is three times the coefficient of x2.
Work out the value of a.
(5 marks)
4
ΑΒ
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8365)
Worksheet 1
Coordinate Geometry Circles
Version 2.0
versions.
You can download this resource from our All About Maths website (http://allaboutmaths.aqa.org.uk/).
1 Coordinate Geometry - Circles
Question 1
Write down the equation of each of these circles.

(a) Centre (0, 3) radius 2 (2 marks)
(b) Centre (1, −5) radius 4 (2 marks)
(c) Centre (−3, 4) radius 7 (2 marks)
(d) Centre (8, 15) radius 17

Does this circle pass through the origin?
Show working to support your answer. (4 marks)
Question 2
Write down the centre and radius of each of these circles.
(a) x 2 + y 2 = 36 (2 marks)
2 2
(b) (x − 3) + (y − 4) = 100 (2 marks)
2 2
(c) (x + 5) + y = 3 (2 marks)
Question 3 (non-calculator)
AB is the diameter of a circle.

A is (−3, 6) and B is (5, 12).
Work out the equation of the circle. (5 marks)
3
PQ is a diameter of a circle, centre C.
y Not drawn
accurately
C (1, 2)
P (−1, 1)
O x
(a) Work out the coordinates of Q. (1 mark)
(b) Work out the equation of the circle. (3 marks)
4
LEVEL 2 CERTIFICATE FURTHER MATHEMATICS
A (12, 6) and B (14, 4) are two points on a circle, centre C (20, 12).
y
Not drawn
accurately
C (20, 12)
A (12, 6)
M
B (14, 4)
O x
(a) Work out the coordinates of the midpoint M, of AB. (2 marks)
(b) Show that the length CM = 7 2 (3 marks)
(c) Work out the radius of the circle. (2 marks)
5
Question 6
(0, −2), (0, 12) and (4, 12) are three points on a circle, centre C.
12 (4, 12) Not drawn

accurately
O x
−2
Work out the coordinates of C. (3 marks)
Question 7
AB is a diameter of the circle ABC.
C (4, 6) Not drawn

accurately
A (−2, 3)
B (6, k)
Work out the value of k. (5 marks)
6
Question 8
A circle has equation (x − 5) 2 + (y − 4) 2 = 100
Show that the point (13, −2) lies on the circle. (2 marks)
Question 9
The point (13, −2) lies on the circle (x − a) 2 + (y − 4) 2 = 100
Work out the two possible values of a. (5 marks)
Question 10
A circle passes through the points (0, 3) and (0, 11) and has centre (6, k)
y Not drawn
accurately
11
(6, k)
O x
(a) Work out the value of k.
(b) Hence find the equation of the circle. (5 marks)
7
2 2
The equation of this circle, centre C, is (x − 3) + (y − 5) = 17
P (4, 1) is a point on the circle.
Not drawn
accurately
P (4, 1)
O x
(a) Show working to explain why OP is a tangent to the circle. (5 marks)
(b) Show that the length OP is equal to the radius of the circle. (3 marks)
8
The equation of this circle is x 2 + y 2 = 20

P (4, 2) is a point on the circle.
y
Not drawn
accurately
P (4, 2)
O x
Work out the equation of the tangent to the circle at P.

Give your answer in the form y = mx + c (3 marks)
Question 13
A (–2, 5) and B (4, 13) are points on a circle.
AB is a diameter.
(a) Work out the equation of the circle.

(4 marks)
(b) Work out the equation of the tangent to the circle passing through A.
Give your answer in the form a x + by + c = 0

(4 marks)
9
ΑΒ
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8360)
Worksheet 2
Geometric Problems and Proof
Version 2.0
versions.
You can get further copies of this Teacher Resource from:

The GCSE Mathematics Department
AQA
Devas Street
Manchester
M16 6EX
Or, you can download a copy from our All About Maths website (http://allaboutmaths.aqa.org.uk/).
2 Geometric Problems and Proof
Question 1
SQ is a tangent to the circle at Q.
PR = QR
Not drawn
R accurately
S
Q
Prove that RQ bisects angle PQS. (3 marks)
Question 2
PQRS is a trapezium.
Angle PSR = angle QRS
Not drawn
accurately
P Q
S R
Prove that PQRS is a cyclic quadrilateral. (3 marks)
3
Question 3
p:q:r=4:6:5
R
S Not drawn
r accurately
s
q
Q
Work out s. (5 marks)
Question 4
O is the centre of the circle.
AOBC and EDC are straight lines.
E Not drawn
accurately
y
x 2x
A O C
B
Prove that 4x + y = 90 (4 marks)
4
Question 5
QS bisects both of the angles PSR and PQR.
P Not drawn
accurately
S
y
y
x
x
Q
R
Prove that QS is a diameter of the circle. (4 marks)
Question 6
RSX is a straight line.
XT is a tangent to the circle at T.
SX = ST
Not drawn
accurately
X
T
Prove that triangle RXT is isosceles. (3 marks)
5
Question 7
AB bisects angle OBC.
Not drawn
accurately
A
O
y
x C
x
B
Prove that y = 90 + x (5 marks)
Question 8
RTQ, RSP and PTV are all straight lines.
PT = PQ
Not drawn
Q accurately
V
x
Prove that PTV is a tangent to circle RST at T (5 marks)
6
Question 9
ABF is a common tangent to the two circles at A and B.
CDE is a straight line.
AC is parallel to BD.
E
Not drawn
accurately
D
A x
B
F
Prove that AD is parallel to BE. (5 marks)
7
ΑΒ
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8360)
Worksheet 3
Algebraic Proof
Version 2.0
versions.

AQA
Devas Street
Manchester
M16 6EX
3 Algebraic Proof
Question 1
Prove that 4(p − 3) − 2(2p − 1) is always a negative integer. (2 marks)
Question 2
Prove that 8(y + 3) + 3(2 − y) is a multiple of 5 when y is a positive integer. (3 marks)
Question 3
a is a positive integer.
Prove that 4a 2(2a + 1) − (2a) 2 is a cube number. (3 marks)
Question 4
a and b are positive integers.
a<b
ax + 3a
Prove that <1 x≠–3 (3 marks)
bx + 3b
Question 5
(a) Express x 2 + 6x + 11 in the form (x + a) 2 + b where a and b are integers. (2 marks)
(b) Hence, prove that x 2 + 6x + 11 is always positive. (2 marks)
Question 6
Prove that, for all values of x, x 2 + 2x + 6 > 0 (4 marks)
Question 7
f(x) = (2x + 3) 2 + 8(x + 2) for all values of x.
Prove that there is exactly one value of x for which f(x) = 0 (4 marks)
3
Question 8
1
The nth term of a sequence is n(n + 1)
2
(a) Work out an expression for the (n − 1)th term of the sequence.
Give your answer in its simplest form. (2 marks)
(b) Hence, or otherwise, prove that the sum of any consecutive pair of terms of the
sequence is a square number. (3 marks)
Question 9
x2 − 4 10 x 2
Prove that × is always positive. (5 marks)
5 x − 10 x+2
Question 10
f(n) = n 2 − n
Prove that f(3n) + f(n + 1) = kn(5n − 1) where k is an integer. (3 marks)
4
ΑΒ
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8365)
Worksheet 4
Trigonometry
Version 2.0
versions.
4 Trigonometry
Work out the exact value of sin 60° + sin 120° + sin 270°.
Give your answer in its simplest form. (3 marks)
Are these statements true or false?
True False
sin 37° = sin 127°
cos 54° = cos 306°
sin 135° = cos 135°
tan 126° = tan 306°

(4 marks)
3
Work out the area of triangle ABC.
Write your answer in its simplest form.
Not drawn
C accurately
45°
6 2
5 cm
A (3 marks)
Question 4 (calculator or non-calculator)

1
Show that tan 2 θ ≡ −1 (3 marks)
cos 2 θ
Question 5 (calculator)
AC is a diameter of the circle.
AC = 5 cm, AD = 4 cm
Not drawn
D accurately
A C
B
Work out angle ABD. (4 marks)
4
A hanging basket is made from a hemisphere and three chains.
The radius of the hemisphere is 10 cm.
Each chain is 30 cm long.
The chains are equally spaced around the rim of the hemisphere.
Work out angle AOB.

O
A
B
(5 marks)
Solve the following equation for 0 < θ < 360°.
2
tan θ = 2 (4 marks)
Solve the following equation for 0 < θ < 360°.
2
3cos θ + 2cos θ − 1 = 0 (5 marks)
5
A cuboid has length 30 cm and width 20 cm
A, B and C are midpoints of three of the edges.
B 20 cm
C
30 cm
(a) Work out the distance BC.

(2 marks)
(b) Given that angle ACB = 59° and AB = 22 cm

work out the size of angle CAB.
(3 marks)
6
ΑΒ
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8360)
Worksheet 5
Matrices 1
Version 2.0
versions.

AQA
Devas Street
Manchester
M16 6EX
5 Matrices 1
Question 1
Work out
(a)  4 2 7 (b) 5 0  − 3 (c)  5 − 2

        2  
 − 3 5   1 0 5  − 4 6 − 3
(d)  1 0  3  (e) − 4 7  (f) 8 4  − 3

    6      
 0 1  − 2   − 1 − 3  4 2  6 
(12 marks)
Question 2
Work out
(a)  2 − 1  0 3  (b)  − 3 − 2  − 2 4 (c) 3 2  5 − 2

           
 1 3   1 − 4  − 1 5   3 4 7 5  − 7 3 
(d) 10 − 7   2 4  (e)  1 − 2  2 3  (f)  2 3   1 − 2

          
9 8   − 2 3  3 − 5  1 4  1 4 3 − 5
(12 marks)
Work out, giving your answers as simply as possible.
(a)  2 1   2 0  (b)  1   − 2 4 (c) 3 2

2
  − − 1    
 −1 3 2 − 3 − 2 2   2 
      7 5
 3  − 1 3
 5  
 2   2 
(d) 3 3 − 4   3 1 (e) 1 1 (f)  2

2
     2 3  2 
 2 3 3   − 4 0 3 2    7
      3 
2 1
   1 4
 
3 4
(17 marks)
3
Question 4
(a) −1 0   p  (b) 3 0  x  (c)  0 1  m 

           
 0 − 1  p + 1 0 3  y   1 0   2m 
(d) 2 0  − a 0 (e)  4t 0 3 0 (f) −1 0  1 0   3 

             
0 2  0 a  0 4t  0 3  0 − 1  0 −1  − 2 
(13 marks)
Question 5
(a)  2 x − 3   x 3x  (b)  a 3a   7 8 (c)  x 0

2
         
 − 5 4x   − 3 0   − 2 1   − 10 11 1 x
(d)  y y 2 3y (e) a +1 a  a +1 −a  (f)  3x − 3 

2
         
− 3 x  0 1   a + 2 a + 1  − a − 2 a + 1  − 9 x + 1
(14 marks)
4
ΑΒ
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8360)
Worksheet 6
Matrices 2
Version 2.0
versions.

AQA
Devas Street
Manchester
M16 6EX
6 Matrices 2
Question 1
 2 − 1 7 4 − 2 3 
A =   B =   C =  
3 4  5 3  1 − 1
Work out
(a) AB (b) BC (c) 3A
(d) BA (e) −C (f)  1 − 4

B  
− 5 7 
(12 marks)
Question 2
 − 2 0 − 4 1   3 
P =   Q =   C =  
 5 1  3 − 2  − 2
Work out
2
(a) P (b) QP (c) 5Q
(d) PC (e) IQ (f) 3I
(12 marks)
Question 3
 − 2 a   3   22 
    =  
 − 4 3 7  9 
Work out the value of a. (2 marks)
Question 4
Work out the values of a, b and c.
 2 a   1 3  12 26 
    =  
 3 1   2 b   c 13 
(3 marks)
3
Question 5
 2 3
Work out the image of the point D (−1, 2) after transformation by the matrix  
 − 1 1
(2 marks)
Question 6
2 3
The point A(m, n) is transformed to the point A′ (−2, 0) by the matrix  
 1 1
Work out the values of m and n.
(4 marks)
Question 7
The matrix A represents a reflection in the line y = x.
Write down the matrix A.
The unit square is transformed by the matrix A and then by rotation through −90° about O.
Work out the matrix representing the combined transformation.
(4 marks)
Question 8
 0 − 1
Describe fully the transformation given by the matrix  
−1 0 
(2 marks)
 h 0
The unit square OABC is transformed by the matrix   to the square OA′B′C′.
0 h
The area of OA′B′C′ is 27.
Work out the exact value of h.

(3 marks)
4
Question 10
3 0  − 1 0
A =   and B =  
0 3  0 1
The point P (2, 7) is transformed by matrix BA to P′.
Show that P lies on the line 7x + 2y = 0
(3 marks)
5
ΑΒ
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8360)
Worksheet 7
Inequalities
Version 2.0
versions.

AQA
Devas Street
Manchester
M16 6EX
7 Inequalities
Question 1
−6 < 3x ≤ 6
x is an integer
Write down all the possible values for x. (2 marks)
Question 2
Solve 6x > 24 − 2x (2 marks)
Question 3
Solve 4(2x − 1) < 2 (3 marks)
Question 4
A rhombus and a rectangle are shown.
The perimeter of the rhombus is greater than the perimeter of the rectangle.
Not drawn
accurately
y+4
2y + 6 2y + 10
Show that y>k where k is an integer. (4 marks)
3
Question 5
p<–1 and q>1
Tick the correct box for each statement.
Always true Sometimes true Never true
5p < 0
p2 < 0
p+q>0
q
–1< <0
p
(4 marks)
4
Question 6
B O A x
y = 16 − x 2
(a) Write down the coordinates of points A and B. (2 marks)
(b) Hence, or otherwise, solve 16 − x 2 ≥ 0 (2 marks)
5
Question 7
(a) Factorise x 2 + 3x (1 mark)
(b) Sketch y = x 2 + 3x
Label the x values of the points of intersection with the x-axis. (2 marks)
(c) Hence, or otherwise, solve x 2 + 3x < 0 (2 marks)
Question 8
Solve (x – 5)(x + 2) ≥ 0 (3 marks)
Question 9
Solve x 2 + 4x − 12 < 0 (4 marks)
Question 10
Solve 2x 2 − x − 3 < 0 (4 marks)
Question 11
Solve 3x 2 > 14x − 8 (4 marks)
Question12
A triangle and a square are shown.
4n − 8 n
Work out the range of values of n for which
area of triangle < area of square

(5 marks)
6
ΑΒ
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8365)
Worksheet 8
Functions
Version 2.1
versions.
8 Functions
3
f(x) = 2x − 250
Work out x when f(x) = 0 (3 marks)
Question 2
2
f(x) = x + ax − 8
f(−3) = 13
Question 3
2
f(x) = x + 3x − 10
Show that f(x + 2) = x(x + 7) (3 marks)
Question 4
Work out the range for each of these functions.
2
(a) f(x) = x + 6 for all x (1 mark)
(b) f(x) = 3x − 5 −2 ⩽ x ⩽ 6 (2 marks)

4
(c) f(x) = 3x x < −2 (1 mark)
Question 5
x+2
(a) The function =
x−3
Give a reason why x > 0 is not a suitable domain for f(x). (1 mark)
(b) Give a possible domain for f(x) = x−5 (1 mark)
3
Question 6
f(x) = 3 − 2x a<x<b
The range of f(x) is −5 < f(x) < 5
Work out a and b. (3 marks)
Question 7
2
Here is a sketch of f(x) = x + 6x + a for all x, where a is a constant
O x
The range of f(x) is f(x) … 11
Question 8
(a) Factorise x 2 − 5x − 14 (2 marks)
2
(b) Sketch the function f(x) = x − 5x − 14 for all x.
Label the points of intersection with the x and y axes. (3 marks)
Question 9
2
f(x) = − x 0⩽x<2
−4 2⩽x<3
2x − 10 3⩽x⩽5
Draw the graph of f(x) for values of x from 0 to 5 (3 marks)
4
Question 10
Here is a sketch of y = f(x) for values of x from 0 to 7.
f(x) = 2x 0⩽x<1
3−x 1⩽x<4
x−7
4⩽x⩽7
3
x
B
Show that
area of triangle A : area of triangle B = 3 : 2 (4 marks)
Question 11
x −a
f( x ) = for x > 0, where a is a positive constant.
2
If f −1(3a ) = 306.25 work out the value of a

(4 marks)
Question 12
2x − 1 5
f(x) = g(x) =
4 x +1
Work out fg(x)

ax + b
Give your answer in the form where a, b, c and d are integers.
cx + d
(2 marks)
5
Question 13
y = f(x) is the graph of a function.
dy
= (x – 5)(2x + 1)
dx
Work out the vales of x for which the function is decreasing.

Give your answer as an inequality.
(2 marks)
6
ΑΒ
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8360)
Worksheet 9
Coordinate Geometry - Calculus
Version 2.0
versions.

AQA
Devas Street
Manchester
M16 6EX
9 Coordinate Geometry - Calculus
Question 1
For each of these straight lines, work out
(i) The gradient of the line (1 mark for each part)
(ii) The gradient of the line that is perpendicular to the given line (1 mark for each part)
(iii) The y-intercept of the line (1 mark for each part)
(a) y = 5x − 4 (b) 3y = 9 − 6x (c) 3y − 12 = 2x
(d) 5x − 2y + 15 = 0 (e) x
−
y
=2
4 3
Question 2
For each of these straight line segments, AB, work out
(i) The mid-point of AB (2 marks for each part)
(ii) The gradient of AB (1 mark for each part)
(iii) The length of AB, giving your answer as an integer or a surd (2 marks for each part)
(a) A = (−3, −4) B = (4, 3) (b) A = (−4, 1) B = (1, 5) (c) A = (5, −2) B = (0, 10)
(d) A = (−2, −6) B = (−6, 0) (e) A = (1, 9) B = (9, −6) (f) A = (7, 1) B = (−5, −3)
3
Question 3
In each of these line segments, B lies between A and C.
Work out the coordinates of C in each case. (2 marks for each part)
(a) A = (−1, 3) B = (1, 1) and AB : BC = 1 : 2
(b) A = (−4, −2) B = (2, −5) and AB : BC = 3 : 1
(c) A = (11, 0) B = (1, −5) and AB : BC = 5 : 3
(d) A = (−6, 2) B = (0, 4) and AB : BC = 2 : 3
(e) A = (2, −9) B = (−3, 1) and AB : BC = 5 : 4
Question 4
2
Work out the coordinates of the points of intersection of the curve y = x + 7 and
the straight line y = 5x + 1 (4 marks)
Question 5
Line L has equation y + 3x = 7
Line N is perpendicular to line L and passes through (3, −1).
Work out the equation of line N.

Give your answer in the form y = ax + b (4 marks)
Question 6
dy
Work out for each of the following
dx
(a) y = 7x + 3 (1 mark) (b) y = 8 − 5x + x 2 (2 marks)
(c) y = 3x 3 + 4x (2 marks) (d) y = x 3 − 7x 2 + 10x − 1 (2 marks)
(e) y = 4x(x 2 + 2x − 3) (3 marks) (f) y = (3x − 5)(x + 8) (3 marks)
(g) y = x(7 − x)(6 − 2x) (3 marks) (h) y = (x + 3)(x − 1)(x − 6) (4 marks)
4
Question 7
A curve has equation y = x 3 + x 2 + 2x − 4
Work out the equation of the tangent to this curve where x = −2

Give your answer in the form y = ax + b (5 marks)
Question 8
A curve has equation y = x 3 + 2x 2 − 9x + 3
Work out the equation of the normal to this curve at the point (1, −3)
Give your answer in the form ax + by + c = 0, where a, b and c are integers. (5 marks)
Question 9
A curve has equation y = x 3 − 6x 2 + 20
dy
(a) Write down an expression for (1 mark)
dx
(b) Work out the coordinates of the points at which the gradient is zero and determine whether
they are maximum or minimum. (5 marks)
(c) Sketch the curve on the axes clearly labelling the maximum and minimum points. (2 marks)
y
O x
5
Question 10
3 2
A curve has equation y = x − x + k x − 2
dy
(a) Write down an expression for (1 mark)
dx
(b) The curve has a minimum point at the point where x = 2
(c) Work out the x coordinate of the maximum point on the curve. (3 marks)
Question 11
1 9 1 2
(a) Show that the line y = x− is the tangent to the curve y= x − x
2 4 4
3
at the point A (3, − ). (4 marks)
4
(b) The point B on the curve is such that the tangent at B is perpendicular to the tangent at A,
as shown in the diagram.
y Not drawn
1 2 accurately
y= x −x
4
Work out the coordinates of B. (4 marks)
6
Question 12
dy
Work out for each of the following
dx
(a) y = 3x -2 + 3 (1 mark) (b) y = 5x -1 + 2x 2 (2 marks)
(c) y = 3x -3 – 4x-5 (2 marks) 5+ x (2 marks)

(d) y=
x2
(e) 1 4 (3 marks) 3x + 2x 6 (3 marks)

y= (x + 2x − 4) (f) y=
x 4x3
Question 13
A pentagon is made from a rectangle and an isosceles triangle.
16x
8x y
8x
6x
(a) The perimeter of the pentagon is 84 cm
Show that y = 42 – 18x

(2 marks)
(b) Show that the area, A cm2, of the pentagon is given by
A = 672x – 240x2
(2 marks)
(c) Using calculus, work out the maximum value of A as x varies.

(3 marks)
Question 13
x 8
The curve y= + has a minimum point
4 x2
Work out this minimum value of y.

(4 marks)
7
ΑΒ
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8365)
Worksheet 10
Factor Theorem
Version 2.0
versions.
10 Factor Theorem
Question 1
(a) Show that x(x + 4)(x − 9) = x 3 − 5x 2 − 36x (1 mark)
(b) Write down the x values of the three points where the graph of y = x 3 − 5x 2 − 36x
crosses the x-axis. (2 marks)
Question 2
f(x) = x 3 + 2x 2 − 5x − 6
(a) Work out f(1) and f(−1) (2 marks)
(b) Work out f(2) and f(−2) (2 marks)
(c) Work out f(3) and f(−3) (2 marks)
(d) Write down the three linear factors of f(x). (1 mark)
Question 3
(a) Show that (x + 5) is a factor of x 3 + 7x 2 + 2x − 40 (2 marks)
(b) Work out the other two linear factors of x 3 + 7x 2 + 2x − 40 (3 marks)
(c) Hence, solve x 3 + 7x 2 + 2x − 40 = 0 (1 mark)
3
Question 4
A sketch of y = x 3 + 5x 2 + 9x + k where k is an integer, is shown.
−2 x
Question 5
(a) (x + 3) is a factor of f(x) = x 3 + x 2 + ax − 72 where a is an integer.
(b) Work out the other linear factors of f(x). (3 marks)
Question 6
(x − 3) and (x + 4) are factors of f(x) = x 3 + ax 2 + bx + 24 where a and b are integers.
(a) Work out the third linear factor of f(x). (2 marks)
(b) Work out the values of a and b. (4 marks)
Question 7
(a) (x − 5) is a factor of f(x) = x 3 + kx 2 + 9x − 20 where k is an integer.
(b) Express f(x) as a product of (x − 5) and a quadratic factor. (2 marks)
(c) Show that (x − 5) is the only linear factor of f(x). (2 marks)
4
Question 8
Solve x 3 − 6x 2 − 25x − 18 = 0 (5 marks)
Question 9
f(x) = x 5 − 2x 4 − 81x + 162 = 0
(a) Use the factor theorem to show that f(x) has a factor of (x – 2)
(1 mark)
(b) Hence work out the integer solutions of f(x) = 0
(4 marks)
Question 10
(a) Use the factor theorem to show that (3x + 2) is a factor of 3x3 + 2x2 – 3x – 2
(2 marks)
(b) Factorise fully 3x + 2x – 3x – 2
3 2
(2 marks)
5
ΑΒ
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8360)
Worksheet 11
Sequences
Version 2.0
versions.

AQA
Devas Street
Manchester
M16 6EX
11 Sequences
Question 1
A linear sequence starts
250 246 242 238 ….…
Which term is the first to have a negative value? (4 marks)
Question 2
Work out the nth term of this quadratic sequence.
8 9 14 23 36 ….…
(4 marks)
Question 3
(a) Show that the nth term of the quadratic sequence
4 10 18 28 …… is n2 + 3 n
(3 marks)
(b) Hence, write down the nth term of these quadratic sequences.
(b) (i) 5 11 19 29 ……
(1 mark)
(b) (ii) 5 12 21 32 ……
(1 mark)
3
Question 4 (non calculator)
(a) Write down the nth term of the linear sequence
4 7 10 13 ……
(1 mark)
(b) Hence, write down the nth term of the quadratic sequence.
16 49 100 169 ……
(1 mark)
(c) For the sequence in part 4(b), show that the 30th term is equal to the product
of the 2nd and 4th terms (3 marks)
Question 5
6 cm
4 cm 5 cm
3 cm 4 cm 5 cm
This pattern of rectangles continues.

Show that the sequence of numbers formed by the areas of these rectangles has nth term
n 2 + 5n + 6 (4 marks)
Question 6
A linear sequence starts
a+b a + 3b a + 5b a + 7b …………..
The 5th and 8th terms have values 35 and 59.
(a) Work out a and b. (4 marks)
(b) Work out the nth term of the sequence. (2 marks)
4
Question 7
3n + 1
A sequence has nth term
n
1
(a) Show that the difference between the nth and (n + 1)th terms is (3 marks)
n ( n + 1)
(b) Which are the first two consecutive terms with a difference less than 0.01? (2 marks)
(c) Write down the limiting value of the sequence as n  ∞ (1 mark)
Question 8
5n + 2
A sequence has nth term
2n
Show that the limiting value of the sequence, S, as n  ∞ is 2.5 (2 marks)
Question 9
Here is the sequence of odd numbers
1 3 5 7 9 ……
A quadratic sequence is formed by multiplying consecutive odd numbers in successive pairs.
3 15 35 63 ……
Work out the nth term of this sequence. (3 marks)
Question 10
2n 2 − 1
The nth term of a sequence is
3n 2 + 2
3
(a) Show that the difference between the first two terms is (3 marks)
10
(b) Write down the limiting value of the sequence as n  ∞ (1 mark)
5
ΑΒ
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8360)
Worksheet 12
Algebraic Problems – including ratio
Version 2.0
versions.

AQA
Devas Street
Manchester
M16 6EX
12 Algebraic Problems - including ratio
Note
x 4
• If x : y = 4 : 7, then =
y 7
• If, in a problem, two numbers are in the ratio 4 : 7, use 4x and 7x as the numbers
(usually leading to a linear equation); otherwise, use x and y as the numbers
(which will lead to simultaneous equations).
• If x : y = 4 : 7, what is x + 2y : 3x?
Think in terms of ‘parts’, ie 4 parts and 7 parts, so x + 2y : 3x = 4 + 14 : 12
= 18 : 12
= 3:2
Question 1
2n − 1
Work out the possible values of if n 2 = 16
3n + 2
Give your answers as fractions in their simplest form. (4 marks)
Question 2
x:y=6:5
(a) Express x in terms of y. (2 marks)
(b) Show that x + 3y : 2x − y = 3 : 1 (2 marks)
3
Question 3
A point P divides XY in the ratio 3 : 7
Not drawn
accurately
Y (6a, 11b)
X (a, b)
Work out the coordinates of P, in terms of a and b. (3 marks)
Question 4
Here is a linear sequence
a+b a + 3b a + 5b a + 7b ……..
Given that
• 2nd term : 4th term = 2 : 5
• 1st term = − 4
Work out a and b. (5 marks)
Question 5
You are given that ab + a = 5 and a:b=4:3
Work out the possible pairs of values of a and b. (7 marks)
Question 6
The sum of the ages of two people is 90 years.
Six years ago, their ages were in the ratio 8 : 5
How old are they now?

Do not use trial and improvement.
You must show your working. (5 marks)
4
Question 7
Not drawn
y accurately
x
Given that x:y=4:5
Work out the value of y.

Do not use trial and improvement.
You must show your working. (7 marks)
Question 8
A rectangular picture is surrounded by a frame of constant width.
All measurements are in centimetres.
a
Not drawn
accurately
7x
9 3x b
Given that a:b=3:2
Work out x. (5 marks)
5
Question 9
If x : y = 3 : 5 and y : z = 10 : 9
Find, in its simplest form
(a) x:z (3 marks)
(b) 10x : 7y (2 marks)
(c) x+y:y (2 marks)
Question 10
A cuboid has dimensions 2n, n and n − 1 cm.
A diagonal has length 2n + 1 cm.
Not drawn
accurately
n−1
2n + 1
2n
Work out n. (6 marks)
6
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8365)
Mark Scheme
Miscellaneous
Version 1.0
Our specification is published on our website (www.aqa.org.uk). We will let centres know in
writing about any changes to the specification. We will also publish changes on our website.
The definitive version of our specification will always be the one on our website, this may differ
from printed versions.
You can download a copy of this resource from our All About Maths website
centres for AQA are permitted to copy material from this specification booklet for their own
internal use.
Glossary for Mark Schemes
These examinations are marked in such a way as to award positive achievement wherever possible. Thus,
for these papers, marks are awarded under various categories.
M Method marks are awarded for a correct method which could lead
to a correct answer.
A Accuracy marks are awarded when following on from a correct

method. It is not necessary to always see the method. This can be
implied.
B Marks awarded independent of method.
M Dep A method mark dependent on a previous method mark being

awarded.
B Dep A mark that can only be awarded if a previous independent mark

has been awarded.
ft Follow through marks. Marks awarded following a mistake in an

earlier step.
SC Special case. Marks awarded within the scheme for a common

misinterpretation which has some mathematical worth.
oe Or equivalent. Accept answers that are equivalent.

1
eg, accept 0.5 as well as
2
3
M Miscellaneous
Question Answer Mark Comments
1 (x = 0, y = 8 ⇒) a = 8 M1
1 = 8 × b –3 M1
b=2 A1
y = 2 3 × 2x A1 must see both lines
x+3
=2
2 3×?×?×? M1
3×4×3×2 or 72 M1
5×4×3×2 or 120 M1 oe
192 A1
3 LHS or LHS numerator M1 Allow one error
6n2 + 17n + 12 +30n
RHS 9n2 + 12n M1
0 = 3n2 – 35n – 12 M1 oe
(3n + 1)(n – 12) M1
12 A1
4
Level 2 Certificate in Further Mathematics: Worksheet 1 – Mark Scheme
Coordinate Geometry – Circles LEVEL 2 CERTIFICATE FURTHER MATHEMATICS
4 4 x5 − 2 x3 − 2 x3 + x M1 At least two terms correct
4 x5 − 4 x3 + x = x + 4 x5 + 108 M1
4 x3 = −108 M1 oe
–3 A1
5 4a3 .5 x and 6a 2 .52 x2 M2 M1 for each
4a3 .5= 3 × 6a 2 .52 M1
22.5 A1
5
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8365)
Mark Scheme
Worksheet 1
Coordinate Geometry Circles
Version 2.0
You can download this resource from our All About Maths website
internal use.

implied.

awarded.

has been awarded.

earlier step.


1
2
3
1 Coordinate Geometry - Circles
1(a) x 2 + (y − 3) 2 = 4 B2 B1 LHS, B1 RHS
2 2
1(b) (x − 1) + (y + 5) = 16 B2 B1 LHS, B1 RHS
2 2
1(c) (x + 3) + (y − 4) = 7 B2 B1 LHS, B1 RHS
2 2
1(d) (x − 8) + (y − 15) = 289 B2 B1 LHS, B1 RHS
2 2
(−8) + (−15) M1 oe
64 + 225 = 289, Yes A1
2(a) (r) = 6 (centre =) (0, 0) B2 B1 For each
2(b) (r) = 10 (centre =) (3, 4) B2 B1 For each
2(c) (r ) = 3 (centre =) (−5, 0) B2 B1 For each
3 −3 + 5 6 + 12 M1
or
2 2
(1, 9) A1
(5 − 1) 2 + (12 − 9) 2 M1 oe
ft Their centre
5 A1
2 2
(x − 1) + (y − 9) = 25 A1 ft ft Their centre and radius
4(a) (3, 3) B1
4(b) 22 + 12 M1 oe
5 A1
2 2
(x − 1) + (y − 2) = 5 B1 ft ft Their radius
4
5(a) 12 + 14 6+4 M1
or
2 2
(13, 5) A1
5(b) (20 − 13)2 + (12 − 5)2 M1 ft Their M
98 A1 72 + 72
49 × 2 = 7 2 A1 72 (1 + 1) = 7 2
5(c) (20 − 12)2 + (12 − 6)2 M1 oe
10 A1
6 − 2 + 12 M1
2
0+4 M1
2
C (2, 5) A1
7 6−3 M1 oe
Gradient AC =
4 − −2
3  1 A1 oe
= = 
6  2
Gradient BC = −2 B1 ft
6−k M1
= −2
4−6
k=2 A1
2 2
8 (13 − 5) + (−2 − 4) M1
64 + 36 = 100 A1
5
Coordinate Geometry – Circles

2 2
9 (13 − a) + (−2 − 4) = 100 M1
2
169 − 13a − 13a + a + 36 (= 100) M1 Allow 1 error
a 2 − 26a + 105 = 0 A1
(a − 5)(a − 21) = 0 M1
a = 5 and a = 21 A1
10(a) 3 + 11 M1 oe eg, 3 + 4
2
k=7 A1
10(b) 6 2 + (7 − 3) 2 M1 oe
ft Their k
52 A1
2 2
(x − 6) + (y − 7) = 52 A1 ft ft Their k and their radius
11(a) C is (3, 5) B1
5 −1 M1
Gradient CP =
3−4
−4 A1
1 B1
Gradient OP =
4
1 A1
−4 × = −1
4
So perpendicular (ie, tangent)
11(b) r = 17 B1
OP = 4 2 + 12 M1
= 17 A1
6
12 2  1 B1
Gradient OP = = 
4  2
Gradient of tangent = −2 B1 ft
y − 2 = −2 (x − 4) M1
y = −2x + 10 A1
13(a) Centre (1, 9) B2 B1 for each coordinate
2
r= 32 + 42 or 2
d= 62 + 82 M1
( x − 1)2 + ( y − 9)2 =
25 A1ft ft their centre
13 − 5
13(b) Grad AB = or using their M1
4+2
8 4
centre with A or B; or or
6 3
3
Grad tangent − or – their grad AB M1
4
3 M1
y −=
5 their − ( x + 2)
4
3 x + 4 y − 14 =
0 A1
7
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8360)
Mark Scheme
Worksheet 2
Geometric Problems and Proof
Version 2.0

AQA
Devas Street
Manchester
M16 6EX
internal use.

implied.

awarded.

has been awarded.

earlier step.


1
2
3
2 Geometric Problems and Proof
1 Let angle SQR = x M1 Any order of angles
∴ angle RPQ = x alternate segment
∴ angle RQP = x isosceles triangle M1
∴ ∠ RQS = ∠ RQP A1 SC2 ‘Correct’ solution without reasons
2 Let angle PSR = x = angle QRS M1 ∠ PQR = 180 − x

∴ ∠ SPQ = 180 − x
Allied angles on parallel lines
∴ ∠ SPQ + ∠ QRS = 180 A1 ∠ PSR + ∠ PQR = 180
PQRS is a cyclic quadrilateral A1 SC2 ‘Correct’ solution without reasons

(converse of) opposite angles add up
to 180°
3 p + r = 180 M1
4x + 5x = 180 M1 oe
(9x = 180) A1
x = 20
6x = 120 M1 ft Their x
s = 60 A1 ft ft Their x
4
4 ∠ BED = x M1
angles in same segment
∠ AEB = 90° A1
angle in semicircle = 90°
In Δ ACE A1
y + x + 2x + x + 90 = 180
angle sum of a triangle = 180
y + 4x = 180 − 90 A1 SC2 ‘Correct’ solution without reasons

= 90
5 2x + 2y = 180 M1
opposite angles of a cyclic
quadrilateral = 180
x + y = 90 A1
∴ ∠ QPS = 90 A1
angle sum of triangle = 180
QS is diameter A1 SC2 ‘Correct’ solution without reasons

(converse of) angle in a semicircle
= 90)
6 Let ∠ SXT = x M1
∴ ∠ STX = x isosceles triangle
∴ ∠ SRT = x alternate segment M1
∴ triangle RXT = is isosceles - A1 SC2 ‘Correct’ solution without reasons

2 base angles equal
5
7 ∠ OAB = x isosceles triangle M1
∠ BOA = 180 − 2x M1
angle sum of triangle = 180
∧
Reflex BOA = 360 − (180 − 2x) M1
(Angles at a point = 360) = 180 + 2x A1
y = 90 + x A1 SC3 ‘Correct’ solution without reasons

Angle at centre = 2 × angle at
circumference
8 ∠ QTP = x isosceles triangle M1
∠ VTR = x vertically opposite angles M1

equal
∠ TQP = x = ∠ RST exterior angle of M1 oe

cyclic quadrilateral = opposite interior
angle
∴ ∠ VTR = ∠ RST A2 SC3 ‘Correct’ solution without reasons

PVT is tangent
(converse of) alternate segment
theorem
9 ∠ EDB = x alternate segment M1
∴ ∠ DCA = x corresponding angles M1

equal
∴ ∠ DAB = x alternate segment M1
ie, ∠ DAB = ∠ EBF A2 SC3 ‘Correct’ solution without reasons

∴ AD is parallel to BE
(converse of) corresponding angles
equal
6
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8360)
Mark Scheme
Worksheet 3
Algebraic Proof
Version 2.0

AQA
Devas Street
Manchester
M16 6EX
internal use.

implied.

awarded.

has been awarded.

earlier step.


1
2
3
3 Algebraic Proof
1 4p − 12 − 4p + 2 M1 4 terms with 3 correct
− 10 A1
2 8y + 24 + 6 − 3y or 5y + 30 M2 M1 4 terms with 3 correct
5y + 30 and 5(y + 6) A1 oe eg, 5y + 30 and states both terms

divisible by 5
3 2 2 3
3 8 a + 4a − 4a or 8a M2 M1 3 terms with 2 correct
3 3 3
8a and (2a) A1 oe eg, 8a and states that 8 is a cube
number
4 a(x + 3) or b(x + 3) M1
a ( x + 3) A1
and cancelling seen
b ( x + 3)
a A1 oe
and explains that as numerator is
b
smaller than denominator value will
be < 1
5(a) a=3 B1
2
b=2 B1 ft ft 11 − their a
2
5(b) (x + 3) ≥ 0 M1 oe Allow their a
Adding 2 means always positive A1 Must have a = 3 and b = 2
4

2
6 (x + 1) B1
2 2
(x + 1) + 5 B1 ft ft Their (x + 1)
2
(x + 1) ≥ 0 M1 oe Allow their 1
2
Adding 5 means always positive A1 Must have (x + 1) + 5
2
7 4x + 6x + 6x + 9 + 8x + 16 or M2 M1 Allow one error in expansions
2
4x + 20x + 25
2 2
4x + 20x + 25 and (2x + 5) A1 oe
2
eg, 4x + 20x + 25 and (2x + 5)(2x + 5)
Explains that only solution is A1 oe

(x = ) − 2.5 eg, explains that because the brackets are
the same there is exactly one solution
8(a) 1
(n − 1)(n − 1 + 1) M1
2
1 A1 1 2
n(n − 1) oe eg, n − 1n
2 2 2
8(b) 1
n(n + 1) + 1
n(n− 1) M1 1
n(n + 1) + their (a)
2 2 2
1 2 1 1 2 M1 Expands brackets
n + n+ n − 1 n
2 2 2 2 ft Their (a)
n2 A1
Alt 8(b) 1 1 M1 oe
n(n + 1) + (n + 1)(n + 1 + 1)
2 2
1 2 1 1 2 1 M1 Expands brackets
n + n + n +n+ n +1
2 2 2 2
oe eg, n 2 + 2n + 1
1
ft Their (n + 1)(n + 1 + 1)
2
2
(n + 1) A1
5
9 ( x + 2)( x − 2) M2 M1 For either numerator or denominator

5( x − 2) factorised correctly
At least one correct cancellation in M1

the product
2
2x A1 10 x 2
oe eg,
5
Explains that 2 > 0 and x2 ≥ 0 so 2x2 A1 oe eg, Explains that 10 > 0 and
always positive
2 10 x 2
5 > 0 and x ≥ 0 so always positive
5
2 2 2 2
10 (3n) − 3n + {(n + 1) − (n + 1)} M1 oe 9n − 3n or n + n + n + 1 − n − 1
2 2 2
9n − 3n + n + n + n + 1 − n − 1 A1 oe eg, 10n − 2n
2 2
10n − 2n and 2n(5n − 1) A1 oe eg 10n − 2n and k = 2
6
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8365)
Mark Scheme
Worksheet 4
Trigonometry
Version 2.0
internal use.

implied.

awarded.

has been awarded.

earlier step.


1
2
3
4 Trigonometry
1 √3/2 + √3/2 − 1 M1 Any 2 values correctly stated in surd form
√3/2 + √3/2 − 1 M1 All 3 values correctly stated in surd form
√3 − 1 A1
2 False A1
True A1
False A1
True A1
3 Evidence that sin 45° = 1/√2 B1
1 M1
Area = × 5 × 6√2 × sin 45°
2
15 A1
4 sin θ M1
tan θ ≡ seen
cos θ
sin2 θ 1 − cos 2 θ M1
≡
cos θ
2
cos 2 θ
1 A1 Accurate method with clear steps is

tan θ ≡ −1 required for all 3 marks
cos 2 θ
Alt 4 1 − cos 2 θ M1 oe
cos 2 θ
sin2 θ M1
cos 2 θ
2
tan θ A1 Accurate method with clear steps is
required for all 3 marks
4
5 Evidence that angle ADC is a right M1

angle
4 M1
sin ACD =
5
ACD = [53.1, 53.13010235] A1 Allow 53 with method seen
Angle ABD = [53.1, 53.13010235] B1 ft ft From 3rd mark their angle ACD
6 A triangle formed with A, B and the M1

centre of the hemisphere with 2
sides of 10 cm and an angle of 120°
2 2 2
(AB =) 10 + 10 − 2 × 10 × 10 × M1 2 × 10 × sin 60
cos 120
(AB =) [17.3, 17.321] A1 oe eg, 300

-1
30 2 + 30 2 − their AB 2 M1 2 × sin (0.5 their AB ÷ 30)
(cos AOB =)
2 × 30 × 30
[33.557, 33.6] A1 ft ft Their AB

Accept 34 with correct method seen
7 tan θ = +√2 or tan θ = −√2 M1
[54.7,54.74] or [125.26,125.3] A1
180 + their [54.7,54.74] or M1

180 + their [125.26,125.3]
[54.7,54.74] and [125.26,125.3] and A1ft All 4 solutions

180 + their [54.7,54.74] and [54.7,54.74] and [125.26,125.3] must be
180 + their [125.26,125.3] correct
ft For other two solutions
5
8 (3cos θ − 1)( cos θ + 1) M2 M2 Fully correct use of quadratic formula

M1 (acos θ + b)(ccos θ + d) where ac = 3
and bd = ± 1 or
quadratic formula with one sign error
cos θ = −1 so θ = 180° A1
1 A1
cos θ = so θ = [70.5, 70.53]
3
θ = 289.50 A1 ft ft 360 − their [70.5, 70.53]
9 (a) 2 2 M1
 20   30 
  +  
 2   2 
325 or 5 13 or 18.0(2…) or 18.03 A1
(b) sin CAB sin 59 M1 oe

=
their 18.03 22
-1  sin 59 
sin  × their 18.03  M1dep
 22 
44.6… A1
6
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FURTHER MATHEMATICS
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Worksheet 5
Matrices 1
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5 Matrices 1
Question 1
Each question 2 marks. M1 for a correct row by column multiplication. A1 for the correct answer.
(a)  30  (b)  − 15  (c) 10 − 4 

     
 − 16   − 20  12 − 6 
(d)  3  (e)  − 24 42  (f) 0

     0 
 − 2  − 6 − 18   
Question 2
(a)  − 1 10  (b)  0 − 20  (c)  1 0

     
 3 − 9 17 16   0 1
(d)  34 19  (e) 0 − 5  (f)  11 − 19 

     
 2 60   1 − 11 13 − 22 
Question 3 (Non-calculator)
3 marks per question. 1 mark for multiplication of row by column, 1 mark for 2 simplified elements,
1 for other 2 elements correct. Part (c) 2 marks.
(a)  −1 − 2 2  (b)  3  (c)  23 16 

  − 5  
 − 10 2 − 12   2   56 39 

 11 
− 21 
 2 
(d)  25 3 3  (e) 7  (f)  16 2 2 + 2 3 

  3 
 − 10 3 2  6  7 2 + 7 3 17 
  
 19 
 3
 12 
4
LEVEL 2 CERTIFICATE FURTHERGeometry
Coordinate MATHEMATICS
– Circles
Question 4
(f) 3 marks. 2 for 1 pair correctly multiplied, 1 for final answer.
(a)  −p  (b)  3x  (c)  2m 

     
 − p − 1 3y  m
(d)  − 2a 0  (e) 12t 0  (f)  − 1 0  3   − 3

 
2a 
   0 1  − 2  =  − 2 
 0  0 12t      
Question 5
(a) to (d) 2 marks each
(e) and (f) 3 marks each, 1 for a correct multiplication, 1 for two elements correct, 1 for all correct.
(a)  2x 2 + 9 6x 2  (b)  − 23a 41a  (c)  x2 0

     
 − 17 x − 15 x   − 24 − 5   2x x 2 
  
(d)  2y 3y2 + y  (e)  1 0 (f)  9 x 2 + 27 − 12 x − 3 

     
− 6 − 9y + x  − 36 x − 9 x 2 + 2 x + 28 
   0 1  
5
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Matrices 2
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6 Matrices 2
Question 1
(a) 9 5 (b)  − 10 17  (c) 6 − 3

     
 41 24   − 7 12   9 12 
(d)  26 9  (e)  2 − 3 (f)  − 13 0 

     
 19 7  −1 1   − 10 1 
Question 2
(a)  4 0 (b)  13 1  (c)  − 20 5 

     
 − 5 1  − 16 − 2   15 − 10 
(d)  − 6 (e) − 4 1  (f) 3 0

   3 − 2   
 13    0 3
Question 3
−6 + 7a = 22 M1
a=4 A1
Question 4
Work out the values of a, b and c.
 2 + 2a 6 + ab  12 26 
  = 
 
 5 9+b   c 13 
a = 5, b = 4, c = 5 B1, B1, B1
4
LEVEL 2 CERTIFICATE FURTHERGeometry
Coordinate MATHEMATICS
– Circles
Question 5
 4
(4, 3) B2 (B1 for (4, ?), (?, 3) or   .
3
Question 6
2m + 3n = −2, m + n = 0 M1 for either, A1 for both
Attempt to solve M1
m = 2, n = −2 A1
Question 7
 0 1
A =   B1
 1 0
 0 1
Rotation   B1
 −1 0
 0 1  0 1  1 0 
Combined     =   M1 Multiplication in correct order.
 − 1 0   1 0   0 −1
1 0 
A1  
 0 − 1
Question 8
Reflection, in the line y = −x B1, B1
Question 9 (Non-calculator)
Vertices of image A′ (h, 0) B′ (h, h) C′(0, h) Any 1 correct B1
Area of OA′B′C′. = h2 M1
h = 3√3 A1
5
Question 10
 − 3 0
BA =   B1
 0 3
 − 3 0  2  − 6
    =   B1
 0 3   7   21 
Show this satisfies 7x + 2y = 0 M1
6
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Inequalities
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7 Inequalities
1 −2 < x ≤ 2 M1
−1 0 1 2 A2 A1 3 correct with none incorrect

or 4 correct with one incorrect
2 6x + 2x > 24 M1 oe
x>3 A1
3 8x − 4 < 2 M1 oe
2
2x − 1 < oe
4
8x < 2 + 4 M1 oe
2
2x < + 1 oe
4
3 A1
x<
4
4 4(2y + 6) > 2y + 10 + 2y + 10 + y + 4 M2 oe eg, 8y + 24 > 6y + 28

+y+4 M1 4(2y + 6) or
2y + 10 + 2y + 10 + y + 4 + y + 4
8y − 6y > 28 − 24 M1 oe
y > 2 or k = 2 A1
5 Always B4 B1 For each correct part

Never
Sometimes
Sometimes
4
6(a) (4, 0) B1
(−4, 0) B1 SC1 4 and −4 seen
6(b) −4 ≤ x ≤ 4 B2 ft ft Their 4 and their −4

B1 ft − 4 < x < 4
Alt 6(b) (4 + x) (4 − x) and −4 and 4 M1
−4 ≤ x ≤ 4 A1
7(a) x(x + 3) B1
7(b) U-shaped parabola M1
0 and − 3 labelled on x-axis A1 ft ft Their factors in (a)
7(c) x < − 3 and x > 0 B2 ft ft Their factors in (a)

B1 ft x ≤ − 3 and x ≥ 0
8 5 and −2 B1
Sketch of graph M1 Sign diagram using their 5 and their −2

y = (x − 5)(x + 2)
x < −2 and x > 5 A1
9 (x + 6)(x − 2) M1 (x + a)(x + b) where ab = ± 12 or

a+b=±4
−6 and 2 A1
Sketch of graph M1 Sign diagram using their −6 and their 2

y = (x + 6)(x − 2)
−6 < x < 2 A1
5
10 (2x − 3)(x + 1) M1 (2x + a)(x + b) where ab = ± 3 or

a + 2b = ± 1
3 A1 oe
and −1
2
Sketch of graph M1 3
Sign diagram using their and their −1
y = (2x − 3)(x + 1) 2
3 A1
−1 < x <
2
11 (3x − 2)(x − 4) M1 (3x + a)(x + b) where ab = ± 8 or

a + 3b = ± 14
2 A1
and 4
3
Sketch of graph M1 2
Sign diagram using their and their 4
y = (3x − 2)(x − 4) 3
2 A1
x< and x>4
3
12 n2 > 1
(4n − 8)n
M1 oe
2
2
0 > n − 4n A1
n(n − 4) M1 Factorises their quadratic expression
Sketch of graph of M1 Sign diagram using their 0 and their 4

y = n(n − 4)
0<n<4 A1
6
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Functions
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1
2
3
8 Functions
3
1 2x − 250 = 0 M1
250 M1 oe
x3 =
2
x=5 A1
2
2 (−3) + a (− 3) − 8 = 13 M1
9 − 8 − 13 = 3a M1 oe Allow 1 error
a = −4 A1
2
3 (x + 2) + 3(x + 2) − 10 M1
x 2 + 2x + 2x + 4 + 3x + 6 − 10 M1 oe Allow 1 error
x 2 + 7x A1
= x(x + 7)
4(a) f(x) … 6 B1
4(b) −11⩽ f(x) ⩽ 13 B1 B1 For −11 or 13 seen
4(c) f(x) > 48 B1
5(a) Not defined when x = 3 B1 oe

or cannot divide by 0 when x = 3
5(b) x … a where a … 5 B1 eg x … 5
or x>6
x > a where a … 5
Allow list of x values if all are … 5
4
6 Either 3 − 2x = −5 M1
or 3 − 2x = 5
a = −1 A1
b=4 A1 SC2 a = 4, b = −1
7 Attempt to complete the square in M1

2
the form (x + 3)
2 A1 oe
(x + 3) − 9 + a
a = 20 A1
8(a) (x + a)(x + b) M1 ab = −14 or a + b = −5
(x − 7)(x + 2) A1
8(b) y B3 B1 Curve through their (7, 0) and (−2, 0)

(from 8(a))
B1 Curve through (0, −14)
B1 Smooth U shape
−2 O 7 x
−14
5
9 y B3 B1 For each part
O 1 2 3 4 5
x
−4
10 (3, 0) and (7, 0) marked or used M1
(1, 2) and (4, −1) marked or used M1
Either of their triangular areas M1

calculated correctly
1 1 A1
× 3 × 2 and ×4×1
2 2
=3:2
11 2y + a = x M1 oe
f −1(=
x ) (2 x + a )2 M1 oe
f −1(3a ) = (7a )2 or 49a 2 = 306.25 M1 oe
2.5 A1
12  5  M1
2  −1
 x + 1
4
9−x −x + 9 A1
or
4x + 4 4x + 4
13 1 B2 B1 for stating must be negative or correct

− < x<5
2 inequality with one or two ≤ symbols in
place of <
6
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Coordinate Geometry - Calculus
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9 Coordinate Geometry - Calculus
1(a) 5 B1
1 B1 ft −1
− ft
5 their 5
−4 B1
1(b) −2 B1
1 B1 ft −1
ft
2 their − 2
3 B1
1(c) 2 B1
3
3 B1 ft −1
− ft
2 2
their
3
4 B1
1(d) 5 B1
2
2 B1 ft −1
− ft
5 5
their
2
15 B1
2
1(e) 3 B1
4
4 B1 ft −1
− ft
3 3
their
4
−6 B1
4
2(a) 1 1 B2 B1 For each coordinate

 ,− 
 2 2 
1 B1
2 2
√(7 + 7 ) M1
√98 or 7√2 A1
2(b) 1 B2 B1 For each coordinate

(−1 , 3)
2
4 B1
5
2 2
√(5 + 4 ) M1
√41 A1
2(c) 1 B2 B1 For each coordinate

(2 , 4)
2
12 B1
−
5
2 2
√(5 + 12 ) M1
13 A1
2(d) (−4, −3) B2 B1 For each coordinate
3 B1
−
2
2 2
√(4 + 6 ) M1
√52 or 2√13 A1
2(e) 1 B2 B1 For each coordinate

(5, 1 )
2
15 B1
−
8
2 2
√(8 + 15 ) M1
17 A1
5
2(f) (1, −1) B2 B1 For each coordinate
1 B1
3
2 2
√(12 + 4 ) M1
√160 or 4√10 A1
3(a) (5, −3) B2 B1 For each coordinate
3(b) (4, −6) B2 B1 For each coordinate
3(c) (−5, −8) B2 B1 For each coordinate
3(d) (9, 7) B2 B1 For each coordinate
3(e) (−7, 9) B2 B1 For each coordinate
4 x 2 + 7 = 5x + 1 M1
or
x 2 − 5x + 6 = 0
(x − 2)(x − 3) = 0 M1 Attempt to factorise the quadratic
(2, 11) or (3, 16) A1 ft ft Their factors
(2, 11) and (3, 16) A1
5 Gradient of L = −3 B1
1 M1
Gradient of N =
3
1 M1
y − (−1) = (x − 3)
3
1 A1
y= x−2
3
6
6(a) dy B1
=7
dx
6(b) dy B2 B1 For each term

= 2x − 5
dx
6(c) dy B2 B1 For each term

= 9x 2 + 4
dx
6(d) dy B2 B1 For two terms correct

= 3x 2 − 14x + 10
dx
6(e) y = 4x 3 + 8x 2 − 12x B1
dy B2 ft B1 For two terms correct

= 12x 2 + 16x − 12
dx
ft Their y = ....
6(f) y = 3x 2 + 19x − 40 B1
dy B2 ft B1 For one term correct

= 6x + 19
dx
ft Their y = ....
6(g) y = 42x − 20x 2 + 2x 3 B1

= 42 − 40x + 6x 2
dx
ft Their y = ….
6(h) y = x 3 − 4x 2 − 15x + 18 B2 B1 For four terms, three of which are

correct

= 3x 2 − 8x − 15x
dx
ft Their y = ....
7 dy M1
= 3x 2 + 2x + 2
dx
(when x = −2) gradient tgt = 10 A1
(when x = −2) y = −12 B1
y − (−12) = 10(x − (−2)) M1 oe
y = 10x + 8 A1 ft ft Their m and c
7
8 dy M1
= 3x 2 + 4x − 9
dx
(when x = 1) gradient tgt = −2 A1
1 A1 ft ft Their −2
(when x = 1) gradient nl =
2
1 M1 oe
y − (−3) = (x − 1)
2
x − 2y − 7 = 0 A1ft ft Their m and c
9(a) dy M1
= 3x 2 − 12x
dx
2
9(b) 3x − 12x = 0 or 3x(x − 4) = 0 M1
x = 0 and x = 4 A1
(0, 20) and (4, −12) A1
dy M1
Testing the sign of for values of
dx
x either side of 0 and 4
Maximum at (0, 20) A1 If previous M1 earned

Minimum at (4, −12)
9(c) y B2 B1 For correct general shape
0, 20 B1 ft For labelling the stationary points
4, −12
8
10(a) dy B1
= 3x 2 − 2x + k
dx
2
10(b) 3(2) − 2(2) + k = 0 M1
k = −8 A1
10(c) 2 M1
3x − 2x − 8 = 0
(3x + 4)(x − 2) = 0 A1
4 A1
Maximum at x = −
3
11(a) dy 1 M1
= x−1
dx 2
dy 3 1 A1
(when x = 3) = −1 =
dx 2 2
3 1 M1
y − (− ) = (x − 3)
4 2
1 1 3 A1 1 9
y= x−1 − Clearly shown since y = x − answer
2 2 4 2 4
given
11(b) Gradient tangent at B = −2 B1
1 M1
x − 1 = −2
2
x = −2 A1 ft ft Their tangent gradient
B = (−2, 3) A1
-3
12(a) –6x B1
-2
12(b) –5x + 4x B2 B1 for each term
-4 -6
12(c) –9x + 20x B2 B1 for each term
-3 -2
12(d) –10x –x B2 B1 for each term
12(e) x3 + 2 – 4x-1 B1
2 -2 B2ft B1ft for each term
3x + 4x
9
12(f) 3 -2 Coordinate
1 Geometry – CirclesB1
x + x3
4 2
3 -3 3 2
– x + x B2ft B1ft for each term
2 2
13(a) Hypotenuse 10x M1
2y = 84 – 36x A1
y = 42 – 18x
13(b) A = 16x(42 – 18x) + ½ × 16x × 6x M1
A = 672x – 288x 2 + 48x 2 A1
= 672x – 240x 2
13(c) dA M1
= 672 – 480x
dx
672
= 0 when x = or 1.4 M1
480
470.4 A1
14 x dy 1 M1
+ 8 x −2 or = ......seen
4 dx 4
dy 1 16 M1 oe
= −
dx 4 x3
1 16
= 0 when = or x3 = 64 or x = 4 M1
4 x3
1.5 A1
10
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Factor Theorem
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1
2
3
10 Factor Theorem
1(a) x(x 2 − 5x − 36) B1
1(b) x = 0, x = −4, x = 9 B2 B1 For two solutions
2(a) f(1) = 1 + 2 − 5 − 6 = −8 B1
f(−1) = −1 + 2 + 5 − 6 = 0 B1
2(b) f(2) = 8 + 8 − 10 − 6 = 0 B1
f(−2) = −8 + 8 + 10 − 6 = 4 B1
2(c) f(3) = 27 + 18 − 15 − 6 = 24 B1
f(−3) = −27 + 18 + 15 − 6 = 0 B1
2(d) (x + 1), (x − 2) and (x + 3) B1
4

3 2
3(a) (−5) + 7(−5) + 2(−5) − 40 M1 oe
−125 + 175 − 10 − 40 = 0 A1 Clearly shown to = 0
3(b) x 3 + 7x 2 + 2x − 40 M1 2
Sight of x and −8 in a quadratic factor
≡ (x + 5)(x 2 + kx − 8)
(x − 2) A1
(x + 4) A1
Alt 1 Substitutes another value into the M1

3(b) expression and tests for ‘= 0’
(x − 2) A1
(x + 4) A1
Alt 2 Long division of polynomials getting as M1

2
3(b) far as x + 2x
(x − 2) A1
(x + 4) A1
3(c) (x =) −5, −4 and 2 B1
3 2
4 (−2) + 5(−2) + 9(−2) + k = 0 M1
−8 + 20 − 18 + k = 0 A1
k=6 A1
5

3 2
5(a) (−3) + (−3) + (−3)a − 72 = 0 M1
−27 + 9 − 3a − 72 = 0 A1
a = −30 A1
2
5(b) x 3 + x 2 − 30x − 72 M1 Sight of x and −24 in a quadratic factor
≡ (x + 3)(x 2 + kx − 24)
(x + 4) A1
(x − 6) A1
Alt 1 Substitutes another value into the M1

5(b) expression and tests for ‘= 0’
(x + 4) A1
(x − 6) A1
Alt 2 Long division of polynomials getting M1

2
5(b) as far as x − 2x
(x + 4) A1
(x − 6) A1
6
6(a) (x − 3)(x + 4)(x + k) M1 or −3 × 4 × k = 24

3 2
≡ x + ax + bx + 24
(x − 2) A1
6(b) (x − 3)(x + 4)(x − 2) M1

2
(x − 3)(x + 2x − 8) M1 oe
x 3 – x 2 − 14x + 24 A1
a = −1 and b = −14 A1 ft ft Their expansion
Alt 6(b) Substitutes any two of M1

x = −4, x = 2 or x = 3
3 2
into x + ax + bx + 24 to create
simultaneous equations
Any two of M1
−64 + 16a − 4b + 24 = 0
or
8 + 4a + 2b + 24 = 0
or
27 + 9a + 3b + 24 = 0
a = −1 A1
b = −14 A1 ft ft Their first solution
7

3 2
7(a) (5) + k(5) + 9(5) − 20 = 0 M1
125 + 25k + 45 − 20 = 0 A1
k = −6 A1
7(b) x 3 − 6x 2 + 9x − 20 M1 2
Sight of x and 4 in a quadratic factor
≡ (x − 5)(x 2 + kx + 4)
(x − 5)(x2 − x + 4) A1
2
7c Tests ‘b − 4ac’ for the quadratic M1 ft Their quadratic
or attempts to solve their quadratic = 0
2
Shows ‘b − 4ac’ = −15 (or < 0) and A1 States 'no solutions' to their quadratic = 0
states no more linear factors
8 Substitutes a value of x into the M1

expression and tests for ‘= 0’
Works out first linear factor A1

(x + 1), (x + 2) or (x − 9)
x 3 − 6x 2 − 25x − 18 M1 Attempts to work out the quadratic factor

2
≡ (x + 1)(x 2 + kx − 18) Sight of x and −18 in a quadratic factor
2
2
or (x + 2)(x + kx − 9) or sight of x and −9 in a quadratic factor
2
2
or (x − 9)(x + kx + 2) or sight of x and 2 in a quadratic factor
2nd and 3rd linear factors A1
−1, −2 and 9 A1
Alt 1 Substitutes a value of x into the M1

8 expression and tests for ‘= 0’

(x + 1), (x + 2) or (x − 9)
Substitutes another value into the M1

expression and tests for ‘= 0’
−1, −2 and 9 A1
8
Alt 2 Substitutes a value of x into the M1

8 expression and tests for ‘= 0’

(x + 1), (x + 2) or (x − 9)
Long division of polynomials getting as M1 Depending on first linear factor

2
far as x − 7x
2
or x − 8x
2
or x + 3x
−1, −2 and 9 A1
9(a) 32 – 32 – 162 + 162 B1
9(b) (x – 2)( x4 – 81) M1
(x – 2)(x2 + 9)(x2 – 9) M1
(x – 2)(x2 + 9)(x + 3)(x – 3) M1
2, –3 and 3 A1
10(a)  2
3
 2
2
 2
3 −  + 2 −  - 3 −  - 2 M1 Oe
 3  3  3
8 8
– + +2–2=0 A1 Clearly shown to = 0
9 9
10(b) (3x + 2)(x2 – 1) M1
(3x + 2)(x + 1)(x – 1) A1
9
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8360)
Mark Scheme
Worksheet 11
Sequences
Version 2.0

AQA
Devas Street
Manchester
M16 6EX
internal use.

implied.

awarded.

has been awarded.

earlier step.


1
2
3
11 Sequences
For the nth terms of quadratic sequences two methods are shown (see example 2).
Other valid methods may be used.
1 −4n M1
254 − 4n A1
254 − 4n < 0 M1 oe
64th A1
2 Method A M1
8 9 14 23 36
1 5 9 13
4 4 4
2
Subtract 2n from sequence A1
6 1 −4 ……
nth term of this sequence is M1
11 − 5n
2
Giving 2n − 5n + 11 A1
Alt 2 Method B M1
2
Using an + bn + c
a+b+c=8
4 a + 2b + c = 9
9a + 3b + c = 14
3a + b = 1 M1 oe
5a + b = 5
a = 2 and b = −5 A1
2
Giving 2n − 5n + 11 A1
4
3(a) Use Method A or B from Q2 3 marks or any other valid method
3(b)(i) n 2 + 3n + 1 B1
3(b)(ii) n 2 + 4n B1
4(a) 3n + 1 B1
2
4(b) (3n + 1) B1 oe
2 2
4(c) 49 × 169 = 7 × 13 B1 oe 8281
2
30th is 91 M1
= (7 × 13) 2 A1 oe 8281
= 7 2 × 13 2
5 nth term of lengths is n+3 M1
nth term of widths is n+2 M1
Area is (n + 3)(n + 2) M1
n 2 + 3n + 2n + 6 A1
2
= n + 5n + 6
Alt 5 nth term of 4 marks or any other valid method

12 20 30
by Method A or Method B
6(a) a + 9b = 35 M1
a + 15b = 59
6b = 24 M1 oe
b=4 A1
a = −1 A1 ft
6(b) … B1 ft
3 11 19
…
8n − 5 B1 ft
5
7(a) 3n + 1 3( n + 1) + 1 M1 oe
−
n n +1 eg subtracts in different order
(3n + 1)( n + 1) − n (3n + 4) M1 oe

n ( n + 1)
3n 2 + n + 3n + 1 − 3n 2 − 4n A1
n ( n + 1)
1
=
n ( n + 1)
Alt 7(a) 3n + 1 1 M1 oe
=3+
n n eg subtracts in different order
1 1 M1 oe
(3 + ) − (3 + )
n n+1
n + 1− n A1
n ( n + 1)
1
=
n ( n + 1)
7(b) Any substitution and evaluation for M1 oe

2
1 „ n „ 10 eg 1 < 0.01n + 0.01n and attempt to solve
1 1
eg =
9 × 10 90
1 1
or =
10 × 11 110
10th and 11th A1
7(c) 3 B1
6
8 5n + 2 5n 2 M1 oe
= +
2n 2n 2n
 5 1
 + 
2 n
1 5 A1
 0 as n  ∝ S = (= 2.5)
n 2
9 Odd number is 2n + 1 or 2n − 1 M1
2n − 1 and 2n + 1 M1
Sequence is (2n − 1)(2n + 1) A1

2
(= 4n − 1)
Alt 9 Using Method A or Method B giving 3 marks or any other valid method
2
4n − 1 eg
1 4 9 16  n2
4 16 36 64  4n 2
3 15 35 63
 4n 2 –
1
10(a) 1 B1
T1 =
5
7 B1 oe
T2 =
14
1
(= )
2
5 2 3 B1 oe
− =
10 10 10
10(b) 2 B1
3
7
AQA Qualifications

FURTHER MATHEMATICS
Level 2 (8360)
Mark Scheme
Worksheet 12
Algebraic Problems – including ratio
Version 2.0

AQA
Devas Street
Manchester
M16 6EX
internal use.

implied.

awarded.

has been awarded.

earlier step.


1
2
3
12 Algebraic Problems – including ratio
1 n=4 M1
1 A1
2
n = −4 M1
9 A1
10
2(a) x 6 M1
=
y 5
6y A1 oe
x=
5
2(b) 6y 15 y 12 y 5y M1 oe 6 + 3 × 5 : 2 × 6 − 5
+ : −
5 5 5 5
21( y ) 7( y ) A1
:
(5 ) (5 )
3 3 3 M1 oe
of (6a − a) or of (11b − b)
10 10
(2.5a, 4b) A2 oe A1 For each coordinate

SC2 (1.5a, 3b)
4 a + 3b 2 M1
=
a + 7b 5
5a + 15b = 2a + 14b M1 Allow one error
3a + b = 0 A1 oe
a + b = −4 A1 ft
2a = 4
a = 2 and b = −6 A1 ft
4
5 a 4 M1 oe
=
b 3
3a A1 4b
b= a=
4 3
3a M1 4b 4b
a× +a=5 × b+ =5
4 3 3
2 2
3a + 4a − 20 = 0 A1 4b + 4b − 15 = 0
(3a + 10)(a − 2) = 0 M1 (2b + 5)(2b − 3)
10 A1 ft 5 3
a=− a=2 b=− b=
3 2 2
5 3 A1 ft 10
b=− b= a=− a=2
2 2 3
6 Let their ages 6 years ago be M1

8x and 5x
8x + 5x = 90 − 12 M1 Allow 90 − 6 for M1
13x = 78 A1
(x = 6)
Their 6 × 8 and their 6 × 5 M1

(48) (30)
54 and 36 A1
Alt 6 x + y = 90 M1
x−6 8 M1
=
y−6 5
18 = 8y − 5x A1
Eliminates a letter M1
(x =) 54 and (y =) 36 A1
5
7 x, x and 180 − 2x M1
seen or on diagram
x 4 M1
=
y 5
4y A1 oe
x=
5
2y = 180 − 2x M1 oe
(or y = 90 − x)
4y M1 oe
y = 90 −
5
9y M1 oe
= 90
5
y = 50 A1
8 a = 7x + 18 or b = 3x + 18 B1 oe
their (7 x + 18) 3 M1
=
their (3 x + 18) 2
14x + 36 = 9x + 54 M1 Rearranging
5x = 18 M1 Solving
x = 3.6 A1
9(a) x : y = 6 : 10 M1 oe
x : y : z = 6 : 10 : 9 M1
x:z=2:3 A1
9(b) 3 × 10 : 7 × 5 M1 oe
6:7 A1
9(c) 3+5:5 M1 x+ y x 3
= + 1 or +1
y y 5
8:5 A1
6

2 2
10 (2n) + n M1 oe
2 2 2 2 M1
(2n) + n + (n − 1) = (2n + 1)
2 2 2 M1 Allow one error
4n + n + n − n − n + 1
= 4n 2 + 2n + 2n + 1
2 2
2n − 6n = 0 M1 Rearranging ; or 2n = 6n
2n(n − 3) = 0 M1 (allow ÷ by n ) 2n = 6
n=3 A1

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AQA Level 2 Certificate

Copyright © 2018 AQA and its licensors. All rights reserved.

A (0, 8) and B (–3, 1) are points on y = abx as shown.

Question 2 (Spec ref 1.2)

Here are five cards.

Question 3 (Spec ref 2.9/2.20)

Question 5 (Spec ref 2.7)

The coefficient of x is three times the coefficient of x2.

Work out the value of a.

AQA Level 2 Certificate

Copyright © 2018 AQA and its licensors. All rights reserved.

Write down the equation of each of these circles.

(b) Centre (1, −5) radius 4 (2 marks)

(c) Centre (−3, 4) radius 7 (2 marks)

(d) Centre (8, 15) radius 17

AB is the diameter of a circle.

PQ is a diameter of a circle, centre C.

(a) Work out the coordinates of Q. (1 mark)

(b) Work out the equation of the circle. (3 marks)

(a) Work out the coordinates of the midpoint M, of AB. (2 marks)

(b) Show that the length CM = 7 2 (3 marks)

(c) Work out the radius of the circle. (2 marks)

12 (4, 12) Not drawn

Work out the coordinates of C. (3 marks)

AB is a diameter of the circle ABC.

C (4, 6) Not drawn

Work out the value of k. (5 marks)

A circle has equation (x − 5) 2 + (y − 4) 2 = 100

The point (13, −2) lies on the circle (x − a) 2 + (y − 4) 2 = 100

Work out the two possible values of a. (5 marks)

(a) Work out the value of k.

(b) Hence find the equation of the circle. (5 marks)

(a) Show working to explain why OP is a tangent to the circle. (5 marks)

The equation of this circle is x 2 + y 2 = 20

Work out the equation of the tangent to the circle at P.

(a) Work out the equation of the circle.

Give your answer in the form a x + by + c = 0

AQA Level 2 Certificate

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Prove that PQRS is a cyclic quadrilateral. (3 marks)

Work out s. (5 marks)

Prove that 4x + y = 90 (4 marks)

Prove that triangle RXT is isosceles. (3 marks)

Prove that y = 90 + x (5 marks)

Prove that PTV is a tangent to circle RST at T (5 marks)

Prove that AD is parallel to BE. (5 marks)

AQA Level 2 Certificate

You can get further copies of this Teacher Resource from:

Copyright © 2012 AQA and its licensors. All rights reserved.

(b) Hence, prove that x 2 + 6x + 11 is always positive. (2 marks)

Prove that f(3n) + f(n + 1) = kn(5n − 1) where k is an integer. (3 marks)

AQA Level 2 Certificate

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Give your answer in its simplest form. (3 marks)

sin 37° = sin 127°

cos 54° = cos 306°

sin 135° = cos 135°

tan 126° = tan 306°