C2 Revision
Workbook
C2 factor and remainder theorem
f(x) = 2x3 + 3x2 – 29x – 60.
1.
(a)
Find the remainder when f(x) is divided by (x + 2).
(2)
(b)
Use the factor theorem to show that (x + 3) is a factor of f(x).
(2)
(c)
Factorise f(x) completely.
(4)
f(x) = x3 + 4x2 + x–6.
2.
(a)
Use the factor theorem to show that (x + 2) is a factor of f(x).
(2)
(b)
Factorise f(x) completely.
(4)
(c)
Write down all the solutions to the equation
x3 + 4x2 + x – 6 = 0.
(1)
f(x) = 3x3 – 5x2 – 16x + 12.
3.
(a)
Find the remainder when f(x) is divided by (x – 2).
(2)
Given that (x + 2) is a factor of f(x),
(b)
factorise f(x) completely.
(4)
4.
(a)
Find the remainder when
x3 – 2x2 – 4x + 8
is divided by
(i)
x – 3,
(ii)
x + 2.
(3)
(b)
Hence, or otherwise, find all the solutions to the equation
x3 – 2x2 – 4x + 8 = 0.
(4)
f(x) = 2x3 – 3x2 – 39x+ 20
5.
(a)
Use the factor theorem to show that (x + 4) is a factor of f (x).
(2)
(b)
Factorise f (x) completely.
(4)
f ( x)  x 4  5 x 3  ax  b ,
6.
where a and b are constants.
The remainder when f(x) is divided by (x – 2) is equal to the remainder when f(x) is divided by
(x + 1).
(a)
Find the value of a.
(5)
Given that (x + 3) is a factor of f(x),
(b)
find the value of b.
(3)
f(x) = (3x – 2)(x – k) – 8
7.
where k is a constant.
(a)
Write down the value of f (k).
(1)
When f (x) is divided by (x – 2) the remainder is 4
(b)
Find the value of k.
(2)
(c)
Factorise f (x) completely.
(3)
f(x) = 2x3 + ax2 + bx – 6
8.
where a and b are constants.
When f(x) is divided by (2x – 1) the remainder is –5.
When f(x) is divided by (x + 2) there is no remainder.
(a)
Find the value of a and the value of b.
(6)
(b)
Factorise f(x) completely.
(3)
Total marks
57
Video solutions can be found in the following links
1.
2.
3.
4.
5.
6.
7.
8.
Edexcel C2 June 2006 Question 4
Edexcel C2 January 2007 Question 5
Edexcel C2 June 2007 Question 2
Edexcel C2 January 2008 Question 1
Edexcel C2 June 2008 Question 1
Edexcel C2 January 2009 Question 6
Edexcel C2 June 2009 Question 3
Edexcel C2 January 2010 Question 3
C2 Coordinate geometry (circles)
1.
y
y = 3x – 4
C
P (2, 2)
Q
O
x
The line y = 3x – 4 is a tangent to the circle C, touching C at the point P(2, 2), as shown in the figure above.
The point Q is the centre of C.
(a)
Find an equation of the straight line through P and Q.
(3)
Given that Q lies on the line y = 1,
(b)
show that the x-coordinate of Q is 5,
(1)
(c)
find an equation for C.
(4)
2.
The line joining the points (–1, 4) and (3, 6) is a diameter of the circle C.
Find an equation for C.
3.
y
B
M
(3, 1)
O
x
P
A
(1, –2)
l
The points A and B lie on a circle with centre P, as shown in the diagram above. The point A has coordinates (1,-2)
and the mid-point M of AB has coordinates (3, 1). The line l passes through the points M and P.
(a)
Find an equation for l.
(4)
Given that the x-coordinate of P is 6,
(b)
use your answer to part (a) to show that the y-coordinate of P is –1,
(1)
(c)
find an equation for the circle.
(4)
4.
A circle C has centre M (6, 4) and radius 3.
(a)
Write down the equation of the circle in the form
(x – a)2 + (y – b)2 = r2.
(2)
y
T
P
(12, 6)
3
Q
M
(6, 4)
C
O
x
The diagram above shows the circle C. The point T lies on the circle and the tangent at T passes through the point
P (12, 6). The line MP cuts the circle at Q.
(b)
Show that the angle TMQ is 1.0766 radians to 4 decimal places.
(4)
The shaded region TPQ is bounded by the straight lines TP, QP and the arc TQ, as shown in the diagram above.
(c)
Find the area of the shaded region TPQ. Give your answer to 3 decimal places.
(5)
5.
The circle C has centre (3, 1) and passes through the point P(8, 3).
(a)
Find an equation for C.
(4)
(b)
Find an equation for the tangent to C at P, giving your answer in the form
ax + by + c = 0, where a, b and c are integers.
(5)
6.
The points P(–3, 2), Q(9, 10) and R(a, 4) lie on the circle C, as shown in the diagram above.
Given that PR is a diameter of C,
(a)
show that a = 13,
(3)
(b)
find an equation for C.
(5)
7.
The circle C has equation
x2 + y2 – 6x + 4y = 12
(a)
Find the centre and the radius of C.
(5)
The point P(–1, 1) and the point Q(7, –5) both lie on C.
(b)
Show that PQ is a diameter of C.
(2)
The point R lies on the positive y-axis and the angle PRQ = 90°.
(c)
Find the coordinates of R.
(4)
8.
The diagram above shows a sketch of the circle C with centre N and equation
(x – 2)2 + (y + 1)2 =
(a)
169
4
Write down the coordinates of N.
(2)
(b)
Find the radius of C.
(1)
The chord AB of C is parallel to the x-axis, lies below the x-axis and is of length 12 units as shown in the
diagram above.
(c)
Find the coordinates of A and the coordinates of B.
(5)
(d)
Show that angle ANB = 134.8°, to the nearest 0.1 of a degree.
(2)
The tangents to C at the points A and B meet at the point P.
(e)
Find the length AP, giving your answer to 3 significant figures.
(2)
Total marks
74
Video solutions can be found in the following links
1.
2.
3.
4.
5.
6.
7.
8.
Edexcel C2 June 2006 Question 7
Edexcel C2 January 2007 Question 3
Edexcel C2 June 2007 Question 7
Edexcel C2 January 2008 Question 8
Edexcel C2 June 2008 Question 5
Edexcel C2 January 2009 Question 5
Edexcel C2 June 2009 Question 6
Edexcel C2 January 2010 Question 8
C2 logarithms
1.
Solve the equation
5x = 17,
giving your answer to 3 significant figures.
(3)
2.
(a)
Find, to 3 significant figures, the value of x for which 8x = 0.8.
(2)
(b)
Solve the equation
2log3 x – log3 7x = 1.
(4)
3.
Given that a and b are positive constants, solve the simultaneous equations
a = 3b,
log3 a + log3 b = 2.
Give your answers as exact numbers.
(6)
4.
(a)
Find, to 3 significant figures, the value of x for which 5x = 7.
(2)
(b)
Solve the equation 52x – 12(5x) + 35 = 0.
(4)
5.
Given that 0 < x < 4 and
log5(4 – x) –2log5 x = 1,
find the value of x.
(6)
6.
(a)
Find the value of y such that
log2 y = –3
(2)
(b)
Find the values of x such that
log 2 32  log 2 16
 log 2 x
log 2 x
(5)
7.
(a)
Find the positive value of x such that
log x 64 = 2
(2)
(b)
Solve for x
log2(11 – 6x) = 2 log2(x – 1) + 3
(6)
Total marks 42
Video solutions can be found in the following links
1.
2.
3.
4.
5.
6.
7.
Edexcel C2 January 2007 (Question 4)
Edexcel C2 June 2007 (Question 6)
Edexcel C2 January 2008 (Question 5)
Edexcel C2 June 2008 (Question 4)
Edexcel C2 January 2009 (Question 4)
Edexcel C2 June 2009 (Question 8)
Edexcel C2 January 2010 (Question 5)
C2 geometric series
1.
A geometric series has first term a and common ratio r.
The second term of the series is 4 and the sum to infinity of the series is 25.
(a)
Show that 25r2 – 25r + 4 = 0.
(4)
(b)
Find the two possible values of r.
(2)
(c)
Find the corresponding two possible values of a.
(2)
(d)
Show that the sum, Sn, of the first n terms of the series is given by
Sn = 25(1 – rn).
(1)
Given that r takes the larger of its two possible values,
(e)
find the smallest value of n for which Sn exceeds 24.
(2)
)
2.
A geometric series is a + ar + ar2 + ...
(a)
Prove that the sum of the first n terms of this series is given by
Sn 


a 1 r n
.
1 r
(4)
(b)
Find
 100 2 .
10
k
k 1
(3)
(c)
Find the sum to infinity of the geometric series
5 5 5
   .....
6 18 54
(3)
(d)
State the condition for an infinite geometric series with common ratio r to be convergent.
(1)
3.
A trading company made a profit of £50 000 in 2006 (Year 1).
A model for future trading predicts that profits will increase year by year in a geometric sequence with
common ratio r, r > 1.
The model therefore predicts that in 2007 (Year 2) a profit of £50 000r will be made.
(a)
Write down an expression for the predicted profit in Year n.
(1)
The model predicts that in Year n, the profit made will exceed £200 000.
(b)
Show that n >
log 4
1 .
log r
(3)
Using the model with r = 1.09,
(c)
find the year in which the profit made will first exceed £200 000,
(2)
(d)
find the total of the profits that will be made by the company over the 10 years from 2006 to 2015
inclusive, giving your answer to the nearest £10 000.
(3)
4.
The fourth term of a geometric series is 10 and the seventh term of the series is 80.
For this series, find
(a)
the common ratio,
(2)
(b)
the first term,
(2)
(c)
the sum of the first 20 terms, giving your answer to the nearest whole number.
(2)
)
5.
A geometric series has first term 5 and common ratio
4
.
5
Calculate
(a)
the 20th term of the series, to 3 decimal places,
(2)
(b)
the sum to infinity of the series.
(2)
Given that the sum to k terms of the series is greater than 24.95,
(c)
show that k 
log 0.002
,
log 0.8
(4)
(d)
find the smallest possible value of k.
(1)
6.
The first three terms of a geometric series are (k + 4), k and (2k – 15) respectively, where k is a positive
constant.
(a)
Show that k2 – 7k – 60 = 0.
(4)
(b)
Hence show that k = 12.
(2)
(c)
Find the common ratio of this series.
(2)
(d)
Find the sum to infinity of this series.
(2)
7.
The third term of a geometric sequence is 324 and the sixth term is 96
(a)
Show that the common ratio of the sequence is
2
3
(2)
(b)
Find the first term of the sequence.
(2)
(c)
Find the sum of the first 15 terms of the sequence.
(3)
(d)
Find the sum to infinity of the sequence.
(2)
8.
A car was purchased for £18 000 on 1st January.
On 1st January each following year, the value of the car is 80% of its value on 1st January in the previous
year.
(a)
Show that the value of the car exactly 3 years after it was purchased is £9216.
(1)
The value of the car falls below £1000 for the first time n years after it was purchased.
(b)
Find the value of n.
(3)
An insurance company has a scheme to cover the maintenance of the car. The cost is £200 for the first year,
and for every following year the cost increases by 12% so that for the 3rd year the cost of the scheme is
£250.88
(c)
Find the cost of the scheme for the 5th year, giving your answer to the nearest penny.
(2)
(d)
Find the total cost of the insurance scheme for the first 15 years.
(3)
Total marks 74
Video solutions can be found in the following links
1.
2.
3.
4.
5.
6.
7.
8.
Edexcel C2 June 2006 (Question 9)
Edexcel C2l January 2007 (Question 10)
Edexcel C2 June 2007 (Question 8)
Edexcel C2 January 2008 (Question 2)
Edexcel C2 June 2008 (Question 6)
Edexcel C2 January 2009 (Question 9)
Edexcel C2 June 2009 (Question 5)
Edexcel C2 January 2010 (Question 6)
C2 differentiation
1.
y
y = x 3 – 8x 2 + 20x
A
B
R
O
N
x
The figure above shows a sketch of part of the curve with equation y = x3 – 8x2 + 20x.
The curve has stationary points A and B.
(a)
Use calculus to find the x-coordinates of A and B.
(4)
(b)
Find the value of
d2 y
at A, and hence verify that A is a maximum.
dx 2
(2)
The line through B parallel to the y-axis meets the x-axis at the point N.
The region R, shown shaded in the figure above, is bounded by the curve, the x-axis and
the line from A to N.
(c)
Find
 (x
3
 8x 2  20 x) dx
(3)
(d)
Hence calculate the exact area of R.
(5)
f(x) = x3+ 3x2 + 5.
2.
Find
(a)
f′′(x),
(3)
(b)
2
 f ( x)dx.
1
(4)
3.
A diesel lorry is driven from Birmingham to Bury at a steady speed of v kilometres per hour. The total cost
of the journey, £C, is given by
C
(a)
1400 2v
 .
v
7
Find the value of v for which C is a minimum.
(5)
(b)
Find
d 2C
and hence verify that C is a minimum for this value of v.
dv 2
(2)
(c)
Calculate the minimum total cost of the journey.
(2)
4.
2x cm
x cm
y cm
The diagram above shows a solid brick in the shape of a cuboid measuring 2x cm by x cm by y cm.
The total surface area of the brick is 600 cm2.
(a)
Show that the volume, V cm3, of the brick is given by
V  200 x 
4x 3
.
3
(4)
Given that x can vary,
(b)
use calculus to find the maximum value of V, giving your answer to the nearest cm3.
(5)
(c)
Justify that the value of V you have found is a maximum.
(2)
5.
x
x
y
The diagram above shows an open-topped water tank, in the shape of a cuboid, which is made of sheet
metal. The base of the tank is a rectangle x metres by y metres. The height of the tank is x metres.
The capacity of the tank is 100 m3.
(a)
Show that the area A m2 of the sheet metal used to make the tank is given by
A
300
 2x2
x
(4)
(b)
Use calculus to find the value of x for which A is stationary.
(4)
(c)
Prove that this value of x gives a minimum value of A.
(2)
(d)
Calculate the minimum area of sheet metal needed to make the tank.
(2)
6.
y
A
R
x
O
The diagram above shows a sketch of part of the curve with equation y = 10 + 8x + x2 – x3.
The curve has a maximum turning point A.
(a)
Using calculus, show that the x-coordinate of A is 2.
(3)
The region R, shown shaded in the diagram, is bounded by the curve, the y-axis and the line from O to A,
where O is the origin.
(b)
Using calculus, find the exact area of R.
(8)
7.
A solid right circular cylinder has radius r cm and height h cm.
The total surface area of the cylinder is 800 cm2.
(a)
Show that the volume, V cm3 , of the cylinder is given by
V = 400r – πr3.
(4)
Given that r varies,
(b)
use calculus to find the maximum value of V, to the nearest cm3.
(6)
(c)
Justify that the value of V you have found is a maximum.
(2)
8.
The diagram above shows a closed box used by a shop for packing pieces of cake. The box is a right prism
of height h cm. The cross section is a sector of a circle. The sector has radius r cm and angle 1 radian.
The volume of the box is 300 cm3.
(a)
Show that the surface area of the box, S cm2, is given by
S r2 
1800
r
(5)
(b)
Use calculus to find the value of r for which S is stationary.
(4)
(c)
Prove that this value of r gives a minimum value of S.
(2)
(d)
Find, to the nearest cm2, this minimum value of S.
(2)
The curve C has equation y = 12 √ ( x ) 
9.
(a)
3
x2
 10, x > 0
Use calculus to find the coordinates of the turning point on C.
(7)
(b)
Find
d2 y
.
dx 2
(2)
(c)
State the nature of the turning point.
(1)
Total marks 99
Video solutions can be found in the following links
1.
2.
3.
4.
5.
6.
7.
8.
9.
Edexcel C2 June 2006 (Question 10(a),(b))
Edexcel C2 January 2007 (Question 1(a))
Edexcel C2 January 2007 (Question 8)
Edexcel C2 June 2007 (Question 10)
Edecxel C2 January 2008 (Question 9)
Edexcel C2 June 2008 (Question 8a)
Edexcel C2 January 2009 (Question 10)
Edexcel C2 June 2009 (Question 9)
Edexcel C2 January 2010 (Question 9)
C2 Integration
1.
Use calculus to find the exact value of

2
1
4
2
 3x  5  2
x


dx

(5)
2.
y
y = x 3 – 8x 2 + 20x
A
B
R
O
N
x
The figure above shows a sketch of part of the curve with equation y = x3 – 8x2 + 20x.
The curve has stationary points A and B.
(a)
Use calculus to find the x-coordinates of A and B.
(4)
(b)
d2 y
Find the value of
at A, and hence verify that A is a maximum.
dx 2
(2)
The line through B parallel to the y-axis meets the x-axis at the point N.
The region R, shown shaded in the figure above, is bounded by the curve, the x-axis and
the line from A to N.
(c)
Find
 (x
3
 8x 2  20 x) dx
(3)
(d)
Hence calculate the exact area of R.
(5)
f(x) = x3+ 3x2 + 5.
3.
Find
(a)
f′′(x),
(3)
(b)
2
 f ( x)dx.
1
(4)
4.
y
O
1
2
5
x
C
The diagram above shows a sketch of part of the curve C with equation
y = x(x – 1)(x – 5).
Use calculus to find the total area of the finite region, shown shaded in the diagram, that is between x = 0
and x = 2 and is bounded by C, the x-axis and the line x = 2.
(9)
5.
Evaluate

8
1
1
dx , giving your answer in the form a + b√2, where a and b are integers.
x
(4)
6.
y
A
R
O
x
The diagram above shows a sketch of part of the curve with equation y = 10 + 8x + x2 – x3.
The curve has a maximum turning point A.
(a)
Using calculus, show that the x-coordinate of A is 2.
(3)
The region R, shown shaded in the diagram, is bounded by the curve, the y-axis and the line from O to A,
where O is the origin.
(b)
Using calculus, find the exact area of R.
(8)
7.
y
L
R
C
O
x
In the diagram above the curve C has equation y = 6x – x2 and the line L has equation y = 2x.
(a)
Show that the curve C intersects the x-axis at x = 0 and x = 6.
(1)
(b)
Show that the line L intersects the curve C at the points (0, 0) and (4, 8).
(3)
The region R, bounded by the curve C and the line L, is shown shaded in the diagram above.
(c)
Use calculus to find the area of R.
(6)
8.
The diagram above shows part of the curve C with equation y  (1  x)(4  x) .
The curve intersects the x-axis at x = –1 and x = 4. The region R, shown shaded in the diagram, is bounded
by C and the x-axis.
Use calculus to find the exact area of R.
(5)
9.
Use calculus to find the value of

4
1
( 2 x  3 x )dx.
(5)
10.
The curve C has equation y = x2 – 5x + 4. It cuts the x-axis at the points L and M as shown in the diagram
above.
(a)
Find the coordinates of the point L and the point M.
(2)
(b)
Show that the point N (5, 4) lies on C.
(1)
(c)
Find
 (x
2
 5 x  4) dx.
(2)
The finite region R is bounded by LN, LM and the curve C as shown in Figure 2.
(d)
Use your answer to part (c) to find the exact value of the area of R.
(5)
(Total 10 marks)
Total marks 80
Video solutions can be found in the following links
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Edexcel C2 June 2006 (Question 2)
Edexcel C2 June 2006 (Question 10 (c),(d))
Edexcel C2 January 2007 (Question 1(b))
Edexcel C2 January 2007 (Question 7)
Edexcel C2 June 2007 (Question 1)
Edexcel C2 January 2008 (Question 7)
Edexcel C2 June 2008 (Question 8)
Edexcel C2 January 2009 (Question 2)
Edexcel C2 June 2009 (Question 1)
Edexcel C2 January 2010 (Question 7)
C2 trigonometric equations
1.
(a)
Given that sin θ = 5cos θ, find the value of tan θ.
(1)
(b)
Hence, or otherwise, find the values of θ in the interval 0  θ < 360° for which
sin θ = 5cos θ,
giving your answers to 1 decimal place.
(3)
2.
Find all the solutions, in the interval 0 ≤ x < 2π, of the equation
2 cos2 x + 1 = 5 sin x,
giving each solution in terms of π.
(6)
3.
(a)
 
Sketch, for 0 ≤ x ≤ 2, the graph of y = sin  x  .
6

(2)
(b)
Write down the exact coordinates of the points where the graph meets the coordinate axes.
(3)
(c)
Solve, for 0 ≤ x ≤ 2, the equation


sin  x    0.65,
6

giving your answers in radians to 2 decimal places
(5)
4.
(a)
Show that the equation
3 sin2θ – 2 cos2θ = 1
can be written as
5 sin2θ = 3.
(2)
(b)
Hence solve, for 0° ≤ θ < 360°, the equation
3 sin2θ – 2 cos2θ = 1,
giving your answers to 1 decimal place.
(7)
5.
(a)
Show that the equation
4 sin2 x + 9 cos x – 6 = 0
can be written as
4 cos2 x – 9 cos x + 2 = 0.
(2)
(b)
Hence solve, for 0  x  720,
4 sin2 x + 9 cos x – 6 = 0,
giving your answers to 1 decimal place.
(6)
6.
(i)
Solve, for –180° ≤ θ < 180°,
(1 + tan θ)(5 sin θ – 2) = 0.
(4)
(ii)
Solve, for 0 ≤ x < 360°,
4sin x = 3tan x.
(6)
7.
(a)
Show that the equation
5 sin x = 1 + 2 cos2 x
can be written in the form
2 sin2 x + 5 sin x – 3 = 0
(2)
(b)
Solve, for 0  x < 360°,
2 sin2 x + 5 sin x – 3 = 0
(4)
8.
(a)
Given that 5sinθ = 2cosθ, find the value of tan θ .
(1)
(b)
Solve, for 0  x < 360°,
5sin 2x = 2cos 2x,
giving your answers to 1 decimal place.
(5)
Total marks 59
Video solutions can be found in the following links
1.
2.
3.
4.
5.
6.
7.
8.
Edexcel C2 June 2006 (Question 6)
Edexcel C2 January 2007 (Question 6)
Edexcel C2 June 2007 (Question 9)
Edexcel C2 January 2008 (Question 4)
Edexcel C2 June 2008 (Question 9)
Edexcel C2 January 2009 (Question 8)
Edexcel C2 June 2009 (Question 7)
Edexcel C2 January 2010 (Question 2)
C2 trigonometry
1.
B
C
2.12 m
A
D
1.86 m
The figure above shows the cross section ABCD of a small shed.
The straight line AB is vertical and has length 2.12 m.
The straight line AD is horizontal and has length 1.86 m.
The curve BC is an arc of a circle with centre A, and CD is a straight line.
Given that the size of BAC is 0.65 radians, find
(a)
the length of the arc BC, in m, to 2 decimal places,
(2)
(b)
the area of the sector BAC, in m2, to 2 decimal places,
(2)
(c)
the size of CAD, in radians, to 2 decimal places,
(2)
(d)
the area of the cross section ABCD of the shed, in m2, to 2 decimal places.
(3)
2.
S
6 3m
P
R
6m
6m
Q
The diagram above shows a plan of a patio. The patio PQRS is in the shape of a sector of a circle with
centre Q and radius 6 m.
Given that the length of the straight line PR is 6√3m,
(a)
find the exact size of angle PQR in radians.
(3)
(b)
Show that the area of the patio PQRS is 12 m2.
(2)
(c)
Find the exact area of the triangle PQR.
(2)
(d)
Find, in m2 to 1 decimal place, the area of the segment PRS.
(2)
(e)
Find, in m to 1 decimal place, the perimeter of the patio PQRS.
(2)
3.
C
5 cm
4 cm
A
6 cm
B
The diagram above shows the triangle ABC, with AB = 6 cm, BC = 4 cm and CA = 5 cm.
(a)
3
Show that cos A  .
4
(3)
(b)
Hence, or otherwise, find the exact value of sin A.
(2)
4.
C
N

B
700 m
500 m
15 
A
The diagram above shows 3 yachts A, B and C which are assumed to be in the same horizontal plane. Yacht
B is 500 m due north of yacht A and yacht C is 700 m from A. The bearing of C from A is 015°.
(a)
Calculate the distance between yacht B and yacht C, in metres to 3 significant figures.
(3)
The bearing of yacht C from yacht B is θ°, as shown in the diagram.
(b)
Calculate the value of θ.
(4)
5.
A circle C has centre M (6, 4) and radius 3.
(a)
Write down the equation of the circle in the form
(x – a)2 + (y – b)2 = r2.
(2)
y
T
P
(12, 6)
3
Q
M
(6, 4)
C
O
x
The diagram above shows the circle C. The point T lies on the circle and the tangent at T passes through the point
P (12, 6). The line MP cuts the circle at Q.
(b)
Show that the angle TMQ is 1.0766 radians to 4 decimal places.
(4)
The shaded region TPQ is bounded by the straight lines TP, QP and the arc TQ, as shown in the diagram above.
(c)
Find the area of the shaded region TPQ. Give your answer to 3 decimal places.
(5)
6.
B
7 cm
R
0.8 rad
A
D
C
The diagram above shows ABC, a sector of a circle with centre A and radius 7 cm.
Given that the size of BAC is exactly 0.8 radians, find
(a)
the length of the arc BC,
(2)
(b)
the area of the sector ABC.
(2)
The point D is the mid-point of AC. The region R, shown shaded in the diagram above, is bounded by CD,
DB and the arc BC.
Find
(c)
the perimeter of R, giving your answer to 3 significant figures,
(4)
(d)
the area of R, giving your answer to 3 significant figures.
(4)
7.
The shape BCD shown above is a design for a logo.
The straight lines DB and DC are equal in length. The curve BC is an arc of a circle with centre A and radius 6 cm.
The size of  BAC is 2.2 radians and AD = 4 cm.
Find
(a)
the area of the sector BAC, in cm2,
(2)
(b)
the size of  DAC, in radians to 3 significant figures,
(2)
(c)
the complete area of the logo design, to the nearest cm2.
(4)
8.
An emblem, as shown in the diagram above, consists of a triangle ABC joined to a sector CBD of a circle
with radius 4 cm and centre B. The points A, B and D lie on a straight line with AB = 5 cm and BD = 4 cm.
Angle BAC = 0.6 radians and AC is the longest side of the triangle ABC.
(a)
Show that angle ABC = 1.76 radians, correct to 3 significant figures.
(4)
(b)
Find the area of the emblem.
(3)
(Total 7 marks)
Total marks 70
Video solutions can be found in the following links
1.
2.
3.
4.
5.
6.
7.
Edexcel C2 June 2006 (Question 8)
Edexcel C2 January 2007 (Question 9)
Edexcel C2 June 2007 (Question 4)
Edexcel C2 January 2008 (Question 6)
Edexcel C2 June 2008 (Question 7)
Edexcel C2 January 2009 (Question 7)
Edexcel C2 January 2010 (Question 4)
C2 Binomial Expansion
1.
Find the first 3 terms, in ascending powers of x, of the binomial expansion of (2 + x)6,
giving each term in its simplest form.
(4)
2.
(a) Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 – 2x)5. Give each term
in its simplest form.
(4)
(b)
If x is small, so that x2 and higher powers can be ignored, show that
(1 + x)(1 – 2x)5 ≈ 1 – 9x.
(2)
3.
(a)
Find the first four terms, in ascending powers of x, in the binomial expansion of
(1 + kx)6,where k is a non-zero constant.
(3)
Given that, in this expansion, the coefficients of x and x2 are equal, find
(b)
the value of k,
(2)
(c)
the coefficient of x3.
(1)
4.
(a)

Find the first 4 terms of the expansion of 1 

simplest form.
10
x

2
in ascending powers of x, giving each term in its
(4)
(b)
Use your expansion to estimate the value of (1.005)10, giving your answer to 5 decimal places.
(3)
5.
(a)
Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 + ax)10, where a is a
non-zero constant. Give each term in its simplest form.
(4)
Given that, in this expansion, the coefficient of x3 is double the coefficient of x3,
(b)
find the value of a.
(2)
6.
Find the first 3 terms, in ascending powers of x, of the binomial expansion of 3  2 x  , giving each term in
its simplest form.
5
(4)
7.
(a)
Find the first 3 terms, in ascending powers of x, of the binomial expansion of
(2 + kx)7
where k is a constant. Give each term in its simplest form.
(4)
Given that the coefficient of x2 is 6 times the coefficient of x,
(b)
find the value of k.
(2)
8.
Find the first 3 terms, in ascending powers of x, of the binomial expansion of
(3 – x)6
and simplify each term.
(4)
Total marks 43
Video solutions can be found in the following links
1.
2.
3.
4.
5.
6.
7.
Edexcel C2 June 2006 (Question 1)
Edexcel C2 January 2007 (Question 2)
Edexcel C2 June 2007 (Question 3)
Edexcel C2 January 2008 (Question 3)
Edexcel C2 June 2008 (Question 3)
Edexcel C2 January 2009 (Question 2)
Edexcel C2 January 2010 (Question 1)
C2 Trapezium Rule
1.
(a)
In the space provided, sketch the graph of y = 3x , x ∈
which the graph meets the y-axis.
, showing the coordinates of the point at
(2)
(b)
Complete the table, giving the values of 3x to 3 decimal places.
x
0
3x
0.2
0.4
0.6
1.246
1.552
0.8
1
3
(2)
(c)
Use the trapezium rule, with all the values from your table, to find an approximation for the value of

1
0
3 x dx .
(4)
2.
The curve C has equation
0  x  2.
y = x(x3 + 1),
(a)
Complete the table below, giving the values of y to 3 decimal places at x = 1 and x = 1.5.
x
0
0.5
y
0
0.530
1
1.5
2
6
(2)
(b)
Use the trapezium rule, with all the y values from your table, to find an approximation for the value
of

2
0


x x 3  1 dx , giving your answer to 3 significant figures.
(4)
y
(2, 6)
l
C
R
x
O
The diagram above shows the curve C with equation y = x√(x3 + 1), 0 ≤ x ≤ 2, and the straight line segment
l, which joins the origin and the point (2, 6). The finite region R is bounded by C and l.
(c)
Use your answer to part (b) to find an approximation for the area of R, giving your answer to 3
significant figures.
(3)
y = √(5x + 2)
3.
(a)
Complete the table below, giving the values of y to 3 decimal places.
x
0
0.5
y
1
1.5
2.646
3.630
2
(2)
(b)
Use the trapezium rule, with all the values of y from your table, to find an approximation for the
2
value of

(5 x  2) dx .
0
(4)
4.
(a)
Complete the table below, giving values of √(2x + 1) to 3 decimal places.
x
0
0.5
1
1.5
√(2x + 1)
1.414
1.554
1.732
1.957
2
2.5
3
3
(2)
The diagram above shows the region R which is bounded by the curve with equation
y = √(2x + 1), the x-axis and the lines x = 0 and x = 3
(b)
Use the trapezium rule, with all the values from your table, to find an approximation for the area of R.
(4)
(c)
By reference to the curve in Figure 1 state, giving a reason, whether your approximation in part (b) is
an overestimate or an underestimate for the area of R.
(2)
Total marks 31
Video solutions can be found in the following links
1.
2.
3.
4.
Edexcel C2 June 2006 (Question 5)
Edexcel C2 June 2007 (Question 5)
Edexcel C2 June 2008 (Question 2)
Edexcel C2 January 2009 (Question 3)
C2 Workbook
Answers
C2 Factor remainder theorem Answers
1.
(a)
f (2) = 2(2)3 + 3(2)2  29(2)  60
M: Attempt f(2) or f(2)
M1
= –16 + 12+ 58  60 = 6
Alternative (long division):
(b)
(c)
f (3) = 2(3)3 + 3(3)2  29(–3)  60
A1
M: Attempt f(3) or f(3)
M1
(= 54 + 27 + 87  60) = 0  (x + 3) is a factor
A1
(x + 3)(2x2  3x  20)
M1 A1
= (x + 3)(2x + 5)(x  4)
Alternative (first 2 marks):
2
M1 A1
2
4
(x + 3)(2x2 + ax + b) = 2x3 + (6 + a)x2 + (3a + b)x + 3b = 0, then
compare coefficients to find values of a and b.
[M1]
a = 3, b = –20
[A1]
2.
(a)
(b)
4.
5.
f(–2) = (–2)3 + 4(–2)2 + (–2) – 6
{= –8 + 16 – 2 – 6}
= 0,  x + 2 is a factor
A1
M1
2
x3 + 4x2 + x – 6 = (x + 2)(x2 + 2x – 3)
= (x + 2)(x + 3)(x – 1)
(c)
–3, –2, 1
3.
(a) f(2) = 24 – 20 – 32 + 12 = –16 (M: Attempt f(2) or f(–2))
(b)
(x + 2)(3x2 – 11x + 6)
(x + 2)(3x – 2)(x – 3)
(a)
(i)
f(3) = 33 – 2 × 32 – 4 × 3 + 8; = 5
(ii)
f(–2) = (– 8 – 8 + 8 + 8) = 0 (B1 on Epen, but A1 in fact)
M1 is for attempt at either f(3) or f(–3) in (i)
or f(–2) or f(2) in (ii).
M1, A1
M1, A1
4
B1
1
M1A1
2
M1A1
M1A1 4
M1; A1
B1
3
(b)
[(x + 2)](x2 – 4x + 4) (= 0 not required) [must be seen or used in (b)]
(x + 2) (x – 2)2 (= 0) ( can imply previous 2 marks)
Solutions: × = 2 or – 2 (both) or (–2, 2, 2) [no wrong working seen]
M1A1
M1
A1
4
(a)
Attempt to find f(–4) or f(4). (f(–4) = 2(–4)3 – 3(–4)2 – 39(–4) + 20)
(= –128 – 48 + 156 + 20) = 0, so (x + 4) is a factor.
M1
A1
Long division scores no marks in part (a)
(b)
2x3 – 3x2 – 39x + 20 = (x + 4)(2x2 – 11x + 5)
.....(2x – 1)(x – 5) (The 3 brackets need not be written together)
so
M1A1
M1A1c
4
2
1

or ......... x  (2 x  10) or equivalent
2

6.
(a)
f (2) = 16 + 40 + 2a + b or f (– 1) = 1 – 5 – a + b
M1 A1
Finds 2nd remainder and equates to 1st  16 + 40 +
2a + b = 1 – 5 – a + b
M1 A1
a = – 20
(b)
A1cso
f (– 3) = (– 3)4 + 5(– 3)3 – 3a + b = 0
M1 A1ft
81 – 135 + 60 + b = 0 gives b = – 6
7.
3
1
f(k) = –8
B1
(b)
f(2) = 4  4 = (6 – 2)(2 – k) – 8
M1
k=–1
A1
(c)
f(x) = 3x2 – (2 + 3k) x + (2k – 8) = 3x2 + x –10
= (3x – 5)(x + 2)
8.
A1 cso
(a)
So
(a)
3
M1
1
4
a  12 b  34 or a  2b  3
f – 2  –16  4a – 2b – 6
A1
M1
f – 2  0  4a – 2b  22
(b)
2
M1
M1A1
1
1
1
f 12   2   a   b  – 6
8
4
2
f  12   –5 
5
A1
Eliminating one variable from 2 linear simultaneous
equations in a and b
M1
a = 5 and b = – 1
A1
2x3 + 5x2 – x – 6 = (x+2)(2x2 + x – 3)
M1
= (x + 2)(2x + 3)(x – 1)
NB x  2x  32 2 x – 2 is A0
is A1
6
But 2 x  2x  32 x – 1
M1A1
3
C2 Coordinate geometry (Circles) Answers
1.
Gradient of PQ is –
(a)
1
3
1
y  2   ( x  2)
3
2.
B1
(3y + x = 8)
3+x=8
M1 A1 3
(b)
y = 1:
x=5
B1
(c)
(“5”  2)2 + (1  2)2
M: Attempt PQ2 or PQ
(x  5)2 + (y  1)2 = 10
M: (x ± a)2 + (y ± b)2 = k
M1 A1
 1 3 6  4 
Centre 
,
 , i e (1, 5)
2 
 2
M1, A1
2
or r = (1 – (–1))2 + (5 – 4)2 or r2 = (3 – 1)2 + (6 – 5)2 o.e.
(x – 1)2 + (y – 5)2 = 5
2
4.
(a)
1  (2) 3  3
 or
3 1
2 2
2
Gradient of l:
= 
M: use of m1m2 = –1, or equiv.
3
y 1
2
2
y – 1 =  ( x  3) or
  [3y = –2x + 9] (Any equiv. form)
3
x 3
3
Gradient of AM:
(b)
x = 6: 3y = –12 + 9 = –3 y = –1 (or show that for y = –1, x = 6)(*)
(A conclusion is not required)
(c)
(r2 =) (6 – 1)2 + (–1 – (–2))2
M: Attempt r2 or r
N.B. Simplification is not required to score M1A1
(x  6)2 + (y  1)2 = k, k  0 (Value for k not needed, could be r2 or r)
(x – 6)2 + (y + 1)2 = 26 (or equiv.)
(a)
M1 A1 4
(3  (1)) 2  (6  4) 2
r=
3.
1
(x – 6)2 + (y – 4)2 =; 32
M1
M1, A1, A1
6
B1
M1
M1A1
4
B1
1
M1A1
M1
A1
4
B1; B1 2
Allow 9 for 32.
(b)
Complete method for MP: =
=
(12  6) 2  (6  4) 2
M1
40 or awrt 6.325
A1
[These first two marks can be scored if seen as part of solution for (c)]
Complete method for cos θ, sin θ or tan θ
MT
3
e.g. cos  =
(= 0.4743) ( = 61.6835°)

MP candidate' s 40
[If TP = 6 is used, then M0]
θ = 1.0766 rad AG
(c)
Complete method for area TMP; e.g. =
=
3
31 (= 8.3516..) allow awrt 8.35
2
1
 3  40 sin 
2
M1
A1
M1
A1
4
Area (sector)MTQ = 0.5 × 32 × 1.0766 (= 4.8446…)
Area TPQ = candidate’s (8.3516.. – 4.8446..)
= 3.507 awrt
[Note: 3.51 is A0]
5.
(a)
(8 – 3)2 + (3 –1)2 or
2
M1
M1
A1
(8  3) 2  (3  1) 2
2
2
M1A1
2
(x ± 3) + (y ± 1) = k or (x ± 1) + (y ± 3) = k (k a positive value)
(x – 3)2 + (y – 1)2 = 29 (Not
(b)
6.
(a)
(a)
PQ: m1 
10 – 2
( 2 )
9 – (– 3) 3
(c)
A1
QR: m2 
and
8
6

 –1
12 9 – a
a = 13
10 – 4
9–a
(*)
Uses (x – a)2 + (y – b)2 = r2 or x2 + y2 + 2gx + 2fy + c = 0 and
substitutes
(–3, 2), (9, 10) and (13, 4) then eliminates one unknown
Eliminates second unknown
a = 5, b = 3, r2 = 65
(x – 3)2 –9 + (y + 2)2 –4 =12
Centre is (3, –2)
(x – 3)2 + (y + 2)2 = 12 + “9” + “4”
(b)
or 5.392)
2
(or exact equiv.) Must be seen or used in (b)
5
5
Gradient of tangent =
(Using perpendicular gradient method)
2
5
y–3=
(x – 8) (ft gradient of radius, dependent upon both M marks)
2
M1A1ft
5x + 2y – 46 = 0 (Or equiv., equated to zero, e.g. 92 – 4y – 10x = 0)
(Must have integer coefficients)
Obtains g= – 5, f = –3, c = –31 or
7.
PQ  (7 – – 1) 2  (– 5 – 1) 2 or
r  12  "9"  "4"  5 (or 25 )
82  6 2
A1
M1
M1 A1
M1
A1, A1, B1cao
5
M1 A1, A1
M1 A1
5
M1
B1
N (2, –1)
3
M1
R must lie on the circle (angle in a semicircle theorem)… often implied
by a diagram with R on the circle or by subsequent working)
y = – 6 or 2 (Ignore y = –6 if seen, and ‘coordinates’ are not required))
(a)
M1
A1
y2 +4y – 12 = 0
(y – 2)(y + 6) = 0 y = ..... (M is dependent on previous M)
4
B1
= 10 = 2 × radius, diam. (N.B. For A1, need a comment or conclusion )
x=0 
8.
M1
Gradient of radius =
m1 m2 = – 1:
(b)
 29 
2
5
2
M1
dM1
A1
4
B1 B1
2
5
169 13
  6.5
4
2
(b)
r
(c)
Complete Method to find x coordinates, x2 – x1 = 12 and
x1  x 2
 2 then solve
2
B1
To obtain x1 = –4,x2= 8
So y2 = y1 = –3.5
Let ANˆ B  2  sin  
6
  67.38......
"6.5"
So angle ANB is 134.8 *
(e)
AP is perpendicular to AN so using triangle
AP
ANP tan  
"6.5"
Therefore
M1
A1ft A1ft
Complete Method to find y coordinates, using
equation of circle or Pythagoras i.e. let d be the distance
below N of A then d2=6.52–62  d=2.5  y =..
(d)
1
AP = 15.6
M1
A1
5
M1
A1
2
M1
A1cao
2
C2 Logarithms Answers
1.
2.
xlog5 = log17
log 17
x=
log 5
= 1.76
or
x = log517
M1
A1
A1
log 0.8
or log8 0.8, = –0.107
log 8
(a)
x=
(b)
2logx = logx2
Allow awrt
3.
M1, A1 2
B1
2
x
7x
“Removes logs” to form equation in x, using the base
x2
correctly:
=3
7x
x = 21
(Ignore x = 0, if seen)
logx2 – log7x = log
3
M1
M1
A1cso
4
(Working with two equations in log3a and log3b)
“Taking logs” of first equation and “separating” log3 a = log3 3 + log3 b
(= 1 + log3b)
Solving simultaneous equations to find log3 a or log3 b
[log3 a = 1½, log3 b = ½]
Using base correctly to find a or b
Correct value for a or b a = 33 or b = 3
Correct method for second answer, dep. on first M; correct second answer
[Ignore negative values]
M1
M1
M1
A1
M1; A1 6
Answers must be exact; decimal answers lose both A marks
4.
(a)
log 7
or x = log5 7 (i.e. correct method up to x =…)
log 5
1.21 Must be this answer (3 s.f.)
x=
M1
A1
2
1.21 with no working: M1 A1 (even if it left as 51.21).
Other answers which round to 1.2 with no working: M1 A0.
(b)
5.
(5x – 7)(5x – 5) Or another variable, e.g. (y – 7)(y – 5),
even (x – 7)(x – 5)
(5x = 7 or 5x = 5)
x = 1.2 (awrt) ft from the answer to (a), if used
x = 1 (allow 1.0 or 1.00 or 1.000)
4– x
M1A1
A1ft
B1
4
2log5 x = log5 (x2),
log5 (4 – x) – log5 (x2) = log5
4– x
log  2   log 5
 x 
5x2 + x – 4 = 0 or 5x2 + x = 4 o.e.
M1 A1
(x = – 1)
dM1 A1
(5x – 4)(x + 1) = 0
x
4
5
x2
B1 M1
6
6.
(a)
(b)

log2y = –3
y = 2–3
M1
y  18 or 0.125
A1
32 = 25 or 16 = 24 or 512 = 29
[or log232 = 5log22 or log216=4log2 2
[or log2 32 =
(a)
M1
or log2 512 9log2 2]
log 10 16
log 10 512
log 10 32
or log2 16 =
or log2 512 =
]
log 10 2
log 10 2
log 10 2
log232 + log216 = 9
A1
(log x)2 = ... or (log x)(log x) = ... (May not be seen explicitly, so
M1 may be implied by later work, and the base may be 10 rather than 2)
log2x = 3  x = 23 = 8
M1
A1
log2 x = –3  x = 2–3 =
7.
1
8
A1ft
logx 64 = 2  64 = x2
A1
log2(11–6x)=log2(x–1)2 + 3
M1
 11 – 6 x 
log 2 
3
2 
  x – 1 
M1
11 – 6 x
x – 1
2
5
M1
So x = 8
(b)
2
=23
{11 – 6x = 8(x2 –2x+1)} and so 0 = 8x2 – 10x – 3
0=(4x+1)(2x–3)  x =...
3  1
x  , – 
2  4
M1
A1
dM1
2
C2 Geometric series
1.
(a)
ar = 4,
a
 25
1 r
a = 25(1  r)
(These can be seen elsewhere)
25r(1  r) = 4
B1, B1
M: Eliminate a
M1
25r2  25r + 4 = 0
A1cso 4
1
5
4
5
M1,A1
2
20 or
5
M1,A1
2
(5r  l)(5r  4) = 0
(c)
r = ... a = ... ,
(d)
Sn 
(e)
25(1  0.8n) > 24 and proceed to n = ..
(or >, or <) with no unsound algebra.
r =... ,
a (1  r n )
a
, but
 25, so Sn = 25(l  rn)
1 r
1 r


log 0.04
 n 
(14.425) 
log 0.8


2.
or
(b)
(a)
M1
n = 15
A1
a (1  r n )
(*) A1cso 4
1 r
a = 200, r = 2, n = 10, S10 =
200 (1  210 )
1 2
M1, A1
= 204,600
(c)
a

A1
5
1
,r 
6
3
a
S =
, S 
1 r
3.
2
{Sn = } a + ar + … + arn – 1 B1
{rSn = } ar + ar2 + … + arn M1
(1 – r)Sn = a(1 – rn)
dM1
Sn =
(b)
B1
3
B1
5
6
M1
1
1
3
5
o.e.
4
A1
(d)
–1 < r < 1 (or r< 1)
(a)
50 000rn–1 (or equiv.) (Allow arn–1 if 50 000rn–1 is seen in (b))
(b)
50 000rn–1 > 200 000
(Using answer to (a), which must include r and n, and 200 000)
(Allow equals sign or the wrong inequality sign)
(Condone ‘slips’ such as omitting a zero)
B1
1
rn–1 > 4  (n – 1)logr > log4
(Introducing logs and dealing correctly with the power)
(Allow equals sign or the wrong inequality sign)
M1
B1
1
M1
3
1
n>
(c)
(d)
4.
(a)
(b)
(c)
5.
log 4
1
log r
log 4
log 4
(n > 17.086...) (Allow equality)
 1 or n  1 
log 1.09
log 1.09
Year 18 or 2023 (If one of these is correct, ignore the other)
r = 1.09: n >
a (1  r n ) 50 000 (1  1.09 10 )

1 r
1  1.09
£760 000 (Must be this answer... nearest £10000)
Sn =
3
M1
A1
2
M1A1
A1
3
M1
Complete method for finding a
[e.g. Substituting value for r into equation of form ark = 10 or 80
and finding a value for a.]
5
1
(8a = 10) a   1 (equivalent single fraction or 1.25)
4
4
M1
Substituting their values of a and r into correct formula for sum.
a ( r n  1) 5 20
 (2  1) (= 1310718.75)
S=
1 310 719 (only this)
r 1
4
M1
4
T20 = 5 ×  
5
(b)
S 
(d)
A1cso
Complete method, using terms of form ark, to find r
[e.g. Dividing ar6 = 80 by ar3 = 10 to find r; r6 – r3 = 8 is M0]
r=2
(a)
(c)
(*)
19
= 0.072 (Accept awrt) Allow 5 ×
A1
A1
A1
4 19
for M1
5
2
2
2
M1 A1 2
5
 25
1  0.8
M1A1 2
5(1  0.8 k )
> 24.95
1  0.8
1 – 0.8k > 0.998 (or equiv., see below)
k log 0.8 < log 0.002 or k > log0.8 0.002
log 0.002
k>
(*)
log 0.8
(Allow with = or <)
M1
(Allow with = or <)
(Allow with = or <)
A1
M1
A1cso 4
k = 28 (Must be this integer value) Not k > 27, or k < 28, or k > 28
B1
1
[9]
6.
(a)
Initial step: Two of: a = k + 4, ar = k, ar2 = 2k – 15
Or one of: r 
2k – 15
2k – 15
k
, r
, r
k 4
k 4
k
M1
Or k  (k  4)(2k – 15) or even k3 = (k + 4)k(2k – 15)
k2 = (k + 4)(2k – 15), so k2 = 2k2 + 8k – 15k – 60
Proceed to k2 – 7k – 60 = 0
(b)
(k – 12)(k + 5) = 0 k = 12
(*)
(*)
M1, A1
A1
4
M1 A1
2
7.
2k – 15 12  3
k

or
   or 0.75 
k 4
k
16  4

(c)
Common ratio:
(d)
a
16

 64
1– r 1
4
M1 A1 2
(a)
324r3 = 96
r3 
 
r
or
8.
8
27
2
M1
A1cso
2
M1, A1
2
M1A1ft,
3
or 80% or equivalent].
B1cso
1
18000 × (0.8) n < 1000
M1
n log(0.8) < log 181 
M1
5
2
2
a   324 or a   96
3
3

729 1 – 23 
15
S15 
(d)
S 
(a)
18000×(0.8)3
1–
2
3
,
729
,
1 – 23
n

a = ...,
729
= 2182.00...
(AWRT 2180)
= 2187 M1, A1 2
=£9216 *
log 181 
log 0.8
[may see
 12.952 ......
u5 =200× (1.12)4,
(c)
(d)
r3 
(*)
(c)
(b)
or
2
3
2
(b)
96
324
M1 A1

1
4
so n = 13.
= £314.70 or £314.71

3
M1 A1
2
M1A1 A1
3

200 1.12 15 – 1
200 1.1215
or
,  7455 .94......
1.12 – 1
1 – 1.12
awrt £7460
S15 
A1 cso
C2 Differentiation Answers
1.
(a)
dy
 3x 2 16 x  20
dx
M1 A1
3x2  16x + 20 = 0
(3x  10)(x  2) = 0
x = ...,
10
and 2
3
dM1
A1
(b)
(c)
4
d2 y
d2 y
At
x
=
2,

6
x

16
 ...
dx 2
dx 2
M1
 4 (or < 0, or both), therefore maximum
A1ft
 x
M1 A1
3

 8 x 2  20 x dx 
x 4 8 x 3 20 x 2


( C )
4
3
2
A1
(d)
4
3
 68 
 
 3 
64
 40
3
Area of  =
1  10

  2  16
2 3

Shaded area =
(a)
(b)
M1
y = 8  32 + 40 = 16
A: x = 2:
2.
68 32 100


3
3
3
(May be scored elsewhere)
1

 ( xB  x A )  y A 
2


B1
 32 
 
 3 
M1
1

  33 
3

f(x) = 3x2 + 6x
f(x) = 6x + 6

M1 A1 5
B1
M1, A1cao
( x 3  3 x 2  5)dx 
x 4 3x 3

 5x
4
3
3
M1, A1
2
 x4

1
3
  x  5 x   4  8  10  (  1  5)
4
 4
1
3
 15 o.e.
4
3.
(a)
(b)
2
dC
2
 1400 v  2 
dv
7
2
–1400v–2 +
=0
7
v2 = 4900
v = 70
d 2C
dv 2
= 2800v–3
v = 70,
M1
A1
4
M1, A1
M1
dM1
A1cso
5
M1
d 2C
> 0 { minimum}
dv 2
2
d 2C
or v = 70,
= 2800 × 70–3 { 
= 0.00816...}{ minimum}
2
245
dv
A1ft
2
(c)
v = 70, C =
1400 2  70

70
7
M1
C = 40
4.
(a)
(b)
(c)
5.
(a)
(b)
(c)
(d)
6.
(a)
(b)
4x2 + 6xy = 600
 600  4 x 2
V = 2x2y = 2 x 2 
6x

A1
M1A1

4x 3
 V = 200x –

3

(*)
dV
= 200 – 4x2
dx
dV
Equate their
to 0 and solve for x2 or x : x2 = 50 or x = 50 (7.07...)
dx
4
Evaluate V: V = 200(50) – (5050) = 943 cm3 Allow awrt
3
d 2V
dx 2
2
= –8x Negative.  Maximum
M1A1cso
4
B1
M1A1
M1A1
5
M1, A1ft 2
(Total area) = 3xy + 2x2
100
100
(Vol:) x2y = 100 (y = 2 , xy 
)
x
x
B1
Deriving expression for area in terms of x only
(Substitution, or clear use of, y or xy into expression for area)
300
(Area =)
AG
 2x 2
x
M1
dA
300
  2  4x
dx
x
dA
Setting
= 0 and finding a value for correct power of x, for cand. M1
dx
[x3 = 75]
x = 4.2172 awrt 4.22 (allow exact 3 75 )
d 2 A 600
 3  4 = positive, > 0; therefore minimum
dx 2
x
Substituting found value of x into (a)
(Or finding y for found × and substituting both in 3xy + 2x2)
100
[y 
 5.6228 ]
4.2172 2
Area = 106.707 awrt 107
 dy 
  8 + 2x – 3x2 (M: xn  xn–1 for one of the terms, not just 10  0)

d
x


2
3x – 2x – 8 = 0 (3x + 4)(x – 2) = 0 x = 2 (Ignore other solution) (*)
1
 2  22 (M: Correct method to find area of triangle)
2
(Area = 22 with no working is acceptable)
8x 2 x 3 x 4
2
2
10

8
x

x

x
d
x

10
x




2
3
4
n
n+1
(M: x  x for one of the terms)
M1A1A1
Only one term correct: M1A0A0
Area of triangle =
B1
A1cso 4
M1A1
A1
4
M1; A1 2
M1
A1
2
M1A1
A1cso 3
M1A1
2 or 3 terms correct: M1A1A0
Integrating the gradient function loses this M mark.
2

8x 2 x 3 x 4 
10
x


   ...... (Substitute limit 2 into a ‘changed function’)

2
3
4 0

8


  20  16   4  (This M can be awarded even if the other limit is wrong)
3


2
38 
2
Area of R = 34  22 
  12  (Or 12. 6 )
3
3 
3
7.
(a)
2πrh + 2πr2 = 800
h
(b)
400 – r 2
,
r
M1A1 8
B1
 400 – r 2
V = πr2 
r


  400 r – r 3


(*)
dV
 400 – 3r 2
dr
400 – 3π r2 = 0
M1
M1, M1 A1
4
M1 A1
r2 = ...,
V = 400r – πr3 = 1737 =
800
3
r
400
3
(= 6.5 (2 s.f.))
400
(cm3 )
3
M1 A1
M1 A1
6
M1 A1
2
(accept awrt 1737 or exact answer)
(c)
8.
(a)
d 2V
 – 6r , Negative,  maximum
dr 2
(Arc length =) rθ = r × 1 = r . Can be awarded by implication from
later work, e.g. 3rh or (2rh + rh) in the S formula.
(Requires use of θ = 1).
1 2
1
r2
r   r 2 1  . Can be awarded by
2
2
2
implication from later work, e.g. the correct volume formula. (Requires
use of θ = 1).
Surface area = 2 sectors + 2 rectangles + curved face
(= r2 + 3rh) (See notes below for what is allowed here)
Volume = 300 = 12 r 2 h
(Sector area =)
Sub for h: S  r 2  3 3
(b)
600 2 1800
r 
r
r
(*)
dS
1800
 2r – 2 or 2r –1800r–2 or 2r + –1800r–2
dr
r
dS
 0  r 3  ....,r  3 900 or AWRT 9.7 (NOT – 9.7 or ± 9.7)
dr
(c)
d2S
d2S
3600
and
consider
sign,

.....
 2  3  0 so point is a
2
2
dr
dr
r
minimum
B1
B1
M1
B1
A1cso
5
M1A1
M1, A1
4
M1, A1ft
2
S min  (9.65...)2 
(d)
1800
9.65...
(Using their value of r, however found, in the given S formula)
M1
= 279.65… (AWRT: 280) (Dependent on full marks in part (b))
9.
(a)
A1
2
 y  12 x 12 – x 32 – 10 


 y  
6x
6
Puts their
x
1
2
–
3 12
x
2
M1 A1
3 12
x 0
2
M1
– 12
–
12
 4 (If x = 0 appears also
3
as solution then lose A1)
So x =
,
M1 A1
3
x = 4,  y = 12×2 – 4 2 – 10,
–
3
2
so y = 6
dM1 A1
7
M1A1
2
B1
1
1
3 –2
x
4
(b)
y"  –3x
(c)
[Since x >0] It is a maximum
–
C2 Integration Answers
1.
 3x
x
2.
3
2

3x 3
4 x 1
 5x 
3
1
 5  4 x  2 dx 
5 x  4 x 1

2
1
(= x3 + 5x  4x1)
M1 A1 A1
= (8 + 10  2)  (1 + 5  4), = 14
M1, A1
5
dy
 3x 2 16 x  20
dx
(a)
(b)
(c)
M1 A1
3x2  16x + 20 = 0
(3x  10)(x  2) = 0
A1
4
x = ...,
10
and 2
3
dM1
d2 y
d2 y
At
x
=
2,

6
x

16
 ...
dx 2
dx 2
M1
 4 (or < 0, or both), therefore maximum
A1ft
 x
M1 A1
3

 8 x 2  20 x dx 
x 4 8 x 3 20 x 2


( C )
4
3
2
A1
4
(d)
3
 68 
 
 3 
64
 40
3
Area of  =
M1
y = 8  32 + 40 = 16
A: x = 2:
1  10

  2  16
2 3

Shaded area =
3.
(May be scored elsewhere)
1

 ( xB  x A )  y A 
2


(b)
B1
 32 
 
 3 
M1
68 32 100 
1


  33 
3
3
3 
3
f(x) = 3x2 + 6x
f(x) = 6x + 6
(a)
M1 A1 5
B1
M1, A1cao
( x 3  3 x 2  5)dx 
x 4 3x 3

 5x
4
3
3
M1, A1
2
 x4

1
3
  x  5 x   4  8  10  (  1  5)
4
 4
1
3
 15 o.e.
4
4.
y = x(x2 – 6x + 5)
= x3 – 6x2 + 5x
 (x
3
2
M1
A1
M1, A1
 6 x 2  5 x ) dx 
4
3
x
6x
5x


4
3
2
3
M1, A1ft
1
 x4
5x 2 
5
3
1
3
  2x 
   2 0 
2 0  4
2
4
 4
M1
2
 x4
5x 2 
3
11
3
  2x 
  (4  16  10)   
2 1
4
4
 4
M1, A1(both)
4
total area =

3 11

4 4
M1
7
o.e.
2
A1cso
9
1
5.
x

1
2 dx
x2

1
 
2
(Or equivalent, such as
1
2x 2
, or 2 x )
M1A1
8
 1 
 2 
 x   2 8  2  2  4 2 [or 42 – 2, or 2(22 – 1), or 2(–1 + 22)]
 1  
  
  2  1
6.
(a)
 dy 
  8 + 2x – 3x2 (M: xn  xn–1 for one of the terms, not just 10  0)

 dx 
3x2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0 x = 2 (Ignore other solution) (*)
The final mark may also be scored by verifying that
(b)
M1A1 4
M1A1
A1cso 3
dy
 0 at x = 2.
dx
1
 2  22 (M: Correct method to find area of triangle)
2
(Area = 22 with no working is acceptable)
8x 2 x 3 x 4
2
2
10

8
x

x

x
d
x

10
x




2
3
4
n
n+1
(M: x  x for one of the terms)
M1A1A1
Only one term correct: M1A0A0
2 or 3 terms correct: M1A1A0
Integrating the gradient function loses this M mark.
Area of triangle =
M1A1
2

8x 2 x 3 x 4 

   ...... (Substitute limit 2 into a ‘changed function’)
10 x 
2
3
4 0

8


  20  16   4  (This M can be awarded even if the other limit is wrong)
3


2
38 
2
Area of R = 34  22 
  12  (Or 12. 6 )
3
3 
3
7.
M1
M1A1 8
(a)
Either solving 0 = x(6 – x) and showing x = 6 (and x = 0)
B1
1
(b)
or showing (6, 0) (and x = 0) satisfies y = 6x – x2
[allow for showing x = 6]
Solving 2x = 6x – x2 (x2 = 4x) to x = ….
x = 4 ( and x = 0)
Conclusion: when x = 4, y = 8 and when x = 0, y = 0,
M1
A1
A1
3
(c)
(Area =)

( 4)
( 0)
(6 x  x 2 )dx Limits not required
x3
(+ c)
3
Correct use of correct limits on their result above (see notes on limits)
x3 4
x3
1
2
["3x 2 
" ]  ["3 x 2 
" ] 0 with limits substitute d [ 48  21  26 ]
3
3
3
3
Area of triangle = 2 × 8 =16 (Can be awarded even if no M scored,
Correct integration 3 x 2 
M1
A1
M1
i.e. B1)
Shaded area = ± (area under curve – area of triangle ) applied correctly
2
2
( 26  16)  10 (awrt 10.7)
3
3
8.
y = (1+ x)(4 – x) = 4 + 3x – x2

(4  3 x – x 2 )dx  4 x 

3x 2 x 3
–
2
3
M: Attempt to integrate
1


 2 x  3x 2  dx =


1


(a)
M1 A1
5
M1 A1A1
4
3
 2

 x  2 x 2   (16  2  8) – (1  2)

 1
= 29
10.
5

  20 
6

3
1
3x
2


2
x
 2 x  3x 2  dx 
 32


2
2


4
A1
M: Expand, giving 3 (or 4) terms
64  
3 1  125

= [.................]4–1 = 16  24 –  –  – 4    
3 
2 3 6

9.
A1
M1
M1
(29 + C scores A0)
A1
Puts y = 0 and attempts to solve quadratic e.g. (x – 4)(x – 1) = 0
Points are (1,0) and (4, 0)
5
M1
A1
2
(b)
x = 5 gives y = 25 – 25 + 4 and so (5, 4) lies on the curve
B1cso
1
(c)
 x
M1A1
2
(d)
Area of triangle  14  4  4  8 or

– 5x  4 dx  13 x 3 – 52 x 2  4 x
2
 c
 x – 1dx 
1
2
x2 – x
with limits 1 and 5 to give 8
B1
5
Area under the curve 

1
 53
3
4
1
3
5

4
–
5

– 52  5 2  4  5   
6

 4 3 – 52  4 2  4  4
8

 – 3 


5
8 11
– 
or equivalent (allow 1.83 or 1.8 here)
6
3 6
Area of R = 8 –
(not 6.17)
M1
M1
A1 cao
11
37
 6 16 or or 6.16 r
6
6
A1 cao 5
6
C2 Trigonometry equations
1.
(a)
tan θ = 5
(b)
tanθ = k
B1
  tan k 
1
 = 78.7,
M1
258.7
(Accept awrt)
A1,
3
A1ft
2.
3.
1
2(1 – sin2x) + 1 = 5sinx
2sin2x + 5sinx – 3 = 0
(2sinx – 1)(sinx + 3) = 0
1
sinx =
2
 5
x= ,
6 6
M1
M1
M1, A1
M1, M1, A1cso
6
(a)
y
1
0.5
x
2

12
2
–0.5
–1
Sine wave (anywhere with at least 2 turning points.
Starting on positive y-axis, going up to a max., then min. below
x-axis, no further turning points in range, finishing above x-axis at
x = 2 or 360°. There must be some indication of scale on the y-axis...
(e.g. 1, –1 or 0.5)
Ignore parts of the graph outside 0 to 2.
(b)
(c)
 1   5   11 
,0 , 
,0 
 0, , 
 2  6   6 
(Ignore any extra solutions) (Not 150°, 330°)
B1, B1, B1
awrt 0.71 radians (0.70758...), or awrt 40.5° (40.5416...) ()

 

( – ) (2.43...) or (180 – ) if  is in degrees.  NOT      
6 


Subtract

from  (or from ( – ))... or subtract 30 if  is in degrees
6
0.18 (or 0.06), 1.91 (or 0.61)
Allow awrt
st
nd
(The 1 A mark is dependent on just the 2 M mark)
4.
(a)
(b)
3 sin2  – 2 cos2  = 1
3 sin2  – 2(1 – sin2 ) = 1
3 sin2  – 2 + 2 sin2  = 1
5 sin2  = 3 cso
A1
2
3
B1
M1
M1
A1, A1
5
(M1: Use of sin2  + cos2  = 1)
M1
AG
A1
3
, so sin  = (±)0.6
5
Attempt to solve both sin = +.. and sin  = – …
(may be implied by later work)
 = 50.7685° awrt  = 50.8° (dependent on first M1 only)
 (= 180º – 50.7685c°); = 129.23…° awrt 129.2º
sin2  =
M1
M1
M1
A1
M1;
2
A1ft
[f.t. dependent on first M and 3rd M]
sin  = –0.6
 = 230.785° and 309.23152° awrt 230.8º, 309.2º (both)
5.
(a)
4(1 – cos2 x) + 9cos x – 6 = 0
(b)
(4cos x – 1)(cos x – 2) = 0 cos x = ...,
x = 75.5
4cos2 x – 9cos x + 2 = 0 (*)
1
4
(i)
(α)
(ii)
M1, M1
A1
θ = –45,
135

θ = 23.6,
156.4
2
5
4 sin x =
(AWRT: 24, 156)
3 sin x
cos x

(AWRT: 41, 319)

2sin2x+ 5sinx–3 = 0
(b)
(a)
(b)
M1
B1, B1ft
6
M1
(*)
A1cso
(2s–1)(s+3)=0 giving s =
M1
[sin x = –3 has no solution] so sin x= 12
A1
 x = 30, 150
8.
4
B1, B1
x = 41.4, 318.6
5sin x =1+ 2 1 – sin 2 x
B1, B1ft
M1
x = 0, 180 seen
(a)
6
B1, B1ft
4sinxcosx=3sinx
 sinx(4cosx – 3)=0
Other possibilities (after squaring): sin2 x(16sin2 x – 7) = 0,
(16cos2x – 9)(cos2x –1) = 0
7.
2
B1
tanθ = –1 
sinθ =
M1 A1
M1 A1
360 – α, 360 + α or 720 – α
284.5, 435.5, 644.5
6.
M1A1 7
2
B1 B1ft
4
tan 
B1
1
awrt 21.8 ( )
B1
2
(or 0.4)
(i.s.w. if a value of  is subsequently found)
5
Requires the correct value with no incorrect working seen.
5
in (a), but no other ft)
2
(This value must be seen in part (b). It may be implied by a correct
solution, e.g. 10.9)
180   ( 201.8) , or 90   / 2  (if division by 2 has
already occurred)
(  found from tan 2 x  ... or tan x  ... or sin 2 x  ...
or cos 2 x  ... )
360   ( 381 .8) , or 180   / 2 
(  found from tan 2 x  ... or sin 2 x  ... or cos 2 x  ... )
OR 540   ( 561 .8) , or 270   / 2 
(  found from tan 2 x  ... )
(Also allow awrt 68.2, ft from tan 
M1
M1
Dividing at least one of the angles by 2
(  found from tan 2 x  ... or sin 2 x  ... or cos 2 x  ... )
x  10.9, 100.9, 190.9, 280.9 (Allow awrt)
M1
A1
5
C2 Trigonometry
1.
(a)
rθ = 2.12 × 0.65
1.38 (m)
(b)
1 2
1
r    2.12 2  0.65
2
2
1.46 (m2)
(c)
(d)

2
 0.65
0.92 (radians)
(a)
cos PQR =
PQR =
(b)
(c)
()
M1 A1
2
M1 A1
2
1
ACD: (2.12)(1.86) sinα (With the value of α from part (c))
2
Area = “1.46” + “1.57”,
2.
M1 A1 2
3.03 (m2)
M1 A1 3
6 2  6 2  (6 3 ) 2 
1
  
266
2

M1, A1
2
3
A1
1 2  2
m
6 
2
3
= 12 m2 (*)
Area =
Area of  =
(e)
A1cso
1
2 2
m
 6  6  sin
2
3
Area of segment = 12 – 9 3 m2
= 22.1 m2
 2 
Perimeter = 6 + 6 + 6   m
3 

= 24.6 m
42 = 52 + 62 – (2 × 5 × 6cos)
52  62  42
cos =
2 5 6
 45  3
  
 60  4
A1cso
2
M1
A1
2
M1
A1ft
2
M1
Correct for their (a)to 1 decimal place or more
(a)
2
M1
6 + 6 + [6 × their (a)].
3.
3
M1
= 9 3 m2
(d)
M1
A1 ft
M1
A1
(*)
A1cso
3
2
(b)
3
sin2A +   = 1
(or equiv. Pythag. method)
4
7
1
7
 2
7 or equivalent exact form, e.g.
, 0.4375
 sin A   sin A 
16 
4
16

M1
A1
2
4.
N
C
B
700m
500m
15º
A
(a)
(b)
BC2 = 7002 + 5002 – 2 × 500 × 700 cos 15°
(= 63851.92…)
BC = 253 awrt
M1A1
A1
sin B
sin 15
M1

700 candidate' s BC
sin B = sin 15 × 700 /253c = 0.716.. and giving an obtuse B
(134.2°) dep on 1st M
M1
 = 180º – candidate’s angle B (Dep. on first M only, B can be acute)
 = 180 – 134.2 = (0)45.8 (allow 46 or awrt 45.7, 45.8, 45.9)
M1
A1
3
4
[46 needs to be from correct working]
5.
(a)
(x – 6)2 + (y – 4)2 =; 32
B1; B1 2
Allow 9 for 32.
(b)
Complete method for MP: =
=
(12  6) 2  (6  4) 2
M1
40 or awrt 6.325
A1
[These first two marks can be scored if seen as part of solution for (c)]
Complete method for cos θ, sin θ or tan θ
MT
3
e.g. cos  =
(= 0.4743) ( = 61.6835°)

MP candidate' s 40
[If TP = 6 is used, then M0]
θ = 1.0766 rad AG
(c)
Complete method for area TMP; e.g. =
1
 3  40 sin 
2
3
31 (= 8.3516..) allow awrt 8.35
2
Area (sector)MTQ = 0.5 × 32 × 1.0766 (= 4.8446…)
Area TPQ = candidate’s (8.3516.. – 4.8446..)
= 3.507 awrt
[Note: 3.51 is A0]
=
M1
A1
4
M1
A1
M1
M1
A1
5
6.
(a)
rθ = 7 × 0.8 = 5.6 (cm)
M1A1 2
(b)
1 2
1
r θ = × 72 × 0.8 = 19.6 (cm2)
2
2
M1A1 2
(c)
(d)
BD2 = 72 + (their AD)2 – (2× 7 × (their AD) × cos 0.8)
BD2 = 72 + 3.52 – (2 ×7 × 3.5 × cos 0.8) (or awrt 46° for the angle)
(BD = 5.21)
Perimeter = (their DC) + “5.6” + “5.21” = 14.3 (cm) (Accept awrt)
∆ ABD =
M1
A1
M1A1 4
1
× 7 × (their AD) × sin 0.8 (or awrt 46° for the angle) (ft their AD)
2
M1A1f
t
(= 8.78…)
1
ab sin C is quoted the use of any
2
two of the sides of ∆ ABD as a and b scores the M mark).
Area = “19.6” – “8.78…” = 10.8 (cm2) (Accept awrt)
(If the correct formula
7.
(a)
1 2
1
r    6 2  2.2  39.6 (cm 2 )
2
2
M1 A1
2
(b)
 2 – 2.2 
  – 1.1  2.04 (rad)

2


M1 A1
2
(c)
ΔDAC =
1
× 6 × 4sin 2.04
2
(≈ 10.7)
Total area = sector + 2 triangles = 61
8.
M1 A1 4
(a)
M1A1ft
(cm2)
M1 A1
(Arc length =) rθ = r × 1 = r . Can be awarded by implication from
later work, e.g. 3rh or (2rh + rh) in the S formula.
(Requires use of θ = 1).
1 2
1
r2
r   r 2 1  . Can be awarded by
2
2
2
implication from later work, e.g. the correct volume formula. (Requires
use of θ = 1).
Surface area = 2 sectors + 2 rectangles + curved face
(= r2 + 3rh) (See notes below for what is allowed here)
Volume = 300 = 12 r 2 h
(Sector area =)
Sub for h: S  r 2  3 3
(b)
600 2 1800
r 
r
r
(*)
dS
1800
 2r – 2 or 2r –1800r–2 or 2r + –1800r–2
dr
r
dS
 0  r 3  ....,r  3 900 or AWRT 9.7 (NOT – 9.7 or ± 9.7)
dr
(c)
d2S
d2S
3600
and
consider
sign,

.....
 2  3  0 so point is a
2
2
dr
dr
r
minimum
4
B1
B1
M1
B1
A1cso
5
M1A1
M1, A1
4
M1, A1ft
2
(d)
S min  (9.65...)2 
1800
9.65...
(Using their value of r, however found, in the given S formula)
= 279.65… (AWRT: 280) (Dependent on full marks in part (b))
9.
M1
A1
2
(a)
Either


sin ACˆ B sin 0.6

5
4
or 42 = b2 + 52 – 2 × b × 5 cos0.6
b
 ACˆ B = arcsin(0.7058...)
= [0.7835.. or 2.358]
Use angles of triangle
ABˆ C    0.6  ACˆ B
CBˆD   – 1.76  1.38 Sector area 
100 cos
2
0.6 – 36

2
= [6.96 or 1.29]
Use sine / cosine rule with value for b
–b2
sin B  sin40.6  b or cos B  2516
40
(But as AC is the longest side so)
ABˆ C = 1.76 (*)(3sf)
(But as AC is the longest side so)
ABˆ C = 1.76 (*)(3sf) [Allow 100.7° → 1.76]
In degrees 0.6 = 34.377°, ACˆ B =44.9
(b)
10 cos 0.6 
M1
M1
M1,
A1
4
1
2
 4 2   – 1.76 
1
= 11.0 ~ 11.1  4 2  79.3 is M0
2
Area of ABC  12  5  4  sin 1.76   9.8or
M1
1
 5  4  sin 101
2
A1
3
C2 Binomial Expansion
1.
(2 + x)6 = 64 ...
B1
6  5 4

(6 × 25 × x) + 
 2  x 2 ,
 2

+192x, +240x2
M1,A1,A1
2.
(a)
(b)
3.
(a)
5 4
5 4 3
(2 x) 2 
(2 x) 3  ...
2!
3!
= 1 – 10x + 40x2 – 80x3 + ...
(1 – 2x)5 = 1 + 5 × (–2x) +
(1 + x)(1 – 2x)5 = (1 + x)(1 – 10x + …)
= 1 + x – 10x + …
 1 – 9x (*)
1 + 6kx [Allow unsimplified versions, e.g. 16 + 6(15)kx, 6C0 + 6C1kx]
65
6 5 4
[See below for acceptable versions]

(kx) 2 
(kx) 3
2
3 2
N.B. THIS NEED NOT BE SIMPLIFIED FOR THE A1 (isw is applied)
4
B1, M1, A1,
A1
4
M1
A1
2
B1
M1A1
3
M1A1cso
2
A1cso
1
The terms can be ‘listed’ rather than added.
2
(or equiv. fraction, or 0.4) (Ignore k = 0, if seen)
5
(b)
6k = 15k2 k =
(c)
65 4  2 
32
c=
  
3 2  5 
25
3
(or equiv. fraction, or 1.28)
(Ignore x3, so
4.
(a)
 1 
1  x 
 2 
10
32 3
x is fine)
25
2
3
10  1  10  1  10  1 
 1    x     x     x 
 1  2   2  2   3  2 
45
(or 11.25)x2 + 15x3
4
(coeffs need to be these, i.e, simplified)
[Allow A1 A0, if totally correct with unsimplified, single
fraction coefficients)
M1A1
= 1 + 5x; 
(b)
(1 +
A1; A1 4
1
45
× 0.01)10 = 1 + 5(0.01) + (
or 11.25) (0.01)2 + 15(0.01)3
2
4
M1A1f
t
= 1 + 0.05 + 0.001125 + 0.000015
= 1.05114 cao
5.
(a)
(b)
(1 + ax)10 = 1 + 10ax....... (Not unsimplified versions)
10  9
10  9  8

(ax) 2 
(ax) 3 Evidence from one of these terms is sufficient
2
6
+ 45(ax)2, + 120(ax)3 or + 45a2x2, + 120a3x3
120a3 = 2 × 45a2 a =
90
3


,0.75  Ignore a = 0, if seen
or equiv.  e.g.
120
4


A1
3
B1
M1
A1, A1 4
M1 A1 2
6.
(3 – 2x)+ = 243,

7.
…… + 5 × (3)4 (–2x) = –810x ……
5 4 3
(3) (– 2 x) 2 
2
(a)
+ 1080x2
(7 × ...× x) or (21× ...× x2) The 7 or 21 can be in ‘unsimplified’ form.
B1 B1
M1 A1
4
M1
7
(2 + kx)7 = 27 + 26 × 7 × kx + 25 ×   k 2 x 2
 2
= 128; + 448kx, +672k2x2 [or 672(kx)2]
(If 672kx2 follows 672(kx)2, isw and
allow A1)
(b)
6 × 448k = 672k2
k=4
8.
3 – x 3
6
6
B1; A1, A1
4
M1
(Ignore k = 0, if seen)
6
2
 35  6  – x   3 4     – x 
2
 
= 729, –1458x, +1215x2
A1
2
M1
B1 A1 A1
[4]
C2 Trapezium rule
1.
(a)
Shape
B1
(0, 1), or just 1 on the y-axis, or seen in table for (b)
Beware the order of marks!
B1
(b)
Missing values:
(c)
1
 0.2, 1  3  2 (1.246  1.552  1.933  2.408)
2
= 1.8278
2.
(a)
(b)
(c)
3.
(a)
(b)
1.933,
2.408
(Accept awrt)
B1, B1
2
2
B1, M1 A1ft
(awrt 1.83)
Beware the order of marks!
A1
1.414 (allow also exact answer 2), 3.137
Allow awrt
1
(0.5)...
2
...{0 + 6 + 2(0.530 + 1.414 + 3.137)}
= 4.04 (Must be 3 s.f.)
4
B1, B1
2
B1
M1A1ft
A1
1
(2 × 6)
2
(Could also be found by integration, or even by the trapezium rule
on y = 3x)
Area required = Area of triangle – Answer to (b) (Subtract
either way round)
6 – 4.04 = 1.96
Allow awrt
(ft from (b), dependent on the B1, and on answer to (b) less than 6)
Area of triangle =
4
B1
M1
A1ft
1.732, 2.058, 5.196 awrt (One or two correct B1 B0, All correct B1 B1)
3
B1B1 2
1
 0.5 .......
2
..........{(1 .732 + 5.196) + 2(2.058 + 2.646 + 3.630)}
B1
M1A1f
t
= 5.899 (awrt 5.9, allowed even after minor slips in values)
4.
(a)
(b)
gives 2.236
(allow AWRT) Accept
x = 2.5
gives 2.580
(allow AWRT) Accept 2.58
1 1
  , [(1.414 + 3) + 2(1.554 + 1.732 + 1.957 + 2.236 + 2.580)]
2 2
= 6.133
(c)
5
x=2
(AWRT 6.13, even following minor slips)
Overestimate
‘Since the trapezia lie above the curve’, or an equivalent explanation, or
sketch of (one or more) trapezia above the curve on a diagram (or
A1
B1
B1
2
B1, [M1A1ft]
A1
4
B1
dB1
2
4