C1 Revision
Workbook
Video solutions can be found in the following link
http://www.examsolutions.co.uk/a-level-maths-papers/Edexcel/Core-Maths/Core-Maths-C1/C1-Papers.php
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1
C1 Algebra
1.
(a)
Express √108 in the form a√3, where a is an integer.
(1)
(b)
Express (2 – √3)2 in the form b + c√3, where b and c are integers to be found.
(3)
(Total 4 marks)
2.
Simplify
5 3
2 3
,
giving your answer in the form a+b√3, where a and b are integers.
(Total 4 marks)
3.
(a)
Write down the value of
1
125 3
.
(1)
(b)
Find the value of 125

2
3
.
(2)
(Total 3 marks)
4.
Write
√(75) – √(27)
in the form k √x, where k and x are integers.
(Total 2 marks)
5.
(a)
Find the value of
4
83
.
(2)
(b)
Simplify
4
15 x 3
3x
.
(2)
(Total 4 marks)
1
6.
(a)
Write down the value of 16 4 .
(1)
(b)
Simplify
3
12 4
(16 x )
.
(2)
(Total 3 marks)
7.
Given that 32 √2 = 2a, find the value of a.
(Total 3 marks)
8.
Factorise completely
x3 – 9x.
(Total 3 marks)
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9.
Solve the simultaneous equations
y – 3x + 2 = 0
y2 – x – 6x2 = 0
(Total 7 marks)
10.
Solve the simultaneous equations
y = x – 2,
y2 + x2 = 10.
(Total 7 marks)
11.
Find the set of values of x for which
(a)
4x – 3 > 7 – x
(2)
(b)
2x2 – 5x – 12 < 0
(4)
(c)
both 4x – 3 > 7 – x and 2x2 – 5x – 12 < 0
(1)
(Total 7 marks)
12.
The equation 2x2 – 3x – (k + 1) = 0, where k is a constant, has no real roots.
Find the set of possible values of k.
(Total 4 marks)
13.
(a)
Show that x2 + 6x + 11 can be written as
(x + p)2 + q
where p and q are integers to be found.
(2)
(b)
In the space at the top of page 7, sketch the curve with equation y = x2 + 6x + 11, showing clearly any
intersections with the coordinate axes.
(2)
(c)
Find the value of the discriminant of x2 + 6x + 11
(2)
(Total 6 marks)
14.
The equation x2 + kx + (k + 3) = 0, where k is a constant, has different real roots.
(a)
Show that k 2 – 4k – 12 > 0.
(2)
(b)
Find the set of possible values of k.
(4)
(Total 6 marks)
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15.
The equation kx2  4 x  (5  k )  0 , where k is a constant, has 2 different real solutions for x.
(a)
Show that k satisfies
k 2  5k  4  0.
(3)
(b)
Hence find the set of possible values of k.
(4)
(Total 7 marks)
f(x) = x2 + 4kx + (3 + 11k),
16.
(a)
where k is a constant.
Express f(x) in the form (x + p)2 + q, where p and q are constants to be found in terms of k.
(3)
Given that the equation f(x) = 0 has no real roots,
(b)
find the set of possible values of k.
(4)
Given that k = 1,
(c)
sketch the graph of y = f(x), showing the coordinates of any point at which the graph crosses a
coordinate axis.
(3)
(Total 10 marks)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
January 2007
January 2008
January 2009
June2010
June 2007
January 2008
June 2009
June 2008
January 2010
January 2007
June 2009
January 2007
June 2007
June 2010
January 2009
January 2010
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Topic
Indices & surds
Indices & surds
Indices & surds
Indices & surds
Indices & surds
Indices & surds
Indices & surds
Algebra
Simultaneous equations
Simultaneous equations
Inequalities
Inequalities
Quadratics
Inequalities
Quadratics
Quadratics
Marks
4 marks
4 marks
3 marks
2 marks
4 marks
3 marks
3 marks
3 marks
7 marks
7 marks
7 marks
4 marks
6 marks
6 marks
7 marks
10 marks
Marks you scored
4
C1 Coordinate geometry
1.
The point A (–6, 4) and the point B (8, –3) lie on the line L.
(a)
Find an equation for L in the form ax + by + c = 0, where a, b and c are integers.
(4)
(b)
Find the distance AB, giving your answer in the form k√5, where k is an integer.
(3)
(Total 7 marks)
2.
The line l1 has equation 3x + 5y – 2 = 0
(a)
Find the gradient of l1.
(2)
The line l2 is perpendicular to l1 and passes through the point (3, 1).
(b)
Find the equation of l2 in the form y = mx + c, where m and c are constants.
(3)
(Total 5 marks)
3.
The line l1 has equation y = 3x + 2 and the line l2 has equation 3x + 2 y – 8 = 0.
(a)
Find the gradient of the line l2.
(2)
The point of intersection of l1 and l2 is P.
(b)
Find the coordinates of P.
(3)
The lines l1 and l2 cross the line y= 1 at the points A and B respectively.
(c)
Find the area of triangle ABP.
(4)
(Total 9 marks)
4.
(a)
Find an equation of the line joining A (7, 4) and B (2, 0), giving your answer in the form ax + by + c =
0, where a, b and c are integers.
(3)
(b)
Find the length of AB, leaving your answer in surd form.
(2)
The point C has coordinates (2, t), where t > 0, and AC = AB.
(c)
Find the value of t.
(1)
(d)
Find the area of triangle ABC.
(2)
(Total 8 marks)
5.
1
The line l1 passes through the point A (2, 5) and has gradient  .
2
(a)
Find an equation of l1, giving your answer in the form y = mx + c.
(3)
The point B has coordinates (–2, 7).
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(b)
Show that B lies on l1.
(1)
(c)
Find the length of AB, giving your answer in the form k 5 , where k is an integer.
(3)
The point C lies on l1 and has x-coordinate equal to p.
The length of AC is 5 units.
(d)
Show that p satisfies
p 2  4 p  16  0.
(4)
(Total 11 marks)
6.
The curve C has equation y 
(a)
3
and the line l has equation y = 2x + 5.
x
On the axes below, sketch the graphs of C and l, indicating clearly the coordinates of any intersections
with the axes.
(3)
(b)
Find the coordinates of the points of intersection of C and l.
y
O
x
(6)
(Total 9 marks)
Question
Topic
Marks
1.
January 2008
Coordinate Geometry
7 marks
2.
January 2010
Straight lines
5 marks
3.
June 2007
Straight lines
9 marks
4.
June 2010
Straight lines
8 marks
5.
January 2009
Coordinate Geometry
11 marks
6.
June 2008
Coordinate Geometry
9 marks
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Marks you
scored
6
C1 differentiation and Integration
1.
Given that
3
y = 4x – 1 +
find
1
2x 2
,
x > 0,
dy
.
dx
(Total 4 marks)
1
2.
dy
.
dx
Given that y = x4 + x 3  3, find
(Total 3 marks)
f(x) = 3x + x3, x > 0.
3.
(a)
Differentiate to find f′ (x).
(2)
Given that f′ (x) = 15,
(b)
find the value of x.
(3)
(Total 5 marks)
f( x) 
4.
(3 – 4 x ) 2
, x>0
x
(a)
Show that f( x)  9 x
–
1
2

1
Ax 2
 B, where A and B are constants to be found.
(3)
(b)
Find f´(x).
(3)
(c)
Evaluate f´(9).
(2)
(Total 8 marks)
5.
The curve C has equation
y = x3 – 2x2 – x + 9,
x>0
The point P has coordinates (2, 7).
(a)
Show that P lies on C.
(1)
(b)
Find the equation of the tangent to C at P, giving your answer in the form y = mx + c, where m and c
are constants.
(5)
The point Q also lies on C.
Given that the tangent to C at Q is perpendicular to the tangent to C at P,
(c)
show that the x-coordinate of Q is
1
(2  6 ).
3
(Total 11 marks)
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6.
The curve C has equation
y
(a)
Find
( x  3)( x  8)
, x>0
x
dy
in its simplest form.
dx
(4)
(b)
Find an equation of the tangent to C at the point where x = 2
(4)
(Total 8 marks)
7.
Given that
y  8x 3  4 x 
find
3x 2  2
, x0
x
dy
.
dx
(Total 6 marks)
8.
Given that y = 3x2 + 4x, x > 0, find
(a)
dy
dx
(2)
(b)
d2 y
,
dx 2
(2)
(c)
 ydx .
(3)
(Total 7 marks)
9.
Find
 (3x
2
 4 x5  7)dx .
(Total 4 marks)
10.
The curve C has equation y = f(x), x ≠ 0, and the point P(2, 1) lies on C. Given that
f ′(x) = 3x2 – 6 –
(a)
8
,
x2
find f(x).
(5)
(b)
Find an equation for the tangent to C at the point P, giving your answer in the form
y = mx + c, where m and c are integers.
(4)
(Total 9 marks)
11.
A curve has equation y = f(x) and passes through the point (4, 22).
Given that
f' ( x)  3x
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1
 3x 2
 7,
8
use integration to find f(x), giving each term in its simplest form.
(Total 5 marks)
12.
The curve C has equation y = f(x), x > 0, and f ( x)  4 x  6 x 
8
x2
.
Given that the point P (4, 1) lies on C,
(a)
find f(x) ands simplify your answer.
(6)
(b)
Find an equation of the normal to C at the point P (4, 1).
(4)
(Total 10 marks)
13.
The curve C with equation y = f(x) passes through the point (5, 65).
Given that f’(x) = 6x2 – 10x –12,
(a)
use integration to find f (x).
(4)
(b)
Hence show that f(x) = x(2x + 3)(x – 4).
(2)
(c)
Sketch C, showing the coordinates of the points where C crosses the x-axis.
(3)
(Total 9 marks)
Question
Topic
Marks
1.
January 2007
Basic differentiation
4 marks
2.
January 2008
Basic differentiation
3 marks
3.
June 2008
Basic differentiation
5 marks
4.
June 2009
Indices & surds
8 marks
5.
June 2009
Straight lines
11 marks
6.
January 2010
Basic differentiation
8 marks
7.
June 2010
Basic differentiation
6 marks
8.
June 2007
Differentiation
7 marks
9.
January 2008
Basic integration
4 marks
10. January 2007
Basic integration
9 marks
11. January 2008
Basic integration
5 marks
12. January 2009
Basic integration
10 marks
13.
Basic integration
9 marks
June 2007
Glyn Technology School
marks you socred
9
C1 Sequences
1.
A girl saves money over a period of 200 weeks. She saves 5p in Week 1, 7p in Week 2, 9p in Week 3, and so
on until Week 200. Her weekly savings form an arithmetic sequence.
(a)
Find the amount she saves in Week 200.
(3)
(b)
Calculate her total savings over the complete 200 week period.
(3)
(Total 6 marks)
2.
A sequence x1, x2, x3, … is defined by
x1 = 1,
xn+1 = axn – 3, n > 1,
where a is a constant.
(a)
Find an expression for x2 in terms of a.
(1)
(b)
Show that x3 = a2 – 3a – 3.
(2)
Given that x3 = 7,
(c)
find the possible values of a.
(3)
(Total 6 marks)
3.
A sequence of positive numbers is defined by
n  1,
a n 1  (a n2  3 ,
a1  2
(a)
Find a2 and a3, leaving your answers in surd form.
(2)
(b)
Show that a5 = 4
(2)
(Total 4 marks)
4.
A sequence a1, a2, a3, … is defined by
a1 = k,
an+1 = 2an – 7,
n ≥ 1,
where k is a constant.
(a)
Write down an expression for a2 in terms of k.
(1)
(b)
Show that a3 = 4k – 21.
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(2)
4
Given that
a
r
 43,
r 1
(c)
find the value of k.
(4)
(Total 7 marks)
5.
A sequence a1, a2, a3 ... , is defined by
a1 = k,
n ≥ 1,
an+1 = 3an + 5,
where k is a positive integer.
(a)
Write down an expression for a2 in terms of k.
(1)
(b)
Show that a3 = 9k + 20.
(2)
4
(c)
(i)
Find
a
in terms of k.
r
r 1
4
(ii)
Show that
a
r
is divisible by 10.
r 1
(4)
(Total 7 marks)
6.
A sequence is given by:
x1 = 1,
xn+1 = xn(p + xn),
where p is a constant (p ≠ 0).
(a)
Find x2 in terms of p.
(1)
(b)
Show that x3=1 + 3p + 2p2.
(2)
Given that x3 = 1,
(c)
find the value of p,
(3)
(d)
write down the value of x2008.
(2)
(Total 8 marks)
7.
Sue is training for a marathon. Her training includes a run every Saturday starting with a run of 5 km on the
first Saturday. Each Saturday she increases the length of her run from the previous Saturday by 2 km.
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(a)
Show that on the 4th Saturday of training she runs 11 km.
(1)
(b)
Find an expression, in terms of n, for the length of her training run on the nth Saturday.
(2)
(c)
Show that the total distance she runs on Saturdays in n weeks of training is n(n + 4) km.
(3)
On the n th Saturday Sue runs 43 km.
(d)
Find the value of n.
(2)
(e)
Find the total distance, in km, Sue runs on Saturdays in n weeks of training.
(2)
(Total 10 marks)
8.
The first term of an arithmetic series is a and the common difference is d.
The 18th term of the series is 25 and the 21st term of the series is 32
(a)
1
.
2
Use this information to write down two equations for a and d.
(2)
(b)
Show that a = –7.5 and find the value of d.
(2)
The sum of the first n terms of the series is 2750.
(c)
Show that n is given by
n 2  15n  55  40.
(4)
(d)
Hence find the value of n.
(3)
(Total 11 marks)
Question
Topic
Marks
1. June 2007
Arithmetic series
6 marks
2.
June 2008
General sequences & series
6 marks
3.
June 2010
General sequences & series
4 marks
4.
June 2009
General sequences & series
7 marks
5.
June 2007
General sequences & series
7 marks
6.
January 2008
General sequences & series
8 marks
7.
June 2008
Arithmetic series
10 marks
8.
January 2009
Simultaneous equations
11 marks
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Marks you socred
12
C1 Transformations of curves
1.
(a)
On the same axes sketch the graphs of the curves with equations
(i)
y = x2(x – 2),
(3)
(ii)
y = x(6 – x),
(3)
and indicate on your sketches the coordinates of all the points where the curves cross the x-axis.
(b)
Use algebra to find the coordinates of the points where the graphs intersect.
(7)
(Total 13 marks)
2.
y
(0, 7)
y = f( x)
O
(7, 0)
x
The diagram above shows a sketch of the curve with equation y = f(x). The curve passes through the point (0,
7) and has a minimum point at (7, 0). On separate diagrams, sketch the curve with equation
(a)
y = f(x) + 3,
(3)
(b)
y = f(2x).
(2)
On each diagram, show clearly the coordinates of the minimum point and the coordinates of the point at which the
curve crosses the y-axis.
(Total 5 marks)
3.
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Above is a sketch of the curve C with equation y = f(x). There is a maximum at (0, 0), a minimum at (2, –1)
and C passes through (3, 0).
On separate diagrams sketch the curve with equation
(a)
y = f(x + 3)
(3)
(b)
y = f(–x).
(3)
On each diagram show clearly the coordinates of the maximum point, the minimum point and any points of
intersection with the x-axis.
(Total 6 marks)
4.
y
(2, 5)
1
4
x
The diagram above shows a sketch of the curve with equation y = f(x). The curve crosses the x-axis at the
points (1, 0) and (4, 0). The maximum point on the curve is (2, 5). In separate diagrams sketch the curves
with the following equations. On each diagram show clearly the coordinates of the maximum point and of
each point at which the curve crosses the x-axis.
(a)
y = 2f(x),
(3)
(b)
y = f(–x).
(3)
The maximum point on the curve with equation y = f(x + a) is on the y-axis.
(c)
Write down the value of the constant a.
(1)
(Total 7 marks)
5.
Given that
(a)
f(x) =
1
,
x
x ≠ 0,
sketch the graph of y = f(x) + 3 and state the equations of the asymptotes.
(4)
(b)
Find the coordinates of the point where y = f(x) + 3 crosses a coordinate axis.
(2)
(Total 6 marks)
6.
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y
O
x
3
The diagram above shows a sketch of the curve with equation y  , x  0.
x
(a)
On a separate diagram, sketch the curve with equation y 
3
, x   2, showing the coordinates of
x2
any point at which the curve crosses a coordinate axis.
(3)
(b)
Write down the equations of the asymptotes of the curve in part (a).
(2)
(Total 5 marks)
7.
(a)
Factorise completely x3 – 6x2 + 9x
(3)
(b)
Sketch the curve with equation
y = x3 – 6x2 + 9x
showing the coordinates of the points at which the curve meets the x-axis.
(4)
Using your answer to part (b), or otherwise,
(c)
sketch, on a separate diagram, the curve with equation
y = (x – 2)3 – 6(x – 2)2 + 9(x – 2)
showing the coordinates of the points at which the curve meets the x-axis.
(2)
(Total 9 marks)
Question
Topic
Marks
1.
January 2007
Transformations & graphs
13 marks
2.
June 2008
Transformations & graphs
5 marks
3.
January 2009
Transformations & graphs
6 marks
4.
January 208
Transformations & graphs
7 marks
5.
January 2007
Transformations & graphs
6 marks
6.
June 2007
Transformations & graphs
5 marks
7.
June 2009
Quadratics
9 marks
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marks you scored
15
C1 Answers
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C1 Algebra
1.
(a)
6√3
(a = 6)
B1
1
Expanding (2 – √3)2 to get 3 or 4 separate terms
7, –4√3
(b = 7, c = –4)
M1
A1, A1
3.
±6√3 also scores B1.
(b)
[4]
2.
(5  3 )

(2  3 )
M1
(2  3 ) (2  3 )

10  2 3  5 3  ( 3 ) 2
...
 10  7 3  3 




...



13  7 3 
( 13  7 3 )  Allow


1


M1
13 (a = 13)
A1
 7 3 (b  7)
A1
4
[4]
3.
(a)
5
(b)
1
 1 
or 

2
( their 5)
 their 5 
=
(± 5 is B0) B1
1
2
M1
1
or 0.04
25
(
1
is A0)
25
A1
2
[3]
4.


75  27  5 3  3 3
M1
= 2 3
A1
[2]
5.
3
8 or
A1
3
(8 4 )
2
(a)
Attempt
= 16
M1
1
1
(b)
5x 3
5, x 3
B1, B1
2
[4]
6.
(a)
2
B1
1
Negative answers:
Allow –2. Allow ±2. Allow ‘2 or –2’.
(b)
x9 seen, or (answer to (a))3 seen, or (2x3)3 seen.
M1
8x9
A1
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2
[3]
7.
2 2
32 = 25 or 2048 = 211,
a
1
1
2
or 2048  (2048 ) 2
1


 or 5 or 5.5 
2


11
2
B1, B1
B1
[3]
8.
x(x2 – 9) or (x ± 0)(x2 – 9) or (x – 3)(x2 + 3x) or (x + 3)(x2 – 3x)
B1
x(x – 3)(x + 3)
M1A1 3.
[3]
9.
(3x – 2)2 – x – 6x2 (= 0)
y = 3x – 2
M1
9x2 – 12x + 4 – x – 6x2 = 0
3x2 – 13x + 4 = 0 (or equiv., e.g. 3x2 = 13x – 4)
(3x –1)(x – 4) = 0
x
x=…
M1 A1cso
1
(or exact
3
equivalent) x = 4
M1 A1
y = –1 y = 10
(Solutions need not be “paired”)
M1 A1
[7]
10.
(x – 2)2 = x2 – 4x + 4
or (y + 2)2 = y2 + 4y + 4 M: 3 or 4 terms
2
2
(x – 2) + x = 10
or y2 + (y + 2)2 = 10
M: Substitute
2x2 – 4x – 6 = 0
or 2y2 + 4y – 6 = 0
Correct 3 terms
(x– 3)(x + 1) = 0, x = ... or (y + 3)(y – 1) = 0, y = ...
(The above factorisations may also appear as (2x – 6)(x + 1) or equivalent).
x = 3 x= –1
y = 1 y= –3
y= –3 y = 1
x= –1 x = 3
6
(Allow equivalent fractions such as: x =
for x = 3).
2
or
or
M1
M1
A1
M1
A1
M1M1
7
[7]
11.
10
2
(a)
5x > 10, x > 2 [Condone x >
= 2 for M1A1]
(b)
(2x + 3)(x – 4) = 0, ‘Critical values’ are –
3
and 4
2
3
–  x 4
2
(c)
2<x<4
M1, A1
2
M1, A1
M1 A1ft
4
B1ft
1
[7]
12.
Use of b2 – 4ac, perhaps implicit (e.g. in quadratic formula)
(–3)2 – 4 × 2 × –(k + 1) < 0
(9 + 8(k + 1) < 0)
M1
A1
8k < –17 (Manipulate to get pk < q, or pk > q, or pk = q)
M1
k< 
1
17 

 Or equiv : k  2 or k  2.125 
8 
8

A1cso
4
[4]
Glyn Technology School
18
13.
(a)
 x  3
2
or
2
p  3 or
6
2
q2
B1
B1
2
(b)
(c)
U shape with min in 2nd quad
(Must be above x-axis and not on y = axis)
B1
U shape crossing y-axis at (0, 11) only
(Condone (11,0) marked on y-axis)
B1
b2  4ac  62  4 11
M1
= 8
A1
2
2
[6]
14.
(a)
(b)
Attempt to use discriminant b2 – 4ac
k2 – 4(k + 3) > 0  k2 – 4k – 12 > 0
(*)
M1
A1cso 2
k2 – 4k – 12 = 0  (k  a)(k  b), with ab = 12
4  4 2  4 12
or (k – 2)2  22 – 12
2
k = –2 and 6
(both)
k < –2, k > 6 or (–, –2);(6, )
M: choosing “outside”
or (k =)
M1
A1
M1 A1ft
4
[6]
15.
(a)
b2 – 4ac > 0  16 – 4k(5 – k) > 0 or equiv., e.g. 16 > 4k(5 – k)
So k2 – 5k + 4 > 0 (Allow any order of terms,
e.g. 4 – 5k + k2 > 0)
(b)
Critical Values
(k – 4)(k – 1) = 0
M1A1
(*)
k=....
k = 1 or 4
A1cso 3
M1
A1
Choosing “outside” region
k < 1 or k > 4
M1
A1
4
[7]
16.
(a)
4k 

(x + 2k)2 or  x 

2 

2
M1
(x ± F)2 ±G ± 3 ± 11k (where F and G are any functions of k,
not involving x)
M1
(x + 2k)2 – 4k2 + (3+11k)
equivalents such as
A1
Glyn Technology School
Accept unsimplified
3
19
2
2
4k   4k 

x
 –    3  11k ,
2   2 

and i.s.w. if necessary.
(b)
Accept part (b) solutions seen in part (a).
“4k 2 –11k – 3”= 0
(4k +1)(k – 3) = 0
k = …,
M1
[Or, ‘starting again’, b2 – 4ac = (4k)2 – 4(3 +11k)
1
and proceed to k = …] –  k  3 (Ignore any inequalities
4
for the first 2 marks in (b)).
A1
Using b2 – 4ac < 0 for no real roots, i.e. “4k2 –11k – 3”< 0,
to establish inequalities involving their two
critical values m and n
M1
(even if the inequalities are wrong, e.g. k < m, k < n).
1
 k  3 (See conditions below) Follow through
4
their critical values.
–
A1ft
4
The final A1ft is still scored if the answer m< k < n
follows k <m, k < n.
(c)
Shape
(seen in (c))
Minimum in correct quadrant, not touching the x-axis, not on the
y-axis, and there must be no other minimum or maximum.
B1
B1
(0, 14) or 14 on y-axis.
Allow (14, 0) marked on y-axis.
n.b. Minimum is at (–2,10), (but there is no mark for this).
B1
3
[10]
Glyn Technology School
20
C1 Coordinate geometry
1.
(a)
(b)
4  (3)
1
3 4
7
7

or
, 
or
  
2
6 8
8  (6)
 14
14

1
1
Equation: y  4   ( x  (6)) or y  (3)   ( x  8)
2
2
x + 2y – 2 = 0 (or equiv. with integer coefficients… must have ‘= 0’)
(e.g. 14y + 7x – 14 = 0 and 14 – 7x – 14y = 0 are acceptable)
m=
M1, A1
M1
A1
4
(–6 – 8)2 + (4 –(–3))2
142 + 72 or (–14)2 + 72 or 142 + (–7)2 (M1 A1 may be implied by 245)
M1
A1
AB = 14 2  7 2 or 7 2 (2 2  12 ) or 245
A1cso 3
7 5
[7]
2. (a) Putting the equation in the form y = mx (+c) and attempting to extract the m or mx (not the c) ,M1
or finding 2 points on the line and using the correct gradient formula.
Gradient = –
(b)
3
(or equivalent)
5
Gradient of perp. line =
A1
1
–1
(Using –
with the m from part (a))
3
m
"–
"
5


2
M1
5
y – 1 "  " x – 3
 3
M1
5
12
5x – 12
5
)
y  – x – 4 (Must be in this form... allow y= x –
but not y=
3
3
3
3
A1
3
This A mark is dependent upon both M marks.
[5]
3.
(a)
3
y =  x(+ 4)
2
(b)
3
3x + 2 =  x + 4
2
Gradient = 
3
2
M1A1
x = ...,
4
9
M1, A1
10 
1
4
y = 3  + 2 =  3 
3
3
9
(c)
Where y = 1, l1: xA = 
1
l2: xB = 2
3
1
(xB – xA)(yP – 1)
2
1 7 7 49
13
2
=   
2 3 3 18
18
A1
M: Attempt one of these
Area =
2
3
M1A1
M1
o.e.
A1
4
[9]
mAB 
4.
(a)
40  4
 
72  5
Glyn Technology School
M1
21
Equation of AB is: y  0 
4
( x  2)
5
or
y4
4
 x  7
5
(o.e.)
4x – 5y – 8 = 0 (o.e.)
(b)
A1
 AB   7  22  4  02
3
M1
= 41
(c)
Using isos triangle with AB = AC then t = 2 × yA = 2 × 4 = 8
(d)
Area of triangle =
=
M1
1 t  (7  2)
2
A1
2
B1
1
M1
20
A1
2
[8]
5.
(a)
1
( x – 2) or equivalent,
2
y –5
1
e.g.
–
x–2
2
y–5= –
y= –
1
x  6 M1A1,A1cao
2
1
(–2) + 6 = 7 (therefore B lies on the line)
2
(b)
x = –2  y = –
(c)
(AB2 =) (2 – –2)2 + (7 – 5)2, = 16 + 4 = 20, AB =
C is (p, –
(d)
3
1
p + 6), so
2
20  2 5
 1

AC2 = (p – 2)2 +  – p  6 – 5 
 2

Therefore 25 = p2 –4p + 4 +
B1
1
M1, A1, A1
3
2
1 2
p –p+1
4
25 = 1.25p2 – 5p + 5 or 100 = 5p2 – 20p + 20 (or better, RHS simplified
to 3 terms) Leading to: 0 = p2 – 4p – 16
(*)
M1
M1
A1
A1cso
4
[11]
6.
(a)
y
5
-2.5
x
B1M1A1
3
B1
for curve of correct shape i.e 2 branches of curve, in correct quadrants, of roughly the correct
shape and no touching or intersections with axes.
M1
for a straight line cutting the positive y-axis and the negative x-axis. Ignore any values.
A1
for (0, 5) and (–2.5, 0) or points correctly marked on axes.
Glyn Technology School
22
Do not give for values in tables.
(b)
3
x
2
2x + 5x = 3[= 0] or 2x2 + 5x = 3
(2x – 1)(x + 3)[= 0]
1
x = –3 or
2
1
3
3
y=
or 2 × (–3) + 5 or y =
or 2 × ( ) + 5
1
2
3
2
2x + 5 =
1
Points are (3,1) and ( ,6) (correct pairings)
2
M1
A1
M1
A1
M1
A1ft
6
[9]
Glyn Technology School
23
C1 Differentiation and Integration
1.
1
2x 2
4x  kx or
3
2
12x2,  x

1
2
 kx

1
2
(k a non–zero constant)
...., (–1  0) A1, A1, B1
M1
4
[4]
2.
x4 → kx3 or x
1
3
–2
 kx
3
or 3  0
(k a non-zero constant)
M1
 dy  3
   4 x .........., with ‘3’ differentiated to zero (or ‘vanishing’)
 dx 
1 –2
 dy 
   ............. + x 3
3
 dx 
1
or equivalent, e.g.
3
3 x
2
or
A1
1
 
33 x
A1
2
[3]
3.
(a)
[f(x)= ] 3 + 3x2
M1A1 2
(b)
3 + 3x2 = 15 and start to try and simplify
x2 =k → x = √k (ignore )
x = 2 (ignore × = –2)
M1
M1
A1
3
[5]
4.
(a)
(3 – 4

x ) 2  9 – 12 x – 12 x  (– 4) 2 x
9x
–
1
2
1
 16 x 2
3
M1
– 24
A1, A1
3
M1 A1, A1ft
3
M1 A1
2
1
9 – 2 16 – 2
x , x
2
2
(b)
f’(x) = –
(c)
9 1 16 1
1 16 5
f ′(9) = –     –  
2 27 2 3
6 6 2
[8]
5.
y = 8 – 8 – 2 + 9 = 7 (*)
(a)
x = 2:
(b)
dy
 3x 2 – 4 x – 1
dx
x = 2:
m–
1
M1 A1
dy
12 – 8 – 1( 3)
dx
y – 7 = 3(x – 2),
(c)
B1
A1ft
y = 3x +1
1
3
M1, A1
(for 
1
with their m)
m
1
4
2
3x 2 – 4 x – 1  – , 9 x 2 – 12 x – 2  0 or x 2 – x –  0 (o.e.)
3
3
9
 12  144  72 
x



18


1
x 2 6
3

Glyn Technology School



216  36 6  6 6 or (3x – 2)2
5
B1ft
M1, A1
6 → 3x = 2  6
M1
A1cso
5
24
[11]
6.
y
(a)
x 2 – 5 x – 24
 x – 5 – 24 x –1
x
dy
 1  24 x – 2
dx
(b)
24
)
x
M1 A1
dy
24
 1 2
dx
x
or
y = –15
x = 2:
(or equiv., e.g. x  3 – 8 –
M1 A1 4
Allow if seen in part (a).
B1
24
 dy 
7
  1
4
 dx 
B1ft
This must be simplified to a “single value”. y +15 = 7(x – 2) (or equiv., e.g. y = 7x – 29)
y  15
7
x–2
Allow
M1 A1 4
[8]
7.
3x 2  2
 3 x  2 x 1
x
 y  
M1 A1
1
24 x 2 , 2 x 2 , 3  2 x 2
1

24 x 2  2 x 2


M1 A1

 3  2 x 2 

A1 A1
[6]
8.
(a)
1
1


4 
 dy 
   6 x 1  x 2 or  6 x  2 x 2

2
 dx 


3
2
or 6  1 x
(b)
6  x
(c)
8
x3  x 2  C
3





3
2
M1A1
2
M1A1ft
2
M1A1A1
3
3
3
3
4x 2
A1: x3 or
A1: both, simplified and +C
3
3
 
2
[7]
9.
3x2  kx3 or 4x5 → kx6 or –7 → kx (k a non-zero constant)
M1
3x 3
4x 6
or
(Either of these, simplified or unsimplified)
3
6
A1
x3 +
2x 6
3x 3 4 x 6

 7 x1
– 7x or equivalent unsimplified, such as
3
3
6
+ C (or any other constant, e.g. + K)
A1
B1
4
[4]
10.
(a)
3x2  cx3 or –6  cx or –8x–2  cx–1
3x 2
8 x 1
8

 6x 
( C )  x 3  6 x  
f(x) =
3
1
x

Substitute x = 2 and y = 1 into a ‘changed function’ to form an
equation in C.
Glyn Technology School
M1
A1A1
M1
25
1 = 8 – 12 + 4 + C
(b)
3 × 22 – 6 –
C=1
A1cso
8
M1
22
=4
Eqn. of tangent: y – 1 = 4(x – 2)
y = 4x – 7 (Must be in this form)
A1
M1
A1 4
[9]
3
11.
3x 3 3x 2
(f(x) =)
–
– 7 x(c)
3
3
2
= x
3
3
– 2x 2
M1
– 7 x ( c)
A1A1
f(4) = 22  22 = 64 – 16 – 28 + c
M1
c=2
A1cso
5
[5]
12.
(a)
4x  kx2 or 6 x  kx
3
3
2
or
8
x2
 kx1 (k a non-zero constant)
, –8x–1 (+ C)(+ C not required)
3
At x = 4, y = 1: 1 = (2 × 16) –  4  4 2   8  4 1  C Must be in part (a)


C=3
f(x) = 2x2,  4x
(b)
2

f(4) = 16 – (6 × 2) +
8 9
 (= m)
16 2

 M : Attempt f (4) with the given f  




Must
be
in
part
(b)



2
1  M : Attempt perp. grad. rule.

Gradient of normal is     
9
m  Dependent on the use of their f ( x) 
2
Eqn. of normal: y – 1 =  ( x  4)
9
y 1
2
or any equiv. form, e.g.
 )
x4
9
M1
A1, A1, A1
M1
A1
6
M1
M1
M1A1
4
[10]
3
13.
(a)
(b)
2
6x
10 x

– 12x (+C)
3
2
x = 5: 250 – 125 – 60 + C = 65
f(x) =
M1A1
C=0
x(2x2 – 5x – 12) or (2x2 + 3x)(x – 4) or (2x + 3)(x2 – 4x)
= x(2x + 3)(x – 4)
(*)
M1A1
4
M1
A1cso
2
(c)
4 y
2
x
–4 –3 –2 –1
1 2
3
4
5
6
–2
–4
–6
Shape
 3 
  ,0  and (4, 0)
 2 
Glyn Technology School
B1
Through origin
B1
B1
3
26
[9]
C1 Sequences
1.
(a)
(b)
Identify a = 5 and d = 2
(u200 =)a + (200 – 1)d
= 430(p) or (£)4.30
May be implied
(= 5 + (200 – 1) × 2)
B1
M1
A1
3
200
200
[2a + (200 – 1)d] or
(a + “their 403”)
2
2
200
200
=
[2 × 5 + (200 – 1) × 2] or
(5 + “their 403”)
2
2
= 40 800 or £408
(S200 =)
M1
A1
A1
3
[6]
2.
(a)
[x2 =] a – 3
B1
(b)
[x3 = ] ax2 – 3 or a(a – 3) – 3
= a(a – 3) – 3 (*)
= a2 – 3a – 3 (*)
M1
(c)
1
A1cso 2
a2 – 3a – 3 = 7
a2 – 3a – 10 = 0 or a2 – 3a = 10
(a – 5)(a + 2) = 0
a = 5 or – 2
M1
dM1
A1
3
[6]
3.
(a)
a2 


43  7
B1
a3  "their 7" 3  10
(b)

a4  10  3  13
B1ft

2
M1
a5  13  3  4 *
A1 cso 2
[4]
4.
(a)
(a2 =)2k – 7
(b)
(a3 =)2(2k – 7) – 7 or 4k – 14 – 7, = 4k – 21
(c)
(a4 =)2(4k – 21) –7 (= 8k – 49)
B1
1
(*)
M1, A1cso
2
M1
4
a
r
= k + “(2k – 7)” + (4k – 21) + “(8k – 49)”
M1
r 1
k + (2k – 7) + (4k – 21) + (8k – 49) = 15k – 77 = 43
k=8
M1 A1
4
[7]
5.
(a)
(a2 =)3k + 5 [must be seen in part (a) or labelled a2 =]
(b)
(a3 =)3(3k + 5) + 5
= 9k + 20
(c)
(i)
a4 = 3(9k + 20) + 5 (= 27k + 65)
B1
(*)
1
M1
A1cso 2
M1
4
a
r
= k + (3k + 5) + (9k + 20) + (27k + 65)
M1
r 1
(ii)
= 40k + 90
= 10(4k + 9) (or explain why divisible by 10)
Glyn Technology School
A1
A1ft
4
27
[7]
6.
(a)
1(p + 1) or p +1
B1
(b)
((a))(p + (a)) [(a) must be a function of p]. [( p +1)(p + p + 1)]
=1+ 3p + 2p2
(*)
M1
A1cso 2
(c)
1+ 3p + 2p2 =1
p(2p + 3) = 0
p = ...
3
p =  (ignore p = 0, if seen, even if ‘chosen’ as the answer)
2
M1
M1
Noting that even terms are the same.
M1
(d)
A1
1
1
This M mark can be implied by listing at least 4 terms, e.g. 1,  , 1, ,...
2
2
1
x2008 = 
2
A1
1
3
2
[8]
7.
(a)
(b)
(c)
(d)
(e)
5, 7, 9, 11 or 5 + 2 + 2 + 2 = 11 or 5 + 6 = 11
use a = 5, d = 2, n = 4 and t4 = 5 + 3 × 2 = 11
B1
1
tn = a + (n – 1)d with one of a = 5 or d = 2 correct
(can have a letter for the other)
= 5 + 2(n – 1) or 2n + 3 or 1 + 2(n + 1)
M1
A1
2
n
n
[2 × 5 + 2(n – 1)] or use of (5 + “their 2n + 3”)
2
2
(may also be scored in (b))
= {n(5 + n – 1)} = n(n + 4) (*)
M1A1
A1cso 3
43 = 2n + 3
[n] = 20
M1
A1
S20 =20 × 24, = 480 (km)
M1A1 2
Sn =
2
[10]
8.
(a)
(b)
(c)
st
a +17d = 25 or equiv. (for 1 B1),
a + 20d = 32.5 or equiv. (for 2nd B1),
Solving (Subtract)
2750 =
B1, B1 2
3d = 7.5
so d = 2.5
a = 32.5 – 20 × 2.5 so a = –17.5 (*)
n
5
[– 35  (n – 1)]
2
2
M1
A1cso
2
M1A1ft
{ 4 × 2750 = n(5n – 75) }
4 × 550 = n(n – 15)
n2 – 15n = 55 × 40
(d)
M1
(*)
n2 – 15n – 55 × 40 = 0
(n – 55)(n + 40) = 0
n = 55 (ignore – 40)A1
A1cso
or n2 – 15n – 2200 = 0
n=...
4
M1
M1
3
[11]
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C1 Transformations of curve
1.
(a)
20 y
10
x
–3
–2
–1
1
2
3
4
5
6
7
–10
–20
(i)
(ii)
(b)
Shape
or
or
B1
Max. at (0, 0).
(2, 0), (or 2 shown on x-axis).
B1
B1
Shape
B1
(It need not go below x-axis)
Through origin.
(6, 0), (or 6 shown on x-axis).
B1
B1
x2(x – 2) = x(6 – x)
x3 – x2 – 6x = 0
Expand to form 3-term cubic (or 3-term quadratic if divided by x),
with all terms on one side. The “= 0” may be implied.
x(x – 3)(x + 2) = 0 x = ... Factor x (or divide by x), and solve quadratic.
x = 3 and x = –2
x = –2 : y = –16
Attempt y value for a non-zero x value by
substituting back into x2 (x – 2) or x(6 – x).
x=3:y=9
Both y values are needed for A1.
(–2,–16) and (3,9)
(0, 0)
This can just be written down. Ignore any ‘method’ shown.
(But must be seen in part (b)).
3
3
M1
M1
M1
M1
A1
B1
7
[13]
2.
(a)
10
(7, 3)
B1B1B1
1st B1
for moving the given curve up. Must be U shaped curve, minimum in first quadrant, not
touching x-axis but cutting positive y-axis. Ignore any values on axes.
2nd B1
for curve cutting y-axis at (0, 10). Point 10(or even (10, 0) marked on positive y-axis is OK)
3rd B1
for minimum indicated at (7, 3). Must have both coordinates and in the right order.
(b)
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3
y
7
x
(3.5, 0)
B1B1 2
1st B1 for U shaped curve, touching positive x-axis and crossing y-axis at (0, 7)[condone (7, 0) if
marked on positive y axis] or 7 marked on y-axis
2nd B1 for minimum at (3.5, 0) or 3.5 or
7
or marked on x-axis.
2
Allow “stopping at” (0, 10) or (0, 7) instead of “cutting”
[5]
3.
(a)
Shape
, touching the x-axis at its maximum.
Through (0, 0) & – 3 marked on x-axis, or (–3,0) seen.
Allow (0, – 3) if marked on the x-axis.
M1
A1
Marked in the correct place, but 3, is A0.
Min at (–1, –1)
A1
Correct shape
(top left – bottom right)
B1
Through – 3 and max at (0, 0). Marked in the correct place, but 3, is B0.
B1
3
(b)
Min at (–2, –1)
B1
3
[6]
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4.
(a)
(2, 10)
1
4
Shape: Max in 1st quadrant and 2 intersections on positive x-axis
B1
1 and 4 labelled (in correct place) or clearly stated as coordinates
B1
(2, 10) labelled or clearly stated
B1
3
(b)
(–2, 5)
–4
(c)
–1
Shape: Max in 2nd quadrant and 2 intersections on negative x-axis
B1
–1 and –4 labelled (in correct place) or clearly stated as coordinates
B1
(–2, 5) labelled or clearly stated
B1
3
B1
1
(a = ) 2
May be implicit, i.e. f (x + 2)
[7]
5.
(a)
8 y
6
4
2
–4
–3
–2
–1
1
2
3
x
4
–2
–4
Shape of f(x)
B1
Moved up ↑
M1
Asymptotes: y = 3 B1
x = 0 (Allow “ y –axis”)
(b)
1
3 = 0
x
1
x =  (or – 0.33 …)
3
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B1
4
No variations accepted.
M1
Decimal answer requires at least 2 d.p.
A1
2
31
This scores B0 (clear overlap with horiz. asymp.) M1 (Upward translation… bod that both
branches have been translated).
(a)
B0 M1
B0 M1
B0 M0
No marks unless the original curve is seen, to show upward translation.
[6]
6.
(a)
6
4
2
–6
–4
–2
2
4
6
–2
–4
–6
Translation parallel to x-axis
Top branch intersects  ve y  axis 

Lower branch has no intersections 

No obvious overlap

3
 3
 0,  or marked on y-axis
2
 2
(b)
x = –2, y = 0
M1
A1
B1
3
B1, B1
2
[5]
7.
(a)
x(x2 – 6x + 9)
= x(x – 3)(x – 3)
B1
M1 A1 3
(b)
Shape
Through origin (not touching)
Touching x-axis only once
Touching at (3, 0), or 3 on x-axis
B1
B1
B1ft
4
M1
A1
2
(c)
Moved horizontally (either way)
(2, 0) and (5, 0), or 2 and 5 on x-axis
Allow a fully correct graph (as shown above) to score M1A1 whatever they have in (b)
[9]
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