FP1 PAST PAPERSwith mark schemesJune 2014 back to January 2010Included:June 2014June 2014 (R)June 2013June 2013 (R)Jan 2013June 2012Jan 2012June 2011Feb 2010Omitted:June 2013 (withdrawn paper)January 2011June 2010
P43153AThis publication may only be reproduced in accordance with Pearson Education Limited copyright policy.©2014 Pearson Education Limited.
Paper Reference(s)
6667/01Edexcel GCEFurther Pure Mathematics FP1
Advanced/Advanced SubsidiaryTuesday 10 June 2014 Morning
Time: 1 hour 30 minutes
Materials required for examination Items included with question papersMathematical Formulae (Pink) Nil
Candidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for symbolicalgebra manipulation or symbolic differentiation/integration, or haveretrievable mathematical formulae stored in them.
Instructions to Candidates
In the boxes above, write your centre number, candidate number, your surname, initials and signature.Check that you have the correct question paper.Answer ALL the questions.You must write your answer for each question in the space following the question.When a calculator is used, the answer should be given to an appropriate degree of accuracy.
Information for Candidates
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for the parts of questions are shown in round brackets, e.g. (2).There are 9 questions in this question paper. The total mark for this paper is 75.There are 28 pages in this question paper. Any blank pages are indicated.
Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.You must show sufficient working to make your methods clear to the Examiner. Answers without working may not gain full credit.
P43153A 2
1. The complex numbers z1 and z2 are given by
z1 = p + 2i and z2 = 1 – 2i
where p is an integer.
(a) Find in the form a + bi where a and b are real. Give your answer in its simplest form in terms of p.
(4)
Given that ,
(b) find the possible values of p.(4)
2.
, x > 0
(a) Show that the equation f(x) = 0 has a root α in the interval [1.1, 1.5].(2)
(b) Find f ʹ(x).(2)
(c) Using x0 = 1.1 as a first approximation to α, apply the Newton-Raphson procedure once to f(x) to find a second approximation to α, giving your answer to 3 decimal places.
(3)
3. Given that 2 and 1 – 5i are roots of the equation
x3 + px2 + 30x + q = 0,
(a) write down the third root of the equation.(1)
(b) Find the value of p and the value of q.(5)
(c) Show the three roots of this equation on a single Argand diagram.(2)
P43152AThis publication may only be reproduced in accordance with Pearson Education Limited copyright policy.©2014 Pearson Education Limited.
4. (i) Given that
and ,
(a) find AB.
(b) Explain why AB ≠ BA.(4)
(ii) Given that
, where k is a real number
find C–1, giving your answer in terms of k.(3)
5. (a) Use the standard results for and to show that
(6)
(b) Hence show that
where a and b are constants to be found.(3)
P43152A 4
6. The rectangular hyperbola H has cartesian equation xy = c2.
The point P , t > 0, is a general point on H.
(a) Show that an equation of the tangent to H at the point P is
t2y + x = 2ct(4)
An equation of the normal to H at the point P is t3x – ty = ct4 – c.
Given that the normal to H at P meets the x-axis at the point A and the tangent to H at P meets the x-axis at the point B,
(b) find, in terms of c and t, the coordinates of A and the coordinates of B.(2)
Given that c = 4,
(c) find, in terms of t, the area of the triangle APB. Give your answer in its simplest form.(3)
7. (i) In each of the following cases, find a 2 × 2 matrix that represents
(a) a reflection in the line y = –x,
(b) a rotation of 135° anticlockwise about (0, 0),
(c) a reflection in the line y = –x followed by a rotation of 135° anticlockwise about (0, 0).
(4)
(ii) The triangle T has vertices at the points (1, k), (3, 0) and (11, 0), where k is a constant. Triangle T is transformed onto the triangle T ʹ by the matrix
Given that the area of triangle T ʹ is 364 square units, find the value of k.(6)
P43152A 5
8. The points P(4k2, 8k) and Q(k2, 4k), where k is a constant, lie on the parabola C with equation y2 = 16x.
The straight line l1 passes through the points P and Q.
(a) Show that an equation of the line l1 is given by
3ky – 4x = 8k2
(4)
The line l2 is perpendicular to the line l1 and passes through the focus of the parabola C.The line l2 meets the directrix of C at the point R.
(b) Find, in terms of k, the y coordinate of the point R.(7)
9. Prove by induction that, for ,
f(n) = 8n – 2n
is divisible by 6.(6)
TOTAL FOR PAPER: 75 MARKS
END
P43152A 6
Paper Reference(s)
6667/01REdexcel GCEFurther Pure Mathematics FP1 (R)
Advanced/Advanced SubsidiaryTuesday 10 June 2014 Morning
Time: 1 hour 30 minutes
Materials required for examination Items included with question papersMathematical Formulae (Pink) Nil
Candidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for symbolicalgebra manipulation or symbolic differentiation/integration, or haveretrievable mathematical formulae stored in them.
Instructions to Candidates
In the boxes above, write your centre number, candidate number, your surname, initials and signature.Check that you have the correct question paper.Answer ALL the questions.You must write your answer for each question in the space following the question.When a calculator is used, the answer should be given to an appropriate degree of accuracy.
Information for Candidates
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for the parts of questions are shown in round brackets, e.g. (2).There are 9 questions in this question paper. The total mark for this paper is 75.There are 28 pages in this question paper. Any blank pages are indicated.
Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.You must show sufficient working to make your methods clear to the Examiner. Answers without working may not gain full credit.
P43152A 7
1. The roots of the equation
2z3 – 3z2 + 8z + 5 = 0
are z1, z2 and z3.
Given that z1 = 1 + 2i, find z2 and z3.(5)
2. f(x) = 3cos 2x + x – 2, –π ≤ x < π
(a) Show that the equation f(x) = 0 has a root α in the interval [2, 3].(2)
(b) Use linear interpolation once on the interval [2, 3] to find an approximation to α.
Give your answer to 3 decimal places.(3)
(c) The equation f(x) = 0 has another root β in the interval [–1, 0]. Starting with this interval, use interval bisection to find an interval of width 0.25 which contains β.
(4)
3. (i)
(a) Describe fully the single transformation represented by the matrix A.(2)
The matrix B represents an enlargement, scale factor –2, with centre the origin.
(b) Write down the matrix B.(1)
(ii)
, where k is a positive constant.
Triangle T has an area of 16 square units.
Triangle T is transformed onto the triangle Tʹ by the transformation represented by the matrix M.
Given that the area of the triangle Tʹ is 224 square units, find the value of k.(3)
P43138AThis publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2013 Edexcel Limited.
P43138A 9
4. The complex number z is given by
where p is an integer.
(a) Express z in the form a + bi where a and b are real. Give your answer in its simplest form in terms of p.
(4)
(b) Given that arg(z) = θ, where tan θ = 1 find the possible values of p.(5)
5. (a) Use the standard results for and to show that
(5)
(b) Calculate the value of .(3)
6.
and
Given that M = (A + B)(2A – B),
(a) calculate the matrix M,(6)
(b) find the matrix C such that MC = A.(4)
P43138A 10
7. The parabola C has cartesian equation y2 = 4ax, a > 0.
The points P(ap2, 2ap) and Pʹ(ap2, –2ap) lie on C.
(a) Show that an equation of the normal to C at the point P is
y + px = 2ap + ap3
(5)
(b) Write down an equation of the normal to C at the point Pʹ.(1)
The normal to C at P meets the normal to C at Pʹ at the point Q.
(c) Find, in terms of a and p, the coordinates of Q.(2)
Given that S is the focus of the parabola,
(d) find the area of the quadrilateral SPQPʹ.(3)
8. The rectangular hyperbola H has equation xy = c2, where c is a positive constant.
The point , t ≠ 0 is a general point on H.
An equation for the tangent to H at P is given by
The points A and B lie on H.
The tangent to H at A and the tangent to H at B meet at the point .
Find, in terms of c, the coordinates of A and the coordinates of B.(5)
P43138A 11
9. (a) Prove by induction that, for ,
(5)
(b) A sequence of numbers is defined by
u1 = 0, u2 = 32,
un+2 = 6un+1 – 8un n ≥ 1
Prove by induction that, for ,
un = 4n+1 – 2n+3
(7)
TOTAL FOR PAPER: 75 MARKS
END
P43138A 12
Paper Reference(s)
6667/01Edexcel GCEFurther Pure Mathematics FP1
Advanced/Advanced SubsidiaryMonday 10 June 2013 Morning
Time: 1 hour 30 minutes
Materials required for examination Items included with question papersMathematical Formulae (Pink) Nil
Candidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for symbolicalgebra manipulation or symbolic differentiation/integration, or haveretrievable mathematical formulae stored in them.
Instructions to Candidates
In the boxes above, write your centre number, candidate number, your surname, initials and signature.Check that you have the correct question paper.Answer ALL the questions.You must write your answer for each question in the space following the question.When a calculator is used, the answer should be given to an appropriate degree of accuracy.
Information for Candidates
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for the parts of questions are shown in round brackets, e.g. (2).There are 9 questions in this question paper. The total mark for this paper is 75.There are 32 pages in this question paper. Any blank pages are indicated.
Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.You must show sufficient working to make your methods clear to the Examiner. Answers without working may not gain full credit.
P43138A 13
1.
M =
Given that the matrix M is singular, find the possible values of x.(4)
2. f(x) = cos(x2) – x + 3, 0 < x < π
(a) Show that the equation f(x) = 0 has a root α in the interval [2.5, 3].(2)
(b) Use linear interpolation once on the interval [2.5, 3] to find an approximation for α, giving your answer to 2 decimal places.
(3)
3. Given that x = is a root of the equation
2x3 – 9x2 + kx – 13 = 0,
find
(a) the value of k,(3)
(b) the other 2 roots of the equation.(4)
4. The rectangular hyperbola H has Cartesian equation xy = 4.
The point lies on H, where t ≠ 0.
(a) Show that an equation of the normal to H at the point P is
ty – t3x = 2 – 2t4
(5)
The normal to H at the point where t = meets H again at the point Q.
(b) Find the coordinates of the point Q.(4)
P42828AThis publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2013 Edexcel Limited.
P42828A 15
5. (a) Use the standard results for and to show that
for all positive integers n.
(6)
(b) Hence show that
where a, b and c are integers to be found.(4)
6. A parabola C has equation y2 = 4ax, a > 0
The points P(ap2, 2ap) and Q(aq2, 2aq) lie on C, where p ≠ 0, q ≠ 0, p ≠ q.
(a) Show that an equation of the tangent to the parabola at P is
py – x = ap2
(4)
(b) Write down the equation of the tangent at Q.(1)
The tangent at P meets the tangent at Q at the point R.
(c) Find, in terms of p and q, the coordinates of R, giving your answers in their simplest form.
(4)
Given that R lies on the directrix of C,
(d) find the value of pq.(2)
P42828A 16
7. z1 = 2 + 3i, z2 = 3 + 2i, z3 = a + bi, a, b
(a) Find the exact value of |z1 + z2|.(2)
Given that w = ,
(b) find w in terms of a and b, giving your answer in the form x + iy, x, y .(4)
Given also that w = ,
(c) find the value of a and the value of b,(3)
(d) find arg w, giving your answer in radians to 3 decimal places.(2)
8.
A =
and I is the 2 × 2 identity matrix.
(a) Prove that
A2 = 7A + 2I(2)
(b) Hence show that
A–1 = (A – 7I)(2)
The transformation represented by A maps the point P onto the point Q.
Given that Q has coordinates (2k + 8, –2k – 5), where k is a constant,
(c) find, in terms of k, the coordinates of P.(4)
P42828A 17
9. (a) A sequence of numbers is defined by
u1 = 8
un + 1 = 4un – 9n, n ≥ 1
Prove by induction that, for n ,
un = 4n + 3n +1(5)
(b) Prove by induction that, for m ,
(5)
TOTAL FOR PAPER: 75 MARKS
END
P42828A 18
Paper Reference(s)
6667/01REdexcel GCEFurther Pure Mathematics FP1 (R)
Advanced/Advanced SubsidiaryMonday 10 June 2013 Morning
Time: 1 hour 30 minutes
Materials required for examination Items included with question papersMathematical Formulae (Pink) Nil
Candidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for symbolicalgebra manipulation or symbolic differentiation/integration, or haveretrievable mathematical formulae stored in them.
This paper is strictly for students outside the UK.
Instructions to Candidates
In the boxes above, write your centre number, candidate number, your surname, initials and signature.Check that you have the correct question paper.Answer ALL the questions.You must write your answer for each question in the space following the question.When a calculator is used, the answer should be given to an appropriate degree of accuracy.
Information for Candidates
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for the parts of questions are shown in round brackets, e.g. (2).There are 10 questions in this question paper. The total mark for this paper is 75.There are 36 pages in this question paper. Any blank pages are indicated.
Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.You must show sufficient working to make your methods clear to the Examiner. Answers without working may not gain full credit.
P42828A 19
1. The complex numbers z and w are given by
z = 8 + 3i, w = –2i
Express in the form a + bi, where a and b are real constants,
(a) z – w,(1)
(b) zw.(2)
2. (i) , where k is a constantGiven that
B = A + 3I
where I is the 2 × 2 identity matrix, find
(a) B in terms of k,(2)
(b) the value of k for which B is singular.(2)
(ii) Given that
, D = (2 –1 5)and
E = CD
find E.(2)
P41485A 20
3. f(x) =
(a) Show that the equation f(x) = 0 has a root α between x = 2 and x = 2.5.(2)
(b) Starting with the interval [2, 2.5] use interval bisection twice to find an interval of width 0.125 which contains α.
(3)
The equation f(x) = 0 has a root β in the interval [–2, –1].
(c) Taking –1.5 as a first approximation to β, apply the Newton-Raphson process once to f(x) to obtain a second approximation to β.Give your answer to 2 decimal places.
(5)
4. f(x) = (4x2 +9)(x2 – 2x + 5)
(a) Find the four roots of f(x) = 0.(4)
(b) Show the four roots of f(x) = 0 on a single Argand diagram.(2)
P41485A This publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2013 Edexcel Limited.
5.
Figure 1
Figure 1 shows a rectangular hyperbola H with parametric equations
x = 3t, y = , t ≠ 0
The line L with equation 6y = 4x – 15 intersects H at the point P and at the point Q as shown in Figure 1.
(a) Show that L intersects H where 4t2 – 5t – 6 = 0.(3)
(b) Hence, or otherwise, find the coordinates of points P and Q.(5)
P41485A 22
6. ,
The transformation represented by B followed by the transformation represented by A is equivalent to the transformation represented by P.
(a) Find the matrix P.(2)
Triangle T is transformed to the triangle T´ by the transformation represented by P.
Given that the area of triangle T´ is 24 square units,
(b) find the area of triangle T.(3)
Triangle T´ is transformed to the original triangle T by the matrix represented by Q.
(c) Find the matrix Q.(2)
P41485A This publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2013 Edexcel Limited.
7. The parabola C has equation y2 = 4ax, where a is a positive constant.
The point P(at2, 2at) is a general point on C.
(a) Show that the equation of the tangent to C at P(at2, 2at) is
ty = x + at2
(4)
The tangent to C at P meets the y-axis at a point Q.
(b) Find the coordinates of Q.(1)
Given that the point S is the focus of C,
(c) show that PQ is perpendicular to SQ.(3)
8. (a) Prove by induction, that for ,
(6)
(b) Hence, show that
where a, b and c are integers to be found.(4)
P41485A 24
9. The complex number w is given by
w = 10 – 5i
(a) Find .(1)
(b) Find arg w, giving your answer in radians to 2 decimal places(2)
The complex numbers z and w satisfy the equation
(2 + i)(z + 3i) = w
(c) Use algebra to find z, giving your answer in the form a + bi,where a and b are real numbers.
(4)
Given that
arg(λ + 9i + w) =
where λ is a real constant,
(d) find the value of λ.(2)
P41485A This publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2013 Edexcel Limited.
10. (i) Use the standard results for and to evaluate
(2)
(ii) Use the standard results for and to show that
for all integers n ≥ 0, where a, b and c are constant integers to be found.(6)
TOTAL FOR PAPER: 75 MARKS
END
P41485A 26
Paper Reference(s)
6667/01Edexcel GCEFurther Pure Mathematics FP1
Advanced Subsidiary/Advanced LevelMonday 28 January 2013 Morning
Time: 1 hour 30 minutes
Materials required for examination Items included with question papersMathematical Formulae (Pink) Nil
Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation, differentiation or integration, or have retrievable mathematical formulae stored in them.
Instructions to Candidates
Write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Further Pure Mathematics FP1), the paper reference (6667), your surname, initials and signature.
Information for Candidates
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for the parts of questions are shown in round brackets, e.g. (2).There are 9 questions in this question paper. The total mark for this paper is 75.
Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.You must show sufficient working to make your methods clear to the Examiner. Answers without working may not gain full credit.
P41485A This publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2013 Edexcel Limited.
1. Show, using the formulae for ∑r = 1
n
r and
∑r = 1
n
r2
, that
∑r = 1
n
3(2r−1)2 = n(2n + 1)(2n – 1), for all positive integers n.
(5)
2. z =
503+4 i .
Find, in the form a + ib where a, b ℝ,
(a) z,(2)
(b) z2.(2)
Find
(c) z,(2)
(d) arg z2, giving your answer in degrees to 1 decimal place.(2)
3. f(x) = 2 x12 + x
−12 − 5, x > 0.
(a) Find f ′(x).(2)
The equation f(x) = 0 has a root in the interval [4.5, 5.5].
(b) Using x0 = 5 as a first approximation to , apply the Newton-Raphson procedure once to f(x) to find a second approximation to , giving your answer to 3 significant figures.
(4)
P41485A 28
4. The transformation U, represented by the 2 2 matrix P, is a rotation through 90° anticlockwise about the origin.
(a) Write down the matrix P.(1)
The transformation V, represented by the 2 × 2 matrix Q, is a reflection in the line y = −x.
(b) Write down the matrix Q.(1)
Given that U followed by V is transformation T, which is represented by the matrix R,
(c) express R in terms of P and Q,(1)
(d) find the matrix R,(2)
(e) give a full geometrical description of T as a single transformation.(2)
5. f(x) = (4x2 + 9)(x2 − 6x + 34).
(a) Find the four roots of f (x) = 0.
Give your answers in the form x = p + iq , where p and q are real.(5)
(b) Show these four roots on a single Argand diagram.(2)
P41485A 29
6. X = (1 a3 2 ) , where a is a constant.
(a) Find the value of a for which the matrix X is singular.(2)
Y = (1 −13 2 )
.(b) Find Y−1.
(2)
The transformation represented by Y maps the point A onto the point B.
Given that B has coordinates (1 – , 7 – 2), where is a constant,
(c) find, in terms of , the coordinates of point A.(4)
7. The rectangular hyperbola, H, has cartesian equation xy = 25.
The point P (5 p , 5
p ) and the point Q (5q , 5
q ), where p, q 0, p q, are points on the rectangular hyperbola H.
(a) Show that the equation of the tangent at point P is
p2y + x = 10p.(4)
(b) Write down the equation of the tangent at point Q.(1)
The tangents at P and Q meet at the point N.
Given p + q 0,
(c) show that point N has coordinates (10 pq
p+q, 10
p+q ) .(4)
The line joining N to the origin is perpendicular to the line PQ.
(d) Find the value of p2q2. (5)
P41485A 30
8. (a) Prove by induction that, for n ℤ+,
∑r = 1
n
r (r+3 ) =
13 n(n + 1)(n + 5).
(6)
(b) A sequence of positive integers is defined by
u1 = 1,
un + 1 = un + n(3n + 1), n ℤ+.
Prove by induction thatun = n2(n – 1) + 1, n ℤ+.
(5)
9.
Figure 1
Figure 1 shows a sketch of part of the parabola with equation y2 = 36x.
The point P (4, 12) lies on the parabola.
(a) Find an equation for the normal to the parabola at P.(5)
This normal meets the x-axis at the point N and S is the focus of the parabola, as shown in Figure 1.
(b) Find the area of triangle PSN. (4)
TOTAL FOR PAPER: 75 MARKS
ENDPaper Reference(s)
6667/01
N26109A 31
Edexcel GCEFurther Pure Mathematics FP1
Advanced SubsidiaryFriday 1 June 2012 Morning
Time: 1 hour 30 minutes
Materials required for examination Items included with question papersMathematical Formulae (Pink) Nil
Candidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for
symbolicalgebra manipulation, differentiation and integration, or have retrievablemathematical formulae stored in them.
Instructions to Candidates
Write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Core Mathematics FP1), the paper reference (6667), your surname, initials and signature.
Information for Candidates
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.There are 10 questions in this question paper. The total mark for this paper is 75.
Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.You must show sufficient working to make your methods clear to the Examiner. Answers without working may not gain full credit.
P40688A 32
1. f(x) = 2x3 – 6x2 – 7x − 4.
(a) Show that f(4) = 0.(1)
(b) Use algebra to solve f(x) = 0 completely.(4)
2. Given that
A = (3 1 34 5 5 ) and B =
(1 11 20 −1 ),
(a) find AB.(2)
Given that
C = (3 28 6 ) and D =
(5 2 k4 k )
, where k is a constant
and
E = C + D,
(b) find the value of k for which E has no inverse.(4)
3. f(x) = x2 +
34 √ x – 3x – 7, x > 0.
A root of the equation f(x) = 0 lies in the interval [3, 5].
Taking 4 as a first approximation to , apply the Newton-Raphson process once to f(x) to obtain a second approximation to . Give your answer to 2 decimal places.
(6)
P40086A 33
4. (a) Use the standard results for ∑r = 1
n
r3
and ∑r = 1
n
r to show that
∑r = 1
n
(r3+6 r−3 ) =
14 n2(n2 + 2n + 13)
for all positive integers n.(5)
(b) Hence find the exact value of
∑r = 16
30
(r3+6 r−3).
(2)
N17584A 34
5.
Figure 1
Figure 1 shows a sketch of the parabola P with equation y2 = 8x. The point P lies on C, where y > 0, and the point Q lies on C, where y < 0. The line segment PQ is parallel to the y-axis.
Given that the distance PQ is 12,
(a) write down the y-coordinate of P,(1)
(b) find the x-coordinate of P.(2)
Figure 1 shows the point S which is the focus of C.
The line l passes through the point P and the point S.
(c) Find an equation for l in the form ax + by + c = 0, where a, b and c are integers.(4)
P40086A This publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2012 Edexcel Limited.
6. f(x) = tan ( x
2 ) + 3x – 6, – < x < .
(a) Show that the equation f(x) = 0 has a root in the interval [1, 2].(2)
(b) Use linear interpolation once on the interval [1, 2] to find an approximation to . Give your answer to 2 decimal places.
(3)
7. z = 2 − i√3.
(a) Calculate arg z, giving your answer in radians to 2 decimal places.(2)
Use algebra to express
(b) z + z2 in the form a + bi√3, where a and b are integers,(3)
(c)
z+7z−1 in the form c + di√3, where c and d are integers.
(4)
Given that
w = – 3i,
where is a real constant, and arg (4 – 5i + 3w) = –
π2 ,
(d) find the value of .(2)
N17584A 36
8. The rectangular hyperbola H has equation xy = c2, where c is a positive constant.
The point P(ct , c
t ), t ≠ 0, is a general point on H.
(a) Show that an equation for the tangent to H at P is
x + t 2 y = 2ct.(4)
The tangent to H at the point P meets the x-axis at the point A and the y-axis at the point B.
Given that the area of the triangle OAB, where O is the origin, is 36,
(b) find the exact value of c, expressing your answer in the form k√2, where k is an integer.(4)
P40086A This publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2012 Edexcel Limited.
9. M = (3 42 −5 ) .
(a) Find det M.(1)
The transformation represented by M maps the point S(2a – 7, a – 1), where a is a constant, onto the point S (25, –14).
(b) Find the value of a.(3)
The point R has coordinates (6, 0).
Given that O is the origin,
(c) find the area of triangle ORS.(2)
Triangle ORS is mapped onto triangle OR 'S ' by the transformation represented by M.
(d) Find the area of triangle OR 'S '.(2)
Given that
A =(0 −11 0 )
(e) describe fully the single geometrical transformation represented by A.(2)
The transformation represented by A followed by the transformation represented by B is equivalent to the transformation represented by M.
(f) Find B.(4)
10. Prove by induction that, for n ℤ+,
f(n) = 22n – 1 + 32n – 1
is divisible by 5.(6)
TOTAL FOR PAPER: 75 MARKS
ENDPaper Reference(s)
N17584A 38
6667/01
Edexcel GCEFurther Pure Mathematics FP1
Advanced SubsidiaryMonday 30 January 2012 Afternoon
Time: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Pink) Nil
Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation, differentiation or integration, or have retrievable mathematical formulae stored in them.
Instructions to Candidates
In the boxes on the answer book, write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Further Pure Mathematics FP1), the paper reference (6667), your surname, initials and signature.When a calculator is used, the answer should be given to an appropriate degree of accuracy.
Information for Candidates
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for individual questions and the parts of questions are shown in round brackets: e.g. (2).There are 9 questions on this paper. The total mark for this paper is 75.
Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.You must show sufficient working to make your methods clear to the Examiner. Answers without working may not gain full credit.
P40086A This publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2012 Edexcel Limited.
1. Given that z1 =1 − i,
(a) find arg (z1 ).(2)
Given also that z2 = 3 + 4i, find, in the form a + ib, a, b ℝ,
(b) z1 z2 ,(2)
(c)
z2
z1 .(3)
In part (b) and part (c) you must show all your working clearly.
2. (a) Show that f(x) = x4 + x − 1 has a real root in the interval [0.5, 1.0].(2)
(b) Starting with the interval [0.5, 1.0], use interval bisection twice to find an interval of width 0.125 which contains .
(3)
(c) Taking 0.75 as a first approximation, apply the Newton Raphson process twice to f(x) to obtain an approximate value of . Give your answer to 3 decimal places.
(5)
3. A parabola C has cartesian equation y2 = 16x. The point P(4t2, 8t) is a general point on C.
(a) Write down the coordinates of the focus F and the equation of the directrix of C.(3)
(b) Show that the equation of the normal to C at P is y + tx = 8t + 4t3.(5)
N26109A 40
4. A right angled triangle T has vertices A(1, 1), B(2, 1) and C(2, 4). When T is transformed by
the matrix P = (0 11 0 ) , the image is T ′.
(a) Find the coordinates of the vertices of T ′.(2)
(b) Describe fully the transformation represented by P.(2)
The matrices Q = (4 −23 −1 ) and R =
(1 23 4 ) represent two transformations. When T is
transformed by the matrix QR, the image is T .
(c) Find QR.(2)
(d) Find the determinant of QR.(2)
(e) Using your answer to part (d), find the area of T .(3)
5. The roots of the equation
z3 − 8z2 + 22z − 20 = 0
are z1 , z2 and z3 .
(a) Given that z1 = 3 + i, find z2 and z3 .(4)
(b) Show, on a single Argand diagram, the points representing z1 , z2 and z3 .(2)
6. (a) Prove by induction
∑r = 1
n
r3
=
14 n2(n + 1)2.
(5)(b) Using the result in part (a), show that
∑r = 1
n
(r3−2) =
14 n(n3 + 2n2 + n – 8).
(3)
P38168A This publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2011 Edexcel Limited.
(c) Calculate the exact value of ∑
r = 20
50
( r3−2). (3)
7. A sequence can be described by the recurrence formula
un + 1 = 2un + 1, n 1, u1 = 1.
(a) Find u2 and u3 .(2)
(b) Prove by induction that un = 2n − 1.(5)
8. A = (0 12 3 ) .
(a) Show that A is non-singular.(2)
(b) Find B such that BA2 = A.(4)
9. The rectangular hyperbola H has cartesian equation xy = 9.
The points P(3 p , 3
p )and Q(3 q , 3
q ) lie on H, where p ≠ ± q.
(a) Show that the equation of the tangent at P is x + p2y = 6p.(4)
(b) Write down the equation of the tangent at Q.(1)
The tangent at the point P and the tangent at the point Q intersect at R.
(c) Find, as single fractions in their simplest form, the coordinates of R in terms of p and q.(4)
TOTAL FOR PAPER: 75 MARKSEND
P38168A 42
Paper Reference(s)
6667/01Edexcel GCE
Further Pure Mathematics FP1
Advanced/ Advanced SubsidiaryWednesday 22 June 2011 Morning
Time: 1 hour 30 minutes
Materials required for examination Items included with question papersMathematical Formulae (Pink) Nil
Candidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for
symbolicalgebra manipulation, differentiation and integration, or have retrievablemathematical formulae stored in them.
Instructions to Candidates
Write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Core Mathematics FP1), the paper reference (6667), your surname, initials and signature.
Information for Candidates
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.There are 9 questions in this question paper. The total mark for this paper is 75.
Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.You must show sufficient working to make your methods clear to the Examiner. Answers without working may not gain full credit.
P38168A This publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2011 Edexcel Limited.
1. f (x) = 3x + 3x − 7
(a) Show that the equation f (x) = 0 has a root between x =1 and x = 2.(2)
(b) Starting with the interval [1, 2], use interval bisection twice to find an interval of width 0.25 which contains .
(3)
2. z1 = − 2 + i
(a) Find the modulus of z1 .(1)
(b) Find, in radians, the argument of z1 , giving your answer to 2 decimal places.(2)
The solutions to the quadratic equation
z2 − 10z + 28 = 0
are z2 and z3 .
(c) Find z2 and z3 , giving your answers in the form p iq, where p and q are integers.(3)
(d) Show, on an Argand diagram, the points representing your complex numbers z1 , z2 and z3 .
(2))
N35143A 44
3. (a) Given that
A = ( 1 √ 2√ 2 −1 ) ,
(i) find A2,
(ii) describe fully the geometrical transformation represented by A2.(4)
(b) Given that
B = ( 0 −1−1 0 )
,
describe fully the geometrical transformation represented by B.(2)
(c) Given that
C = (k+1 12
k 9 ),
where k is a constant, find the value of k for which the matrix C is singular.(3)
4. f(x) = x2 +
52 x – 3x – 1, x 0.
(a) Use differentiation to find f ′(x).(2)
The root of the equation f(x) = 0 lies in the interval [0.7, 0.9].
(b) Taking 0.8 as a first approximation to , apply the Newton-Raphson process once to f(x) to obtain a second approximation to . Give your answer to 3 decimal places.
(4)
N35143A This publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2010 Edexcel Limited.
5. A = (−4 a
b −2 ), where a and b are constants.
Given that the matrix A maps the point with coordinates (4, 6) onto the point with coordinates (2, −8),
(a) find the value of a and the value of b.(4)
A quadrilateral R has area 30 square units.It is transformed into another quadrilateral S by the matrix A.Using your values of a and b,
(b) find the area of quadrilateral S.(4)
6. Given that z = x + iy, find the value of x and the value of y such that
z + 3iz* = −1 + 13i
where z* is the complex conjugate of z.(7)
7. (a) Use the results for ∑r = 1
n
rand
∑r = 1
n
r2
to show that
∑r = 1
n
(2 r−1 )2 =
13 n(2n + 1)(2n – 1)
for all positive integers n.(6)
(b) Hence show that
∑r = n + 1
3 n
(2 r−1 )2 =
23 n(an2 + b)
where a and b are integers to be found.(4)
N17584A 46
8. The parabola C has equation y2 = 48x.
The point P(12t 2, 24t) is a general point on C.
(a) Find the equation of the directrix of C.(2)
(b) Show that the equation of the tangent to C at P(12t 2, 24t) is
x − ty + 12t 2 = 0.(4)
The tangent to C at the point (3, 12) meets the directrix of C at the point X.
(c) Find the coordinates of X.(4)
9. Prove by induction, that for n ℤ+,
(a)(3 06 1 )
n
= ( 3n 03 (3n−1) 1 )
(6)
(b) f(n) = 72n − 1 + 5 is divisible by 12.(6)
TOTAL FOR PAPER: 75 MARKS
END
N35143A This publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2010 Edexcel Limited.
Paper Reference(s)
6667/01
Edexcel GCEFurther Pure Mathematics FP1
Advanced/Advanced SubsidiaryMonday 1 February 2010 Afternoon
Time: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Orange) Nil
Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation, differentiation or integration, or have retrievable mathematical formulae stored in them.
Instructions to Candidates
In the boxes on the answer book, write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Further Pure Mathematics FP1), the paper reference (6667), your surname, initials and signature.When a calculator is used, the answer should be given to an appropriate degree of accuracy.
Information for Candidates
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for individual questions and the parts of questions are shown in round brackets: e.g. (2).There are 9 questions on this paper. The total mark for this paper is 75.
Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.You must show sufficient working to make your methods clear to the Examiner. Answers without working may not gain full credit.
1. The complex numbers z1 and z2 are given by
z1 = 2 + 8i and z2 = 1 – i
Find, showing your working,
(a)
z1
z2 in the form a + bi, where a and b are real,(3)
N17584A 48
(b) the value of |
z1
z2|,
(2)
(c) the value of arg
z1
z2 , giving your answer in radians to 2 decimal places.(2)
2. f(x) = 3x2 –
11x2
.
(a) Write down, to 3 decimal places, the value of f(1.3) and the value of f(1.4).(1)
The equation f(x) = 0 has a root α between 1.3 and 1.4
(b) Starting with the interval [1.3, 1.4], use interval bisection to find an interval of width 0.025 which contains α.(3)
(c) Taking 1.4 as a first approximation to α, apply the Newton-Raphson procedure once to f(x) to obtain a second approximation to α, giving your answer to 3 decimal places.(5)
3. A sequence of numbers is defined by
u1 = 2,
un+1 = 5un – 4, n 1.
Prove by induction that, for n ℤ, un = 5n – 1 + 1.(4)
N35143A This publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2010 Edexcel Limited.
4.
Figure 1
Figure 1 shows a sketch of part of the parabola with equation y2 = 12x .
The point P on the parabola has x-coordinate
13 .
The point S is the focus of the parabola.
(a) Write down the coordinates of S.(1)
The points A and B lie on the directrix of the parabola.The point A is on the x-axis and the y-coordinate of B is positive.
Given that ABPS is a trapezium,
(b) calculate the perimeter of ABPS.(5)
N17584A 50
5. A = (a −52 a+4 ) , where a is real.
(a) Find det A in terms of a.(2)
(b) Show that the matrix A is non-singular for all values of a.(3)
Given that a = 0,
(c) find A–1.(3)
6. Given that 2 and 5 + 2i are roots of the equation
x3 − 12x2 + cx + d = 0, c, d ∈ℝ,
(a) write down the other complex root of the equation.(1)
(b) Find the value of c and the value of d.(5)
(c) Show the three roots of this equation on a single Argand diagram.(2)
7. The rectangular hyperbola H has equation xy = c2, where c is a constant.
The point P(ct , c
t ) is a general point on H.
(a) Show that the tangent to H at P has equation
t2y + x = 2ct.(4)
The tangents to H at the points A and B meet at the point (15c, –c).
(b) Find, in terms of c, the coordinates of A and B.(5)
GCE Further Pure Mathematics FP1 (6667) January 2010 51
8. (a) Prove by induction that, for any positive integer n,
∑r = 1
n
r3
=
14 n2(n + 1)2.
(5)
(b) Using the formulae for ∑r = 1
n
r and
∑r = 1
n
r3
, show that
∑r = 1
n
(r3+3 r+2)=
14 n(n + 2)(n2 + 7).
(5)
(c) Hence evaluate ∑
r = 15
25
(r 3+3 r+2 ).
(2)
9. M = (
1√ 2
− 1√ 2
1√ 2
1√ 2
).
(a) Describe fully the geometrical transformation represented by the matrix M.(2)
The transformation represented by M maps the point A with coordinates (p, q) onto the point B with coordinates (3√2, 4√2).
(b) Find the value of p and the value of q.(4)
(c) Find, in its simplest surd form, the length OA, where O is the origin.(2)
(d) Find M2.(2)
The point B is mapped onto the point C by the transformation represented by M2.
(e) Find the coordinates of C.(2)
TOTAL FOR PAPER: 75 MARKSEND
N17584A 52
JUNE 2014 MARK SCHEMEQuestionNumber Scheme Marks
1.(a) M1
M1
A1, A1
(4)
(b)`
M1
dM1
dM1A1
OR
M1
oedM1
oedM1A1
(4)Total 8
QuestionNumber Scheme Marks
2.
(a) M1
Sign change (and is continuous) therefore a root / α is between and
A1
GCE Further Pure Mathematics FP1 (6667) January 2010 53
(2)
(b) M1A1
(2)
(c) M1
M1
A1(3)
Total 7
Question Number Scheme Marks
3.(a) B1
(1)
(b) M1A1
M1
A1, A1
OR f(1+5i)=0 or f(1-5i)=0 M1
and A1
M1A1, A1
(5)
N17584A 54
Question Number Scheme Marks
(c)
B1
B1
(2)Total 8
Question Number Scheme Marks
4.A = , B =
(i)(a) M1A2
(b) B1
(4)(ii) M1
M1A1
(3)Total 7
5.(a) B1Proof by induction will usually score no more marks without use of standard results
GCE Further Pure Mathematics FP1 (6667) January 2010
21
5
O
55
M1A1B1
M1
A1
(6)
(b) M1
A1
A1
(3)Total 9
Question Number Scheme Marks
6.
(a)
M1
or equivalent expressions
A1
dM1
A1*(4)
(b) B1
B1(2)
(c) or PA= and PB=
M1
N17584A 56
Area APB = M1
=
A1
(3)Total 9
Question Number Scheme Marks
7.(i)(a) B1
(b) B1
(c) M1A1
(4)
(ii)Area triangle T =
M1A1
M1A1
Area triangle T = M1
A1
(6)Total
10
8.(a) M1
or
orM1A1
or A1*
(4)GCE Further Pure Mathematics FP1 (6667) January 2010 57
(b) (Focus) (4, 0) B1(Directrix) x = -4 B1
Gradient of l2 is M1
M1, A1
M1
A1(7)
Total 11
Question Number Scheme Marks
9. is divisible by 6.B1
Assume that for is divisible by 6.M1
M1A1
A1
If the result is true for then it is now true for As the result
has been shown to be true for then the result is true for all A1cso
(6)Total 6
JUNE 2014(R) MARK SCHEMEQuestio
n Number
Scheme Marks
1.B1M1A1
M1
A1
(5)Total 5
2.
(a) f(2) = -1.9609......f(3) = 3.8805...... M1
Sign change (and is continuous) therefore a root A1
N17584A 58
Question
NumberScheme Marks
is between and (2)
(b) M1
If any “negative lengths” are used, score M0
A1ft
A1(3)
(c) f(0) = +(1) or f(-1) = -(4.248) B1
f(-0.5) (= -0.879.....) M1
f(-0.25) (= 0.382....) M1
A1(4)
Total 9
Question
NumberScheme Marks
3.(i)(a) Rotation of 45 degrees anticlockwise, about the origin B1B1
(2)
(b) B1
(1)
(ii) B1
M1
A1(3)
Total 6
4.(a) M1
GCE Further Pure Mathematics FP1 (6667) January 2010 59
M1
A1, A1
(4)
(b) M1
M1
A1
M1A1
(5)Total 9
Question
NumberScheme Marks
5.(a) B1
M1A1
M1
A1
(5)
(b) M1
N17584A 60
A1
= 1621800 - 1890
= 1619910 A1
(3)Total 8
6.(a) M1A1
M1A1
M1A1
(6)(b) B1
M1
dM1
A1
(4)Total
10
Question
NumberScheme Marks
7.(a) M1
A1
At P, gradient of normal = -p A1
M1A1*
(5)
GCE Further Pure Mathematics FP1 (6667) January 2010 61
(b) B1
(1)
(c) M1A1
(2)(d) S is (a, 0) B1
Area SPQP’ = M1
A1
(3)Total 11
8.
M1
A1
M1
M1A1
(5)Total 5
Question
NumberScheme Marks
9.(a)When n = 1, rhs = lhs = 2
B1
M1A1
A1If the result is true for n = k then it has been shown true for n = k + 1. As it is true for n = 1 then it is true for all n (positive integers.)
A1
(5)
N17584A 62
(b)When n = 1 B1
When n = 2 B1True for n = 1 and n = 2
Assume and M1A1
M1
So A1If the result is true for n = k and n = k + 1 then it has been shown true for n = k + 2. As it is true for n = 1 and n = 2 then it is true for all n (positive integers.)
A1
(7)Total 12
JUNE 2013 MARK SCHEMEQuestio
n Number
Scheme Marks
1.
detM = x(4x – 11) – (3x – 6)(x – 2) M1
x2 + x – 12 (=0) A1(x + 4)(x – 3) (= 0 ) x = ... M1
A1[4]
2
(a)f(2.5) = 1.499.....
f(3) = -0.9111.....
M1
Sign change (positive, negative) (and is continuous) therefore root or
equivalent.A1
Use of degrees gives f(2.5) = 1.494 and f(3) = 0.988 which is awarded M1A0 (2)(b)
M1 A1ft
GCE Further Pure Mathematics FP1 (6667) January 2010 63
(2d.p.) A1 cao(3)[5]
Question Number
Scheme Marks
3(a) Ignore part labels and mark part (a) and part (b) together.
M1
dM1
k = 30 A1 caoAlternative using long division:
M1
dM1
A1Alternative by inspection:
M1dM1
k = 30 A1(3)
(b)
M1
or A1
or equivalentM1
A1 oe
(4)[7]
N17584A 64
Question Number
Scheme Marks
4(a)
M1
or equivalent expressionsA1
M1
M1
A1* cso(5)
(b) M1
or or
.
M1
or
or M1
A1
(4)[9]
Question Number
Scheme Marks
5(a) B1
M1,B1ft
M1 A1
GCE Further Pure Mathematics FP1 (6667) January 2010 65
A1*cso
(6)5(b)
M1A1
3f(n) – f(n or n+1) is M0
dM1
A1
(4)[10]
Question Number
Scheme Marks
6(a)
M1
or
A1
M1
py – x = ap2 * A1 cso(4)
(b) qy – x = aq2 B1
(1)
N17584A 66
(c)qy – aq2= py – ap2 M1
M1
A1,A1
(4)(d)
M1
A1(2)
[11]
Question Number
Scheme Marks
7(a) M1
A1 cao(2)
(b)
M1
B1
dM1A1
(4)(c)
M1
dM1
a = 1, b = -1 A1
(3)(d)
M1
A1
(2)GCE Further Pure Mathematics FP1 (6667) January 2010 67
[11]
Question Number
Scheme Marks
8(a)
M1A1
OR
(2)(b)
M1
*A1* cso
Numerical approach award 0/2.
(2)(c)
B1
M1
A1,A1
Or:
B1
M1
A1,A1
(4)[8]
Question Number
Scheme Marks
9(a) B1
N17584A 68
Assume true for n = k so that
M1
A1A1
If true for n = k then true for n = k + 1 and as true for n = 1 true for all n A1 cso
(5)(b)
Condone use of n here.
B1
M1
A1
A1
If true for m = k then true for m = k + 1 and as true for m = 1 true for all m A1 cso
(5)[10]
JUNE 2013 (R) MARK SCHEME
Question
Number
Scheme Marks
1.(a) B1
(1)
(b) M1A1
(2)[3]
GCE Further Pure Mathematics FP1 (6667) January 2010 69
2.
(i)(a) M1
A1
(2) (b) B is singular
M1 A1cao
(2)
(ii)
M1
A1
(2)[6]
Question
NumberScheme Marks
3.
(a)M1
Sign change (and is continuous) therefore a root exists between and A1
(2)(b) B1
M1A1
(3)
(c) M1A1B1
M1
A1 cao
(5)[10]
N17584A 70
Question
NumberScheme Marks
4.
(a) M1A1
M1
A1(4)
(b)
B1ft
B1ft
(2)[6]
Ignore part labels and mark part (a) and part (b) together
5.
(a)M1 A1
* A1 cso(3)
(b) M1A1
When
When
M1
A1
A1
(5) [8]
GCE Further Pure Mathematics FP1 (6667) January 2010
O x
y
71
Question
NumberScheme Marks
6.
(a) M1
A1
(2)(b) M1
dM1A1ft
(3)(c)
M1A1ft
(2)[7]
7.(a)
or (implicitly)
or (chain rule)
M1
When
or
A1
T: M1
T:
T: A1 cso *
(4)
(b)At Q,
B1
(1)(c)
M1
A1
N17584A 72
A1 cso (3) [8]
Question
NumberScheme Marks
8. (a)
B1As the summation formula is true for Assume that the summation formula is true for
With terms the summation formula becomes:
M1
dM1
A1
dM1
If the summation formula is true for then it is shown to be true for
As the result is true for , it is now also true for all and by mathematical induction.
A1 cso
(6)
8. (b)
M1A1
dM1
GCE Further Pure Mathematics FP1 (6667) January 2010 73
A1
(4)[10]
Question
Number
Scheme Marks
9.
(a) 11.1803... B1
(b) M1A1 oe
(2)
(c) B1
M1
M1
(Note: A1(4)
(d)
M1
So, A1(2)[9]
10.
(i) M1
A1 cao(2)
(ii)
M1N17584A 74
A1B1B1
M1
A1
(6)[8]
GCE Further Pure Mathematics FP1 (6667) January 2010 75
JANUARY 2013 MARK SCHEME
Question Number Scheme Marks
M1
=
A1, B1
M1
A1 cso
[5]
2.(a)
M1 A1cao
(2)
(b) = = M1 A1 (2)
(c) =10 M1 A1ft
(2)
(d) M1
so or A1 cao (2)
[8]
3. (a) M1 A1(2)
(b) f(5) = - 0.0807
B1
M1
M1
=5.2(0)
A1
(4)
[6]
N17584A 76
5
5
3 i2
3 i2
3 5i
3 5i
O
Question Number Scheme Marks
4.
(a) B1 (1)
(b) B1 (1)
(c) B1 (1)
(d) M1 A1 cao
(2)
(e) Reflection in the y axis B1 B1 (2)[7]
5.
(a) , or equivalent M1, A1
Solving 3-term quadratic by formula or completion of the square
or ( x−3 )2−9+34=0
M1
A1 A1ft (5)(b)
Two roots on imaginary axis B1ft
Two roots – one the conjugate of the other B1ft
Accept points or vectors
(2)[7]
GCE Further Pure Mathematics FP1 (6667) January 2010 77
Question Number Scheme Marks
6.(a) Determinant: 2 – 3a = 0 and solve for a = M1
So or equivalent A1(2)
(b) Determinant:
M1A1
(2)
(c)
M1depM1A1A1
(4)[8]
Question Number Scheme Marks
7.
(a) M1
A1
(*)M1 A1
(4)
(b) only B1
(1)
(c) so M1 A1cso
= M1 A1 cso
(4)
(d) Line PQ has gradient
5p− 5
q5 p−5 q (¿− 1
pq ) M1 A1
ON has gradient
10p+q
10 pqp+q
(¿ 1pq )
or
−1−1pq
(¿ pq )
could be as unsimplified
B1
N17584A 78
Question Number Scheme Marks
equivalents seen anywhere
As these lines are perpendicular so
OR for ON
with gradient (equivalent to) pq and sub in points O
AND N to give OR for PQ
with gradient (equivalent to) –pq and sub in points P
AND Q to give
M1 A1
(5)[14]
Question Number Scheme Marks
8.
(a) If n =1, and ,
B1
(so true for n = 1. Assume true for n = k)
So
M1
= =
A1
= which implies is true for
dA1
As result is true for this implies true for all positive integers and so result is true by induction
dM1A1cso
(6)
(b) B1(so true for . Assume true for n = k)
M1,
which implies is true for n = k + 1 A1
As result is true for n = 1 this implies true for all positive integers and so result is true by induction
M1A1cso(5)
GCE Further Pure Mathematics FP1 (6667) January 2010 79
[11]
N17584A 80
Question Number Scheme Marks
9.
(a) so
M1
Gradient when x = 4 is and gradient of normal is M1 A1
So equation of normal is (or ) M1 A1
(5)(b) S is at point (9,0) B1
N is at (22,0), found by substituting y = 0 into their part (a) B1ft
Both B marks can be implied or on diagram.
So area is M1 A1 cao
(4)
[9]
JUNE 2012 MARK SCHEMEQuestion Number Scheme Marks
1.
B1(a)[1]
(b)
M1 A1
So,
M1
A1
[4]
2. (a)
GCE Further Pure Mathematics FP1 (6667) January 2010 81
M1
A1
[2]
where k is a constant,(b)
M1
E does not have an inverse
M1M1
A1 oe[4]
6 marks
Question Number Scheme Marks
3.
M1A1
B1
M1
M1
N17584A 82
A1 cao
[6]
6 marks
Question Number Scheme Marks
4. (a)
M1A1B1
dM1
(AG)A1 *
[5]
(b)
M1
A1 cao
[2]7 marks
Question Number Scheme Marks
5.
(a) B1
[1]
(b)M1
A1 oe
GCE Further Pure Mathematics FP1 (6667) January 2010 83
(So P has coordinates )
[2]
(c) Focus B1
Gradient M1
Either
;M1
or and
;
l: A1
[4]7 marks
Question Number Scheme Marks
6.
(a)M1
Sign change (and is continuous) therefore a rootis between and
A1
[2]
orM1
(b)
A1
A1[3]
5 marks
N17584A 84
Question Scheme Marks
7. (a) M1
A1[2]
(b)
M1
(Note:
M1A1
[3]
M1(c)
dM1
M1
(Note: A1
[4](d)
, and
So real part of = 0 or M1
So, A1[2]11
marks
Question
NumberScheme Marks
8.(a)
M1
GCE Further Pure Mathematics FP1 (6667) January 2010 85
or equivalent expressions
A1
M1
A1 *
[4](b) B1
B1
Area M1
A1
[4]8 marks
Question Number Scheme Marks
9. (a) B1[1]
(b)
Therefore, M1
Either, or
or
A1
giving A1
[3]
(c) M1A1
N17584A 86
[2](d) M1
A1[2]
(e) Rotation; anti-clockwise (or clockwise) about B1;B1
[2](f) M1
A1
M1
A1
[4]14
marks
Question
NumberScheme Marks
is divisible by 5.10.
B1
Assume that for
is divisible by 5 for
M1A1
M1
GCE Further Pure Mathematics FP1 (6667) January 2010 87
A1
If the result is true for then it is now true for As the result has shown to be true for then the result is true for all n.
A1 cso
[6]6 marks
JANUARY 2012 MARK SCHEME
Question Number
Scheme NotesMarks
1(a) or or M1
or -45 or awrt -0.785 (oe e.g )A1
Correct answer only 2/2 (2)(b) At least 3 correct terms (Unsimplified) M1
cao A1(2)
(c)Multiply top and bottom by (1 + i) M1
A1
or A1 (3)
Correct answers only in (b) and (c) scores no marks (7 marks)
Question Number
Scheme NotesMarks
2
(a) f(0.5) = -0.4375 (- ) f(1) = 1
Either any one of f(0.5) = awrt -0.4 or f(1) = 1 M1
N17584A 88
Sign change (positive, negative) (and is continuous) therefore (a root) is between and
f(0.5) = awrt -0.4 and f(1) = 1, sign change and conclusion
A1
(2)
(b) f(0.75) = 0.06640625( )Attempt f(0.75) M1
f(0.625) = -0.222412109375( )f(0.75) = awrt 0.07 and f(0.625) = awrt -0.2 A1
or
or or equivalent in words.
A1
In (b) there is no credit for linear interpolation and acorrect answer with no working scores no marks.
(3)
(c) Correct derivative (May be implied later by e.g. 4(0.75)3 + 1) B1
Attempt Newton-Raphson M1
Correct first application – a correct numerical expression e.g.
or awrt 0.725 (may
be implied)
A1
Awrt 0.724 A1
cao A1
A final answer of 0.724 with evidence of NR applied twice with no incorrect work should score 5/5 (5)
(10 marks)
Question Number
Scheme NotesMarks
3(a) Focus B1
Directrix x + “4” = 0 or x = - “4” M1
x + 4 = 0 or x = - 4 A1(3)
GCE Further Pure Mathematics FP1 (6667) January 2010 89
(b)
or
their
M1
Correct differentiation A1
At P, gradient of normal = -t Correct normal gradient with no errors seen. A1
Applies
or using
in an attempt to find c. Their mN must be different from their mT and must be a function of t.
M1
cso **given answer** A1
Special case – if the correct gradient is quoted could score M0A0A0M1A1 (5)
(8 marks)
Question Number
Scheme NotesMarks
4(a) Attempt to multiply the right way round with at least 4 correct elements M1
has coordinates (1,1), (1,2) and (4,2)
or NOT just Correct coordinates or vectors A1
(2)(b)
Reflection in the line y = xReflection B1
y = x B1
Allow ‘in the axis’ ‘about the line’ y = x etc. Provided both features are
mentioned ignore any reference to the origin unless there is a clear
contradiction.
N17584A 90
(2)(c) 2 correct elements M1
Correct matrix A1
Note that scores M0A0 in (c) but allow all the marks in (d) and (e)
(2)(d) “-2”x”2” – “0”x”0” M1
-4 A1
Answer only scores 2/2
scores M0
(2)
(e)Area of T =
Correct area for T B1
Area of
Attempt at M1
6 or follow through their det(QR) x Their triangle area provided area > 0
A1ft
(3)(11
marks)
Question Number
Scheme NotesMarks
5(a) B1
Attempt to expand or any valid
method to establish the quadratic factor e.g.
Sum of roots 6, product of roots 10
M1
Attempt at linear factor with their cd in M1
GCE Further Pure Mathematics FP1 (6667) January 2010 91
Argand Diagram
3, 1
3, -1
2, 0
-1.5
-1
-0.5
0
0.5
1
1.5
0 0.5 1 1.5 2 2.5 3 3.5 Re
Im
Or
Or attempts f(2)A1
Showing that f(2) = 0 is equivalent to scoring both M’s so it is possible to
gain all 4 marks quite easily e.g. B1, shows f(2) = 0 M2, A1.Answers only can score 4/4
(4)
5(b)
First B1 for plotting (3, 1) and (3, -1) correctly with an indication of scale or labelled with coordinates (allow points/lines/crosses/vectors etc.) Allow i/-i for 1/-1 marked on imaginary axis.Second B1 for plotting (2, 0) correctly relative to the conjugate pair with an indication of scale or labelled with coordinates or just 2
B1B1
(2)(6 marks)
N17584A 92
GCE Further Pure Mathematics FP1 (6667) January 2010 93
Question Number
Scheme Notes Marks
6(a) Shows both LHS = 1 and RHS = 1 B1
Assume true for n = kWhen n = k + 1
Adds (k + 1)3 to the given result M1
Attempt to factorise out
dM1
Correct expression with factorised out.
A1
Must see 4 things: true for n = 1, assumption true for n = k, said true for n = k + 1 and therefore true for all n
Fully complete proof with no errors and comment. All the previous marks must have been scored.
A1cso
See extra notes for alternative approaches (5)
(b)Attempt two sums M1
is M0
Correct expression A1
Completion to printed answer with no errors seen. A1
(3)(c) Attempt S50 – S20 or S50 – S19 and
substitutes into a correct expression at least once.
M1
Correct numerical expression (unsimplified) A1
= 1 589 463 cao A1(3)(11
marks)
Question Number
Scheme NotesMarks
7(a)B1, B1
(2)(b) At n =1,
and so result true for n = 1B1
Assume true for n = k;
and so Substitutes uk into uk+1 (must see this line) M1
Correct expression A1
Correct completion to A1
Must see 4 things: true for n = 1, assumption true for n = k, said true for n = k + 1 and therefore true for all n
Fully complete proof with no errors and comment. All the previous marks in (b) must have been scored.
A1cso
Ignore any subsequent attempts e.g. etc. (5)
Total 7
N17584A 94
Question Number
Scheme NotesMark
s
8(a) Correct attempt at the determinant M1
(so A is non singular)det(A) = -2 and some reference to zero A1
scores M0
(2)
(b) Recognising that A-1 is required M1
At least 3 correct terms in M1
B1ft
Fully correct answer A1
Correct answer only score 4/4(4)
Ignore poor matrix algebra notation if the intention is clear(6 marks)
Question Number
Scheme NotesMarks
9 (a)
M1
Correct use of product rule. The sum
of two terms, one of which is correct.
or their
Correct differentiation. A1
Applies M1GCE Further Pure Mathematics FP1 (6667) January 2010 95
or using
in an attempt to find c. Their m must be a function of p and come from their dy/dx.
* Cso **given answer** A1
Special case – if the correct gradient is quoted could score M0A0M1A1 (4)
(b) Allow this to score here or in (c) B1
(1)(c) Attempt to obtain an equation in
one variable x or y M1
Attempt to isolate x or y – must reach x or y = f(p, q) or f(p) or f(q)
M1
One correct simplified coordinate A1
Both coordinates correct and simplified A1
(4)(9 marks)
JUNE 2011 MARK SCHEME
Question Number Scheme Notes Marks
1.
(a) Either any one of f(1) = -1 or f(2) = 8. M1
Sign change (positive, negative) (and is continuous) therefore (a root) is between and
Both values correct, sign change and conclusion A1
(2)
(b) (or truncated to 2.6)
B1
Attempt to find . M1
with or
A1 (3)
N17584A 96
or or equivalent in words.
5
GCE Further Pure Mathematics FP1 (6667) January 2010 97
Question Number Scheme Notes Marks
2. (a) or awrt 2.24 B1(1)
(b) or or
or
or or M1
awrt 2.68 A1 oe
(2)
(c)
An attempt to use the quadratic formula (usual rules) M1
Attempt to simplify their
in terms of i,. e.g. i or i M1
So, . A1 oe (3)
(d) Note that the points are
and
The point plotted correctly on the Argand diagram with/without label.
B1
The distinct points and plotted correctly and symmetrically about the x-axis on the Argand diagram with/without label.
B1 (2)
8
N17584A 98
x
y
Question Number Scheme
Notes Marks
3. (a)A =
( 1 √ 2√ 2 −1 )
(i)A2 =
( 1 √ 2√ 2 −1 )( 1 √ 2
√ 2 −1 )
= ( 1+2 √ 2−√ 2√ 2−√ 2 2+1 )
A correct method to multiply out two matrices. Can be implied by two out of four correct elements.
M1
Correct answer A1
(2)
(ii) Enlargement; scale factor 3, centre
Enlargement; B1;
scale factor 3, centre (0, 0) B1
(2)
(b)
Reflection; in the line Reflection; B1;
y = –x B1 (2)
(c) k is a constant.
C is singular (Can be implied) B1
Applies M1
A1 (3)
9
GCE Further Pure Mathematics FP1 (6667) January 2010 99
Question Number Scheme Notes Marks
4.
(a)
At least two of the four terms differentiated correctly. M1
Correct differentiation. (Allow any correct unsimplified form)
A1
(2)
(b) A correct numerical
expression for f(0.8)B1
Attempt to insert x = 0.8 into their f’(x). Does not require an evaluation.(If is incorrect for their derivative and there is no working score M0)
M1
Correct application of Newton-Raphson using their values. Does not require an evaluation.
M1
0.869 A1 cao (4)
6
N17584A 100
Question Number Scheme
Notes Marks
5. where a and b are constants.
(a)
Therefore,
Using the information in the question to form the matrix equation. Can be implied by both correct equations below.
M1
So, and
Allow
Any one correct equation.
Any correct horizontal line M1
giving and Any one of or A1
Both and A1(4)
(b)Finds determinant by applying M1
A1
or M1
150 or ft answer A1 (4)
8
GCE Further Pure Mathematics FP1 (6667) January 2010 101
Question Number Scheme Notes Marks
6.
B1
Substituting and
their z* into M1
Correct equation in x and y with i2 = -1. Can be implied.
A1
An attempt to equate real and imaginary parts. M1
Correct equations. A1
Attempt to solve simultaneous equations to find one of x or y. At least one of the equations must contain both x and y terms.
M1
Both and A1(7)
7
N17584A 102
Question Number Scheme
Notes Marks
7.
(a)Multiplying out brackets and an attempt to use at least one of the two standard formulae correctly. M1
First two terms correct. A1B1
Attempt to factorise out M1
Correct expression with factorised out with no errors seen.
A1
Correct proof. No errors seen. A1 *(6)
(b)
Use of or with the result from (a) used at least once.
M1
Correct unsimplified expression.E.g. Allow 2(3n) for 6n. A1
Factorising out ( or ) dM1
A1(4)
10GCE Further Pure Mathematics FP1 (6667) January 2010 103
N17584A 104
Question Number Scheme
Notes Marks
8. with general point
(a) Using to find a. M1
So, directrix has the equation A1 oe (2)
(b)
or (implicitly)
or (chain rule) their
M1
When
or
A1
T:
Applies
or
using
in an attempt to find c. Their mT must be a function of t .
M1
T:
T: Correct solution. A1 cso*(4)
(c) Compare with gives B1
NB with x = 3 and y = 12 gives
into T gives Substitutes their t into T. M1
At X, Substitutes their x from (a) into T. M1
So,
GCE Further Pure Mathematics FP1 (6667) January 2010 105
So the coordinates of X are A1 (4)
10
N17584A 106
Question Number Scheme
Notes Marks
9.(a)
Check to see that the result is true for
B1As the matrix result is true for
Assume that the matrix equation is true for
With the matrix equation becomes
by
M1
Correct unsimplified matrix with no errors seen.
A1
Manipulates so that on at least one term.
dM1
Correct result with no errors seen with some working between this and the previous A1
A1
If the result is true for then it is now true for Correct A1 csoGCE Further Pure Mathematics FP1 (6667) January 2010 107
(2) As the result has shown to be true for then the result is true for all n. (4)
conclusion with all previous marks earned
(6)
Question Number Scheme Notes Marks
(b) Shows that B1
{which is divisible by 12}.{ is divisible by 12 when }
Assume that for
is divisible by 12 for
So,
Correct unsimplified expression for B1
giving,
Applies
No simplification is necessary and condone missing brackets.
M1
Attempting to isolate
M1
A1cso
, which is divisible by
12 as both and are both divisible by 12.(1) If the result is true for (2) then it is now true for (3) As the result has shown to be true for then the result is true for all n. (5).
Correct conclusion with no incorrect work. Don’t condone missing brackets.
A1 cso (6)
12
N17584A 108
JANUARY 2010Mark Scheme
Question
Number
Scheme Marks
Q1(a)
z1
z2=2+8i
1−i×1+ i
1+ i
=2+2 i+8i−8
2=−3+5 i
M1
A1 A1(3)
(b) |z1
z2|=√(−3 )2+52=√34
(or awrt 5.83)M1 A1ft
(2)
(c) tan α=−5
3or 5
3
arg
z1
z2=π−1 . 03 .. .=2 .11
M1
A1(2)[7]
Notes
(a) ¿ 1+i
1+i and attempt to multiply out for M1-3 for first A1, +5i for second A1(b) Square root required without i for M1|z1||z2| award M1 for attempt at Pythagoras for both numerator and denominator
(c) tan or tan−1, ±5
3 or ±3
5 seen with their 3 and 5 award M12.11 correct answer only award A1
GCE Further Pure Mathematics FP1 (6667) January 2010 109
Question Number
Scheme Marks
Q2 (a) f (1. 3)=−1. 439 and f (1. 4 )=0 .268 (allow awrt)
B1(1)
(b) f (1. 35)<0 (−0 .568 . .. )⇒1.35<α<1.4
f (1. 375)<0 (−0 .146 . .. )⇒1. 375<α<1. 4
M1 A1
A1(3)
(c) f'( x )=6 x+22 x−3
x1=x0−
f ( x0)
f '( x0)=1. 4− 0 . 268
16 . 417,=1 .384
M1 A1
M1 A1, A1(5)
[9]Notes
(a) Both answers required for B1. Accept anything that rounds to 3dp values above.
(b) f(1.35) or awrt -0.6 M1(f(1.35) and awrt -0.6) AND (f(1.375) and awrt -0.1) for first A11 .375<α<1 . 4 or expression using brackets or equivalent in words for second A1(c) One term correct for M1, both correct for A1Correct formula seen or implied and attempt to substitute for M1awrt 16.4 for second A1 which can be implied by correct final answerawrt 1.384 correct answer only A1
Q3 For n = 1: u1=2, u1=50+1=2
Assume true for n = k:
B1
M1 A1P43153A 110
uk+1=5 uk−4=5 (5k−1+1)−4=5k+5−4=5k+1
True for n = k + 1 if true for n = k.
True for n = 1,
true for all n. A1 cso
[4]
Notes
Accept u1=1+1=2 or above B15(5k−1+1 )−4 seen award M15k+1 or 5
(k+1 )−1+1 award first A1All three elements stated somewhere in the solution award final A1
Q4 (a) (3, 0)
cao
B1(1)
(b) P: x=1
3⇒ y=2
A and B lie on x=−3
PB=PS or a correct method to find both PB and PS
B1
B1
M1
GCE Further Pure Mathematics FP1 (6667) January 2010 111
Perimeter = 6+2+3 1
3+3 1
3=14 2
3M1 A1
(5)[6]
Notes(b) Both B marks can be implied by correct diagram with lengths labelled or coordinates of vertices stated.Second M1 for their four values added together.
14 23 or awrt 14.7 for final A1
Q5 (a) det A = a (a+4 )−(−5×2)=a2+4 a+10 M1 A1(2)
(b) a2+4 a+10=(a+2)2+6
Positive for all values of a, so A is non-singular
M1 A1ft A1cso
(3)
(c) A−1= 1
10 ( 4 5−2 0 ) B1 for
110 B1 M1 A1
(3)[8]
N17584A 112
Notes(a) Correct use of ad−bc for M1(b) Attempt to complete square for M1Alt 1Attempt to establish turning point (e.g. calculus, graph) M1Minimum value 6 for A1ftPositive for all values of a, so A is non-singular for A1 csoAlt 2Attempt at b
2−4ac for M1. Can be part of quadratic formulaTheir correct -24 for first A1No real roots or equivalent, so A is non-singular for final A1cso(c) Swap leading diagonal, and change sign of other diagonal, with numbers or a for M1
Correct matrix independent of ‘their
110 award’ final A1
Q6 (a) 5−2 i is a root B1 (1)
(b) ( x−(5+2 i)) ( x−(5−2 i ))=x2−10 x+29
x3−12 x2+cx+d=(x2−10x+29 ) ( x−2 )
c=49 , d=−58
M1 M1
M1
A1, A1(5)
(c)
Conjugate pair in 1st and 4th quadrants(symmetrical about real axis)
Fully correct, labelled
B1
B1(2)
[8]
GCE Further Pure Mathematics FP1 (6667) January 2010 113
x
y
(b) 1st M: Form brackets using ( x−α ) ( x−β ) and expand. 2nd M: Achieve a 3-term quadratic with no i's.
(b) Alternative: Substitute a complex root (usually 5+2i) and expand brackets M1
(5+2 i)3−12(5+2i )2+c (5+2 i)+d=0 (125+150 i−60−8 i)−12(25+20 i−4 )+(5c+2ci )+d=0 M1 (2nd M for achieving an expression with no powers of i) Equate real and imaginary parts M1 c=49 , d=−58
A1, A1
Q7(a)
y= c2
xdydx
=−c2 x−2
dydx
=− c2
(ct )2=− 1
t2
without
y− c
t=− 1
t2( x−ct )⇒ t2 y+x=2ct
(*)
B1
M1
M1 A1cso
(4)
(b) Substitute (15 c , −c ): −ct 2+15 c=2ct
t2+2 t−15=0
( t+5)( t−3 )=0⇒ t=−5 t=3
Points are (−5 c , − c
5 ) and (3 c , c
3 ) both
M1
A1
M1 A1
A1(5)[9]
N17584A 114
Notes
(a) Use of y− y1=m( x−x1 )where m is their gradient expression in terms of c and / or t only for second M1. Accept y=mx+k and attempt to find k for second M1.(b) Correct absolute factors for their constant for second M1.Accept correct use of quadratic formula for second M1.
Alternatives:
(a)
dxdt
=c and
dydt
=−ct−2
B1
dydx
=dydt
÷dxdt
=− 1t 2
M1, then as in main scheme.
(a) y+x dy
dx=0
B1
dydx
=− yx=− 1
t 2 M1, then as in main scheme.
Q8
(a) ∑r=1
1
r 3=13=1 and
14×12×22=1
Assume true for n = k :
∑r=1
k+1
r 3= 14
k2(k+1 )2+(k+1)3
14( k+1)2 [k 2+4 (k+1 )]=1
4( k+1)2( k+2)2
True for n = k + 1 if true for n = k.True for n = 1,true for all n.
B1
B1
M1 A1
A1cso(5)
(b) ∑ r3+3∑ r+∑ 2=1
4n2(n+1)2+3 (12 n(n+1 )) , +2n B1, B1
GCE Further Pure Mathematics FP1 (6667) January 2010 115
= 1
4n [n (n+1)2+6(n+1 )+8 ]
= 1
4n [n3+2n2+7 n+14 ]=1
4n(n+2)( n2+7 )
(*)
M1
A1 A1cso
(5)
(c) ∑15
25
=∑1
25
−∑1
14
with attempt to sub in answer to
part (b)
= 1
4(25×27×632)− 1
4(14×16×203)=106650−11368=95282
M1
A1(2)
[12]
Notes
(a) Correct method to identify (k+1 )2as a factor award M114( k+1)2( k+2)2
award first A1All three elements stated somewhere in the solution award final A1(b) Attempt to factorise by n for M114 and n
3+2 n2+7 n+14 for first A1(c) no working 0/2
Q9(a) 45 or
π4 rotation (anticlockwise), about the origin
B1, B1(2)
(b) (
1√ 2
− 1√ 2
1√ 2
1√ 2
)( pq )=(3 √ 2
4√ 2)
p−q=6 and p+q=8
or equivalent
p = 7 and q = 1
both correct
M1
M1 A1
A1(4)
(c) Length of OA (= length of OB) = √72+12 , =√50=5√2 M1, A1(2)
(d)
M 2=(1
√ 2− 1
√ 21
√ 21
√ 2)(
1√ 2
− 1√ 2
1√ 2
1√ 2
)=(0 −11 0 ) M1 A1
(2)
N17584A 116
(e) (0 −11 0 )¿ (3√2 ¿ )¿
¿¿ so coordinates are (−4 √ 2 , 3 √ 2)
M1 A1(2)
[12]NotesOrder of matrix multiplication needs to be correct to award Ms(a) More than one transformation 0/2(b) Second M1 for correct matrix multiplication to give two equationsAlternative:
(b)
M−1=(1
√ 21
√ 2
− 1√ 2
1√ 2
) First M1 A1
(
1√ 2
1√ 2
− 1√ 2
1√ 2
)(3 √ 24 √ 2)=(71 )
Second M1 A1(c) Correct use of their p and their q award M1(e) Accept column vector for final A1.
GCE Further Pure Mathematics FP1 (6667) January 2010 117