Paper Reference(s) - Web viewItems included with question papers. Mathematical Formulae ... Paper Reference(s) Last modified by: Martin Thomas Company: The City of London of Academy

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2010Included:June 2014June 2014 (R)June 2013June 2013 (R)Jan 2013June 2012Jan 2012June 2011Feb 2010Omitted:June 2013 (withdrawn paper)January 2011June 2010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










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)

(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)


END

P43138A 12

Paper Reference(s)


Advanced/Advanced SubsidiaryMonday 10 June 2013 Morning










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


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)


END

P42828A 18

Paper Reference(s)

6667/01REdexcel GCEFurther Pure Mathematics FP1 (R)

Advanced/Advanced SubsidiaryMonday 10 June 2013 Morning




This paper is strictly for students outside the UK.







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)


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)


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)


END

P41485A 26

Paper Reference(s)


Advanced Subsidiary/Advanced LevelMonday 28 January 2013 Morning



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.


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.


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.




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)


ENDPaper Reference(s)

6667/01

N26109A 31

Edexcel GCEFurther Pure Mathematics FP1

Advanced SubsidiaryFriday 1 June 2012 Morning



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.


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.


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.



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)


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)


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)


ENDPaper Reference(s)

N17584A 38

6667/01


Advanced SubsidiaryMonday 30 January 2012 Afternoon

Time: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Pink) Nil



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.


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.




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)


(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



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.


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.


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.




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)


END


Paper Reference(s)

6667/01


Advanced/Advanced SubsidiaryMonday 1 February 2010 Afternoon

Time: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Orange) Nil



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.


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.



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)


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


(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


(c)

B1

B1

(2)Total 8


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


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


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


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


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)


(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


(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


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]


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


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]


JANUARY 2013 MARK SCHEME


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


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]



6.(a) Determinant: 2 – 3a = 0 and solve for a = M1

So or equivalent A1(2)

(b) Determinant:

M1A1

(2)

(c)

M1depM1A1A1

(4)[8]


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


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]


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)


[11]

N17584A 80


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)


M1

A1

[2]

where k is a constant,(b)

M1

E does not have an inverse

M1M1

A1 oe[4]

6 marks


3.

M1A1

B1

M1

M1

N17584A 82

A1 cao

[6]

6 marks


4. (a)

M1A1B1

dM1

(AG)A1 *

[5]

(b)

M1

A1 cao

[2]7 marks


5.

(a) B1

[1]

(b)M1

A1 oe


(So P has coordinates )

[2]

(c) Focus B1

Gradient M1

Either

;M1

or and

;

l: A1

[4]7 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


or equivalent expressions

A1

M1

A1 *

[4](b) B1

B1

Area M1

A1

[4]8 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


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)


(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


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


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



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



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


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



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


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


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,


So the coordinates of X are A1 (4)

10

N17584A 106


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)


(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


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


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]


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


= 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.


Documents

Paper Reference(s) - Web viewItems included with question papers. Mathematical Formulae ... Paper Reference(s) Last modified by: Martin Thomas Company: The City of London of Academy