Extra FP3 past paper - A - WordPress.com · mark scheme for the question can be located in the mark...

Extra FP3 papers - 1

Extra FP3 past paper - A This paper is compiled from the list of sample questions which Edexcel circulated in 2009, when they changed the syllabus. It may not have exactly the same balance as an actual past paper, but it is a good approximation. Each numbered question carries an extra number in square brackets [...] That is the number of the question in Edexcel's 2009 list, and the number by which the mark scheme for the question can be located in the mark scheme document. Maximum marks for this paper are 82.

1.[1] An ellipse has equation 16

2x+

9

2y = 1.

(a) Sketch the ellipse.

(1) (b) Find the value of the eccentricity e.

(2) (c) State the coordinates of the foci of the ellipse.

(2)

[P5 June 2002 Qn 1]

3. [3] Solve the equation

l0 cosh x + 2 sinh x = 11.

Give each answer in the form ln a where a is a rational number. (7)

[P5 June 2002 Qn 3]

4. [4] In = xxx n dcos2

0

, n 0.

(a) Prove that In = n

2

n(n – 1)In – 2 , n 2.

(5)

(b) Find an exact expression for I6. (4)

[P5 June 2002 Qn 4]

Mark schemes for these "Extra FP3" papers at https://mathsmartinthomas.files.wordpress.com/2014/12/extra_fp3_markscheme.pdf

More FP3 practice papers, with mark schemes, compiled from Solomon and AQA papers, at the end of this booklet


5. [5] (a) Given that y = arctan 3x, and assuming the derivative of tan x, prove that

x

y

d

d=

291

3

x.

(4)

(b) Show that

33

0

d3arctan6 xxx = 91 (4 33).

(6)

[P5 June 2002 Qn 6]

6. [6] Figure 1

y

O x

The curve C shown in Fig. 1 has equation y2 = 4x, 0 x 1.

The part of the curve in the first quadrant is rotated through 2 radians about the x-axis. (a) Show that the surface area of the solid generated is given by

4

1

0

.d)1( xx

(4) (b) Find the exact value of this surface area.

(3)

(c) Show also that the length of the curve C, between the points (1, 2) and (1, 2), is given by


2

1

0

.d1

xx

x

(3) (d) Use the substitution x = sinh2 to show that the exact value of this length is

2[2 + ln(1 + 2)]. (6)

[P5 June 2002 Qn 8]

7. [7] Prove that sinh (i ) = sinh . (4)

[P6 June 2002 Qn 1]

8. [8] A =

344

450

401

.

(a) Verify that

1

2

2

is an eigenvector of A and find the corresponding eigenvalue.

(3)

(b) Show that 9 is another eigenvalue of A and find the corresponding eigenvector. (5)

(c) Given that the third eigenvector of A is

2

1

2

, write down a matrix P and a

diagonal matrix D such that PTAP = D.

(5)

[P6 June 2002 Qn 5]


9. [9] The plane Π passes through the points

A(1, 1, 1), B(4, 2, 1) and C (2, 1, 0).

(a) Find a vector equation of the line perpendicular to Π which passes through the point D (1, 2, 3).

(3)

(b) Find the volume of the tetrahedron ABCD. (3)

(c) Obtain the equation of Π in the form r.n = p. (3)

The perpendicular from D to the plane Π meets Π at the point E. (d) Find the coordinates of E.

(4)

(e) Show that DE = 35

3511.

(2)

The point D is the reflection of D in . (f) Find the coordinates of D.

(3)

[P6 June 2002 Qn 7]

End of paper


Extra FP3 past paper - B This paper is compiled from the list of sample questions which Edexcel circulated in 2009, when they changed the syllabus. It may not have exactly the same balance as an actual past paper, but it is a good approximation. Each numbered question carries an extra number in square brackets [...] That is the number of the question in Edexcel's 2009 list, and the number by which the mark scheme for the question can be located in the mark scheme document. Maximum marks for this paper are 71.

1. [10] Find the values of x for which

4 cosh x + sinh x = 8,

giving your answer as natural logarithms. (6)

[P5 June 2003 Qn 1]

2. [11] (a) Prove that the derivative of artanh x, 1 < x < 1, is 21

1

x.

(3)

(b) Find xx d artanh .

(4)

[P5 June 2003 Qn 2]


3. [12] Figure 1

Figure 1 shows the cross-section R of an artificial ski slope. The slope is modelled by the curve with equation

5.0 ,)94(

102

xx

y

Given that 1 unit on each axis represents 10 metres, use integration to calculate the

area R. Show your method clearly and give your answer to 2 significant figures. (7)

[P5 June 2003 Qn 3]

4. [15] In =

1

0

andde xx xn Jn =

1

0

.0,de nxx xn

(a) Show that, for ,1 n

.e 1 nn nII

(2)

(b) Find a similar reduction formula for Jn . (3)

(c) Show that .e

522 J

(3)

(d) Show that

1

0

dcosh xxx n = 21 (In + Jn ). (1)

(e) Hence, or otherwise, evaluate

1

0

2 ,dcosh xxx giving your answer in terms of e.

(4)

[P5 June 2003 Qn 7]

x

y

O

R

5


5. [16] The hyperbola C has equation .12

2

2

2

b

y

a

x

(a) Show that an equation of the normal to C at the point P (a sec t, b tan t) is

ax sin t + by = (a2 + b2) tan t. (6)

The normal to C at P cuts the x-axis at the point A and S is a focus of C. Given that the eccentricity of C is 2

3 , and that OA = 3OS, where O is the origin,

(b) determine the possible values of t, for .20 t

(8)

[P5 June 2003 Qn 1] 6. [17] Referred to a fixed origin O, the position vectors of three non-collinear points

A, B and C are a, b and c respectively. By considering AB AC , prove that the area of ABC can be expressed in the form accbba 2

1 .

(5)

[P6 June 2003 Qn 1]

7. [18]

96

54M

(a) Find the eigenvalues of M.

(4)

A transformation T: ℝ2 ℝ2 is represented by the matrix M. There is a line through the origin for which every point on the line is mapped onto itself under T. (b) Find a cartesian equation of this line.

(3)

[P6 June 2003 Qn 3]


8. [20] The plane 1Π passes through the P, with position vector i + 2j – k, and is

perpendicular to the line L with equation

r = 3i – 2k +(i + 2j + 3k).

(a) Show that the Cartesian equation of 1Π is x – 5y – 3z = 6.

(4)

The plane 2Π contains the line L and passes through the point Q, with position vector

i + 2j + 2k. (b) Find the perpendicular distance of Q from 1Π .

(4)

(c) Find the equation of 2Π in the form r = a + sb + tc.

(4)

[P6 June 2003 Qn 7]

End of paper

5


Extra FP3 past paper - C This paper is compiled from the list of sample questions which Edexcel circulated in 2009, when they changed the syllabus. It may not have exactly the same balance as an actual past paper, but it is a good approximation. Each numbered question carries an extra number in square brackets [...] That is the number of the question in Edexcel's 2009 list, and the number by which the mark scheme for the question can be located in the mark scheme document. Maximum marks for this paper are 79.


1. [13] Figure 2

A rope is hung from points P and Q on the same horizontal level, as shown in Fig. 2. The curve formed by the rope is modelled by the equation

, ,cosh kaxkaa

xay

where a and k are positive constants. (a) Prove that the length of the rope is 2a sinh k.

(5)

Given that the length of the rope is 8a, (b) find the coordinates of Q, leaving your answer in terms of natural logarithms and

surds, where appropriate. (4) [P5 June 2003 Qn 5]

2. [14] The curve C has equation

y = arcsec ex, x > 0, 0 < y < 21 .

(a) Prove that )1e(

1

d

d2

xx

y.

(5)

(b) Sketch the graph of C.

(2)

The point A on C has x-coordinate ln 2. The tangent to C at A intersects the y-axis at the point B. (c) Find the exact value of the y-coordinate of B. (4) [P5 June 2003 Qn 6]

x

y

ka -ka

Q P

O


3. [19] .1 ,

35

111

113

u

u

A

(a) Show that det A =2(u – 1). (2)

(b) Find the inverse of A . (6)

The image of the vector

a

b

c

when transformed by the matrix

3 1 1 3

1 1 1 is 1 .

5 3 6 6

(c) Find the values of a, b and c.

(3)

[P6 June 2003 Qn 6]

4. [23] An ellipse, with equation 49

22 yx = 1, has foci S and S.

(a) Find the coordinates of the foci of the ellipse.

(4)

(b) Using the focus-directrix property of the ellipse, show that, for any point P on the ellipse,

SP + SP = 6. (3)

[P5 June 2004 Qn 3]


5. [25] Figure 1

Figure 1 shows the curve with parametric equations

x = a cos3 , y = a sin3 , 0 < 2. (a) Find the total length of this curve.

(7)

The curve is rotated through radians about the x-axis. (b) Find the area of the surface generated.

(5)

[P5 June 2004 Qn 7] 6. [26] The points A, B and C lie on the plane Π and, relative to a fixed origin O,

they have position vectors a = 3i j + 4k, b = i + 2j, c = 5i 3j + 7k

respectively.

(a) Find AB AC . (4)

(b) Find an equation of Π in the form r.n = p. (2)

The point D has position vector 5i + 2j + 3k.

(c) Calculate the volume of the tetrahedron ABCD. (4)

[P6 June 2004 Qn 3]


7. [27] The matrix M is given by

M =

cba

p03

141

,

where p, a, b and c are constants and a > 0. Given that MMT = kI for some constant k, find

(a) the value of p, (2)

(b) the value of k, (2)

(c) the values of a, b and c, (6)

(d) det M. (2)

[P6 June 2004 Qn 5]


8. [31] Figure 1 Figure 1 shows a sketch of the curve with parametric equations

x = a cos3 t, y = a sin3 t, 0 t 2

,

where a is a positive constant. The curve is rotated through 2 radians about the x-axis. Find the exact value of the area of the curved surface generated.

(7) [FP2/P5 June 2005 Qn 3]

End of paper

y

O x


Extra FP3 past paper - D This paper is compiled from the list of sample questions which Edexcel circulated in 2009, when they changed the syllabus. It may not have exactly the same balance as an actual past paper, but it is a good approximation. Each numbered question carries an extra number in square brackets [...] That is the number of the question in Edexcel's 2009 list, and the number by which the mark scheme for the question can be located in the mark scheme document. Maximum marks for this paper are 78.

1. [21] Using the definitions of cosh x and sinh x in terms of exponentials,

(a) prove that cosh2 x sinh2 x = 1, (3)

(b) solve cosech x 2 coth x = 2,

giving your answer in the form k ln a, where k and a are integers. (4)

[P5 June 2004 Qn 1]

2. [22] 4x2 + 4x + 17 (ax + b)2 + c, a > 0.

(a) Find the values of a, b and c.

(3)

(b) Find the exact value of

5.1

5.02 1744

1

xx dx.

(4)

[P5 June 2004 Qn 2]

3. [24] Given that y = sinhn 1 x cosh x,

(a) show that x

y

d

d = (n 1) sinhn 2 x + n sinhn x.

(3)


The integral In is defined by In =

1arsinh

0

sinhn x dx, n 0.

(b) Using the result in part (a), or otherwise, show that

nIn = 2 (n 1)In 2, n 2

(2)

(c) Hence find the value of I4. (4)

[P5 June 2004 Qn 5] 4. [28] The transformation R is represented by the matrix A, where

31

13A .

(a) Find the eigenvectors of A.

(5)

(b) Find an orthogonal matrix P and a diagonal matrix D such that A = PDP1.

(5)

(c) Hence describe the transformation R as a combination of geometrical transformations, stating clearly their order.

(4)

[P6 June 2004 Qn 6]

5. [32] In =

.0,de2 nxx xn

(a) Prove that, for n 1,

In = 2

1(xn

e2x – nIn – 1).

(3)

(b) Find, in terms of e, the exact value of

1

0

22 de xx x . (5) [FP2/P5 June 2005 Qn 4]


6. [35] (a) (i) Explain why, for any two vectors a and b, a.b a = 0. (2)

(ii) Given vectors a, b and c such that a b = a c, where a 0 and b c, show

that

b – c = a, where is a scalar. (2)

(b) A, B and C are 2 2 matrices. (i) Given that AB = AC, and that A is not singular, prove that B = C.

(2)

(ii) Given that AB = AC, where A =

21

63 and B =

10

51, find a matrix C

whose elements are all non-zero. (3)

[FP3/P6 June 2005 Qn 2]

7. [39] The hyperbola H has equation 16

2x –

4

2y = 1.

Find (a) the value of the eccentricity of H,

(2)

(b) the distance between the foci of H. (2)

The ellipse E has equation 16

2x +

4

2y = 1.

(c) Sketch H and E on the same diagram, showing the coordinates of the points

where each curve crosses the axes. (3) [FP2/P5 January 2006 Qn 2]

8. [40] A curve is defined by

x = t + sin t, y = 1 – cos t,

where t is a parameter.

Find the length of the curve from t = 0 to t = 2

, giving your answer in surd form.

(7) [FP2/P5 January 2006 Qn 3]


9. [72]

M =

12

30

21

p

q

p

,

where p and q are constants.

Given that

1

2

1

is an eigenvector of M,

(a) show that q = 4p.

(3) Given also that λ = 5 is an eigenvalue of M, and p < 0 and q < 0, find (b) the values of p and q,

(4)

(c) an eigenvector corresponding to the eigenvalue λ = 5. (3)

[FP3 June 2008 Qn 2]

End of paper


Extra FP3 past paper - E This paper is compiled from the list of sample questions which Edexcel circulated in 2009, when they changed the syllabus. It may not have exactly the same balance as an actual past paper, but it is a good approximation. Each numbered question carries an extra number in square brackets [...] That is the number of the question in Edexcel's 2009 list, and the number by which the mark scheme for the question can be located in the mark scheme document. Maximum marks for this paper are 73.

1. [29] (a) Find

)41(

12x

x dx.

(5)

(b) Find, to 3 decimal places, the value of

3.0

02 )41(

1

x

x dx.

(2)

(Total 7 marks)

[FP2/P5 June 2005 Qn 1]

2. [30] (a) Show that, for x = ln k, where k is a positive constant,

cosh 2x = 2

4

2

1

k

k .

(3)

Given that f(x) = px – tanh 2x, where p is a constant, (b) find the value of p for which f(x) has a stationary value at x = ln 2, giving your

answer as an exact fraction. (4)

(Total 7 marks)

[FP2/P5 June 2005 Qn 2]


3. [33] Figure 2

Figure 2 shows a sketch of the curve with equation

y = x arcosh x, 1 x 2.

The region R, as shown shaded in Figure 2, is bounded by the curve, the x-axis and the line x = 2. Show that the area of R is

4

7 ln (2 + 3) –

2

3.

(Total 10 marks)

[FP2/P5 June 2005 Qn 6] 4. [36] The line l1 has equation

r = i + 6j – k + (2i + 3k)

and the line l2 has equation

r = 3i + pj + (i – 2j + k), where p is a constant.

The plane 1Π contains l1 and l2.

(a) Find a vector which is normal to 1Π .

(2)

(b) Show that an equation for 1Π is 6x + y – 4z = 16.

(2)

2 1

y

O x

R


(c) Find the value of p. (1)

The plane 2Π has equation r.(i + 2j + k) = 2.

(d) Find an equation for the line of intersection of 1Π and 2Π , giving your answer in

the form

(r – a) b = 0. (5)

[FP3/P6 June 2005 Qn 3]

5. [37] A =

k24

202

423

.

(a) Show that det A = 20 – 4k.

(2)

(b) Find A–1. (6)

Given that k = 3 and that

1

2

0

is an eigenvector of A,

(c) find the corresponding eigenvalue.

(2)

Given that the only other distinct eigenvalue of A is 8, (d) find a corresponding eigenvector.

(4)

[FP3/P6 June 2005 Qn 7]


6. [42] Given that

In =

4

0

,0,d)4( nxxx n

(a) show that In = 32

8

n

nIn – 1 , n 1.

(6)

Given that

4

0 3

16d)4( xx ,

(b) use the result in part (a) to find the exact value of

4

0

2 d)4( xxx .

(3)

[FP2/P5 January 2006 Qn 7]

7. [48] The point S, which lies on the positive x-axis, is a focus of the ellipse with

equation 4

2x + y2 = 1.

Given that S is also the focus of a parabola P, with vertex at the origin, find (a) a cartesian equation for P,

(4)

(b) an equation for the directrix of P. (1)



8. [50] Figure 1 The curve C, shown in Figure 1, has parametric equations x = t – ln t,

y = 4t, 1 t 4. (a) Show that the length of C is 3 + ln 4.

(7)

The curve is rotated through 2 radians about the x-axis.

(b) Find the exact area of the curved surface generated. (4)


End of paper

y

C

x O


Extra FP3 past paper - F This paper is compiled from the list of sample questions which Edexcel circulated in 2009, when they changed the syllabus. It may not have exactly the same balance as an actual past paper, but it is a good approximation. Each numbered question carries an extra number in square brackets [...] That is the number of the question in Edexcel's 2009 list, and the number by which the mark scheme for the question can be located in the mark scheme document. Maximum marks for this paper are 69.

1. [34] (a) Show that, for 0 < x 1,

ln

x

x )1(1 2

= –ln

x

x )1(1 2

.

(3)

(b) Using the definition of cosh x or sech x in terms of exponentials, show that, for 0 < x 1,

arsech x = ln

x

x )1(1 2

.

(5)

(c) Solve the equation

3 tanh2 x – 4 sech x + 1 = 0,

giving exact answers in terms of natural logarithms. (5)

(Total 13 marks)

[FP2/P5 June 2005 Qn 8]

2. [38] Evaluate

4

12 )172(

1

xx dx, giving your answer as an exact logarithm.

(5)



3. [41] (a) Using the definition of cosh x in terms of exponentials, prove that

4 cosh3 x – 3 cosh x = cosh 3x. (3)

(b) Hence, or otherwise, solve the equation

cosh 3x = 5 cosh x,

giving your answer as natural logarithms. (4)


4. [44] A transformation T : ℝ2 ℝ2 is represented by the matrix

A = 2 2

2 1

, where k is a constant.

Find (a) the two eigenvalues of A,

(4)

(b) a cartesian equation for each of the two lines passing through the origin which are invariant under T.

(3) [*FP3/P6 January 2006 Qn 3]

5. [46] The plane passes through the points

P(–1, 3, –2), Q(4, –1, –1) and R(3, 0, c), where c is a constant. (a) Find, in terms of c, RP RQ .

(3)

Given that RP RQ = 3i + dj + k, where d is a constant, (b) find the value of c and show that d = 4,

(2)

(c) find an equation of in the form r.n = p, where p is a constant. (3)

The point S has position vector i + 5j + 10k. The point S is the image of S under

reflection in . (d) Find the position vector of S . (5) [FP3/P6 January 2006 Qn 7]


6. [51] Figure 2 Figure 2 shows a sketch of part of the curve with equation

y = x2 arsinh x.

The region R, shown shaded in Figure 2, is bounded by the curve, the x-axis and the line x = 3.

Show that the area of R is

9 ln (3 + 10) – 91 (2 + 710).

(10)


7. [53] The ellipse E has equation 2

2

a

x +

2

2

b

y = 1 and the line L has equation y = mx +

c, where m > 0 and c > 0. (a) Show that, if L and E have any points of intersection, the x-coordinates of these

points are the roots of the equation

(b2 + a2m2)x2 + 2a2mcx + a2(c2 – b2) = 0. (2)

Hence, given that L is a tangent to E, (b) show that c2 = b2 + a2m2.

(2)

The tangent L meets the negative x-axis at the point A and the positive y-axis at the point B, and O is the origin.

R

x

y

O 3


(c) Find, in terms of m, a and b, the area of triangle OAB. (4)

(d) Prove that, as m varies, the minimum area of triangle OAB is ab. (3)

(e) Find, in terms of a, the x-coordinate of the point of contact of L and E when the area of triangle OAB is a minimum.

(3)


End of paper


Extra FP3 past paper - G This paper is compiled from the list of sample questions which Edexcel circulated in 2009, when they changed the syllabus. It may not have exactly the same balance as an actual past paper, but it is a good approximation. Each numbered question carries an extra number in square brackets [...] That is the number of the question in Edexcel's 2009 list, and the number by which the mark scheme for the question can be located in the mark scheme document. Maximum marks for this paper are 68.

1. [43] (a) Show that artanh

4sin

= ln (1 + 2).

(3)

(b) Given that y = artanh (sin x), show that x

y

d

d = sec x.

(2)

(c) Find the exact value of xxx d)(sinartanhsin4

0

.

(5)


2. [45] A =

019

10

21

k

k

, where k is a real constant.

(a) Find values of k for which A is singular.

(4)

Given that A is non-singular, (b) find, in terms of k, A–1.

(5)




5 cosh x – 2 sinh x = 11,

giving your answers as natural logarithms. (6) [FP2 June 2006 Qn 1]

4. [49] The curve with equation

y = –x + tanh 4x, x 0,

has a maximum turning point A. (a) Find, in exact logarithmic form, the x-coordinate of A.

(4)

(b) Show that the y-coordinate of A is 41 {23 – ln(2 + 3)}. (3) [FP2 June 2006 Qn 5]

5. [52] In =

xxx n dcosh , n 0.

(a) Show that, for 2n ,

12sinh cosh 1n n

n nI x x nx x n n I

.

(4)

(b) Hence show that

I4 = f(x) sinh x + g(x) cosh x + C,

where f(x) and g(x) are functions of x to be found, and C is an arbitrary constant. (5)

(c) Find the exact value of xxx dcosh1

0

4

, giving your answer in terms of e.

(3)



6. [54] A =

100

110

211

.

Prove by induction, that for all positive integers n,

An =

100

10

)3(1 221

n

nnn

.

(5)


7. [58] The ellipse D has equation 25

2x +

9

2y = 1 and the ellipse E has equation

4

2x +

9

2y = 1.

(a) Sketch D and E on the same diagram, showing the coordinates of the points

where each curve crosses the axes. (3)

The point S is a focus of D and the point T is a focus of E. (b) Find the length of ST.

(5)



y = 41 (2x2 – ln x), x > 0.

Find the length of C from x = 0.5 to x = 2, giving your answer in the form a + b ln 2, where a and b are rational numbers.

(7)


9. [65] Show that

xd

d[ln(tanh x)] = 2 cosech 2x, x > 0. (4) [FP2 June 2008 Qn 1]

End of paper


Extra FP3 past paper - H

This paper is compiled from the list of sample questions which Edexcel circulated in 2009, when they changed the syllabus. It may not have exactly the same balance as an actual past paper, but it is a good approximation. Each numbered question carries an extra number in square brackets [...] That is the number of the question in Edexcel's 2009 list, and the number by which the mark scheme for the question can be located in the mark scheme document. Maximum marks for this paper are 78.

1. [55] The eigenvalues of the matrix M, where

M =

11

24,

are 1 and 2, where 1 < 2. (a) Find the value of 1 and the value of 2.

(3)

(b) Find M–1. (2)

(c) Verify that the eigenvalues of M–1 are 1–1 and 2

–1. (3)

A transformation T : ℝ2 ℝ2 is represented by the matrix M. There are two lines, passing through the origin, each of which is mapped onto itself under the transformation T.

(d) Find cartesian equations for each of these lines. (4)



2. [56] The points A, B and C lie on the plane 1Π and, relative to a fixed origin O, they

have position vectors

a = i + 3j – k, b = 3i + 3j – 4k and c = 5i – 2j – 2k

respectively. (a) Find (b – a) (c – a).

(4)

(b) Find an equation for 1Π , giving your answer in the form r.n = p.

(2)

The plane 2Π has cartesian equation x + z = 3 and 1Π and 2Π intersect in the line l.

(c) Find an equation for l, giving your answer in the form (r p) q = 0.

(4)

The point P is the point on l that is the nearest to the origin O.

(d) Find the coordinates of P. (4)


3. [57] Evaluate

3

12 )54(

1

xx dx, giving your answer as an exact logarithm.

(5)


4. [60] (a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that

cosh(A – B) = cosh A cosh B – sinh A sinh B. (3)

(b) Hence, or otherwise, given that cosh (x – 1) = sinh x, show that

tanh x = 1e2e

1e2

2

.

(4)



5. [61] Given that In =

8

0

3

1

d)8( xxx n , n 0,

(a) show that In = 43

24

n

nIn – 1, n 1.

(6)

(b) Hence find the exact value of

8

0

3

1

d)8)(5( xxxx .

(6)


6. [62]

Figure 1

Figure 1 shows part of the curve C with equation y = arsinh (x), x 0. (a) Find the gradient of C at the point where x = 4.

(3) The region R, shown shaded in Figure 1, is bounded by C, the x-axis and the line x = 4. (b) Using the substitution x = sinh2

, or otherwise, show that the area of R is

k ln (2 + 5) – 5, where k is a constant to be found.

(10)



7. [73]

Figure 1

Figure 1 shows a pyramid PQRST with base PQRS. The coordinates of P, Q and R are P (1, 0, –1), Q (2, –1, 1) and R (3, –3, 2). Find (a) PQ PR

(3)

(b) a vector equation for the plane containing the face PQRS, giving your answer in the form r . n = d.

(2)

The plane Π contains the face PST. The vector equation of Π is r . (i – 2j – 5k) = 6. (c) Find cartesian equations of the line through P and S.

(5)

(d) Hence show that PS is parallel to QR. (2)

Given that PQRS is a parallelogram and that T has coordinates (5, 2, –1), (e) find the volume of the pyramid PQRST.

(3)


End of paper


Extra FP3 past paper - I

This paper is compiled from the list of sample questions which Edexcel circulated in 2009, when they changed the syllabus. It may not have exactly the same balance as an actual past paper, but it is a good approximation. Each numbered question carries an extra number in square brackets [...] That is the number of the question in Edexcel's 2009 list, and the number by which the mark scheme for the question can be located in the mark scheme document. Maximum marks for this paper are 80.

1. [64] The points A, B and C have position vectors, relative to a fixed origin O, a = 2i – j, b = i + 2j + 3k, c = 2i + 3j + 2k,

respectively. The plane Π passes through A, B and C.

(a) Find AB AC .(4)

(b) Show that a cartesian equation of Π is 3x – y + 2z (2)

The line l has equation (r – 5i – 5j – 3k) × (2i – j – 2k) = 0. The line l and the plane Π

intersect at the point T. (c) Find the coordinates of T.

(5)

(d) Show that A, B and T lie on the same straight line. (3)



8 cosh x – 4 sinh x = 13, giving your answers as natural logarithms. (6) [FP2 June 2008 Qn 2]


3. [67] Show that 6

25

3

( 9)

x

x

dx = 3 ln

3

32 + 33 – 4.

(7)



y = arsinh (x3), x 0. The point P on C has x-coordinate √2. (a) Show that an equation of the tangent to C at P is

y = 2x – 2√2 + ln (3 + 2√2). (5)

The tangent to C at the point Q is parallel to the tangent to C at P. (b) Find the x-coordinate of Q, giving your answer to 2 decimal places.

(5)


5. [69] Given that

In =

0

dsine xxnx , n 0,

(a) show that, for n 2,

In = 1

)1(2

n

nnIn – 2 .

(8)

(b) Find the exact value of I4. (4)



6. [70]

Figure 1

Figure 1 shows the curve C with equation

y = 10

1cosh x arctan (sinh x), x 0.

The shaded region R is bounded by C, the x-axis and the line x = 2.

(a) Find

xxx d)(sinharctancosh .

(5)

(b) Hence show that, to 2 significant figures, the area of R is 0.34. (2)



7. [71] The hyperbola H has equation

16

2x –

9

2y = 1.

(a) Show that an equation for the normal to H at a point P (4 sec t, 3 tan t) is

4x sin t + 3y = 25 tan t. (6)

The point S, which lies on the positive x-axis, is a focus of H. Given that PS is parallel to the y-axis and that the y-coordinate of P is positive, (b) find the values of the coordinates of P.

(5) Given that the normal to H at this point P intersects the x-axis at the point R, (c) find the area of triangle PRS. (3) [FP2 June 2008 Qn 7]

8. [63] Given that

1

1

0

is an eigenvector of the matrix A, where

A

311

41

43

q

p

(a) find the eigenvalue of A corresponding to

1

1

0

,

(2)

(b) find the value of p and the value of q. (4)

The image of the vector

n

m

l

when transformed by A is

3

4

10

.

(c) Using the values of p and q from part (b), find the values of the constants l, m and n. (4) [FP3 June 2007 Qn 3]

End of paper

Solomon Press P6A page 2

1. With respect to a fixed origin O, the lines l1 and l2 are given by the equations l1 : [r – (– 3i + 2j – k)] × (i + k) = 0, l2 : [r – (i + j + 4k)] × (2i – j – 2k) = 0.

(a) Find (i + k) × (2i – j – 2k). (3 marks)

(b) Find the shortest distance between l1 and l2. (3 marks) 2. Prove by induction that, for all n ∈ +,

1

n

r=∑ (r2 + 1)r! = n(n + 1)! (6 marks)

3. (a) Solve the equation z3 + 27 = 0,

giving your answers in the form reiθ where r > 0, – π < θ ≤ π. (5 marks) (b) Show the points representing your solutions on an Argand diagram. (2 marks)

4. A =

b

a

2

2.

The matrix A has eigenvalues λ1 = – 2 and λ2 = 3.

(a) Find the value of a and the value of b. (4 marks)

Using your values of a and b,

(b) for each eigenvalue, find a corresponding eigenvector, (3 marks)

(c) find a matrix P such that PTAP =

−30

02. (2 marks)

www.XtremePapers.net

FP3-type paper compiled from Solomon P6 papers A and B. Total: 73 marks

A

A

http://www.xtremepapers.net

Solomon Press P6A page 3

5. (1 + x2)2x

y

d

d2

+ 4xx

y

d

d + 2y = 0 and y = 1,

x

y

d

d = 1 at x = – 1.

Find a series solution of the differential equation in ascending powers of (x + 1) up to and including the term in (x + 1)4.

(11 marks) 6. The variable y satisfies the differential equation

2

2

d

d

y

x = x

d

d

y

x + y2 with y = 1.2 at x = 0.1 and y = 0.9 at x = 0.2

Use the approximations 0d

d

x

y ≈ h

yy

211 −−

and 0

2

d

d

2x

y ≈ 20 11

2

h

yyy −+− with a step length

of 0.1 to estimate the values of y at x = 0.3 and x = 0.4 giving your answers to 3 significant figures.

(11 marks)

7. M =

− 21

34

112

k

k .

(a) Find the determinant of M in terms of k. (2 marks) (b) Prove that M is non-singular for all real values of k. (2 marks) (c) Given that k = 3, find M −1, showing each step of your working. (4 marks)

When k = 3 the image of the vector

c

b

a

when transformed by M is the vector

5

3

0

.

(d) Find the values of a, b and c. (3 marks)

Turn over


A


Solomon Press P6B page 2

1. Given that x is so small that terms in x3 and higher powers of x may be neglected, find the values of the constants a and b for which

bx

ax

1

) ln(1

++

= 3x + 32 x2. (5 marks)

2. Given that

| z + 1 – 4i | = 1, (a) sketch, in an Argand diagram, the locus of z, (2 marks) (b) find the maximum value of arg z in degrees to one decimal place. (3 marks) 3. (a) Show that

cosh ix = cos x where x ∈ . (2 marks) (b) Hence, or otherwise, solve the equation cosh ix = eix for 0 ≤ x < 2π. (3 marks) 4. Given that un+2 = 5un+1 – 6un for n ≥ 1, u1 = 2 and u2 = 4,

prove by induction that un = 2n for all integers n, n ≥ 1. (6 marks)


B



5. M =

−−−

13

410

121

x

.

(a) Given that λ = – 1 is an eigenvalue of M, find the value of x. (3 marks) (b) Show that λ = – 1 is the only real eigenvalue of M. (6 marks) (c) Find an eigenvector corresponding to the eigenvalue λ = – 1. (2 marks) 6. A student is looking at different methods of solving the differential equation

x

y

d

d= xy with y = 1 at x = 0.2

The first method the student tries is to use the approximation 0d

d

x

y ≈ h

yy01 −

twice with a

step length of 0.1 to obtain an estimate for y at x = 0.4 (a) Find the value of the student’s estimate for y at x = 0.4 (6 marks) The student then realises that the exact value of y at x = 0.4 can be found using integration. (b) Use integration to find the exact value of y at x = 0.4 (4 marks) (c) Find, correct to 1 decimal place, the percentage error in the estimated value in part (a).

(2 marks)

Turn over


B



7. (a) Given that z = cosθ + i sinθ, show that

n

n

zz

1+ = 2 cos nθ and n

n

zz

1− = 2i sin nθ,

where n is a positive integer. (3 marks) (b) Given that cos4θ + sin4θ = A cos 4θ + B, find the values of the constants A and B. (8 marks) (c) Hence find the exact value of

π8

0∫ cos4θ + sin4θ dθ . (3 marks)

8. The points A, B, C and D have coordinates (3, – 1, 2), (– 2, 0, – 1), (1, 2, 6) and (– 1, – 5, 8)

respectively, relative to the origin O.

(a) Find AB × AC . (5 marks) (b) Find the volume of the tetrahedron ABCD. (3 marks) The plane Π contains the points A, B and C. (c) Find a vector equation of Π in the form r.n = p. (3 marks) The perpendicular from D to Π meets the plane at the point E. (d) Find the coordinates of E. (6 marks)

END


B

B


Solomon Press P6A MARKS page 2

P6 Paper A – Marking Guide

1. (a) 1 0 1

2 1 2− −

i j k

= i(0 + 1) − j(−2 − 2) + k(−1 − 0) = i + 4j − k M1 A2

(b) d = ( 4 5 ) ( 4 )

1 16 1

− + − + −+ +

i j k . i j k M1 A1

= 4 4 5

18

− + + = 5

18 or 5

6 √2 A1 (6)

2. assume true for n = k ∴ 1

k

r =∑ (r2 + 1)r! = k(k + 1)!

∴ 1

1

k

r

+

=∑ (r2 + 1)r! = k(k + 1)! + [(k + 1)2 + 1](k + 1)! M1 A1

= (k + 1)!(k + k2 + 2k + 2) = (k + 1)!(k2 + 3k + 2) M1 = (k + 1)!(k + 2)(k + 1) = (k + 1)[(k + 1) + 1]! A1

∴ true for n = k + 1 if true for n = k

if n = 1 1

n

r =∑ (r2 + 1)r! = 2 × 1! = 2; n(n + 1)! = 1 × 2! = 2 ∴ true for n = 1 B1

∴ by induction true for n ∈ + A1 (6)

3. (a) z3 = −27 ∴ (reiθ )3 = 27eiπ M1 r3 = 27 so r = 3 A1 3θ = 2nπ + π M1

n = −1, 0, 1 gives θ = π3− , π3 , π A1

∴ z1 = π3i

3e−

, z2 = π3i

3e , z3 = 3eiπ A1 (b) Im(z) z2 α

z3 α O α 3 Re(z)

z1 α = 2π3 B2 (7)

4. (a) 2

2

a

b

λλ

−−

= 0 ∴ (2 − λ)(b − λ) − 2a = 0 M1 A1

λ1 = −2 gives 4b − 2a + 8 = 0 λ2 = 3 gives −b − 2a + 3 = 0 M1 solve simul. giving a = 2, b = −1 A1

(b) λ1 = −2, 4 2 0

2 1 0

x

y

=

,

4 2 0

2

x y

y x

+ =∴ = −

∴ eigenvector 1

2k

−

M1 A1

λ1 = 3, 1 2 0

2 4 0

x

y

− = −

, 2 0

2

x y

x y

− + =∴ =

∴ eigenvector 2

1k

A1

(c) P = 1

5

1 2

2 1

−

M1 A1 (9)


A

A


Solomon Press P6A MARKS page 4

7. (a) det M = 2(8 − 3k) − 1(2k + 3) + 1(k2 + 4) = k2 − 8k + 17 M1 A1

(b) det M = (k − 4)2 − 16 + 17 = (k − 4)2 + 1 M1 (k − 4)2 ≥ 0 ∴ det M > 0 ∴ M non-singular for all real k A1

(c) k = 3, M =

2 1 1

3 4 3

1 3 2

−

, det M = 2 B1

matrix of cofactors:

1 9 13

1 5 7

1 3 5

− − − − −

M1 A1

∴ M−1 = 1

2

1 1 1

9 5 3

13 7 5

− − − − −

A1

(d)

a

b

c

= M−1

0

3

5

= 1

2

1 1 1

9 5 3

13 7 5

− − − − −

0

3

5

= 1

2

2

0

4

−

=

1

0

2

−

a = −1, b = 0, c = 2 M1 A2 (11)

8. (a) w(iz − 1) = z + 1; iwz − w = z + 1 M1

z(iw − 1) = w + 1 ∴ z = 1

i 1

w

w

+−

A1

| z | = 1 ∴ | w + 1 | = | iw − 1 | M1 | w + 1 | = | i(w + i) | = | i | | w + i | = | w + i | ∴ perp. bisector of −1 and −i ∴ u = v M1 A1

(b) Im z = 0 ∴ y = 0 so u + iv = 1

i 1

x

x

+−

M1

(u + iv)(−1 + ix) = x + 1 −u − vx + i(ux − v) = x + 1 M1

−u − vx = x + 1 ∴ x = 1

1

u

v

− −+

; ux − v = 0 ∴ x = v

u A1

∴ 1

1

u

v

− −+

= v

u; −u2 − u = v + v2 M1

giving (u + 12 )2 + (v + 1

2 )2 = 12 A1

∴ circle, centre − 12 − 1

2 i, radius 12

A1

(c) Im(z)

12− L

C 12− Re(z) B3 (14)

Total (75)

O


A


Solomon Press P6B MARKS page 2

P6 Paper B – Marking Guide 1. ln (1 + ax) × (1 + bx)−1 = (ax − 1

2 a2x2 + …)(1 − bx + …) B1

= ax − abx2 − 12 a2x2 + ... M1 A1

∴ ax + (−ab − 12 a2)x2 = 3x + 3

2 x2

∴ a = 3 and −ab − 12 a2 = 3

2 M1

giving −3b − 92 = 3

2 so a = 3, b = −2 A1 (5) 2. (a) Im(z) 4 circle centre −1 + 4i, radius 1 B2 -1 O Re(z) (b) Im(z) 4 -1 O Re(z)

tan α = 14 , α = 14.036...° M1

max. arg z = 90° + 2α = 118.1° (1dp) M1 A1 (5)

3. (a) cosh ix = 12 (eix + e−ix) = 1

2 [cos x + isin x + cos (−x) + isin (−x)] M1

= 12 [ cos x + isin x + cos x − isin x] = cos x A1

(b) cosh ix = eix ∴ cos x = cos x + isin x M1 ∴ sin x = 0 giving x = 0, π M1 A1 (5)

4. assume true for n = k and n = k + 1 ∴ uk = 2k, uk+1 = 2k+1 M1 ∴ uk+2 = 5(2k+1) − 6(2k) M1 = 10(2k) − 6(2k) = 4(2k) = 2k+2 M1 A1 ∴ true for n = k + 2 if true for n = k and n = k + 1 if n = 1, u1 = 21 = 2; if n = 2, u2 = 22 = 4 ∴ true for n = 1 and n = 2 B1 ∴ by induction true for integer n, n ≥ 1 A1 (6)

α


B



5. (a) λ = −1,

2 2 1

0 2 4

3 0x

−− = 0 M1

∴ 2(0 + 12) − 2(0 + 4x) − 1(0 − 2x) = 0 M1 24 − 8x + 2x = 0 so x = 4 A1

(b)

1 2 1

0 1 4

4 3 1

λλ

λ

− −− −

− − = 0 M1

(1 − λ)[(1 − λ)(−1 − λ) + 12] − 2(0 + 16) − 1[0 − 4(1 − λ)] = 0 A1 −(1 + λ)(1 − λ)2 + 12 − 12λ − 32 + 4 − 4λ = 0 M1 −(1 + λ)(1 − λ)2 − 16λ − 16 = 0 (1 + λ)(1 − λ)2 + 16(1 + λ) = 0 (1 + λ)[(1 − λ)2 + 16] = 0 A1 λ = −1; or (1 − λ)2 = −16, not poss. for real λ M1 ∴ λ = −1 is only real eigenvalue A1

(c)

2 2 1 0

0 2 4 0

4 3 0 0

x

y

z

− − =

2 2 0

2 4 0

2

x y z

y z

y z

+ − =− =

∴ =

2 4 0

2 3 0

2 3

x z z

x z

x z

∴ + − =+ =

∴ = − ∴ eigenvector

3

4

2

k

−

M1 A1 (11)

6. (a) 1 0

0.1

y y− ≈ x0y0 ∴ y1 ≈ 0.1x0y0 + y0 M1 A1

x0 = 0.2, x1 = 0.3, y0 = 1 ∴ y1 ≈ 1.02 M1 A1 y2 ≈ 0.1x1y1 + y1 M1 x1 = 0.3, x2 = 0.4, y1 = 1.02 ∴ y2 ≈ 1.0506 A1

(b)

1

y

∫ 1y dy =

0.4

0.2∫ x dx M1

[ ln | y | ] 1y

= [ 12 x2 ] 0.4

0.2 A1

ln | y | − ln 1 = 0.08 − 0.02 giving y = e0.06 M1 A1

(c) % error = 0.06

0.06e 1.0506

e− × 100% = 1.1% (1dp) M1 A1 (12)


B



7. (a) zn + 1nz

= cos nθ + isin nθ + cos (−nθ) + isin (−nθ) M1

= cos nθ + isin nθ + cos nθ − isin nθ = 2cos nθ A1

zn − 1nz

= cos nθ + isin nθ − cos nθ + isin nθ = 2isin nθ A1

(b) (z + 1

z)4 = z4 + 4z3 1

z + 6z2

2

1

z + 4z

3

1

z +

4

1

z M1

(2cos θ)4 = 2cos 4θ + 4(2cos 2θ) + 6 M1 A1 16cos4 θ = 2cos 4θ + 8cos 2θ + 6 A1

(z − 1

z)4 = z4 − 4z2 + 6 −

2

4

z +

4

1

z M1

(2isin θ)4 = 2cos 4θ − 8cos 2θ + 6 16sin4 θ = 2cos 4θ − 8cos 2θ + 6 A1 ∴ 16(cos4 θ + sin4 θ) = 4cos 4θ + 12 M1

cos4 θ + sin4 θ = 14 cos 4θ + 3

4 so A = 14 , B = 3

4 A1

(c) I = π8

0∫ 14 cos 4θ + 3

4 dθ

= [ 116 sin 4θ + 3

4 θ ]π80 M1 A1

= 116 + 3

32 π A1 (14)

8. (a) ABuuur

= −5i + j − 3k, ACuuur

= −2i + 3j + 4k M1 A1

ABuuur

× ACuuur

= 5 1 3

2 3 4

− −−

i j k

= i(4 + 9) − j(−20 − 6) + k(−15 + 2) = 13i + 26j − 13k M1 A2

(b) ADuuur

= −4i − 4j + 6k B1

volume = 16 (13i + 26j − 13k).(−4i − 4j + 6k) M1

= 136 −4 − 8 − 6 = 39 units3 A1

(c) n = i + 2j − k B1 r.n = (3i − j + 2k).(i + 2j − k) = 3 − 2 − 2 = −1 M1 ∴ eqn. is r.(i + 2j − k) = −1 A1

(d) line through DE has eqn. r = −i − 5j + 8k + λ(i + 2j − k) M1 A1 at intersection [(−1 + λ)i + (−5 + 2λ)j + (8 − λ)k].(i + 2j − k) = −1 M1 −1 + λ − 10 + 4λ − 8 + λ = −1 giving λ = 3 M1 A1 ∴ E is (2, 1, 5) A1 (17)

Total (75)


B

B


FP3-type paper compiled from Solomon papers C, D, E for P6. 72 marks.

C

D

MARK SCHEME

C

D

FP3-type paper from AQA papers. 80 marks

AQA FP2 January 2006

AQA FP2 June 2006

AQA FP2 February 2007


AQA FP2 June 2008


MARK SCHEME

Jan 2006

June 2006

February 2007

January 2013

June 2008

June 2006

January 2009

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