1. Find the values of x for which

5 cosh x – 2 sinh x = 11,

giving your answers as natural logarithms.(Total 6 marks)

2. A = k

k1 2

0 19 1 0

−−

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟ , 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)

(Total 9 marks)

3. 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 14 {2√3 – ln(2 + √3)}.

(3)

(Total 7 marks)

4. M =−−

⎛

⎝⎜

⎞

⎠⎟

4 56 9

(a) Find the eigenvalues of M.(4)

A transformation T : 2 → 2 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)

(Total 7 marks)

Practice Paper APractice papers A and B, produced by Edexcel in 2009, with mark schemes

5. In = xn xe2⌠

⌡

⎮⎮

dx, n 0.

(a) Prove that, for n 1,In =

12

(xn e2x – nIn – 1).(3)

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

x xx2 2

0

1

e d⌠

⌡

⎮⎮

.

(5)

(Total 8 marks)

6.

Figure 1

The curve C, shown in Figure 1, has parametric equations

x = t – ln t,

y = 4√t, 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)

(Total 11 marks)

y

C

xO

7. The plane Π 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 Π 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 Π.

(d) Find the position vector of S ′.(5)

(Total 13 marks)

8. (a) Show that the normal to the rectangular hyperbola xy = c2, at the point P ct ct

,⎛⎝⎜

⎞⎠⎟ , t ≠ 0 has

equation

y = t2x + ct

– ct3.(5)

The normal to the hyperbola at P meets the hyperbola again at the point Q.

(b) Find, in terms of t, the coordinates of the point Q.(5)

Given that the mid-point of PQ is (X, Y) and that t ≠ ±1,

(c) show that XY

= – t12 ,

(2)

(d) show that, as t varies, the locus of the mid-point of PQ is given by the equation

4xy + c2 yx

xy

−⎛

⎝⎜

⎞

⎠⎟

2

= 0.(2)

(Total 14 marks)

TOTAL FOR PAPER: 75 MARKS

END

1. The transformation R is represented by the matrix A, where

A =⎛

⎝⎜

⎞

⎠⎟

3 11 3

.

Find the eigenvectors of A.

(Total 5 marks)

2. Find the values of x for which

4 cosh x + sinh x = 8,

giving your answer as natural logarithms.

(Total 6 marks)

3. 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

14 4 172

0 5

1 5

x x+ +−

⌠

⌡⎮⎮

.

.

dx.

(4)

(Total 7 marks)

4. (a) Show that, for x = ln k, where k is a positive constant,

cosh 2x = kk

4

2

12+ .

(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)

Practice paper B

5. Given that y = sinhn − 1 x cosh x,

(a) show that ddyx

= (n − 1) sinhn − 2 x + n sinhn x.(3)

The integral In is defined by In = 0

arsinh 1⌠

⌡

⎮⎮

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)

(Total 9 marks)

6. The matrix M is given by

M = 1 4 13 0

−⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

pa b c

,

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)

(Total 12 marks)

7. The hyperbola C has equation xa

yb

2

2

2

2 1− = .

(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 3

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

(b) determine the possible values of t, for 0 t < 2 .(8)

(Total 14 marks)

8. 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 = 11 3535

.(2)

(Total 15 marks)

TOTAL FOR PAPER: 75 MARKS

END

GCE Further Pure Mathematics and Decision Mathematics mock paper mark schemes – UA019582 1

Further Pure Mathematics FP3 (6669)

Practice paper A mark scheme

Question number

Scheme Marks

1. 112

ee2

2

ee5

xxxx

B1

07e22e3 2 xx M1 A1

7e,3

1e0)7e)(1e3( xxxx M1 A1

7ln3ln or 3

1ln xx A1 (6)

(6 marks)

2. (a) det A = 1892 kk M1 A1

Setting to zero and solving for k [ ]0)3(6 kk M1

6,3 kk A1 (4)

(b) Cofactors

kkk

k

kk

22

9182

99

(B1 for each row or column) B3

A–1 =

kk

kk

kk

99

189

22

det

1 2 M1 A1ft

(5)

(9 marks)

2 GCE Further Pure Mathematics and Decision Mathematics mock paper mark schemes – UA019582

Question number

Scheme Marks

3. (a) 14hsec4d

d 2 xx

y B1

Put 24cosh44cosh0d

d 2 xxx

y M1

)32ln(4 x or )347ln(8 x

or 32e4 x or 347e8 x

A1

)32ln(4

1x or )347ln(

8

1x (or equiv.) A1 (4)

(b) tanh(....))32ln(4

1y (Substitute for x) M1

2

34tanh,4tanh1

2

14hsec 2 xxx M1

32ln324

1)32ln(

4

1

2

3y () A1 (3)

(7 marks)

4. (a) Deriving characteristic equation (4 – )(–9 – ) + 30 = 0 M1 A1

2 + 5 –6 = 0 ( + 6)( – 1) = 0 = –6, = 1 M1 A1 (4)

(b) Stating, implying or showing = 1 associated with point invariant line

B1

y

x

y

x

96

54

Equation is 4x – 5y = x 3x – 5y = 0 (any equiv form) M1 A1 (3)

(7 marks)

GCE Further Pure Mathematics and Decision Mathematics mock paper mark schemes – UA019582 3

Question number

Scheme Marks

5. (a) 2 1 21e e d

2 2n x n x

n

nI x x x , 2

1

1e

2n x

n nI x nI () M1 A1 A1 (3)

(b) 11

2 2 2 2 22 1 1

00

1 1e d e e

2 2x xx x I x I I

(one correct statement) M1

I1 = 02

0

1

0

2

2

1

2

1

2

1e

2

1IeIx x

(linking all three) M1 A1

1

0

21

0

220 2

1e

2

1e

2

1de xx xI

2 2 2

22

e e e 1 1e 1

2 2 4 4 4I

M1 A1 (5)

(8 marks)

6. (a) 2

1

2d

d11

d

d

tt

y

tt

x B1 B1

tttt

t

11

121,2

11

2

2

2

12

or t

t 1

M1, A1

4ln314ln4lnd

11Length

4

1

4

1

ttt

t

()

M1 M1 A1 (7)

(b) Surface area =

tttttt

t d8d21

142

4

1

2

1

2

14

1

2

2

12

M1

3

1602

3

24

3

16)8(2

3

2)8(

4

1

2

12

3

tt

M1 M1 A1

(4)

(11 marks)

4 GCE Further Pure Mathematics and Decision Mathematics mock paper mark schemes – UA019582

Question number

Scheme Marks

7. (a)

1

1

1

RQ

c

,

4

3

2

RP

c

(both) B1

4 3 2

1 1 1

RP RQ c

c

i j k

= (–5 – 4c) i – (6 + 5c) j + k M1 A1ft (3)

(b) c = –2 A1ft

d = –6 – 5c = 4 () A1 cso (2)

(c) r. p

1

4

3

M1

Substituting point in plane to give p, r.

3

4 7

1

M1 A1 (3)

(d) Equation of normal to plane through S : r =

1

4

3

10

5

1

t B1

Meets plane where 7

1

4

3

.

10

45

31

t

t

t

t –1 M1 A1ft

S has position vector

1

4

3

2

10

5

1

t =

8

3

5

M1 A1 (5)

(13 marks)

GCE Further Pure Mathematics and Decision Mathematics mock paper mark schemes – UA019582 5

Question number

Scheme Marks

8.

(a) 2

2

dd 1d

ddd

y cy t t

xx c tt

M1 A1

The normal to the curve has gradient 2t B1

The equation of the normal is 2 ( )c

y t x ctt

M1

The equation may be written 2 3cy t x ct

t () A1 (5)

(b) Let Q be the point (cq, c/q)

Then 2 3c ccqt ct

q t and so 2( )

( )c t q

ct q tqt

M1 A1

Attempt to find q, e.g. 2( )( 1) 0q t t qt or quadratic formula

M1

3

1q t or

t

A1

So Q has coordinates 33

,c

ctt

A1 (5)

(c) 33

1 1( ), ( )

2 2

c cX t Y t

t t M1

4

3 4 2

1 1

1

X t t

Y t t t

() A1 (2)

(d) 2

2 22

1( )

4

cXY t

t

M1

22

4

c X YXY

Y X

2

24 0y x

xy cx y

() A1 (2)

(14 marks)

GCE Further Pure Mathematics and Decision Mathematics mock paper mark schemes – UA019582 1

Further Pure Mathematics FP3 (6669)

Practice paper B mark scheme

Question number

Scheme Marks

1. det A = 0 01)3( 2 M1

4,20)4)(2(0862 A1

1

1r Eigenvecto,0,0

11

11:2 yx

y

x

M1 A1

1

1r Eigenvecto,0,0

11

11:4 yx

y

x

A1 (5)

(5 marks)

2. 2

ee

2

ee4

xxxx

= 8 M1

5e2x – 16ex + 3 = 0 M1 A1

(5ex – 1)(ex – 3) = 0 A1

ex = 51 , 3

x = ln( 51 ), ln 3 accept – ln 5 M1 A1 (6)

(6 marks)

3. (a) 2, 1, 16a b c B1 B1 B1 (3)

(b) xx

d16)12(

15.1

5.02

M1

1.5

0.5

1 2 1arc tan

8 4

x

M1 A1

32

B1 (4)

(7 marks)

2 GCE Further Pure Mathematics and Decision Mathematics mock paper mark schemes – UA019582

Question number

Scheme Marks

4. (a) cosh 2x = 2

ee 22 xx =

2

ee ln2ln2 kk (or use ex = k) M1

2 2 4

2

1

2 2

k k k

k

() M1 A1 (3)

(b) 2f ( ) 2sech 2x p x M1 A1

For ln 2x 4

2

2 1 17cosh 2

2 2 8x

B1

2

20

cosh 2p

x ,

64 1282

289 289p A1 (4)

(7 marks)

5. (a) Using product rule 2 2d( 1)sinh cosh sinh

dn ny

n x x xx

M1

Using 2 2cosh 1 sinhx x in derived expression M1

to obtain 2 2d( 1)sinh (1 sinh ) sinh

dn ny

n x x xx

and 2d( 1)sinh sinh

dn ny

n x n xx

() A1 (3)

(b)

1arsinh

0

1 coshsinh

xxn

=

1arsinh

0

2 dsinh)1( xxn n +

1arsinh

0

dsinh xxn n

So 2cosh( sinh1) ( 1) n nar n I nI M1

If sinh =1 then cosh = 21 sinh = 2

22 ( 1)n nnI n I () A1 (2)

(c) 0 ar sinh1I B1

2 02 2I I M1

4 24 2 3I I and use with previous results to obtain M1

18 (3arsinh1 2) = 0.154 (either answer acceptable) A1 (4)

(9 marks)

GCE Further Pure Mathematics and Decision Mathematics mock paper mark schemes – UA019582 3

Question number

Scheme Marks

6. (a)

k

k

k

cp

b

a

cba

p

00

00

00

1

04

31

03

141TMM

300

3

141

p

p

M1 A1 (2)

(b) 18

1

4

1

141

kk (ft on their p, if used) M1 A1 (2)

(c) 2 equations: a + 4b – c = 0 3a + 3c = 0 M1

a and b in terms of c (or equiv.): a = – c b =

12 c (ft on

their p)

M1 A1ft

Using )18(18 222

cba

c

b

a

c

b

a

. M1

a = 22, b = –2, c = –22 A2, 1, 0 (6)

(d) det M= (32) – 4(–122) –1(–32) = 542 M1

A1 cso (2)

(12 marks)

4 GCE Further Pure Mathematics and Decision Mathematics mock paper mark schemes – UA019582

Question number

Scheme Marks

7. (a) t

x

d

d = a sec t tan t,

t

y

d

d = b sec2t M1 A1

x

y

d

d =

ta

b

tta

tb

sin

tansec

sec 2

M1 A1

gradient of normal is –b

ta sin

y – b tan t = –b

ta sin(x – a sec t) M1

ax sin t + by = (a2 + b2) tan t () A1 cso (6)

(b) y = 0 x =

ta

ba

ta

tba

cos

sin

tan)( 2222

B1

b2 = a2(e2 – 1) b2 = 4

5 2a M1

OS = ae and OA = 3OS M1

a2 + 4

5 2a = 3a2 ×

2

3× cos t

cos t = 21 M1 A1

t = 3

5,

3

A1

By symmetry or (as OA = ta

ba

cos

22 ) –

ta

ba

cos

22 = 3ae

t = 3

4,

3

2 M1 A1 (8)

(14 marks)

GCE Further Pure Mathematics and Decision Mathematics mock paper mark schemes – UA019582 5

Question number

Scheme Marks

8. (a) AB 5i + 3j AC 3i + 2j – k or BC –2i – j – k

5 3 0

3 2 1

AB AC

i j k

= –3i + 5j + k M1 A1

r = i + 2j + 3k + (–3i + 5j + k) B1ft (3)

(b) Volume = 1

.( )6

AD AB AC AD =2i + 3j + 2k B1

= 16 (2i + 3j + 2k).(–3i + 5j + k) M1

= 11

6 A1 (3)

(c) r.(–3i + 5j + k) = (2i + j).(–3i + 5j + k) M1 A1ft

= –1 A1 (3)

(d) [i.(1 – 3) + j(2 + 5) + k(3 + )].(–3i + 5j + k) = –1 M1 A1ft

–3 + 9 + 10 + 25 + 3 + = –1

35 + 10 = – 1 = – 3511 M1

E is

35

94,

35

15,

35

68 A1 (4)

(e) Distance = 11 11 35

3 535 35

i j k () M1 A1 (2)

(15 marks)