Examination Style Paper
1
An ellipse E has a focus at (6, 0) and the corresponding directrix has equation x = 12, find
a
the exact value of the eccentricity of E,
(4)
b a Cartesian equation of E.
cosh 2 x
,
1  sinh 2 x
f ( x) 
2
(2)
x0
Find the x- coordinate of the stationary point of f(x), giving your answer to 3 significant
figures.
(7)
3
a Find

1  6x
1  9 x2
dx
(5)
1
6
b Find the exact value of

0
1  6x
I n   x n e 3 x dx ,
4
a
dx
1  9 x2
(3)
n0
Prove that, for n > 1,


1 n 3x
x e  nI n 1
3
b Find, in terms of e, the exact value of
In 
2 3 3x
0 x e
5
(3)
(5)
dx
The line l1 has Cartesian equations x – 2 = 4y = z + 3 and the line l2 has equation
 4   1
   
r = 5  s 2 
1  0 
   
The plane 1 contains l1 and l2 .
a Find a vector which is normal to 1 .
(3)
b Show that an equation of 1 is 8x + 4y – 9z = 43
(2)
c
Find the shortest distance of the point D (1, 1, 1) from 1 .
(3)
© Edexcel Limited 2009
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6
Figure 1
y
Q
P
x
Figure 1 shows a sketch of the curve with parametric equations

x  a cos3 t , y  a sin 3 t , 0  t 
2
where a is a positive constant.
The curve cuts the x-axis at the point P and the y axis at the point Q.
Find the perimeter of the shaded region bounded by the arc PQ and the chord PQ.
7
8
(8)
The point P( ap 2 ,2ap) lies on the parabola M with equation y 2 =4ax, where a is a positive
constant.
a Show that an equation of the tangent to M at P is
py = x + ap 2
(3)
The point Q (9 ap 2 , 6ap) also lies on M.
b Write down an equation of the tangent to M at Q.
(2)
The tangent at P and the tangent at Q intersect at the point V.
c Find, as p varies, the locus of V.
(4)
a Using the definition of sinhx or cosechx in terms of exponentials,
show that, for x > 0
 1  1  x2 

arcosechx = ln 


x


b Solve the equation
3coth 2 x = 7cosechx + 1
giving your answers in terms of natural logarithms.
(5)
(5)
© Edexcel Limited 2009
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9
3 0 0 


A =  k 2 5  where k and l are constants
 l 2 1 


0
 
a Show that  5  is an eigenvector of A and find the corresponding eigenvalue.
 2
 
Given that -3 is also an eigenvalue of A
b find the corresponding eigenvector.
 1
 
Given that 1 is also an eigenvector of A
 1
 
c find the corresponding eigenvalue, the value of k and the value of l.
(2)
(4)
(5)
© Edexcel Limited 2009
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Worked Solutions
1
a
ae = 6 and
a
=12
e
Use formula from booklet for
focus and directrix
a
 72
e
6
1
e

72
2
Solving a 2  ae 
So
b
2
M1
A1
b 2  72 1  12   36
So equation of E is
f ( x) 
f ( x) 

Use b 2  a 2 1  e2
x2 y 2

1
72 36
and

Use the standard equation of an
x2 y2
ellipse 2  2  1
a
b
1  sinh 2 x  2sinh 2 x  cosh 2 x  12 1  sinh 2 x 
1
2
M1
A1
M1A1A1
 2 cosh 2 x
1  sinh 2 x 
2sinh 2 x 1  sinh 2 x   cosh 2 2 x
3
1  sinh 2 x  2
f ( x)  0  0  sinh 2 2 x  2 sinh 2 x  1
So
B1
B1
sinh 2 x  1  2
1
x  arsinh( 2  1)
2
= 0.201599… = 0.202 (3sf)
Differentiate using the quotient rule
and then simplify. Then set
derivative = 0
Use cosh 2 A  sinh 2 A  1
Solve using the quadratic
formula or complete the
square
© Edexcel Limited 2009
M1
A1
M1
A1
4
3
I 
a
1
1 9x
2

dx   6 x 1  9 x 2


1
2
I  arcsin 3 x  1  9 x 2
3
3
So
1
2
b I =  arcsin 3 x  1  9 x 2
3
3


1
2

1
2
Split the integral into two parts
dx
+c
First integral is based on arcsinx
given in formula booklet.
For second identify f ( x)f ( x) or
use substitution

0
Substitute limits and recall that
  1
sin   
6 2
1
 
2
2
9
=  arcsin  63   1  36
0 


3
3
3
 
 2
3 2
=
 

18 3 2 3
1
1
 1
I n   x n d  e3 x   e3 x x n   e3 x nx n 1 dx
3
3
 3
1
I n  x n e3 x  nI n 1
3

M1
A1
A1

3 2
=


18 3 3
a
M1A1
M1A1
1
1 6
2
1
2
4
M1

Use integration by parts
M1A1
Simplify to given answer
A1cso
b
1 3 3x
x e  I2
3
1
2
I 2  x 2 e3 x  I1
3
3
1
1
I1  xe3 x  I 0
3
3
I3 
Use the reduction
formula from (a)
Use integration to find
the I 0 term
1
I 0   e dx  e 3 x + c
3
1
1
2
2 1
So I 3  x3e3 x  x 2 e3 x  xe3 x   e3 x
3
3
9
9 3
3x
2 3 3x
xe
0

2
1
2
2 1 
1
dx   x3e3 x  x 2 e3 x  xe3 x   e3 x 
3
9
9 3 0
3
M1
B1
Link the terms together
M1
Apply the limits
M1
4
4
2   2 
8
  e 6  e 6  e6  e 6     
3
9
27   27 
3
46 6 2
=
e 
27
27
A1
© Edexcel Limited 2009
5
5
 2   4
   
a l1 has vector equation r   0   t  1 
 3   4 
   
 4   1   8 
     
Perpendicular vector to plane is  1    2    4 
 4  0   9 
     
b Since plane contains both lines:
 x   8   8   2 
       
 y    4    4    0   43
 z   9   9   3 
       
So -8x - 4y + 9z = -43
Or 8x + 4y - 9z = 43
 1
 
c Vector from 1 to
 1
 
2 1
   
 0  is  1 
 3   4 
   
 1   8 
   
Distance is +  1    4  
   
 4   9 
So shortest distance is
6
B1for
1:0.25:1 ratio
Use the vector
product
Use r.n=a.n
formula for a
plane
Find a suitable vector
and then scalar
product with n̂

M1
A1cso
M1
1
M1
 8 2   4 2  92
40
161
A1
P (a, 0) and Q (0, a) so chord PQ = a 2
Arc PQ =
M1A1
x 2  y 2 dt
x  3a cos 2 t    sin t 
B1
Use formula from formula
booklet
y  3a sin t   cos t 
M1
A1
So arc =  3a sin t cos t sin 2 t  cos 2 t dt
M1
2
3a
=  sin 2t dt
2

2
 3a
=   cos 2t 
 4
0
 3a   3a 
=   
 4   4 
3a
So perimeter =
a 2
2
Use sin2A formula to help
integrate
M1
A1
M1
A1
© Edexcel Limited 2009
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7
a
dy y 2a
1
 

dx x 2ap p
Equation is
So
y  2ap 
Differentiate using chain
rule to find gradient

1
x  ap 2
p
py  x  ap

Let Q be ( aq 2 , 2aq ) and then
compare with given point
So equation is:
3 py  x  9ap 2
Solve simultaneously to find the
coordinates of V
y 2  16a 2 p 2 and 3 y 2  48a 2 p 2
a Let y = arcosechx
Then x = cosechy
1
So sinhy =
x
y
y
e e
1

2
x
2y
xe  x  2e y or xe2 y  2e y  x  0
2  4  4 x2 1  1  x2
ey 
=
2x
x
y
Since e >0 then we must choose the “+” sign
 1  1  x2 

So
y = arcosechx = ln 


x


2
b 3 + 3cosech x = 7cosechx + 1
3c 2 -7c + 2 = 0
(3c - 1)(c - 2) = 0
cosechx = 2 or 13
 1 5 
So x = ln 
 or ln 3  10
2



M1
A1
So locus of V is: 3 y 2  16ax
8
M1
A1
c Subtracting equations
2 py  8ap 2
So
y = 4ap
And
x = 3ap 2
Therefore
M1
A1cso
2
b Let q = 3p
M1

Eliminate p to find
locus
M1
A1
M1
Use the definition of
sinhx
Rearrange to form quadratic
in e y
Use the quadratic
formula
Take logs
M1
A1
M1
A1cso
Use coth 2 x = 1 + cosech 2 x
M1
Solve quadratic in cosechx
M1A1
Use formula from part (a)
© Edexcel Limited 2009
M1A1
7
9
a
0  0 
 0
   
 
A  5    20   4   5 
 2  8 
 2
   
 
Multiply A times the vector
and use the definition of an
eigenvector: Ax = x
A1
So the vector is an eigenvector with eigenvalue of 4
 x
 x
 
 
b A  y   3  y  
z
z
 
 
3x = -3x and so x = 0
kx + 2y + 5z = -3y
lx + 2y - z = -3z
Solving: y = - z
0
 
So the eigenvector is: p  1  , for some constant p
 1 
 
 1
 1
 
 
c A  1    1 
 1
 1
 
 
M1
Again use the basic definition
of an eigenvector
M1
B1
Solve to find a solution
M1
A1
M1A1
3 =  and so the eigen value is 3
k + 7 = 3 so k = -4
l + 1 = 3 so l = 2
M1A1A1
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