INTERNAL MOCK MARKING GUIDE
1.

2.

dy 4 1  y 2

1 x 

If y  tan  2 tan  , show that
2
dx
4  x2

2 . 12
du
4
1 x


Let u  2 tan
,
2
4  x2
2 dx 1  x
4
dy
 sec 2 u  1  y 2
For y  tan u ,
du
4
dy 4 1  y 2
dy dy du dy
2
 1 y 
 
 
So
as required
4  x2
dx
4  x2
dx du dx dx
Comment:

 



Solve the simultaneous equations x 2  y 2  25r 2 and 2 y  x  10r
giving the answers in terms of r .
2
x  10r  2 y so 10r  2 y   y 2  25r 2
100r 2  40ry  4 y 2  y 2  25r 2 , 5 y 2  40ry  75r 2  0
y 2  8ry  15r 2  0 ,   y  3r  y  5r   0
y  3r , y  5r
x  4r ,
2012
x0
M1 diff arc tan
A1 for answer
M1 diff. tan
M1 chain rule
B1 for proof
M1sub & exp.
A1for quad. eqn
M1 factorising
A1 for both y
A1for both x
Comment:
3.


Solve the equation 2 cos2  x    3 cos x    1  0 for 0  x  2 .

2

2

Let  x   be y , 2 cos2 y  3 cos y  1  0

2
2 cos y  1cos y  1   0 ,
cos y 
1
 5
, y ,
2
3 3
M1factorising
A1for both cos
A1
A1
A1for all angles
cos y  1, y  0,  , 2 thus
5
 3
   5

, x , x ,
x    ,
6
2 2
2 3 3

Comment:
4.
The angle between r1 and r2 is cos 1  214  . If r1  6i  3 j  2k and
r2   2i  j  4k , find the possible values of  .
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cos  
r1 . r2 6i  3 j  2k  .  2i  j  4k  4


r1 r2
36  9  4 4  2  16 21
3  4
20  

2


652  216  176  0
A1 for
quadratic
652  44  260  176  0
  465  44  0
  4,   
M1for dotting
M1for modulus
4
3
93  4  16 20  2
2
2012
M1for
factorising
A1 for both
44
65
Comment:
5
Find the coordinates of the points of intersection of the curve
x  2t 2  1, y  3t  1 and the straight line 3 x  4 y  3 .
Substituting for x and y in the equation of the line, we get
3(2t 2  1)  12(t  1)  3
6t 2  12t  18  0
6(t-3)(t+1)=0
t  3, t  1.
i.e
The parameters of the points on the curve at its
intersection with the line are t  3andt  1.
Thus points of intersection are (17,12) and (1,0).
Comment:
6
Prove that if x  cos 2 , then
1

0
M1substituting
A1for quadratic
A1for both t
A1
A1for each
coordinate
1 x

dx   1
1 x
2
1  cos 2
.  2 sin 2 d
4 1  cos 2
x  cos 2 , dx   2 sin 2 d
x  0,    x  1,   0
4

2 sin 2 
 4 4
. sin  cos  d
0
2 cos 2 
0
 
© Prepared by Ssekyewa Edward GHS
M1for changing
limits
M1for even
2
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2012
power

 4 4 sin 2  d
0
 2

4
0
1  cos 2  d
M1for integratn
 2  12 sin 2 0 4

M1for limits
  1 

 2    0
 4 2 


 1
2
A1for proof
ALTERNATIVELY:
1  cos 2 1  cos 2
.
 2 sin 2 d
1  cos 2
4 1  cos 2
x  cos 2 , dx   2 sin 2 d
x  0,    x  1,   0
4
0
(1  cos 2 ) 2
 
.  2 sin 2 d
1  cos 2 2
4
0
 
 2

0
4
1  cos 2  d
 2  12 sin 2 0 4

  1 

 2    0
 4 2 



2
1
Comment:
7
Find the values of  for which 10 x 2  4 x  1  2x2  x  has equal
roots.
10 x  4 x  1  4x  2x
 (10  2 ) x  (4  4 ) x  1  0
For equal roots: b  4ac  0 , so
2
(4  4 )  4(10  2 )  0
162  40  24  0
 22  5  3  0
   3,    12
(  3)( 2  1)  0
2
2
2
2
ALT; Let the roots be  & 
2 
 4  4 
10  2
where  
© Prepared by Ssekyewa Edward GHS
M1collecting
terms
M1use of
discriminant.
A1for quadratic
M1for
factorising
A1for both
 1
1
and  2 
 5
10  2
3
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INTERNAL MOCK MARKING GUIDE
2012
2
  1
1
 
So, 
, to get 22  5  3  0
2  5
  5
   3,    12
(  3)( 2  1)  0
Comment:
8
d  3 2  x 3
ln
dx 
2  x3
Show that
2

  4x
.
 4  x6


2  x3  1

3
Let y  ln
 In 2  x3  In 2  x3
3 

2 x  3

dy 1   3 x 2    3 x 2  


 
dx 3   2  x 3   2  x3  
 
dy 3x 2  2  x 3  2  x 3


dx
3  2  x 3 2  x 3
dy
4 x2

dx
4  x6






M1splitting
M1diffg
M1diffg




M1symplifying
A1for answer
Comment:
9
Given that f ' ' x   2 
x
f ' ' x   2 
2
x
3
, f ' 1  0 , f 1  8 , find f x  .
2
3
 2  2x
f ' x    2  2 x
3
2
3
2
dx  2 x  4 x
1
2
 c , since f ' 1  0 ,
2  4  c  0 c   6
f ' x   2 x  4 x
f x    2 x  4 x
1
1
2
2
6
 6 dx  x 2  8 x
1
2
 6 x  C for f 1  8
f 1  1  8  6  C  8 C  5
Thus f x   x 2  8 x
Comment:
1
2
 6x  5
b) Find and classify the nature of the stationary point on the curve
x  4  t 3 and y  t 2  2t .
© Prepared by Ssekyewa Edward GHS
M1Integrating
M1for initial
values
A1for arbt. cons
M1Integrating
A1for constant
A1for constant
M1differentiatg
A1for value of t
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dx
dy
  3t 2 ,
 2t  2
dt
dt
dy
1
dy
 2t  2  2 at the stationary point,
0
Thus
dx
3t
dx
2  2t
 0 ,  t  1 thus the coordinates are 3,  1
So
3t 2
d 2 y d  2  2 2 1    1 
2   1 
4
  t  t  2    3  2  2 
2
3t   3t 
dx  3
3   3t 
dx
 3t
2
d y   4 2   1  2
For t  1

       0 thus 3,  1 min .
3  3  9
dx 2  3
10
2012
M1coordinates
M1for 2nd
derive.
M1for checking.
A1for max.
The position vector of points P and Q are 2i  3 j  4k and
3i  7 j  12k respectively.
Determine;
i)
the size of PQ.
ii)
The Cartesian equation of PQ.
iii)
The position vector of the point of intersection and the
plane 4 x  5 y  2 z  5 .
iv)
Angle between PQ and plane in (iii) above.
i)
  2  3   1 
     
PQ =  3     7     4 
  4   12   8 
     
 PQ
ii)
= (12  (4) 2  8 2 = 9
Vector equation PQ is rPQ
 2 
 1 
 
 
   3     4 
 4 
 8 
 
 
 x  2   
  

  y     3  4 
 z   4  8 
  

(iii)
 Cartesian equation is;
x2 y3 z4


(  )
1
4
8
from (ii) above, put x, y, z into 4 x  5 y  2 z  5
 4(2   )  5(3  4 )  2(4  8 )  5
5
 32  20    
8
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2012
5 11
20
1
40
 , y  3 
  , z  4
 1
8 8
8
2
8
11 1
 position vector =
i jk
8
2
 x  2
iv)
Let  be the angle required,
 (i  4 j  8k ).( 4i  5 j  2k )  81x 45 sin 
 32
   sin 1
9 x3 5
Since dot product is negative, such an angle cannot be
between a line and a plane.
Comment:
11
a) Prove that, if tan A 
tan( A  B)
is independent of x.
3  4x
6  7x
and tan B 
, the value of
4  3x
7  6x
tan A  tan B
1  tan A tan B
3  4x 6  7x

4  3x 7  6 x

 3  4 x  6  7 x 


1  
 4  3x  7  6 x 
tan  A  B  


3  4 x 7  6 x   6  7 x 4  3x 
4  3x 7  6 x   3  4 x 6  7 x 
21  18 x  28 x  24 x 2  24  18 x  28 x  21x 2
28  24 x  21x  18 x 2  18  21x  28 x 2


 3x 2  3
46 x 2  46
 3( x 2  1)
3
thus the value is independent of x .

2
46
46( x  1)
Comment:
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2012
b)
Find the maximum and minimum values of the function
6 cos   8 sin  cos  and the corresponding value of  , hence solve
6 cos 2   8 sin  cos   4 .
3(2 cos 2  )  4(2 sin  cos  ) = 3(1  cos 2 )  4 sin 2
= 3cos 2  4 sin 2  3
Let 3cos 2  4 sin 2  3  R cos 2 cos  R sin 2 sin 
So, R cos 2  3 , R sin 2  4
2
R sin  3
   53.1o

R cos  4
R 2 cos 2   R 2 sin 2   3 2  4 2  R  5
 3 cos 2  4 sin 2  3  5 cos(2  53.1o )  3
Max. Value = 8 when 2  53.1o  0o    116.55 o
Min. Value =  2 when 2  53.1o  180o 
  116.55 o
Thus: 6 cos 2   8 sin  cos   4
5 cos(2  53.1o )  3  4
1
 cos( 2  53.1o ) 
2  53.1o  78.5o ,281.5o
5
o
o
   65.8 ,167.3
 tan  
Comment:
12a) Express  1  3i 6 in the form x  iy .
Let z   1  3i , z   12   3   2
2
 3
2
 
arg z  tan 1 

3
 1 
2
2
Thus  1  3i  2 cos   i sin   
3
3 

 1  3i   2 cos 4  i sin  4 
 1  3i   641  0i   64  0i
6
6
6
Comment:
b) w is the complex number such that w 
p
q

where p, q
2  i 1  3i
are real. Given that argw  2 , and w  7 , find p and q .
© Prepared by Ssekyewa Edward GHS
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2012
p2  i 
q1  3i 

2  i 2  i  1  3i 1  3i 
p2  i  q1  3i 
w

5
10
4 p  q 2 p  3q i
w

10
10
1
4 p  q 2  2 p  3q 2  7
w  7 , so
10
2 p  3q 
2 p  3q 1
arg w  tan 1
 , thus
 , so q   4 p
4p  q 0
4p  q 2
1
4 p   4 p 2  2 p  12 p 2  7
10
196 p 2  4900 ,
p 2  25
thus p   5 , q   20
w
Comment:
13
Express 5  2 x  3x 2 in the form a  b( x  c) 2 . Deduce the maximum
or
minimum value of the expression.
5  2 x  3x 2 = 5  3x 2  23 x 

= 5  3 x 2  23 x  13 2  13 2

= 5  13  3x  13 2
=
16
3
 3x  13 
2
For greatest value occurs when x  13  0 , i.e x   13
Greatest value =
16
3
Comment:
b) Show that the gradient of the curve y  xx  32 is zero at the
point P1, 4 , and sketch the curve. The tangent at P cuts the
curve again at Q . Calculate the area contained between the
chord PQ and the curve.
© Prepared by Ssekyewa Edward GHS
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2012
y  x3  6 x 2  9 x
dy
dy
 3 x 2  12 x  9 , for turning points,
 0 , thus 3x 2  12 x  9  0 ,
dx
dx
dy
So when x  1 ,  3  12  9  0 thus the gradient at
dx
P1, 4 is 0
3x  1x  3  0 , so, x  1, x  3 and y  4, y  0 , the
other turning point is 3, 0
d2y
d2y
,
so
when

6
x

12
x

1
,
 0 1, 4 max
dx 2
dx 2
d2y
when x  3,
 0 3, 0 min
dx 2
Intercepts are 0, 0 & 3, 0
y  4.
Grad of tangent at P is 0, so equation of tangent is
At Q , y  4 , so,  x 3  6 x 2  9 x  4  0
To get x  1x  1x  4  0 , thus
x  1, x  4
Area  1 4  x3  6 x2  9 x  dx
4
4

x4
9x2 
 4 x 
 2 x3 

4
2 1

1
9
3

 16  64  128  12   4   2    6 sq. units
4
2
4

Comment:
14a) The gradient of the side PQ of the rectangle PQRS is 3 4 . The
coordinates
of the opposite corners Q , S are respectively 6, 3 and  5, 1 .
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INTERNAL MOCK MARKING GUIDE
2012
Find the equation of PR .
y3 3
 ,  4 y  3x  6
x6 4
__
y 1
4
  ,  3 y   4 x  17 since they
Find the eqn of RS 
x5
3
__
Find the eqn of PQ 
are perpendicular.
Solving the two simultaneously, we get the coordinates of P
as P 2,  3
__
Let M be the midpoint of QS , thus the coordinates are
M  12 , 2 .
__
M is also the midpoint of PR , gradient of
__
3 2
PR 
2
 2  12
__
y3
 Equation of PR :
 2 , giving y  2 x  1
x2
Comment:
b)
Find the equation of the line through the intersection of the lines
3x  4 y  6  0 and 5 x  y  13  0 and
a) passes through the point 2, 4.
The equation of any line through the points of intersection of
3x  4 y  6  k (5 x  y  13)  0
two lines may be written as:
So if the line passes through (2,4) , then,
3.2  4.4  6  k (5.2  4  13)  0
4
k
27
Thus equation becomes: 3x  4 y  6  274 (5x  y  13)  0
101x  104 y  214  0
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INTERNAL MOCK MARKING GUIDE
2012
ALT:
Points of intersection y   5 x  13 , 3x  4 5x  13  6  0
58
9
, y
23
23
101
y  4 101


, equation is
, 104 y  101x  214
104
x  2 104
Thus 23x   58 , x  
m
4   239
2   58
23
b)
making an angle of 60o with the x-axis.
If line makes gradient 60o with x-axis, then its gradient is
tan 60 o  3 .
3  5k
 3
4k
3  4 3 23 3  27
k 

22
5 3
23 3  27
(5 x  y  13)  0
Thus required equation is 3x  4 y  6 
22
But gradient is given by
ALT:
tan 60o  3 , thus equation is
y
x
9
23
58
23
 3 , 23 y  23 3 x  9  58 3  0
Comment:
15a) Using the substitution y  x 2  x to solve the equation
x 4  2 x 3  7 x 2  8x  12  0 .
y  x 2  x , y 2  x 4  2 x3  x 2
x
4
 

 2 x3  x 2  8 x 2  x  12  0
y 2  8 y  12  0 ,  y  2 y  6  0 so, y  2, y  6
x 2  x  2  0 , x  2x  1  0 , x  2, x   1
And x2  x  6  0 , x  3x  2  0 , x  3, x   2
b)
M1for substg
A1for quadratic
M1for
factorising
A1for both y.
A1for both x
A1for both x .
Solve for x and y :
x log 4 128  y log 8 2  6
log 2 x  13 log 2 y 3  2 log 4 6
x log 2 128 y log 2 2

6
log 2 4
log 2 8
log 2 xy  log 2 6
7x y
  6 …..(i)
we get;
2 3
© Prepared by Ssekyewa Edward GHS

xy  6 …….(ii)
M1for change of
base
M1for equation
M1for equation
A1for quadratic.
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7 6 y
.  6
2 y 3
( y  21)( y  3)  0
So
y  3, x  2
y  21, x  
Comment:
16a) Find


y 2  18 y  63  0
2012
A1for both y
A1for both x .
2
7
 x sin x cos x dx
u  x,
1
x sin 2 x dx
2
Let
dv
 sin 2 x
dx
du
 1,
v   12 cos 2 x
dx
x cos 2 x 
1 cos 2 x
 
dx

2 
2
2
1
1
x sin 2 x dx  

2
2
1
  2 x cos 2 x  sin 2 x   c
8
M1double angle
M1for parts.
M1for
integration
A1for answer.
Comment:
16b) Express 6  9 x in partial fractions and hence find
3
27 x  8
6  9x
6  9x



Let
27 x  8 3x  2 9 x 2  6 x  4
6  9x
A
Bx  C

 2
2
3x  29 x  6 x  4 3x  2 9 x  6 x  4
3

6  9x
dx
3
8
 27 x

M1factorising
6  9 x  A 9 x2  6 x  4  Bx  C 3x  2
x 2 : 9 A  3B  0 , x :  6 A  2 B  3C   9 , x o : 4 A  2C  6
Solve to get; A  1, B   3, C  1
6  9x
 3x  1
1
Thus 
dx

dx

 3x  2
 9 x 2  6x  4 dx
27 x 3  8
1
1
 In3x  2   In9 x 2  6 x  4  c
3
6
Or In
9 x
3x  2
2
1
3
 6x  4

1
c
6
M1for partial fr
M1solving
A1for all values
M1for substg.
M1 M1for
integration
A1for answer.
Comment:
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12
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INTERNAL MOCK MARKING GUIDE
© Prepared by Ssekyewa Edward GHS
2012
13
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