DUNMAN HIGH SCHOOL
H2 Mathematics
2009 Yr 6 Prelim Paper 1 Suggested Solns
1
Let x, y, z be the number of B1, B2, B3 bears made in the month.
x  y  z  180
5 x  4 y  3z  650
20 x  12 y  9 z  2100
 1 1 1 180 
Augmented matrix M =  5 4 3 650 
 20 12 9 2100 


 1 0 0 30 


RREF(M) =  0 1 0 50 
 0 0 1 100 


Thus x = 30, y = 50, z = 100
2
From GC, a = 0
tan 1 x  y
x  tan y
x   tan y
Since x  0 and 0  y 
f 1 : x

, therefore x   tan y
2

 tan x , 0  x 
2
3(a)
9 x 2  4 y 2  36 x  0
9  x  2   4 y 2  36
2
 x  2
2

4
y2
 1 i.e Hyperbola
9
Asymptotes:
 x  2
2
4
3
y   x3
2
y2
3
3
 y   x  2  or y    x  2
9
2
2
3
y
y   x  2
2

2
(b)(i)
(ii)
Asymptotes: y 
4
x
and x  1
2
dy 1
A
 
0
dx 2  x  12
 x  1
2
 2A
Therefore, for C not to have stationary points, A < 0.
4(a)
f ( x)  tan  eg( x )   f '( x)  sec 2  eg( x )  g '( x)eg( x )
x

f '(5)  sec2 (e)(3)e  9.81
(b)
x2
dy (2 x  1)(2 x)  x 2 (2)


2 x  1 dx
(2 x  1)2
2 x2  2 x

(2 x  1)2
dy
2 x2  2 x
Function is increasing 
0
0
dx
(2 x  1)2
2 x( x  1)
 0  x  0 or x  1
(2 x  1)2
y
5(a)
 2 5
   
 3   2  1  4 
 4   2   
By ratio theorem, OP        1 
3
3
 
 0 
(b)
 11   3   8 
    

AB  EC  OE       4      4 
14   6   8 
    

 2 
 
AD   1 
 1
 
(c)
 3   2 
   
 4    1 
 6  1
Angle required = cos 1      102
61 6
(d)
 11   5   6 
3
    

 
BC  k AB  BC       1     1  k  4 
14   2   12 
6
    

 
 k  2 &   8  1  9
6(i)
Set

(ii)





(iii)
dm
 R  Bm, where R and B are positive constants.
dt
dm
2
3
 R  B( )  0, B  R.
dt
3
2
dm
3
 R  Rm
dt
2
3
 R(1  m)
2
R
 (2  3m)  k (2  3m)
2
dm
 k (2  3m)
dt
1
dm
k
(2  3m) dt
1
 ln | 2  3m | kt  c
3
ln | 2  3m | 3kt  3c
dm
3
R
 (2  3m)  e 3kt 3c (
 R  Rm  (2  3m)  0)
dt
2
2
2
m   Ae 3kt
3
Any positive value of k
But A must be a positive value STRICTLY LESS than
1
1
e.g. k=1, A= , then m  [2  e3t ]
3
3
m
2/3
2/3-A
0
t
2
3
7
(a)
r4
2
3
1
 

r  r  1 r  2  r r  1 r  2
2
3
1


r r 1 r  2
1
  r 
 2r  3(1  r )   2 1   
  2 
47
Coefficient of r3 = 16
n
n
2
r4
3
1 




 

r  r  1 r  2  r 1  r (r  1) (r  2) 
r 1
1
(b)
1
2
(1)
2

(2)
2

(3)
3
(2)
3

(3)
3

(4)

1
(3)
1

(4)
1

(5)


2
3
1


(n  2) ( n  1)
( n)
2
3
1



(n  1)
( n)
(n  1)
2
3
1



n
(n  1)
(n  2)
3
1
3
1
 


2 (n  1) (n  1) (n  2)



3
2
1


2 (n  1) (n  2)

3  2n  4  (n  1)  3
n3

 

2  (n  1)(n  2)  2  n  1 n  2 
(shown)
n
r4
3
 r  r  1 r  2  2
r 1
n

r 1
n

r 1
n
r4

(r  2)  r  2  r  2  r 1
r4
 r  2
n
3

r 1
r4
 r  2
r4
3

r  r  1 r  2  2
3
8(a)
2
0
2
x  4 x  3 dx 
1
0
x
2
 4 x  3  dx 
1
(b)


1 x
2
 4 x  3  dx
2
 x3
  x3

2
   2 x  3x    2 x2  3x
3
0  3
1
1
 8
 1

   2  3    8  6     2  3
3
 3
 3

2
 1 
tan 1 x dx  x tan 1 x  x 
dx
2 
 1 x 
1
2x
 x tan 1 x 
dx
2 1  x2
1
 x tan 1 x  ln(1  x 2 )  c
2
1
1
dx 
sec 2 tdt
2
2
2
tan t sec t
x x 1


(c)
2

cos t

 sin t dt
2
  (sin t )1  c

x2  1
c
x
9(a)
Common ratio =
S n   S  S n 
q
p
2 S n  S
  q n 
p 1    
  p 
 p
2 
q
q
1
1
p
p
  q n 
2 1      1
  p 


n
q
2   1
 p
p n  2q n (shown)
(b)(i) First term of each row: 2, 4, 8, 14….
Difference between the first terms: 2, 4, 6…. is AP with a = 2, d
= 2.
Sum of the first (n – 1) differences
n 1
 2  2    n  2  2   n 2  n
=
2
(ii)
First term of nth row = Sum of the first (n – 1) differences + 1st
term = n 2  n  2
Total no. of terms from 1st row to (n – 1)th row = 1 + 2 + 3 …+
(n – 1)

 n  1 1 
2

 n  1 
n  n  1
2
Sum of all terms from 1st row to (n – 1)th row
n  n  1
2
2   n2  n  2  2 



2
n  n  1  n 2  n  2 

4
10(i)
(ii)
(iii)
At point A, 7 1  2 8  8  1  RHS
 1   0   14 
     
Normal vector of 2 =  4    2    4 
1  4  2 
     
Let  be the angle required.
 14   7 
   
 4    2 
 2 1
108

   
sin  

 1    90 or
2
216 54
11664
 7
 
Normal vector of 1=  2  is parallel to direction vector of l
 1
 
Since l is both perpendicular to 1 and 2, 1 and 2 must be
parallel.
Alternative:
(iv)
 7
 14 
  1 
Normal vector of 1=  2  =  4 
 1  2 2 
 
 
Since the normal vectors of 1 and 2 are parallel, 1 and 2 are
also parallel.
Observe that 2 contains origin,
Shortest distance required =
1
72   2   12
2

1
54

1
54
Alternative:
Let B   0,1, 2
 0   1   1 
     
AB   1    8    7 
 2   8   6 
     
Shortest distance required =
 1   7 
   
 7    2 
 6   1 
   
7 2   2   12
2
(v)
Let M be the foot of perpendicular of P on 1.
  3   7    7 
    
  7    2    2 
7
  8   1    1 
1 
PM  PA  nˆ nˆ  
   2 

2 
54

 54
1






OP '  PP '  OP


 7   4   3 
 1      
 2     2    1    3 
 2      
 1   0   1 
Alternative:
Let l2 be the line passing through point P and parallel to normal
vector of 1.
 4  7 
   
l2 : r   1   s  2 
0  1 
   
 4   7    7 
      
 1   s  2     2   1
 0   1    1 
1
28  49 s  2  4 s  s  1  s  
2
Let M be the foot of perpendicular of P on 1 and P’ be the
point of reflection required.
 1 
a
 2 
 
OM   2  & OP '   b 


c
 1 
 
2

0c
1
 4  a 1 1 b
 ,
 2,
 
Using midpoint theorem: 
2 2
2
2
 2
 3 
 
Thus OP '   3 
 1 
 
11(a)
(b)
dC
 2π
dr
dA
A  πr 2 
 2πr
dr
dC dC dA 1



dA dr dr r
When r  k cm,
dC dC dA 2


dt dA dt k
dC 2
 k 3
dt 3
2

1 1   1
2  xy  2 xy  π  x    (πx)(2 x)  800π
2  2   2

C  2πr 
1


2 3 xy  πx 2   πx 2  800π
8


5
6 xy  πx 2  800π
4
2
5πx  24 xy  3200π
1


V   xy  πx 2  (2 x)
8


1
 2 x 2 y  πx 3
4
2
1 3
2  3200π  5πx 
 2x 
  πx
24 x

 4
800
1
πx  πx 3
3
6
1
 πx 1600  x 2 
6

For max. V,
dV 1
 π(1600  3x 2 )  0
dx 6
 x  23.094 (reject negative ans.)
y = 3.0230
Thus the values of x and y are 23.1 and 3.02cm resply.
d 2V 1
 π(6 x)   πx  0
dx 2 6
Thus volume is maximum.
12(a)
i
i

1  e  e (e
2
i

2
i

 e ) = 2 cos
2
1  e   1  e 
i
4
 i

2
i

e 2 ,  .
4
 

 i 2  
   i  2  
  2 cos e    2 cos    e


2
 2

 

4
 2 cos
4
4
 24 cos 4
 16 cos 4

2

2

2
e
i
4
2
 2 cos
4
4

2
e
i
4
4
2
(ei2  e  i2 )
(2i sin 2 )

(b)
 32i cos 4 ( ) sin(2 ).
2
1


arg  iz    
2
4

1 


 arg  i[ z  ]   
2i 
4

i


 arg(i)  arg  z    
2
4

i
3

 L : arg  z    
2
4

L
h

 cos
1
4
|   (1) |
2
 min | z  i | h 
1
2 2
From the graph above, observe that in order for the two loci
1


arg  iz     and z  i  c to have 2 intersection points,
2
4

1
1
1
1
we have
 c  |   (1) |
c
2
2
2 2
2 2