7
Serangoon Garden Secondary School
Mid-Year Exam 2010
Sec 3E Add Maths
QN
1
SOLUTIONS
MARKS
4 x  6 y  34 ----------------------- (1)
2
2x  3 y  5
----------------------- (2)
e
f
a
3 y  2 x  5 -------------------- (3)
From (2),
M1
4 x 2  2(3 y )  34
Subst (3) into (1)
4 x 2  2(2 x  5)  34
4 x 2  4 x  24  0
x2  x  6  0
x  2x  3  0
s
h
x  2 or x  3
When x  2 , y  3
When x  3 ,
t
a
M
y
1
3
A is (-2, -3) and B (3,
2
 2 2

A  
 3 8
e
h
A 1 
T
1
)
3
C
M1
A1, A1
Q1: 5 marks
8  2
1


(2)(8)  (2)(3)  3 2 
1 8  2


=
22  3 2 
B1
2 y  2 x  3 ---------------- (1)
8 y  3x  10 ----------------- (2)
From (1)
2x  2 y  3
From (2)
 3x  8 y  10
 2 2  x   3 

   

  3 8  y    10 
AddMaths/3E/MYE2010
M1
M1
8
 x  1  8  2  3 
  



 y  22  3 2   10 
M1
 x  1  44 
  


 y  22   11
 x   2 
1
  


y
   2 
x  2, y  
A1, A1
x 2  x  1  q  x 2  x  (1  q )  0
3(a)
e
f
a
1
2
s
h
C
b 2  4ac > 0 since the roots are real and distinct
 12  411  q  > 0
t
a
M
1  4  4q > 0
3
q>
4
Q 2: 5 marks
M1
A1
y  2kx  10
(b)
y  x2  2
e
h
 2kx  10  x 2  2
T
x 2  2kx  8  0
M1
Since the line does not intersect the curve,
b 2  4ac < 0

2k

2
 4(1)(8) < 0
M1
2k 2  32 < 0
k 2  16 < 0
k  4k  4 < 0
- 4< k < 4
M1
A1
Q3: 6 marks
AddMaths/3E/MYE2010
9
y  2qx  3 px  8q
4.
2
b 2  4ac  0   3 p   42q 8q   0
2
M1
9 p 2  64q 2  0
9 p 2  64q 2
2
 p
64
  
9
q
M1
e
f
a
 p
64
   
9
q
M1
 p
8
   
3
q
 p
2
   2
3
q
5(a)
s
h
t
a
M
5 x 2  2 x  15  0     
2
;   3
5
2
4
2
 2 =    2(3)  6
25
5
(i)
       
(ii)
   2  ( 2   2 )  2
(iii)
Subst x   and x   into 5 x 2  2 x  15  0
2
2
2
e
h
C
 p
2
or    2 (reject)
3
q
6
4
4
 2(3)  12
25
25
A1
Q4: 4 marks
M1
A1
A1
5 2  2  15  0 -------------- (1)
5 2  2   15  0 -------------- (2)
T
(1)  
5 3  2 2  15  0 ---------------- (3)
(1)  
5 3  2  2  15  0 --------------- (4)
(3) + (4) 5 3  5 3  2 2  2  2  15  15  0
5( 3   3 )  2( 2   2 )  15(   )  0
5( 3   3 )  2(6
M1
4
2
)  15( )  0
25
5
( 3   3 )  3
AddMaths/3E/MYE2010
M1
83
125
A1
5(b)
10
1 1
1
 
  
1 1  
 
;
 

2
= 5
3
= 
M1 for both
parts
1
3
=
2
15
e
f
a
 2  1 
Equation is x 2     x  
0
 15    3 
x2 
s
h
6.
C
2
1
x   0 or 15 x 2  2 x  5  0
15
3
t
a
M
A1
If expression is
given instead of
equation, no
mark given
Q5: 8 marks
2 x 2  px  6  0
Let the two roots be  and   2
e
h
2  2 
p
----------- (1) ;  2  2  3 ----------------- (2)
2
M1 for both
eqns
Form (2),  2  2  3  0
T
  3  1  0
  3 (NA since  is positive ) or   1
Hence, the two roots are 1 and 3.
From (1)
4
M1
A1
p
2
p 8
A1
Q6: 4 marks
AddMaths/3E/MYE2010
11
7(i)
Area of ABC =
=

1
 2  11  5
2
 11  5 
M1
 11   5 
2
2
= 11 – 5
= 6 cm 2
(ii)
A1
 11  5    11  5 
= 11  2 11 5  5  11  2
= 16  2 55   16  2 55 
2
AC 2 
M1
11 5  5

= 32cm 2
8.
2m
p
pm
=
T
s
h
t
a
M
2m
p
pm
2m


p
 p  
2m p 
=
=
e
h
e
f
a
2
2
p
pm
p
pm
pm
2m
= 2


p  pm
m
p  pm
Q7: 5 marks
M1
2
p   p  m
=
C
A1


p  m
pm
M1
M1


M1
= 2 pm 2 p
= 2( p  m 
p ) (shown)
A1
Q8: 4 marks
AddMaths/3E/MYE2010
12
9(a)
8x  4 y 
1
---------------- (1)
32
9y
 1 -------------- (2)
3 x7
From (1)
2 3 x  2 2 y  2 5
M1
3x  2 y  5 ------------- (3)
From (2)
e
f
a
32 y
 30
3 x 7
2y  x  7  0
 x  2 y  7 ------------- (4)
4 x  12
(3) - (4)
x  3
Subst x  3 into (4)
t
a
M
1
5 x  2  25 4 x 
(b)
125 2 x 3
5
T
A1
3( 2 x  3)
M1
1
5 x  2 8 x 
5
59 x 2  5


1
3( 2 x  3) 2
1
6 x  9 
2
9 x  2  3x  4.5
x
C
A1
1
5 x  2  (5 2 ) 4 x 
e
h
s
h
y2
M1
M1
M1
5
or -0.2083
24
x
5
or -0.208
24
A1
Q9: 8 marks
AddMaths/3E/MYE2010
13
2 log 3 a  log 3 6 y  3
10(a)
a2
3
6y
log 3
M1
a2
 33
6y
M1
a 2  27  6 y
y
e
f
a
a2
162
A1
log 3  x  6  log 3  x  4   log 9 x 2
(b)
log 3  x  6  log 3  x  4  
log 3 x 2
log 3 9
log 3  x  6  log 3  x  4  
2 log 3 x
2
s
h
t
a
M
x  6  x  4x
x 2  5x  6  0
x  1x  6  0
x  1 (rej) or x  6
C
M1
M1
A1, A1
Q10: 7 marks
f  x   6 x 3  x 2  11x  m
e
h
11 (a)
2 x 2  x  3  2 x  3 x  1
(b)(i)
T
f  1  6(1) 3  (1) 2  11(1)  m  0
m  6
M1
A1
1
g x    x 3  4 x 2
2
1
g  2n   (2n) 3  4 (2n) 2  2  3n
2
M1
1
 (8n 3 )  4 (4n 2 )  2  3n
2
8n 3  18n 2  3n  2  0 (shown)
AddMaths/3E/MYE2010
A1
14
Let h(n)  8n  18n  3n  2
(b) (ii)
3
2
h(2)  8(2) 3  18(2) 2  3(2)  2  0
Hence, n  2  is a factor.

M1

8n 3  18n 2  3n  2  n  2 8n 2  An  1
18n 2  An 2  16n 2
A2

e
f
a

8n 3  18n 2  3n  2  n  2  8n 2  2n  1
= n  2 4n  12n  1
Hence,
n  24n  12n  1  0
1 1
n  2, ,
2 4
12.
M1
A
B
C
= 

x ( x  2)  x  2 2
4  x  Ax  2  B( x)( x  2)  C ( x)
e
h
2
s
h
t
a
M
4 x
4 x

3
2
x  4 x  4 x x  x  2 2
C
A1, A1, A1
Q11: 9 marks
M1
M1
When x  0 , 4 = A(2) 2 , A = 1
T
When x  2 , 6  2C , C  3
When x  1 , 3 = A(9) +B(1)(3)+C(1)
3 = 9 + 3B – 3
3B = -3
B = -1
4 x
1
1
3
 

2
x  4 x  4 x x  x  2   x  2 2
3
A1 for each
correct part
Q12: 5 marks
AddMaths/3E/MYE2010
15
13(a)
8  (1) 9
1
 1
3  (3) 6
2
Gradient of AB =
1
2

1
3
1
2
Gradient of BC = 
Equation of BC is y  8  
M1
2
x  3
3
3 y  24  2 x  3
e
f
a
A1
3 y  2 x  30
(b)
Equation of AB is y  1  1
1
 x  3
2
2 y  2  3( x  3)
2 y  3x  7
s
h
Let E be (0, y) and subst E (0, y) into 2 y  3 x  7
y  3.5
t
a
M
Hence, E is (0, 3.5).
C
M1
A1
Let F be (x, 0) and subst F (x, 0) into 3 y  2 x  30
x  15
Hence, F is (15, 0).
(c)
e
h
A1
Let C be ( x, y ) . F is midpoint of B (3, 8) and C.
3 x 8 y
(15,0)  
,

2 
 2
T
M1
x  27, y  8
C is (27, -8)
(d)
Area of ABC =
A1
1 3  3 27 3
2 8 1  8 8
M1
1
237    75
2
M1
=
=
1
312
2
= 156 cm 2
A1
Q13: 10 marks
AddMaths/3E/MYE2010