Download 2016-17 F4 Math E1 P..

MUNSANG COLLEGE
2016-2017 First Term Examination
F. 4 Mathematics
Paper 1
Class : _______
Name : _____________________
Time allowed :
1 hour 30 minutes
Full mark
70
:
Class Number : _____
This question-answer book consists of 12 printed pages.
1. This paper must be answered in English with a blue / black ball pen.
2. Write your name, class and class number in the space provided on this cover.
3. This paper consists of THREE sections, A(1), A(2) and B.
Section A(1) carries 23 marks, Section A(2) carries 23 marks and Section B carries 24 marks.
4. Answer ALL questions in this paper. Write your answers in the spaces provided in this Question-Answer
Book. Do not write in the margins. Answers written in the margins will not be marked.
5. Graph paper and supplementary answer sheets will be supplied on request. Write your name, class and
class number on each sheet, and fasten them INSIDE this book.
6. Unless otherwise specified, all working must be clearly shown.
7. The diagrams in this paper are not necessarily drawn to scale.
8. Unless otherwise specified, numerical answers must be exact or correct to 3 significant figures.
9. Calculator pad printed with the “HKEA Approved” / “HKEAA Approved” label is allowed.
Remove the calculator cover / jacket.
Answers written in the margins will not be marked.
1
Answers
Answerswritten
writtenininthe
themargins
marginswill
willnot
notbebemarked.
marked.
Answers written in the margins will not be marked.
Instructions to candidates:
Section A(1) (23 marks)
1. Simplify (2 xy 2 )3  (3x2 y3 )2 and express your answer with positive indices.
(3 marks)
( 2 xy 2 )3  (3 x 2 y 3 ) 2
( 2 xy 2 )3
(3 x 2 y 3 ) 2
1M : -ve index, 1A: x4-3 or y6-6, 1A: answer
8 x 3 y 6
9 x4 y6
8

9x
Answers written in the margins will not be marked.

2. Make x the subject of the formula z 
x
 xy 4
3
x  3 xy 4
z
3
x(1  3 y 4 )
z
3
3 z  x(1  3 y 4 )
x
 xy 4 .
3
(3 marks)
z
1M: common denominator + 1A: Factorization
3z
x
1  3y4
x
3z
1  3y4
1A
3. Mr Chan buys a car for $40 000. He sells it to Mr Cheng at a loss of 30%. Mr Cheng then sells it to
Mr Lee at a loss of $11 200. How much does Mr Lee pay?
(3 marks)
Amount Mr Lee pays  $[40 000  (1  30%)  11 200]
 $(40 000  0.7  11 200)
 $(28 000  11 200)
 $16 800
Answers written in the margins will not be marked.
2
1M: 40 000  (1  30%)  1A
1A
Answers
Answerswritten
writtenininthe
themargins
marginswill
willnot
notbebemarked.
marked.

5
4. In the figure, the graph of y  x 2  x  6 cuts the x-axis at two points A and B.
2
Find the area of △ABC.
(4 marks)
5
x60
2
2 x 2  5 x  12  0
x2 
1M
(2 x  3)( x  4)  0
1 
2 
3
2
or
4
1A
 3 
 2 
Area of △ABC =  4      [0  (7)] sq. units
Answers written in the margins will not be marked.
=
77
sq. units
4
1A
1A
5. It is given that f ( x)  2 x  7 and g ( x)  x 2  3x  k . If g (5)  f (2) , solve the equation
g ( x)  f ( x) .
∵
(3 marks)
g(5) = f(2)
52  3(5)  k  2(2)  7
25  15  k  3
k  7
1M
g(x) = f(x)
x  3x  7  2 x  7
2
1M
x  5 x  14  0
2
( x  7)( x  2)  0
x  7 or x   2
Answers written in the margins will not be marked.
1A
3
Answers
Answerswritten
writtenininthe
themargins
marginswill
willnot
notbebemarked.
marked.
x
6. Find the equation of the straight line L2 which is parallel to L1: 4 x  y  5  0 and passes through
A(1, 2).
(4 marks)
∵ L2 is parallel to L1.
∴
Slope of L2  slope of L1

1M
4
(1)
4
1A
L2:
y  2  4[ x  (1)]
1M
y  2  4( x  1)
y  2  4x  4
1A
or y = 4x + 6
7. When ( x  3)( x  2)  2 is divided by x  k , the remainder is k 2 . Find the value of k.
(3 marks)
(k  3)(k  2)  2  k 2
1M+1A
k2  k  6  2  k2
k4
1A
Answers written in the margins will not be marked.
4
Answers
Answerswritten
writtenininthe
themargins
marginswill
willnot
notbebemarked.
marked.
Answers written in the margins will not be marked.
4x  y  6  0
Section A(2) (23 marks)
8. (a) Factorize 49  p 2  7q  pq .
(3 marks)
(b) Hence, factorize 49  (a  b)  14a  2a(a  b) .
2
(3 marks)
(a) 49  p 2  7q  pq
 (7  p)(7  p)  q(7  p)
1M (any one of the factorization)+ 1A
 (7  p)(7  p  q)
1A
 49  (a  b)2  7(a)  (a  b)(a)
1M: follow the pattern
 (7  (a  b))(7  (a  b)  a)
1M: using result of (a)
 (7  a  b)(7  b  2a)
1A
9. It is given that  and  are the roots of the quadratic equation 2 x2  k  2(2  k ) x , where k is a
real number.
(a) Express    and  in terms of k.
(3 marks)
(b) It is given that (   )  76 . Find the value(s) of k.
(3 marks)
2
2 x 2  k  2(2  k ) x
(a)
2 x 2  k  (4  2k ) x
2 x 2  (2k  4) x  k  0
1A
2k  4
2
 2k
  
 
(b)
(   ) 2  76
(   ) 2  4  76
1M
k
(2  k ) 2  4    76
2
4  4k  k 2  2k  76
k 2  6k  72  0
1A
(k  6)(k  12)  0
k   6 or k  12
1A: both
Answers written in the margins will not be marked.
5
k
2
1A+1A
Answers
Answerswritten
writtenininthe
themargins
marginswill
willnot
notbebemarked.
marked.
Answers written in the margins will not be marked.
(b) 49  (a  b)2  7a  a(a  b)
10. The figure shows the graph of y  x 2  6 x  c . Its axis of
symmetry is x = 3, and the minimum value of y is 5.
(a) Find the value of c.
(b) If the graphs of y  2( x  1)2  k and y  x 2  6 x  c
have the same y-intercept, find
(i)
the value of k,
(ii) the optimum value of the function y  2( x  1)2  k .
(5 marks)
(a)
∵
Coordinates of the vertex = (3, 5)
∴
By substituting x = 3 and y = 5 into
(b)
(i)
∵
∴
 5  (3) 2  6(3)  c
1M
 5  9 18  c
c4
1A
The two graphs have the same y-intercept.
By substituting x = 0 and y = 4 into
y  2( x  1) 2  k , we have
4  2(0  1) 2  k
4  2  k
k 6
(ii) ∵
∴
1M
1A
Coefficient of x2 = 2 < 0
The maximum value of y is 6.
Answers written in the margins will not be marked.
1A
6
Answers
Answerswritten
writtenininthe
themargins
marginswill
willnot
notbebemarked.
marked.
Answers written in the margins will not be marked.
y  x 2  6 x  c , we have
2 x  y  32 x 2

11. Solve 
.
2 x 3 y
 90

3
(6 marks)
2 x  y  32 x  2  (1)
 2 x 3 y
3
 9 0  (2)
From (1),
2 x  y  25( x 2)
1M
x  y  5 x  10
∴
4 x  y  10
(3)
1A
From (2),
32 x 3 y  90
32 x 3 y  30
2x  3y  0
(4)
1A
(3)  (4)  2 :
(4 x  y )  2(2 x  3 y )  10
5 y  10
y2
1A
By substituting y  2 into (4), we have
2 x  3(2)  0
x3
Answers written in the margins will not be marked.
1A
7
Answers
Answerswritten
writtenininthe
themargins
marginswill
willnot
notbebemarked.
marked.
Answers written in the margins will not be marked.
∴
1A
Section B (24 marks)
12. (a) Express
1
in the form of a + bi, where a and b are real numbers.
2  3i
(2 marks)
1
is a root of the quadratic equation mx 2  nx  1  0 , where m and n are
2  3i
(b) It is given that
real numbers.
(i)
Find the values of m and n.
(4 marks)
(ii) Find the range of values of p such that the graph of y  mx  nx  1  p has two
2
x-intercepts.
(b)
1
1
2  3i


2  3i 2  3i 2  3i
2  3i
 2
2  (3i ) 2
2 3
  i
13 13
(i)
1M
1A
∵
1
2 3
  i is a root of the quadratic equation mx 2  nx  1  0 .
2  3i 13 13
∴
2 3
 i is another root.
13 13
Sum of roots =
4
13
Product of roots =
x2 
1A
4
9
13
1



169 169 169 13
1A
4
1
x 0
13
13
13x 2  4 x  1  0
1M
1A
∴ m  13 and n  4
(ii) ∵
The graph of y  mx2  nx  1  p has two x-intercepts.
0
∴
(4)  4(13)(1  p)  0
2
16  52  52 p  0
1M+1A
52 p  36
p
∴
9
13
The range of values of p is p  
Answers written in the margins will not be marked.
8
9
.
13
1A
Answers
Answerswritten
writtenininthe
themargins
marginswill
willnot
notbebemarked.
marked.
Answers written in the margins will not be marked.
(a)
(3 marks)
13. In the figure, L1 and L2 are straight lines. L1 intersects the x-axis
and the y-axis at A and B(0, 2) respectively. L2 is perpendicular
to L1 and intersects the x-axis at the same point A as L1. Suppose
the slope of L1 is
2
.
5
(a) Find the equations of L1 and L2.
(4 marks)
(b) Another straight line L3: 8x + 9y + k = 0 intersects L1 and L2
at B and C respectively,
(i) Find the value of k.
(1 mark)
(ii) Find the coordinates of C.
(2 marks)
(iii) D is a point on BC such that the area of △ABD : the area of △ACD = 2 : 3. Find the
(a)
y
L1:
(3 marks)
2
x2
5
1A
or 2x – 5y – 10 = 0
Sub. y = 0 into equation of L1,
0
2
x2
5
x=5
∴
The coordinates of A are (5, 0).
1A
Product of slopes = 1
Slope of L2 = 
L2 :
5
2
5
y  0   ( x  5)
2
1M
5x + 2y  25 = 0
(b)
(i)
1A
Sub. (0, –2) into 8 x  9 y  k  0 ,
8(0)  9(2)  k  0
k =18
Answers written in the margins will not be marked.
1A
9
5
25
or y   x 
2
2
Answers
Answerswritten
writtenininthe
themargins
marginswill
willnot
notbebemarked.
marked.
Answers written in the margins will not be marked.
coordinates of D.
(ii)
5 x  2 y  25  0 ......(1)

8 x  9 y  18  0 ......(2)
From (1),
5 x  2 y  25  0
y
25  5 x
......(3)
2
Sub. (3) into (2),
8x 
9(25  5 x)
 18  0
2
1M (can be absorbed)
261  29x = 0
x=9
Sub. x = 9 into (3), we have
∴
(iii) ∵
∴
25  5(9)
 10
2
The coordinates of C are (9, –10).
1A
△ABD and △ADC has the same height.
BD : DC
= area of △ABD : area of △ACD
=2:3
1A
Coordinates of D
 0  3  9  2 2  3  (10)  2 

,

23
 23

26 
 18
 ,  
5 
5
Answers written in the margins will not be marked.
1M
1A
10
Answers
Answerswritten
writtenininthe
themargins
marginswill
willnot
notbebemarked.
marked.
Answers written in the margins will not be marked.
y
14. Let f ( x)  x3  2 x 2  5x  6 .
(a) Factorize f ( x) .
(b) Simplify
(2 marks)
x2  1
4  2x
.
 2
3
2
x  2 x  5x  6 x  2 x  3
(3 marks)
(a) f ( x)  x3  2 x2  5x  6
f (2)  23  2(2)2  5(2)  6  0
∴ ( x  2) is a factor of f ( x) .
1M
f (1)  (1)3  2(1)2  5(1)  6  0
∴ f ( x)  ( x  1)( x  2)( x  3)
1A
(b)
x2  1
4  2x
 2
3
2
x  2x  5x  6 x  2x  3
( x  1)( x  1)
2( x  2)


( x  1)( x  2)( x  3) ( x  3)( x  1)
2

( x  3)( x  3)

1M+1A
2
( x  3) 2
1A
End of Paper
Answers written in the margins will not be marked.
11
Answers
Answerswritten
writtenininthe
themargins
marginswill
willnot
notbebemarked.
marked.
Answers written in the margins will not be marked.
∴ ( x  1) is a factor of f ( x) .