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Transcript
Open-ended Question Zone
163
Introduction
Open-ended questions are questions allowing a variety of correct answers or solutions. These
questions require students to understand, exercise reasoning and apply their knowledge under
unconventional situations. For example, the question ‘write down an equation with a root 2 and
explain briefly how you obtain the answer’ is an open-ended question, which is different from
the conventional ones like ‘solve the equation 2x  4  0’.
Characteristics of open-ended questions
Open-ended questions help students recollect their knowledge from perspectives other than the
conventional ones. They also train up the higher-order thinking skills of students in
communication, problem-solving and mathematical reasoning. Consider the questions mentioned
above:
1. Solve the equation 2x  4  0.
2. Write down an equation with a root 2 and explain briefly how you obtain the answer.
To answer question 1, students can simply obtain the answer by performing mechanical
operations without knowing a more abstract concept about ‘ the relation between equations and
their roots’. But for question 2, no specified equation is provided for students as a starting point.
Therefore, students need to analyze how they can obtain an equation with a root 2, which
gradually leads students to the key concept about ‘the relation between equations and their roots’.
As students are asked to use simple and clear mathematical language to explain the concepts
involved in the thinking process, this provides students an opportunity to express abstract
concepts explicitly and thus students can understand the concepts more thoroughly.
Skills in handling open-ended questions
During the process of problem solving, open-ended questions act like mirrors, which reflect
students’ ability in communication, problem-solving and mathematical reasoning. When dealing
with an open-ended question, students should first analyze it, then determine what knowledge is
relevant and adopt appropriate strategies to solve the problem. Finally, students should write
down the solution to the question systematically by using simple language to ensure others can
understand it easily. The following are examples and exercises that help students tackle
open-ended questions better.
164
New Trend Mathematics S4A — Supplement
Numbers and Functions
E XAMPLE 1
Explain how to find two values of a so that the values of the expression a 2  14a  49 are
odd numbers.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
S UGGESTED SOLUTION
[ Analysis: There are three terms in the given expression. As the value of each term will affect the value
of the whole expression, we shall try to break down the expression and consider its value part by part. ]
Method 1:
 14a is an even number and 49 is an odd number.
 14a  49 is an odd number.
 For the value of a 2  (14a  49) to be an odd number, a 2 must be an even number, i.e.
a must be an even number.

When the value of a is 2 or 6, the value of a 2  14a  49 is an odd number.
Method 2:
a 2  14a  49  (a  7) 2
 If a 2  14a  49 is an odd number, i.e. (a  7) 2  (a  7)(a  7) is an odd number, then
a  7 is an odd number, i.e. a must be an even number.
 When the value of a is 2 or 6, the value of a 2  14a  49 is an odd number.
S OLUTION OF A STUDENT
Comments
This student can factorize the expression as the product of two identical factors a + 7 and deduce that the factor a + 7 is
an odd number. However, the student cannot further deduce that a is an even number. If the answer was verified, the
student could have found that a  7 was wrong.
Open-ended Question Zone
165
E XAMPLE 2
Frankie says, ‘For any positive number, when it is multiplied by another positive number,
the product must be greater than the former positive number. Also, when a positive number
is divided by another positive number, the quotient must be smaller than the former
positive number.’
Do you agree with what Frankie says? Explain briefly.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
S UGGESTED SOLUTION
[ Analysis: Although the example involves inequalities, Frankie’s opinion cannot be proved or disproved by
just considering the properties of inequalities. As we do not have a concrete starting point, we may
consider different positive numbers to help make reasonable judgement. ]
1
, the product
2
becomes half the former positive number, i.e. the product is smaller than the former
positive number.
When a positive number is multiplied by a proper positive fraction l ike

The first half of Frankie’s opinion is incorrect.
1
, the quotient
2
becomes twice the former positive number, i.e. the quotient is greater than the former
positive number.
When a positive number is divided by a proper positive fracti on like

The second half of Frankie’s opinion is also incorrect.
S OLUTION OF A STUDENT
Comments
This student knows that Frankie’s opinion is wrong and is able to give a correct counter example for the first half of
Frankie’s opinion. However, a wrong example has been given to support the incorrectness of the second half of Frankie’s
opinion. Here are two suggestions to the student:
1. The solution should be further elaborated to let others understand its meaning thoroughly.
7
2. Fractions and decimals are not necessarily smaller than one.
and 2.14 are examples of fractions and decimals
3
greater than one.
166
New Trend Mathematics S4A — Supplement
E XAMPLE 3
(a) Simon thinks that every function f (x) has the property that f (ab)  f (a)  f (b). Do you
agree with his opinion?
(b) Can you give a function which has the above property?
8 marks: (a) 4 marks for clear explanation. (b) 2 marks for correct answer; 2 marks for correct verification.
S UGGESTED SOLUTION
[ Analysis: The knowledge we have learned in function does not indicate explicitly whether all functions
have the above property. As we do not have a concrete starting point, we may consider different
functions to help make reasonable judgement. ]
(a) Consider the function f (x)  2x,
f (2  3)  12
f (2)  4 and f (3)  6, i.e. f (2) f (3)  4  6  24

Obviously, f (2  3)  f (2) f (3)

Not all functions have the property f (ab)  f (a)  f (b).
(b) Consider f (x)  x 2 ,
f (ab)  a 2b 2
f (a)  f (b)  (a 2 )(b 2 )
 f (ab)
i.e.
f (ab)  f (a)  f (b)
 The function f (x)  x 2 has the required property.
S OLUTION OF A STUDENT
Open-ended Question Zone
167
Comments
This student can argue that f (ab) = f (a)  f (b) is not a common property for any func tion f (x) and can give appropriate
examples to support his/her idea in both parts (a) and (b). However, two conceptual mistakes have been made in
verifying f (ab) = f (a)  f (b) for f (x) = x :
1. The student should verify the statement f (1  1) = f (1)  f (1) instead of using it to develop the verification.
2. The student has only verified that f (x) = x has the property f (ab) = f (a)  f (b) for a specific case (a  1 and b  1), but
not generally for all values of a and b.
E X E RC I S E 1
Level 1
1. Find two pairs of integral values of a and b so that
how you get the answer.
1 a 1
  holds and explain briefly
3 b 2
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
2. Explain how to find two values of a so that the values of the expression a 2  18a  81
are even numbers.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
3. Find two values of m so that
m m m m
m
are integers. Explain briefly how
, , ,
and
2 3 4 5
6
you obtain the answers.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
4. Find two pairs of positive integral values of a and b so that a  2 b holds and explain
briefly why your answers satisfy the equality.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
5. Find two rational numbers between
answers.
2 and
3 , and explain briefly how you obtain the
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
6. Find two irrational numbers between 4 and 5, and explain briefly how you obtain the
answers.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
7. Give a function y  f (x) such that f (0)  2 and f (2)  0.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
Level 2
8. (a) Find the value of 9  4a 2  12a for
(i) a  2.
(ii) a  3.
(b) Ann thinks that the value of 9  4a 2  12a is a complete square for any number a.
Is Ann’s opinion correct? Explain briefly.
5 marks: (a) 2 marks for correct answer. (b) 3 marks for clear explanation.
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New Trend Mathematics S4A — Supplement
9. Is it true that the square root of any positive number is smaller than the positive
number? Explain your answer briefly. 5 marks: 2 marks for correct answer; 3 marks for clear explanation.
10. Explain how to find two pairs of positive integral values of a and b so that
6 marks: 2 marks for correct answer; 4 marks for clear explanation.
3  a  b holds.
11. Without using a calculator, find a pair of rational numbers p and q so that
p  11  14  q holds for p  6 and q  8 . Explain briefly.
6 marks: 2 marks for correct answer; 4 marks for clear explanation.
12. Find two sets of integral values of a, b, c and d so that
how you obtain the answer.
a c ac
holds, and explain
 
b d bd
6 marks: 2 marks for correct answer; 4 marks for clear explanation.
13. The definition of a function f (x) is given as follows:
When x > 0, f (x)  x;
when x  0, f (x)  0;
when x < 0, f (x)  x.
(a) Find the values of f (x) when x  2, 1, 1 and 2.
(b) Write down two properties of f (x).
(c) Find two pairs of values of a and b so that f (a  b)  3 holds.
8 marks: (a) 2 marks for correct answer.
(b) 2 marks for correct answer.
(c) 2 marks for correct answer; 2 marks for clear explanation.
Open-ended Question Zone
169
Quadratic Equations
E XAMPLE 1
Find two values of c so that the equation x 2  x  c  0 does not have real roots. Under
what conditions does the equation have no real roots? E xplain briefly.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
S UGGESTED SOLUTION
[ Analysis: Since we know the relation between the value of discriminant and the number of roots, it is
easier for us to answer the second part of the question first, i.e. the condition under which the equation
has no real roots. Then we can use the information to help us find two values of c. ]

The equation x 2  x  c  0 < 0 does not have real solutions when  < 0.
i.e.   1 2  4(1)(c)  1  4c < 0.
1
 c
4
In other words, if we take any value of c greater than
1
, then the equation has no real
4
roots.
5
Thus when c = 1 or 3 , x 2  x  c  0 does not have real roots.
7
S OLUTION OF A STUDENT
Comments
This student knows that a quadratic equation does not have any real roots when the value of its discriminant is smaller
than zero. However, no possible values of c are provided as requested. Besides, something about graphs of equations is
mentioned, which is in fact irrelevant to this question.
170
New Trend Mathematics S4A — Supplement
E XAMPLE 2
Find two pairs of values of a and b so that the equation x 2  ax  b  0 has two negative
roots. Write down the characteristics of a and b and explain briefly.
6 marks: 2 marks for correct answer; 4 marks for clear explanation.
S UGGESTED SOLUTION
[ Analysis: The knowledge we have learned does not indicate explicitly the relation between the signs of
roots and the values of a and b. As we do not have a definite starting point, we may work backwards by
writing down some quadratic equations with negative roots and then observe the characteristics of a and
b. ]
Consider a quadratic equation with roots 1 and 2,
i.e. (x  1)(x  2)  0
x 2  3x  2  0

When a  3 and b  2 , x 2  ax  b  0 has two negative roots.
1
Similarly, consider a quadratic equation with roots 2 and  ,
3
1
i.e. ( x  2)( x  )  0
3
7
2
x2  x   0
3
3
7
2
and b  , x 2  ax  b  0 has two negative roots.
3
3
We can see that a and b are positive numbers in the above two cases.
Apparently, if the roots of the equation (x  p)(x  q)  0 are p and q, where p and q are
negative numbers, then we will have

When a 

a  p  q > 0 and b  pq > 0 as we express the equation in the form of x 2  ax  b  0.
a and b are positive numbers.
S OLUTION OF A STUDENT
Open-ended Question Zone
171
Comments
This student is able to write down two pairs of correct values of a and b and claim that the roots are negative if both a
and b are positive. However, no explanation has been pr ovided.
E X E RC I S E 2
Level 1
1. Mandy thinks that the root of the equation ax  b  0 is a rational number only if a and
b are rational numbers where a  0. Do you agree with her opinion? Explain briefly.
3 marks: 1 mark for correct answer; 2 marks for clear explanation.
2. Find two integral values of k so that 6x 2  13x  k can be factorized and then factorize
6x 2  13x  k for each value of k.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
3. Explain how to obtain two quadratic equations both with 1 as one of its roots.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
4. Find two values of c so that the equation x 2  x  c  0 has two unequal real roots.
Under what conditions does the equation have two unequal real roots? Explain briefly.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
5. Find two pairs of values of b and c so that the equation x 2  bx  c  0 has two equal
real roots. Under what conditions does the equat ion have two equal real roots? Explain
briefly.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
6. If a, b, c, f, g and h are six numbers not all identical to each other, may the quadratic
equations ax 2  bx  c  0 and fx 2  gx  h  0 have the same roots? Explain briefly.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
Level 2
7. Find two pairs of values of a and b so that the product of the roots of the equation
x 2  ax  b  0 is 1. Write down some characteristics of a and b and explain briefly.
6 marks: 2 marks for correct answer; 4 marks for clear explanation.
8. Find two pairs of values of a and b so that the sum of the roots of the equation
x 2  ax  b  0 is 0. Write down some characteristics of a and b and explain briefly.
6 marks: 2 marks for correct answer; 4 marks for clear explanation.
9. Find two pairs of values of a and b so that the equation x 2  (b  a)x  ab  0 has two
equal real roots. Under what conditions does x 2  (b  a)x  ab  0 have two equal real
roots? Explain briefly.
6 marks: 2 marks for correct answer; 4 marks for clear explanation.
172
New Trend Mathematics S4A — Supplement
10. Put 1, 2, 3, 4, 5, 6, 7, 8 and 9 into the following equations without repetition.
 x2   x    0
 x2   x    0
 x2   x    0
..................... (1)
..................... (2)
..................... (3),
so that equation (1) has two equal real roots, equation (2) has two unequal real roots
and equation (3) does not have real roots. Explain briefly.
6 marks: 3 marks for correct answer; 3 marks for clear explanation.
Open-ended Question Zone
173
Equations and Graphs
E XAMPLE 1
Find two pairs of values of a and b so that x  1, y  2 is a solution of the equation
ax  by  4. Explain how you find the relation between a and b.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
S UGGESTED SOLUTION
[ Analysis: We may consider the relation between the given equation and its solution to see whether the
relation between a and b can be obtained. ]
Given that x  1, y  2 is a solution of ax  by  4,

a(1)  b(2)  4, i.e. a  2b  4 represents the relation between a and b.
a  2b  4
a  4  2b
When b  1, a  4  2(1)  2
When b  2, a  4  2(2)  0
 a  0, b  2 and a  2, b  1 are two pairs of the required values of a and b.
S OLUTION OF A STUDENT
Comments
This student knows that if x = 1, y = 2 is a solution of the equation, then x = 1, y = 2 can be substituted into the equation
to obtain an equation of a and b. However, the student fails to understand what should be done in the last part of the
question. Instead of explaining the method of finding out the relation between a and b as requested, the student only
tries to explain ‘the relation between a and b’.
174
New Trend Mathematics S4A — Supplement
E XAMPLE 2
If one of the x-intercepts of the graph of the quadratic equation y  x 2  bx  c is 1, find
two pairs of values of b and c and explain how you find the relation between b and c.
5 marks: 2 marks for correct answer; 3 marks for clear explanation.
S UGGESTED SOLUTION
[ Analysis: Since the given information involves an equation and the x-intercepts of its graph, we may
start from the relation between an equation and the points on its graph to see whether the relation
between b and c can be obtained. ]
Method 1:

One of the x-intercepts of the graph of the equation y  x 2  bx  c is 1, i.e. the graph
passes through (1, 0).

1 is a root of the corresponding equation x 2  bx  c  0.
If the other root is also 1, then x 2  2x  1  0 (i.e. (x  1)(x  1) 0) is a quadratic
equation with double root 1, i.e. y  x 2  2x  1 is a possible equation.
Similarly, if the other root is 2, then x 2  3x  2  0 (i.e. (x  1)(x  2)  0)) is a quadratic
equation with roots 1 and 2, i.e. y  x 2  3x  2 is another possible equation.

b  2, c  1 and b  3, c  2 are two pairs of the required values of b and c.
On the other hand, it is given that the graph of y  x 2  bx  c passes through (1, 0). Thus
x  1, y  0 satisfies the equation.

0  (1) 2  b(1)  c
i.e. b + c  1 represents the relation between b and c.
Method 2:
It is given that the graph of the equation y  x 2 + bx  c passes through (1, 0). Thus x  1,
y  0 satisfies the equation.

0  (1) 2 + b(1)  c
i.e. b  c  1 represents the relation between b and c.
b  c  1
c  1  b
When b  2, c  1  (2)  1
When b  3, c  1  (3)  2

b  2, c  1 and b  3, c  2 are two pairs of required values of b and c.
Open-ended Question Zone
175
S OLUTION OF A STUDENT
Comments
This student is able to find two pairs of correct values of b and c but no explanation has been provided on how to find
the relation observed. Besides, ‘one of the root of x 2  3x  2  y is 1’ should be written as ‘one of the x-intercepts of the
graph of x 2  3x  2  y is 1’ or ‘one of the solutions of x 2  3x  2  y is x  1, y  0’.
E X E RC I S E 3
Level 1
1. Explain how to find one pair of simultaneous linear equations with no solutions.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
2. Explain how to find one pair of simultaneous linear equations with more than one
solution.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
3. Find two pairs of values of a and b so that the graph of the equation y  ax 2  bx  36
has integral x-intercepts. List some limitations of a and b.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
4. If the graph of the equation y  ax 2  2x  c cuts the x-axis at one point, find two pairs
of values of a and c. Explain how you find the relation between a and c.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
5. If the graph of the equation y  x 2  bx  c passes through P(1, 3), find two pairs of
values of b and c. Explain how you find the relation between b and c.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
176
New Trend Mathematics S4A — Supplement
6. Explain how to find two quadratic equations y  ax 2 + bx  c of which the x-intercepts
of their graphs are 2 and 3.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
7. Figure A and Figure B are sketches of the equations y  x 3  2 x 2  3 x and y  3x 2  9 x
from x  0 to x  3 respectively. Jack finds out that these two sketches both pass
through (0, 0), (2, 6) and (3, 0), and he claims that these two graphs are the same
within the range 0  x  3 . Do you agree with his opinion? Explain briefly.
y
y
y  x 3  2x 2  3x
3
O
y  3x 2  9x
x
O
(2, 6)
3
x
(2, 6)
Figure A
Figure B
4 marks: 1 mark for correct answer; 3 marks for clear explanation.
Level 2
8. Explain how to find a quadratic equation y  ax 2  bx  c such that its graph passes
through (2, 3) and does not intersect the x-axis.
5 marks: 2 marks for correct answer; 3 marks for clear explanation.
 y  x 2  bx  c
9. Find a pair of simultaneous equations 
such that the x-coordinates of
 y  mx  k
the points of intersection of their graphs are 0 and 2 . Find the relation among b, c, m
and k.
6 marks: 4 marks for correct answer; 2 marks for clear explanation.
10. The figure shows the graphs of y  x 2  4x  c and y  2x  k.
y
y  x 2  4x  c
O
x
y  2x  k
Find two pairs of values of c and k such that the graphs intersect at one point and find
the relation between c and k.
6 marks: 4 marks for correct answer; 2 marks for clear explanation.
Open-ended Question Zone
177
 y  x 2  8
 y  x 2  8
11. (a) Solve two pairs of simultaneous equations 
and 
.
 y  2 x
 y  7 x
 y  x 2  8
(b) It is given the simultaneous equations 
where k is an integer. Mandy
 y  k x
thinks that ‘if k 2  32 is a perfect square, then the solutions of the simultaneous
equations are integers’. Do you agree with her opinion? Explain briefly.
8 marks: (a) 4 marks for correct answer. (b) 2 marks for correct answer; 2 marks for clear explanation.
178
New Trend Mathematics S4A — Supplement
Graphs of Functions
E XAMPLE 1
Draw two figures such that when each of them reflects along the x-axis, the image will
overlap the original figure. What property do these two figures have?
3 marks: 2 marks for correct answer; 1 mark for clear explanation.
S UGGESTED SOLUTION
[ Analysis: We may work backwards by first analyzing the common property of the required figures, then
drawing the figures based on the property obtained. ]
y
O
x
Any figure which is symmetrical about the x-axis will have an image overlap it when it is
reflected along the x-axis. Therefore, the circle and the square drawn above are the
required figures.
S OLUTION OF A STUDENT
Comments
This student is able to draw two correct figures but the proper ty stated is not correct. A correct property is that the
y-intercepts of the figure above and below the x-axis are equal in magnitude but different in sign.
Open-ended Question Zone
179
E XAMPLE 2
It is given that the graph C 1 of the quadratic function y  x 2  bx intersects the x-axis at A
and the origin O. If C 1 is reflected along the x-axis and then translated 6 units upwards to
obtain the image C 2 , explain how to find two values of b so that the vertex of C 2 lies below
the x-axis.
6 marks: 4 marks for correct answer; 2 marks for clear explanation.
S UGGESTED SOLUTION
[ Analysis: This question involves the transformation of graphs of functions. From the knowledge we have
learned, we can find the function represented by the image after the required transformations. We can
then find the vertex of the graph of the quadratic function and hence find out for what values of b the
vertex will lie below the x-axis. ]
The function represented by the image after reflecting C 1 along the x-axis is
y  (x 2  bx)  x 2  bx;
the function represented by the image after translating the graph of y  x 2  bx 6 units
upwards is y  x 2  bx  6.
The coordinates of the vertex of the graph of y  x 2  bx  6 are
 (b) 4(1)(6)  (b) 2
b 24  b 2
).
(
,
) , i.e. ( ,
2
4
2(1)
4(1)
24  b 2
0
If the vertex lies below the x-axis, then
4
24  b 2  0
b 2  24

When b  5 or
26 , the vertex of C 2 lies below the x-axis.
S OLUTION OF A STUDENT
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New Trend Mathematics S4A — Supplement
Comments
This student knows how to obtain the range of possible values of b but the following mistakes have been made in the
solution:
1. Only one value of b answered is correct, the other one does not lie in the range required.
2. The equation of C 2 is wrongly stated as y  x 2  bx  6.
3. A wrong conclusion ‘the value of b must be greater than 24 ’ has been made from an invalid deduction ‘if b 2 > 24,
then b  24 ’. When the deduction is corrected to ‘if b 2 > 24, then b  24 or b   24 ’, the student should be able
to reach a correct conclusion like ‘the va lue of b can be greater than 24 ’.
E X E RC I S E 4
Level 1
1. Draw two figures such that when each of them reflects along the y-axis, the image
overlaps the original figure. What property do these two figures have?
3 marks: 2 marks for correct answer; 1 mark for clear explanation.
2. It is known that by applying certain transformations on the line segment AB in the
figure, an image AC would be obtained where AB  AC and ABC  90. Give one set
of such transformations and hence find the coordinates of C.
y
B(3, 5)
A(1, 2)
O
x
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
3. Give two quadratic functions such that the vertices of their graphs are the same but the
directions of their openings are different. Give also one similar ity and one distinction
between them.
4 marks: 3 marks for correct answer; 1 mark for clear explanation.
4. Give two sets of values of a, b and c so that the graphs of quadratic functions
y  x 2  ax  1 and y  x 2  bx  c have the same vertex. Find the relation among a, b
and c.
4 marks: 3 marks for correct answer; 1 mark for correct explanation.
5. Explain how to find two pairs of integral values of b and c so that the minimum value
of the quadratic function y  x 2  bx  c is 15.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
6. Give two quadratic functions such that their graphs are symmetrical along x  3, and
write down the similarities between these functions.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
Open-ended Question Zone
181
7. Write down four common features of the two quadratic functions y  x 2  4x and
y  2x 2  8x.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
Level 2
8. The graphs of two quadratic functions pass through (0, 3) an d (3, 0). Do these graphs
intersect at (0, 3) and (3, 0) only? Explain briefly.
6 marks: 2 marks for correct answer; 4 marks for clear explanation.
9. Given that a > 0, b < 0 and c > 0, sketch the graphs of the quadratic functions
y  ax 2  bx  c, y  bx 2  cx  a and y  cx 2  ax  b.
6 marks: 3 marks for correct answer; 3 marks for clear explanation.
10. Fanny thinks that the images obtained by transforming the graph of the quadratic
function y = f (x) = x 2  4x  5 according to the following two sets of transformation
never overlap each other.
Set A: Reflect the graph along the y-axis and then translate its image.
Set B: Translate the graph in the same way as set A and then reflect its image along the
y-axis.
May does not agree with Fanny’s opinion. She thinks that whether the two images will
overlap each other depends on the translation applied in both sets of transformation.
Determine whose opinions are correct and explain briefly.
8 marks: 2 marks for correct answer; 6 marks for clear explanation.
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New Trend Mathematics S4A — Supplement
Properties of Circles
E XAMPLE 1
In the figure, O is the centre and ADE is a straight line. Point out the contradictions in the
figure and eliminate these contradictions.
C
B
95
166
O
190
83
A
D
97
E
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
S UGGESTED SOLUTION
[ Analysis: Contradictions occur because the angles marked do not satisfy the theorems we have learned.
Thus we can use appropriate theorems to check which angles are not reasonable. ]
According to ‘opp. s, cyclic quad.’, ABC  ADC should be 180. However, a
contradiction occurs in the figure where ABC  ADC  95  83  178  180.
Let ABC  95 be true,
then ADC  180  ABC
 180  95
 85 instead of 83
Thus according to ‘ext. s, cyclic quad.’,
CDE  ABC  95 instead of 97;
and according to ‘ at centre  2 at ☉ ce ’,
AOC  2ADC
 2  85
 170 instead of 166
S OLUTION OF A STUDENT
Open-ended Question Zone
183
Comments
This student is able to point out the contradictions in the figure but fails to correct the given data whic h lead to the
contradictions.
E XAMPLE 2
In the figure, O is the centre and PQ is the tangent to the circle at T. Give one set of
possible values of a, b, c, d and e.
A
B
e
c
O
d
b
P
T
a
Q
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
S UGGESTED SOLUTION
[ Analysis: We can first assign a reasonable value to a and see if we can find the remaining angles by
applying appropriate theorems. ]
Let a  48,
a  b  90 (tangent  radius)
b  90  48
 42
ea
( in alt. segment)
 48
d  2e ( at centre  2 at ☉ ce )
 2  48
 96
 OT  OB (radii)
 cb
(base s, isos. )
 42
S OLUTION OF A STUDENT
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New Trend Mathematics S4A — Supplement
Comments
This student is able to apply appropriate theorems to give one set of possible values of a, b, c, d and e.
E X E RC I S E 5
Level 1
1. In the figure, CDE is a straight line parallel to AB. BCD is an isosceles triangle where
BC  BD. Give one set of possible values of a, b, c, d and e, and explain how you
obtain the answer.
B
A
a
b
e
E
d
D
c
C
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
2. The figure shows a rectangle ABCD where the length of AD is longer than twice the
length DC. If E is the mid-point of AD, is it possible for BEC to be a right angle?
Explain briefly.
E
A
B
D
C
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
3. In the figure, O is the centre. AB and AC are the tangents to the circle at B and C
respectively. Point out the contradictions in the figure and eliminate these
contradictions.
B
124
61
A
O
63
C
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
Open-ended Question Zone
185
4. In the figure, chords AC and BD intersect at E. Give two sets of possible values of a
and b. Explain how you find the relation between a and b.
B
a
A
b
E
C
D
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
5. In the figure, O is the centre. ABCD is a cyclic quadrilateral. Chord AC and diameter
BD intersect at Q. ACB  35 and CBD  50. If CB is produced to P, under what
conditions is APC equal to 70?
A
D
O
35
Q
50
P
B
C
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
6. In the figure, O is the centre and the radius of the circle is 10 cm.
O
10 cm
A
B
If we have to find the distance between chord AB and the centre, write down a piece of
information we have to know. Hence find the distance between AB and the centre.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
7. In the figure, O is the centre of circle C 3 . A, C, D and F are points on the
circumference of C 3 . AOD and COF are straight lines. The circumference of circle C 1
passes through A, C and O, and the circumference of circle C 2 passes through F, D and
O. AB and BC are tangents to C 1 , EF and ED are tangents to C 2 .
C3
A
B
C1
C2
F
E
O
C
D
If we have to find ABC and FED, write down a condition we have to know. Hence
find ABC and FED.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
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New Trend Mathematics S4A — Supplement
Level 2
8. Give three equalities (e.g. lengths, areas) which can be obtained in the figure and
explain your answers.
D
C
E
A
B
6 marks: 3 marks for correct answer; 3 marks for clear explanation.
9. Below is the solution done by Tony to the given question. He has found that the length
of OE is longer than the radius. Explain whether Tony’s solution is correct. How can
we modify the question so that the length of OE is shorter than the radius?
Question: In the figure, chords AB and CD intersect at E perpendicularly. Given that
the radius of the circle is 8, AB  10 and CD  12, find the length of OE.
C
E
A
B
O
D
Solution:
Construct OM and ON where M and N are points on AB and CD
respectively such that AB  OM and CD  ON.
C
M
A
B
E
N
O
D
 AB  OM
1
 AM  AB
2
5
( from centre bisects chord)
OA2  AM 2  OM 2
(Pyth. theorem)
OM  OA2  AM 2
 82  52
 39
Open-ended Question Zone
 CD  ON
1
 DN  CD
2
6
( from centre bisects chord)
OD 2  ON 2  DN 2
ON  OD  DN
2
187
(Pyth. theorem)
2
 82  6 2
 28
 OMEN is a rectangle.
 EN  OM  39
OE 2  ON 2  EN 2
(Pyth. theorem)
OE  ( 39 ) 2  ( 28 ) 2
 67 , i.e. OE  8
6 marks: 2 marks for correct answer; 4 marks for clear explanation.
10. In the figure, the centre O of circle C 2 lies on the circumference of circle C 1 . A lies on
the circumference of C 1 . The circumferences of C 1 and C 2 intersect at B and D. AC is
the tangent to C 2 at B and AE is the tangent to C 2 at D.
C1
C
B
A
f
a
e
d
c
O
C2
b
D
E
(a) Which of the unknowns a, b, c, d, e and f are constants? Explain briefly.
(b) Give one set of possible values of a, b, c, d, e and f.
6 marks: (a) 2 marks for correct answer; 1 mark for clear explanation;
(b) 2 marks for correct answer; 1 mark for clear explanation.
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New Trend Mathematics S4A — Supplement
More about Polynomials
E XAMPLE 1
Given that f (x) and g (x) are two different polynomials of degree two, explain how to find
f (x) and g (x) so that f (3)  g (3).
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
S UGGESTED SOLUTION
[ Analysis: From the knowledge we have learned, we know that f (3) and g (3) are the remainders of f (x)
and g (x) respectively when each of them is divided by x  3. According to these facts, we can try to
construct f (x) and g (x). ]
Let f (3)  g (3)  1.
According to the remainder theorem and the division algorithm,
f (x)  (x  3) h 1 (x)  1 and g (x)  (x  3) h 2 (x)  1
where h 1 (x) and h 2 (x) are polynomials of degree one or above.


f (x) and g (x) are different polynomials of degree two.
We can let h 1 (x)  x and h 2 (x)  x  2,
i.e. f (x)  (x  3)x  1  x 2  3x  1,
g (x)  (x  3)(x  2)  1  x 2  5x  7
S OLUTION OF A STUDENT
Comments
This student gives two correct polynomials first and describes correctly the way to find the answers at the end of the
solution. The presentation of the solution would be better if the calculation was accompanied by the explanation during
the process, instead of having it separated from the explanation.
Open-ended Question Zone
189
E XAMPLE 2
It is given that f (x)  x 3  kx 2  2kx  2k 2 , where k is an integer, has three different linear
factors. Find one possible value of k and factorize f (x) accordingly.
6 marks: 4 marks for correct answer; 2 marks for clear explanation.
S UGGESTED SOLUTION
[ Analysis: If f (x) can be factorized easily, we can simply obtain a value of k from the result; if it is
hard to factorize f (x), we may use the remainder theorem or let f (x)  (x  a)(x  b)(x  c) to see
whether we can find a value of k accordingly. ]
f (x)  x 3  kx 2  2kx  2k 2  x 2 (x  k)  2k(x  k)  (x  k)(x 2  2k)
Since f (x) has three different linear factors and x  k is one of the factors, the other two
factors should be found by factorizing x 2  2k.
Observe x 2  2k and take k  8, we have
x 2  16  (x  4)(x  4) and f (x)  (x  8)(x  4)(x  4)
S OLUTION OF A STUDENT
Comments
This student knows how to handle this question by using the factor theorem and by working backwards. However, the
value of k obtained is incorrect. This mistake could have been avoided if the factors of f (x) were verified.
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New Trend Mathematics S4A — Supplement
E X E RC I S E 6
Level 1
1. (a) When a number is divided by 4, the remainder is 3. Find two possible values of this
number.
(b) When a number is divided by 4, the remainder is 3. When this number is divided by
7, the remainder is 1. Find two possible values of this number.
4 marks: (a) 1 mark for correct answer; 1 mark for clear explanation;
(b) 1 mark for correct answer; 1 mark for clear explanation.
2. If (x  1)(x  p) 2  x 3  ax 2  bx  c, give two sets of values of a, b, c and p.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
3. Based on the identity x 3  4x 2  1  (x 2  1)(x  4)  (x  3), Mabel thinks that the
quotient and remainder of x 3  4x 2  1 when it is divided by x  4 are x 2  1 and x  3
respectively. Determine whether Mabel’s opinion is correct and explain briefly.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
4. Given that f (x) and g (x) are two different polynomials, explain how to find f (x) and
g (x) so that f (2)  g (2).
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
5. Given a polynomial of degree greater than two such that the remainder is 3 when it is
divided by x  4, find two possible polynomials and explain how to obtain them.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
6. When a polynomial f (x) is divided by another polynomial, the quotient is 3x  1 and
the remainder is x 2  x  1. Explain how to find two possible f (x).
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
7. Find two polynomials which cannot be factorized by using the factor theorem but can
be factorized by other methods. Explain briefly.
4 marks: 2 marks for correct answer; 2 marks for clear explanation.
Level 2
8. Given that f (x)  2x 3  x 2  k(x  1)  1 has three different linear factors where k is an
integer, find a possible value of k and factorize f (x) accordingly.
6 marks: 4 marks for correct answer; 2 marks for clear explanation.
9. Given that f (x)  (x  a)(x  b) Q(x)  mx  n where a  b, find a pair of values of a
and b so that when the polynomial f (x) is divided by x  a and x  b, the remainders
are the same. Hence find a possible polynomial f (x).
6 marks: 3 marks for correct answer; 3 marks for clear explanation.
10. Explain how to find a polynomial such that when it is divided by x  1 and x  1, the
remainders are 5 and 7 respectively. 6 marks: 3 marks for correct answer; 3 marks for clear explanation.