Comparison between HKDSEand HKCEE Syllabuses

1. Topics removed from and added to the syllabus

Section Topics removed Topics added

Number and Algebra Strand

Quadratic equations in one unknown

• Sum of roots and product of roots• Operations of complex numbers

Functions and graphs • Concepts of domains and co-domains of functions

More about graphs of functions • Enlargement and reduction

Exponential and logarithmic functions • Change of base

More about polynomials • G.C.D. and L.C.M. of polynomials

• Operations of rational functions

More about equations • Using a given quadratic graph to solve another quadratic equation

Arithmetic and geometric sequences and their summations

• Properties of arithmetic and geometric sequences

Inequalities and linear programming

• Solving quadratic inequalities in one unknown by the algebraic method

• Solving compound linear inequalities involving ‘or’

Measures, Shape and Space Strand

Locus • Describing the locus of points with algebraic equations

Equations of straight lines and circles

• Possible intersection of two straight lines

• Intersection of a straight line and a circle

Data Handling Strand

Permutation and combination • Concepts and notations of permutation and combination

II

Public Assessment of the HKDSE Mathematics Examination1.Public AssessmentThe mode of public assessment of the Hong Kong Diploma of Secondary Education (HKDSE) Mathematics Exam in the Compulsory Part is shown below.

Component Weighting DurationWeighting Duration

Public Examination Paper 1 Conventional questionsPaper 2 Multiple-choice questions

65%35%

21 4 hours11 4 hours

The mode of public assessment of the Hong Kong Diploma of Secondary Education (HKDSE) Mathematics Exam in Module 1 (Calculus and Statistics) is shown below.

Component Weighting DurationWeighting DurationPublic Examination Conventional questions 100% 21 2 hours

The mode of public assessment of the Hong Kong Diploma of Secondary Education (HKDSE) Mathematics Exam in Module 2 (Algebra and Calculus) is shown below.

Component Weighting DurationWeighting DurationPublic Examination Conventional questions 100% 21 2 hours

2.Standards-referenced ReportingThe HKDSE makes use of standards-referenced reporting, which means candidates’ levels of performance will be reported with reference to a set of standards as defined by cut scores on the variable or scale for a given subject. The following diagram represents the set of standards for a given subject:

Cut scores

U 1 2 3 4 5

Variable/scale

Within the context of the HKDSE there will be five cut scores, which will be used to distinguish five levels of performance (1–5), with 5 being the highest. The Level 5 candidates with the best performance will have their results annotated with the symbols ∗∗ and the next top group with the symbol ∗. A performance below the threshold cut score for Level 1 will be labelled as ‘Unclassified’ (U).

IV

Exam StrategiesA. General Strategies

1. Before the start of the examination

• Make sure that the time on your watch matches with that of the examination centre.

• Listen carefully to the invigilator for any errors and changes in the examination papers.

• Read carefully the instructions on the cover of the question-answer book or question book.

• Check carefully whether there are any omitted or blank pages in the examination paper according to the invigilator’s instruction.

2. During the examination

• Use proper stationery.

– Paper 1: use a pen mainly, but an HB pencil for drawing.

– Paper 2: use an HB pencil.

• Show your work clearly and neatly.Show your work clearly and neatly.Show your work clearly and neatly

• Do not get stuck on any one of the questions. Skip it and go on to another one.

3. After answering all the questions

• Do not be tempted to leave early.Do not be tempted to leave early.Do not be tempted to leave early

• Check whether there are any questions that were missed.

• Go back to questions skipped earlier.Go back to questions skipped earlier.Go back to questions skipped earlier

• Check whether there are any careless mistakes or not.

• Do not cross out anything unless you have enough time to replace it correctly.

• Make sure you write your candidate number on the answer book, supplementary answer sheets and multiple-choice answer sheet.

B. Specific Strategies

1. Paper 1 (214

hours)

• Allocate a reasonable proportion of time to each section.

Sections SuggestedSections Suggested Time Allocation ApprAllocation Approximate Time per QuestionA (1) 40 minutes 3 − 5 minutesA (2) 40 minutes 5 – 10 minutes

B 50 minutes 5 – 15 minutes

– In general, spend 5 minutes for every 4 marks.

– Allow 5 minutes for final checking.V

10

Mathematics: Conventional Questions Compulsory Part Book 1

10.If 9x2 + kxkxk + 16 = 0 has one double ppositive real root, find the value of k.(5 marks)

Suggested Solution:

D = (k)2 - 4(9)(16) 1M

= k2 - 576Since the equation has one double real root, D = 0.

k2 - 576= 0 1Mk = ±24 1A

Remember to check the given condition for the values of k found. k found. kThe root that corresponds to each value of k is calculated so as to k is calculated so as to kknow which value of k gives a positive root.k gives a positive root.k

D = (k)2 - 4(9)(16) 1M

= k2 - 576Since the equation has one double real root, D = 0.

k2 - 576= 0 1Mk = ±24 1A

When k = 24,

9x2 + 24x + 16= 0

(3x + 4)2 = 0

x = - 43

(double root)

When k = -24,

9x2 - 24x + 16= 0

(3x - 4)2 = 0

x = 43

(double root) 1M

\ k = -24 1A

Note tha t the re a re two

conditions:

1. double root

2. positive root

The candidate is able to find

the value of the unknown, but

not able to justify the answer

with respect to the given

conditions.

The candidate is able to find

the value of the unknown as

well as justify the answer by

using mathematical language.

Section A (2)Hint 1 Factorize x2 - 4.

Hint 2 Consider (x3)2 + 7(x3) - 8 = 0. We can let y = x3.

Hint 3 x x43

23 2= ( )

Hint 4 x x x=32

Hint 5 9 3x x=

Hint 6 4 2x x=

Hint 7 16x = 42x

Hint 8 Consider (5x)2 - 7(5x) + 10 = 0. We can let y = 5x.

Hint 9 9x = 32x

Hint 10 Note that log 1 = 0.

Hint 11 Combine the two given equations to get a new equation (*). The discriminant of (*) < 0.

Hint 12 Combine the two given equations to get a new equation (*). The discriminant of (*) > 0.

Section BHint 13 First combine the two given equations to get a new equation (*). The discriminant of (*) = 0.

Hint 14 Let the slope be m. Write down the equation of the line in point-slope form. Then combine the equationof the line with that of the given parabola to get a new equation (*). The discriminant of (*) = 0.

Hint 15 The four right-angled triangles are congruent. The hypotenuse = x.

Hint 16 By considering the number of calculators produced per day, we have6000

56000

40x x−

− = .

Hint 17 The length of the tunnel + the length of the train = the speed of the train × the time taken to pass throughthe tunnel completely.

Hint 18 The shaded region consists of four right-angled triangles which are congruent. The hypotenuse = 10.

Hint 19 The distance travelled by the sphere = 2 × the maximum height reached by the sphere.

Hint 20 The y-intercept of the line = 30 and the slope of the line = -12.

161

More about Equations

42

Mathematics: Conventional Questions Compulsory Part Book 1

41.Henry holds a ball just above his head and then throws it upwards. After t seconds, the ball is h m

above the ground and h = -8t2 + 8t + 32

.

(a) Find Henry’s height. Hint 13

(b) When will the ball be 2.5 m above the ground?

(c) When will the ball be 0.5 m below Henry’s head?

(d) When will the ball reach the ground?

(e) What is the maximum height that the ball can reach?

42.An object A is thrown vertically upwards. The height hA (in m) of object A after t seconds is given byhA = 40t - 5t2.

(a) Find the maximum height that object A reaches.

(b) After how many seconds will object A reach a height of 40 m? (Give the answer correct to1 decimal place.)

(c) Another object B, which is 30 m away from object A horizontally, is also thrown verticallyupwards. The height hB (in m) of object B after t seconds is given by hB = 30t - 5t2.

(i) After how many seconds will object B return to the ground?

(ii) If the two objects are thrown at the same time, express the difference in the vertical distancesof the two objects in terms of t. Hint 14

(iii)Tommy claims that the two objects will be 50 m apart 4 seconds later. Do your agree? Explain.(iii)Tommy

333 Exponential and Logarithmic Functions

Laws of Rational IndicesFor non-zero real numbers a and b, and rational b, and rational bnumbers m and n, we have:

1. a a am n m n× = + 2. a aa

aam n

m

nm n÷ = = −

3. ( )a am n mn= 4. ( )ab a bn n n=

5.ab

a

b

n n

n

= 6. a 0 1=

7. aa

nn

− = 18. a an n

1

=

9. a a amn n m m

n= =( ) if n is a positive integer

Exponential FunctionsA function in the form y = axaxa for x for x a > 0 and a ≠ 1 is called an exponential function, where a is the base and x is the exponent.x is the exponent.x

Graphs of Exponential FunctionsFor a > 1, we have:

Fig. 3.1

Exponential EquationsAn equation in the form axaxa = b, where b, where b a and bare non-zero constants and a ≠ 1, is called an exponential equation.

Exponential and

Exponential and Logarithmic Functions

Graphs of Logarithmic FunctionsFor a > 1, we have:

Fig. 3.2

Logarithmic Functions

Logarithmic FunctionsIf a number y(> 0) can be expressed as y = axaxa , where a > 0 and a ≠ 1, then x is called the x is called the xlogarithm of y to the base y to the base y a and denoted by x = loga y.

Logarithmic EquationsAny equation containing the logarithm of one or more variables is called a logarithmic equation.

Properties of Logarithmic FunctionsFor a > 0, a ≠ 1, M > 0 and N > 0, we have:1. loga a = 1

2.loga 1 = 0

3.a Ma Mlog =4.loga (MN (MN ( ) = loga M + loga N

5. log log loga a aMN

M N

= −

6.loga (M (M ( nMnM ) = n loga M

7. logloglogb

a

aM

Mb

= , where b > 0 and b ≠ 1.

(The change of base formula)

Relationship between the Graphs of Exponential and Logarithmic FunctionsThe graph of y = loga x is the image of the graph of x is the image of the graph of xy = axaxa when it is reflected about the line x when it is reflected about the line x y = x.

Applications of Logarithms1. Loudness of Sound2.Magnitude of Earthquakes3.Logarithmic TransformationLogarithmic TransformationLogarithmic T

Intersections of Two Straight Lines

Given two straight lines L1: y = m1x + c1 and L2: y = m2m2m x + c2, either one of

the following three cases may happen.

1. The two straight lines cut at one point.

The two lines have different slopes, i.e., m1 ≠ m2.

2. The two straight lines do not cut each other.

There is no point of intersection. Both lines have the same slope but

different y-intercepts, i.e., y-intercepts, i.e., y m1 = m2 and c1 ≠ c2.

3. The two straight lines cut at infinitely many points.

The two lines overlap each other. Both lines have the same slope and

y-intercepts, i.e., y-intercepts, i.e., y m1 = m2 and c1 = c2.

D

Fig. 4.14

Fig. 4.15

Fig. 4.16

4444Determine whether each of the following statements is true or false.

1. The equation of the vertical line passing through (-8, 0) is y = 0.

2. The equation of the straight line passing through (0, 7) with slope 3 is 3xough (0, 7) with slope 3 is 3xough (0, 7) with slope 3 is 3 + y + 7 = 0.

3. The slope of the line 9xThe slope of the line 9xThe slope of the line 9 - 8y 8y 8 = 0 is 0.

4. The x-intercept of the line 2x - 5y 5y 5 + 6 = 0 is -3.

5. The equation of the horizontal line passing through (-2, 5) is y = 5.

6. The lines y = 4x 4x 4 + 6 and y = 4x 4x 4 - 6 have no point of intersection.(For answers, see the bottom of the page.)

4444

Suggested Answers (Check Your Progress 4)

1. F (The equation is x = -8.) 2. F (The line is y = 3x 3x 3 + 7, i.e., 3x 7, i.e., 3x 7, i.e., 3 - y - 7 = 0.) 3. F (The slope is 98

.)

4. T (xT (xT ( -intercept = - 6

2= -3 ) 5. T 6. T (The lines have the same slope but

different y-intercepts.)74

Mathematics: Conventional Questions Compulsory Part Book 1

Polynomials in One Variable

A polynomial in one variable is a sum of terms. Each term is a product of a

variable of a non-negative integral exponent and a real number, called the

coefficient. The degree of the polynomial is the degree of the term with the

highest degree.

For example,

13x4 + 12x3 - 17x2 + 18x + 19 19

degree of polynomial = 4 variable = x

constant termcoefficients

number of terms = 5

Division of Polynomials

We can perform division of polynomials by the following ways:

1. Cancelling common factors

For example, ( )

( )

9 12 3

9 123

3 3 43

3 4

2

2

x x x

x xx

x xx

x

− ÷

= −

= −

= −

2. Long division

For example,

3x − 2 6x 3 + 2x 2 − x + 5

6x 3 − 4x 2

2x 2 + 2x + 1

2 6

6x 2 −

6x 2 − 4x + 5

3x + 5

3x − 2

7

x + 5

Divisor

Quotient

Dividend

Remainder

A

B

3x3x3 is the common factor.x is the common factor.x

Remainder should be of degree

less than or equal to that of the

divisor.

92

Mathematics: Conventional Questions Compulsory Part Book 1

This means P(a(a( ) = a2. Use this

information to solve for a.

• Unless otherwise specified, numerical answers should be either exactor correct to 3 significant figures.

• The diagrams are not necessarily drawn to scale.

Section A(1)

1. Find the remainder when x4 + 3x3 - 8x2 + x + 4 is divided by x + 1.(3 marks)

Suggested Solution Remainder

= (-1)4 + 3(-1)3 - 8(-1)2 + (-1) + 4 2M= 1 - 3 - 8 - 1 + 4= -7 1A

2. Given that f (x) = (x - 2)(2x + 5)(3x - 4). Find the remainder when f (x)is divided by (2x + 3). (3 marks)

Suggested Solution Remainder

= − −

−

+

−

−

32

2 232

5 332

4 2M

= −

−

72

2172

( )

=119

21A

3. When (x + 1)(x - 2) + 3 is divided byy x - a, the remainder is a2. Findthe value(s) of a.

(3 marks)

Suggested SolutionLet P(x) = (x + 1)(x - 2) + 3.

P(a)= a2

(a + 1)(a - 2) + 3= a2 2A

a2 - a - 2 + 3= a2

-a + 1= 0a = 1 1A

It is not necessary to expand

f (x).

97

More about Polynomials

87

Basic Properties of Circles

(c) FE2FE2FE = CE × DE 322 = (CD + DE) × DE 9= (8 + DE) × DE 9= 8DE + DE2DE2DE

DE2DE2DE + 8DE - 9= 0 (DE + 9)(DE - 1) = 0

DE=DE=DE -9(rejected) or 1 CE = CD + DE

= 8 + 1 = 9

\ Radius = =92

4 5� .

The length of DE cannot be negative.DE cannot be negative.DE

29. (a) ∠BAE = ∠DAC (common) ∠AEB = ∠ACD (ext. ∠, cyclic quad.) ∠ABE = ∠ADC (ext. ∠, cyclic quad.) \ ∆AEB\ ∆AEB\ ∆ ∼ ∆ACD∆ACD∆ (AAA) (b) Since ∆AEB∆AEB∆ ∼ ∆ACD∆ACD∆ ,

AEAC

EBCD

AE

=

=6

25�cm

(corr. sides, ~∆s)

\ AE = 2.4 cm ∠BEA = ∠DCA = 50° (proved in (a))

Area of ∆ = ∠

=

AEB AE EB AEB12122 4 2 50

( )( )sin

( . )( )sin �° cmm

cm

2

21 84= . �

30. (a) Reflex ∠BOD = 180° + y Reflex ∠BOD = 2∠BCD

180° + y = 2∠BCD

∠ = +BCD y1802°

(cor. to 3 sig. fig)

(∠ at centretwice ∠ at Bce)

∠BCD + ∠BED = 180°

∠ = − −

∠ = −

BED y

BED y

180 1802

1802

° °

°

Since AOD is a straight line, ∠AOD∠AOD∠ = 180°.

(b) In ∆ACD,∆ACD,∆∠ + ∠ + ∠ =

+ + + ∠ =

∠ =

CAD ACD CDA

x y CDA

CDA

180180

2180

180

°° °

° −− −22x y

31. In ∆OAB,

tan�

.

..

∠ =

=

∠ =

BOA ABOA

BOA

6 84 357 7°

Since ∠BOC = ∠BOA,∠BOC = 57.7° (cor. to 3 sig. fig.)

Since ∠OAB = 90° (tangent ⊥ radius),

∆OAB is a right-angled triangle.

32. Since BD is a tangent to the circle,∠CBD = ∠BAC = 35° (∠ in alt. segment)

Since BC = DC,∠CDB = ∠CBD = 35° (base∠s, isos. ∆)

In ∆BCD∆BCD∆ ,∠BCD + ∠CBD + ∠CDB = 180°(∠ sum of ∆) ∠BCD + 35° + 35°= 180°

∠BCD = 110°

(opp. ∠s, cyclicquad.)

(∠ sum of ∆)

(cor.(cor.(cor to 3 sig. fig.)(tangent properties)