Further Mathematics T Paper 1 讨论专区

由 **SpeedMaths** 9/6/2009, 00:53

STPM Mathematics T (also known as Pure Mathematics) Syllabus
STPM Mathematics S (also known as Statistical Mathematics) Syllabus
STPM Further Mathematics Syllabus

我手中有历年的Further Mathematics T Paper 1考卷。
1979-1988 Model Answer
1991年至2000年； 2005年
得空的时候，我会Scan出来给大家做。
我会附上解答。

希望这些可以帮到考生们。

P/S:
若有人想放出 FM 的练习题或考题，欢迎欢迎~
但是，我将不去解答，因为我工作了，没时间。
希望有同志们可以帮到你们。

即将登场~

由 **SpeedMaths** 9/6/2009, 00:53

Please take note that the first eight topics (Paper 1) of Maths T and Maths S are the same. Besides, Maths T and Maths S are mutually exclusive. In other words, a STPM candidate cannot take both subjects at the same time. Maths T is taken by most science stream stuedents whereas Maths S is taken by some art stream students. Meanwhile, Further Maths is taken as the optional fifth subject by some science stream students.

STPM Mathematics T (also known as Pure Mathematics) Syllabus

Numbers and Sets
Real numbers
Exponents and logarithms
Complex numbers
Sets
Polynomials
Polynomials
Equations and inequalities
Partial fractions
Sequences and Series
Sequences
Series
Binomial expansions
Matrices
Matrices
Inverse matrices
System of linear equations
Coordinate Geometry
Cartesian coordinates in a plane
Straight lines
Curves
Functions
Functions and graphs
Composite functions
Inverse functions
Limit and continuity of a function
Differentiation
Derivative of a function
Rules for differentiation
Derivative of a function defined implicitly or parametrically
Applications of differentiation
Integration
Integral of a function
Integration techniques
Definite integrals
Applications of integration
Differential Equations
Differential equations
First order differential equations with separable variables
First order homogeneous differential equations
Trigonometry
Solution of a ***
Trigonometric formulae
Trigonometric equations and inequalities
Deductive Geometry
Axioms
Polygons
Circles
Vectors
Vectors
Applications of vectors
Data Description
Representation of data
Measures of location
Measures of dispersion
Probability
Techniques of counting
Events and probabilities
Mutually exclusive events
Independent and conditional events
Discrete Probability Distributions
Discrete random variables
Mathematical expectation
The binomial distribution
The Poisson distribution
Continuous Probability Distributions
Continuous random variable
Probability density function
Mathematical expectation
The normal distribution

由 **SpeedMaths** 9/6/2009, 00:55

STPM Mathematics S (also known as Statistical Mathematics) Syllabus

* Numbers and Sets
Real numbers
Exponents and logarithms
Complex numbers
Sets
* Polynomials
Polynomials
Equations and inequalities
Partial fractions
* Sequences and Series
Sequences
Series
Binomial expansions
* Matrices
Matrices
Inverse matrices
System of linear equations
* Coordinate Geometry
Cartesian coordinates in a plane
Straight lines
Curves
* Functions
Functions and graphs
Composite functions
Inverse functions
Limit and continuity of a function
* Differentiation
Derivative of a function
Rules for differentiation
Derivative of a function defined implicitly or parametrically
Applications of differentiation
* Integration
Integral of a function
Integration techniques
Definite integrals
Applications of integration
* Linear Programming
* Network Planning
* Data Description
* Probability
* Probability Distributions
* Sampling and Estimation
* Correlation and Regression
* Time Series and Index Number

由 **SpeedMaths** 9/6/2009, 00:56

心血来潮。。。
今天我来介绍Remainder Theorem。
很多的Math S or T 生只学到一个Remainder Theorem..
^^
我教多你们一个Theorem.
你们学过第一个，但是没学过第二个。

Remainder Theorem
1.
If a polynomial f(x) is divided by x -a, the remainder is f(a).
i.e. f(x) = (x - a)q(x) + f(a)

As a consequence of this theorem,
it follows that if f(a) = 0, then (x - a) is a factor of f(x).

2.
If a polynomial f(x) is divided by (x - a)^2,
the remainder is f'(a)(x -a) + f(a).
i.e. f(x) = q(x)(x - a)^2 + f'(a)(x - a) + f(a)

If f'(a) = f(a) = 0, then (x - a) is a repeated factor of f(x).

由 **SpeedMaths** 9/6/2009, 00:57

这里要update一些资料：

政府考生若要报考STPM Further Mathematics T, 你得是理科生。
私人重考生若要报考STPM Futher Mathematics T, 你得以前是理科生，
重考的同一时，你得多报考 Maths T, 才能允许你报考 Further Maths T.

由 **SpeedMaths** 9/6/2009, 00:58

如题：

∫ ln x dx 該怎麽solve?

你可以用Integration by part.

我额外补充一下。当你们用integration by part时，
你们得知道一些 rule,就是选出 first function当成你们的fixed function.
例如：

∫ f(x). g(x) dx = f(x) ∫ g(x) dx - ∫ [ ∫ g(x) dx ] . f '(x) dx

In words, this formula states that

The integral of the product of two functions

= first function x integral of second - integral of ( integral of second x diff. coefficient of first )

所以你们先在两个functions中锁定其中一个function为你的first function(我习惯用Fixed这个字眼).

其实，这里有个我个人的口诀。

By using integration by part,

Fixed . Integrate - ∫ Integrate . differentiate Fixed dx

接下来，我教教你们如何辨认哪个 functions 为你们的 first function.

这里就是要obey the rules 来找出 first function。

The first function is chosen as the function which comes first in the word ILATE, where

I - stands for Inverse function (Exp: f(x) = sin^(-1) x; cos^(-1) )
L - stands for Logarithmic function (Exp: f(x) = log x; ln x )
A - stands for Algebraic function (Exp: f(x) = 2x+3 )
T - stands for Trigonometric function (Exp: f(x) = sin x; cos x )
E - stands for Exponential function (Exp: f(x) = e^(x) )

通常STPM的level, 只有以下的rules:

lnx
x
e^x

你们根据以上的rules就可以找出fixed 就是 (first function)。

回刚才的问题：

∫ ln x dx

这个问题有两个functions就是

∫ ln x dx
= ∫ 1 . ln x dx

设 f(x) = ln x 和 g(x) = 1

你现在可以很清楚地辨认哪一个是 fixed 吗？
就是用刚才我所说的rules来找出fixed.

其实 ln x 是属于 Logarithmic function，
而 1 是属于 Algebraic function

根据rules:

lnx
x
e^x

你看到lnx 排在 x 的上面，所以 lnx 就是设为 Fixed.

然后根据这个

∫ f(x). g(x) dx = f(x) ∫ g(x) dx - ∫ [ ∫ g(x) dx ] . f '(x) dx

或

∫ f(x). g(x) dx = Fixed . Integrate - ∫ Integrate . differentiate Fixed dx

Solution：
设
f(x) = ln x 为 Fixed
和
g(x) = 1 为 second

∫ ln x dx
= ∫ 1 . ln x dx
= ln x . x - ∫ x . 1/ x dx {因为 Fixed . Integrate - ∫ Integrate . differentiate Fixed dx }
= x.lnx - ∫ 1 dx
= x.lnx - x + c
= x ln (x/e) + c

由数学白痴 9/13/2009, 14:53

If f'(a) = f(a) = 0, then (x - a) is a repeated factor of f(x)

可以有example看吗？

由 **SpeedMaths** 10/11/2009, 17:48

数学白痴写道:If f'(a) = f(a) = 0, then (x - a) is a repeated factor of f(x)

可以有example看吗？

a)
Proof : Suppose (x - a)^2 is a factor of f(x), then
f(x) = Q(x)(x-a)^2
=> f(x) = Q'(x).(x-a)^2 + 2(x-a).Q(x)
=> (x - a) is a factor of f'(x). (proved)

现在我们看看以下的converse is false.
Let f(x) = (1/2) (x -7)(x+3)
= (x^2)/2 - 2x - 21/2
Then f'(x) = x - 2

我们清楚地知道 (x- 2)^2 is not a factor of f(x).

Examples
i) 3x^3 + 29x^2 + 65x - 25 = 0 given that two of its roots are repeated.

Solution:
i)
Let f(x) = 3x^3 + 29x^2 + 65x - 25

Suppose f'(x) = 0

Then 9x^2 + 58x + 65 = 0
By quadratic formula, x = [-58 ± √ [58^2 -4(9)(65)]]/[2(9)]
x = -5 or -13/9

f(-5) = 0 but f(-13/9) ≠ 0

Since f'(-5) = f(-5) = 0, then (x+5) is a repeated factor of f(x).
Hence, -5 is the repeated root of f(x).

By long division f(x) = (x+5)^2 . (3x - 1)
Therefore, the solution to f(x) = 0 are x = -5, -5, 1/3.

由 **SpeedMaths** 10/20/2009, 03:48

考生面对的一个问题就是不知道什么时候该用哪一个formula来做积分。

其实很容易又简单，只要你依照以下的步骤就可以了：

当你无法使用step 1的其中一种fomula,
你就尝试使用step 2的fomula,
若step 2无法使用，就以此类推，直到step 4为止。

Step 1:
i) Integrate ax^n dx = [ax^(n+1)]/(n+1) + c
ii) Integrate f'(x) [f(x)]^n dx = { [f(x)]^(n+1) } / (n+1) + c
iii) Integrate f'(x)/ f(x) dx = ln | f(x) | + c
iv) Integrate f'(x) e^f(x) dx = e^f(x) + c

Step 2:
Integration by Partial Fraction

Step 3:
Integration by Parts

Step 4:
Integration by Substitution

注意：
当然，有些题目是可同时用两个method找出答案，
如：Integration by Parts 和 Integration by Substitution都可同时用.
但是若可以的话尽量用Integration by Parts，
而 Integration by Substitution是最后的选择，
因为避免当你用 Integration by Substitution时，
你会substitute错variable或浪费时间。

由 Log 7/9/2010, 23:02

SpeedMaths 写道:心血来潮。。。
今天我来介绍Remainder Theorem。
很多的Math S or T 生只学到一个Remainder Theorem..
^^
我教多你们一个Theorem.
你们学过第一个，但是没学过第二个。

Remainder Theorem
1.
If a polynomial f(x) is divided by x -a, the remainder is f(a).
i.e. f(x) = (x - a)q(x) + f(a)

As a consequence of this theorem,
it follows that if f(a) = 0, then (x - a) is a factor of f(x).

2.
If a polynomial f(x) is divided by (x - a)^2,
the remainder is f'(a)(x -a) + f(a).
i.e. f(x) = q(x)(x - a)^2 + f'(a)(x - a) + f(a)

If f'(a) = f(a) = 0, then (x - a) is a repeated factor of f(x).

If a polynomial f(x) is divided by (x - a)^2,
the remainder is f'(a)(x -a) + f(a).
i.e. f(x) = q(x)(x - a)^2 + f'(a)(x - a) + f(a)
可以看下它的proof吗？ study

Further Mathematics T Paper 1 讨论专区

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