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Further Mathematics T Paper 1 讨论专区

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Further Mathematics T Paper 1 讨论专区

帖子 由 SpeedMaths 于 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~
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回复: Further Mathematics T Paper 1 讨论专区

帖子 由 SpeedMaths 于 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
  1. Numbers and Sets
    Real numbers
    Exponents and logarithms
    Complex numbers
    Sets
  2. Polynomials
    Polynomials
    Equations and inequalities
    Partial fractions
  3. Sequences and Series
    Sequences
    Series
    Binomial expansions
  4. Matrices
    Matrices
    Inverse matrices
    System of linear equations
  5. Coordinate Geometry
    Cartesian coordinates in a plane
    Straight lines
    Curves
  6. Functions
    Functions and graphs
    Composite functions
    Inverse functions
    Limit and continuity of a function
  7. Differentiation
    Derivative of a function
    Rules for differentiation
    Derivative of a function defined implicitly or parametrically
    Applications of differentiation
  8. Integration
    Integral of a function
    Integration techniques
    Definite integrals
    Applications of integration
  9. Differential Equations
    Differential equations
    First order differential equations with separable variables
    First order homogeneous differential equations
  10. Trigonometry
    Solution of a ***
    Trigonometric formulae
    Trigonometric equations and inequalities
  11. Deductive Geometry
    Axioms
    Polygons
    Circles
  12. Vectors
    Vectors
    Applications of vectors
  13. Data Description
    Representation of data
    Measures of location
    Measures of dispersion
  14. Probability
    Techniques of counting
    Events and probabilities
    Mutually exclusive events
    Independent and conditional events
  15. Discrete Probability Distributions
    Discrete random variables
    Mathematical expectation
    The binomial distribution
    The Poisson distribution
  16. Continuous Probability Distributions
    Continuous random variable
    Probability density function
    Mathematical expectation
    The normal distribution
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回复: Further Mathematics T Paper 1 讨论专区

帖子 由 SpeedMaths 于 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
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回复: Further Mathematics T Paper 1 讨论专区

帖子 由 SpeedMaths 于 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).
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回复: Further Mathematics T Paper 1 讨论专区

帖子 由 SpeedMaths 于 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.
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回复: Further Mathematics T Paper 1 讨论专区

帖子 由 SpeedMaths 于 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


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

Solution:

f(x) = ln x 为 Fixed

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
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回复: Further Mathematics T Paper 1 讨论专区

帖子 由 数学白痴 于 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?

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回复: Further Mathematics T Paper 1 讨论专区

帖子 由 SpeedMaths 于 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.

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人生难免充满挑战,只有梦想才能给与我们能量。
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回复: Further Mathematics T Paper 1 讨论专区

帖子 由 SpeedMaths 于 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。

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人生难免充满挑战,只有梦想才能给与我们能量。
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回复: Further Mathematics T Paper 1 讨论专区

帖子 由 Log 于 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
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