STPM MATHEMATICS T SYLLABUS 

Chapter  1 Functions

1.1 Functions

(a) state the domain and range of a function, and find composite functions;

The domain of a function  is all the possible input values, and the range is all possible output values.Domain: set of all possible input values (usually x)Range: set of all possible output values (usually y)

a) State the domain and range of the following relation{(3, –2), (5, 4), (1, –1), (2, 6)}

The domain is all the x-values, and the range is all the y-values       Domain: {3, 5, 1, 2}     Range:{–2, 4, -1, 6}

a) State the domain and range of the following relation{(3, 5), (5, 5), (1, 5), (2, 5)}       Domain: {3, 5, 1, 2}       Range: {5}

What is a Function? A function, f(x) relates an input to an output. Each input is related to exactly one output.This is a function. There is only one y for each x

This is a function. There is only one arrow coming from each x; there is only one y for each x

This is not a function. 5 is in the domain, but it has no range element that corresponds to it

This is not a function. 3 is associated with two different range elements

Functions: Domain and Range

The function composition of two functions takes the output of one function as the input

of a second one

Aninverse function for f, denoted by f-1, is a function in the opposite direction.- See more at: http://coolmathsolutions.blogspot.com/2013/02/what-is-function.html#sthash.t3Rri8gK.dpuf

(b) determine whether a function is one-to-one, and find the inverse of a one-to-one function;

A one-to-one function is a function in which every element in the range of the function corresponds with one and only one element in the domain.

A function, f(x), has an inverse function if f(x) is one-to-one.

The Horizontal Line Test: If you can draw a horizontal line so that it hits the graph in more than one spot, then it is NOT one-to-one.

a) Is below function one-to-one?

f(x)=x3

f(x)=x3 is one-to-one function and has inverse function. (The horizontal line cuts the graph of function f at 1 point, therefore f is  a one-to-one function)

b)  Is below function one-to-one? 

f(x)=x2

f(x)=x2 is NOT one-to-one function and does NOT has an inverse function. (The horizontal line cuts the graph of function f at 2 points, therefore f is NOT a one-to-one function)

If we restricted x greater than or equal to 0. The horizontal line cuts the graph of function f once and f(x)=x2 is one-to-one function and has an inverse function.

(c) sketch the graphs of simple functions, including piecewise-defined functions;

How to find the inverse of one-to-one function below?

f(x)=3x−4Draw the graph of f(x)=3x-4

The horizontal line cuts the graph of function f once, therefore f is  a one-to-one function and it has has inverse function.

Find the inverse function

f(y)=x3y−4=xy=x+43

f−1(x)=x+43

The graph of an inverse relation is the reflection of the original graph over the identity line, y = x.

Piecewise-defined function is a function which is defined by multiple sub-functions, each sub-function applying to a certain interval of the main function's domain.

Example 1

f(x)={2x−1,x<1x2−3,x≤1

Example 2

f(x)=⎧⎩⎨⎪⎪3x−2, x<−3 x2,  −3≤x≤52x+5, x>51.2 Polynomial and rational functions

A polynomial function is a function that can be written in the form

f(x)=anxn+an−1xn−1+a1x+a0

where an, an-1,.... a0  are real numbers and n is a nonnegative integer.

Some example of polynomials:

f(x)=4x+1f(x)=2x3−4x2+5x+11f(x)=x7−6x4+3x2

Rational function is division of two polynomial functions.

f(x)=P(x)Q(x) where P and Q are polynomial functions in x.

y=2x+5x−1f(x) is not defined at x=1.

We can't have x = 1, and therefore f(x) has a vertical asymptote (a line that a curve approaches as it heads towards infinity) at

x=1

Horizontal asymptote will be the result of dividing the leading coefficients.

y=21=2

Graph of Rational Functions of f(x).

(d) use the factor theorem and the remainder theorem;

The Factor TheoremIf the remainder of a  polynomial, f(x), when divide by (x-a) is zero, then (x-a) is a factor.Show that (x+3) is a factor of

x3+5x2+5x−3f(−3)=(−3)3+5(−3)2+5(−3)−3

f(−3)=0

By factor theorem (x+3) is  factor of x3+5x2+5x-3Remainder TheoremIf a polynomial f(x) is divided by (x − r) and a remainder R is obtained, then f(r) = R.

f(x)=(x−r)⋅q(x)+R

Example: f(x)=x4+3x3-2x2+x-7 divide by (x-1)

After dividing by x-1, there is a remainder of -4. We can write:

f(x)=(x−1)(x3+4x2+2x+3)−4

(e) solve polynomial and rational equations and inequalities;

Solving Rational Inequalities

A rational function is a quotient of two polynomials.

x2+3x+2x2−16≥0x2+3x+2x2−16=(x+2)(x+1)(x−4)(x+4)

as an example for an inequality involving a rational function.

This polynomial fraction will be zero wherever numerator is zero 

(x+2)(x+1)=0x=−2,x=−1

The fraction will be undefined wherever the denominator is zero

(x−4)(x+4)=0x=4,x=−4

Consequently, the set 

(−∞,−4),[−2,−1],(4,+∞)

Solution in "inequality" notation 

x<−4,−2≤x≤−1,x>4

Find the interval with f(x) > 0

Rational equations and inequalities 02

Solving Rational Inequalities

A rational function is a quotient of two polynomials.

x2+3x+2x2−16≥0x2+3x+2x2−16=(x+2)(x+1)(x−4)(x+4)

as an example for an inequality involving a rational function.

This polynomial fraction will be zero wherever numerator is zero 

(x+2)(x+1)=0x=−2,x=−1

The fraction will be undefined wherever the denominator is zero

(x−4)(x+4)=0x=4,x=−4

Consequently, the set 

(−∞,−4),[−2,−1],(4,+∞)

Solution in "inequality" notation 

x<−4,−2≤x≤−1,x>4

Find the interval with f(x) > 0

(f) solve equations and inequalities involving modulus signs in simple cases;

Solve the inequality |x − 2| ≤ |x + 1|

Square both sides of each equation to omit the modulus signs. Rearrange the inequality into an equivalent form.

So the solution set is

 

(g) decompose a rational expression into partial fractions in cases where the denominator has two distinct linear factors, or a linear factor and a prime quadratic factor;

A rational function P(x)/Q(x) can be rewritten using what is known as partial fraction decomposition.

A1ax+b+A1(ax+b)2+...+Am(ax+b)m

  Write the partial fraction decomposition of 

11x+21(2x−3)(x+6)=A2x−3+Bx+6

Distribute A and B

11x+21=(x+6)A+(2x−3)BIf x=-6

11(−6)+21=[2(−6)−3]BB=3

If x=0,21=6A−3B

A=5

The partial fraction 

11x+21(2x−3)(x+6)=52x−3+3x+6

This is the Exponential Function: f(x)=ax

 a is any value greater than 0

For 0 < a < 1:

As x increases, f(x) heads to 0 As x decreases, f(x) heads to infinity It is a Strictly Decreasing function  It has a Horizontal Asymptote along the x-axis (y=0).

For a > 1:

As x increases, f(x) heads to infinity As x decreases, f(x) heads to 0 it is a Strictly Increasing function  It has a Horizontal Asymptote along the x-axis (y=0).

The Natural Exponential Function is the function f(x)=ex

where e is the number (approximately 2.718281828)

1.3 Exponential and logarithmic functions

(h) relate exponential and logarithmic functions, algebraically and graphically;

This is the Exponential Function: f(x)=ax

 a is any value greater than 0

For 0 < a < 1:

As x increases, f(x) heads to 0 As x decreases, f(x) heads to infinity It is a Strictly Decreasing function  It has a Horizontal Asymptote along the x-axis (y=0).

For a > 1:

As x increases, f(x) heads to infinity As x decreases, f(x) heads to 0 it is a Strictly Increasing function  It has a Horizontal Asymptote along the x-axis (y=0).

The Natural Exponential Function is the function f(x)=ex

where e is the number (approximately 2.718281828)

The logarithmic function is the function y = logax, where a is any number such that a > 0, a ≠ 1, and x > 0. 

y = logax  is equivalent to x = ay

The inverse of an exponential function is a logarithmic function.

Since y = log2x  is a one-to-one function, we know that its inverse will also be a function. When we graph the inverse of the logarithmic function, we notice that we obtain the exponential function, y=2x

Comparison of Exponential and Logarithmic Functions

(i) use the properties of exponents and logarithms;

1. Indicesan,a≠0a0=1,a≠0a−n=1/an

2. Law of Indicesam×an=am+n

am÷ an=am−n

(am)n=am×n

am×bm=(a×b)mam÷bm=(ab)m

3. LogarithmsIfa=bc,then logba=cloga1=0logaa=1alogab=1

4. Law of Logarithmslogaxy= logax+logayloga(xy)= logax−logaylogaxn= nlogax

5 Change of base of Logarithmslogab=1logbalogab=lognblogna

(j) solveequations and inequalities involving exponential or logarithmic expressions;

Solving Rational Inequalities

A rational function is a quotient of two polynomials.

x2+3x+2x2−16≥0x2+3x+2x2−16=(x+2)(x+1)(x−4)(x+4)

as an example for an inequality involving a rational function.

This polynomial fraction will be zero wherever numerator is zero 

(x+2)(x+1)=0x=−2,x=−1

The fraction will be undefined wherever the denominator is zero

(x−4)(x+4)=0x=4,x=−4

Consequently, the set 

(−∞,−4),[−2,−1],(4,+∞)

Solution in "inequality" notation 

x<−4,−2≤x≤−1,x>4

Find the interval with f(x) > 0

1.4 Trigonometric functions

(k) relate the periodicity and symmetries of the sine, cosine and tangent functions to their graphs, and identify the inverse sine, inverse cosine and inverse tangent functions and their graphs;

(l) use basic trigonometric identities and the formulae for sin (A ± B), cos (A ± B) and tan (A ± B), including sin 2A, cos2A and tan 2A; 

(m) express a sin θ + b cos θ in the forms r sin ( θ   ± α) and r cos ( θ   ± α) ; (n) find the solutions, within specified intervals, of trigonometric equations and

inequalities.

Trigonometric Identities

sin(A+B)=sinAcosB+cosAsinBsin(A−B)=sinAcosB−cosAsinBcos(A+B)=cosAcosB−sinAsinBcos(A−B)=cosAcosB+sinAsinB

tan(A+B)=tanA+tanB1−tanAtanBtan(A−B)=tanA−tanB1+tanAtanB

sin(2A)=2sinAcosA

cos(2A)=cos2(A)−sin2(A)=2cos2(A)−1=1−2sin2(A)

tan2A=2tanA1−tan2A

Express a sin θ ± b cos θ in the form R sin (θ ± α)

Express a sin θ ± b cos θ in the form R sin(θ ± α), where a, b, R and α are positive constants.

Let asinθ+bcosθ≡Rsin(θ+α)

Expand R sin(θ ± α) as followRsin(θ+α)≡R(sinθcosα+cosθsinα)Rsin(θ+α)≡Rsinθcosα +Rcosθsinα

Apply Trigonometric Identities as belowsin(A+B)=sinAcosB+cosAsinB

Soasinθ+bcosθ≡ Rcosαsinθ+Rsinαcosθ

Equating the coefficients of sin θ and cos θ in this identity, we have:For sin θ:    a = R cos α ----(1)For cos θ:    b = R sin α ----(2)

ba=RsinαRcosα=tanα

So α=tan−1ba

(α is a positive acute angle and a and b are positive)a2+b2=R2cos2α+R2sin2α

=R2(cos2α+sin2α)=R2

R=a2+b2−−−−−−√Then we have expressed a sin θ + b cos θ in the form required

asinθ+bcosθ≡Rsin(θ+α)

Similar to minus case,asinθ−bcosθ≡Rsin(θ−α)

Express Rsin(θ−α)≡Rcosαsinθ−Rsinαcosθ

ExampleExpress 4 sin θ +3 cos θ  in the form R sin(θ + α)

R=42+32−−−−−−√=25−−√=5 

α=tan−1 34=36.87∘So

4sinθ+3cosα=5sin(θ+36.87∘)Summary of the expressions and conditions

Express a sin θ ± b cos θ in the form R cos (θ ± α)

Express a sin θ ± b cos θ in the form R cos(θ ± α), where a, b, R and α are positive constants.

asinθ+bcosθ≡Rcos(θ−α)Expanding R cos (θ − α), Trigonometric Identities 

Rcos(θ−α)=Rcosθcosα+Rsinθsinαasinθ+bcosθ≡Rcosαcosθ+Rsinαsinθ

Equating the coefficients of sin θ and cos θ in this identity, we have:For sin θ:    a = R sin α ----(1)For cos θ:    b = R cos α ----(2)

ab=RsinαRcosα=tanα

So α=tan−1ab

R=a2+b2−−−−−−√

Then we have expressed a sin θ + b cos θ in the form required asinθ+bcosθ≡Rcos(θ−α)

Summary of the expressions and conditions

Chapter 2

Sequences and Series

2.1 Sequences

(a) use an explicit formula and a recursive formula for a sequence; (b) find the limit of a convergent sequence;

Convergent Sequence

A sequence is said to be convergent if it approaches some limit. Formally, a sequence converges to the limit  

limn→∞Sn=S

Example

The sequence 2.1, 2.01, 2.001, 2.0001, . . . has limit 2, so the sequence converges to 2. The sequence 1, 2, 3, 4, 5, 6, . . . has a limit of infinity (∞). This is not a real number, so

the sequence does not converge. It is a divergent sequence.

Convergent sequences have a finite limit.

1,12, 13,14,15,16....,1nLimit=0

1,12, 23,34,45,56....,nn+1Limit=1

2.2 Series

(c) use the formulae for the nth term and for the sum of  the first n terms of an arithmetic series and of a geometric series;

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

The nth term of the sequence (an) is given by

an=a1+(n−1)d

The sum of the sequence of the first n terms is then given by

Sn=n2[2a+(n−1)d]or

Sn=n2(a1+a2)

Geometric Series is a series with a constant ratio between successive terms.

an=arn−1The sum of the first n terms of a geometric series is:

a+ar+ar2+ar3+.....+arn−1=∑k=0n−1ark=a1−rn1−rDerive the sum of the first n terms of a geometric series as follows:Let  

s=a+ar+ar2+ar3+.....+arn−1rs=ar+ar2+ar3+.....+arn

s−rs=a−arns(1−r)=a(1−rn)

Sos=a1−rn1−r

As n approaches infinity, the absolute value of r must be less than one for the series to converge.

a+ar+ar2+ar3+ar4.....=∑k=0∞ark=a1−rWhen a=1

1+r+r2+r3+....=11−rLet  

s=1+r+r2+r3+.....rs=r+r2+r3+.....

s−rs=1s(1−r)=1

Sos=11−r

Proof of convergence

1+r+r2+r3+....=limn→∞(1+r+r2+r3+....+rn)

Since (1 + r + r2 + ... + rn)(1−r) = 1−rn+1 and rn+1 → 0 for | r | < 1 

1+r+r2+r3+....=limn→∞1−rn+11−r

(d) identify the condition for the convergence of a geometric series, and use the formula for the sum of a convergent geometric series;

(e) use the method of differences to find the nth partial sum of a series, and deduce the sum of the series in the case when it is convergent;

2.3 Binomial expansions

(f) expand (a+b) n  , where n∈ZBinomial theorem describes the algebraic expansion of powers of a binomial

(x+y)0=1(x+y)2=x2+2xy+y2(x+y)3=x3+3x3y+3xy2+y3(x+y)4=x4+4x3y+6x2y2+4xy3+y4

It is possible to expand any power of x + y into a sum of the form

(x+y)n=(n0)xny0+(n1)xn−1y1+(n2)xn−2y2+...+(nn−1)x1yn−1+(nn)x0yn

 It can be written as

(x+y)n=∑k=0n(nk)xn−kyk

Binomial formula : involves only a single variable 

(1+x)n=(n0)x0+(n1)x1+(n2)x2+...+(nn−1)xn−1+(nn)xn

(1+x)n=∑k=0n(nk)xk

According to the ratio test for series convergence a series converges when:

if  limk→∞(ak+1ak)<1,  series  convergesif  limk→∞(ak+1ak)>1,  series  divergesiflimk→∞(ak+1ak)=1,  result  is indeterminate

The Binomial Theorem converges when |x|<1

(g) expand   (1+x) n  , where n∈Q, and identify the condition | x | < 1 for the validity of this expansion;

(h) use binomial expansions in approximations.

Chapter 3

Matrices

3.1 Matrices

(a) identify null, identity, diagonal, triangular and symmetric matrices;

Inverse of Diagonal Matrix

Diagonal matrix is a matrix in which the entries outside the main diagonal are all zero.

A=⎡⎣ a11000  a22 0 00 a33⎤⎦A−1=⎡⎣⎢⎢⎢⎢⎢⎢⎢1a11000 1a22 0 00 1a33⎤⎦⎥⎥⎥⎥⎥⎥⎥

Example

A=⎡⎣ 5000 3 0 00 1⎤⎦A−1=⎡⎣⎢⎢⎢⎢15000 13 0 001⎤⎦⎥⎥⎥⎥

(b) use the conditions for the equality of two matrices;

(c) perform scalar multiplication, addition, subtraction and multiplication of matrices with at  most three rows and three columns;

What is Triangular Matrix

A matrix is a triangular matrix if all the elements either above or below the diagonal are zero. A matrix that is both upper and lower triangular is a diagonal matrix.

Properties of Triangular Matrix

The determinant of a triangular matrix is the product of the diagonal elements. The inverse of a triangular matrix is a triangular matrix. The product of two (upper or lower) triangular matrices is a triangular matrix.

Upper Triangular Matrix 

A=∣∣∣∣313025001∣∣∣∣=3.2.1=6

A−1=∣∣∣∣1/3−1/6−1/601/2−5/25001∣∣∣∣Lower Triangular Matrix

B=∣∣∣∣2003−10461∣∣∣∣=2.−1.1=−2B−1=∣∣∣∣1/2003/2−10−1161∣∣∣∣

(d) use the properties of matrix operations;

Properties of Matrices

AB≠BAA+B=B+A

A(B+C)=AB+AC(A+B)C=AC+BCA(BC)=(AB)CAI=IA=AA0=I

Scalar Multiplicationλ(AB)=(λA)B(AB)λ=A(Bλ)

Transpose(AB)T=BTAT

Properties of Inverse MatrixAA−1=A−1A=I(A−1)−1=A

(AT)−1=(A−1)T(An)−1=(A−1)n

(cA)−1=c−1A−1=1cA−1

(e) find the  inverse of a non-singular matrix using elementary row operations ;

Find the inverse of a non-singular matrix using elementary row operations4shirts +2pants +2pairofshoes=$190

3shirts +4pants +3pairofshoes=$2952shirts +4pants +2pairofshoes=$190

Let x = price of shirt , y= price of pant, z= price for a pair of shoe⎡⎣4322 4 4 232⎤⎦⎡⎣xyz⎤⎦=⎡⎣190295250⎤⎦Let

A=⎡⎣4322 4 4 232⎤⎦,B=⎡⎣190295250⎤⎦A⎡⎣xyz⎤⎦=⎡⎣190295250⎤⎦

AA−1=A−1A=IA−1A⎡⎣xyz⎤⎦=A−1⎡⎣190295250⎤⎦⎡⎣xyz⎤⎦=⎡⎣0.50−0.5−0.5 −0.5 1.50.250.75−1.2

5⎤⎦⎡⎣190295250⎤⎦⎡⎣xyz⎤⎦=⎡⎣104035⎤⎦Alternative,

a1x+b1y+c1z=d1a2x+b2y+c2z=d2a3x+b3y+c3z=d3

Mode > EQN > Unknowns=3, input data below a1=4,b1=2,c1=2,d1=190a2=3,b2=4,c2=3,d2=295a3=2,b3=4,c3=2,d3=250

x=10,y=40,z=35

(f) evaluate the determinant of a matrix;

Inverse of a Matrices 3X3

A=⎡⎣adgbeh cfi⎤⎦1. Determinant of 3x3 Matrices∥A∥=∥∥∥∥adgbehcfi∥∥∥∥=a∥∥∥ e hfi∥∥∥−b∥∥∥dgfi∥∥∥+c∥∥∥dgeh∥∥∥

=a(ei−fh)−b(di−fg)+c(dh−eg)

2. Inverse of 3x3 Matrices

A−1=⎡⎣adgbe h cfi⎤⎦−1=1∥A∥⎡⎣A D GBEHCFI⎤⎦T=1∥A∥⎡⎣A BCDEFGHI⎤⎦

A=∥∥∥ehfk∥∥∥,B=−∥∥∥dgfk∥∥∥,C=∥∥∥dgeh∥∥∥D=−∥∥∥bhck∥∥∥,E=∥∥∥agck∥∥∥,F=−∥∥∥agbh∥∥∥G=∥∥∥becf∥∥∥,H=−∥∥∥adcf∥∥∥,K=∥∥∥adbe∥∥∥⎡⎣+−+− +−+−+⎤⎦(g) use the properties of determinants;

Properties of Determinant

1. |A| = 0 if  it has two equal line ∣∣∣∣3132 22313∣∣∣∣=02. |A| = 0 if all elements of a line are zero ∣∣∣∣303202303∣∣∣∣=03. |A| = 0 if the elements of a line are a linear combination of the others. (row 3 = row 1 + row 2)  ∣∣∣∣213325246∣∣∣∣=04. The determinant of matrix A and its transpose A are equal. |AT|=|A|

A=∣∣∣∣213121342∣∣∣∣,  |AT|=∣∣∣∣213124312∣∣∣∣|A|=|AT|=−5

5. A triangular determinant is the product of the diagonal elements.

A=∣∣∣∣3130 25001∣∣∣∣=3.2.1=66. The determinant of a product equals the product of the determinants.

|A.B|=|A|.|B|

7. If a determinant switches two parallel lines its determinant changes sign.∣∣∣∣213121342∣∣∣∣=−5,  ∣∣∣∣123211432∣∣∣∣=53.2 Systems of linear equations

(h)  reduce an augmented matrix to row-echelon form , and determine whether a system of linear equations has a unique solution, infinitely many solutions or no solution;

(i) apply the Gaussian elimination to solve a system of linear equations;

Gaussian elimination is a method for solving matrix equations of the formAx=b

Gaussian elimination starting with the system of equations

Compose the "augmented matrix equation"

Perform elementary row operations to put the augmented matrix into the upper triangular form

A matrix that has undergone Gaussian elimination is said to be in echelon formGiven below matrix equation

In augmented form, this becomes                                                                 Switching the first and second rows (without switching the elements in the right-hand column vector) gives

First row times 3 and minus third row gives

First row times 2 and minus second row gives

Finally, second row times -7/3 and minus third row gives (augmented matrix has been reduced to row-echelon form)

which can be solved immediately to give  , back-substituting to obtain and then again back-substituting to find  ∴x3=3,  x2=1,  x1=−2

(j) find the unique solution of a system of linear equations using the inverse of a matrix.

Use result in 24a to solve simultaneous equations below.

20x−10z=−10020y+10z=300

−10x+10y+20z=200⎡⎣200−1002010−101020⎤⎦⎡⎣xyz⎤⎦=⎡⎣−100300200⎤⎦10⎡⎣20−1021 −112⎤⎦⎡⎣xyz⎤⎦=⎡⎣−100300200⎤⎦⎡⎣xyz⎤⎦=⎡⎣20−1021 −112⎤⎦−1⎡⎣−103020⎤⎦⎡⎣xyz⎤⎦=14⎡⎣3−12−1 3−2 2−24⎤⎦⎡⎣−103020⎤⎦⎡⎣xyz⎤⎦=⎡⎣−5150⎤⎦∴x=−5,    y=15,   z=0

Alternative, a1x+b1y+c1z=d1a2x+b2y+c2z=d2a3x+b3y+c3z=d3

Mode > EQN > Unknowns=3, input data below a1=20,b1=0,c1=−10,d1=−100a2=0,b2=20,c2=10,d2=300

a3=−10,b3=10,c3=20,d3=200x=−5,y=15,z=0

Chapter 4

Complex Numbers

4 Complex Numbers

(a) identify the real and imaginary parts of a complex number;

The real numbers include all integer, rational number (number that can be expressed as the quotient or fraction p/q of two integers, with the denominator q not equal to zero) and irrational numbers (numbers cannot be represented as a simple fraction)

Intergers...−4,−3,−2−1,0,1,2,3,4....Rational Number 12,−12,1537,....Irrational Number 2√,3√3,π,e,log23,....

Imaginary number is a number than can be written as a real number multiplied by the imaginary unit ii=−1−−−√i2=−1

A complex number is a number that can be put in the form a+bi

wherea and b are real numbers and i is the square root of -1, the imaginary unit i=−1−−−√

i2=−1

The absolute value or modulus of a complex number z = a + bi is|z|=a2+b2−−−−−−√

(b) use the conditions for the equality of two complex numbers;

(c) find the modulus and argument of a complex number in cartesian form and express the  complex number in polar form;

What is Argument of Complex Number

Complex Number, z=a+biModulus is the length of the line segment, that is OP (modulus of z can find using Pythagoras’ theorem)

|z|=a2+b2−−−−−−√|z|=42+32−−−−−−√

|z|=25−−√=5

The angle theta is called the argument of the complex number, ztanΘ=ba

tanΘ=34Θ=tan−134

arg z=0.644 rad

How to change theta=36.870 to radian

rad=Θ∗π1800rad=Θ∗π1800

36.870∗π1800=0.644 rad

(d) represent a complex number geometrically by means of an Argand diagram;

Argand Diagram

The Argand diagram is used to represent complex numbers.Argand diagram form by y-axis which represents  the imaginary part and x-axis represent the real part.

Example:Z1=3+2iZ2=−4+i

Conjugate of Z∗1=3−2iZ∗2=−4−i

Z1+Z2=−1+3i

(e) find the complex roots of a polynomial equation with real coefficients; 

(f) perform elementary operations on two complex numbers expressed in cartesian form;

(g) perform multiplication and division of two complex numbers expressed in polar form;

(h) usede Moivre’s theorem to find the powers and roots of a complex number

De Moivre's formula - STPM Mathematics

De Moivre's 

De Moivre's formula states that for any real number x and any integer n, (cosx+isinx)n=cos(nx)+isin(nx)

From Euler's formula 

eix=cosx+isinx(eix)n=einx

ei(nx)=cos(nx)+isin(nx)

De Moivre's Theorem in Complex Number

If z = x + yi = reiθ, and n is a natural number. Then zn=(x+yi)n=(reiθ)n

Example 1: Use De Moivre’s theorem to find (1+i)10. Write the answer in exact rectangular form.

r=12+12−−−−−−√=2√θ=tan−1(11)

(1+i)10=(2√e45∘i)10=(2√)10e450∘i

=32(cos450∘+isin450∘)=32(0+i)=32i

Example 2: Use De Moivre’s theorem to find (√3 + i ) 7 . Write the answer in exact rectangular form.

r=(3√)2+12−−−−−−−−−√=4√=2sinθ=12, θ=30∘cosθ=3√2, θ=30∘

(3√+i)7=27[cos(7.30∘)+isin(7.30∘)]=128(cos210∘+isin210∘)

=128(−3√2−12i)=643√−64i

Chapter 5 Analytic Geometry

5 Analytic Geometry

(a) transform a given equation of a conic into the standard form;

Transform Conic Section into Standard Form

A conic section is the intersection of a plane and a double right circular cone. By changing the angle and location of the intersection, we can produce different types of conics, such as circles, ellipses, hyperbolas andparabolas.

Equation of a Circle in Standard Form

(x−h)2+(y−k)2=r2Equation of an Ellipse in Standard Form

(x−h)2a2+(y−k)2b2=1Equation of a Hyperbolas in Standard Form

(x−h)2a2−(y−k)2b2=1Equation of a Parabolas in Standard Form

a(x−h)2+k

(b) find the vertex, focus and directrix of a parabola;  

Vertex, Focus and Directrix of a Parabola

A parabola is the set of all points P in the plane that are equidistant from a fixed point F (focus) and a fixed line d(directrix).The vertex of the parabola is at equal distance between focus and the directrix.

Parabolas are frequently encountered as graphs of quadratic functions, such as 

y=x2

(c) find the vertices, centre and foci of an ellipse;

Vertices, Centre and Foci of an Ellipse

Ellipse is a planar curve which in some Cartesian system of coordinates is described by the equation:

(x−h)2a2+(y−k)2b2=1

Vertex of an ellipse.The points at which an ellipse makes its sharpest turns. The vertices are on the major axis.Centre of an ellipse A point inside the ellipse which is the midpoint of the line segment linking the two foci.Foci of an ellipse - The foci, c is always lie on the major (longest) axis, spaced equally each side of the centre. 

c2=a2−b2

Example16x2+25y2=400

16x2400+25y2400=1

x225+y216=1x252+y242=1

(x2−0)252+(y2−0)242=1

Vertex is (-5,0) and (5,0), the major (longest) axis. Co-vertex is (0,-4), (0, 4) minor (shortest) axisThe centre is at (h,k)=(0,0)Foci of an ellipse. The foci are three unit to either side of the centre, at (-3,0) and (3,0)

c2=a2−b2c2=52−42c2=±3

Find Vertices, Centre, and Foci of Ellipse 01 Find Vertices, Centre, and Foci of Ellipse 02 Find Vertices, Centre, and Foci of Ellipse 03

(d) find the vertices, centre, foci and asymptotes of a hyperbola;

(e) find the equations of parabolas, ellipses and hyperbolas satisfying prescribed conditions (excluding eccentricity);

(f) sketch conics;

(g) find the cartesian equation of a conic defined by parametric equations;

(h) use the parametric equations of conics.

Chapter 6 Vectors

6.1 Vectors in two and three dimensions

(a) use unit vectors and position vectors; 

Unit Vectors and Position Vectors

Unit Vectors

A unit vector, or direction vector is a vector which has length of 1 or magnitude of 1.

(b) uˆ=u∥u∥

(c)(d)

Position Vectors

A vector that starts from the origin (O) is called a position vector. 

Point A has the position vector a, if

A=(35) Point B has the position vector b, if

B=(62)

(e)

(b) performscalar multiplication, addition and subtraction of vectors;

 Scalar Multiplication of Vector

Multiplying a vector by a scalar is called scalar multiplication.c[a1a2]=[ca1ca2]

Example: Multiply the vector u = < 2, 3 > of by the scalars 2, –3, and 1/2

2u ⃗ =2[23]=[46]12u ⃗ =12[23]=[13/2]−3u ⃗ =−3[23]=[−6−9]

(c) find the scalar product of two vectors, and determine the angle between two vectors; 

The scalar product, or dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number.

Properties of the Scalar Product

u.v=v.u u(v+w)=uv+uw u.u=||u||^{2} c(u.v)=cu.v=u.cv

Scalar product can be obtained by formula 

a.b=|a||b|cosθ|a| means the magnitude (length) of vector a. and 

a.b=axbx+ayby

Calculate the scalar product of vectors a and b, given θ = 59.500 a=[−68],b=[512]

|a|=(−6)2+82−−−−−−−−−√=10|b|=52+122−−−−−−−√=13

a.b=|a||b|cosθa.b=10∗12cos59.5∘

ora.b=axbx+ayby

a.b=(−6)(5)+(8)12)=66

Find the angle between the two vectors.

a=⎡⎣234⎤⎦,b=⎡⎣1−23⎤⎦a.b=axbx+aybya.b=|a||b|cosθ

a.b=(2)(1)+(3)(−2)+(4)(3)=8|a|=22+32+42−−−−−−−−−−√=29−−√

|a|=12+(−2)2+32−−−−−−−−−−−−√=14−−√8=29−−√14−−√cosθ

cosθ=0.397θ=66.6∘

(d) find the vector product of two vectors, and determine the area a parallelogram and of a triangle;

Vector Product of Two Vectors

The Vector Product, or Cross Product of two vectors is another vector that is at right angles to both. 

The magnitude of the vector product of vectors a and b is 

|a∗b|=|a||b|sinθ n|a| is the magnitude (length) of vector a|b| is the magnitude (length) of vector bθ is the angle between a and bn is the unit vector at right angles to both a and bVector Product, C, 

a ⃗ ∗b ⃗ =∣∣∣∣ia1b1ja2b2 ka3b3∣∣∣∣a ⃗ ∗b ⃗ =∣∣∣a2b2a3b3∣∣∣i−∣∣∣a1b1a3b3∣∣∣j+∣∣∣a1b1a2b2∣∣∣kWhat is the vector product of a = (2,3,4) and b = (5,6,7)

a ⃗ ∗b ⃗ =∣∣∣∣i25j36k47∣∣∣∣a ⃗ ∗b ⃗ =∣∣∣3647∣∣∣i−∣∣∣2547∣∣∣j+∣∣∣2536∣∣∣k

=−3i+6j−3ku ⃗ =⟨−3,6,−3⟩

Vector product, c = a x b = (-3, 6, -3)

6.2 Vector geometry

(e) find and use the vector and cartesian equations of lines;

Vector and Cartesian Equations of Lines

Vector Equations of Lines

The vector equation of the line through points A and B is given by

r=OA+λABor

r=a+λt,   t=b−a

Example: Compute the vector equation of a straight line through point A and B with position vector:

a=⎛⎝2−13⎞⎠,b=⎛⎝13−2⎞⎠,r=⎛⎝xyz⎞⎠b−a=⎛⎝13−2⎞⎠−⎛⎝2−13⎞⎠=⎛⎝−14−5⎞⎠r=a+λtr=⎛⎝2−13⎞⎠+λ⎛⎝−14−5⎞⎠Cartesian Equations of Lines

a=⎛⎝axayaz⎞⎠,b=⎛⎝⎜bxbybz⎞⎠⎟,r=⎛⎝xyz⎞⎠Cartesian Equation will be

x−axbx−ax=y−ayby−ay=z−azbz−azExample: Compute the cartesian equation of a straight line through point A and B with position vector:

a=⎛⎝1−32⎞⎠,b=⎛⎝3−15⎞⎠,r=⎛⎝xyz⎞⎠x−13−1=y−(−3)−1−(−3)=z−25−2x−12=y+32=z−23

(f) find and use the vector and cartesian equations of planes;

Vector Equations of Planes

The vector equation of the plane r=a+λu+μv

Example: Compute the vector equation of a plane through point A, B and C with position vector:

a=⎛⎝2−13⎞⎠,b=⎛⎝14−1⎞⎠,c=⎛⎝0−21⎞⎠u=b−a=⎛⎝14−1⎞⎠−⎛⎝2−13⎞⎠=⎛⎝−15−4⎞⎠v=c−a=⎛⎝0−21⎞⎠−⎛⎝2−13⎞⎠=⎛⎝−2−1−2⎞⎠

r=a+λu+μvr=⎛⎝2−13⎞⎠+λ⎛⎝−15−4⎞⎠+μ⎛⎝−2−1−2⎞⎠

Cartesian Equations of Planes

Formula for Cartesian Equations of Planes

n=(b−a)∗(c−a)r.n=a.n

a=⎛⎝2−13⎞⎠,b=⎛⎝14−1⎞⎠,c=⎛⎝0−21⎞⎠,r=⎛⎝xyz⎞⎠n=(b−a)∗(c−a)

n=⎛⎝−15−4⎞⎠⎛⎝−2−1−2⎞⎠=⎛⎝−14611⎞⎠r.n=a.n⎛⎝xyz⎞⎠.⎛⎝−14611⎞⎠=⎛⎝2−13⎞⎠.⎛⎝−14611⎞⎠

−14x+6y+11z=(2)(−14)+(−1)(6)+(3)(11)Cartesian Equations of Planes

−14x+6y+11z=−1

(g) calculate the angle between two lines, between a line and a plane, and between two planes;

Angle Between Two Lines

Angle Between Two Lines

θ=β−αtanθ=tan(β−α)=tanβ−tanα1+tanβtanα=m1−m21+m1m2

For two line of gradient m1, m2te acute angle between them is always positive

tanθ=∣∣∣m1−m21+m1m2∣∣∣ m1m2 ≠ -1, this formula doesn't work for perpendicular lines.

Example 1: Find the acute angle between the lines y = 3x - 1 and y = -2x + 3.

tanθ=∣∣∣m1−m21+m1m2∣∣∣tanθ=∣∣∣3−(−2)1+(3)(−2)∣∣∣tanθ=|−1|=1θ=tan−1(1)=45∘Example 2: Find the acute angle between the lines 6x - y + 8 = 0 and -3x -11y +10 = 0Rearrange the equation

y=6x+8y=−311x−1011

m1=6, m2=-3/11

tanθ=∣∣∣m1−m21+m1m2∣∣∣tanθ=∣∣∣∣6−(−311)1+(6)(−311)∣∣∣∣tanθ=∣∣∣−697∣∣∣θ=tan−1697=84.2∘

(h) find the point of intersection of two lines, and of a line and a plane;

Find the Point of Intersection Between Two Lines

At the point of intersecting lines, the points are equal.

Example: Find the point of intersection between lines y = 3x - 7 and y = -2x+3.

y=3x−7−−−−−(1)y=−2x+3−−−−−(1)Substitute (1) into (2)

3x−7=−2x+35x=10x=2x=2,

y=3(2)−7y=−1

Hence, the intersecting point is (2, -1)

 5 0 0Google - See more at: http://coolmathsolutions.blogspot.com/2013/03/find-point-of-intersection-

between-two.html#sthash.hyvNVGN1.dpuf

(i) find the line of intersection of two planes.

Line of Intersection of Two Planes

 Line of Intersection of Two Planes

Find the vector equation of the line in which the 2 planes 2x - 5y + 3z = 12 and 3x + 4y - 3z = 6 meet.The normal vector of first plane is <2, -5, 3> and normal vector of second plane it is <3, 4, -3>

To determine whether these two planes parallel

Two planes are parallel if n1=cn2

 Therefore, <2, -5, 3> and <3, 4, -3> not parallel to each others.

Find the vector/cross product of these normal vectors 

n1→∗n2→=∣∣∣∣i23j−54k3−3∣∣∣∣n1→∗n2→=∣∣∣−543−3∣∣∣i−∣∣∣233−3∣∣∣j+∣∣∣23−54∣∣∣k

=3i+15j+23kVector product is <3, 15, 23>

Find the position vector from the origin

Find some point which lies on both the planes because then it must lie on their line of intersection. Any point which lies on both planes will do. Could be plane-xy, yz, xz.If x=0,

−5y+3z=124y−3z=6

y=−18,z=−26Point with position vector (0, -18, -26) lies on the line of intersection.

The equation of the line of intersection is r=(0,−18,−26)+t(3,15,23)

To check that point that we get does really lie on both planes and so on their line of intersection. If t=1

r=(3,−3,−3)Substitute into the planes equations

2(3)−5(−3)+3(−3)=123(3)+4(−3)−3(−3)=6

If y=0

2x+3z=123x−3z=6

x=3.6,z=1.6Point with position vector (0, 3.6, 1.6) lies on the line of intersection.

The equation of the line of intersection is r=(3.6,0,1.6)+t(3,15,23)

To check that point that we get does really lie on both planes and so on their line of intersection. If t=1

r=(6.6,15,24.6)Substitute into the planes equations

2(6.6)−5(15)+3(24.6)=123(6.6)+4(15)−3(24.6)=6

Chapter 7 Limits and Continuity

7.1 Limits

(a) determine the existence and values of the left-hand limit, right-hand limit and limit of a function;

Left-hand Limit and Right-hand Limit

A limit is the value that a function or sequence "approaches" as the input or index approaches some value.

The limit of f(x) as x approaches a from the right. 

limx→a+f(x)

The limit of f(x) as x approaches a from the left.

limx→a−f(x)

Example: Find

limx→2−(x3−1)limx→2−(x3−1)=limx→2−(23−1)=7As x approaches 2 from the left, x3pproaches 8 and x3-1 approaches 7.

Find

limx→2+(x3−1)limx→2+(x3−1)=limx→2+(23−1)=7

(b) use the properties of limits;

Properties of limits

The limit of a constant is the constant itself.

limx→ak=k

The limit of a function multiplied by a constant is equal to the value of the function multiplied by the constant.

limx→ak.f(x)=k.limx→af(x)

The limit of a sum (or difference) of the functions is the sum (or difference) of the limits of the individual functions.

limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)limx→a[f(x)−g(x)]=limx→af(x)−limx→ag(x)

The limit of a product is the product of the limits.

limx→af(x).g(x)=limx→af(x).limx→ag(x)

The limit of a quotient is the quotient of the limits

limx→af(x)g(x)=limx→af(x)limx→ag(x)

The limit of a power is the power of the limit.

limx→axn=an

7.2 Continuity

(c) determine the continuity of a function at a point and on an interval;

Continuity of a Function at A Point and On An Interva

Continuity at a Point

A function f (x) is continuous at a if the following three conditions are valid:

i) The function is de fined at a: That is, a is in the domain of definition of f (x)

ii) if limx→af(x)

exists.

iii) iflimx→af(x)=f(a)

If any of the three conditions in the definition of continuity fails when x = c, the function is discontinuous at that point.

Continuity on An Interval

A function which is continuous at every point of an open interval I is called continuous on I.

(d) use the intermediate value theorem.

Let f (x) be a continuous function on the interval [a b]

If d is between f(a) and f(b), then a corresponding c between a and b, exists, so that f(c)=d

Chapter 8 Differentiation

8.1 Derivatives

(a) identify the derivative of a function as a limit;

Derivative of a Function as a Limit

For a function f(x), its derivative is defined as

f′(x)=lim△x→0f(x+△x)−f(x)△x

Example 1: Compute the derivative of f(x) by using limit definition

f(x)=13x−25f′(x)=lim△x→0f(x+△x)−f(x)△xf′(x)=lim△x→013(x+△x)−25−{13x−25}△x=lim△x→013x+13△x−25−13x+25△x=lim△x→013△x△x=13

Example 1: Compute the derivative of f(x) by using limit definition

f(x)=5−x+2−−−−√f′(x)=lim△x→0f(x+△x)−f(x)△x=lim△x→0{5−(x+△x)+2−−−−−−−−−−−√}−{5−x+2−−−−√}△x=lim△x→0x+2−−−−√−x+△x+2−−−−−−−−−√△x=lim△x→0x+2−−−−√−x+△x+2−−−−−−−−−√△x.x+2−−−−√+x+△x+2−−−−−−−−−√x+2−−−−√+x+△x+2−−−−−−−−−√=lim△x→0(x+3)−(x+△x+3)△x{x+3−−−−√+x+△x+3−−−−−−−−−√}=lim△x→0−△x△x{x+3−−−−√+x+△x+3−−−−−−−−−√}=lim△x→0−1x+3−−−−√+x+△x+3−−−−−−−−−√=−1x+3−−−−√+x+3−−−−√=−12x+3−−−−√

(b) find the derivatives ofx n   ( n ∈ Q),   e x , ln x ,   sin x, cos   x,   tan x,   sin -1 x,   cos -1 x, tan -1 x , with constant multiples, sums, differences, products, quotients and composites;

Common Derivatives

Table of Derivativesddx(x)=1ddx(xn)=nxn−1ddx(ex)=exddx(lnx)=1x,x>0ddx(sinx)=cosxddx(cosx)=−sinxddx(tanx)=sec2xddx(sin−1x)=11−x2−−−−−√ddx(cos−1x)=−11−x2−−−−−√ddx(secx)=secxtanxddx(cscx)=−cscxcotxddx(cotx)=−csc2x

(c) performimplicit differentiation;

Explicit and Implicit Differentiation

There are two ways to define functions, implicitly and explicitly. Most of the equations we have dealt with have been explicit equations, such as y = 3x-2. This is called explicit because given an x, you can directly get f(x).

The technique of implicit differentiation allows you to find the derivative of y with respect to x without having to solve the given equation for y. Given

x2+y2=10

Performing a chain rule to get dy/dx, solve dy/dx in terms of x and y.Find dy/dx for

x2+y2=10

2x+2ydydx=02ydydx=−2xdydx=−xy

(d) find the first derivatives of functions defined parametrically;

First Derivative of Parametric Functions

Parametric derivative is a derivative in calculus that is taken when both the x and y variables (independent and dependent, respectively) depend on an independent third variable t, usually thought of as "time".

The first derivative of the parametric equations is

dydx=dydtdxdt

dydx=dydt.dtdxx=f(t),y=g(t)

Example:  Find the first derivative, given  x=t+costy=sint

dydx=dydtdxdt=cost1−sint

Example:  Find the first derivative, given x=t4−4t2y=t3

dydx=dydtdxdt=2t23t3−8t

8.2 Applications of differentiation

(e) determine where a function is increasing, decreasing, concave upward and concave downward

Where is a function increasing or decreasing?

Increasing function: if f '(x)>0, function is increasing.

Decreasing function: if f '(x)<0,  function is decreasing

Where is function concave up or concave down?

concave down: if the second derivative is negative, f ''(x) < 0 the function curves downward, convex down.

concave upward: if the second derivative is negative, f ''(x)> 0 the function curves upward.

a)  f(x) is increasing and concave up if f'(x) is positive and f"(x) is positive.

b)  f(x) is increasing and concave down if f'(x) is positive and f"(x) is negative.

c) f(x) is decreasing and concave up if f'(x) is negative and f"(x) is positive.

d) f(x) is decreasing and concave down if f'(x) is negative and f"(x) is negative.

(f) determine the stationary points, extremum points and points of inflexion;

Stationary Points

A stationary point is an input to a function where the derivative is zero. At stationary points,

dydx=0

Extremum Points

Maxima and minima are points where a function reaches a highest or lowest value, respectively

(g)

(h)(i)

Second DerivativeIf f'''(x) is positive, then it is a minimum point.If f'''(x) is negative, then it is a maximum point.If f'''(x)= zero, then it could be a maximum, minimum or point of inflexion.

Point of InflexionAn inflection point is a point on a curve at which the sign of the curvature changes.

d2ydx2=0

Third Derivative

If f'''(x) ≠ 0 There is an inflexion point

(g) sketch the graphs of functions, including asymptotes parallel to the coordinate axes;

(h) find the equations of tangents and normals to curves, including parametric curves;

Tangents and Normals to a Curve

At point (x1,y1) on the curve y=f(x) the equation of tangent is y−y1=m1(x−x1)

wherem1=f′(x)=dydx

the gradient to the function of f(x).

Tangent is perpendicular to normal, thus m1m2=−1

Example: Find the equations of the tangent line and the normal line for the curve at t=1.

x=t2y=2t+1

dxdt=2tdydt=2

dydx=dydt.dtdx=2t2

t=1 dydx=2

Since t = 1, x = 1, y = 3 Equation of tangent  y−y1=m1(x−x1)y−3=1(x−1)

y=x+2Equation of normal

m1m2=−1m2=−1

y−3=−1(x−1)y=−x+4

(i) solve problems concerning rates of change, including related rates;

Rates of Change and Related Rates

Related rates problems involve finding a rate at which a quantity changes by relating that quantity to other quantities whose rates of change are known. The rate of change is usually with respect to time.

Example:  Air is being pumped into a spherical balloon such that its radius increases at a rate of .80 cm/min. Find the rate of change of its volume when the radius is 5 cm.

The volume ( V) of a sphere with radius r is

V=43πr3Differentiating above equation with respect to t

dVdt=43π.3r2.drdtdVdt=4πr2.drdtThe rate of change of the radius dr/dt = 0.80 cm/min because the radius is increasing with respect to time.

dVdt=4π(5)2(0.80)dVdt=80π cm3/min

Hence, the volume is increasing at a rate of 80π cm3/min when the radius has a length of 5 cm.

(j) solveoptimisation problems .

The differentiation and its applications can be used to solve practical problems. This include minimizing costs, maximizing areas, minimizing distances and so on.

1. Diagram-  Draw a diagram.2. Goal - Maximize or minimize which unknown?3. Data - Introduce variable names. Which values are given?4. Equation-  Express the unknown as a function of a single variable.5. Differentiate-  Find first and second derivatives.6. Extrema -

Find critical points  Use the first or second derivative test to determine whether the critical points are local

maxima, local minima, or neither.  Check end points of f, if applicable.

Chapter 9 Integration

9.1 Indefinite integrals

(a) identify integration as the reverse of differentiation;

(b) integrate   x n     ( n ∈ Q),   e x   ,   sin x, cos   x,   sec 2 x ,  with constant multiples, sums and differences;

Integral Common Function

Constant

∫adx=ax+C

Variable

∫xdx=x22+C

Power

∫xndx=xn+1n+1+C

Reciprocal

∫1xdx=ln|x|+C

Exponential

∫exdx=ex+C∫axdx=axlna+C∫ln(x)dx=x(ln(x)−1)+C

Trigonometry

∫cos(x)dx=sin(x)+C∫sin(x)dx=−cos(x)+C∫sec2(x)dx=tan(x)+C

(c) integrate rational functions by means of decomposition into partial fractions;

Integration by Partial Fractions Decomposition

How to integrate the rational function, quotient of two polynomials

f(x)=P(x)Q(x)  A rational function can be integrated into 4 steps

1. Reduce the fraction if it is improper (i.e degree of P(x) is greater than degree of Q(x).2. Factor Q(x) into linear and/or quadratic (irreducible) factors.3. Find the partial fraction decomposition.4. Integrate the result in step 3.

Example 1: Evaluate the indefinite integral ∫1x2+5x+6dx

Factor Q(x) into linear and/or quadratic (irreducible) factors &  find the partial fraction decomposition

1(x+2)(x+3)=Ax+2+Bx+31=A(x+3)+B(x+2)

x=−2,A=1x=−3,B=−1

1(x+2)(x+3)=1x+2−1x+3Integrate the result

∫1x2+5x+6dx=∫1(x+2)(x+3)dx=∫1x+2−1x+3dx

=ln|x+2|−ln|x+3|+C

Example 2: Evaluate the indefinite integral 1(x−1)(x+2)(x+1)=Ax−1+Bx+2+Cx+1

Decomposition of rational functions into partial fractions 1=A(x+2)(x+1)+B(x−1)(x+1)+C(x−1)(x+2)

x=1,6A=1,A=16x=−1−2C=1,C=−12x=−23B=1,B=13

1(x−1)(x+2)(x+1)=16(1x−1)+13(1x+2)−12(1x+1)

Integrate the result∫1(x−1)(x+2)(x+1)dx

=∫16(1x−1)+13(1x+2)−12(1x+1)

=16ln|x−1|+13ln|x+2|−12ln|x+1|+C

(d) use trigonometric identities to facilitate the integration of trigonometric functions;

(e) use algebraic and trigonometric substitutions to find integrals;

Integration by Trigonometric Substitution & Identities

Substitute one of the following to simplify the expressions to be integrated

For a2−x2−−−−−−√, use  x=asinθFor a2+x2−−−−−−√, use  x=atanθFor x2−a2−−−−−−√, use  x=asecθ

Basic Trigonometric Identities

csc2θ=1sin2θsec2θ=1cos2θ cot2θ=1tan2θ

Pythagorean Identities

cos2θ+sin2θ=1csc2θ=1+cot2θsec2θ=1+tan2θ

(f) perform integration by parts;

9.2 Definite integrals

(g) identify a definite integral as the area under a curve;

(h) use the properties of definite integrals;

Properties of Integrals

Additive Properties

Split a definite integral up into two integrals with the same integrand but different limits∫baf(x)dx+∫cbf(x)dx=∫caf(x)dx

If the upper and lower bound are the same, the area is 0.∫aaf(x)dx=0

If an interval is backwards, the area is the opposite sign.∫baf(x)dx=−∫abf(x)dx

Integral of Sum

The integral of a sum can be split up into two integrands ∫ba[f(x)+g(x)]dx=∫baf(x)dx+∫bag(x)dx

Scaling by a constant

Constants can be distributed out of the integrand and multiplied afterwards.∫bacf(x)dx=c∫baf(x)dx

Total Area Within an Interval

∫baf(x)dx=F(b)−F(a)∫ba|f(x)|dx=F(b)+F(a)

Integral inequalities

If f(x)≥0  and  a<b, then   ∫baf(x)dx≥0

f(x)≤ g(x)  and  a<b, then  ∫baf(x) dx≤ ∫bag(x) dx

(i) evaluate definite integrals;

(j) calculate the area of a region bounded by a curve (including a parametric curve) and lines parallel to the coordinate axes, or between two curves;

Area Under a Curve for Definite Integrals

Area Under a Curve

If y = f (x ) is continuous and non-negative on [a, b], then the area under the curve of f from a to b is:

Area=∫baf(x)dx

If y =  f(x) is continuous and f(x) < 0 on [a, b], then the area under the curve from a to b is:

Area=−∫baf(x)dx

If y = f(x) is continuous and f(x) < 0 and f(x) > 0 

Area=∫baf(x)d(x)+∣∣∣∫cbf(x)d(x)∣∣∣+∫dcf(x)d(x)

If x = g (y ) is continuous and non-negative on [c, d], then the area under the curve of g from c to d is:

Area=∫dcg(y)dy

(k) calculate volumes of solids of revolution about one of the coordinate axes.

Chapter 10

Differential Equations

(a) find the general solution of a first order differential equation with separable variables; (b) find the general solution of a first order linear differential equation by means of an

integrating factor; (c) transform, by a given substitution, a first order differential equation into one with

separable  variables or one which is linear; (d) use a boundary condition to find a particular solution; (e) solve problems, related to science and technology, that can be modelled by

differential equations.

Chapter 11

Maclaurin Series

(a) find the Maclaurin series for a function and the interval of convergence;

Maclaurin Series

Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point.Taylor Series

f(x)=∑x=0∞fn(a)n!(x−a)n=f(a)+f′(a)(x−a)+f′(a)2!(x−a)2+f′′(a)3!(x−a)3+...+f(n)(a)n!(x−a)n+...

MacLaurin series is the Taylor series of the function about x=0.

f(x)=∑x=0∞fn(0)n!xnf(x)=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3+...+f(n)(0)n!xn+...

Example : Find the Maclaurin Series expansion of e5x

f(x)=e5x

f′(x)=5e5x

f′′(x)=52e5x

f′′′(x)=53e5x

f(4)(x)=54e5x

f(0)=1

f′(0)=5

f′′

(0)=52

f′′′

(0)=53

f(4)=54

f(x)=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3+...+f(n)(0)n!xn+...

e5x=1+5x1!+52x22!+53x33!+...+5nxnn!

=∑n=0∞5nxnn!

Interval of Convergence

The set of points where the series converges is called the interval of convergenceFinding interval of convergence1) perform ratio test to test for the convergence of a series.2) Check endpoint

Ratio test

L=limn→∞∣∣∣an+1an∣∣∣if L < 1 then the series converges.if L > 1 then the series does not converge;if L = 1 or the limit fails to exist, then the test is inconclusive

Example:  Determine the interval of convergence for the series ∑n=1∞(x−2)nn.5n

Apply ratio test

limn→∞∣∣∣an+1an∣∣∣=∣∣∣(x−2)n+1(n+1).5n+1.n.5n(x−2)n∣∣∣=limn→∞∣∣∣x−25.nn+1∣∣∣

=15| x−2|limn→∞∣∣∣nn+1∣∣∣=15| x−2|

As n approcaches infinity, n/n+1 aprpoaches 1

limn→∞∣∣∣nn+1∣∣∣≈ 1 The series converges for  

15| x−2|<1−5<x−2<5−3<x<7

Check end point x=7, ∑n=1∞(5)nn.5n=∑n=1∞1n

This is the harmonic series, and it diverges.Check end point x=-3,

∑n=1∞(−5)nn.5n=∑n=1∞(−1)n1nThe series converges by the Alternating Series Test.

The interval of convergence is  −3≤x<7

or[−3,7)

(b) usestandard series to find the series expansions of the sums, differences, products, quotients and composites of functions;

Standard Maclaurin Series

Standard Maclaurin Series

11−x=∑n=0∞xn=1+x+x2+x3+....

ex=∑n=0∞xnn!=1+x+x22!+x33!+....

ln(x+1)=∑n=0∞(−1)nxn+1n+1=x−x22+x33

(1+x)k=∑n=0∞xn(kn)xn=1+kx+k(k−1)2!x2+k(k−1)(k−2)3!x3+...

Power Series Expansions 08

Find a power series for x4+x2

Rewrite the function asx4+x2=x(14+x2)=x4⎛⎝11+x24⎞⎠

=x4⎛⎝⎜11−(−x24)⎞⎠⎟This is the sum of the infinite geometric series   with the first term x/4 and ratio x2/4

=x4∑n=0∞(−x24)=x4∑n=0∞(−1)nx2n4n=∑n=0∞(−1)nx2n+14n+1

(c) perform differentiation and integration of a power series;

(d) use series expansions to find the limit of a function.

Chapter 12 Numerical Method

12.1 Numerical solution of equations

(a) locate a root of an equation approximately by means of graphical considerations and by searching for a sign change;

(b) use an iterative formula of the form xn+1 =f(xn) to find a root of an equation to a prescribed degree of accuracy;

(c) identify an iteration which converges or diverges;

(d) use the Newton-Raphson method;

12.2 Numerical integration

(e) use the trapezium rule;

(f) use sketch graphs to determine whether the trapezium rule gives an over-estimate or an under-estimate in simple cases.

Chapter 13

Data Description

(a) identify discrete, continuous, ungrouped and grouped data;

Discrete, Continuous, Ungrouped and Grouped Data

Discrete Data

Discrete Data is counted and can only take certain values (whole numbers).Eg Number of students in a class, Number of children in a playground, etc.

Continuous Data

Continuous Data is data that can take any value (within a range)Eg. Height of children, weights of car, etc.

Grouped Data

Data that has been organized into groups (into a frequency distribution).Eg. 

Class0−56−10

11−1516−20

Frequency1020174

Ungrouped Data

Data that has not been organized into groups10 1 2 4 80 45 67 7850 1 11 23 6 9 8 15

(b) construct and interpret stem-and-leaf diagrams, box-and-whisker plots, histograms and cumulative frequency curves;

Stem-and-Leaf Diagrams

A stem-and-leaf diagrams presents quantitative data in a graphical format, similar to a histogram, to assist in visualizing the shape of a distribution, giving the reader a quick overview of distribution.

45 46 48 49 63 65 66 68 68 73 73 75 76 81 85 88 107

4  | 5 6 8 9 5  |6  | 3 5 6 8 87  | 3 3 5 68  | 1 5 810 | 7key 4 | 5 means 45

Box-and-Whisker Plots

Box-and-Whisker Plots displays of the spread of a set of data through five-number summaries: the minimum, lower quartile (Q1), median (Q2), upper quartile(Q3), and maximum.

The first and third quartiles are at the ends of the box,  The median is indicated with a vertical line in the interior of the box

Ends of the whiskers indicated the maximum and minimum.

Histograms and Cumulative Frequency Histograms

Histograms and Cumulative Frequency Histograms

A histogram is constructed from a frequency tableThe cumulative frequency is the running total of the frequencies.

(c) state the mode and range of ungrouped data; 

Mode of ungrouped data

An observation occurring most frequently in the data is called mode of the data

Example: Find the median of the following observations

10, 14, 16, 20, 24, 28, 28, 30, 32, 40

In the given data, the observation 28 occurs maximum. So the mode is 28.

Range of ungrouped data

Range = Highest Value – Lowest value

The range is the simplest measure of dispersion; it only takes into account the highest and lowest values.

The range of above ungrouped data = 40 - 10 = 30

(d) determine the median and interquartile range of ungrouped and grouped data;

Median and Interquartile Range of Ungrouped Data

Interquartile range

Q2 (the middle quartile) is the median.Q1 (the lower quartile) is the median of the numbers to the left of, or below Q2.Q3 (the upper quartile) is the median of the numbers to the right of, or above Q2.Interquartile Range = Q3 -Q1 

Interquartile range for Ungrouped Data

Q2=(n+12)th  observationQ1=(n+14)th  observation

Q3=(3(n+1)4)th  observation

Example 1: Find the median, lower quartile and upper quartile of the following numbers.

12, 5, 22, 30, 7, 38, 16, 42, 15, 43, 35

Rearrange the data in ascending order:

Q1   == 1/4 (11+1) = 3th observation, Q1=12Q2   == 1/2 (11+1) = 6th observation, Q2=22Q3   == 3/4 (11+1) = 9th observation, Q3=38

Interquartile Range = Q3 -Q1  = 38 - 12 = 26Range = Largest value - smallest value = 43 -5 = 38

Example 2: Find the median, lower quartile and upper quartile of the following numbers.

12, 5, 22, 30, 7, 38, 16, 42, 15, 43, 35, 50

Rearrange the data in ascending order:

Q1=(12+152)=13.5Q2=(22+302)=26Q3=(38+422)=40

Interquartile Range = Q3 -Q1   = 40 - 13.5 = 26.5Range = Largest value - smallest value = 50 -5 = 45

Median and Interquartile Range of Grouped Data

Median=Lm+(n2−Ffm)C

n = the total frequency Lm = the lower boundary of the class medianF = the cumulative frequency before class medianf = the frequency of the class medianfm= the lower boundary of the class medianC= the class width

Q1=LQ1+(n4−FfQ1)CQ3=LQ3+⎛⎝3n4−FfQ3⎞⎠C

Example : Find the median and interquartile range of below data.

Q2=20.5+⎛⎝502−2312⎞⎠10=22.167Q1=10.5+⎛⎝504−1010⎞⎠10=13

Q3=30.5+⎛⎝3(50)4−358⎞⎠10=33.625 (e) calculate the mean and standard deviation of ungrouped and grouped data, from raw data and from given totals such as

∑i=1n(xi−a)  and  ∑i=1n(xi−a)2

Mean and Standard Deviation of Ungrouped and Grouped Data

Ungrouped DataMean

x¯=∑xn

Standard Deviation s=∑(x−x¯)2n−−−−−−−−−√

s=∑x2n−(∑xn)2−−−−−−−−−−−−−−⎷

Grouped Data

Mean x¯=∑fx∑f

Standard Deviation s=∑f(x−x¯)2n−−−−−−−−−−√

s=∑fx2∑f−(∑fx∑f)2−−−−−−−−−−−−−−−−⎷

(f) select and use the appropriate measures of central tendency and measures of dispersion;

Measures of Central Tendency

Measure of central tendency is an average of a set of measurements.

Mode -  the number that occurs most frequently. Median - the value of the middle item in a set of observations. Mean - average value of the distribution.

Measures of Dispersion

Measures of Dispersion is group of analytical tools that describes the spread or variability of a data set.

Range - Difference between the largest and smallest sample values. Variance/ Standard Deviation - Measures the dispersion around an average. Coefficient of variation - Expressed in a relative value.

(g) calculate the Pearson coefficient of skewness;

Pearson coefficient of skewness

Skewness

Skewness is a measure of the asymmetry of the probability distribution. Skewness value can be positive or negative.

Negative skew: The left tail is longer; the mass of the distribution is concentrated on the right of the figure.

Positive skew: The right tail is longer; the mass of the distribution is concentrated on the left of the figure.

Pearson coefficient of skewness

Pearson coefficient of skewness is based on arithmetic mean, mode, median and standard deviation.

Pearson's mode or first skewness coefficient: S=mean − modestandard deviation

Pearson's median or second skewness coefficient: S=3(mean − median)standard deviation

If Sk = 0, then the frequency distribution is normal and symmetrical. If Sk> 0, then the frequency distribution is positively skewed. If Sk< 0, then the frequency distribution is negatively skewed.

Example: Calculate the coefficient of skewness of the following data by using Karl Pearson's method.

1, 2, 3, 3, 4, 4, 4

mean=3, standard deviation = 1.07Coefficient of skewness,

Sk=3−41.07=−0.93Therefore, the distribution is negatively skewed.

(h) describe the shape of a data distribution.

Chapter 14

Probability

(a) apply the addition principle and the multiplication principle;

Addition Principle

If we want to find the probability that event A happens or event B happens, we should add the probability that A happens to the probability that B happens. 

Addition Rule:P(A or B) = P(A) + P(B)

Example: A single 6-sided die is rolled. What is the probability of rolling a 1 or a 6P(1 or 6)=P(1)+P(6)

=16+16

=13

Multiplication Principle 

When two compound independent events occur, we use multiplication to determine their probability. To find the probability that event A happens and event B happens, we should multiply the probability that A happens times the probability that B happens.

Multiplication Rule: P(A and B) = P(A) x P(B)

Example: A pizza corner sells pizza in 3 sizes with 6 different toppings. If Peter wants to pick one pizza with one topping, there is a possibility of 18 combinations as the total number of outcomes.

(b) use the formulae for combinations and permutations in simple cases;

Permutations

A permutation is the choice of r things from a set of n things without replacement and where the order matters

Permutation Formula

nPr=n!(n−r)!

Combinations

A combination is the choice of r things from a set of n things without replacement and where order does NOT matter.

Combination Formula

nCr=(nr)=nPrr!=n!r!(n−r)!

(c) identify a sample space, and calculate the probability of an event;

Sample Space - The set of all the possible outcomes in an experiment.

Event - Any subset E of the sample space S, specific outcome of an experiment.

Probability - The measure of how likely an event is.

Example 1 Tossing a coin. The sample space is S = {Head, Tail}, E = {Head} is an event.The probability of getting 'Head' is 1/2

Example 2 Tossing a die. The sample space is S = {1,2,3,4,5,6}, E ={1,3,5} is an event, which can be described in words as "the number is odd".

The probability of getting odd numbers is 1/2.

(d) identify complementary, exhaustive and mutually exclusive events;

Complementary, Exhaustive and Mutually Exclusive Events

Complementary Event

The complement of any event A, A' is the event that A does not occur.

Exhaustive Event

A set of events is collectively exhaustive if at least one of the events must occur. For example, when rolling a six-sided die, the outcomes 1, 2, 3, 4, 5, and 6 are collectively exhaustive, because they include the entire range of possible outcomes. Thus, all sample spaces are collectively exhaustive.

P(A∪B)=1Mutually Exclusive Event

Two events are 'mutually exclusive' if they cannot occur at the same time. Example, tossing a coin once, which can result in either heads or tails, but not both.

P(A∩B)=0P(A∪B)=P(A)+P(B)

(e) use the formula      

   Independent Events Conditional Probabilities

Formula 

P(A∪B)=P(A)+P(B)−P(A∩B)Independent Events

Two events are independent means that the occurrence of one does not affect the probability of the other. Two events, A and B, are independent if and only if

P(A∩B)=P(A).P(B)

Conditional Probability

Conditional Probability is the probability that an event will occur, when another event is known to occur or to have occurred.

Given two events A and B, , the conditional probability of A given B is defined as the quotient of the joint probability of A and B, and the probability of B

P(A|B)=P(A∩B)P(B)P(A∩B)=P(A|B).P(B)

                           

         P(A ∪ B) = P(A) + P(B) - P(A ∩ B);

(f) calculate conditional probabilities, and identify independent events; (g) use the formulae                                                

          P(A ∩ B) = P(A) x P(B|A) = P(B) x P(A |B);

(h) usethe rule of total probability.

The Fundamental Laws of Set Algebra

Commutative laws

A∪B=B∪AA∩B=B∩A

Associative Laws

(A∪B)∪C=A∪(B∪C)(A∩B)∩C=A∩(B∩C)

Distributive Laws

A∪(B∩C)=(A∪B)∩(A∪C)

A∩(B∪C)=(A∩B)∪(A∩C)Impotent Laws

A∩A=AA∪A=ADomination Laws

A∪U=UA∩∅=∅Absorption Laws

A∪(A∩B)=AA∩(A∪B)=AInverse Laws

A∪A′=UA∩A′=∅

Law of Complement

(A′)′=AU′=∅∅′=U

DeMorgan's Law

(A∪B)′=A′∩B′(A∩B)′=A′∪B′

Relative Complement of B in A

A−B=A∩B′

Chapter 15 Probability Distributions

15.1 Discrete random variables

Var(X)=E(X2)−[E(X)]2Var(X)=σ2x=∑[x2i∗P(xi)]−μ2x

(a) identify discrete random variables; (b) construct a probability distribution table for a discrete random variable; (c) use the probability function and cumulative distribution function of a discrete random

variable; (d) calculate the mean and variance of a discrete random variable;

Discrete random variables

A discrete variable is a variable which can only take a countable number of values.

A discrete random variable X is uniquely determined by

Its set of possible values X Its probability density function (pdf): A real-valued function f (·) defined for each x ∈ X as the probability that X has the value x.  Probability Density Function 

f(x)=Pr(X=x)

∑x=inf(xi)=1

Cumulative Distribution Function F(x) is defined to be

F(x)=P(X≤x)

Discrete Probability Distribution

The probability that random variable X will take the value xi is denoted by p(xi)  where P(xi)=P(X=xi)

Mean and Variance of a Discrete Random Variable

Mean  

E(X)=μx=∑[xi∗P(xi)]Variance

15.2 Continuous random variables

(e) identify continuous random variables; (f) relate the probability density function and cumulative distribution function of a

continuous random variable; (g) use the probability density function and cumulative distribution function of a

continuous random variable; (h) calculate the mean and variance of a continuous random variable;

Continuous Random Variables

A random variable X is continuous if its set of possible values is an entire interval of numbers

Probability Density Function

Let X be a continuous random variables. Then a probability distribution or probability density function (pdf) of X is a function f (x) such that for any two numbers a and b, 

P(a≤X≤b)=∫baf(x)dx

The graph of f is the density curve.

Area of the region between the graph of f and the x – axis is equal to 1.

The Cumulative Distribution Function

The cumulative distribution function, F(x) for a continuous random variables X is defined for every number x by

F(x)=P(X≤x)=∫x−∞f(y)dyP(a≤X≤b)=F(b)−F(a)

For each x, F(x) is the area under the density curve to the left of x. 

Expected Value 

The expected or mean value of a continuous random variables X with pdf f(x) is E(X)=μx=∫∞−∞x.f(x)dx

Variance 

The variance of continuous random variables X with pdf f(x) is σ2x=Var(X)=∫∞−∞(x−μ)2.f(x)dx

E[(x−μ)2]Var(X)=E(X2)−[E(x)]2

15.3 Binomial distribution

(i) use the probability function of a binomial distribution, and find its mean and variance; (j) use the binomial distribution as a model for solving problems related to science and

technology;

Binomial Distribution

The binomial distribution is used when there are exactly two mutually exclusive outcomes of a trial. These outcomes are appropriately labeled "success" and "failure".

If the random variable X follows the binomial distribution with parameters n and p, X ~ (B(n, P), the probability of getting exactly k successes in n trials is given by the probability mass function

f(k;n,p)=Pr(X=k)=(nk)pk(1−p)n−kIf X ~ B(n, p), X is a binomially distributed random variable, then the expected value of X is

Mean = npand the variance is

Variance = np(1−p)

n number of successesp is the probability of success in Binomial Distribution, assumes that p is fixed for all trials.

15.4 Poisson distribution

(k) use the probability function of a Poisson distribution, and identify its mean and variance;

(l) use the Poisson distribution as a model for solving problems related to science and technology;

Poisson Distribution

Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate and independently of the time since the last event.

A discrete stochastic variable X is said to have a Poisson distribution with parameter λ > 0, if for k = 0, 1, 2, ... the probability mass function of X is given by

f(k;λ)=Pr(X=k)=λke−λk!

K is the number of occurrences of an event; the probability of which is given by the function λ is a positive real number.

Mean and Variance

The expected value and variance of a Poisson-distributed random variable is equal to λ.

15.5 Normal distribution

(m) identify the general features of a normal distribution, in relation to its mean and standard deviation;

(n) standardise a normal random variable and use the normal distribution tables; (o) use the normal distribution as a model for solving problems related to science and

technology; (p) use the normal distribution, with continuity correction, as an approximation to the

binomial distribution, where appropriate.

Normal distribution

The normal distribution is a continuous probability distribution, defined by the formula f(x)=1σ2π−−√e−(x−μ)22σ2

The normal distribution is also often denoted

X∼ N(μ,σ2)

Standard Normal Distribution

If μ = 0 and σ = 1, the distribution is called the standard normal distribution.

Formula for z-score: z=x−μσ

z is the "z-score" (Standard Score) x is the value to be standardized μ is the mean σ is the standard deviation

Normal Approximations

Binomial Approximation

The normal distribution can be used as an approximation to the binomial distribution, under certain circumstances, namely:

If X ~ B(n, p) and if n is large and/or p is close to ½, then X is approximately N~(np, npq)

Poisson Approximation

The normal distribution can also be used to approximate the Poisson distribution for large values of λ  (the mean of the Poisson distribution).

If X ~ Po(λ) then for large values of l, X ~ N(λ, λ) approximately.

Chapter 16

Sampling and Estimation

16.1 Sampling

(a) distinguish between a population and a sample, and between a parameter and a statistic;

(b) identify a random sample; (c) identify the sampling distribution of a statistic;  (d) determine the mean and standard deviation of the sample mean; (e) use the result that  X has a normal distribution if X has a normal distribution; (f) use the central limit theorem; (g) determine the mean and standard deviation of the sample proportion; (h) use the approximate normality of the sample proportion for a sufficiently large sample

size;

Sampling

Sampling

Sampling is concerned with the selection of a subset of individuals from within a statistical population to estimate characteristics of the whole population.

Population and Sample

A population includes each element from the set of observations that can be made.

A sample consists only of observations drawn from the population.

Depending on the sampling method, a sample can have fewer observations than the population, the same number of observations, or more observations. More than one sample can be derived from the same population.

Parameter and Statistic

Aa measurable characteristic of a population, such as a mean or standard deviation, is called a parameter; but a measurable characteristic of a sample is called a statistic.

Random Sample

A simple random sample is a subset of a sample chosen from aa population. Each individual is chosen randomly and entirely by equal chance (same probability of being chosen at any stage during the sampling process).

Central Limit Theorem

The central limit theorem states that even if a population distribution is strongly non-normal, its sampling distribution of means will be approximately normal for large sample sizes (n over 30). The central limit theorem makes it possible to use probabilities associated with the normal curve to answer questions about the means of sufficiently large samples.

Mean and Variance of Sampling Distribution

According to the central limit theorem, the mean of a sampling distribution of means is an unbiased estimator of the population mean

μx¯=μSimilarly, the standard deviation of a sampling distribution of means is

σx¯=σn√

The larger the sample, the less variable the sample mean

Example: 

The population has a mean of 12 and a standard deviation of 4. The sample size of a sampling distribution is N=20. What is the mean  and standard deviation of the sampling distribution?

μx¯=12σx¯=420−−√=0.8944

Mean and Variance of Sample Proportion

Expected value of a sample proportion μp^=p

Standard deviation of a sample proportionσp^=p(1−p)n−−−−−−−√

Example: 

Suppose the true value of the president's approval rating is 53%. What is the mean  and standard deviation of the sample proportion with sample of 800 people?

μp^=0.53σp^=0.53(1−0.53)800−−−−−−−−−−−−√=0.0176

16.2 Estimation

(i) calculate unbiased estimates for the population mean and population variance; (j) calculate an unbiased estimate for the population proportion; (k) determine and interpret a confidence interval for the population mean based on a

sample from a normally distributed population with known variance; (l) determine and interpret a confidence interval for the population mean based on a large

sample; (m) find the sample size for the estimation of population mean; (n) determine and interpret a confidence interval for the population proportion based on a

large 

sample; (o) find the sample size for the estimation of population proportion.

Unbiased Estimator

Unbiased Estimator of Population Mean

If the mean value of an estimator equals the true value of the quantity it estimates, then the estimator is called an unbiased estimator.  Using the Central Limit Theorem, the mean value of the sample means equals the population mean. Therefore, the sample mean is an unbiased estimator of the population mean.

Unbiased Estimator of Population Variance

This makes the sample variance an unbiased estimator for the population variance.

s2=∑ni=1(xi−x¯)2n−1

Although using (n-1)  as the denominator makes the sample variance, an unbiased estimator of the population variance,the sample standard deviation, , still remains a biased estimator of the population standard deviation. For large sample sizes this bias is negligible.

Confidence Interval

A confidence interval (CI) is a type of interval estimate of a population parameter and is used to indicate the reliability of an estimate.

α: between 0 and 1A confidence level: 1 - α or 100(1 - α)%. E.g. 95%. This is the proportion of times that the confidence interval actually does contain the population parameter, assuming that the estimation process is repeated a large number of times.

The central limit theorem states that when the sample size is large, approximately 95% of the sample means will fall within 1.96 standard errors of the population mean,

μ±1.96(σn√)Stated another way

X¯−1.96(σn√)<μ<X¯+1.96(σn√)

Example:

A survey which estimate the average age of the students presently enrolled. From past studies, the standard deviation is known to be 3 years. A sample of 40 students is selected, and the mean is found to be 25 years. Find the 95% confidence interval of the population mean.

X¯−1.96(σn√)<μ<X¯+1.96(σn√)25−1.96(340−−√)<μ<25+1.96(340−−√)

25−0.93<μ<25+0.9324.07<μ<25.93

Confidence Interval for the Population Proportion

The confidence interval for a proportion isp±zα/2σp

A 1-α confidence interval to population proportion.

p^±zα/2(p^(1−p^)n)−−−−−−−−−−⎷

Chapter 17

Hypothesis Testing

(a) explain the meaning of a null hypothesis and an alternative hypothesis;

(b) explain the meaning of the significance level of a test; (c) carry out a hypothesis test concerning the population mean for a normally distributed

population with known variance; (d) carry out a hypothesis test concerning the population mean in the case where a large

sample is used; (e) carry out a hypothesis test concerning the population proportion by direct evaluation

of binomial probabilities; (f) carry out a hypothesis test concerning the population proportion using a normal

approximation.

Chapter 18

Chi-squared Tests

(a) identify the shape, as well as the mean and variance, of a chi-squared distribution with a given number of degrees of freedom;

(b) use the chi-squared distribution tables; (c) identify the chi-squared statistic; (d) use the result that classes with small expected frequencies should be combined in a

chisquared test; (e) carry out goodness-of-fit tests to fit prescribed probabilities and probability

distributions with known parameters; (f) carry out tests of independence in contingency tables (excluding Yates correction).