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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−4
Draw 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)=x
3y−4=x
y=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>5
1.2 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+1
f(x)=2x3−4x2+5x+11
f(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−1
f(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 Theorem
If 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−3
f(−3)=(−3)3+5(−3)2+5(−3)−3
f(−3)=0
By factor theorem (x+3) is factor of x3+5x2+5x-3
Remainder Theorem
If 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≥0
x2+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)=0
x=−2,x=−1
The fraction will be undefined wherever the denominator is zero
(x−4)(x+4)=0
x=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≥0
x2+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)=0
x=−2,x=−1
The fraction will be undefined wherever the denominator is zero
(x−4)(x+4)=0
x=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)B
If x=-6
11(−6)+21=[2(−6)−3]B
B=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. Indices
an,a≠0
a0=1,a≠0
a−n=1/an
2. Law of Indices
am×an=am+n
am÷ an=am−n
(am)n=am×n
am×bm=(a×b)m
am÷bm=(ab)m
3. Logarithms
Ifa=bc,then logba=c
loga1=0
logaa=1
alogab=1
4. Law of Logarithms
logaxy= logax+logay
loga(xy)= logax−logay
logaxn= nlogax
5 Change of base of Logarithms
logab=1logba
logab=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≥0
x2+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)=0
x=−2,x=−1
The fraction will be undefined wherever the denominator is zero
(x−4)(x+4)=0
x=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+cosAsinB
sin(A−B)=sinAcosB−cosAsinB
cos(A+B)=cosAcosB−sinAsinB
cos(A−B)=cosAcosB+sinAsinB
tan(A+B)=tanA+tanB1−tanAtanB
tan(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 follow
Rsin(θ+α)≡R(sinθcosα+cosθsinα)
Rsin(θ+α)≡Rsinθcosα +Rcosθsinα
Apply Trigonometric Identities as below
sin(A+B)=sinAcosB+cosAsinB
So
asinθ+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
Example
Express 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+Rsinθsinα
asinθ+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....,1n
Limit=0
1,12, 23,34,45,56....,nn+1
Limit=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−1
The sum of the first n terms of a geometric series is:
a+ar+ar2+ar3+.....+arn−1=∑k=0n−1ark=a1−rn1−r
Derive the sum of the first n terms of a geometric series as follows:
Let
s=a+ar+ar2+ar3+.....+arn−1
rs=ar+ar2+ar3+.....+arn
s−rs=a−arn
s(1−r)=a(1−rn)
So
s=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−r
When a=1
1+r+r2+r3+....=11−r
Let
s=1+r+r2+r3+.....
rs=r+r2+r3+.....
s−rs=1
s(1−r)=1
So
s=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∈Z
Binomial 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 writtenas
(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
Accordingto the ratiotest forseriesconvergence aseriesconvergeswhen:
if limk→∞(ak+1ak)<1, series converges
if limk→∞(ak+1ak)>1, series diverges
iflimk→∞(ak+1ak)=1, result is indeterminate
The Binomial Theoremconvergeswhen|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=−2
B−1=∣∣∣∣1/2003/2−10−1161∣∣∣∣
(d) use the properties of matrix operations;
Properties of Matrices
AB≠BA
A+B=B+A
A(B+C)=AB+AC
(A+B)C=AC+BC
A(BC)=(AB)C
AI=IA=A
A0=I
Scalar Multiplication
λ(AB)=(λA)B
(AB)λ=A(Bλ)
Transpose
(AB)T=BTAT
Properties of Inverse Matrix
AA−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 operations
4shirts +2pants +2pairofshoes=$190
3shirts +4pants +3pairofshoes=$295
2shirts +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=I
A−1A⎡⎣xyz⎤⎦=A−1⎡⎣190295250⎤⎦
⎡⎣xyz⎤⎦=⎡⎣0.50−0.5−0.5 −0.5 1.50.250.75−1.25⎤⎦⎡⎣190295250⎤⎦
⎡⎣xyz⎤⎦=⎡⎣104035⎤⎦
Alternative,
a1x+b1y+c1z=d1
a2x+b2y+c2z=d2
a3x+b3y+c3z=d3
Mode > EQN > Unknowns=3, input data below
a1=4,b1=2,c1=2,d1=190
a2=3,b2=4,c2=3,d2=295
a3=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∣∣∣∣=0
2. |A| = 0 if all elements of a line are zero
∣∣∣∣303202303∣∣∣∣=0
3. |A| = 0 if the elements of a line are a linear combination of the others. (row 3 = row 1 +
row 2)
∣∣∣∣213325246∣∣∣∣=0
4. 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=6
6. 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∣∣∣∣=5
3.2 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 form
Ax=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 form
Given 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=−100
20y+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=d1
a2x+b2y+c2z=d2
a3x+b3y+c3z=d3
Mode > EQN > Unknowns=3, input data below
a1=20,b1=0,c1=−10,d1=−100
a2=0,b2=20,c2=10,d2=300
a3=−10,b3=10,c3=20,d3=200
x=−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 i
i=−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+bi
Modulus 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, z
tanΘ=ba
tanΘ=34
Θ=tan−134
arg z=0.644 rad
How to change theta=36.870 to radian
rad=Θ∗π1800
rad=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+2i
Z2=−4+i
Conjugate of
Z∗1=3−2i
Z∗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√=2
sinθ=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 sectionisthe intersectionof aplane anda double rightcircularcone. By changingthe
angle andlocationof the intersection,we canproduce differenttypesof conics,suchas circles,
ellipses,hyperbolasandparabolas.
Equation of a Circle in Standard Form
(x−h)2+(y−k)2=r2
Equation of an Ellipse inStandard Form
(x−h)2a2+(y−k)2b2=1
Equation of a Hyperbolas in Standard Form
(x−h)2a2−(y−k)2b2=1
Equation of a Parabolas inStandard 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
Example
16x2+25y2=400
16x2400+25y2400=1
x225+y216=1
x252+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) axis
The 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−b2
c2=52−42
c2=±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 directionvector isa vectorwhichhaslengthof 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.
PointA has the positionvector a, if
A=(35)
PointB has the positionvector 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
2u⃗ =2[23]=[46]
12u⃗ =12[23]=[13/2]
−3u⃗ =−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|cosθ
a.b=10∗12cos59.5∘
or
a.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+ayby
a.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=0.397
θ=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 b
θ is the angle between a and b
n is the unit vector at right angles to both a and b
Vector Product, C,
a⃗ ∗b⃗ =∣∣∣∣ia1b1ja2b2 ka3b3∣∣∣∣
a⃗ ∗b⃗ =∣∣∣a2b2a3b3∣∣∣i−∣∣∣a1b1a3b3∣∣∣j+∣∣∣a1b1a2b2∣∣∣k
What is the vector product of a = (2,3,4) and b = (5,6,7)
a⃗ ∗b⃗ =∣∣∣∣i25j36k47∣∣∣∣
a⃗ ∗b⃗ =∣∣∣3647∣∣∣i−∣∣∣2547∣∣∣j+∣∣∣2536∣∣∣k
=−3i+6j−3k
u⃗ =⟨−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 equationof the line throughpointsA andB isgivenby
r=OA+λAB
or
r=a+λt, t=b−a
Example:Compute the vectorequationof astraightline throughpointA and B withposition
vector:
a=⎛⎝2−13⎞⎠,b=⎛⎝13−2⎞⎠,r=⎛⎝xyz⎞⎠
b−a=⎛⎝13−2⎞⎠−⎛⎝2−13⎞⎠=⎛⎝−14−5⎞⎠
r=a+λt
r=⎛⎝2−13⎞⎠+λ⎛⎝−14−5⎞⎠
Cartesian Equationsof Lines
a=⎛⎝axayaz⎞⎠,b=⎛⎝⎜bxbybz⎞⎠⎟,r=⎛⎝xyz⎞⎠
CartesianEquationwill be
x−axbx−ax=y−ayby−ay=z−azbz−az
Example:Compute the cartesianequationof astraightline throughpointA andB withposition
vector:
a=⎛⎝1−32⎞⎠,b=⎛⎝3−15⎞⎠,r=⎛⎝xyz⎞⎠
x−13−1=y−(−3)−1−(−3)=z−25−2
x−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+μv
r=⎛⎝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−α
tanθ=tan(β−α)
=tanβ−tanα1+tanβtanα
=m1−m21+m1m2
For twoline of gradientm1,m2te acute angle betweenthemisalwayspositive
tanθ=∣∣∣m1−m21+m1m2∣∣∣
m1m2 ≠ -1, thisformuladoesn'tworkforperpendicular lines.
Example 1: Findthe acute angle betweenthe linesy= 3x - 1 and y = -2x + 3.
tanθ=∣∣∣m1−m21+m1m2∣∣∣
tanθ=∣∣∣3−(−2)1+(3)(−2)∣∣∣
tanθ=|−1|=1
θ=tan−1(1)=45∘
Example 2: Findthe acute angle betweenthe lines6x - y + 8 = 0 and -3x -11y +10 = 0
Rearrange the equation
y=6x+8
y=−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 pointof intersectinglines,the pointsare equal.
Example:Findthe pointof intersectionbetweenlinesy=3x - 7 and y = -2x+3.
y=3x−7−−−−−(1)
y=−2x+3−−−−−(1)
Substitute (1) into(2)
3x−7=−2x+3
5x=10
x=2
x=2,
y=3(2)−7
y=−1
 Hence,the intersectingpointis(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+23k
Vector 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=12
4y−3z=6
y=−18,z=−26
Point 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)=12
3(3)+4(−3)−3(−3)=6
If y=0
2x+3z=12
3x−3z=6
x=3.6,z=1.6
Point 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)=12
3(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 limitisthe value thata functionorsequence "approaches"asthe inputorindex approaches
some value.
The limitof f(x) asx approachesa fromthe right.
limx→a+f(x)
The limitof f(x) asx approachesa fromthe left.
limx→a−f(x)
Example:Find
limx→2−(x3−1)
limx→2−(x3−1)=limx→2−(23−1)=7
As x approaches2 fromthe left,x3
pproaches8andx3
-1 approaches7.
Find
limx→2+(x3−1)
limx→2+(x3−1)=limx→2+(23−1)=7
 (b) use the properties of limits;
Properties of limits
The limitof a constant isthe constant itself.
limx→ak=k
The limitof a functionmultipliedbyaconstant isequal to the value of the functionmultipliedby
the constant.
limx→ak.f(x)=k.limx→af(x)
The limitof a sum (or difference) of the functionsisthe sum(ordifference)of the limitsof the
individualfunctions.
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 limitof a product isthe productof the limits.
limx→af(x).g(x)=limx→af(x).limx→ag(x)
The limitof a quotientisthe quotientof the limits
limx→af(x)g(x)=limx→af(x)limx→ag(x)
The limitof a poweris the powerof 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) if
limx→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 functionf(x),itsderivativeisdefinedas
f′(x)=lim△x→0f(x+△x)−f(x)△x
Example 1: Compute the derivative of f(x)byusinglimitdefinition
f(x)=13x−25
f′(x)=lim△x→0f(x+△x)−f(x)△x
f′(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)byusinglimitdefinition
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 ofxn (n∈Q), ex , ln x, sin x, cos x, tan x, sin-1x, cos-1x, tan-1x, with
constant multiples, sums, differences, products, quotients and composites;
Common Derivatives
Table of Derivatives
ddx(x)=1
ddx(xn)=nxn−1
ddx(ex)=ex
ddx(lnx)=1x,x>0
ddx(sinx)=cosx
ddx(cosx)=−sinx
ddx(tanx)=sec2x
ddx(sin−1x)=11−x2−−−−−√
ddx(cos−1x)=−11−x2−−−−−√
ddx(secx)=secxtanx
ddx(cscx)=−cscxcotx
ddx(cotx)=−csc2x
(c) performimplicit differentiation;
Explicit and Implicit Differentiation
There are two waysto define functions, implicitlyandexplicitly.Mostof the equationswe have
dealtwithhave beenexplicitequations,suchasy = 3x-2. Thisis called explicit because given an
x, you can directly get f(x).
The technique of implicitdifferentiation allowsyoutofindthe derivativeof ywithrespecttox
withouthavingtosolve the givenequationfory.Given
x2+y2=10
Performingachainrule to get dy/dx,solve dy/dx intermsof x and y.
Finddy/dx for
x2+y2=10
2x+2ydydx=0
2ydydx=−2x
dydx=−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.dtdx
x=f(t),y=g(t)
Example: Find the first derivative, given
x=t+cost
y=sint
dydx=dydtdxdt=cost1−sint
Example: Find the first derivative, given
x=t4−4t2
y=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 stationarypointisan inputtoa functionwhere the derivative iszero. Atstationarypoints,
dydx=0
Extremum Points
Maximaand minimaare pointswhere afunctionreachesahighestor lowestvalue,
respectively
(g)
(h)
(i)
SecondDerivative
If f'''(x) is positive,thenitisa minimumpoint.
If f'''(x) is negative,thenitisa maximum point.
If f'''(x)= zero,thenitcouldbe a maximum, minimumorpointof inflexion.
Point of Inflexion
An 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 isan inflexionpoint
(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)
where
m1=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=t2
y=2t+1
dxdt=2t
dydt=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+2
Equation of normal
m1m2=−1
m2=−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
quantitytootherquantitieswhose ratesof change are known. The rate of change is usually
with respect to time.
Example: Air isbeingpumpedintoaspherical balloonsuchthatitsradiusincreasesata
rate of .80 cm/min.Findthe rate of change of itsvolume whenthe radiusis5 cm.
The volume ( V) of a sphere withradiusris
V=43πr3
Differentiatingabove equationwithrespecttot
dVdt=43π.3r2.drdt
dVdt=4πr2.drdt
The rate of change of the radiusdr/dt= 0.80 cm/minbecause the radiusisincreasingwith
respectto time.
dVdt=4π(5)2(0.80)
dVdt=80π cm3/min
Hence,the volume isincreasingatarate of 80π cm3
/minwhenthe radiushasa lengthof 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 xn (n∈Q), ex , sin x, cos x, sec2x, 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+3
1=A(x+3)+B(x+2)
x=−2,A=1
x=−3,B=−1
1(x+2)(x+3)=1x+2−1x+3
Integrate 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=16
x=−1−2C=1,C=−12
x=−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=asinθ
For a2+x2−−−−−−√, use x=atanθ
For x2−a2−−−−−−√, use x=asecθ
Basic Trigonometric Identities
csc2θ=1sin2θ
sec2θ=1cos2θ
cot2θ=1tan2θ
Pythagorean Identities
cos2θ+sin2θ=1
csc2θ=1+cot2θ
sec2θ=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 seriesis a representationof afunctionasan infinite sumof termsthatare calculated
fromthe valuesof the function'sderivativesata 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 seriesisthe Taylorseriesof the functionabout x=0.
f(x)=∑x=0∞fn(0)n!xn
f(x)=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3+...+f(n)(0)n!xn+...
Example : Findthe MaclaurinSeriesexpansionof 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 convergence
Finding interval of convergence
1) 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)n1n
The 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 as
x4+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.
Class
0−5
6−10
11−15
16−20
Frequency
10
20
17
4
Ungrouped Data
Data that has not been organized into groups
10 1 2 4 80 45 67 78
50 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 8
7 | 3 3 5 6
8 | 1 5 8
10 | 7
key 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 table
The 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 observation
Q1=(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=12
Q2 == 1/2 (11+1) = 6th observation, Q2=22
Q3 == 3/4 (11+1) = 9th observation, Q3=38
Interquartile Range = Q3 -Q1 = 38 - 12 = 26
Range = 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)=13.5
Q2=(22+302)=26
Q3=(38+422)=40
Interquartile Range = Q3 -Q1 = 40 - 13.5 = 26.5
Range = 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
median
f = the frequency of the class median
fm= the lower boundary of the class median
C= the class width
Q1=LQ1+(n4−FfQ1)C
Q3=LQ3+⎛⎝3n4−FfQ3⎞⎠C
Example : Find the median and interquartile range of below data.
Q2=20.5+⎛⎝502−2312⎞⎠10=22.167
Q1=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 Data
Mean
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 = 1.07
Coefficient of skewness,
Sk=3−41.07=−0.93
Therefore, 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 6
P(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 permutationisthe choice of r thingsfrom a setof n thingswithoutreplacementandwhere
the order matters
PermutationFormula
nPr=n!(n−r)!
Combinations
A combinationisthe choice of r thingsfroma setof n thingswithoutreplacementandwhere
orderdoesNOT matter.
CombinationFormula
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 complementof anyeventA,A'is the eventthatA doesnotoccur.
Exhaustive Event
A setof eventsiscollectively exhaustive if at least one of the events must occur. For example,
whenrollingasix-sideddie,the outcomes1,2, 3, 4, 5, and 6 are collectivelyexhaustive,because
they include the entire range of possible outcomes. Thus, all sample spaces are collectively
exhaustive.
P(A∪B)=1
MutuallyExclusive 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)=0
P(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∪A
A∩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)
ImpotentLaws
A∩A=A
A∪A=A
Domination Laws
A∪U=U
A∩∅=∅
AbsorptionLaws
A∪(A∩B)=A
A∩(A∪B)=A
Inverse Laws
A∪A′=U
A∩A′=∅
 Law of Complement
(A′)′=A
U′=∅
∅′=U
 DeMorgan'sLaw
(A∪B)′=A′∩B′
(A∩B)′=A′∪B′
 Relative ComplementofB in A
A−B=A∩B′
Chapter 15 Probability Distributions
15.1 Discrete random variables
Var(X)=E(X2)−[E(X)]2
Var(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)
Meanand 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)dy
P(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−k
If X ~ B(n, p), X is a binomially distributed random variable, then the expected value of
X is
Mean = np
and the variance is
Variance = np(1−p)
n number of successes
p 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
givennumberof eventsoccurring 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 stochasticvariable Xissaidtohave a Poissondistributionwithparameterλ> 0, if for
k = 0, 1, 2, ...the probabilitymassfunctionof Xisgivenby
f(k;λ)=Pr(X=k)=λke−λk!
K isthe numberof occurrencesof an event;the probability of whichisgivenbythe function
λ is a positive real number.
Meanand 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
Samplingis concernedwiththe selectionof asubsetof individualsfromwithinastatistical
populationtoestimate characteristicsof 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.
Dependingonthe samplingmethod,asample canhave fewerobservationsthanthe population,
the same numberof observations,ormore observations.More thanone 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 randomsample isa subsetof a sample chosenfromaa population.Eachindividualis
chosenrandomlyandentirelybyequal chance (same probabilityof beingchosenatanystage
duringthe samplingprocess).
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.
Meanand 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¯=μ
Similarly, 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¯=12
σx¯=420−−√=0.8944
Meanand Variance of Sample Proportion
Expected value of a sample proportion
μp^=p
Standard deviation of a sample proportion
σp^=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^=0.53
σp^=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 1
A 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+0.93
24.07<μ<25.93
Confidence Interval for the Population Proportion
The confidence interval for a proportion is
p±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).

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237654933 mathematics-t-form-6

  • 1. Get Homework/Assignment Done Homeworkping.com Homework Help https://www.homeworkping.com/ Research Paper help https://www.homeworkping.com/ Online Tutoring https://www.homeworkping.com/ click here for freelancing tutoring sites 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
  • 2. {(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
  • 3. 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
  • 4. 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−4 Draw the graph of f(x)=3x-4
  • 5. 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)=x 3y−4=x y=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
  • 6. f(x)={2x−1,x<1x2−3,x≤1 Example 2 f(x)=⎧⎩⎨⎪⎪3x−2, x<−3 x2, −3≤x≤52x+5, x>5 1.2 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+1 f(x)=2x3−4x2+5x+11 f(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−1 f(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
  • 7. Graph of Rational Functions of f(x).  (d) use the factor theorem and the remainder theorem; The Factor Theorem If 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−3 f(−3)=(−3)3+5(−3)2+5(−3)−3 f(−3)=0 By factor theorem (x+3) is factor of x3+5x2+5x-3 Remainder Theorem If 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)
  • 8. 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≥0 x2+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)=0 x=−2,x=−1
  • 9. The fraction will be undefined wherever the denominator is zero (x−4)(x+4)=0 x=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≥0 x2+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)=0 x=−2,x=−1 The fraction will be undefined wherever the denominator is zero
  • 10. (x−4)(x+4)=0 x=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
  • 11. (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)B If x=-6 11(−6)+21=[2(−6)−3]B B=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
  • 12. 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
  • 13. 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
  • 14. 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. Indices an,a≠0 a0=1,a≠0 a−n=1/an
  • 15. 2. Law of Indices am×an=am+n am÷ an=am−n (am)n=am×n am×bm=(a×b)m am÷bm=(ab)m 3. Logarithms Ifa=bc,then logba=c loga1=0 logaa=1 alogab=1 4. Law of Logarithms logaxy= logax+logay loga(xy)= logax−logay logaxn= nlogax 5 Change of base of Logarithms logab=1logba logab=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≥0 x2+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)=0 x=−2,x=−1 The fraction will be undefined wherever the denominator is zero
  • 16. (x−4)(x+4)=0 x=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+cosAsinB sin(A−B)=sinAcosB−cosAsinB cos(A+B)=cosAcosB−sinAsinB cos(A−B)=cosAcosB+sinAsinB tan(A+B)=tanA+tanB1−tanAtanB tan(A−B)=tanA−tanB1+tanAtanB
  • 17. 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 follow Rsin(θ+α)≡R(sinθcosα+cosθsinα) Rsin(θ+α)≡Rsinθcosα +Rcosθsinα Apply Trigonometric Identities as below sin(A+B)=sinAcosB+cosAsinB So asinθ+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
  • 18. asinθ+bcosθ≡Rsin(θ+α) Similar to minus case, asinθ−bcosθ≡Rsin(θ−α) Express Rsin(θ−α)≡Rcosαsinθ−Rsinαcosθ Example Express 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+Rsinθsinα asinθ+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)
  • 19. 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
  • 20. 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....,1n Limit=0 1,12, 23,34,45,56....,nn+1 Limit=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;
  • 21. 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−1 The sum of the first n terms of a geometric series is: a+ar+ar2+ar3+.....+arn−1=∑k=0n−1ark=a1−rn1−r Derive the sum of the first n terms of a geometric series as follows: Let s=a+ar+ar2+ar3+.....+arn−1 rs=ar+ar2+ar3+.....+arn s−rs=a−arn s(1−r)=a(1−rn) So s=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−r When a=1 1+r+r2+r3+....=11−r Let s=1+r+r2+r3+..... rs=r+r2+r3+..... s−rs=1 s(1−r)=1 So s=11−r
  • 22. 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∈Z Binomial 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 writtenas (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
  • 23. Accordingto the ratiotest forseriesconvergence aseriesconvergeswhen: if limk→∞(ak+1ak)<1, series converges if limk→∞(ak+1ak)>1, series diverges iflimk→∞(ak+1ak)=1, result is indeterminate The Binomial Theoremconvergeswhen|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.
  • 24. 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∣∣∣∣
  • 25. Lower Triangular Matrix B=∣∣∣∣2003−10461∣∣∣∣=2.−1.1=−2 B−1=∣∣∣∣1/2003/2−10−1161∣∣∣∣ (d) use the properties of matrix operations; Properties of Matrices AB≠BA A+B=B+A A(B+C)=AB+AC (A+B)C=AC+BC A(BC)=(AB)C AI=IA=A A0=I Scalar Multiplication λ(AB)=(λA)B (AB)λ=A(Bλ) Transpose (AB)T=BTAT Properties of Inverse Matrix AA−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 operations 4shirts +2pants +2pairofshoes=$190 3shirts +4pants +3pairofshoes=$295 2shirts +4pants +2pairofshoes=$190
  • 26. 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=I A−1A⎡⎣xyz⎤⎦=A−1⎡⎣190295250⎤⎦ ⎡⎣xyz⎤⎦=⎡⎣0.50−0.5−0.5 −0.5 1.50.250.75−1.25⎤⎦⎡⎣190295250⎤⎦ ⎡⎣xyz⎤⎦=⎡⎣104035⎤⎦ Alternative, a1x+b1y+c1z=d1 a2x+b2y+c2z=d2 a3x+b3y+c3z=d3 Mode > EQN > Unknowns=3, input data below a1=4,b1=2,c1=2,d1=190 a2=3,b2=4,c2=3,d2=295 a3=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∥∥∥
  • 27. 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∣∣∣∣=0 2. |A| = 0 if all elements of a line are zero ∣∣∣∣303202303∣∣∣∣=0 3. |A| = 0 if the elements of a line are a linear combination of the others. (row 3 = row 1 + row 2) ∣∣∣∣213325246∣∣∣∣=0 4. 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=6 6. 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∣∣∣∣=5 3.2 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;
  • 28. (i) apply the Gaussian elimination to solve a system of linear equations; Gaussian elimination is a method for solving matrix equations of the form Ax=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 form Given 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=−100 20y+10z=300 −10x+10y+20z=200 ⎡⎣200−1002010−101020⎤⎦⎡⎣xyz⎤⎦=⎡⎣−100300200⎤⎦ 10⎡⎣20−1021 −112⎤⎦⎡⎣xyz⎤⎦=⎡⎣−100300200⎤⎦
  • 29. ⎡⎣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=d1 a2x+b2y+c2z=d2 a3x+b3y+c3z=d3 Mode > EQN > Unknowns=3, input data below a1=20,b1=0,c1=−10,d1=−100 a2=0,b2=20,c2=10,d2=300 a3=−10,b3=10,c3=20,d3=200 x=−5,y=15,z=0
  • 30. 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 i i=−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−−−−−−√
  • 31. (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+bi Modulus 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, z tanΘ=ba tanΘ=34 Θ=tan−134 arg z=0.644 rad How to change theta=36.870 to radian rad=Θ∗π1800 rad=Θ∗π1800 36.870∗π1800=0.644 rad
  • 32. (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+2i Z2=−4+i Conjugate of Z∗1=3−2i Z∗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;
  • 33. (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.
  • 35. 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 sectionisthe intersectionof aplane anda double rightcircularcone. By changingthe angle andlocationof the intersection,we canproduce differenttypesof conics,suchas circles, ellipses,hyperbolasandparabolas. Equation of a Circle in Standard Form (x−h)2+(y−k)2=r2 Equation of an Ellipse inStandard Form (x−h)2a2+(y−k)2b2=1 Equation of a Hyperbolas in Standard Form (x−h)2a2−(y−k)2b2=1 Equation of a Parabolas inStandard 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
  • 36. 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 Example
  • 37. 16x2+25y2=400 16x2400+25y2400=1 x225+y216=1 x252+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) axis The 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−b2 c2=52−42 c2=±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.
  • 38. 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 directionvector isa vectorwhichhaslengthof 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. PointA has the positionvector a, if A=(35) PointB has the positionvector b, if B=(62)
  • 39. (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 2u⃗ =2[23]=[46] 12u⃗ =12[23]=[13/2] −3u⃗ =−3[23]=[−6−9]
  • 40. (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|cosθ a.b=10∗12cos59.5∘ or a.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+ayby a.b=|a||b|cosθ a.b=(2)(1)+(3)(−2)+(4)(3)=8
  • 41. |a|=22+32+42−−−−−−−−−−√=29−−√ |a|=12+(−2)2+32−−−−−−−−−−−−√=14−−√ 8=29−−√14−−√cosθ cosθ=0.397 θ=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 b θ is the angle between a and b n is the unit vector at right angles to both a and b Vector Product, C, a⃗ ∗b⃗ =∣∣∣∣ia1b1ja2b2 ka3b3∣∣∣∣ a⃗ ∗b⃗ =∣∣∣a2b2a3b3∣∣∣i−∣∣∣a1b1a3b3∣∣∣j+∣∣∣a1b1a2b2∣∣∣k What is the vector product of a = (2,3,4) and b = (5,6,7) a⃗ ∗b⃗ =∣∣∣∣i25j36k47∣∣∣∣ a⃗ ∗b⃗ =∣∣∣3647∣∣∣i−∣∣∣2547∣∣∣j+∣∣∣2536∣∣∣k =−3i+6j−3k u⃗ =⟨−3,6,−3⟩ Vector product, c = a x b = (-3, 6, -3)
  • 42. 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 equationof the line throughpointsA andB isgivenby r=OA+λAB or r=a+λt, t=b−a Example:Compute the vectorequationof astraightline throughpointA and B withposition vector: a=⎛⎝2−13⎞⎠,b=⎛⎝13−2⎞⎠,r=⎛⎝xyz⎞⎠ b−a=⎛⎝13−2⎞⎠−⎛⎝2−13⎞⎠=⎛⎝−14−5⎞⎠ r=a+λt r=⎛⎝2−13⎞⎠+λ⎛⎝−14−5⎞⎠ Cartesian Equationsof Lines a=⎛⎝axayaz⎞⎠,b=⎛⎝⎜bxbybz⎞⎠⎟,r=⎛⎝xyz⎞⎠ CartesianEquationwill be x−axbx−ax=y−ayby−ay=z−azbz−az Example:Compute the cartesianequationof astraightline throughpointA andB withposition vector: a=⎛⎝1−32⎞⎠,b=⎛⎝3−15⎞⎠,r=⎛⎝xyz⎞⎠ x−13−1=y−(−3)−1−(−3)=z−25−2 x−12=y+32=z−23
  • 43. (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+μv r=⎛⎝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
  • 44. θ=β−α tanθ=tan(β−α) =tanβ−tanα1+tanβtanα =m1−m21+m1m2 For twoline of gradientm1,m2te acute angle betweenthemisalwayspositive tanθ=∣∣∣m1−m21+m1m2∣∣∣ m1m2 ≠ -1, thisformuladoesn'tworkforperpendicular lines. Example 1: Findthe acute angle betweenthe linesy= 3x - 1 and y = -2x + 3. tanθ=∣∣∣m1−m21+m1m2∣∣∣ tanθ=∣∣∣3−(−2)1+(3)(−2)∣∣∣ tanθ=|−1|=1 θ=tan−1(1)=45∘ Example 2: Findthe acute angle betweenthe lines6x - y + 8 = 0 and -3x -11y +10 = 0 Rearrange the equation y=6x+8 y=−311x−1011 m1=6, m2=-3/11 tanθ=∣∣∣m1−m21+m1m2∣∣∣ tanθ=∣∣∣∣6−(−311)1+(6)(−311)∣∣∣∣ tanθ=∣∣∣−697∣∣∣ θ=tan−1697=84.2∘
  • 45. (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 pointof intersectinglines,the pointsare equal. Example:Findthe pointof intersectionbetweenlinesy=3x - 7 and y = -2x+3. y=3x−7−−−−−(1) y=−2x+3−−−−−(1) Substitute (1) into(2) 3x−7=−2x+3 5x=10 x=2 x=2, y=3(2)−7 y=−1  Hence,the intersectingpointis(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>
  • 46. 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+23k Vector 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=12 4y−3z=6 y=−18,z=−26 Point 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)=12 3(3)+4(−3)−3(−3)=6 If y=0 2x+3z=12 3x−3z=6
  • 47. x=3.6,z=1.6 Point 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)=12 3(6.6)+4(15)−3(24.6)=6
  • 48. 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 limitisthe value thata functionorsequence "approaches"asthe inputorindex approaches some value. The limitof f(x) asx approachesa fromthe right. limx→a+f(x) The limitof f(x) asx approachesa fromthe left. limx→a−f(x) Example:Find limx→2−(x3−1) limx→2−(x3−1)=limx→2−(23−1)=7 As x approaches2 fromthe left,x3 pproaches8andx3 -1 approaches7. Find limx→2+(x3−1) limx→2+(x3−1)=limx→2+(23−1)=7  (b) use the properties of limits; Properties of limits The limitof a constant isthe constant itself. limx→ak=k The limitof a functionmultipliedbyaconstant isequal to the value of the functionmultipliedby the constant. limx→ak.f(x)=k.limx→af(x)
  • 49. The limitof a sum (or difference) of the functionsisthe sum(ordifference)of the limitsof the individualfunctions. 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 limitof a product isthe productof the limits. limx→af(x).g(x)=limx→af(x).limx→ag(x) The limitof a quotientisthe quotientof the limits limx→af(x)g(x)=limx→af(x)limx→ag(x) The limitof a poweris the powerof 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) if limx→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.
  • 50. 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
  • 51. 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 functionf(x),itsderivativeisdefinedas f′(x)=lim△x→0f(x+△x)−f(x)△x Example 1: Compute the derivative of f(x)byusinglimitdefinition f(x)=13x−25 f′(x)=lim△x→0f(x+△x)−f(x)△x f′(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)byusinglimitdefinition 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−−−−√
  • 52.  (b) find the derivatives ofxn (n∈Q), ex , ln x, sin x, cos x, tan x, sin-1x, cos-1x, tan-1x, with constant multiples, sums, differences, products, quotients and composites; Common Derivatives Table of Derivatives ddx(x)=1 ddx(xn)=nxn−1 ddx(ex)=ex ddx(lnx)=1x,x>0 ddx(sinx)=cosx ddx(cosx)=−sinx ddx(tanx)=sec2x ddx(sin−1x)=11−x2−−−−−√ ddx(cos−1x)=−11−x2−−−−−√ ddx(secx)=secxtanx ddx(cscx)=−cscxcotx ddx(cotx)=−csc2x (c) performimplicit differentiation; Explicit and Implicit Differentiation There are two waysto define functions, implicitlyandexplicitly.Mostof the equationswe have dealtwithhave beenexplicitequations,suchasy = 3x-2. Thisis called explicit because given an x, you can directly get f(x). The technique of implicitdifferentiation allowsyoutofindthe derivativeof ywithrespecttox withouthavingtosolve the givenequationfory.Given x2+y2=10 Performingachainrule to get dy/dx,solve dy/dx intermsof x and y. Finddy/dx for x2+y2=10 2x+2ydydx=0 2ydydx=−2x dydx=−xy
  • 53. (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.dtdx x=f(t),y=g(t) Example: Find the first derivative, given x=t+cost y=sint dydx=dydtdxdt=cost1−sint Example: Find the first derivative, given x=t4−4t2 y=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
  • 54. 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 stationarypointisan inputtoa functionwhere the derivative iszero. Atstationarypoints, dydx=0
  • 55. Extremum Points Maximaand minimaare pointswhere afunctionreachesahighestor lowestvalue, respectively (g) (h) (i) SecondDerivative If f'''(x) is positive,thenitisa minimumpoint. If f'''(x) is negative,thenitisa maximum point. If f'''(x)= zero,thenitcouldbe a maximum, minimumorpointof inflexion. Point of Inflexion An 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 isan inflexionpoint
  • 56. (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) where m1=f′(x)=dydx the gradient to the function of f(x). Tangent is perpendicular to normal, thus m1m2=−1
  • 57. Example: Find the equations of the tangent line and the normal line for the curve at t=1. x=t2 y=2t+1 dxdt=2t dydt=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+2 Equation of normal m1m2=−1 m2=−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 quantitytootherquantitieswhose ratesof change are known. The rate of change is usually with respect to time.
  • 58. Example: Air isbeingpumpedintoaspherical balloonsuchthatitsradiusincreasesata rate of .80 cm/min.Findthe rate of change of itsvolume whenthe radiusis5 cm. The volume ( V) of a sphere withradiusris V=43πr3 Differentiatingabove equationwithrespecttot dVdt=43π.3r2.drdt dVdt=4πr2.drdt The rate of change of the radiusdr/dt= 0.80 cm/minbecause the radiusisincreasingwith respectto time. dVdt=4π(5)2(0.80) dVdt=80π cm3/min Hence,the volume isincreasingatarate of 80π cm3 /minwhenthe radiushasa lengthof 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.
  • 59. Chapter 9 Integration 9.1 Indefinite integrals (a) identify integration as the reverse of differentiation; (b) integrate xn (n∈Q), ex , sin x, cos x, sec2x, 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
  • 60. (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+3 1=A(x+3)+B(x+2) x=−2,A=1 x=−3,B=−1 1(x+2)(x+3)=1x+2−1x+3 Integrate 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=16 x=−1−2C=1,C=−12 x=−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)
  • 61. =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=asinθ For a2+x2−−−−−−√, use x=atanθ For x2−a2−−−−−−√, use x=asecθ Basic Trigonometric Identities csc2θ=1sin2θ sec2θ=1cos2θ cot2θ=1tan2θ Pythagorean Identities cos2θ+sin2θ=1 csc2θ=1+cot2θ sec2θ=1+tan2θ (f) perform integration by parts; 9.2 Definite integrals (g) identify a definite integral as the area under a curve;
  • 62. (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
  • 63. (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)
  • 64. 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.
  • 65. 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.
  • 66. Chapter 11 Maclaurin Series (a) find the Maclaurin series for a function and the interval of convergence; Maclaurin Series Taylor seriesis a representationof afunctionasan infinite sumof termsthatare calculated fromthe valuesof the function'sderivativesata 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 seriesisthe Taylorseriesof the functionabout x=0. f(x)=∑x=0∞fn(0)n!xn f(x)=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3+...+f(n)(0)n!xn+... Example : Findthe MaclaurinSeriesexpansionof 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!
  • 67. =∑n=0∞5nxnn! Interval of Convergence The set of points where the series converges is called the interval of convergence Finding interval of convergence 1) 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)n1n The series converges by the Alternating Series Test. The interval of convergence is −3≤x<7 or [−3,7)
  • 68. (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 as x4+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.
  • 69. 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.
  • 70. 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. Class 0−5 6−10 11−15 16−20 Frequency 10 20 17 4 Ungrouped Data Data that has not been organized into groups 10 1 2 4 80 45 67 78 50 1 11 23 6 9 8 15
  • 71. (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 8 7 | 3 3 5 6 8 | 1 5 8 10 | 7 key 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.
  • 72. Histograms and Cumulative Frequency Histograms Histograms and Cumulative Frequency Histograms A histogram is constructed from a frequency table The cumulative frequency is the running total of the frequencies.
  • 73. (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 observation Q1=(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.
  • 74. 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=12 Q2 == 1/2 (11+1) = 6th observation, Q2=22 Q3 == 3/4 (11+1) = 9th observation, Q3=38 Interquartile Range = Q3 -Q1 = 38 - 12 = 26 Range = 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)=13.5 Q2=(22+302)=26 Q3=(38+422)=40 Interquartile Range = Q3 -Q1 = 40 - 13.5 = 26.5 Range = 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 median f = the frequency of the class median fm= the lower boundary of the class median C= the class width Q1=LQ1+(n4−FfQ1)C
  • 75. Q3=LQ3+⎛⎝3n4−FfQ3⎞⎠C Example : Find the median and interquartile range of below data. Q2=20.5+⎛⎝502−2312⎞⎠10=22.167 Q1=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 Data Mean x¯=∑xn Standard Deviation s=∑(x−x¯)2n−−−−−−−−−√ s=∑x2n−(∑xn)2−−−−−−−−−−−−−− ⎷ Grouped Data Mean
  • 76. 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.
  • 77. 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 = 1.07 Coefficient of skewness, Sk=3−41.07=−0.93 Therefore, the distribution is negatively skewed. (h) describe the shape of a data distribution.
  • 78. 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 6 P(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;
  • 79.  Permutations A permutationisthe choice of r thingsfrom a setof n thingswithoutreplacementandwhere the order matters PermutationFormula nPr=n!(n−r)! Combinations A combinationisthe choice of r thingsfroma setof n thingswithoutreplacementandwhere orderdoesNOT matter. CombinationFormula 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.
  • 80.  (d) identify complementary, exhaustive and mutually exclusive events; Complementary, Exhaustive and Mutually Exclusive Events Complementary Event The complementof anyeventA,A'is the eventthatA doesnotoccur. Exhaustive Event A setof eventsiscollectively exhaustive if at least one of the events must occur. For example, whenrollingasix-sideddie,the outcomes1,2, 3, 4, 5, and 6 are collectivelyexhaustive,because they include the entire range of possible outcomes. Thus, all sample spaces are collectively exhaustive. P(A∪B)=1 MutuallyExclusive 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)=0 P(A∪B)=P(A)+P(B)
  • 81. (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);
  • 82.  (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∪A A∩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) ImpotentLaws A∩A=A A∪A=A Domination Laws A∪U=U A∩∅=∅ AbsorptionLaws A∪(A∩B)=A A∩(A∪B)=A Inverse Laws A∪A′=U A∩A′=∅
  • 83.  Law of Complement (A′)′=A U′=∅ ∅′=U  DeMorgan'sLaw (A∪B)′=A′∩B′ (A∩B)′=A′∪B′  Relative ComplementofB in A A−B=A∩B′
  • 84. Chapter 15 Probability Distributions 15.1 Discrete random variables Var(X)=E(X2)−[E(X)]2 Var(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)
  • 85. Meanand 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.
  • 86. 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)dy P(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".
  • 87. 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−k If X ~ B(n, p), X is a binomially distributed random variable, then the expected value of X is Mean = np and the variance is Variance = np(1−p) n number of successes p 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 givennumberof eventsoccurring 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 stochasticvariable Xissaidtohave a Poissondistributionwithparameterλ> 0, if for k = 0, 1, 2, ...the probabilitymassfunctionof Xisgivenby f(k;λ)=Pr(X=k)=λke−λk! K isthe numberof occurrencesof an event;the probability of whichisgivenbythe function λ is a positive real number. Meanand Variance The expected value and variance of a Poisson-distributed random variable is equal to λ.
  • 88. 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
  • 89. 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)
  • 90. 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.
  • 91. 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 Samplingis concernedwiththe selectionof asubsetof individualsfromwithinastatistical populationtoestimate characteristicsof 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. Dependingonthe samplingmethod,asample canhave fewerobservationsthanthe population, the same numberof observations,ormore observations.More thanone sample can be derived from the same population.
  • 92. 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 randomsample isa subsetof a sample chosenfromaa population.Eachindividualis chosenrandomlyandentirelybyequal chance (same probabilityof beingchosenatanystage duringthe samplingprocess). 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. Meanand 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¯=μ Similarly, the standard deviation of a sampling distribution of means is σx¯=σn√ The larger the sample, the less variable the sample mean Example:
  • 93. 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¯=12 σx¯=420−−√=0.8944 Meanand Variance of Sample Proportion Expected value of a sample proportion μp^=p Standard deviation of a sample proportion σp^=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^=0.53 σp^=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.
  • 94. 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 1 A 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.
  • 95. 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+0.93 24.07<μ<25.93 Confidence Interval for the Population Proportion The confidence interval for a proportion is p±zα/2σp
  • 96. A 1-α confidence interval to population proportion. p^±zα/2(p^(1−p^)n)−−−−−−−−−− ⎷
  • 97. 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.
  • 98. 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).