1. Index number is a measurement used to show the average changes in a certain quality with time. An example of index number is price index.
2. Price index
I = P1/P0 X 100
P0 = the price of an item at base time
P1 = the price of the item at a specific time
3. Composite index number
(T) = (total)(I)(W) / (total)(W)
I = Index number
W = Weightage
Example :
1. In the year 2002, the price and price index of a kilogram of onions are RM 3.60 and 120
respectively. Using the year 2000 as the base year, calculate the of a kilogram of onions in the year 2000.
P(2000) / P(2002) X 100 = 120
RM 3.60 / P(2002) X 100 = 120
P(2002) = (RM 3.60 x 100) / 120
= RM 3.00
Jumaat, 21 November 2008
Chapter 10 Solution of Triangles
1. Sin rule :
a/sin A = b/sin B = c/sin C
2. An ambiguous case will arise if given the two sides a and b , where a is shorter than b, and the acute angle A is not the included angle.
3. Cosine rule :
a^2 = b^2 + c^2 - 2bc cos A
b^2 = a^2 + c^2 - 2ac cos B
c^2 = a^2 + b^2 - 2ab cos C
4. Area of triangle ABC
= 1/2 ab sin C
= 1/2 ac sin B
= 1/2 bc sin A
a/sin A = b/sin B = c/sin C
2. An ambiguous case will arise if given the two sides a and b , where a is shorter than b, and the acute angle A is not the included angle.
3. Cosine rule :
a^2 = b^2 + c^2 - 2bc cos A
b^2 = a^2 + c^2 - 2ac cos B
c^2 = a^2 + b^2 - 2ab cos C
4. Area of triangle ABC
= 1/2 ab sin C
= 1/2 ac sin B
= 1/2 bc sin A
Khamis, 20 November 2008
Chapter 9 Differentiation
(Q) = small change
1. If y = f (x) , that is a function in terms of x, therefore dy/dx = f' (x).
2. If y = ax^n , therefore dy/dx = nax^(n - 1) , a is a constant.
a) If y = x^n , therefore dy/dx = nx^(-1)
b) If y = k , therefore dy/dx = 0 , k is a constant.
3. The gradient of curve y = f (x) at a certain point is the derivative of y with respect to x , that is dy/dx or f' (x).
4. a) Equation of tangent :
y - b = m1 (x - a) , m1 = f'(a)
b) Equation of normal :
y - b = m2 (x - a) , m2 = -1/m2
5. Differentiation of a product ( Product Rule )
y = uv
dy/dx = (u) dy/dx + (v) du/dx
6. Differentiation of a quotient ( Quotient Rule )
y = u/v
dy/dx = ((v)du/dx - (u)dv/dx) / v^2
7. Differentiation of a composite function (Chain Rule )
y = f (u) and u = g (x) , therefore :
dy/dx = dy/du X du/dx
8. At the turning point or stationary point , dy/dx = 0.
a) Maximum point if d^2y/dx^2 < 0
b) Minimum point if d^2y/dx^2 > 0
9. The related rate of change :
dy/dt = dy/dx X dx/dt
10. Small changes and approximation :
Qy = dy/dx X Qx
11. Second order differentiation :
d^2y/dx^2 = d/dx (dy/dx) = f'' (x)
1. If y = f (x) , that is a function in terms of x, therefore dy/dx = f' (x).
2. If y = ax^n , therefore dy/dx = nax^(n - 1) , a is a constant.
a) If y = x^n , therefore dy/dx = nx^(-1)
b) If y = k , therefore dy/dx = 0 , k is a constant.
3. The gradient of curve y = f (x) at a certain point is the derivative of y with respect to x , that is dy/dx or f' (x).
4. a) Equation of tangent :
y - b = m1 (x - a) , m1 = f'(a)
b) Equation of normal :
y - b = m2 (x - a) , m2 = -1/m2
5. Differentiation of a product ( Product Rule )
y = uv
dy/dx = (u) dy/dx + (v) du/dx
6. Differentiation of a quotient ( Quotient Rule )
y = u/v
dy/dx = ((v)du/dx - (u)dv/dx) / v^2
7. Differentiation of a composite function (Chain Rule )
y = f (u) and u = g (x) , therefore :
dy/dx = dy/du X du/dx
8. At the turning point or stationary point , dy/dx = 0.
a) Maximum point if d^2y/dx^2 < 0
b) Minimum point if d^2y/dx^2 > 0
9. The related rate of change :
dy/dt = dy/dx X dx/dt
10. Small changes and approximation :
Qy = dy/dx X Qx
11. Second order differentiation :
d^2y/dx^2 = d/dx (dy/dx) = f'' (x)
Chapter 6 Coordinate Geometry
(~) square root
1. The distance between point A (x1 , y1) and point B (x2 , y2) , is :
AB = ~((x2 - x1)^2 + (y2 - y1)^2)
2. Midpoint of AB :
AB = ( (x1 + x2)/2 , (y1 + y2)/2 )
3. The point that divides a straight line line AB in the ratio m : n is :
( (nx1 + mx2) /(m + n) , (ny1 + my2) / (m + n) )
a) Area of a triangle
A = 1/2 |x1 x2 x3 x1|
|y1 y2 y3 y1|
= 1/2 | (x1)(y2) + (x2)(y3) + (x3)(y1) - (y1)(x2) - (y2)(x3) - (y3)(x1) |
b) Area of a quadrilateral
= 1/2 |x1 x2 x3 x4 x1|
|y1 y2 y3 y4 y1|
4. Gradient of a straight line
m = (y2 - y1) / (x2 - x1)
m = - (y - intercept) / (x - intercept)
5. Equation of a straight line
a) y = mx + c , m = gradient and c = y - intercept
b) y - y1 = m(x - x1)
c) (y - y1) = (y2 - y1)
(x - x1) (x2 - x1)
d) x/a + y/ b ＝ 1 ， a = x- intercept and b = y - intercept
6. Straight lines which are parallel have the same gradient, that is m1 = m2 and vice versa.
7. Two straight lines are perpendicular to each other if m1 m2 = -1 and vice versa.
1. The distance between point A (x1 , y1) and point B (x2 , y2) , is :
AB = ~((x2 - x1)^2 + (y2 - y1)^2)
2. Midpoint of AB :
AB = ( (x1 + x2)/2 , (y1 + y2)/2 )
3. The point that divides a straight line line AB in the ratio m : n is :
( (nx1 + mx2) /(m + n) , (ny1 + my2) / (m + n) )
a) Area of a triangle
A = 1/2 |x1 x2 x3 x1|
|y1 y2 y3 y1|
= 1/2 | (x1)(y2) + (x2)(y3) + (x3)(y1) - (y1)(x2) - (y2)(x3) - (y3)(x1) |
b) Area of a quadrilateral
= 1/2 |x1 x2 x3 x4 x1|
|y1 y2 y3 y4 y1|
4. Gradient of a straight line
m = (y2 - y1) / (x2 - x1)
m = - (y - intercept) / (x - intercept)
5. Equation of a straight line
a) y = mx + c , m = gradient and c = y - intercept
b) y - y1 = m(x - x1)
c) (y - y1) = (y2 - y1)
(x - x1) (x2 - x1)
d) x/a + y/ b ＝ 1 ， a = x- intercept and b = y - intercept
6. Straight lines which are parallel have the same gradient, that is m1 = m2 and vice versa.
7. Two straight lines are perpendicular to each other if m1 m2 = -1 and vice versa.
Chapter 5 Indices And Logarithms
(~) = square root
(*) = divide
1. a^n = a x a x a x a ....................... x a
a is the base and n is the index.
2 (a) Zero index, a^0 = 1
(b) Negative integer index, a^(-n) = 1/(a^n)
(c) Fractional index, a^(1/n) = n~(a)
a^(m/n) = (n~(a^m)) =(n~(a))^m
3. Laws of indeces:
(a) a^m x a^n = a^(m + n)
(b) a^m * a^n = a^(m - n)
(c) (a^m)^n = a^(m x n)
4. If y = a^x, then log(a)y = x.
(a) If a^0 = 1 therefore log(a)1 = 0
(b) If a^1 = a therefore log(a) a = 1
5. Laws of logarithms:
(a) log (a) xy = log (a) x + log (a) y
(b) log (a) (x/y) = log (a) x - log (a) y
(c) log (a) x^n = n log (a) x
6. Changing the base of logarithms:
(a) log (a) b = (log (c) b) / (log (c) a)
(b) log (a) b = 1/(log(b) a) , when c = b
Example:
a) 27^x / 81^ 2x = 1/ 243
3^3x / (3^4)^2x = 1/3^5
3^(3x-8x) = 3^(-5)
-5x = -5
x = 1
b) log (10)(x - 5) = log (10) (x - 1) + 2
log (10) (x-5) - log (10) (x - 1) = 2
log (10) (x - 5/x - 1) = 2
(x - 5/x - 1) = 10^2
x - 5 = 100 (x - 1)
x - 5 = 100 x - 100
99 x = 95
x = 95/99
(*) = divide
1. a^n = a x a x a x a ....................... x a
a is the base and n is the index.
2 (a) Zero index, a^0 = 1
(b) Negative integer index, a^(-n) = 1/(a^n)
(c) Fractional index, a^(1/n) = n~(a)
a^(m/n) = (n~(a^m)) =(n~(a))^m
3. Laws of indeces:
(a) a^m x a^n = a^(m + n)
(b) a^m * a^n = a^(m - n)
(c) (a^m)^n = a^(m x n)
4. If y = a^x, then log(a)y = x.
(a) If a^0 = 1 therefore log(a)1 = 0
(b) If a^1 = a therefore log(a) a = 1
5. Laws of logarithms:
(a) log (a) xy = log (a) x + log (a) y
(b) log (a) (x/y) = log (a) x - log (a) y
(c) log (a) x^n = n log (a) x
6. Changing the base of logarithms:
(a) log (a) b = (log (c) b) / (log (c) a)
(b) log (a) b = 1/(log(b) a) , when c = b
Example:
a) 27^x / 81^ 2x = 1/ 243
3^3x / (3^4)^2x = 1/3^5
3^(3x-8x) = 3^(-5)
-5x = -5
x = 1
b) log (10)(x - 5) = log (10) (x - 1) + 2
log (10) (x-5) - log (10) (x - 1) = 2
log (10) (x - 5/x - 1) = 2
(x - 5/x - 1) = 10^2
x - 5 = 100 (x - 1)
x - 5 = 100 x - 100
99 x = 95
x = 95/99
Chapter 4 Simultaneous Equations
Simultaneous equations in two unknowns, one linear equation and one non-linear equation, can be solved using the sudstitution method by following the steps below:
Step 1: arrange the linear equation so that one of the two unknowns become the subject of the equation.
Step 2: substitude the equation from step 1 into the non-linear equation, simplify and express it
in the form of :
ax^2 + bx + c = 0.
Step 3: Solve the quadratic equation using factorisation, completing the square, or using the
formula.
Step 4: Substitude the value of the unknown obtained in step 3 into the linear equation to find the value of the other unknown.
Example:
(1) y - 2x = 0
(2) x^2 + xy + 5x = 0
From (1) y = 2x ..................................................................(3)
Substitude (3) into (2) :
x^2 + x(2x) + 5x = 0
x^2 + 2x^2 + 5x = 0
3x^2 + 5x - 8 = 0
(3x + 8)(x - 1) = 0
3x + 8 = 0 or x - 1 = 0
x = -8/3 or x = 1
substitude x = -8/3 into (3) , y = 2 (-8/3)
= -16/3
substitude x = 1 into (3) , y = 2 (1)
= 2
the solution is: x = -8/3 , y = -16/3
x = 1 , y = 2
Step 1: arrange the linear equation so that one of the two unknowns become the subject of the equation.
Step 2: substitude the equation from step 1 into the non-linear equation, simplify and express it
in the form of :
ax^2 + bx + c = 0.
Step 3: Solve the quadratic equation using factorisation, completing the square, or using the
formula.
Step 4: Substitude the value of the unknown obtained in step 3 into the linear equation to find the value of the other unknown.
Example:
(1) y - 2x = 0
(2) x^2 + xy + 5x = 0
From (1) y = 2x ..................................................................(3)
Substitude (3) into (2) :
x^2 + x(2x) + 5x = 0
x^2 + 2x^2 + 5x = 0
3x^2 + 5x - 8 = 0
(3x + 8)(x - 1) = 0
3x + 8 = 0 or x - 1 = 0
x = -8/3 or x = 1
substitude x = -8/3 into (3) , y = 2 (-8/3)
= -16/3
substitude x = 1 into (3) , y = 2 (1)
= 2
the solution is: x = -8/3 , y = -16/3
x = 1 , y = 2
Chapter 2 Quadratic Equations
1. The general form of a quadratic equation is ax^2 + bx + c = 0, where a, b and c are constant and a is not equal to 0.
2. The roots of a quadratic equation are the values of the unknown that satisfy the equation.
3. A quadratic equation ax^2 + bx + c = 0 can be solved by :
a) factorisation
b) completing the square
c) (-b ((+/-) (b^2 - 4 a c) ^(1/2))) / 2a
4. If a and b are roots of a quadratic equation, then the quadratic equation is (x - a) (x - b) = 0,
that is x^2 - (a + b)x + ab = 0.
In general, a quadratic equation is of the form:
x^2 - (sum of roots)x + (product of roots) = 0
5. Types of roots of a quadratic equation ax^2 + bx + c = 0 :
a) If b^2 - 4ac > 0, the equation has two different real roots.
b) If b^2 - 4ac = 0, the equation has two equal real roots.
c) If b^2 - 4ac < 0, the equation has no real roots.
2. The roots of a quadratic equation are the values of the unknown that satisfy the equation.
3. A quadratic equation ax^2 + bx + c = 0 can be solved by :
a) factorisation
b) completing the square
c) (-b ((+/-) (b^2 - 4 a c) ^(1/2))) / 2a
4. If a and b are roots of a quadratic equation, then the quadratic equation is (x - a) (x - b) = 0,
that is x^2 - (a + b)x + ab = 0.
In general, a quadratic equation is of the form:
x^2 - (sum of roots)x + (product of roots) = 0
5. Types of roots of a quadratic equation ax^2 + bx + c = 0 :
a) If b^2 - 4ac > 0, the equation has two different real roots.
b) If b^2 - 4ac = 0, the equation has two equal real roots.
c) If b^2 - 4ac < 0, the equation has no real roots.
Jumaat, 14 November 2008
Chapter 1 Functions
1. A relation can be represended in three ways.
a) Arrow diagram
b) Graph
c) Ordered pairs
2. Terminology
a) domain
b) codomain
c) objects
d) images
e) range
3. There's 4 types of relation, which is:
a) one to one relation
b) one to many relation
c) many to one relation
d) many to many relation
4. A funtion is said to exist if every object in the relation has one and only one image.
5. It means that only one to one and many to one relation can be said as
functions.
6. A function f, which maps x to px + q can be written as f : x = px + q . x is the
object and f(x) or px + q is the image of x under the function f.
7. A composite function is made up of two or more functions. In the diagram below, gf(x) is a
composite function where the function f maps set A to set B folows by the function g which
maps set B to set C.
8. The inverse of a funtion f, f-1, is a relation which maps every image of the function to its
object.
Example:
1) Given that f(x)=2x - 3, find the images of -2 and 1/2.
f(x) = 2x-3
f(-2) = 2(-2) -3
= - 4 - 3
= - 7
f(x) = 2x - 3
f(1/2) = 2(1/2) - 3
= 1 - 3
= - 2
a) Arrow diagram
b) Graph
c) Ordered pairs
2. Terminology
a) domain
b) codomain
c) objects
d) images
e) range
3. There's 4 types of relation, which is:
a) one to one relation
b) one to many relation
c) many to one relation
d) many to many relation
4. A funtion is said to exist if every object in the relation has one and only one image.
5. It means that only one to one and many to one relation can be said as
functions.
6. A function f, which maps x to px + q can be written as f : x = px + q . x is the
object and f(x) or px + q is the image of x under the function f.
7. A composite function is made up of two or more functions. In the diagram below, gf(x) is a
composite function where the function f maps set A to set B folows by the function g which
maps set B to set C.
8. The inverse of a funtion f, f-1, is a relation which maps every image of the function to its
object.
Example:
1) Given that f(x)=2x - 3, find the images of -2 and 1/2.
f(x) = 2x-3
f(-2) = 2(-2) -3
= - 4 - 3
= - 7
f(x) = 2x - 3
f(1/2) = 2(1/2) - 3
= 1 - 3
= - 2
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here, you can learn about the methods and ways to solve the add maths question...
lets start!
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