Arimethic Progression
a , a + d , a + 2d , a + 3d , ……..
a = first term
d = common terms
Formula
Tn = a (n-1)d
Sn = n/2 [ 2a(n-1)d ]
Geometric Progression
a , ar , ar^2 , ar^3 , …….
Formula
Tn = a(r^n-1)/(r-1) , when r > 1
Tn = a(1-r^n)/(1-r) , when r < 1
S infinity = a / r-1
Example
1. The third and the eighth term of an arithmetic progression are 22 and 57 respectively. Find
the first term and the common difference.
Tn = a + (n-1) d
T3 = a + (3-1) d = 22
= a + 2d = 22
= a = 22 - 2d - eq 1
T8 = a + (8-1) d = 57
= a + 7d = 57 - eq 2
replace eq 1 into eq 2
(22 - 2d) + 7d = 57
5d = 35
d = 7
when d = 7
a = 22 - 2 (7)
a = 8
2. Calculate the sum of all the terms for each of the arithmetic progressions below.
2, 6, 10, 14, ...... 50
Tn = 50
a + (n-1) = 50
2 + (n-1)4 = 50
4n - 4 = 48
4n = 52
n = 13
Sn = n/2(a + l)
S13 = 13/2 (2 + 50)
= 338
Ahad, 11 Januari 2009
Jumaat, 21 November 2008
Chapter 11 Index Number (Social Science Application Pakage)
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
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
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
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