KOLEKSI SOALAN PERCUBAAN SPM 2012 MATEMATIK TAMBAHAN DAN MATEMATIK NEGERI-NEGERI
Mohon laluan ke blog berikut.. sila klik link di bawah....Good Luck
Sunday, July 22, 2012
Wednesday, July 18, 2012
Permutation And Combination
Combinations and Permutations
What's the Difference?
In English we use the word "combination" loosely, without thinking if the order of things is important. In other words: |
"My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad. |
|
"The combination to the safe was 472". Now we do care about the order. "724" would not work, nor would "247". It has to be exactly 4-7-2. |
 |
If the order doesn't matter, it is a Combination. |
|
If the order does matter it is a Permutation. |
 |
So, we should really call this a "Permutation Lock"! |
A Permutation is an ordered Combination.
 |
To help you to remember, think "Permutation ... Position" |
Permutations
There are basically two types of permutation:- Repetition is Allowed: such as the lock above. It could be "333".
- No Repetition: for example the first three people in a running race. You can't be first and second.
1. Permutations with Repetition
These are the easiest to calculate.When you have n things to choose from ... you have n choices each time!
When choosing r of them, the permutations are:
n × n × ... (r times)
(In other words, there are n possibilities for the first choice, THEN there are n possibilites for the second choice, and so on, multplying each time.)Which is easier to write down using an exponent of r:
n × n × ... (r times) = nr
Example: in the lock above, there are 10 numbers to choose from (0,1,..9) and you choose 3 of them:
So, the formula is simply:
10 × 10 × ... (3 times) = 103 = 1,000 permutations
| nr |
| where n is the number of things to choose from, and you choose r of them (Repetition allowed, order matters) |
2. Permutations without Repetition
In this case, you have to reduce the number of available choices each time. |
For example, what order could 16 pool balls be in? After choosing, say, number "14" you can't choose it again. |
16 × 15 × 14 × 13 × ... = 20,922,789,888,000
But maybe you don't want to choose them all, just 3 of them, so that would be only:
16 × 15 × 14 = 3,360
In other words, there are 3,360 different ways that 3 pool balls could be selected out of 16 balls.But how do we write that mathematically? Answer: we use the "factorial function"
 |
The factorial function (symbol: !) just means to multiply a series of descending natural numbers. Examples:
|
| Note: it is generally agreed that 0! = 1. It may seem funny that multiplying no numbers together gets you 1, but it helps simplify a lot of equations. | |
16! = 20,922,789,888,000
But if you wanted to select just 3, then you have to stop the
multiplying after 14. How do you do that? There is a neat trick ... you
divide by 13! ...
16 × 15 × 14 × 13 × 12 ...
|
= 16 × 15 × 14 = 3,360 | |
13 × 12 ...
|
The formula is written:
 |
| where n is the number of things to choose from, and you choose r of them (No repetition, order matters) |
Examples:
Our "order of 3 out of 16 pool balls example" would be:| 16! | = | 16! | = | 20,922,789,888,000 | = 3,360 |
| (16-3)! | 13! | 6,227,020,800 |
(which is just the same as: 16 × 15 × 14 = 3,360)
How many ways can first and second place be awarded to 10 people?| 10! | = | 10! | = | 3,628,800 | = 90 |
| (10-2)! | 8! | 40,320 |
(which is just the same as: 10 × 9 = 90)
Notation
Instead of writing the whole formula, people use different notations such as these:
Example: P(10,2) = 90
Combinations
There are also two types of combinations (remember the order does not matter now):- Repetition is Allowed: such as coins in your pocket (5,5,5,10,10)
- No Repetition: such as lottery numbers (2,14,15,27,30,33)
1. Combinations with Repetition
Actually, these are the hardest to explain, so I will come back to this later.2. Combinations without Repetition
This is how lotteries work. The numbers are drawn one at a time, and if you have the lucky numbers (no matter what order) you win!The easiest way to explain it is to:
- assume that the order does matter (ie permutations),
- then alter it so the order does not matter.
We already know that 3 out of 16 gave us 3,360 permutations.
But many of those will be the same to us now, because we don't care what order!
For example, let us say balls 1, 2 and 3 were chosen. These are the possibilites:
| Order does matter | Order doesn't matter |
| 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1 |
1 2 3 |
In fact there is an easy way to work out how many ways "1 2 3" could be placed in order, and we have already talked about it. The answer is:
3! = 3 × 2 × 1 = 6
(Another example: 4 things can be placed in 4! = 4 × 3 × 2 × 1 = 24 different ways, try it for yourself!)So, all we need to do is adjust our permutations formula to reduce it by how many ways the objects could be in order (because we aren't interested in the order any more):
 |
| where n is the number of things to choose from, and you choose r of them (No repetition, order doesn't matter) |
It is often called "n choose r" (such as "16 choose 3")
And is also known as the "Binomial Coefficient"
And is also known as the "Binomial Coefficient"
Notation
As well as the "big parentheses", people also use these notations:
Example
| 16! | = | 16! | = | 20,922,789,888,000 | = 560 |
| 3!(16-3)! | 3!×13! | 6×6,227,020,800 |
| 16×15×14 | = | 3360 | = 560 |
| 3×2×1 | 6 |
So remember, do the permutation, then reduce by a further "r!"
... or better still ...
Remember the Formula!
It is interesting to also note how this formula is nice and symmetrical:
| 16! | = | 16! | = | 16! | = 560 |
| 3!(16-3)! | 13!(16-13)! | 3!×13! |
Pascal's Triangle
You can also use Pascal's Triangle to find the values. Go down to row "n" (the top row is 0), and then along "r" places and the value there is your answer. Here is an extract showing row 16:1 14 91 364 ... 1 15 105 455 1365 ... 1 16 120 560 1820 4368 ...
1. Combinations with Repetition
OK, now we can tackle this one ... |
Let us say there are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be? Let's use letters for the flavors: {b, c, l, s, v}. Example selections would be
|
(And just to be clear: There are n=5 things to choose from, and you choose r=3 of them.
Order does not matter, and you can repeat!)
Now, I can't describe directly to you how to calculate this, but I can show you a special technique that lets you work it out. Order does not matter, and you can repeat!)
 |
Think about the ice cream being in boxes, you could say "move past the first box, then take 3 scoops, then move along 3 more boxes to the end" and you will have 3 scoops of chocolate! |
| So, it is like you are ordering a robot to get your ice cream, but it doesn't change anything, you still get what you want. |
(arrow means move, circle means scoop).In fact the three examples above would be written like this:
| {c, c, c} (3 scoops of chocolate): | |
| {b, l, v} (one each of banana, lemon and vanilla): |  |
| {b, v, v} (one of banana, two of vanilla): |  |
Notice that there are always 3 circles (3 scoops of ice cream) and 4 arrows (you need to move 4 times to go from the 1st to 5th container).
So (being general here) there are r + (n-1) positions, and we want to choose r of them to have circles.
This is like saying "we have r + (n-1) pool balls and want to choose r of them". In other words it is now like the pool balls problem, but with slightly changed numbers. And you would write it like this:
 |
| where n is the number of things to choose from, and you choose r of them (Repetition allowed, order doesn't matter) |
| (5+3-1)! | = | 7! | = | 5040 | = 35 |
| 3!(5-1)! | 3!×4! | 6×24 |
In Conclusion
Phew, that was a lot to absorb, so maybe you could read it again to be sure!But knowing how these formulas work is only half the battle. Figuring out how to interpret a real world situation can be quite hard.
But at least now you know how to calculate all 4 variations of "Order does/does not matter" and "Repeats are/are not allowed".
Monday, June 18, 2012
TRIGONOMETRIC FUNCTIONS
5.TRIGONOMETRIC
FUNCTIONS
To sketch the graph of sine or cosine function , students
are encouraged to follow the steps below.
1. Determine the
angle to be labeled on x-axis.
eg : Function angle
y = sin x x = 90o
y
= cos 2x 2x = 90o
x = 45o
y
= sin 
= 90o
= 90o
x = 60o
2. Calculate the values of y for each value of x by using calculator
eg : y = 1 – 2 cos 2x
|
x
|
0
|
45
|
90
|
135
|
180
|
225
|
270
|
315
|
360
|
|
y
|
-1
|
1
|
3
|
1
|
-1
|
1
|
3
|
1
|
-1
|
P/Sbab ni mesti score sbb sama mcm bab Matematik Form 4 shj....ok..all da best guys..
Monday, May 21, 2012
Multiplying Matrices
Sunday, May 6, 2012
Example of Solutions for Paper 1
|
1
|
Given f(x) = x
+ 3 and g(x) = 6x2 + 3.
Diberi f(x) = x + 3 dan g(x) = 6x2 + 3. gf(x) = g(f(x)) = g(x + 3) = 6(x + 3)2 + 3 = 6(x2 + 6x + 9) + 3 = 6x2 + 36x + 57 gf(−3) = 6(−3)2 + 36(−3) + 57 = 3 |
|
2
|
α + β = −4
αβ = −4 New equation: Sum of roots, (1 − 9α) + (1 − 9β) = −9(α + β) + 1 = −9(−4) + 1 = 37 Product of roots, (1 − 9α)(1 − 9β) = 81αβ − 9α − 9β + 1 = 81(−4) − 9(−4) + 1 = −287 Equation: x2 − 37x − 287 = 0 |
|
3
|
−6x2 − nx
− 6 = x
−6x2 + (−n − 1)x − 6 = 0 The equation has two equal roots b2 − 4ac = 0 (−n − 1)2 − 4(−6)(−6) = 0 n2 + 2n + 1 − 144 = 0 n2 + 2n − 143 = 0 (n − 11)(n + 13) = 0 n = 11 or n = −13 |
|
4
|
f(x) = y = 4x2 + 8x
+ k + 2
The equation does not have real roots b2 − 4ac < 0 (8)2 − 4(4)(k + 2) < 0 64 − 32 − 16k < 0 −16k < −32 k > 2 |
|
5
|
y = x2 − 2hx − 2h
= x2 − 2hx + ()2 − ()2 − 2h = (x − h)2 − h2 − 2h Minimum value = − h2 − 2h −h2 − 2h = −3 h2 + 2h − 3 = 0 (h − 1)(h + 3) = 0 h = 1 or h = −3 |
|
6
|
−3 logx a
− 5 logx b + logx ab
= −logx a3 − logx b5 + logx ab = logx = logx |
|
7
|
logx 128 = −
128 = x x = 128 = 2 = 4 |
|
9
|
PQ = 5
= 5 = 5 Squaring both sides, (6a − 4)2 + 42 = 52 (6x − 4)2 = 25 − 16 (6x − 4)2 = 9 6x − 4 = ±3 a = or a = |
|
10
|
Gradient of EF
= = 1 (1)m2 = −1 m2 = −1 Midpoint of EF = (, ) = (−6, 6) Equation: y − 6 = −(x + 6) −y + 6 = x + 6 x + y = 0 |
|
11
|
Rearrange the data,
41 59 62 78 85 87 89 First quartile = 59 Third quartile = 87 Interquartile range = 87 − 59 = 28 |
|
12
|
(3x2 + 2x + 4) = f(x)
= [3x2 + 2x + 4]
= 3(2)2 + 2(2) + 4 − [3(2)2 + 2(2) + 4] = 20 − (20) = 0 |
|
13
|
30° 54'
= 30.9 × = 0.5394 rad. |
|
14
|
0.6π rad.
= 0.6π × = 108° |
|
15
|
Step 1:
Langkah 1: Let δx be a small increment in x and δy be the corresponding small increment in y. Biar δx menjadi tokokan kecil pada x dan δy menjadi tokokan yang sepadan pada y. Step 2: Langkah 2: y = x2 + 7 ----- (1) y + δy = (x + δx)2 + 7 y + δy = x2 + 2x(δx) + (δx)2 + 7 ----- (2) Step 3: Langkah 3: (2) − (1), δy = 2x(δx) + (δx)2 Step 4: Langkah 4: = 2x + (δx) = (2x + (δx)) = 2x Therefore, Maka, = 2x |
|
16
|
y = −x4 + 2x3
= −4x3 + 6x2 At point (2, 0), = −4(2)3 + 6(2)2 = −8 Therefore, the gradient of the tangent at point (2, 0) = −8 |
|
17
|
|
|
18
|
|
|
19
|
Price index, I = × 100
178.3 = × 100 P1985 = = RM2300 |
|
20
|
Composite index,
= = = 152.3 |
|
21
|
|
|
22
|
|
|
23
|
= + 8
= 8() − 8 X = Y = |
|
24
|
|
|
25
|
OH = 3HF
OH→ = OF→ = (OE→ + OG→) = (8x~ + 10y~) = 6x~ + y~ |
Subscribe to:
Posts (Atom)