Nota: Mathematical Reasoning

Tips SPM:

1. Soalan untuk bab ini akan keluar dalam Kertas 2 sahaja

2. Fokus SPM:

- menentukan ayat yang diberi adalah pernyataan/statement atau tidak
- menentukan pernyataan yang diberi benar atau palsu
- melakukan operasi 'semua', 'sebilangan', 'tidak', 'bukan', 'dan', 'atau' untuk pernyataan yang diberikan
- operasi 'jika...maka..' dan 'jika dan hanya jika'
- melengkapkan penghujahan yang diberikan - Premis dan kesimpulan
- deduction dan induction

3. Semua senarai diatas boleh disoal dalam setiap soalan kertas 2

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Pernyataan (Statement)

ialah satu ayat yang benar atau palsu tetapi bukan kedua-duanya.

 

Contoh:

  9+ 6 = 15                             :Pernyataan palsu

  10 adalah nombor genap     :Pernyataan benar

Jenis ayat seperti soalan, arahan dan seruan bukanlah pernyataan:

 

Contoh:

a) Siapa nama kamu?

b) Jangan bising!

c) 8xy

Hujah(Argument)

Satu proses membuat  kesimpulan berdasarkan pernyataan yang diberi beberapa.
Hujah terdiri daripada premis dan kesimpulan.
Premis = kenyataan yang diberikan

Mengenal pasti premis dan kesimpulan dalam setiap hujah berikut:

Jika x ialah nombor genap, maka 3x ialah nombor genap. 14 adalah nombor genap. Oleh itu, 42 ialah nombor genap.
Premis 1: Jika x ialah nombor genap, maka 3x ialah nombor genap
Premis 2: 14 adalah nombor genap
Kesimpulan: 42 adalah nombor genap.

Heksagon A mempunyai 6 sisi. ABCDEF adalah heksagon. Oleh itu, ABCDEF mempunyai 6 sisi

Premis 1: Sebuah heksagon mempunyai 6 sisi.
Premis 2: ABCDEF adalah heksagon.
Kesimpulan: ABCDEF mempunyai 6 sisi.

Normal Distribution (Part 5)

Normal Distribution (Part 4)

Arithmetic Progression and Sums

Sequence

A Sequence is a set of things (usually numbers) that are in order.

Arithmetic Progression

In an Arithmetic Progression the difference between one term and the next is a constant.
In other words, we just add the same value each time ... infinitely.

Example:

1, 4, 7, 10, 13, 16, 19, 22, 25, ...

This sequence has a difference of 3 between each number.

In General we could write an arithmetic sequence like this:

{a, a+d, a+2d, a+3d, ... }

where:

• a is the first term, and
• d is the difference between the terms (called the "common difference")

Example: (continued)

1, 4, 7, 10, 13, 16, 19, 22, 25, ...

Has:

• a = 1 (the first term)
• d = 3 (the "common difference" between terms)

And we get:

{a, a+d, a+2d, a+3d, ... }

{1, 1+3, 1+2×3, 1+3×3, ... }

{1, 4, 7, 10, ... }

Rule

We can write an Arithmetic Sequence as a rule:

T_n = a + (n-1)d

Example 1: Calculate the 4th term for

3, 8, 13, 18, 23, 28, 33, 38, ...

This sequence has a difference of 5 between each number.

The values of a and d are:

• a = 3 
• d = 5

So, the 4th term can be calculated:
 

T_n = a + (n-1)d

T4 = a + (4-1)d 

     = a + 3d

     = 3 + 3(5)

     =18

 

 

Example 2: If T₃ = 8 and T₆ = 17, find T₁₄.

Solution:

T_{3 = a + 2d}

   _{8 = a + 2d -------①}

T_{6  = a +5d} 

_{17 = a + 5d --------②}

 

_{② - ①}

_{17-8  = a-a  + (5d-2d)}

_{9 = 3d}

 _d=3

_{substitute d=3 into ①}

_{8 = a + 2(3)}

_{a = 2}

 

T_{14  = a +13d}

         _{= 2 + 13(3)}

_{T14  = 41}

 

Summing an Arithmetic Series

To sum up the terms of this arithmetic sequence:

a + (a+d) + (a+2d) + (a+3d) + ...

use this formula:

  

Example1: 

Add up the first 10 terms of the arithmetic sequence:

{ 1, 4, 7, 10, 13, ... }

The values of a, d and n are:

• a = 1
• d = 3
• n = 10 (how many terms to add up)
• S10 = 10/2 [2(1) + ( 10-1) 3]
•        =  5 [ 2+9(3)]
•        = 5[29]
•        = 145

Example2:

0.6, 1.7, 2.8, ……………… to 100 terms

Solution:
a = 0.6, d = 1.1 and n = 100
Sum of n terms can be given as follows:

Thus, sum of the 100 term,
S₁₀₀= 5505

Problem 1:
The first term of an arithmetic sequence is equal to 200 and the common difference is equal to
-10. Find the value of the 20 th term

 

Solution to Problem 1:

Use the value of the common difference d = -10 and the first term a₁ = 200 in the formula for the n th term given above and then apply it to the 20 th term

T₂₀ = 200 +(20 - 1 ) (-10) 

      = 10

   

Problem 2:
An arithmetic sequence has a common difference equal to 10 and its 6 th term is equal to 52. Find its 15 th term.

 

Solution to Problem 2:

• We use the n th term formula for the 6 th term, which is known, to write
  T₆ = 52    
•     = a +(6 - 1 ) (10)
• 52 = a +50
• a   = 2
  
  
• The above equation allows us to calculate a.
  
  a = 2
  
  
• Now that we know the first term and the common difference, we use the n th term formula to find the 15 th term as follows.
  
  T₁₅ = 2 + (15 - 1)(10) = 142

Problem 3:
Find the sum of all the integers from 1 to 1000.

 

Solution to Problem 3:

• The sequence of integers starting from 1 to 1000 is given by
  
  1 , 2 , 3 , 4 , ... , 1000
  
  
• The above sequence has 1000 terms. The first term is 1 and the last term is 1000 and the common difference is equal to 1. We have the formula that gives the sum of the first n terms of an arithmetic sequence knowing the first and last term of the sequence and the number of terms (see formula above).
  
  s₁₀₀₀ = 1000 (1 + 1000) / 2 = 500500

Problem 4:

Find the sum of all positive integers, from 5 to 1555 inclusive, that are divisible by 5.

 

Solution to Problem 4:

• The first few terms of a sequence of positive integers divisible by 5 is given by
  
  5 , 10 , 15 , ...
  
  
• The above sequence has a first term equal to 5 and a common difference d = 5. We need to know the rank of the term 1555. We use the formula for the n th term as follows
  
  1555 = a₁ + (n - 1 )d
  
  
• Substitute a₁ and d by their values
  
  1555 = 5 + 5(n - 1 )
  
  
• Solve for n to obtain
  
  n = 311
  
  
• We now know that 1555 is the 311 th term, we can use the formula for the sum as follows
  
  s₃₁₁ = 311 (5 + 1555) / 2 = 242580 

We can use this fomula to calculate the sum of the AP :

 where a = first term,  l = last term

 Problem 5:

Find the sum of the first 50 even positive integers. 

Solution to Problem 5:

• The sequence of the first 50 even positive integers is given by
  
  2 , 4 , 6 , ...
  
  
• The above sequence has a first term equal to 2 and a common difference d = 2. We use the n th term formula to find the 50 th term
  T₅₀ = 2 +  49(2) = 100
  
  
• We now the first term and last term and the number of terms in the sequence, we now find the sum of the first 50 terms
  
  s₅₀ = 50 (2 + 100) / 2 = 2550

 

Problems on Operation on Sets

1. Let A = {x : x is a natural number and a factor of 18}

B = {x : x is a natural number and less than 6}

Find A ∪ B and A ∩ B.

Solution:

A = {1, 2, 3, 6, 9, 18}

B = {1, 2, 3, 4, 5}

Therefore, A ∩ B = {1, 2, 3}


2. If P = {multiples of 3 between 1 and 20} and Q = {even natural numbers upto 15}. Find the intersection of the two given set P and set Q.
Solution:
P = {multiples of 3 between 1 and 20}
So, P = {3, 6, 9, 12, 15, 18}
Q = {even natural numbers upto 15}
So, Q = {2, 4, 6, 8, 10, 12, 14}
Therefore, intersection of P and Q is the largest set containing only those elements which are common to both the given sets P and Q
Hence, P ∩ Q = {6, 12}.

Sets: Notes

• The intersection of two sets can be represented by Venn diagram, with the shaded region representing A ∩ B.

A ∩ B when A ⊂ B, i.e., A ∩ B = A

A ∩ B when neither A ⊂ B nor B ⊂ A

A ∩ B = ϕ No shaded part



• The union of two sets can be represented by Venn diagrams by the shaded region, representing A ∪ B.

A ∪ B when A ⊂ B

A ∪ B when neither A ⊂ B nor B ⊂ A

A ∪ B when A and B are disjoint sets



Relationship between the three Sets using Venn Diagram
• If ξ represents the universal set and A, B, C are the three subsets of the universal sets. Here, all the three sets are overlapping sets.

Let us learn to represent various operations on these sets.

A ∪ B ∪ C

A ∩ B ∩ C

A ∪ (B ∩ C)

A ∩ (B ∪ C)










Observe the Venn diagrams. The shaded portion represents the following sets.

(a) A’ (A prime)

(b) A ∪ B (A union B)

(c) A ∩ B (A intersection B)

(d) (A ∪ B)’ (A union B dash)

(e) (A ∩ B)’ (A intersection B dash)

(f) B’ (B prime)

For example;
Use Venn diagrams in different situations to find the following sets.

(a) A ∪ B

(b) A ∩ B

(c) A'

(e) (A ∩ B)'

(f) (A ∪ B)'

Solution:

ξ = {a, b, c, d, e, f, g, h, i, j}

A = {a, b, c, d, f}

B = {d, f, e, g}

A ∪ B = {elements which are in A or in B or in both}

         = {a, b, c, d, e, f, g}

A ∩ B = {elements which are common to both A and B}

        = {d, f}

A' = {elements of ξ, which are not in A}

    = {e, g, h, i, j}


(A ∩ B)' = {elements of ξ which are not in A ∩ B}

            = {a, b, c, e, g, h, i, j}

(A ∪ B)' = {elements of ξ which are not in A ∪ B}

             = {h, i, j}

Chapter 3: Extra Exercise

Practice Test on Operations on Sets

1. If A = {2, 3, 4, 5}     B = {4, 5, 6, 7}     C = {6, 7, 8, 9}     D = {8, 9, 10, 11}, find

(a) A ∪ B

(b) A ∪ C

(c) B ∪ C

(d) B ∪ D

(e) (A ∪ B) ∪ C

(f) A ∪ (B ∪ C)

(g) B ∪ (C ∪ D)


2. If A = {4, 6, 8, 10, 12} B = {8, 10, 12, 14} C = {12, 14, 16} D = {16, 18}, find

(a) A ∩ B

(b) B ∩ C

(c) A ∩ (C ∩ D)

(d) A ∩ C

(e) B ∩ D

(f)(A ∩ B) ∪ C

(g) A ∩ (B ∪ D)

(h) (A ∩ B) ∪ (B ∩ C)

(i) (A ∪ D) ∩ (B ∪ C)


3. Let ξ = {1, 2, 3, 4, 5, 6, 7} and A = {1, 2, 3, 4, 5} B = {2, 5, 7} show that

(a) (A ∪ B)' = A' ∩ B'

(b) (A ∩ B)' = A' ∪ B'

(c) (A ∩ B) = B ∩ A

(d) (A ∪ B) = B ∪ A



4. Let P = {a, b, c, d}   Q = {b, d, f}   R = {a, c, e} verify that

(a) (P ∪ Q) ∪ R = P ∪ (Q ∪ R)

(b) (P ∩ Q) ∩ R = P ∩ (Q ∩ R)


Answers for practice test on operations on sets are given below to check the correct answers

Answers:

1. (a) {2, 3, 4, 5, 6, 7}
(b) {2, 3, 4, 5, 6, 7, 8, 9}
(c) {4, 5, 6, 7, 8, 9}
(d) {4, 5, 6, 7, 8, 9, 10, 11}
(e) {2, 3, 4, 5, 6, 7, 8, 9}
(f) {2, 3, 4, 5, 6, 7, 8, 9}
(g) {4, 5, 6, 7, 8, 9, 10, 11}

2. (a) {8, 10, 12}
(b) {12, 14}
(c) ∅
(d) {12}
(e) d
(f) {8, 10, 12, 14, 16}
(g) {8}
(h) {8, 10, 12, 14}
(i) {8, 10, 12, 16}


3. (a) L.H.S. = R. H. S = {6}
(b) L.H.S. = R. H. S = {1, 3, 4, 6, 7}
(c) {2, 5}
(d) {1, 2, 3, 4, 5, 7}


4. (a) {a, b, c, d, e, f}
(b) d

Chap3: Quiz 1

Quiz 1

Chapter 2: Notes

• Solve (x + 1)(x – 3) = 0 

          (x + 1)(x – 3) = 0

x + 1 = 0  or  x – 3 = 0
x = –1  or  x = 3
The solution is  x = –1, 3 

• Solve x² – 3x – 4 = 0.

x² – 3x – 4 = 0
(x + 1)(x – 4) = 0
x + 1 = 0  or  x – 4 = 0
x = –1  or  x = 4 
The solution is x = –1, 4

• Solve x² – 4 = 0

          x² – 4 = 0
      (x + 2)(x – 2) = 0
      x + 2 = 0  or  x – 2 = 0
      x = –2  or  x = 2

        The solution is x = ± 2

• Solve x² – 7x = 0.

x² – 7x = 0
x(x – 7) = 0
x = 0  or  x – 7 = 0
x = 0  or  x = 7 
The solution is x = 0, 7

 

Chapter 2 : Test Yourself!!

Try to do it!!! then only check the answer~~~~^^

REAdy ~~~~Go!

Normal Distribution (Part 3)

Normal Distribution (Part 2)

Normal Distribution Worksheet Answer (Part 1)

Answer for Normal Distribution ①