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3.2 Subset, Universal Set and the Complement of a Set

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3.2b Universal Set
1. Universal set is a set that contains all the elements under consideration.
2. In set notation, the symbol ξ denotes a universal set.

Example:
Given that the universal set, ξ = {whole numbers less than 9}, A = {prime number} and B = {multiple of 4}.
(a) List all the elements of set A and set B.
(b) Illustrate the relationships between the following sets using Venn diagrams.
(i) ξ and A
(ii) ξ, A and B

Solution:
(a) ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8}
 A = {2, 3, 5, 7}
 B = {4, 8}
 
(b)(i)








(b)(ii)

3.4 SPM Practice (Short Questions)

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Question 1:

List all the subsets of set P = {r, s}.

Solution:

There are 2 elements, so the number of subsets of set P is 2n = 22 = 4.

Set P = {r, s}
Therefore subsets of set P = {r}, {s}, {r, s}, {}

Question 2:


Diagram above shows a Venn diagram with the universal set,  ξ = Q ∪ P. List all the subset of set P.

Solution:

Set P has 3 elements, so the number of subsets of set P is 2n = 23 = 8.

Set P = {2, 3, 5}
Therefore subsets of set P = {}, {2}, {3}, {5}, {2, 3}, {2, 5}, {3, 5}, {2, 3, 5}.

Question 3:

It is given that the universal set, ξ = {x : 30 ≤ x < 42, x is an integer} and set P = {x : x is a number such that the sum of it its two digits is an even number}.
Find set P’.

Solution:

ξ = {30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41}
P = {31, 33, 35, 37, 39, 40}

Therefore P’ = {30, 32, 34, 36, 38, 41}

Question 4:

Given that universal set ξ = {x : 3 < x ≤ 16, x is an integer},
Set A = {4, 11, 13, 16},
Set B = {x : x is an odd number} and
Set C = {x : x is a multiple of 3}.
The elements of the set (A ∪ C)’ ∩ B are

Solution:

ξ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}
A = {4, 11, 13, 16}
B = {5, 7, 9, 11, 13, 15}
C = {6, 9, 12, 15}

(A∪C)' = {5, 6, 7, 8, 10, 14}
Therefore  (A∪C)’ ∩ B = {5, 7}

 

 

 

3.4 SPM Practice (Short Questions)

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Question 5:

Diagram below shows a Venn diagram with the number of elements of set P, set Q and set R.

It is given that the universal set, ξ=P∪Q∪R and n(Q‘)=n(Q∩R).
Find the value of x.

Solution:

n(Q)’ = n(Q ∩ R)
3 + 8 + 5 = x– 3 + 9
16 = x + 6
x = 10

Question 6:

Diagram below is a Venn diagram showing the number of quiz participants in set P, set Q and set R.
It is given that the universal set, ξ=P∪Q∪R  , set P = {Science quiz participants}, set Q = {Mathematics quiz participants} and set R = {History quiz participants}.

If the number of participants who participate in only one quiz is 76, find the total number of the participants.

Solution:

Number of participants who participate in only one quiz = 76
(5x – 2) + (x + 6) + (2x + 8) = 76
8x + 12 = 76
8x = 64
x = 8
Total number of the participants
= 76 + 7 + 4 + 5 + 3(8)
= 116

Question 7:

Diagram below is a Venn diagram showing the number of students in set K, set L and set M.
It is given that the universal set, ξ=K∪L∪M  , set K = {Karate Club}, set L = {Life Guards Club} and set M = {Martial Arts Club}.

If the number of students who join both the Life Guards Club and the Martial Arts Club is 8, find the number of students who join only two clubs.

Solution:

Number of students who join both the Life Guards Club and the Martial Arts Club = n(L ∩ M) = 2 + 2x
2 + 2x = 8
2x = 6
x = 3
Number of students who join only two clubs
= x + 4 + 2x
= 3 + 4 + 2(3)
= 13

 

 

 

 

 

3.4 SPM Practice (Long Questions)

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Question 1:

The Venn diagrams in the answer space shows sets X, Y and Z such that the universal set, ξ = X ∪ Y ∪ Z
On the diagrams in the answer space, shade

( a ) X’ ∩ Y , ( b ) ( X ∪ Y’ ) ∩ Z
 

 Solution:

  1. X’ ∩ Y means the intersection of the region outside X with the region Y.
    • Find the region of (X ∪ Y’) first.
    • (X ∪ Y’) means the union of the region X and the region outside Y.
    • The region then intersects with region Z to give the result of  (X ∪ Y’) ∩ Z.

Question 2:

The Venn diagrams in the answer space shows sets P, Q and R such that the universal set,
On the diagrams in the answer space, shade

  1. Q ∩ R,
  2. (P’ ∩ R) ∪ Q.

Solution:

  1. Q ∩ R means the intersection of the region Q and the region R.
    • Find the region of (P’ ∩ R) first.
    • (P’ ∩ R) means the region that is outside P and is inside R.
    • The union of this region with region Q give the result of (P’ ∩ R) ∪ Q.

 

 

 

3.3 Operations on Sets

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3.3b Union of Sets (Part 1)
 
1. The union of set A and set B, denoted by A υ B is the set consisting of all elements in set A or set B or both the sets.
The Venn diagram of A υ B is illustrated as below:

 

2. The union of set A, set B and set C, denoted by A υ B υ C is the set consisting of all elements in set A, set B or set C or all the three sets.
The Venn diagram of A υ B υ C is illustrated as below:


 

2.4 Roots of Quadratic Equations

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2.4 Roots of Quadratic Equations
1. A root of quadratic equation is the value of the unknown which satisfies the quadratic equation.

2. Roots of an equation are also called the solution of an equation.

3. To solve a quadratic equation by the factorisation method, follow the steps below:

Step 1: Express the quadratic equation in general form ax2 + bx + c = 0.
Step 2: Factorise the quadratic expression ax2 + bx + c = 0 as the product of two linear expressions, that is, (mx+ p) (nx + q) = 0.
Step 3: Equate each factor to zero and obtain the roots or solutions of the quadratic equation.

mx+p=0      or      nx+q=0     x=− p m       or          x=− q n


Example 1:
Solve the quadratic equation  2 x 2 − 5 3 = 3 x

Solution:
2 x 2 − 5 3 = 3 x
2x2 – 5 = 9x
2x2 – 9x – 5 = 0
(x – 5)(2x + 1) = 0
x – 5 = 0, x = 5
or 2x + 1 = 0
x = − 1 2  
Therefore, x = 5 and x = ½ are roots or solutions of the quadratic equation.


Example 2:
Solve the quadratic equation 4x2 – 12 = –13x
 
Solution:
4x2 – 12 = –13x
4x2 + 13x – 12 = 0
(4x – 3)(x + 4) = 0
4x – 3 = 0,  x = 3 4
or x + 4 = 0
x = –4


Example 3:
Solve the quadratic equation 5x2 = 3 (x + 2) – 4

Solution:
5x2 = 3 (x + 2) – 4
5x2 = 3x + 6 – 4
5x2 – 3x – 2 = 0
(5x + 2)(x – 1) = 0
5x + 2 = 0,  x = − 2 5
or x – 1 = 0
x = 1


Example 4:
Solve the quadratic equation 
3 x ( x − 3 ) 4 = − x + 3.
 
Solution:
3 x ( x − 3 ) 4 = − x + 3
3x2 – 9x = – 4x + 12
3x2 – 5x – 12 = 0
(3x + 4)(x – 3) = 0
3x + 4 = 0,  x = − 4 3
or x – 3 = 0
x = 3

2.1 Quadratic Expression (Sample Questions)

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Example 1:
Form a quadratic expression by multiplying each of the following.
(a) (6p – 2)(2p – 1)
(b)   (m + 5)(4 – 7m)
(c)    (x + 2) (2x – 3)

Solution:
(a) (6p – 2)(2p – 1)
= (6p)(2p) + (6p)(-1) + (-2)(2p) +(-2)(-1)
= 12p2 – 6p – 4p + 2
= 12p2 – 10p + 2

(b) (m + 5)(4 – 7m)
= (m)(4) + (m)(-7m) + (5)(4) + (5)(-7m)
= 4m – 7m2 + 20 – 35m
= – 7m2 – 31m+ 20

(c) (x + 2) (2x – 3)
= (x)(2x) + (x)(-3) + (2)(2x) + (2)(-3)
= 2x2-3x + 4x – 6
= 2x2 + x - 6

Quadratic Equations Long Questions (Question 8 – 10)

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Question 8:
Solve the following quadratic equation:
4x (x + 4) = 9 + 16x

Solution:
4x( x+4 )=9+16x 4 x 2 +16x=9+16x    4 x 2 −9=0 ( 2x ) 2 − 3 2 =0 ( 2x+3 )( 2x−3 )=0 2x+3=0     or     2x−3=0      2x=−3                2x=3        x=− 3 2                  x= 3 2    


Question 9:
Solve the following quadratic equation:
(x + 2)2 = 2x + 7

Solution:
( x+2 ) 2 =2x+7 x 2 +4x+4=2x+7 x 2 −2x−3=0 ( x+1 )( x−3 )=0 x+1=0     or     x−3=0      x=−1                 x=3


Question 10:
Solve the following quadratic equation:
− 2 3x−5 = x 3x−1

Solution:
              − 2 3x−5 = x 3x−1          −2( 3x−1 )=x( 3x−5 )               −6x+2=3 x 2 −5x 3 x 2 −5x+6x−2=0          3 x 2 +x−2=0    ( 3x−2 )( x+1 )=0 3x−2=0     or     x+1=0      3x=2                  x=−1        x= 2 3                  x=−1

2.3 Quadratic Equations

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2.3 Quadratic Equations
1. Quadratic equations are equations which fulfil the following characteristics:
(a) Have an equal ‘=’ sign
(b) Contain only one unknown
(c) Highest power of the unknown is 2.
 
For example,

 
2. The general form of a quadratic equation is written as:
(a) 
ax2 + bx + c = 0,
where a ≠ 0, b ≠ 0 and c ≠ 0,
example: 4x2+ 13x – 12 = 0

(b) 
ax2 + bx = 0,
where a ≠ 0, b ≠ 0 but c = 0,
example: 7x2+ 9x = 0

(c) 
ax2 + c = 0,
where a ≠ 0, c ≠ 0 but b = 0,
example: 9x2– 3 = 0



Example 1:
Write each quadratic equation in the general form.
(a) x2 – 5x = 12
(b) -2 + 5x2– 6x = 0
(c) 7p2 – 3p = 4p2+ 4p – 3
(d) (x – 2)(x + 6) = 0
(e) 3 – 13x = 4 (x2 + 2)
(f)   2 − y = 1 − 3 y y
(g)  p 4 = 2 p 2 − 3 10
(h) y 2 + 5 4 = y − 1 2
(i) 4 p 7 = p ( 7 p − 6 )


Solution:
A quadratic equation in the general form is written as ax2 + bx + c = 0
(a) x2 – 5x = 12
x2 – 5x -12 = 0

(b) –2 + 5x2– 6x = 0
5x2 – 6x –2 = 0


(c) 7p2 – 3p = 4p2+ 4p – 3
7p2 – 3p – 4p2– 4p + 3 = 0
3p2 – 7p  + 3 = 0

(d) (x – 2)(x + 6) = 0
x2 + 6x – 2x– 12 = 0
x2 + 4x – 12 = 0

(e) 3 – 13x = 4 (x2 + 2)
3 – 13x = 4x2 + 8
–4x2 – 8 + 3 – 13x = 0
–4x2 – 13x – 5 = 0
4x2 + 13x + 5 = 0

(f) 2 − y = 1 − 3 y y
2y – y2 = 1 – 3y
2y – y2 – 1 + 3y = 0
 – y2 + 5y – 1 = 0
y2 – 5y + 1 = 0


(g)  p 4 = 2 p 2 − 3 10
10p = 8p2 – 12
–8p2 + 10p +12 = 0
8p2 – 10p – 12 = 0

(h) y 2 + 5 4 = y − 1 2
2y2 + 10 = 4y – 4
2y2 – 4y + 10 + 4 = 0
2y2 – 4y + 14 = 0

(i)  4 p 7 = p ( 7 p − 6 )
4p = 7p (7p– 6)
4p = 49p2 – 42p
– 49p2 + 42p + 4p  = 0
49p2 – 46p = 0

Quadratic Equations Long Questions (Question 5 – 7)

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Question 5:
Solve the equation:
(m + 2)(m – 4) = 7(m – 4).
 
Solution:
(m + 2)(m – 4) = 7(m – 4)
m2– 4m + 2m – 8 = 7m – 28
m2– 9m + 20 = 0
(m – 5)(m – 4) = 0
m = 5 or m = 4   


Question 6:
Solve the equation:
− 6 y − 2 y = 7 1 − y

Solution:
−6y−2 y = 7 1−y ( −6y−2 )( 1−y )=7y −6y+6 y 2 −2+2y−7y=0 6 y 2 −11y−2=0 ( 6y+1 )( y−2 )=0 6y+1=0        or        y=2 y=− 1 6


Question 7:
Solve the equation:
4 m 7 = m ( 8 m − 9 )

Solution:
4m 7 =m( 8m−9 ) 4m=7m( 8m−9 ) 4m=56 m 2 −63m 56 m 2 −63m−4m=0 56 m 2 −67m=0 m(56m−67)=0 m=0        or        56m−67=0                                        m= 67 56