Question #172049

A. List the members of the following sets

1. {x| x is real numbers and x2 = 1}

2. {x| x is an integer and -4 < x ≤ 3}


B. Use set builder notation to give description of each of these sets.

1. {a, e,i ,o, u}

2. {=2, -1, 0, 1, 2}

C. Let A= (a, b, c), B = (x, y) and C = (0, 1)

Find:

1. A U C

2. C x B

3. B – A

4. (A ∩ C) U B


D. Find these terms of the sequence (An}, where An = 2(3)n + 5

1. A0

2. A5

3. A3

4. 8th term

5. 2nd term

6. Sum of the sequence


E. Given the following set:

2. X = {-1, 0, 1, 2, 3, 4, 5} defined by the rule (x, y) ∈R if x ≤ y

F. List the elements of R

G. Find the domain of R

H. Find the range of R

I. Draw the digraph

J. Properties of the Relation


1
Expert's answer
2021-03-18T15:47:47-0400

A. (1). {1,1-1,1 }

(2). {3,2,1,0,1,2,3-3,-2,-1,0,1,2,3 }


B. (1). {xxx|x is a vowel}

(2). {xxx|x is an integer and 2x2-2\leq x\le 2 }


C.(1) AC=A\cup C = {a,b,c,0,1a,b,c,0,1 }

(2) C×B=C×B= {(0,x),(0,y),(1,x),(1,y)(0,x),(0,y),(1,x),(1,y) }

(3) BA=B-A= { x,yx,y }

(4) (AC)B=(A\cap C)\cup B= ϕB=B=\phi \cup B =B= {x,yx,y } [ Since AA and C have no common element therefore AC=ϕA\cup C=\phi ]


D. Given sequence is An=23.n+5A_n=2^3.n+5

(1) A0=5A_0=5

(2) A5=(23.5+5)=45A_5=( 2^3.5+5)=45

(3) A3=(23.3+5)=29A_3=( 2^3.3+5)=29

(4) 8 th term =A7=(23.7+5)=61=A_7=( 2^3.7+5)=61

(5) 2nd term =A1=(23.1+5)=13=A_1=( 2^3.1+5)=13

(6) Sum of the sequence =Sn=(A1+A2+A3+......+An)=S_n=(A_1+A_2+A_3+......+A_n)

=23(1+2+3+....+n)+5n=2^3(1+2+3+....+n)+5n

=23.n.(n+1)2+5n=2^3.\frac{n.(n+1)}{2}+5n

=n[4n+9]=n[4n+9]


E. (F). R=R= {(1,1),(1,0),(1,1),(1,2),(1,3),(1,4),(1,5),(0,0),(0,1),(0,2),(0,3),(0,4),(0,5),(1,1)(-1,-1),(-1,0),(-1,1),(-1,2),(-1,3),(-1,4),(-1,5),(0,0),(0,1),(0,2),(0,3),(0,4),(0,5),(1,1)

(1,2),(1,3),(1,4),(1,5),(2,2)(2,3),(2,4),(2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,5)(1,2),(1,3),(1,4),(1,5),(2,2)(2,3),(2,4),(2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,5) }


(G)Domain of RR == {1,0,1,2,3,4,5-1,0,1,2,3,4,5 }

(H) Range of R=R= {1,0,1,2,3,4,5-1,0,1,2,3,4,5 }

(I) Digraph of R given below



(J) As we seen above that their is a loop in each point. Therefore RR is reflexive.

But not symmetric as (0,2)R(0,2)\in R but (2,0)R(2,0)\notin R ,

Also the relation is transitive .


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