Question #172398

Self - Assessment

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
2021-03-18T15:48:58-0400

Solution:

(A):

1: {1,1}\{ 1,-1\}

2: {3,2,1,0,1,3}\{ -3,-2,-1,0,1,3\}

(B):

1: {x| x is a vowel }

2: {x| x is an integer and 2x2-2\le x\le2 }

(C):

1: AC={a,b,c,0,1}A\cup C=\{ a,b,c,0,1\}

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

3: BA={x,y}B-A=\{ x,y\}

4: (AC)B={ϕ}{x,y}={x,y}(A\cap C)\cup B=\{\phi\}\cup \{x,y\}=\{x,y\}

(D):

An=2(3)n+5A_n=2(3)^n+5

(1): Put n = 0

A0=2(3)0+5=2(1)+5=7A_0=2(3)^0+5=2(1)+5=7

(2): Put n = 5

A5=2(3)5+5=2(243)+5=491A_5=2(3)^5+5=2(243)+5=491

(3): Put n = 3

A3=2(3)3+5=2(27)+5=59A_3=2(3)^3+5=2(27)+5=59

(4): For 8th term, put n = 8

A8=2(3)8+5=2(6561)+5=13127A_8=2(3)^8+5=2(6561)+5=13127

(5): For 2nd term, put n = 2

A2=2(3)2+5=2(9)+5=23A_2=2(3)^2+5=2(9)+5=23

(6): Sn=[2(3)n+5]=23n+51S_n=\sum [2(3)^n+5]=2\sum 3^n+5\sum 1

=2(30+31+...+3n)+5n=2[1(3n11)31]+5n [Using GP]=3n1+5n1=2(3^0+3^1+...+3^n)+5n \\=2[\dfrac{1(3^{n-1}-1)}{3-1}]+5n \ [\text{Using GP}] \\=3^{n-1}+5n-1

(E):

(F): 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),(1,2),(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 R = {-1, 0, 1, 2, 3, 4, 5}

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

(I): Digraph:



(J): This relation is reflexive and transitive but not symmetric as-

Reflexive: (xx)(x\le x), this is true

Transitive: (xy) &(yz)(xz)\\(x\le y)\ \& (y\le z)\Rightarrow (x\le z), this is true

Symmetric: (xy)(yx)(x\le y)\Rightarrow (y\le x), this is not true.


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