**A. SET. Let A, B and C are sets and U be universal set. U = {-1, 0, 1, 2, 3, 4, 5, 6, a, b, c, d, e} A = {-1, 1, 2, 4} B = {0, 2, 4, 6} C = {b, c, d} Find for the following. Show complete solutions. 1. π΅ βͺ πΆ 2. π΄ β π΅ π₯ πΆ 3. πππ€ππ π ππ‘ ππ πΆ 4. |π(π΅)| B. SEQUENCES. Consider the sequence {Sn} defined by Sn = 2n β 5, where π β₯ βπ. Find for: 1. β1π=β1 ππ 2. β ππ 4π=2 C. RELATION. Consider X = {-3, -2, -1, 0, 1} defined by (x,y) β R if x β₯ y. Find for: 1. Elements of R (3 pts) 2. Domain and Range of R (2 pts) 3. Draw the digraph (3 pts) 4. Identify the properties of R (2pts)**

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Solution:

(A):

U = {-1, 0, 1, 2, 3, 4, 5, 6, a, b, c, d, e}

A = {-1, 1, 2, 4}

B = {0, 2, 4, 6}

C = {b, c, d}

1. π΅ βͺ πΆ = {0, 2, 4, 6, b, c, d}

2.Β $π΅ \times πΆ=\{(0,b),(0,c),(0,d),(2,b),(2,c),(2,d),(4,b),(4,c),(4,d),(6,b),(6,c),(6,d)\}$

Now,Β $A-B\times C=\{-1, 1, 2, 4\}$

3. πππ€ππ π ππ‘ ππ πΆ$=\{\phi,\{b\}, \{c\},\{d\},\{b,c\},\{c,d\},\{b,d\},\{b,c,d\}\}$

4. |π(π΅)|$=2^n$Β , where n is the number of elements in set B.

$|P(B)|=2^4=16$

(B):

$S_n=2n-5,n\ge-1$

$\Sigma_{-1}^1 S_i=S_{-1}+S_0+S_1 \\=2(-1)-5+2(0)-5+2(1)-5 \\=-2+0+2-15 \\=-15$

$\Pi_2^4S_i=S_2\times S_3 \times S_4=(4-5)(6-5)(8-5)=(-1)(1)(3)=-3$

(C):

Consider X = {-3, -2, -1, 0, 1} defined by (x,y)Β βΒ R if xΒ β₯Β y.

1.

$R=\{(-3,-3),(-2,-2),(-1,-1),(0,0),(1,1),(-2,-3),(-1,-3),\\(0,-3),(1,-3),(-1,-2),(0,-2),(1,-2),(0,-1),(1,-1)(1,0)\}$

2. Domain of R$=\{-3,-2,-1,0,1\}$

And range of R$=\{-3,-2,-1,0,1\}$

3. Digraph of R:

4.

Reflexive:

It is clearly reflexive asΒ $(a,a)\in R, \forall a\in X$

Symmetric:

It is clearly not symmetric asΒ $a\ge b$Β butΒ $b\ge a$Β is not true,Β $\forall a,b \in X$

Moreover,Β $(0,-2)\in R$Β butΒ $(-2,0)\not\in R$

Transitive:

$a\ge b, b\ge c \Rightarrow a\ge c$Β which is trueΒ $\forall a,b,c \in X$

Hence, it is transitive.