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## Here's the Solution to this Question

A. (1). {$-1,1$ }

(2). {$-3,-2,-1,0,1,2,3$ }

B. (1). {$x|x$ is a vowel}

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

C.(1) $A\cup C =$ {$a,b,c,0,1$ }

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

(3) $B-A=$ { $x,y$ }

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

D. Given sequence is $A_n=2^3.n+5$

(1) $A_0=5$

(2) $A_5=( 2^3.5+5)=45$

(3) $A_3=( 2^3.3+5)=29$

(4) 8 th term $=A_7=( 2^3.7+5)=61$

(5) 2nd term $=A_1=( 2^3.1+5)=13$

(6) Sum of the sequence $=S_n=(A_1+A_2+A_3+......+A_n)$

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

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

$=n[4n+9]$

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 of R given below

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

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

Also the relation is transitive .