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## Here's the Solution to this Question

1 . Factors of $231$ are $1,3,7,11,21,33,77$

In set notation$,A=$ {$1,3,7,11,21,33,77$}

1 . These numbers are in succesive powers of 3.

So,in set builder form,$A=$ {$x\in 3^n | 6\geqslant n \geqslant 0 \ \ \& \ n\neq1$ }

2 .A={$x\in Q|-4.1 }

3 .As the given numbers are negative even numbers,this means they must be divisible by 2,So $A=$ {$7n\ | n<0\ \&\ n=0\, mod\, 2$ }

4 .Given,$A =$ {$a, b, c, 1, 2, 3, q, r, s$ }. B = {a, 1, r}, and C = {a, 3, q, x, y, z}

1 . Yes $3 \in A$

2 . False

3 . {a,1,r} $\in$ $A$ .So,Yes,B ⊂A

4 . False,As we know $1\in A,3\in A$ but {1,3} does not belong to A.

5 . Yes {${1,3}$ } as a set is a subset of A .So,{1,3}⊂A

6 . Every set is a not a subset,but a proper subset of itself .Hence,it is not true.

7 . Every set is a proper subset of itself.So,B ⊆B is true.

8 .$\phi$ is a subset of every set.Hence,∅ ⊆C

Let A, B, and C be as in 4 and let U = {1, 2, 3, 5, 7, 8, 9, a, b, c, d, e, f, g, x, y, z}.

1. A ∩B={a,1,r}

2 .A ∩C={a,3,q}

3 .A U B={a, b, c, 1, 2, 3, q, r, s}

4 .A U C={a, b, c, 1, 2, 3, q, r, s,x,y,z}

5 .A-B={ b, c, 2, 3, q,s}

6 .A -C={ b, c, 1, 2, r, s}

7 .B-A=$\phi$

8 .C-A={x,y,z}

9 .$A^c=$ {5,7,8,9,d,e,f,x,y,z}

10 .A⨁C$=(A∩C^c)U(C∩A^c)$ ={b,c,1,2,r,s,x,y,z}