Exemplar Solution

$\begin{array}{l}L.H.S.=A\cup \left(B-A\right)=A\cup \left(B\cap A\text{'}\right)\left[?A-B=A\cap B\text{'}\right]\\ =\left(A\cup B\right)\cap \left(A\cup A\text{'}\right)\\ =A\cup B\cap U\left[?A\cup A\text{'}=U\right]\\ =A\cup B=R.H.S.\left[?A\cup U=A\right]\\ Hence,thegivenstatementisproved.\end{array}$

$\begin{array}{l}Letx\in A\cup B\\ \Rightarrow x\in Aorx\in B\\ \Rightarrow x\in Corx\in C\left[?A\subset CandB\subset C\right]\\ \Rightarrow x\in C\Rightarrow A\cup B\subset C\\ Hence,thegivenstatementis\text{'}True\text{'}.\end{array}$

$\begin{array}{l}Letx\in A\cup C\\ \Rightarrow x\in Aorx\in C\\ \Rightarrow x\in Borx\in C\left[?A\subset B\right]\\ \Rightarrow x\in \left(B\cup C\right)\Rightarrow \left(A\cup C\right)\subset \left(B\cup C\right)\\ Hence,thegivenstatementis\text{'}True\text{'}.\end{array}$

$\begin{array}{l}Letx\in A\cap C\\ \Rightarrow x\in Aandx\in C\\ \Rightarrow x\in Bandx\in C\\ \Rightarrow x\in \left(B\cap C\right)\Rightarrow \left(A\cap C\right)\subset \left(B\cap C\right)\\ Hence,thegivenstatementis\text{'}True\text{'}.\end{array}$

$\begin{array}{l}L.H.S.=\left(A-B\right)\cup \left(A\cap B\right)\\ =\left[\left(A-B\right)\cup A\right]\cap \left[\left(A-B\right)\cup B\right]\\ =A\cap \left(A\cup B\right)=A=R.H.S.\\ Hence,thegivenstatementis\text{'}True\text{'}.\end{array}$

$\begin{array}{l}Toprove\left(A-B\right)\cap \left(A-C\right)=A-\left(B\cup C\right)\\ L.H.S.Letx\in \left(A-B\right)\cap \left(A-C\right)\\ \Rightarrow x\in \left(A-B\right)andx\in \left(A-C\right)\\ \Rightarrow \left(x\in Aandx\notin B\right)and\left(x\in Aandx\notin C\right)\\ \Rightarrow x\in Aand\left(x\notin Bandx\notin C\right)\\ \Rightarrow x\in A-\left(B\cup C\right)\\ So,\left(A-B\right)\cap \left(A-C\right)\subset A-\left(B\cup C\right)\dots \left(i\right)\\ R.H.S.Lety\in A-\left(B\cup C\right)\\ \Rightarrow y\in Aandy\notin \left(B\cup C\right)\\ \Rightarrow y\in Aand\left(y\notin Bandy\notin C\right)\\ \Rightarrow \left(y\in Aandy\notin B\right)and\left(y\in Aandy\notin C\right)\\ \Rightarrow y\in \left(A-B\right)andy\in \left(A-C\right)\\ So,A-\left(B\cup C\right)\subset \left(A-B\right)\cap \left(A-C\right)\dots \left(ii\right)\\ Fromeq.\left(i\right)and\left(ii\right),weget\\ A-\left(B\cup C\right)=\left(A-B\right)\cap \left(A-C\right)\end{array}$

Kindly go through the solution

$\begin{array}{l}Giventhat:Y=\left\{1,2,3,\dots ,10\right\}\\ \left(i\right)\left\{a\in Ybut{a}^2\notin Y\right\}=\left\{4,5,6,7,8,9,10\right\}\\ \left(ii\right)\left\{a+1=6,a\in Y\right\}=\left\{5\right\}\\ \left(iii\right)\left\{a<6anda\in Y\right\}=\left\{1,2,3,4,5\right\}\end{array}$

$\begin{array}{l}Wearegiventhat:X=\left\{1,2,3\right\}\\ \left(i\right)\left\{4n|n\in X\right\}=\left\{4,8,12\right\}\left(ii\right)\left\{n+6|n\in X\right\}=\left\{7,8,9\right\}\\ \left(iii\right)\left\{\frac{n}{2}|n\in X\right\}=\left\{\frac{1}{2},1,\frac{3}{2}\right\}\left(iv\right)\left\{\left(n-1\right)|n\in X\right\}=\left\{0,1,2\right\}\end{array}$

^{$\begin{array}{l}Wearegiventhat:N=\left\{1,2,3,4,5,\dots ,100\right\}\\ \left(i\right)Requiredsubsetwhoseelementsareeven\\ =\left\{2,4,6,8,\dots ,100\right\}\\ \left(ii\right)Requiredsubsetwhoseelementsareperfectsquares\\ =\left\{1,4,9,16,25,36,\dots ,100\right\}\end{array}$}