\displaystyle \textbf{Question 1. }\text{If }A=\{1,2,4\},\ B=\{2,4,5\},\ C=\{2,5\},\text{ then } \\ (A-B)\times(B-C)\text{ is}
\displaystyle \text{(a) }\{(1,2),(1,5),(2,5)\} \qquad \text{(b) }\{(1,4)\} \qquad \text{(c) }(1,4) \qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle A-B=\{1,2,4\}-\{2,4,5\}=\{1\}
\displaystyle B-C=\{2,4,5\}-\{2,5\}=\{4\}
\displaystyle \therefore (A-B)\times(B-C)=\{1\}\times\{4\}=\{(1,4)\}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 2. }\text{If }R\text{ is a relation on the set }A=\{1,2,3,4,5,6,7,8,9\}\text{ given by } \\ xRy\Leftrightarrow y=3x,
\displaystyle \text{then }R=
\displaystyle \text{(a) }\{(3,1),(6,2),(8,2),(9,3)\} \qquad \text{(b) }\{(3,1),(6,2),(9,3)\}
\displaystyle \text{(c) }\{(3,1),(2,6),(3,9)\} \qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle y=3x
\displaystyle \text{For }x=1,\ y=3 \Rightarrow (1,3)\in R
\displaystyle \text{For }x=2,\ y=6 \Rightarrow (2,6)\in R
\displaystyle \text{For }x=3,\ y=9 \Rightarrow (3,9)\in R
\displaystyle \text{For }x\geq4,\ y\notin A
\displaystyle \therefore R=\{(1,3),(2,6),(3,9)\}
\displaystyle \therefore \text{Correct option is (d) none of these.}

\displaystyle \textbf{Question 3. }\text{Let }A=\{1,2,3\},\ B=\{1,3,5\}.\text{ If relation }R\text{ from }A\text{ to }B\text{ is given by } \\ R=\{(1,3),(2,5),(3,3)\}. \text{ Then, }R^{-1}\text{ is}
\displaystyle \text{(a) }\{(3,3),(3,1),(5,2)\} \qquad \text{(b) }\{(1,3),(2,5),(3,3)\}
\displaystyle \text{(c) }\{(1,3),(5,2)\} \qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle R=\{(1,3),(2,5),(3,3)\}
\displaystyle \therefore R^{-1}=\{(3,1),(5,2),(3,3)\}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 4. }\text{If }A=\{1,2,3\},\ B=\{1,4,6,9\}\text{ and }R\text{ is a relation from }A\text{ to }B\text{ defined by }x \\ \text{ is greater than }y. \text{ The range of }R\text{ is}
\displaystyle \text{(a) }\{1,4,6,9\} \qquad \text{(b) }\{4,6,9\} \qquad \text{(c) }\{1\} \qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle R=\{(2,1),(3,1)\}
\displaystyle \therefore \text{Range of }R=\{1\}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 5. }\text{If }R=\{(x,y):x,y\in Z,\ x^2+y^2\leq4\}\text{ is a relation on }Z,\text{ then domain of }R\text{ is}
\displaystyle \text{(a) }\{0,1,2\} \qquad \text{(b) }\{0,-1,-2\} \qquad \text{(c) }\{-2,-1,0,1,2\} \qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle x^2+y^2\leq4
\displaystyle \text{Possible values of }x\text{ are }-2,-1,0,1,2
\displaystyle \therefore \text{Domain of }R=\{-2,-1,0,1,2\}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 6. }\text{A relation }R\text{ is defined from }\{2,3,4,5\}\text{ to }\{3,6,7,10\}\text{ by } \\ xRy\Leftrightarrow x\text{ is relatively prime to }y. \text{ Then, domain of }R\text{ is}
\displaystyle \text{(a) }\{2,3,5\} \qquad \text{(b) }\{3,5\} \qquad \text{(c) }\{2,3,4\} \qquad \text{(d) }\{2,3,4,5\}
\displaystyle \text{Answer:}
\displaystyle 2\text{ is relatively prime to }3\Rightarrow2\in\text{Domain}
\displaystyle 3\text{ is relatively prime to }7\Rightarrow3\in\text{Domain}
\displaystyle 4\text{ is relatively prime to }3\Rightarrow4\in\text{Domain}
\displaystyle 5\text{ is relatively prime to }3\Rightarrow5\in\text{Domain}
\displaystyle \therefore \text{Domain of }R=\{2,3,4,5\}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 7. }\text{A relation }\phi\text{ from }C\text{ to }R\text{ is defined by }z\ \phi\ y\Leftrightarrow |z|=y.\text{ Which one is correct?}
\displaystyle \text{(a) }(2+3i)\ \phi  \ 13 \qquad \text{(b) } \ 3 \ \phi  \ (-3) \qquad \text{(c) } \ (1+i) \ \phi  \ 2 \ \qquad \text{(d)} \ i \ \phi \ \ 1
\displaystyle \text{Answer:}
\displaystyle |i|=1
\displaystyle \therefore i\ \phi\ 1
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 8. }\text{Let }R\text{ be a relation on }N\text{ defined by }x+2y=8.\text{ The domain of }R\text{ is}
\displaystyle \text{(a) }\{2,4,8\} \qquad \text{(b) }\{2,4,6,8\} \qquad \text{(c) }\{2,4,6\} \qquad \text{(d) }\{1,2,3,4\}
\displaystyle \text{Answer:}
\displaystyle x+2y=8
\displaystyle \text{Possible ordered pairs are }(2,3),(4,2),(6,1)
\displaystyle \therefore \text{Domain of }R=\{2,4,6\}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 9. }\text{R is a relation from }\{11,12,13\}\text{ to }\{8,10,12\}\text{ defined by }y=x-3.\text{ Then, }R^{-1}\text{ is}
\displaystyle \text{(a) }\{(8,11),(10,13)\} \qquad \text{(b) }\{(11,8),(13,10)\}
\displaystyle \text{(c) }\{(10,13),(8,11),(12,10)\} \qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle R=\{(11,8),(13,10)\}
\displaystyle \therefore R^{-1}=\{(8,11),(10,13)\}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 10. }\text{If the set }A\text{ has }p\text{ elements, }B\text{ has }q\text{ elements, then the number of elements in } \\ A\times B\text{ is}
\displaystyle \text{(a) }p+q \qquad \text{(b) }p+q+1 \qquad \text{(c) }pq \qquad \text{(d) }p^2
\displaystyle \text{Answer:}
\displaystyle n(A\times B)=n(A)\times n(B)=pq
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 11. }\text{Let }R\text{ be a relation from a set }A\text{ to a set }B,\text{ then}
\displaystyle \text{(a) }R=A\cup B \qquad \text{(b) }R=A\cap B \qquad \text{(c) }R\subseteq A\times B \qquad \text{(d) }R\subseteq B\times A
\displaystyle \text{Answer:}
\displaystyle \text{A relation from }A\text{ to }B\text{ is a subset of }A\times B
\displaystyle \therefore R\subseteq A\times B
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 12. }\text{If }R\text{ is a relation from a finite set }A\text{ having }m\text{ elements to a finite set } \\ B\text{ having }n\text{ elements, then the number of relations from }A\text{ to }B\text{ is}
\displaystyle \text{(a) }2^{mn} \qquad \text{(b) }2^{mn}-1 \qquad \text{(c) }2mn \qquad \text{(d) }m^n
\displaystyle \text{Answer:}
\displaystyle n(A\times B)=mn
\displaystyle \therefore \text{Number of relations}=2^{mn}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 13. }\text{If }R\text{ is a relation on a finite set having }n\text{ elements, then the number of relations on }A\text{ is}
\displaystyle \text{(a) }2^n \qquad \text{(b) }2^{n^2} \qquad \text{(c) }n^2 \qquad \text{(d) }n^n
\displaystyle \text{Answer:}
\displaystyle n(A\times A)=n^2
\displaystyle \therefore \text{Number of relations on }A=2^{n^2}
\displaystyle \therefore \text{Correct option is (b).}

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