\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

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

\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}
\displaystyle \text{by }xRy\Leftrightarrow y=3x,\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 xRy\Leftrightarrow y=3x
\displaystyle \text{For }x=1,\ y=3;\qquad x=2,\ y=6;\qquad x=3,\ y=9.
\displaystyle \text{For }x\geq4,\ y=3x>9,\text{ so }y\notin A.
\displaystyle \therefore R=\{(1,3),(2,6),(3,9)\}
\displaystyle \text{This relation is not among options (a), (b) and (c).}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Let }A=\{1,2,3\},\ B=\{1,3,5\}.\text{ If relation }R\text{ from }A\text{ to }B
\displaystyle \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 \text{The inverse relation is obtained by interchanging the components of each ordered pair.}
\displaystyle R^{-1}=\{(3,1),(5,2),(3,3)\}
\displaystyle =\{(3,3),(3,1),(5,2)\}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\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
\displaystyle \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\}
\displaystyle \text{(d) none of these.}
\displaystyle \text{Answer:}
\displaystyle R=\{(x,y):x\in A,\ y\in B\text{ and }x>y\}
\displaystyle \text{For }x=2,\ y=1,\text{ and for }x=3,\ y=1.
\displaystyle \therefore R=\{(2,1),(3,1)\}
\displaystyle \therefore \text{Range}(R)=\{1\}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }R=\{(x,y):x,y\in Z,\ x^2+y^2\leq4\}\text{ is a relation on }Z,
\displaystyle \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\}
\displaystyle \text{(d) none of these.}
\displaystyle \text{Answer:}
\displaystyle x^2+y^2\leq4
\displaystyle \text{Since }y^2\geq0,\text{ we must have }x^2\leq4.
\displaystyle \therefore -2\leq x\leq2
\displaystyle \text{For each }x=-2,-1,0,1,2,\text{ there exists an integer }y\text{ satisfying the relation.}
\displaystyle \therefore \text{Domain}(R)=\{-2,-1,0,1,2\}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A relation }R\text{ is defined from }\{2,3,4,5\}\text{ to }\{3,6,7,10\}\text{ by:}
\displaystyle 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\}
\displaystyle \text{(d) }\{2,3,4,5\}.
\displaystyle \text{Answer:}
\displaystyle \gcd(2,3)=1,\qquad \gcd(3,7)=1
\displaystyle \gcd(4,3)=1,\qquad \gcd(5,3)=1
\displaystyle \text{Thus, each element of }\{2,3,4,5\}\text{ is related to at least one element of the second set.}
\displaystyle \therefore \text{Domain}(R)=\{2,3,4,5\}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A relation }\phi\text{ from }C\text{ to }R\text{ is defined by }x\phi y\Leftrightarrow |x|=y.
\displaystyle \text{Which one is correct?}
\displaystyle \text{(a) }(2+3i)\phi13\qquad\text{(b) }3\phi(-3)\qquad\text{(c) }(1+i)\phi2\qquad\text{(d) }i\phi1
\displaystyle \text{Answer:}
\displaystyle x\phi y\Leftrightarrow |x|=y
\displaystyle |2+3i|=\sqrt{2^2+3^2}=\sqrt{13}\ne13
\displaystyle |3|=3\ne-3
\displaystyle |1+i|=\sqrt{1^2+1^2}=\sqrt2\ne2
\displaystyle |i|=1
\displaystyle \therefore i\phi1
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\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\Rightarrow x=8-2y
\displaystyle \text{Since }x,y\in N,\text{ take }y=1,2,3.
\displaystyle y=1\Rightarrow x=6,\qquad y=2\Rightarrow x=4,\qquad y=3\Rightarrow x=2
\displaystyle \therefore \text{Domain}(R)=\{2,4,6\}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{R is a relation from }\{11,12,13\}\text{ to }\{8,10,12\}\text{ defined by}
\displaystyle 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 y=x-3
\displaystyle x=11\Rightarrow y=8,\qquad x=12\Rightarrow y=9,\qquad x=13\Rightarrow y=10
\displaystyle \text{Since }9\notin\{8,10,12\},\ (12,9)\text{ does not belong to }R.
\displaystyle \therefore R=\{(11,8),(13,10)\}
\displaystyle \therefore R^{-1}=\{(8,11),(10,13)\}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If the set }A\text{ has }p\text{ elements, }B\text{ has }q\text{ elements, then the number}
\displaystyle \text{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)
\displaystyle =p\times q=pq
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\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
\displaystyle \text{(d) }R\subseteq B\times A.
\displaystyle \text{Answer:}
\displaystyle \text{By definition, 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{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }R\text{ is a relation from a finite set }A\text{ having }m\text{ elements to a}
\displaystyle \text{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)=n(A)n(B)=mn
\displaystyle \text{Every subset of }A\times B\text{ defines a relation from }A\text{ to }B.
\displaystyle \therefore \text{Number of relations}=2^{mn}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }R\text{ is a relation on a finite set having }n\text{ elements, then the number}
\displaystyle \text{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(A)n(A)=n^2
\displaystyle \text{A relation on }A\text{ is any subset of }A\times A.
\displaystyle \therefore \text{Number of relations on }A=2^{n^2}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{If }A=\{1,2,4\},\ B=\{2,4,5\}\text{ and }C=\{2,5\},\text{ write}
\displaystyle (A-C)\times(B-C).
\displaystyle \text{Answer:}
\displaystyle A-C=\{1,4\}\text{ and }B-C=\{4\}
\displaystyle \therefore (A-C)\times(B-C)=\{(1,4),(4,4)\}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }n(A)=3,\ n(B)=4,\text{ then write }n(A\times A\times B).
\displaystyle \text{Answer:}
\displaystyle n(A\times A\times B)=n(A)\times n(A)\times n(B)
\displaystyle =3\times3\times4=36
\displaystyle \therefore n(A\times A\times B)=36
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }R\text{ is a relation defined on the set }Z\text{ of integers by the rule}
\displaystyle (x,y)\in R\Leftrightarrow x^2+y^2=9,\text{ then write domain of }R.
\displaystyle \text{Answer:}
\displaystyle x^2+y^2=9,\qquad x,y\in Z
\displaystyle \text{The possible ordered pairs are }(\pm3,0)\text{ and }(0,\pm3).
\displaystyle \therefore \text{Domain}(R)=\{-3,0,3\}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }R=\{(x,y):x,y\in Z,\ x^2+y^2\leq4\}\text{ is a relation defined}
\displaystyle \text{on the set }Z\text{ of integers, then write domain of }R.
\displaystyle \text{Answer:}
\displaystyle x^2+y^2\leq4,\qquad x,y\in Z
\displaystyle \text{Since }y^2\geq0,\text{ we have }x^2\leq4.
\displaystyle \therefore x\in\{-2,-1,0,1,2\}
\displaystyle \therefore \text{Domain}(R)=\{-2,-1,0,1,2\}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }R\text{ is a relation from set }A=\{11,12,13\}\text{ to set}
\displaystyle B=\{8,10,12\}\text{ defined by }y=x-3,\text{ then write }R^{-1}.
\displaystyle \text{Answer:}
\displaystyle y=x-3
\displaystyle x=11\Rightarrow y=8,\qquad x=12\Rightarrow y=9,\qquad x=13\Rightarrow y=10
\displaystyle \text{Since }9\notin B,\quad R=\{(11,8),(13,10)\}.
\displaystyle \therefore R^{-1}=\{(8,11),(10,13)\}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Let }A=\{1,2,3\}\text{ and }R=\{(a,b):|a^2-b^2|\leq5,\ a,b\in A\}.
\displaystyle \text{Then write }R\text{ as set of ordered pairs.}
\displaystyle \text{Answer:}
\displaystyle |1^2-3^2|=|3^2-1^2|=8>5
\displaystyle \text{All the other ordered pairs in }A\times A\text{ satisfy }|a^2-b^2|\leq5.
\displaystyle \therefore R=\{(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)\}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Let }R=\{(x,y):x,y\in Z,\ y=2x-4\}.\text{ If }(a,-2)\text{ and}
\displaystyle (4,b^2)\in R,\text{ then write the values of }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle \text{Since }(a,-2)\in R,\quad -2=2a-4.
\displaystyle \therefore 2a=2\Rightarrow a=1
\displaystyle \text{Since }(4,b^2)\in R,\quad b^2=2(4)-4=4.
\displaystyle \therefore b=\pm2
\displaystyle \therefore a=1\text{ and }b=\pm2
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }R=\{(2,1),(4,7),(1,-2),\ldots\},\text{ then write the linear relation}
\displaystyle \text{between the components of the ordered pairs of the relation }R.
\displaystyle \text{Answer:}
\displaystyle \text{Let the linear relation be }y=mx+c.
\displaystyle \text{Using }(2,1)\text{ and }(4,7),\quad m=\frac{7-1}{4-2}=3.
\displaystyle \therefore y=3x+c
\displaystyle \text{Using }(2,1),\quad 1=3(2)+c\Rightarrow c=-5.
\displaystyle \therefore y=3x-5
\displaystyle \text{Also, }(1,-2)\text{ satisfies }y=3x-5.
\displaystyle \therefore \text{The required linear relation is }y=3x-5.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }A=\{1,3,5\}\text{ and }B=\{2,4\},\text{ list the elements of }R,\text{ if}
\displaystyle R=\{(x,y):x,y\in A\times B\text{ and }x>y\}.
\displaystyle \text{Answer:}
\displaystyle A\times B=\{(1,2),(1,4),(3,2),(3,4),(5,2),(5,4)\}
\displaystyle \text{The ordered pairs satisfying }x>y\text{ are }(3,2),(5,2)\text{ and }(5,4).
\displaystyle \therefore R=\{(3,2),(5,2),(5,4)\}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }R=\{(x,y):x,y\in W,\ 2x+y=8\},\text{ then write the domain}
\displaystyle \text{and range of }R.
\displaystyle \text{Answer:}
\displaystyle 2x+y=8\Rightarrow y=8-2x
\displaystyle \text{Since }x,y\in W,\quad x=0,1,2,3,4.
\displaystyle \therefore R=\{(0,8),(1,6),(2,4),(3,2),(4,0)\}
\displaystyle \therefore \text{Domain}(R)=\{0,1,2,3,4\}
\displaystyle \text{Range}(R)=\{0,2,4,6,8\}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Let }A\text{ and }B\text{ be two sets such that }n(A)=3\text{ and }n(B)=2.
\displaystyle \text{If }(x,1),(y,2),(z,1)\text{ are in }A\times B,\text{ write }A\text{ and }B.
\displaystyle \text{Answer:}
\displaystyle (x,1),(y,2),(z,1)\in A\times B
\displaystyle \therefore x,y,z\in A\text{ and }1,2\in B.
\displaystyle \text{Since }n(A)=3\text{ and }n(B)=2,
\displaystyle \therefore A=\{x,y,z\}\text{ and }B=\{1,2\}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Let }A=\{1,2,3,5\},\ B=\{4,6,9\}\text{ and }R\text{ be a relation from }A
\displaystyle \text{to }B\text{ defined by }R=\{(x,y):x-y\text{ is odd}\}.\text{ Write }R\text{ in roster form.}
\displaystyle \text{Answer:}
\displaystyle x-y\text{ is odd when one of }x,y\text{ is odd and the other is even.}
\displaystyle \therefore R=\{(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)\}
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.