\displaystyle \textbf{Question 1: }\ \text{Show that the relation }R\text{ on the set }A=\{1,2,3,4,5\},\text{ given by}
\displaystyle R=\{(a,b):|a-b|\text{ is even}\},\text{ is an equivalence relation.} \text{Show that all the elements of } \\ \{1,3,5\}\text{ are related to each other and all the elements of }\{2,4\}\text{ are related to each} 
\displaystyle \text{other. But, no element of }\{1,3,5\}\text{ is related to any element of }\{2,4\}. \text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle R=\{(a,b):|a-b|\text{ is even}\},\text{ where }a,b\in A=\{1,2,3,4,5\}.
\displaystyle \text{We observe the following properties of relation }R.
\displaystyle \textbf{Reflexivity:}\ \text{For any }a\in A,\text{ we have}
\displaystyle |a-a|=0,\text{ which is even}
\displaystyle \therefore (a,a)\in R\text{ for all }a\in A
\displaystyle \text{So, }R\text{ is reflexive.}
\displaystyle \textbf{Symmetry:}\ \text{Let }(a,b)\in R.\text{ Then,}
\displaystyle (a,b)\in R
\displaystyle \Rightarrow |a-b|\text{ is even}
\displaystyle \Rightarrow |b-a|\text{ is even}
\displaystyle \Rightarrow (b,a)\in R
\displaystyle \text{Thus, }(a,b)\in R\Rightarrow (b,a)\in R
\displaystyle \text{So, }R\text{ is symmetric.}
\displaystyle \textbf{Transitivity:}\ \text{Let }(a,b)\in R\text{ and }(b,c)\in R.\text{ Then,}
\displaystyle (a,b)\in R\text{ and }(b,c)\in R
\displaystyle \Rightarrow |a-b|\text{ is even and }|b-c|\text{ is even}
\displaystyle \Rightarrow (a\text{ and }b\text{ both are even or both are odd})\text{ and }(b\text{ and }c\text{ both are even or both are odd})
\displaystyle \text{Now two cases arise:}
\displaystyle \textbf{CASE I:}\ \text{When }b\text{ is even}
\displaystyle \text{In this case,}
\displaystyle (a,b)\in R\text{ and }(b,c)\in R
\displaystyle \Rightarrow |a-b|\text{ is even and }|b-c|\text{ is even}
\displaystyle \Rightarrow a\text{ is even and }c\text{ is even}\qquad[\because b\text{ is even}]
\displaystyle \Rightarrow |a-c|\text{ is even}
\displaystyle \Rightarrow (a,c)\in R
\displaystyle \textbf{CASE II:}\ \text{When }b\text{ is odd}
\displaystyle \text{In this case,}
\displaystyle (a,b)\in R\text{ and }(b,c)\in R
\displaystyle \Rightarrow |a-b|\text{ is even and }|b-c|\text{ is even}
\displaystyle \Rightarrow a\text{ is odd and }c\text{ is odd}\qquad[\because b\text{ is odd}]
\displaystyle \Rightarrow |a-c|\text{ is even}
\displaystyle \Rightarrow (a,c)\in R
\displaystyle \text{Thus, }(a,b)\in R\text{ and }(b,c)\in R\Rightarrow (a,c)\in R
\displaystyle \text{So, }R\text{ is transitive.}
\displaystyle \text{Hence, }R\text{ is an equivalence relation.}
\displaystyle \text{We know that the difference of any two odd (even) natural numbers is always an} \\ \text{even natural number.}\text{Therefore, all the elements of set }\{1,3,5\}\text{ are related to each} \\ \text{other and all the elements of }\{2,4\}\text{ are related to each other.}\text{We know that the} \\ \text{difference of an even natural number and an odd natural number is an odd natural} \\ \text{number. Therefore, no element of }\{1,3,5\}\text{ is related to any element of }\{2,4\}.

\displaystyle \textbf{Question 2: }\ \text{Show that the relation }R\text{ on the set }A=\{x\in Z:0\leq x\leq12\},\text{ given by }R=\{(a,b):|a-b|\text{ is a multiple of }4\}\text{ is an equivalence relation.}
\displaystyle \text{Find the set of all elements related to }1\text{ i.e. equivalence class }[1]. \ \ \   \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle\text{We have,}
\displaystyle R=\{(a,b):|a-b|\text{ is a multiple of }4\},\text{ where }a,b\in A=\{x\in Z:0\leq x\leq12\}=\{0,1,2,\ldots,12\}.
\displaystyle \text{We observe the following properties of relation }R.
\displaystyle \textbf{Reflexivity:}\ \text{For any }a\in A,\text{ we have}
\displaystyle |a-a|=0,\text{ which is a multiple of }4.
\displaystyle \Rightarrow (a,a)\in R
\displaystyle \text{Thus, }(a,a)\in R\text{ for all }a\in A.
\displaystyle \text{So, }R\text{ is reflexive.}
\displaystyle \textbf{Symmetry:}\ \text{Let }(a,b)\in R.\text{ Then,}
\displaystyle (a,b)\in R
\displaystyle \Rightarrow |a-b|\text{ is a multiple of }4
\displaystyle \Rightarrow |a-b|=4\lambda\text{ for some }\lambda\in N
\displaystyle \Rightarrow |b-a|=4\lambda\text{ for some }\lambda\in N\qquad[\because |a-b|=|b-a|]
\displaystyle \Rightarrow (b,a)\in R
\displaystyle \text{So, }R\text{ is symmetric.}
\displaystyle \textbf{Transitivity:}\ \text{Let }(a,b)\in R\text{ and }(b,c)\in R.\text{ Then,}
\displaystyle (a,b)\in R\text{ and }(b,c)\in R
\displaystyle \Rightarrow |a-b|\text{ is a multiple of }4\text{ and }|b-c|\text{ is a multiple of }4
\displaystyle \Rightarrow |a-b|=4\lambda\text{ and }|b-c|=4\mu\text{ for some }\lambda,\mu\in N
\displaystyle \Rightarrow a-b=\pm4\lambda\text{ and }b-c=\pm4\mu
\displaystyle \Rightarrow a-c=\pm4\lambda\pm4\mu
\displaystyle \Rightarrow a-c\text{ is a multiple of }4
\displaystyle \Rightarrow |a-c|\text{ is a multiple of }4
\displaystyle \Rightarrow (a,c)\in R
\displaystyle \text{Thus, }(a,b)\in R\text{ and }(b,c)\in R\Rightarrow (a,c)\in R
\displaystyle \text{So, }R\text{ is transitive.}
\displaystyle \text{Hence, }R\text{ is an equivalence relation.}
\displaystyle \text{Let }x\text{ be an element of }A\text{ such that }(x,1)\in R.\text{ Then,}
\displaystyle |x-1|\text{ is a multiple of }4
\displaystyle \Rightarrow |x-1|=0,4,8,12
\displaystyle \Rightarrow x-1=0,4,8,12
\displaystyle \Rightarrow x=1,5,9
\displaystyle \text{Hence, the set of all elements of }A\text{ which are related to }1\text{ is }\{1,5,9\}\text{ i.e. }[1]=\{1,5,9\}.

\displaystyle \textbf{Question 3: }\ \text{Prove that the relation }R\text{ on the set }N\times N\text{ defined by}
\displaystyle (a,b)\ R\ (c,d)\iff a+d=b+c\text{ for all }(a,b),(c,d)\in N\times N \text{is an equivalence}
\displaystyle \text{relation. Also, find the equivalence classes }[(2,3)]\text{ and }[(1,3)]. \ \ \ \ \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{We observe the following properties of relation }R.
\displaystyle \textbf{Reflexivity:}\ \text{Let }(a,b)\text{ be an arbitrary element of }N\times N.\text{ Then,}
\displaystyle (a,b)\in N\times N
\displaystyle \Rightarrow a,b\in N
\displaystyle \Rightarrow a+b=b+a\qquad[\text{By commutativity of addition on }N]
\displaystyle \Rightarrow (a,b)\ R\ (a,b)
\displaystyle \text{Thus, }(a,b)\ R\ (a,b)\text{ for all }(a,b)\in N\times N.\text{ So, }R\text{ is reflexive on }N\times N.
\displaystyle \textbf{Symmetry:}\ \text{Let }(a,b),(c,d)\in N\times N\text{ be such that }(a,b)\ R\ (c,d).\text{ Then,}
\displaystyle (a,b)\ R\ (c,d)
\displaystyle \Rightarrow a+d=b+c
\displaystyle \Rightarrow c+b=d+a\qquad[\text{By commutativity of addition on }N]
\displaystyle \Rightarrow (c,d)\ R\ (a,b)\qquad[\text{By definition of }R]
\displaystyle \text{Thus, }(a,b)\ R\ (c,d)\Rightarrow (c,d)\ R\ (a,b)\text{ for all }(a,b),(c,d)\in N\times N.
\displaystyle \text{So, }R\text{ is symmetric on }N\times N.
\displaystyle \textbf{Transitivity:}\ \text{Let }(a,b),(c,d),(e,f)\in N\times N\text{ such that }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f).\text{ Then,}
\displaystyle (a,b)\ R\ (c,d)\Rightarrow a+d=b+c\Rightarrow (a+d)+(c+f)=(b+c)+(d+e)
\displaystyle (c,d)\ R\ (e,f)\Rightarrow c+f=d+e
\displaystyle \Rightarrow a+f=b+e\Rightarrow (a,b)\ R\ (e,f)
\displaystyle \text{Thus, }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f)\Rightarrow (a,b)\ R\ (e,f)\text{ for all }(a,b),(c,d),(e,f)\in N\times N.
\displaystyle \text{So, }R\text{ is transitive on }N\times N.
\displaystyle \text{Hence, }R,\text{ being reflexive, symmetric and transitive, is an equivalence relation on }N\times N.
\displaystyle [(2,3)]=\{(x,y)\in N\times N:(x,y)\ R\ (2,3)\}
\displaystyle \Rightarrow [(2,3)]=\{(x,y)\in N\times N:x+3=y+2\}=\{(x,y)\in N\times N:x-y=1\}
\displaystyle =\{(x,y)\in N\times N:y=x+1\}
\displaystyle =\{(x,x+1):x\in N\}
\displaystyle =\{(1,2),(2,3),(3,4),(4,5),\ldots\}
\displaystyle [(7,3)]=\{(x,y)\in N\times N:(x,y)\ R\ (7,3)\}
\displaystyle =\{(x,y)\in N\times N:x+3=y+7\}
\displaystyle =\{(x,y)\in N\times N:y=x-4\}
\displaystyle =\{(x,x-4)\in N\times N:x\in N\}
\displaystyle =\{(5,1),(6,2),(7,3),(8,4),(9,5),\ldots\}

\displaystyle \textbf{Question 4: }\ \text{Let }A=\{1,2,3,\ldots,9\}\text{ and }R\text{ be the relation on }A\times A \text{ defined by} \\ (a,b)\ R\ (c,d)\text{ if}  a+d=b+c\text{ for all }(a,b),(c,d)\in A\times A.\text{ Prove that }R\text{ is an} \\ \text{equivalence} \text{ relation and also obtain } \text{the equivalence class }[(2,5)]. \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{We observe the following properties of relation }R.
\displaystyle \textbf{Reflexivity:}\ \text{Let }(a,b)\text{ be an arbitrary element of }A\times A.\text{ Then,}
\displaystyle (a,b)\in A\times A
\displaystyle \Rightarrow a,b\in A
\displaystyle \Rightarrow a+b=b+a\qquad[\text{By commutativity of addition on }N]
\displaystyle \Rightarrow (a,b)\ R\ (a,b)
\displaystyle \text{Thus, }(a,b)\ R\ (a,b)\text{ for all }(a,b)\in A\times A.\text{ So, }R\text{ is reflexive on }A\times A.
\displaystyle \textbf{Symmetry:}\ \text{Let }(a,b),(c,d)\in A\times A\text{ be such that }(a,b)\ R\ (c,d).\text{ Then,}
\displaystyle (a,b)\ R\ (c,d)
\displaystyle \Rightarrow a+d=b+c
\displaystyle \Rightarrow c+b=d+a\qquad[\text{By commutativity of addition on }N]
\displaystyle \Rightarrow (c,d)\ R\ (a,b)
\displaystyle \text{Thus, }(a,b)\ R\ (c,d)\Rightarrow (c,d)\ R\ (a,b)\text{ for all }(a,b),(c,d)\in A\times A.
\displaystyle \text{So, }R\text{ is symmetric on }A\times A.
\displaystyle \textbf{Transitivity:}\ \text{Let }(a,b),(c,d),(e,f)\in A\times A\text{ such that }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f).\text{ Then,}
\displaystyle (a,b)\ R\ (c,d)\Rightarrow a+d=b+c
\displaystyle (c,d)\ R\ (e,f)\Rightarrow c+f=d+e\Rightarrow (a+d)+(c+f)=(b+c)+(d+e)
\displaystyle \Rightarrow a+f=b+e
\displaystyle \Rightarrow (a,b)\ R\ (e,f)
\displaystyle \text{Thus, }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f)\Rightarrow (a,b)\ R\ (e,f)\text{ for all }(a,b),(c,d),(e,f)\in A\times A.
\displaystyle \text{So, }R\text{ is a transitive relation on }A\times A.
\displaystyle \text{Hence, }R\text{ is an equivalence relation on }A\times A.
\displaystyle \text{Now,}
\displaystyle [(2,5)]=\{(x,y)\in A\times A:(x,y)\ R\ (2,5)\}
\displaystyle =\{(x,y)\in A\times A:x+5=y+2\}=\{(x,y)\in A\times A:y=x+3\}
\displaystyle =\{(x,x+3):x\in A\text{ and }x+3\in A\}=\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}

\displaystyle \textbf{Question 5: }\ \text{Let }N\text{ denote the set of all natural numbers and }R\text{ be the relation on }\\ N\times N\text{ defined by}
\displaystyle (a,b)\ R\ (c,d)\iff ad(b+c)=bc(a+d).\text{ Check whether }R\text{ is an equivalence relation on } \\ N\times N. \ \ \ \  \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{We observe the following properties of relation }R.
\displaystyle \textbf{Reflexivity:}\ \text{Let }(a,b)\text{ be an arbitrary element of }N\times N.\text{ Then,}
\displaystyle (a,b)\in N\times N
\displaystyle \Rightarrow a,b\in N
\displaystyle \Rightarrow ab(b+a)=ba(a+b)\qquad[\text{By commutativity of addition and multiplication on }N]
\displaystyle \Rightarrow (a,b)\ R\ (a,b)
\displaystyle \text{Thus, }(a,b)\ R\ (a,b)\text{ for all }(a,b)\in N\times N.\text{ So, }R\text{ is reflexive on }N\times N.
\displaystyle \textbf{Symmetry:}\ \text{Let }(a,b),(c,d)\in N\times N\text{ be such that }(a,b)\ R\ (c,d).\text{ Then,}
\displaystyle (a,b)\ R\ (c,d)
\displaystyle \Rightarrow ad(b+c)=bc(a+d)
\displaystyle \Rightarrow cb(d+a)=da(c+b)\qquad[\text{By commutativity of addition and multiplication on }N]
\displaystyle \Rightarrow (c,d)\ R\ (a,b)
\displaystyle \text{Thus, }(a,b)\ R\ (c,d)\Rightarrow (c,d)\ R\ (a,b)\text{ for all }(a,b),(c,d)\in N\times N.
\displaystyle \text{So, }R\text{ is symmetric on }N\times N.
\displaystyle \textbf{Transitivity:}\ \text{Let }(a,b),(c,d),(e,f)\in N\times N\text{ such that }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f).\text{ Then,}
\displaystyle (a,b)\ R\ (c,d) \\ \\ \Rightarrow ad(b+c)=bc(a+d) \\ \\ \Rightarrow \frac{b+c}{bc}=\frac{a+d}{ad}=\frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}\qquad\text{(i)}
\displaystyle \text{and,}
\displaystyle (c,d)\ R\ (e,f) \\ \\ \Rightarrow cf(d+e)=de(c+f) \\ \\ \Rightarrow \frac{d+e}{de}=\frac{c+f}{cf}=\frac{1}{d}+\frac{1}{e}=\frac{1}{c}+\frac{1}{f}\qquad\text{(ii)}
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle \left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{d}+\frac{1}{e}\right)=\left(\frac{1}{a}+\frac{1}{d}\right)+\left(\frac{1}{c}+\frac{1}{f}\right)
\displaystyle \Rightarrow \frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f} \\ \\ \Rightarrow \frac{b+e}{be}=\frac{a+f}{af} \\ \\ \Rightarrow af(b+e)=be(a+f) \\ \\ \Rightarrow (a,b)\ R\ (e,f)
\displaystyle \text{Thus, }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f) \\ \\ \Rightarrow (a,b)\ R\ (e,f)\text{ for all }(a,b),(c,d),(e,f)\in N\times N.
\displaystyle \text{So, }R\text{ is transitive on }N\times N.
\displaystyle \text{Hence, }R,\text{ being reflexive, symmetric and transitive, is an equivalence relation on }N\times N.

\displaystyle \textbf{Question 6: }\ \text{Let }R\text{ be the equivalence relation in the set }A=\{0,1,2,3,4,5\}\text{ given by} 
\displaystyle R=\{(a,b):2\text{ divides }(a-b)\}.\text{ Write the equivalence class }[0]. \ \ \ \  \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Clearly, the equivalence class }[0]\text{ is the set of those elements in }A\text{ which are related}
\displaystyle \text{to }0\text{ under the relation }R\text{ i.e. }[0]=\{(a,0)\in R:a\in A\}.
\displaystyle \text{Now,}\ (a,0)\in R
\displaystyle \Rightarrow a-0\text{ is divisible by }2\text{ and }a\in A
\displaystyle \Rightarrow a\in A\text{ such that }2\text{ divides }a
\displaystyle \Rightarrow a=0,2,4
\displaystyle \text{Thus, }[0]=\{0,2,4\}.

\displaystyle \textbf{Question 7: } \text{Show that the function }f:R\to R\text{ given by } \\ f(x)=ax+b,\text{ where }a,b\in R,\,a\neq 0\text{ is a bijection.} \hspace{2.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle  \text{Injectivity : Let }x,y\text{ be any two real numbers. Then,}
\displaystyle f(x)=f(y)\Rightarrow ax+b=ay+b\Rightarrow ax=ay\Rightarrow x=y
\displaystyle \text{Thus, }f(x)=f(y)\Rightarrow x=y\text{ for all }x,y\in R\text{(domain).}
\displaystyle \text{So, }f\text{ is an injection.}
\displaystyle \text{Surjectivity : Let }y\text{ be an arbitrary element of }R\text{(co-domain). Then,}
\displaystyle f(x)=y\Rightarrow ax+b=y\Rightarrow x=\frac{y-b}{a}
\displaystyle \text{Clearly, }x=\frac{y-b}{a}\in R\text{ (domain) for all }y\in R\text{ (co-domain). Thus, for all }y\in R\text{ (co-domain) there} \text{exists }x=\frac{y-b}{a}\in R\text{ (domain) such that}
\displaystyle f(x)=f\left(\frac{y-b}{a}\right)=a\left(\frac{y-b}{a}\right)+b=y.
\displaystyle \text{This shows that every element in co-domain has its pre-image in domain. So, }f\text{ is a surjection.}
\displaystyle \text{Hence, }f\text{ is a bijection.}

\displaystyle \textbf{Question 8: } \text{Let }f:X\to Y\text{ be a function. Define a relation }R\text{ on }X\text{ given by } \\ R=\{(a,b):f(a)=f(b)\}. \text{ Show that }R\text{ is an equivalence relation on }X. \hspace{0.5cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle  \text{We observe the following properties of relation }R.
\displaystyle \text{Reflexivity: For any }a\in X,\text{ we have}
\displaystyle f(a)=f(a)\Rightarrow (a,a)\in R\Rightarrow R\text{ is reflexive.}
\displaystyle \text{Symmetry: Let }a,b\in X\text{ be such that }(a,b)\in R.\text{ Then,}
\displaystyle (a,b)\in R\Rightarrow f(a)=f(b)\Rightarrow f(b)=f(a)\Rightarrow (b,a)\in R
\displaystyle \text{So, }R\text{ is symmetric.}
\displaystyle \text{Transitivity: Let }a,b,c\in X\text{ be such that }(a,b)\in R\text{ and }(b,c)\in R.\text{ Then,}
\displaystyle (a,b)\in R\text{ and }(b,c)\in R
\displaystyle \Rightarrow f(a)=f(b)\text{ and }f(b)=f(c)
\displaystyle \Rightarrow f(a)=f(c)
\displaystyle \Rightarrow (a,c)\in R
\displaystyle \text{So, }R\text{ is transitive.}
\displaystyle \text{Hence, }R\text{ is an equivalence relation.}

\displaystyle \textbf{Question 9: } \text{Show that }f:N\to N\text{ defined by}
\displaystyle f(n)=\begin{cases}\frac{n+1}{2},&\text{if }n\text{ is odd}\\\frac{n}{2},&\text{if }n\text{ is even}\end{cases} 
\displaystyle \text{is many-one onto function.} \hspace{2.0cm} \text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle  \text{We observe that}
\displaystyle f(1)=\frac{1+1}{2}=1\text{ and }f(2)=\frac{2}{2}=1.
\displaystyle \text{Thus, }1,2\in N\text{ such that }1\neq 2\text{ but }f(1)=f(2).\text{ So, }f\text{ is a many-one function.}
\displaystyle \text{Surjectivity: Let }n\text{ be an arbitrary element of }N.
\displaystyle \text{If }n\text{ is an odd natural number, then }2n-1\text{ is also an odd natural number such that}
\displaystyle f(2n-1)=\frac{2n-1+1}{2}=n
\displaystyle \text{If }n\text{ is an even natural number, then }2n\text{ is also an even natural number such that}
\displaystyle f(2n)=\frac{2n}{2}=n.
\displaystyle \text{Thus, for every }n\in N\text{ (whether even or odd) there exists its pre-image in }N.\text{ So, }f\text{ is a surjection.}
\displaystyle \text{Hence, }f\text{ is a many-one onto function.}

\displaystyle \textbf{Question 10: } \text{If the function }f:R\to R\text{ be given by }f(x)=x^{2}+2\text{ and } \\ g:R-\{1\}\to R\text{ be given by } g(x)=\frac{x}{x-1}.\text{ Find }fog\text{ and }gof. \hspace{2.0cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Clearly, range }f=\text{domain }g\text{ and, range }g=\text{domain }f.\text{ So, }fog\text{ and }gof\text{ both exist.}
\displaystyle \text{Now, }(fog)(x)=f(g(x))=f\left(\frac{x}{x-1}\right)=\left(\frac{x}{x-1}\right)^{2}+2=\frac{x^{2}}{(x-1)^{2}}+2
\displaystyle \text{and, }(gof)(x)=g(f(x))=g(x^{2}+2)=\frac{x^{2}+2}{(x^{2}+2)-1}=\frac{x^{2}+2}{x^{2}+1}
\displaystyle \text{Hence, }gof:R\to R\text{ and }fog:R-\{1\}\to R\text{ are given by}
\displaystyle (gof)(x)=\frac{x^{2}+2}{x^{2}+1}\text{ and }(fog)(x)=\frac{x^{2}}{(x-1)^{2}}+2

\displaystyle \textbf{Question 11: } \text{If }f(x)=e^{x}\text{ and }g(x)=\log_{e}x\,(x>0),\text{ find }fog\text{ and }gof. \\ \text{Is }fog=gof? \hspace{8.0cm} \text{[CBSE 2002]}
\displaystyle \text{Answer:}
\displaystyle  \text{We observe that}
\displaystyle \text{Domain }(f)=R,\text{ Range }(f)=(0,\infty),\text{ Domain }(g)=(0,\infty)\text{ and, Range }(g)=R.
\displaystyle \text{Computation of }fog:\text{ We observe that}
\displaystyle \text{Range }(g)=\text{Domain }(f)
\displaystyle \therefore fog\text{ exists and }fog:\text{ Domain }(g)\to R\text{ i.e. }fog:(0,\infty)\to R\text{ such that}
\displaystyle fog(x)=f(g(x))=f(\log_{e}x)=e^{\log_{e}x}=x
\displaystyle \text{Thus, }fog:(0,\infty)\to R\text{ is defined as }fog(x)=x.
\displaystyle \text{Computation of }gof:\text{ We have,}
\displaystyle \text{Range }(f)=(0,\infty)=\text{Domain }(g)
\displaystyle \therefore gof\text{ exists and }gof:\text{ Domain }(f)\to R\text{ i.e. }gof:R\to R\text{ such that}
\displaystyle gof(x)=g(f(x))=g(e^{x})=\log_{e}e^{x}=x\log_{e}e=x
\displaystyle \text{Thus, }gof:R\to R\text{ is defined as }gof(x)=x.
\displaystyle \text{We observe that Domain }(gof)\neq \text{Domain }(fog).
\displaystyle \therefore gof\neq fog.

\displaystyle \textbf{Question 12: } \text{If }f(x)=\sqrt{x}\,(x>0)\text{ and }g(x)=x^{2}-1\text{ are two real functions, find } \\ fog\text{ and }gof.\text{ Is }fog=gof? \hspace{7.0cm} \text{[CBSE 2002]}
\displaystyle \text{Answer:}
\displaystyle  \text{We observe that}
\displaystyle \text{Domain }(f)=[0,\infty),\text{ Range }(f)=[0,\infty),\text{ Domain }(g)=R
\displaystyle \text{and, Range }(g)=[-1,\infty)\quad [\because x^{2}\ge 0\text{ for all }x\in R\ \therefore x^{2}-1\ge -1\text{ for all }x\in R]
\displaystyle \text{Computation of }gof:\text{ We observe that Range }(f)=[0,\infty)\subseteq \text{Domain }(g).
\displaystyle \therefore gof\text{ exists and }gof:[0,\infty)\to R\text{ such that}
\displaystyle gof(x)=g(f(x))=g(\sqrt{x})=(\sqrt{x})^{2}-1=x-1
\displaystyle \text{Thus, }gof:[0,\infty)\to R\text{ is defined as }gof(x)=x-1.
\displaystyle \text{Computation of }fog:\text{ We observe that Range }(g)=[-1,\infty)\nsubseteq \text{Domain }(f)
\displaystyle \therefore \text{Domain }(fog)=\{x:x\in \text{Domain }(g)\text{ and }g(x)\in \text{Domain }(f)\}
\displaystyle \Rightarrow \text{Domain }(fog)=\{x:x\in R\text{ and }g(x)\in [0,\infty)\}
\displaystyle \Rightarrow \text{Domain }(fog)=\{x:x\in R\text{ and }x^{2}-1\in [0,\infty)\}
\displaystyle \Rightarrow \text{Domain }(fog)=\{x:x\in R\text{ and }x^{2}-1\ge 0\}
\displaystyle \Rightarrow \text{Domain }(fog)=\{x:x\le -1\text{ or }x\ge 1\}
\displaystyle \Rightarrow \text{Domain }(fog)=(-\infty,-1]\cup [1,\infty)
\displaystyle \text{Also, }fog(x)=f(g(x))=f(x^{2}-1)=\sqrt{x^{2}-1}
\displaystyle \text{Thus, }fog:(-\infty,-1]\cup [1,\infty)\to R\text{ is defined as }fog(x)=\sqrt{x^{2}-1}.
\displaystyle \text{We find that }fog\text{ and }gof\text{ have distinct domains. Also, their formulas are not same.}
\displaystyle \text{Hence, }fog\neq gof.

\displaystyle \textbf{Question 13: } \text{Let }f:N\cup\{0\}\to N\cup\{0\}\text{ be defined by}
\displaystyle f(n)=\begin{cases}n+1,&\text{if }n\text{ is even}\\n-1,&\text{if }n\text{ is odd}\end{cases}
\displaystyle \text{Show that }f\text{ is invertible and }f=f^{-1}. \hspace{7.0cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle   f\text{ is a bijection. So, it is invertible.}
\displaystyle \text{In order to find }f^{-1},\text{ let }n,m\in N\cup\{0\}\text{ such that}
\displaystyle f(n)=m
\displaystyle \Rightarrow n+1=m,\text{ if }n\text{ is even}
\displaystyle \Rightarrow n-1=m,\text{ if }n\text{ is odd}
\displaystyle \Rightarrow n=\begin{cases}m-1,&\text{if }m\text{ is odd}\\m+1,&\text{if }m\text{ is even}\end{cases}
\displaystyle \text{[If }n\text{ is even, then }n+1=m\text{ is odd}\quad \text{If }n\text{ is odd, then }n-1=m\text{ is even]}
\displaystyle \Rightarrow f^{-1}(m)=\begin{cases}m-1,&\text{if }m\text{ is odd}\\m+1,&\text{if }m\text{ is even}\end{cases}
\displaystyle \text{Hence, }f^{-1}(n)=\begin{cases}n+1,&\text{if }n\text{ is even}\\n-1,&\text{if }n\text{ is odd}\end{cases}
\displaystyle \text{Clearly, }f=f^{-1}.

\displaystyle \textbf{Question 14: } \text{Let }f:R\to R\text{ be defined as }f(x)=10x+7.\text{ Find the function } \\ g:R\to R\text{ such that} gof=fog=I_{R}. \hspace{7.0cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have,}
\displaystyle fog=I_{R}
\displaystyle \Rightarrow fog(x)=I_{R}(x)\text{ for all }x\in R
\displaystyle \Rightarrow f(g(x))=x\text{ for all }x\in R
\displaystyle \Rightarrow 10g(x)+7=x\text{ for all }x\in R
\displaystyle \Rightarrow g(x)=\frac{x-7}{10}\text{ for all }x\in R

\displaystyle \textbf{Question 15: } \text{Let }f:N\to R\text{ be a function defined as }f(x)=4x^{2}+12x+15.\text{ Show that }f:N\to \text{Range}(f)\text{ is invertible. Find the inverse of }f. \hspace{2.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{In order to prove that }f\text{ is invertible, it is sufficient to show that }f:N\to \text{Range}(f)\text{ is a bijection.}
\displaystyle f\text{ is one-one: For any }x,y\in N,\text{ we find that}
\displaystyle f(x)=f(y)
\displaystyle \Rightarrow 4x^{2}+12x+15=4y^{2}+12y+15
\displaystyle \Rightarrow 4(x^{2}-y^{2})+12(x-y)=0
\displaystyle \Rightarrow (x-y)(4x+4y+3)=0
\displaystyle \Rightarrow x-y=0\quad [\because 4x+4y+3\neq 0\text{ for any }x,y\in N]
\displaystyle \Rightarrow x=y
\displaystyle \text{So, }f:N\to \text{Range}(f)\text{ is one-one.}
\displaystyle \text{Obviously, }f:N\to \text{Range}(f)\text{ is onto. Hence, }f:N\to \text{Range}(f)\text{ is invertible.}
\displaystyle \text{Let }f^{-1}\text{ denote the inverse of }f.\text{ Then,}
\displaystyle f(f^{-1}(x))=x\text{ for all }x\in \text{Range}(f)
\displaystyle \Rightarrow 4(f^{-1}(x))^{2}+12f^{-1}(x)+15=x\text{ for all }x\in \text{Range}(f)
\displaystyle \Rightarrow 4(f^{-1}(x))^{2}+12f^{-1}(x)+15-x=0
\displaystyle \Rightarrow f^{-1}(x)=\frac{-12\pm \sqrt{144-16(15-x)}}{8}
\displaystyle \Rightarrow f^{-1}(x)=\frac{-12\pm \sqrt{16x-96}}{8}=\frac{-3\pm \sqrt{x-6}}{2}
\displaystyle \Rightarrow f^{-1}(x)=\frac{-3+\sqrt{x-6}}{2}\quad [\because f^{-1}(x)\in N\therefore f^{-1}(x)>0]

\displaystyle \textbf{Question 16. }\text{Select the correct option out of the four given options. Let }R
\displaystyle \text{be a relation in the set }N\text{ given by }R=\{(a,b):a=b-2,\ b>6\}.\text{ Then,}
\displaystyle \hspace{0.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }(8,7)\in R\qquad \text{(b) }(6,8)\in R
\displaystyle \text{(c) }(3,8)\in R\qquad \text{(d) }(2,4)\in R
\displaystyle \text{Answer:}
\displaystyle \text{Given, }R=\{(a,b):a=b-2,\ b>6\}
\displaystyle \text{Since }b>6,\text{ possible values of }b\text{ are }7,8,9,\ldots
\displaystyle \text{When }b=7,\text{ then }a=7-2=5
\displaystyle \text{When }b=8,\text{ then }a=8-2=6
\displaystyle \therefore (5,7)\in R\text{ and }(6,8)\in R
\displaystyle \text{Hence, the correct option is }(b).
\\

\displaystyle \textbf{Question 17. }\text{Let }A=\{3,5\}.\text{ Then, number of reflexive relations on }A\text{ is}
\displaystyle \hspace{0.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }2\qquad \text{(b) }4\qquad \text{(c) }0\qquad \text{(d) }8
\displaystyle \text{Answer:}
\displaystyle \text{Here, }n(A)=2
\displaystyle \text{Number of reflexive relations on a set with }n\text{ elements is }2^{n^{2}-n}
\displaystyle \therefore \text{Number of reflexive relations}=2^{2^{2}-2}=2^{4-2}=2^{2}=4
\displaystyle \text{Hence, the correct option is }(b).
\\

\displaystyle \textbf{Question 18. }\text{A relation }R\text{ in set }A=\{1,2,3\}\text{ is defined as}
\displaystyle R=\{(1,1),(1,2),(2,2),(3,3)\}.\text{ Which of the following ordered pair}
\displaystyle \text{in }R\text{ shall be removed to make it an equivalence relation in }A?
\displaystyle \hspace{0.2cm} \text{[CBSE 2022 (Term I)]}
\displaystyle \text{(a) }(1,1)\qquad \text{(b) }(1,2)\qquad \text{(c) }(2,2)\qquad \text{(d) }(3,3)
\displaystyle \text{Answer:}
\displaystyle \text{We have }A=\{1,2,3\}
\displaystyle R=\{(1,1),(1,2),(2,2),(3,3)\}
\displaystyle \text{For equivalence relation, }R\text{ must be reflexive, symmetric and transitive.}
\displaystyle \text{The pair }(1,2)\text{ makes }R\text{ non-symmetric, since }(2,1)\notin R.
\displaystyle \text{Thus, removing }(1,2)\text{ gives }\{(1,1),(2,2),(3,3)\},
\displaystyle \text{which is an equivalence relation on }A.
\displaystyle \text{Hence, the correct option is }(b).
\\

\displaystyle \textbf{Question 19. }\text{Let }A=\{1,2,3\},\ B=\{4,5,6,7\}\text{ and let}
\displaystyle f=\{(1,4),(2,5),(3,6)\}\text{ be a function from }A\text{ to }B.\text{ Based on the}
\displaystyle \text{given information, }f\text{ is best defined as} \hspace{0.2cm} \text{[CBSE Sample}
\displaystyle \text{Paper 2022 (Term I)]}
\displaystyle \text{(a) surjective function}
\displaystyle \text{(b) injective function}
\displaystyle \text{(c) bijective function}
\displaystyle \text{(d) None of the above}
\displaystyle \text{Answer:}
\displaystyle A=\{1,2,3\},\ B=\{4,5,6,7\}
\displaystyle \text{and }f=\{(1,4),(2,5),(3,6)\}
\displaystyle \text{All elements of }A\text{ have distinct images in }B
\displaystyle \therefore f\text{ is an injective (one-one) function.}
\displaystyle \text{Range of }f=\{4,5,6\}
\displaystyle \text{Codomain}=\{4,5,6,7\}
\displaystyle \text{Since Range}\neq\text{Codomain, }f\text{ is not surjective.}
\displaystyle \text{Hence, the correct option is }(b).
\\

\displaystyle \textbf{Question 20. }\text{A relation in a set }A\text{ is called ........ relation, if each}
\displaystyle \text{element of }A\text{ is related to itself.} \hspace{0.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{If }R\text{ is a relation on a set }A\text{ such that }(a,a)\in R,\ \forall a\in A,
\displaystyle \text{then }R\text{ is called a reflexive relation.}
\\

\displaystyle \textbf{Question 21. }\text{If }R=\{(a,a^3):a\text{ is a prime number less than }5\}
\displaystyle \text{be a relation, then find the range of }R. \hspace{0.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }R=\{(a,a^{3}):a\text{ is a prime number less than }5\}
\displaystyle \text{The prime numbers less than }5\text{ are }2\text{ and }3
\displaystyle \therefore R=\{(2,2^{3}),(3,3^{3})\}=\{(2,8),(3,27)\}
\displaystyle \text{Hence, the range of }R\text{ is }\{8,27\}
\\

\displaystyle \textbf{Question 22. }\text{Let }R\text{ be the equivalence relation in the set}
\displaystyle A=\{0,1,2,3,4,5\}\text{ given by }R=\{(a,b):2\text{ divides }(a-b)\}.
\displaystyle \text{Write the equivalence class }[0]. \hspace{0.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }R=\{(a,b):2\text{ divides }(a-b)\}
\displaystyle \text{and }A=\{0,1,2,3,4,5\}
\displaystyle [0]=\{b\in A:(0,b)\in R\}
\displaystyle =\{b\in A:2\text{ divides }(0-b)\}
\displaystyle =\{b\in A:2\text{ divides }(-b)\}
\displaystyle =\{0,2,4\}
\displaystyle \text{Hence, the equivalence class }[0]=\{0,2,4\}
\\

\displaystyle \textbf{Question 23. }\text{If }R=\{(x,y):x+2y=8\}\text{ is a relation on }N,\text{ then write}
\displaystyle \text{the range of }R. \hspace{0.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }R=\{(x,y):x+2y=8\}\text{ on }N
\displaystyle \text{From }x+2y=8,\text{ we get }y=\frac{8-x}{2}
\displaystyle \text{Taking natural number values of }x,
\displaystyle \text{for }x=2,\ y=\frac{8-2}{2}=3
\displaystyle \text{for }x=4,\ y=\frac{8-4}{2}=2
\displaystyle \text{for }x=6,\ y=\frac{8-6}{2}=1
\displaystyle \therefore R=\{(2,3),(4,2),(6,1)\}
\displaystyle \text{Hence, the range of }R\text{ is }\{3,2,1\}
\\

\displaystyle \textbf{Question 24. }\text{If }A=\{1,2,3\},\ B=\{4,5,6,7\}\text{ and}
\displaystyle f=\{(1,4),(2,5),(3,6)\}\text{ is a function from }A\text{ to }B,\text{ state whether}
\displaystyle f\text{ is one-one or not.} \hspace{0.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle A=\{1,2,3\},\ B=\{4,5,6,7\}
\displaystyle \text{and }f:A\to B\text{ is defined as }f=\{(1,4),(2,5),(3,6)\}
\displaystyle \text{i.e. }f(1)=4,\ f(2)=5\text{ and }f(3)=6
\displaystyle \text{The images of distinct elements of }A\text{ are distinct.}
\displaystyle \therefore f\text{ is a one-one function.}
\\

\displaystyle \textbf{Question 25. }\text{A function }f:A\to B\text{ defined by }f(x)=2x\text{ is both}
\displaystyle \text{one-one and onto. If }A=\{1,2,3,4\},\text{ then find the set }B.
\displaystyle \hspace{0.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f:A\to B\text{ such that }f(x)=2x
\displaystyle \text{and }A=\{1,2,3,4\}
\displaystyle f(1)=2\times1=2
\displaystyle f(2)=2\times2=4
\displaystyle f(3)=2\times3=6
\displaystyle f(4)=2\times4=8
\displaystyle \text{Since }f\text{ is onto, Codomain}=\text{Range}
\displaystyle \therefore B=\{2,4,6,8\}
\\

\displaystyle \textbf{Question 26. }\text{Prove that the function }f\text{ is surjective, where}
\displaystyle f:N\to N\text{ such that}
\displaystyle f(n)=\begin{cases}\frac{n+1}{2}, & \text{if }n\text{ is odd} \\ \frac{n}{2}, & \text{if }n\text{ is even}\end{cases}
\displaystyle \text{Is the function injective? Justify your answer.} \hspace{0.2cm} \text{[CBSE Sample}
\displaystyle \text{Paper 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }m\in N\text{ be any element in the codomain.}
\displaystyle \text{Case I: If }m\text{ is odd, then }m=2r+1\text{ for some }r\in N
\displaystyle \text{Consider }n=4r+1
\displaystyle f(4r+1)=\frac{4r+1+1}{2}=\frac{4r+2}{2}=2r+1=m
\displaystyle \text{Case II: If }m\text{ is even, then }m=2r\text{ for some }r\in N
\displaystyle \text{Consider }n=4r
\displaystyle f(4r)=\frac{4r}{2}=2r=m
\displaystyle \text{Thus, every element of codomain has a pre-image in domain.}
\displaystyle \therefore f\text{ is surjective.}
\displaystyle \text{Now, }f(1)=\frac{1+1}{2}=1\text{ and }f(2)=\frac{2}{2}=1
\displaystyle \text{Since }f(1)=f(2)\text{ but }1\neq2,\text{ the function is not injective.}
\displaystyle \therefore f\text{ is not one-one.}
\\

\displaystyle \textbf{Question 27. }\text{Check if the relation }R\text{ in the set }R\text{ of real numbers}
\displaystyle \text{defined as }R=\{(a,b):a<b\}\text{ is}
\displaystyle \text{(i) symmetric}\qquad \text{(ii) transitive} \hspace{0.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }R=\{(a,b):a<b\}\text{ on the set of real numbers}
\displaystyle \text{Symmetric: Let }(a,b)\in R
\displaystyle \Rightarrow a<b
\displaystyle \Rightarrow b\nless a
\displaystyle \Rightarrow (b,a)\notin R
\displaystyle \text{For example, }4<5\text{ but }5\nless4
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{Transitive: Let }(a,b)\in R\text{ and }(b,c)\in R
\displaystyle \Rightarrow a<b\text{ and }b<c
\displaystyle \Rightarrow a<c
\displaystyle \Rightarrow (a,c)\in R
\displaystyle \therefore R\text{ is transitive.}
\\

\displaystyle \textbf{Question 28. }\text{Check if the relation }R\text{ on the set }A=\{1,2,3,4,5,6\}
\displaystyle \text{defined as }R=\{(x,y):y\text{ is divisible by }x\}\text{ is}
\displaystyle \text{(i) symmetric}\qquad \text{(ii) transitive} \hspace{0.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Symmetric: We observe that }6\text{ is divisible by }2
\displaystyle \Rightarrow (2,6)\in R\text{ but }(6,2)\notin R
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{(ii) Transitive: Let }(x,y)\in R\text{ and }(y,z)\in R
\displaystyle \Rightarrow y\text{ is divisible by }x\text{ and }z\text{ is divisible by }y
\displaystyle \Rightarrow z\text{ is divisible by }x
\displaystyle \Rightarrow (x,z)\in R
\displaystyle \therefore R\text{ is transitive.}
\\

\displaystyle \textbf{Question 29. }\text{Check whether the relation }R\text{ defined on the set}
\displaystyle A=\{1,2,3,4,5,6\}\text{ as }R=\{(a,b):b=a+1\}\text{ is reflexive,}
\displaystyle \text{symmetric or transitive.} \hspace{0.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{The relation }R\text{ on }A=\{1,2,3,4,5,6\}\text{ is defined by}
\displaystyle (a,b)\in R\text{ iff }b=a+1
\displaystyle \therefore R=\{(1,2),(2,3),(3,4),(4,5),(5,6)\}
\displaystyle \text{Since }(a,a)\notin R\text{ for any }a\in A,\text{ }R\text{ is not reflexive.}
\displaystyle \text{Also, }(1,2)\in R\text{ but }(2,1)\notin R
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{Further, }(1,2)\in R\text{ and }(2,3)\in R\text{ but }(1,3)\notin R
\displaystyle \therefore R\text{ is not transitive.}
\\

\displaystyle \textbf{Question 30. }\text{Show that the relation }R\text{ on }R\text{ defined as}
\displaystyle R=\{(a,b):a\leq b\}\text{ is reflexive and transitive but not symmetric.}
\displaystyle \hspace{0.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }R=\{(a,b):a\leq b\}\text{ on the set of real numbers}
\displaystyle \text{Reflexive: Since }a\leq a\text{ for every }a\in R,
\displaystyle \therefore (a,a)\in R,\ \forall a\in R
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Transitive: Let }(a,b)\in R\text{ and }(b,c)\in R
\displaystyle \Rightarrow a\leq b\text{ and }b\leq c
\displaystyle \Rightarrow a\leq c
\displaystyle \Rightarrow (a,c)\in R
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Symmetric: }(2,3)\in R\text{ since }2<3
\displaystyle \text{but }(3,2)\notin R\text{ since }3\nleq2
\displaystyle \therefore R\text{ is not symmetric.}
\\

\displaystyle \textbf{Question 31. }\text{If }R\text{ is a relation defined on the set of natural numbers}
\displaystyle N\text{ as }R=\{(x,y):x\in N,\ y\in N\text{ and }2x+y=24\},\text{ then find}
\displaystyle \text{the domain and range of }R.\text{ Also, find whether }R\text{ is an}
\displaystyle \text{equivalence relation or not.} \hspace{0.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }R=\{(x,y):x\in N,\ y\in N\text{ and }2x+y=24\}
\displaystyle \Rightarrow y=24-2x
\displaystyle \text{For }x=1,\ y=22;\quad x=2,\ y=20;\quad x=3,\ y=18
\displaystyle x=4,\ y=16;\quad x=5,\ y=14;\quad x=6,\ y=12
\displaystyle x=7,\ y=10;\quad x=8,\ y=8;\quad x=9,\ y=6
\displaystyle x=10,\ y=4;\quad x=11,\ y=2
\displaystyle \therefore R=\{(1,22),(2,20),(3,18),(4,16),(5,14),(6,12),
\displaystyle (7,10),(8,8),(9,6),(10,4),(11,2)\}
\displaystyle \text{Domain of }R=\{1,2,3,\ldots,11\}
\displaystyle \text{Range of }R=\{2,4,6,8,10,12,14,16,18,20,22\}
\displaystyle \text{Reflexive: }(1,1)\notin R
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle \text{Symmetric: }(1,22)\in R\text{ but }(22,1)\notin R
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{Transitive: }(7,10)\in R\text{ and }(10,4)\in R
\displaystyle \text{but }(7,4)\notin R
\displaystyle \therefore R\text{ is not transitive.}
\displaystyle \text{Hence, }R\text{ is not an equivalence relation.}
\\

\displaystyle \textbf{Question 32. }\text{If }A=R-\{3\}\text{ and }B=R-\{1\}.\text{ Consider the function}
\displaystyle f:A\to B\text{ defined by }f(x)=\frac{x-2}{x-3},\text{ for all }x\in A.\text{ Then, show}
\displaystyle \text{that }f\text{ is bijective.} \hspace{0.2cm} \text{[CBSE 2014C], [CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f:A\to B,\text{ where }A=R-\{3\}\text{ and }B=R-\{1\},
\displaystyle \text{defined by }f(x)=\frac{x-2}{x-3}
\displaystyle \text{One-One: Let }x_{1},x_{2}\in A\text{ such that }f(x_{1})=f(x_{2})
\displaystyle \Rightarrow \frac{x_{1}-2}{x_{1}-3}=\frac{x_{2}-2}{x_{2}-3}
\displaystyle \Rightarrow (x_{1}-2)(x_{2}-3)=(x_{2}-2)(x_{1}-3)
\displaystyle \Rightarrow x_{1}x_{2}-3x_{1}-2x_{2}+6=x_{1}x_{2}-3x_{2}-2x_{1}+6
\displaystyle \Rightarrow -3x_{1}-2x_{2}=-3x_{2}-2x_{1}
\displaystyle \Rightarrow -(x_{1}-x_{2})=0
\displaystyle \Rightarrow x_{1}=x_{2}
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Onto: Let }y\in B=R-\{1\}
\displaystyle \text{Then }f(x)=y
\displaystyle \Rightarrow \frac{x-2}{x-3}=y
\displaystyle \Rightarrow x-2=xy-3y
\displaystyle \Rightarrow x-xy=2-3y
\displaystyle \Rightarrow x(1-y)=2-3y
\displaystyle \Rightarrow x=\frac{2-3y}{1-y}=\frac{3y-2}{y-1}
\displaystyle \text{Since }y\neq1,\ x=\frac{3y-2}{y-1}\in R
\displaystyle \text{Also, }\frac{3y-2}{y-1}\neq3
\displaystyle \left[\because \frac{3y-2}{y-1}=3\Rightarrow 3y-2=3y-3\Rightarrow 2=3,\text{ absurd}\right]
\displaystyle \therefore x=\frac{3y-2}{y-1}\in A
\displaystyle \text{Now, }f\left(\frac{3y-2}{y-1}\right)=\frac{\left(\frac{3y-2}{y-1}\right)-2}{\left(\frac{3y-2}{y-1}\right)-3}
\displaystyle =\frac{3y-2-2y+2}{3y-2-3y+3}=\frac{y}{1}=y
\displaystyle \therefore f\text{ is onto.}
\displaystyle \text{Hence, }f\text{ is bijective.}
\\

\displaystyle \textbf{Question 33. }\text{If }A=\{1,2,3,\ldots,9\}\text{ and }R\text{ is the relation in }A\times A
\displaystyle \text{defined by }(a,b)R(c,d)\text{ if }a+d=b+c\text{ for }(a,b),(c,d)\in A\times A,
\displaystyle \text{prove that }R\text{ is an equivalence relation. Also, obtain the}
\displaystyle \text{equivalence class }[(2,5)]. \hspace{0.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, relation }R\text{ on }A\times A,\text{ where }A=\{1,2,3,\ldots,9\},
\displaystyle \text{defined by }(a,b)\ R\ (c,d)\text{ if }a+d=b+c
\displaystyle \text{Reflexive: Let }(a,b)\in A\times A
\displaystyle \text{Since }a+b=b+a,\text{ we get }(a,b)\ R\ (a,b)
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Symmetric: Let }(a,b)\ R\ (c,d)
\displaystyle \Rightarrow a+d=b+c
\displaystyle \Rightarrow c+b=d+a
\displaystyle \Rightarrow (c,d)\ R\ (a,b)
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{Transitive: Let }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f)
\displaystyle \Rightarrow a+d=b+c\text{ and }c+f=d+e
\displaystyle \text{Adding, we get }a+d+c+f=b+c+d+e
\displaystyle \Rightarrow a+f=b+e
\displaystyle \Rightarrow (a,b)\ R\ (e,f)
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Hence, }R\text{ is an equivalence relation.}
\displaystyle \text{Now, }[(2,5)]=\{(c,d)\in A\times A:(2,5)\ R\ (c,d)\}
\displaystyle \Rightarrow 2+d=5+c
\displaystyle \Rightarrow d-c=3
\displaystyle \therefore [(2,5)]=\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}
\\

\displaystyle \textbf{Question 34. }\text{If }A=R-\{2\},\ B=R-\{1\}\text{ and }f:A\to B\text{ is a function}
\displaystyle \text{defined by }f(x)=\frac{x-1}{x-2},\text{ then show that }f\text{ is one-one and onto. }
\displaystyle \text{[CBSE 2013C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\frac{x-1}{x-2},\ x\neq 2
\displaystyle \text{To show that }f\text{ is one-one, let }f(x_{1})=f(x_{2})
\displaystyle \Rightarrow \frac{x_{1}-1}{x_{1}-2}=\frac{x_{2}-1}{x_{2}-2}
\displaystyle \Rightarrow (x_{1}-1)(x_{2}-2)=(x_{2}-1)(x_{1}-2)
\displaystyle \Rightarrow x_{1}x_{2}-2x_{1}-x_{2}+2=x_{1}x_{2}-2x_{2}-x_{1}+2
\displaystyle \Rightarrow -2x_{1}-x_{2}=-2x_{2}-x_{1}
\displaystyle \Rightarrow x_{2}=x_{1}
\displaystyle \therefore x_{1}=x_{2}
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Now, let }y\in B
\displaystyle \Rightarrow y=\frac{x-1}{x-2}
\displaystyle \Rightarrow y(x-2)=x-1
\displaystyle \Rightarrow yx-2y=x-1
\displaystyle \Rightarrow yx-x=2y-1
\displaystyle \Rightarrow x(y-1)=2y-1
\displaystyle \Rightarrow x=\frac{2y-1}{y-1}
\displaystyle \text{Since }y\neq 1,\ x\text{ exists uniquely.}
\displaystyle \text{Also, }x\neq 2,\text{ because}
\displaystyle \frac{2y-1}{y-1}=2
\displaystyle \Rightarrow 2y-1=2y-2
\displaystyle \Rightarrow -1=-2,\text{ which is not possible.}
\displaystyle \therefore x\in A
\displaystyle \text{Hence, for every }y\in B,\ \exists x\in A\text{ such that }f(x)=y
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is one-one and onto.}
\\

\displaystyle \textbf{Question 35. }\text{Show that the function }f\text{ in}
\displaystyle A=R-\left\{\frac{2}{3}\right\}\text{ defined by }f(x)=\frac{4x+3}{6x-4}\text{ is one-one and onto.}
\displaystyle \hspace{0.2cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f:A\to A,\text{ where }A=R-\left\{\frac{2}{3}\right\}\text{ and}
\displaystyle f(x)=\frac{4x+3}{6x-4}
\displaystyle \text{One-One: Let }x_{1},x_{2}\in A\text{ such that }f(x_{1})=f(x_{2})
\displaystyle \Rightarrow \frac{4x_{1}+3}{6x_{1}-4}=\frac{4x_{2}+3}{6x_{2}-4}
\displaystyle \Rightarrow (4x_{1}+3)(6x_{2}-4)=(4x_{2}+3)(6x_{1}-4)
\displaystyle \Rightarrow 24x_{1}x_{2}-16x_{1}+18x_{2}-12
\displaystyle =24x_{1}x_{2}-16x_{2}+18x_{1}-12
\displaystyle \Rightarrow -34x_{1}=-34x_{2}
\displaystyle \Rightarrow x_{1}=x_{2}
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Onto: Let }y\in A\text{ be any arbitrary element}
\displaystyle \text{Then }f(x)=y
\displaystyle \Rightarrow \frac{4x+3}{6x-4}=y
\displaystyle \Rightarrow 4x+3=6xy-4y
\displaystyle \Rightarrow 4x-6xy=-(4y+3)
\displaystyle \Rightarrow x(4-6y)=-(4y+3)
\displaystyle \Rightarrow x=\frac{4y+3}{6y-4}
\displaystyle \text{Clearly, }x=\frac{4y+3}{6y-4}\in R\text{ for all }y\neq\frac{2}{3}
\displaystyle \text{Also, }\frac{4y+3}{6y-4}\neq\frac{2}{3}
\displaystyle \left[\because \frac{4y+3}{6y-4}=\frac{2}{3}\Rightarrow 12y+9=12y-8,\text{ absurd}\right]
\displaystyle \therefore x=\frac{4y+3}{6y-4}\in A
\displaystyle \text{Now, }f\left(\frac{4y+3}{6y-4}\right)
\displaystyle =\frac{4\left(\frac{4y+3}{6y-4}\right)+3}{6\left(\frac{4y+3}{6y-4}\right)-4}
\displaystyle =\frac{16y+12+18y-12}{24y+18-24y+16}
\displaystyle =\frac{34y}{34}=y
\displaystyle \therefore f\text{ is onto.}
\displaystyle \text{Hence, }f\text{ is bijective.}
\\

\displaystyle \textbf{Question 36. }\text{Show that }f:N\to N,\text{ given by}
\displaystyle f(x)=\begin{cases}x+1, & \text{if }x\text{ is odd} \\ x-1, & \text{if }x\text{ is even}\end{cases}
\displaystyle \text{is bijective (both one-one and onto).} \hspace{0.2cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f:N\to N\text{ such that}
\displaystyle f(x)=\begin{cases}x+1, & \text{if }x\text{ is odd} \\ x-1, & \text{if }x\text{ is even}\end{cases}
\displaystyle \text{One-One: Case I: Let }x_{1},x_{2}\text{ be even}
\displaystyle f(x_{1})=f(x_{2})
\displaystyle \Rightarrow x_{1}-1=x_{2}-1
\displaystyle \Rightarrow x_{1}=x_{2}
\displaystyle \text{Case II: Let }x_{1},x_{2}\text{ be odd}
\displaystyle f(x_{1})=f(x_{2})
\displaystyle \Rightarrow x_{1}+1=x_{2}+1
\displaystyle \Rightarrow x_{1}=x_{2}
\displaystyle \text{Case III: Let }x_{1}\text{ be odd and }x_{2}\text{ be even}
\displaystyle \text{Then }f(x_{1})\text{ is even and }f(x_{2})\text{ is odd}
\displaystyle \Rightarrow f(x_{1})\neq f(x_{2})
\displaystyle \text{Similarly, if }x_{1}\text{ is even and }x_{2}\text{ is odd, then}
\displaystyle f(x_{1})\neq f(x_{2})
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Onto: Let }y\in N
\displaystyle \text{If }y\text{ is odd, then }y+1\text{ is even and}
\displaystyle f(y+1)=(y+1)-1=y
\displaystyle \text{If }y\text{ is even, then }y-1\text{ is odd and}
\displaystyle f(y-1)=(y-1)+1=y
\displaystyle \therefore \text{Every element of codomain has a pre-image in domain.}
\displaystyle \therefore f\text{ is onto.}
\displaystyle \text{Hence, }f\text{ is bijective.}
\\

\displaystyle \textbf{Question 37. }\text{If }f:R\to R\text{ is the function defined by }f(x)=4x^3+7,
\displaystyle \text{then show that }f\text{ is a bijection.} \hspace{0.2cm} \text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f:R\to R\text{ defined by }f(x)=4x^{3}+7
\displaystyle \text{To prove }f\text{ is bijective, we show that it is one-one and onto.}
\displaystyle \text{One-One: Let }x_{1},x_{2}\in R\text{ such that }f(x_{1})=f(x_{2})
\displaystyle \Rightarrow 4x_{1}^{3}+7=4x_{2}^{3}+7
\displaystyle \Rightarrow x_{1}^{3}-x_{2}^{3}=0
\displaystyle \Rightarrow (x_{1}-x_{2})(x_{1}^{2}+x_{1}x_{2}+x_{2}^{2})=0
\displaystyle \Rightarrow (x_{1}-x_{2})\left[\left(x_{1}+\frac{x_{2}}{2}\right)^{2}+\frac{3x_{2}^{2}}{4}\right]=0
\displaystyle \text{Since }\left(x_{1}+\frac{x_{2}}{2}\right)^{2}+\frac{3x_{2}^{2}}{4}\neq0
\displaystyle \Rightarrow x_{1}-x_{2}=0
\displaystyle \Rightarrow x_{1}=x_{2}
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Onto: Let }y\in R
\displaystyle \text{Then }f(x)=y
\displaystyle \Rightarrow 4x^{3}+7=y
\displaystyle \Rightarrow x^{3}=\frac{y-7}{4}
\displaystyle \Rightarrow x=\left(\frac{y-7}{4}\right)^{1/3}
\displaystyle \text{which is a real number.}
\displaystyle \therefore \text{For every }y\in R,\text{ there exists }x=\left(\frac{y-7}{4}\right)^{1/3}\in R
\displaystyle \text{such that}
\displaystyle f\left[\left(\frac{y-7}{4}\right)^{1/3}\right]
\displaystyle =4\left[\left(\frac{y-7}{4}\right)^{1/3}\right]^{3}+7
\displaystyle =4\left(\frac{y-7}{4}\right)+7=y
\displaystyle \therefore f\text{ is onto.}
\displaystyle \text{Hence, }f\text{ is bijective.}
\\

\displaystyle \textbf{Question 38. }\text{Show that the relation }S\text{ in the set }R\text{ of real numbers}
\displaystyle \text{defined as }S=\{(a,b):a,b\in R\text{ and }a\leq b^3\}\text{ is neither}
\displaystyle \text{reflexive nor symmetric nor transitive.} \hspace{0.2cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }S=\{(a,b):a,b\in R\text{ and }a\leq b^{3}\}
\displaystyle \text{Reflexive: For reflexivity, we must have }(a,a)\in S,\ \forall a\in R
\displaystyle \text{Take }a=\frac{1}{2}
\displaystyle \text{Then }\frac{1}{2}\leq\left(\frac{1}{2}\right)^{3}\text{ is false}
\displaystyle \therefore \left(\frac{1}{2},\frac{1}{2}\right)\notin S
\displaystyle \therefore S\text{ is not reflexive.}
\displaystyle \text{Symmetric: We have }-2\leq 3^{3}
\displaystyle \Rightarrow (-2,3)\in S
\displaystyle \text{But }3\nleq(-2)^{3}
\displaystyle \Rightarrow (3,-2)\notin S
\displaystyle \therefore S\text{ is not symmetric.}
\displaystyle \text{Transitive: We have }3\leq\left(\frac{3}{2}\right)^{3}
\displaystyle \text{and }\frac{3}{2}\leq\left(\frac{4}{3}\right)^{3}
\displaystyle \Rightarrow \left(3,\frac{3}{2}\right)\in S\text{ and }\left(\frac{3}{2},\frac{4}{3}\right)\in S
\displaystyle \text{But }3\nleq\left(\frac{4}{3}\right)^{3}
\displaystyle \Rightarrow \left(3,\frac{4}{3}\right)\notin S
\displaystyle \therefore S\text{ is not transitive.}
\displaystyle \text{Hence, }S\text{ is neither reflexive nor symmetric nor transitive.}
\\

\displaystyle \textbf{Question 39. }\text{Show that the relation }S\text{ defined on }N\times N\text{ by}
\displaystyle (a,b)S(c,d)\Rightarrow a+d=b+c\text{ is an equivalence relation.}
\displaystyle \hspace{0.2cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }(a,b)S(c,d)\iff a+d=b+c
\displaystyle \text{Reflexive: For any }(a,b)\in N\times N,
\displaystyle a+b=b+a
\displaystyle \Rightarrow (a,b)S(a,b)
\displaystyle \therefore S\text{ is reflexive.}
\displaystyle \text{Symmetric: Let }(a,b)S(c,d)
\displaystyle \Rightarrow a+d=b+c
\displaystyle \Rightarrow c+b=d+a
\displaystyle \Rightarrow (c,d)S(a,b)
\displaystyle \therefore S\text{ is symmetric.}
\displaystyle \text{Transitive: Let }(a,b)S(c,d)\text{ and }(c,d)S(e,f)
\displaystyle \Rightarrow a+d=b+c\qquad \cdots(i)
\displaystyle \Rightarrow c+f=d+e\qquad \cdots(ii)
\displaystyle \text{Adding Eqs. }(i)\text{ and }(ii),
\displaystyle a+d+c+f=b+c+d+e
\displaystyle \Rightarrow a+f=b+e
\displaystyle \Rightarrow (a,b)S(e,f)
\displaystyle \therefore S\text{ is transitive.}
\displaystyle \text{Hence, }S\text{ is an equivalence relation.}
\\

\displaystyle \textbf{Question 40. }\text{If }f:X\to Y\text{ is a function, define a relation }R\text{ on }X
\displaystyle \text{by }R=\{(a,b):f(a)=f(b)\}.\text{ Show that }R\text{ is an equivalence}
\displaystyle \text{relation on }X. \hspace{0.2cm} \text{[CBSE 2010C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }R=\{(a,b):f(a)=f(b)\}
\displaystyle \text{Reflexive: For every }x\in X,
\displaystyle f(x)=f(x)
\displaystyle \Rightarrow (x,x)\in R
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Symmetric: Let }(x,y)\in R
\displaystyle \Rightarrow f(x)=f(y)
\displaystyle \Rightarrow f(y)=f(x)
\displaystyle \Rightarrow (y,x)\in R
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{Transitive: Let }(x,y)\in R\text{ and }(y,z)\in R
\displaystyle \Rightarrow f(x)=f(y)\text{ and }f(y)=f(z)
\displaystyle \Rightarrow f(x)=f(z)
\displaystyle \Rightarrow (x,z)\in R
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Hence, }R\text{ is an equivalence relation on }X.
\\

\displaystyle \textbf{Question 41. }\text{Show that a function }f:R\to R\text{ given by}
\displaystyle f(x)=ax+b,\ a,b\in R,\ a\neq0\text{ is bijective.}
\displaystyle \hspace{0.2cm} \text{[CBSE 2010C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f:R\to R\text{ defined by }f(x)=ax+b,\ a\neq0
\displaystyle \text{One-One: Let }x_{1},x_{2}\in R\text{ such that }f(x_{1})=f(x_{2})
\displaystyle \Rightarrow ax_{1}+b=ax_{2}+b
\displaystyle \Rightarrow ax_{1}=ax_{2}
\displaystyle \Rightarrow x_{1}=x_{2}\qquad [\because a\neq0]
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Onto: Let }y\in R
\displaystyle \text{Then }f(x)=y
\displaystyle \Rightarrow ax+b=y
\displaystyle \Rightarrow x=\frac{y-b}{a}
\displaystyle \text{Since }a\neq0,\ x\in R
\displaystyle \text{Also, }f\left(\frac{y-b}{a}\right)=a\left(\frac{y-b}{a}\right)+b=y
\displaystyle \therefore f\text{ is onto.}
\displaystyle \text{Hence, }f\text{ is bijective.}
\\

\displaystyle \textbf{Question 42. }\text{A relation }R\text{ is defined on the set of real numbers as}
\displaystyle R=\{(x,y):x\cdot y\text{ is an irrational number}\}. \text{Check whether }R
\displaystyle \text{is reflexive, symmetric and transitive or not.} \hspace{0.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }R=\{(x,y):xy\text{ is an irrational number}\}
\displaystyle \text{Reflexive: Take }x=\sqrt{2}
\displaystyle \sqrt{2}\cdot\sqrt{2}=2,\text{ which is rational}
\displaystyle \Rightarrow (\sqrt{2},\sqrt{2})\notin R
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle \text{Symmetric: If }(x,y)\in R,\text{ then }xy\text{ is irrational}
\displaystyle \Rightarrow yx\text{ is irrational}
\displaystyle \Rightarrow (y,x)\in R
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{Transitive: Consider }2,\sqrt{5}\text{ and }3
\displaystyle 2\cdot\sqrt{5}=2\sqrt{5}\text{ is irrational}
\displaystyle \sqrt{5}\cdot3=3\sqrt{5}\text{ is irrational}
\displaystyle \text{But }2\cdot3=6\text{ is rational}
\displaystyle \Rightarrow (2,\sqrt{5})\in R,\ (\sqrt{5},3)\in R
\displaystyle \text{but }(2,3)\notin R
\displaystyle \therefore R\text{ is not transitive.}
\\

\displaystyle \textbf{Question 43. }\text{Show that a function }f:R\to R\text{ defined by}
\displaystyle f(x)=\frac{5x-3}{4}\text{ is both one-one and onto.} \hspace{0.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f:R\to R\text{ defined by }f(x)=\frac{5x-3}{4}
\displaystyle \text{One-One: Let }x_{1},x_{2}\in R\text{ such that }f(x_{1})=f(x_{2})
\displaystyle \Rightarrow \frac{5x_{1}-3}{4}=\frac{5x_{2}-3}{4}
\displaystyle \Rightarrow 5x_{1}-3=5x_{2}-3
\displaystyle \Rightarrow x_{1}=x_{2}
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Onto: Let }y\in R
\displaystyle \text{Then }y=\frac{5x-3}{4}
\displaystyle \Rightarrow 4y=5x-3
\displaystyle \Rightarrow x=\frac{4y+3}{5}
\displaystyle \text{Clearly, }x\in R
\displaystyle \text{Also, }f\left(\frac{4y+3}{5}\right)=\frac{5\left(\frac{4y+3}{5}\right)-3}{4}=y
\displaystyle \therefore f\text{ is onto.}
\displaystyle \text{Hence, }f\text{ is both one-one and onto.}
\\

\displaystyle \textbf{Question 44. }\text{Let }f:R-\left\{-\frac{4}{3}\right\}\to R\text{ be a function defined as}
\displaystyle f(x)=\frac{4x}{3x+4}.\text{ Show that }f\text{ is a one-one function. Also,}
\displaystyle \text{check whether }f\text{ is an onto function or not.} \hspace{0.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f:R-\left\{-\frac{4}{3}\right\}\to R\text{ defined by}
\displaystyle f(x)=\frac{4x}{3x+4}
\displaystyle \text{One-One: Let }x_{1},x_{2}\in R-\left\{-\frac{4}{3}\right\}\text{ such that}
\displaystyle f(x_{1})=f(x_{2})
\displaystyle \Rightarrow \frac{4x_{1}}{3x_{1}+4}=\frac{4x_{2}}{3x_{2}+4}
\displaystyle \Rightarrow x_{1}(3x_{2}+4)=x_{2}(3x_{1}+4)
\displaystyle \Rightarrow 3x_{1}x_{2}+4x_{1}=3x_{1}x_{2}+4x_{2}
\displaystyle \Rightarrow 4x_{1}=4x_{2}
\displaystyle \Rightarrow x_{1}=x_{2}
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Onto: Let }y=f(x)=\frac{4x}{3x+4}
\displaystyle \Rightarrow 3xy+4y=4x
\displaystyle \Rightarrow 4x-3xy=4y
\displaystyle \Rightarrow x(4-3y)=4y
\displaystyle \Rightarrow x=\frac{4y}{4-3y}
\displaystyle \text{Clearly, }x\text{ is not defined when }4-3y=0
\displaystyle \Rightarrow y=\frac{4}{3}
\displaystyle \therefore \text{Range of }f=R-\left\{\frac{4}{3}\right\}
\displaystyle \text{Since Range}\neq\text{Codomain},\ f\text{ is not onto.}
\\

\displaystyle \textbf{Question 45. }\text{A function }f:[-4,4]\to[0,4]\text{ is given by}
\displaystyle f(x)=\sqrt{16-x^2}.\text{ Show that }f\text{ is an onto function but not}
\displaystyle \text{a one-one function. Further, find all possible values of }a\text{ for}
\displaystyle f(a)=\sqrt{7}. \hspace{0.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f:[-4,4]\to[0,4]\text{ defined by}
\displaystyle f(x)=\sqrt{16-x^{2}}
\displaystyle \text{One-One: Let }x_{1},x_{2}\in[-4,4]\text{ such that}
\displaystyle f(x_{1})=f(x_{2})
\displaystyle \Rightarrow \sqrt{16-x_{1}^{2}}=\sqrt{16-x_{2}^{2}}
\displaystyle \Rightarrow 16-x_{1}^{2}=16-x_{2}^{2}
\displaystyle \Rightarrow x_{1}^{2}=x_{2}^{2}
\displaystyle \Rightarrow (x_{1}-x_{2})(x_{1}+x_{2})=0
\displaystyle \Rightarrow x_{1}=x_{2}\text{ or }x_{1}=-x_{2}
\displaystyle \therefore f\text{ is not one-one.}
\displaystyle \text{Onto: Let }y\in[0,4]
\displaystyle \text{Then }y=f(x)=\sqrt{16-x^{2}}
\displaystyle \Rightarrow y^{2}=16-x^{2}
\displaystyle \Rightarrow x^{2}=16-y^{2}
\displaystyle \Rightarrow x=\pm\sqrt{16-y^{2}}
\displaystyle \text{Since }0\leq y\leq4,\ x\in[-4,4]
\displaystyle \therefore \text{For every }y\in[0,4],\text{ there exists }x\in[-4,4]
\displaystyle \text{such that }f(x)=y
\displaystyle \therefore f\text{ is onto.}
\displaystyle \text{Now, }f(a)=\sqrt{7}
\displaystyle \Rightarrow \sqrt{16-a^{2}}=\sqrt{7}
\displaystyle \Rightarrow 16-a^{2}=7
\displaystyle \Rightarrow a^{2}=9
\displaystyle \Rightarrow a=\pm3
\\

\displaystyle \textbf{Question 46. }\text{If }N\text{ denotes the set of all natural numbers and }R\text{ be}
\displaystyle \text{the relation on }N\times N\text{ defined by }(a,b)R(c,d)\text{ if}
\displaystyle ad(b+c)=bc(a+d).\text{ Show that }R\text{ is an equivalence relation.}
\displaystyle \hspace{0.2cm} \text{[CBSE 2023], [CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given relation }R\text{ on }N\times N\text{ defined by}
\displaystyle (a,b)\ R\ (c,d)\iff ad(b+c)=bc(a+d)
\displaystyle \text{Reflexive: Let }(a,b)\in N\times N
\displaystyle \text{Then }ab(b+a)=ba(a+b),\text{ which is true}
\displaystyle \Rightarrow (a,b)\ R\ (a,b)
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Symmetric: Let }(a,b)\ R\ (c,d)
\displaystyle \Rightarrow ad(b+c)=bc(a+d)\qquad \cdots(i)
\displaystyle \Rightarrow da(c+b)=cb(d+a)
\displaystyle \Rightarrow cb(d+a)=da(c+b)
\displaystyle \Rightarrow (c,d)\ R\ (a,b)
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{Transitive: Let }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f)
\displaystyle \Rightarrow ad(b+c)=bc(a+d)\qquad \cdots(ii)
\displaystyle \Rightarrow cf(d+e)=de(c+f)\qquad \cdots(iii)
\displaystyle \text{From Eq. }(ii),
\displaystyle \frac{b+c}{bc}=\frac{a+d}{ad}
\displaystyle \Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}\qquad \cdots(iv)
\displaystyle \text{From Eq. }(iii),
\displaystyle \frac{d+e}{de}=\frac{c+f}{cf}
\displaystyle \Rightarrow \frac{1}{d}+\frac{1}{e}=\frac{1}{c}+\frac{1}{f}\qquad \cdots(v)
\displaystyle \text{Adding Eqs. }(iv)\text{ and }(v),
\displaystyle \left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{d}+\frac{1}{e}\right)
\displaystyle =\left(\frac{1}{a}+\frac{1}{d}\right)+\left(\frac{1}{c}+\frac{1}{f}\right)
\displaystyle \Rightarrow \frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f}
\displaystyle \Rightarrow \frac{e+b}{be}=\frac{f+a}{af}
\displaystyle \Rightarrow af(e+b)=be(f+a)
\displaystyle \Rightarrow af(b+e)=be(a+f)
\displaystyle \Rightarrow (a,b)\ R\ (e,f)
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Thus, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation.}
\\

\displaystyle \textbf{Question 47. }\text{Show that the function }f:R\to R\text{ defined by}
\displaystyle f(x)=\frac{x}{x^2+1}\text{ is neither one-one nor onto.} \hspace{0.2cm} \text{[CBSE 2023],}
\displaystyle \text{[CBSE 2020], [CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f:R\to R,\text{ defined by }f(x)=\frac{x}{x^{2}+1},\ \forall x\in R
\displaystyle \text{Let }x_{1},x_{2}\in R\text{ such that }f(x_{1})=f(x_{2})
\displaystyle \Rightarrow \frac{x_{1}}{x_{1}^{2}+1}=\frac{x_{2}}{x_{2}^{2}+1}
\displaystyle \Rightarrow x_{1}x_{2}^{2}+x_{1}=x_{2}x_{1}^{2}+x_{2}
\displaystyle \Rightarrow x_{1}x_{2}^{2}-x_{2}x_{1}^{2}+x_{1}-x_{2}=0
\displaystyle \Rightarrow x_{1}x_{2}(x_{2}-x_{1})-(x_{2}-x_{1})=0
\displaystyle \Rightarrow (x_{2}-x_{1})(x_{1}x_{2}-1)=0
\displaystyle \Rightarrow x_{2}=x_{1}\text{ or }x_{1}x_{2}=1
\displaystyle \Rightarrow x_{1}=x_{2}\text{ or }x_{1}=\frac{1}{x_{2}}
\displaystyle \therefore f\text{ is not one-one, since for }x_{1}=3\text{ and }x_{2}=\frac{1}{3},
\displaystyle f(3)=\frac{3}{10}=f\left(\frac{1}{3}\right),\text{ but }3\neq\frac{1}{3}
\displaystyle \text{Now, let }k\in R\text{ be any arbitrary element and let }f(x)=k
\displaystyle \Rightarrow \frac{x}{x^{2}+1}=k \qquad \left[\because f(x)=\frac{x}{x^{2}+1}\right]
\displaystyle \Rightarrow kx^{2}+k=x
\displaystyle \Rightarrow kx^{2}-x+k=0
\displaystyle \Rightarrow x=\frac{1\pm\sqrt{1-4k^{2}}}{2k}
\displaystyle \text{For real values of }x,\ 1-4k^{2}\geq0
\displaystyle \Rightarrow (1-2k)(1+2k)\geq0
\displaystyle \Rightarrow -\frac{1}{2}\leq k\leq\frac{1}{2}
\displaystyle \text{Thus, for }k>\frac{1}{2}\text{ or }k<-\frac{1}{2},\text{ no real }x\text{ exists.}
\displaystyle \therefore f\text{ is not onto.}
\displaystyle \text{Hence, }f\text{ is neither one-one nor onto.}
\\

\displaystyle \textbf{Question 48. }\text{Prove that the relation }R\text{ on }Z,\text{ defined by}
\displaystyle R=\{(x,y):(x-y)\text{ is divisible by }5\}\text{ is an equivalence relation.}
\displaystyle \hspace{0.2cm} \text{[CBSE 2020], [CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }R=\{(x,y):(x-y)\text{ is divisible by }5\}\text{ and }Z\text{ is the set}
\displaystyle \text{of integers.}
\displaystyle \text{Reflexive: Let }x\in Z\text{ be any arbitrary element.}
\displaystyle \text{Then }x-x=0,\text{ which is divisible by }5
\displaystyle \Rightarrow (x,x)\in R
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Symmetric: Let }x,y\in Z\text{ such that }(x,y)\in R
\displaystyle \Rightarrow 5\text{ divides }(x-y)
\displaystyle \Rightarrow 5\text{ divides }[-(x-y)]
\displaystyle \Rightarrow 5\text{ divides }(y-x)
\displaystyle \Rightarrow (y,x)\in R
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{Transitive: Let }x,y,z\in Z\text{ such that }(x,y)\in R\text{ and }(y,z)\in R
\displaystyle \Rightarrow x-y\text{ and }y-z\text{ are both divisible by }5
\displaystyle \Rightarrow (x-y)+(y-z)\text{ is divisible by }5
\displaystyle \Rightarrow (x-z)\text{ is divisible by }5
\displaystyle \Rightarrow (x,z)\in R
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Thus, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \text{Hence, }R\text{ is an equivalence relation.}
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\displaystyle \textbf{Question 49. }\text{Let }A=\{x\in Z:0\leq x\leq12\}.\text{ Show that}
\displaystyle S=\{(a,b):a,b\in A,\ |a-b|\text{ is divisible by }4\}\text{ is an equivalence}
\displaystyle \text{relation. Find the set of all elements related to }1\text{ and also write}
\displaystyle \text{the equivalence class }[2]. \hspace{0.2cm} \text{[CBSE 2018], [CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }S=\{(a,b):|a-b|\text{ is divisible by }4,\ a,b\in A\}
\displaystyle \text{and }A=\{x:x\in Z\text{ and }0\leq x\leq12\}
\displaystyle \text{Thus, }A=\{0,1,2,3,\ldots,12\}
\displaystyle \text{Reflexive: For any }x\in A,\ |x-x|=0,\text{ which is divisible by }4
\displaystyle \Rightarrow (x,x)\in S,\ \forall x\in A
\displaystyle \therefore S\text{ is reflexive.}
\displaystyle \text{Symmetric: Let }(x,y)\in S
\displaystyle \Rightarrow |x-y|\text{ is divisible by }4\qquad [\text{by definition of }S]
\displaystyle \Rightarrow |x-y|=4\lambda,\text{ for some }\lambda\in Z
\displaystyle \Rightarrow |y-x|=4\lambda
\displaystyle \Rightarrow (y,x)\in S
\displaystyle \therefore S\text{ is symmetric.}
\displaystyle \text{Transitive: Let }(x,y)\in S\text{ and }(y,z)\in S
\displaystyle \Rightarrow |x-y|\text{ and }|y-z|\text{ are divisible by }4
\displaystyle \Rightarrow |x-y|=4\lambda\text{ and }|y-z|=4\mu,\text{ for some }\lambda,\mu\in Z
\displaystyle \text{Now, }x-z=(x-y)+(y-z)
\displaystyle \Rightarrow x-z=\pm4\lambda\pm4\mu=\pm4(\lambda+\mu)
\displaystyle \Rightarrow |x-z|\text{ is divisible by }4
\displaystyle \Rightarrow (x,z)\in S
\displaystyle \therefore S\text{ is transitive.}
\displaystyle \text{Since }S\text{ is reflexive, symmetric and transitive, it is an equivalence}
\displaystyle \text{relation.}
\displaystyle \text{The set of all elements related to }1\text{ is }\{1,5,9\}
\displaystyle [2]=\{a\in A:|2-a|\text{ is divisible by }4\}=\{2,6,10\}
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\displaystyle \textbf{Question 50. }\text{Show that the relation }R\text{ on the set }Z\text{ of all integers}
\displaystyle \text{defined by }(x,y)\in R\Leftrightarrow (x-y)\text{ is divisible by }3\text{ is an equivalence}
\displaystyle \text{relation.} \hspace{0.2cm} \text{[CBSE 2018C]}
\displaystyle \text{Answer:}
\displaystyle \text{The given relation is }R=\{(x,y):x,y\in Z\text{ and }x-y\text{ is divisible by }3\}.
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}
\displaystyle \text{Reflexive: For any }x\in Z,\text{ we have }x-x=0,\text{ which is divisible by }3.
\displaystyle \Rightarrow (x-x)\text{ is divisible by }3\Rightarrow (x,x)\in R,\ \forall x\in Z
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Symmetric: Let }(x,y)\in R,\text{ where }x,y\in Z.
\displaystyle \Rightarrow (x-y)\text{ is divisible by }3\qquad [\text{by definition of }R]
\displaystyle \Rightarrow x-y=3\lambda\text{ for some }\lambda\in Z
\displaystyle \Rightarrow y-x=3(-\lambda)\Rightarrow (y-x)\text{ is also divisible by }3
\displaystyle \Rightarrow (y,x)\in R
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{Transitive: Let }(x,y)\in R\text{ and }(y,z)\in R,\text{ where }x,y,z\in Z.
\displaystyle \Rightarrow x-y=3A\text{ and }y-z=3B\text{ for some }A,B\in Z
\displaystyle \text{Adding, we get }(x-y)+(y-z)=3A+3B
\displaystyle \Rightarrow x-z=3(A+B)
\displaystyle \Rightarrow (x-z)\text{ is divisible by }3,\text{ since }(A+B)\in Z
\displaystyle \Rightarrow (x,z)\in R
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Thus, }R\text{ is reflexive, symmetric and transitive. Hence, }R\text{ is an}
\displaystyle \text{equivalence relation.}
\displaystyle \text{Note: If at least one of the conditions namely reflexive, symmetric or}
\displaystyle \text{transitive is not satisfied, then the relation is not an equivalence relation.}
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\displaystyle \textbf{Question 51. }\text{Show that the relation }R\text{ in the set }A=\{1,2,3,4,5\}
\displaystyle \text{given by }R=\{(a,b):|a-b|\text{ is divisible by }2\},\text{ is an equivalence relation.}
\displaystyle \text{Write all the equivalence classes of }R.\ \text{[CBSE 2015C]}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }A=\{1,2,3,4,5\}
\displaystyle R=\{(a,b):|a-b|\text{ is divisible by }2\}
\displaystyle \text{For reflexive, let }a\in A
\displaystyle |a-a|=0,\text{ which is divisible by }2
\displaystyle \therefore (a,a)\in R,\ \forall a\in A
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{For symmetric, let }(a,b)\in R
\displaystyle \Rightarrow |a-b|\text{ is divisible by }2
\displaystyle \text{But }|a-b|=|b-a|
\displaystyle \Rightarrow |b-a|\text{ is divisible by }2
\displaystyle \Rightarrow (b,a)\in R
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{For transitive, let }(a,b)\in R\text{ and }(b,c)\in R
\displaystyle \Rightarrow |a-b|\text{ is divisible by }2\text{ and }|b-c|\text{ is divisible by }2
\displaystyle \Rightarrow a-b\text{ is even and }b-c\text{ is even}
\displaystyle \Rightarrow (a-b)+(b-c)=a-c\text{ is even}
\displaystyle \Rightarrow |a-c|\text{ is divisible by }2
\displaystyle \Rightarrow (a,c)\in R
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Hence, }R\text{ is an equivalence relation.}
\displaystyle \text{Now, elements related to }1\text{ are }1,3,5
\displaystyle \Rightarrow [1]=\{1,3,5\}
\displaystyle \text{Elements related to }2\text{ are }2,4
\displaystyle \Rightarrow [2]=\{2,4\}
\displaystyle \text{Thus, equivalence classes are }\{1,3,5\}\text{ and }\{2,4\}.
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