Class 12: Binary Operations – CBSE Board Problems

$\displaystyle \textbf{Question 1: }\text{Let }\ast\text{ be a binary operation on the set }\text{I}\text{ of integers, defined by }a\ast b=2a+b-3.$
$\displaystyle \text{Find the value of }3\ast 4.\hspace{0.2cm}\text{[CBSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\ast\text{ is a binary operation on the set I of integers.}$
$\displaystyle \text{The operation is defined by }a\ast b=2a+b-3.$
$\displaystyle \text{We need to find the value of }3\ast 4.$
$\displaystyle \text{Since }3\text{ and }4\text{ belong to the set of integers, we can use the binary operation.}$
$\displaystyle \Rightarrow 3\ast 4=(2\times 3)+4-3$
$\displaystyle \Rightarrow 3\ast 4=6+1$
$\displaystyle \Rightarrow 3\ast 4=7$
$\displaystyle \text{Therefore, the value of }3\ast 4\text{ is }7.$
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$\displaystyle \textbf{Question 2: }\text{The binary operation }\ast:R\times R\rightarrow R\text{ is defined as }a\ast b=2a+b. \text{Find }(2\ast 3)\ast 4.\hspace{0.2cm}\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\ast\text{ is an operation from }R\times R\rightarrow R\text{ and is defined by }a\ast b=2a+b.$
$\displaystyle \text{We need to find the value of }(2\ast 3)\ast 4.$
$\displaystyle \Rightarrow (2\ast 3)\ast 4=((2\times 2)+3)\ast 4$
$\displaystyle \Rightarrow (2\ast 3)\ast 4=(4+3)\ast 4$
$\displaystyle \Rightarrow (2\ast 3)\ast 4=7\ast 4$
$\displaystyle \Rightarrow (2\ast 3)\ast 4=(2\times 7)+4$
$\displaystyle \Rightarrow (2\ast 3)\ast 4=14+4$
$\displaystyle \Rightarrow (2\ast 3)\ast 4=18$
$\displaystyle \text{Therefore, the value of }(2\ast 3)\ast 4\text{ is }18.$
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$\displaystyle \textbf{Question 3: }\text{Let }\ast\text{ be a binary operation on }N\text{ given by }a\ast b=\mathrm{LCM}(a,b)$
$\displaystyle \text{for all }a,b\in N.\text{ Find }5\ast 7.\hspace{0.2cm}\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }\ast\text{ is an operation on the set of natural numbers }N\text{ and is defined by }a\ast b=\mathrm{LCM}(a,b).$
$\displaystyle \text{We need to find the value of }5\ast 7.$
$\displaystyle \Rightarrow 5\ast 7=\mathrm{LCM}(5,7)$
$\displaystyle \text{We know that the LCM of two prime numbers is the product of the two prime numbers.}$
$\displaystyle \Rightarrow 5\ast 7=5\times 7$
$\displaystyle \Rightarrow 5\ast 7=35$
$\displaystyle \text{Therefore, the value of }5\ast 7\text{ is }35.$
$\\$

$\displaystyle \textbf{Question 4: }\text{Let }S\text{ be the set of all rational numbers except }1\text{ and }\ast\text{ be defined on }S$
$\displaystyle \text{by }a\ast b=a+b-ab,\text{ for all }a,b\in S.\text{ Prove that:}$
$\displaystyle \text{(i) }\ast\text{ is a binary operation on }S$
$\displaystyle \text{(ii) }\ast\text{ is commutative as well as associative.}\hspace{0.2cm}\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Sum, difference and product of rational numbers is a unique rational number.}$
$\displaystyle \text{Let }a,b\in S.\text{ Then }a,b\in Q\text{ and }a\neq 1,\ b\neq 1.$
$\displaystyle a\ast b=a+b-ab=1-(1-a)(1-b)$
$\displaystyle \text{Since }a\neq 1\text{ and }b\neq 1,\text{ we have }(1-a)(1-b)\neq 0.$
$\displaystyle \Rightarrow 1-(1-a)(1-b)\neq 1$
$\displaystyle \Rightarrow a\ast b\neq 1$
$\displaystyle \therefore a\ast b\in S$
$\displaystyle \text{Hence, }\ast\text{ is a binary operation on }S.$
$\displaystyle \text{(ii) }a\ast b=a+b-ab=b+a-ba=b\ast a$
$\displaystyle \text{Therefore, }\ast\text{ is commutative.}$
$\displaystyle a\ast(b\ast c)=a\ast(b+c-bc)$
$\displaystyle =a+(b+c-bc)-a(b+c-bc)$
$\displaystyle =a+b+c-bc-ab-ac+abc\hspace{1.0cm}\ldots(i)$
$\displaystyle (a\ast b)\ast c=(a+b-ab)\ast c$
$\displaystyle =(a+b-ab)+c-(a+b-ab)c$
$\displaystyle =a+b-ab+c-ac-bc+abc\hspace{1.0cm}\ldots(ii)$
$\displaystyle \text{From }(i)\text{ and }(ii),\text{ we get }a\ast(b\ast c)=(a\ast b)\ast c.$
$\displaystyle \text{Therefore, }\ast\text{ is associative.}$
$\\$

$\displaystyle \textbf{Question 5: }\text{If the binary operation }\ast\text{ on the set }Z\text{ is defined by }a\ast b=a+b-5, \text{then find the identity element with respect to }\ast.\hspace{0.2cm}\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that the binary operation }\ast\text{ is defined on }Z\text{ by }a\ast b=a+b-5,$
$\displaystyle \text{for all }a,b\in Z.$
$\displaystyle \text{Let }a\in Z\text{ and let the identity element be }e\in Z.$
$\displaystyle \text{We know that the identity property is defined as follows:}$
$\displaystyle \Rightarrow a\ast e=e\ast a=a$
$\displaystyle \Rightarrow a+e-5=a$
$\displaystyle \Rightarrow e-5=a-a$
$\displaystyle \Rightarrow e=5$
$\displaystyle \text{Therefore, the required identity element with respect to }\ast\text{ is }5.$
$\\$

$\displaystyle \textbf{Question 6: }\text{On the set }Z\text{ of integers, if the binary operation }\ast\text{ is defined by}$
$\displaystyle a\ast b=a+b+2,\text{ then find the identity element.}\hspace{0.2cm}\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that the binary operation }\ast\text{ is defined on }Z\text{ by }a\ast b=a+b+2,$
$\displaystyle \text{for all }a,b\in Z.$
$\displaystyle \text{Let }a\in Z\text{ and let the identity element be }e\in Z.$
$\displaystyle \text{We know that the identity property is defined as follows:}$
$\displaystyle \Rightarrow a\ast e=e\ast a=a$
$\displaystyle \Rightarrow a+e+2=a$
$\displaystyle \Rightarrow e+2=a-a$
$\displaystyle \Rightarrow e=-2$
$\displaystyle \text{Therefore, the required identity element with respect to }\ast\text{ is }-2.$
$\\$

$\displaystyle \textbf{Question 7: }\text{Let }A=R\times R\text{ and }\ast\text{ be a binary operation on }A\text{ defined by}$
$\displaystyle (a,b)\ast(c,d)=(a+c,b+d). \text{ Show that }\ast\text{ is commutative and associative.}$
$\displaystyle \text{Find the identity element for }\ast\text{ on }A,\text{ if any.}\hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle A=R\times R$
$\displaystyle \text{For }\ast\text{ to be commutative, }p\ast q=q\ast p\text{ must be true for all }p,q\in A.$
$\displaystyle \text{Let }p=(a,b)\text{ and }q=(c,d).$
$\displaystyle p\ast q=(a,b)\ast(c,d)=(a+c,b+d)$
$\displaystyle q\ast p=(c,d)\ast(a,b)=(c+a,d+b)=(a+c,b+d)$
$\displaystyle \Rightarrow p\ast q=q\ast p$
$\displaystyle \therefore \ast\text{ is commutative.}$
$\displaystyle \text{For }\ast\text{ to be associative, }p\ast(q\ast r)=(p\ast q)\ast r\text{ must hold for all }p,q,r\in A.$
$\displaystyle \text{Let }r=(e,f).$
$\displaystyle p\ast(q\ast r)=(a,b)\ast((c,d)\ast(e,f))$
$\displaystyle =(a,b)\ast(c+e,d+f)$
$\displaystyle =(a+c+e,b+d+f)$
$\displaystyle (p\ast q)\ast r=((a,b)\ast(c,d))\ast(e,f)$
$\displaystyle =(a+c,b+d)\ast(e,f)$
$\displaystyle =(a+c+e,b+d+f)$
$\displaystyle \Rightarrow p\ast(q\ast r)=(p\ast q)\ast r$
$\displaystyle \therefore \ast\text{ is associative.}$
$\displaystyle \text{Identity element: Let }e=(x,y)\in A.$
$\displaystyle \text{For }e\text{ to be the identity element,}$
$\displaystyle (a,b)\ast(x,y)=(a,b)$
$\displaystyle \Rightarrow (a+x,b+y)=(a,b)$
$\displaystyle \Rightarrow a+x=a,\ b+y=b$
$\displaystyle \Rightarrow x=0,\ y=0$
$\displaystyle \therefore \text{The identity element is }(0,0).$
$\displaystyle \text{Inverse element: Let }(r,s)\text{ be the inverse of }(a,b).$
$\displaystyle (r,s)\ast(a,b)=(0,0)$
$\displaystyle \Rightarrow (r+a,s+b)=(0,0)$
$\displaystyle \Rightarrow r+a=0,\ s+b=0$
$\displaystyle \Rightarrow r=-a,\ s=-b$
$\displaystyle \therefore (-a,-b)\text{ is the inverse of }(a,b).$
$\\$

$\displaystyle \textbf{Question 8: }\text{Consider the binary operations }\ast:R\times R\rightarrow R\text{ and }\circ:R\times R\rightarrow R$
$\displaystyle \text{defined as }a\ast b=|a-b|\text{ and }a\circ b=a\text{ for all }a,b\in R.\text{ Show that }\ast\text{ is commutative}$
$\displaystyle \text{but not associative, }\circ\text{ is associative but not commutative. Further, show that}$
$\displaystyle \ast\text{ is distributive over }\circ.\text{ Does }\circ\text{ distribute over }\ast\text{? Justify your answer.}$
$\displaystyle \hspace{11.0cm}\textbf{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For any }a,b\in R,\text{ we have}$
$\displaystyle a\ast b=|a-b|\text{ and }b\ast a=|b-a|$
$\displaystyle \because |a-b|=|b-a|\text{ for all }a,b\in R$
$\displaystyle \therefore a\ast b=b\ast a\text{ for all }a,b\in R$
$\displaystyle \text{So, }\ast\text{ is commutative on }R.$
$\displaystyle ((-2)\ast 3)\ast 4=|-2-3|\ast 4=5\ast 4=|5-4|=1$
$\displaystyle (-2)\ast(3\ast 4)=(-2)\ast|3-4|=(-2)\ast 1=|-2-1|=3$
$\displaystyle \therefore ((-2)\ast 3)\ast 4\neq (-2)\ast(3\ast 4)$
$\displaystyle \text{So, }\ast\text{ is not associative on }R.$
$\displaystyle 2\circ 3=2\text{ and }3\circ 2=3$
$\displaystyle \therefore 2\circ 3\neq 3\circ 2$
$\displaystyle \text{So, }\circ\text{ is not commutative on }R.$
$\displaystyle \text{For any }a,b,c\in R,\text{ we have}$
$\displaystyle (a\circ b)\circ c=a\circ c=a$
$\displaystyle a\circ(b\circ c)=a\circ b=a$
$\displaystyle \therefore (a\circ b)\circ c=a\circ(b\circ c)\text{ for all }a,b,c\in R$
$\displaystyle \text{So, }\circ\text{ is associative on }R.$
$\displaystyle \text{For any }a,b,c\in R,\text{ we have}$
$\displaystyle a\ast(b\circ c)=a\ast b=|a-b|$
$\displaystyle (a\ast b)\circ(a\ast c)=|a-b|\circ|a-c|=|a-b|$
$\displaystyle \therefore a\ast(b\circ c)=(a\ast b)\circ(a\ast c)\text{ for all }a,b,c\in R$
$\displaystyle \text{So, }\ast\text{ is distributive over }\circ.$
$\displaystyle \text{Further, for any }a,b,c\in R,\text{ we have}$
$\displaystyle a\circ(b\ast c)=a\circ|b-c|=a$
$\displaystyle (a\circ b)\ast(a\circ c)=a\ast a=|a-a|=0$
$\displaystyle \therefore a\circ(b\ast c)\neq(a\circ b)\ast(a\circ c)$
$\displaystyle \text{So, }\circ\text{ is not distributive over }\ast.$
$\\$

$\displaystyle \textbf{Question 9: }\text{If }\ast\text{ is defined on the set }R_0\text{ of all non-zero real numbers by}$
$\displaystyle a\ast b=\frac{3ab}{7},\text{ find the identity element in }R_0\text{ for the binary operation }\ast.\hspace{0.2cm}\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }e\text{ be the identity element in }R_0\text{ for the binary operation }\ast.$
$\displaystyle \text{Then, }a\ast e=a=e\ast a\text{ for all }a\in R_0.$
$\displaystyle \Rightarrow a\ast e=a\text{ and }e\ast a=a\text{ for all }a\in R_0$
$\displaystyle \Rightarrow \frac{3ae}{7}=a\text{ and }\frac{3ea}{7}=a\text{ for all }a\in R_0$
$\displaystyle \Rightarrow e=\frac{7}{3}$
$\displaystyle \text{Hence, }\frac{7}{3}\text{ is the identity element in }R_0.$
$\\$

$\displaystyle \textbf{Question 10: }\text{Let }\ast\text{ be a binary operation on the set }Q-\{1\}\text{ defined by}$
$\displaystyle a\ast b=a+b-ab\text{ for all }a,b\in Q-\{1\}.\text{ Find the identity element with respect to}$
$\displaystyle \ast\text{ on }Q-\{1\}.\text{ Also, prove that every element of }Q-\{1\}\text{ is invertible.}\hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the identity element }e\text{ exist in }Q-\{1\}\text{ with respect to }\ast.$
$\displaystyle \text{Then, }a\ast e=a=e\ast a\text{ for all }a\in Q-\{1\}.$
$\displaystyle \Rightarrow a\ast e=a\text{ for all }a\in Q-\{1\}$
$\displaystyle \Rightarrow a+e-ae=a$
$\displaystyle \Rightarrow e(1-a)=0$
$\displaystyle \Rightarrow e=0\hspace{1.0cm}[\because a\in Q-\{1\}\Rightarrow a\neq 1]$
$\displaystyle \text{Thus, }0\text{ is the identity element for }\ast\text{ on }Q-\{1\}.$
$\displaystyle \text{Let }a\text{ be an arbitrary element of }Q-\{1\}\text{ and let }b\text{ be the inverse of }a.$
$\displaystyle \text{Then, }a\ast b=0=b\ast a.$
$\displaystyle \Rightarrow a\ast b=0$
$\displaystyle \Rightarrow a+b-ab=0$
$\displaystyle \Rightarrow b(1-a)=-a$
$\displaystyle \Rightarrow b=\frac{a}{a-1}\hspace{1.0cm}[\because a\in Q-\{1\}\Rightarrow a-1\neq 0]$
$\displaystyle \text{Since }a\in Q-\{1\},\text{ therefore }b=\frac{a}{a-1}\in Q-\{1\}.$
$\displaystyle \text{Thus, every element of }Q-\{1\}\text{ is invertible and the inverse of }a\text{ is }\frac{a}{a-1}.$
$\\$

$\displaystyle \textbf{Question 11: }\text{On the set }R-\{-1\},\text{ a binary operation }\ast\text{ is defined by}$
$\displaystyle a\ast b=a+b+ab\text{ for all }a,b\in R-\{-1\}.\text{ Prove that }\ast\text{ is commutative}$
$\displaystyle \text{as well as associative on }R-\{-1\}.\text{ Find the identity element and prove that every element}$
$\displaystyle \text{of }R-\{-1\}\text{ is invertible.}\hspace{0.2cm}\text{[CBSE 2015, 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We observe the following properties of }\ast\text{ on }R-\{-1\}.$
$\displaystyle \text{Commutativity: For any }a,b\in R-\{-1\},\text{ we have}$
$\displaystyle a\ast b=a+b+ab\text{ and }b\ast a=b+a+ba$
$\displaystyle \because a+b+ab=b+a+ba$
$\displaystyle \Rightarrow a\ast b=b\ast a$
$\displaystyle \text{Hence, }\ast\text{ is commutative on }R-\{-1\}.$
$\displaystyle \text{Associativity: For any }a,b,c\in R-\{-1\},\text{ we have}$
$\displaystyle (a\ast b)\ast c=(a+b+ab)\ast c$
$\displaystyle =(a+b+ab)+c+(a+b+ab)c$
$\displaystyle =a+b+c+ab+bc+ac+abc\hspace{1.0cm}\ldots(i)$
$\displaystyle a\ast(b\ast c)=a\ast(b+c+bc)$
$\displaystyle =a+(b+c+bc)+a(b+c+bc)$
$\displaystyle =a+b+c+ab+bc+ac+abc\hspace{1.0cm}\ldots(ii)$
$\displaystyle \text{From }(i)\text{ and }(ii),\text{ we have }(a\ast b)\ast c=a\ast(b\ast c).$
$\displaystyle \text{Hence, }\ast\text{ is associative on }R-\{-1\}.$
$\displaystyle \text{Existence of identity: Let }e\text{ be the identity element. Then,}$
$\displaystyle a\ast e=a=e\ast a\text{ for all }a\in R-\{-1\}$
$\displaystyle \Rightarrow a+e+ae=a$
$\displaystyle \Rightarrow e(1+a)=0$
$\displaystyle \Rightarrow e=0$
$\displaystyle \text{Also, }0\in R-\{-1\}.$
$\displaystyle \text{So, }0\text{ is the identity element for }\ast\text{ on }R-\{-1\}.$
$\displaystyle \text{Existence of inverse: Let }a\in R-\{-1\}\text{ and let }b\text{ be the inverse of }a.$
$\displaystyle \Rightarrow a\ast b=0$
$\displaystyle \Rightarrow a+b+ab=0$
$\displaystyle \Rightarrow b(1+a)=-a$
$\displaystyle \Rightarrow b=\frac{-a}{a+1}$
$\displaystyle \text{Now, }a\in R-\{-1\}\Rightarrow a\neq -1\Rightarrow a+1\neq 0$
$\displaystyle \Rightarrow b=\frac{-a}{a+1}\in R$
$\displaystyle \text{Also, }\frac{-a}{a+1}=-1\Rightarrow -a=-a-1\Rightarrow 0=-1,\text{ which is not possible.}$
$\displaystyle \therefore \frac{-a}{a+1}\in R-\{-1\}$
$\displaystyle \text{Hence, every element of }R-\{-1\}\text{ is invertible and the inverse of }a\text{ is }\frac{-a}{a+1}.$
$\\$

$\displaystyle \textbf{Question 12: }\text{Consider the infimum binary operation }\wedge\text{ on the set }S=\{1,2,3,4,5\}$
$\displaystyle \text{defined by }a\wedge b=\text{ minimum of }a\text{ and }b.\text{ Write the composition table of the operation }\wedge.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle 1\wedge 1=\text{ minimum of }1\text{ and }1=1$
$\displaystyle 1\wedge 2=\text{ minimum of }1\text{ and }2=1$
$\displaystyle 4\wedge 3=\text{ minimum of }4\text{ and }3=3$
$\displaystyle \text{So, we have the following composition table for }\wedge\text{ on }S.$
$\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \wedge & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 1 & 1 & 1 & 1 & 1 \\ \hline 2 & 1 & 2 & 2 & 2 & 2 \\ \hline 3 & 1 & 2 & 3 & 3 & 3 \\ \hline 4 & 1 & 2 & 3 & 4 & 4 \\ \hline 5 & 1 & 2 & 3 & 4 & 5 \\ \hline \end{array}$
$\\$

$\displaystyle \textbf{Question 13: }\text{Define a binary operation }\ast\text{ on the set }\{0,1,2,3,4,5\}\text{ as}$
$\displaystyle a\ast b=\begin{cases}a+b, & \text{if }a+b<6 \\ a+b-6, & \text{if }a+b\geq 6\end{cases}$
$\displaystyle \text{Show that }0\text{ is the identity for this operation and each element }a\neq 0\text{ of the set is invertible}$
$\displaystyle \text{with }6-a\text{ being the inverse of }a.\hspace{0.2cm}\text{[CBSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }X=\{0,1,2,3,4,5\}.$
$\displaystyle \text{The operation }\ast\text{ on }X\text{ is defined as}$
$\displaystyle a\ast b=\begin{cases}a+b, & \text{if }a+b<6 \\ a+b-6, & \text{if }a+b\geq 6\end{cases}$
$\displaystyle \text{An element }e\in X\text{ is the identity element for the operation }\ast\text{ if }a\ast e=a=e\ast a$
$\displaystyle \text{for all }a\in X.$
$\displaystyle \text{For }a\in X,\text{ we have}$
$\displaystyle a\ast 0=a+0=a\hspace{1.0cm}[a\in X\Rightarrow a+0<6]$
$\displaystyle 0\ast a=0+a=a\hspace{1.0cm}[a\in X\Rightarrow 0+a<6]$
$\displaystyle \text{Therefore, }a\ast 0=a=0\ast a\text{ for all }a\in X.$
$\displaystyle \text{Thus, }0\text{ is the identity element for the given operation }\ast.$
$\displaystyle \text{An element }a\in X\text{ is invertible if there exists }b\in X\text{ such that }a\ast b=0=b\ast a.$
$\displaystyle \text{i.e. }\begin{cases}a+b=0=b+a, & \text{if }a+b<6 \\ a+b-6=0=b+a-6, & \text{if }a+b\geq 6\end{cases}$
$\displaystyle \Rightarrow a=-b\text{ or }b=6-a$
$\displaystyle \text{But }X=\{0,1,2,3,4,5\}\text{ and }a,b\in X.\text{ Then }a\neq -b\text{ for }a\neq 0.$
$\displaystyle \text{Therefore, }b=6-a\text{ is the inverse of }a\text{ for all }a\in X,\ a\neq 0.$
$\displaystyle \text{Hence, the inverse of an element }a\in X,\ a\neq 0,\text{ is }6-a,\text{ i.e., }a^{-1}=6-a.$
$\\$