\displaystyle \textbf{Question 1. } \text{Let }L\text{ be a set of all straight lines in a plane. The relation} \ \
\displaystyle  R\text{ on }L\text{ defined as 'perpendicular to' is \hspace{4.2cm} ISC 2024} \ \
\displaystyle \text{(a) Symmetric and Transitive} \ \
\displaystyle \text{(b) Transitive} \ \
\displaystyle \text{(c) Symmetric} \ \
\displaystyle \text{(d) Equivalence} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, the relation } R \text{ is defined as } R = \{(L_1, L_2) : L_1 \perp L_2\}
\displaystyle \text{Now, } R \text{ is not reflexive as every line is not perpendicular to itself.}
\displaystyle \text{Since, } L_1 \perp L_2, \text{ therefore } L_2 \text{ is also perpendicular to } L_1.
\displaystyle \text{Thus, relation is symmetric.}
\displaystyle \text{Also, } L_1 \perp L_2 \text{ and } L_2 \perp L_3.
\displaystyle \text{Then, } L_1 \text{ may or may not be perpendicular to } L_3.
\displaystyle \text{So, } R \text{ is not transitive.}
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\displaystyle \textbf{Question 2. } \text{Let }A\text{ be a non-empty set.} \ \
\displaystyle \text{Statement I }\text{Identity relation on }A\text{ is reflexive.} \ \
\displaystyle \text{Statement II }\text{Every reflexive relation on }A\text{ is an}   \text{identity relation. \hspace{0.2cm} ISC 2024} \ \
\displaystyle \text{(a) Both the statements are true.} \ \
\displaystyle \text{(b) Both the statements are false.} \ \
\displaystyle \text{(c) Statement I is true and Statement II is false.} \ \
\displaystyle \text{(d) Statement I is false and Statement II is true.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{ (c) Let } A \text{ be a non-empty set.}
\displaystyle \text{Statement I: Identity relation on } A \text{ is reflexive.}
\displaystyle \text{Consider an example.}
\displaystyle \text{Let } A = \{1, 2, 3\}
\displaystyle \text{Now, a relation } R \text{ defined as}
\displaystyle R = \{(1,1), (2,2), (3,3)\} \text{ is an identity relation which is reflexive.}
\displaystyle \text{But } R' = \{(1,1), (1,3), (3,3), (2,2), (2,3)\} \text{ is a reflexive relation}
\displaystyle \text{but not an identity relation.}
\displaystyle \text{So, Statement II is not necessarily true.}
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\displaystyle \textbf{Question 3. } \text{Which one of the following graphs is a function of }x\text{? \hspace{0.2cm} ISC 2024} \ \
\displaystyle \text{Graph A \qquad Graph B} \ \
\displaystyle \text{Answer:}
\displaystyle \text{ By the definition of function}
\displaystyle \text{If } A \text{ and } B \text{ are any two sets, then a function } f : A \to B
\displaystyle \text{is a rule that assigns to each element of } A \text{ exactly one element of } B.
\displaystyle \text{So, Graph A represents a function of } x \text{ but Graph B is not a function}
\displaystyle \text{because at } x = 0, \text{ we get three values of } y.
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\displaystyle \textbf{Question 4. } \text{A relation }R\text{ on }\{1,2,3\}\text{ is given by }R=\{(1,1),(2,2),  \ \
\displaystyle (1,2),(3,3),(2,3)\}\text{. Then, the relation }R\text{ is \hspace{3.2cm} ISC 2023} \ \
\displaystyle \text{(a) reflexive} \ \      \displaystyle \text{(b) symmetric} \ \      \displaystyle \text{(c) transitive} \ \      \displaystyle \text{(d) symmetric and transitive} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, } R = \{(1,1), (2,2), (1,2), (3,3), (2,3)\}
\displaystyle \text{Since, } (1,1), (2,2) \text{ and } (3,3) \in R
\displaystyle \text{Hence, } R \text{ is reflexive.}
\displaystyle (1,2) \in R \text{ but } (2,1) \notin R
\displaystyle \text{Hence, } R \text{ is not symmetric.}
\displaystyle (1,2) \in R, (2,3) \in R \text{ but } (1,3) \notin R
\displaystyle \text{Hence, } R \text{ is not transitive.}
\\
\displaystyle \textbf{Question 5. } \text{Let }f(x)=x^{3}\text{ be a function with domain }\{0,1,2,3\}\text{.} \ \
\displaystyle \text{Then, domain of }f^{-1}\text{ is \hspace{7.2cm} ISC 2023} \ \
\displaystyle \text{(a) }\{3,2,1,0\} \ \       \displaystyle \text{(b) }\{0,-1,-2,-3\} \ \      \displaystyle \text{(c) }\{0,1,8,27\} \ \      \displaystyle \text{(d) }\{0,-1,-8,-27\} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(c) } f(x) = x^3
\displaystyle f(0) = 0, \; f(1) = 1, \; f(2) = 8, \; f(3) = 27
\displaystyle f^{-1}(0) = 0, \; f^{-1}(1) = 1, \; f^{-1}(8) = 2, \; f^{-1}(27) = 3
\displaystyle \therefore \text{Domain of } f^{-1} = \{0, 1, 8, 27\}
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\displaystyle \textbf{Question 6. } \text{Let }f:R-\left\{\frac{-1}{3}\right\}\rightarrow R-\{0\}\text{ be defined as }f(x)=\frac{5}{3x+1} \text{ is invertible.}
\displaystyle \text{Find }f^{-1}(x)\text{. \hspace{8.2cm} ISC 2024} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } f : R - \left\{-\frac{1}{3}\right\} \to R - \{0\} \text{ be defined as }
\displaystyle f(x) = \frac{5}{3x+1}
\displaystyle \text{Assume that } y = \frac{5}{3x+1}
\displaystyle \Rightarrow 3xy + y = 5
\displaystyle \Rightarrow 3xy = 5 - y
\displaystyle \Rightarrow x = \frac{5-y}{3y} = f^{-1}(y)
\displaystyle \text{Thus, } f^{-1}(x) = \frac{5-x}{3x}
\displaystyle \text{which is the required solution.}
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\displaystyle \textbf{Question 7. } \text{If }f:R\rightarrow R\text{ is defined by }f(x)=\frac{2x-7}{4}\text{, show that } f(x)\text{ is one-one }
\displaystyle \text{and onto. \hspace{8.2cm} ISC 2024} \ \
\displaystyle \text{Answer:}
\displaystyle \text{A mapping } f : R \to R \text{ is defined by}
\displaystyle f(x) = \frac{2x-7}{4}
\displaystyle \text{Let } x_1, x_2 \in R, \text{ then for one-one}
\displaystyle f(x_1) = f(x_2)
\displaystyle \Rightarrow \frac{2x_1-7}{4} = \frac{2x_2-7}{4}
\displaystyle \Rightarrow 2x_1 - 7 = 2x_2 - 7 \Rightarrow x_1 = x_2
\displaystyle \text{Thus, } f \text{ is one-one.}
\displaystyle \text{Now, let } y = \frac{2x-7}{4}
\displaystyle \text{Then, } 4y = 2x - 7
\displaystyle \Rightarrow 2x = 4y + 7
\displaystyle \Rightarrow x = \frac{4y+7}{2} = f^{-1}(y)
\displaystyle \text{Clearly, for each } y \in R, \text{ there exists } x = \frac{4y+7}{2} \in R
\displaystyle \text{such that } f\left(\frac{4y+7}{2}\right) = \frac{1}{4}\left[2\left(\frac{4y+7}{2}\right) - 7\right]
\displaystyle = \frac{1}{4}[4y + 7 - 7] = y
\displaystyle \Rightarrow f \text{ is onto.}
\displaystyle \text{Hence, } f \text{ is one-one and onto i.e. } f \text{ is bijective.}
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\displaystyle \textbf{Question 8. } \text{If }f(x)=\left[4-(x-7)^{3}\right]^{\frac{1}{5}}\text{ is a real invertible function, } \text{then find }f^{-1}(x).
\displaystyle \text{\hspace{9.2cm} ISC 2023} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } f(x) = [4 - (x - 7)^3]^{1/5}
\displaystyle \text{We know that } f(f^{-1}(x)) = x
\displaystyle \therefore f(f^{-1}(x)) = [4 - (f^{-1}(x) - 7)^3]^{1/5}
\displaystyle \Rightarrow x = [4 - (f^{-1}(x) - 7)^3]^{1/5}
\displaystyle \Rightarrow x^5 = 4 - (f^{-1}(x) - 7)^3
\displaystyle \Rightarrow (f^{-1}(x) - 7)^3 = 4 - x^5
\displaystyle \Rightarrow f^{-1}(x) - 7 = (4 - x^5)^{1/3}
\displaystyle \Rightarrow f^{-1}(x) = 7 + (4 - x^5)^{1/3}
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\displaystyle \textbf{Question 9. } \text{Let }A=R-\{2\}\text{ and }B=R-\{1\}\text{. If }f:A\rightarrow B\text{ is a function defined} \ \
\displaystyle \text{by }f(x)=\frac{x-1}{x-2}\text{ then show that }f\text{ is } \text{a one-one and onto function. \hspace{0.2cm} ISC 2023} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f(x) = \frac{x-1}{x-2}
\displaystyle \text{For one-one function}
\displaystyle \text{Let } x_1, x_2 \in R
\displaystyle \text{Now, } f(x_1) = \frac{x_1-1}{x_1-2} \text{ and } f(x_2) = \frac{x_2-1}{x_2-2}
\displaystyle 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_1x_2 - 2x_1 - x_2 + 2 = x_1x_2 - x_1 - 2x_2 + 2
\displaystyle \Rightarrow x_1 = x_2
\displaystyle \text{Hence, } f \text{ is one-one function.}
\displaystyle \text{For onto function, let } f(x) = y
\displaystyle \therefore y = \frac{x-1}{x-2} \Rightarrow xy - 2y = x - 1
\displaystyle \Rightarrow xy - x = 2y - 1 \Rightarrow x = \frac{2y-1}{y-1}, \; y \neq 1
\displaystyle \text{Thus, for every } y \neq 1, \text{ there exists } x \in R
\displaystyle \text{Now, substituting back }
\displaystyle f\left(\frac{2y-1}{y-1}\right) = \frac{\frac{2y-1}{y-1} - 1}{\frac{2y-1}{y-1} - 2}
\displaystyle = \frac{\frac{2y-1 - (y-1)}{y-1}}{\frac{2y-1 - 2(y-1)}{y-1}}
\displaystyle = \frac{\frac{y}{y-1}}{\frac{1}{y-1}} = y
\displaystyle \Rightarrow f(x) = y
\displaystyle \therefore f \text{ is onto function.}
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\displaystyle \textbf{Question 10. } \text{If }f:A\rightarrow A\text{ and }A=R-\left\{\frac{8}{5}\right\}\text{, show that the function} \ \
\displaystyle f(x)=\frac{8x+3}{5x-8}  \text{ is one-one and onto. Hence, find }f^{-1}\text{. \hspace{3.2cm} ISC 2019} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } f : A \to A, \; A = R - \left\{\frac{8}{5}\right\}
\displaystyle \text{and } f(x) = \frac{8x+3}{5x-8}
\displaystyle \text{For one-one Let } x_1, x_2 \in A \text{ such that}
\displaystyle f(x_1) = f(x_2)
\displaystyle \Rightarrow \frac{8x_1+3}{5x_1-8} = \frac{8x_2+3}{5x_2-8}
\displaystyle \Rightarrow (8x_1+3)(5x_2-8) = (5x_1-8)(8x_2+3)
\displaystyle \Rightarrow 40x_1x_2 - 64x_1 + 15x_2 - 24 = 40x_1x_2 + 15x_1 - 64x_2 - 24
\displaystyle \Rightarrow 15x_1 + 64x_1 = 15x_2 + 64x_2
\displaystyle \Rightarrow 79x_1 = 79x_2
\displaystyle \Rightarrow x_1 = x_2
\displaystyle \therefore f(x) \text{ is one-one function.}
\displaystyle \text{For onto Let } y \text{ be an arbitrary element of } A \text{ (codomain).}
\displaystyle \text{Then, } f(x) = y
\displaystyle \Rightarrow \frac{8x+3}{5x-8} = y
\displaystyle \Rightarrow 5xy - 8y = 8x + 3
\displaystyle \Rightarrow x(5y-8) = 8y + 3
\displaystyle \Rightarrow x = \frac{8y+3}{5y-8}, \; y \neq \frac{8}{5} \qquad (i)
\displaystyle \text{Clearly, } x = \frac{8y+3}{5y-8} \in R \text{ for all } y \neq \frac{8}{5}
\displaystyle \text{Thus, for every } y \in A \text{ there exists } x = \frac{8y+3}{5y-8} \in A
\displaystyle \text{such that } f\left(\frac{8y+3}{5y-8}\right) = \frac{8\left(\frac{8y+3}{5y-8}\right)+3}{5\left(\frac{8y+3}{5y-8}\right)-8}
\displaystyle = \frac{\frac{64y+24}{5y-8} + 3}{\frac{40y+15}{5y-8} - 8}
\displaystyle = \frac{\frac{64y+24 + 15y - 24}{5y-8}}{\frac{40y+15 - 40y + 64}{5y-8}}
\displaystyle = \frac{\frac{79y}{5y-8}}{\frac{79}{5y-8}} = y
\displaystyle \Rightarrow f(x) \text{ is onto function.}
\displaystyle \text{Since, } f(x) \text{ is bijective function, so its inverse exists.}
\displaystyle \text{From Eq. (i), we get } x = \frac{8y+3}{5y-8}
\displaystyle \Rightarrow f^{-1}(y) = \frac{8y+3}{5y-8} \qquad [\because y = f(x) \Rightarrow x = f^{-1}(y)]
\displaystyle \therefore f^{-1}(x) = \frac{8x+3}{5x-8}, \; x \neq \frac{8}{5}
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\displaystyle \textbf{Question 11. } \text{If the function }f(x)=\sqrt{2x-3}\text{ is invertible, then find its inverse} \ \
\displaystyle \text{ \hspace{9.2cm} ISC 2018} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f(x) = \sqrt{2x-3} \text{ is invertible function.}
\displaystyle \text{Let } f(x) = y
\displaystyle \Rightarrow f^{-1}(y) = x
\displaystyle \therefore y = \sqrt{2x-3}
\displaystyle \Rightarrow y^2 = 2x - 3
\displaystyle \Rightarrow x = \frac{y^2 + 3}{2}
\displaystyle \Rightarrow f^{-1}(y) = \frac{y^2 + 3}{2}
\displaystyle \Rightarrow f^{-1}(x) = \frac{x^2 + 3}{2}
\displaystyle \therefore \text{Inverse of } f(x) = \frac{x^2 + 3}{2}
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\displaystyle \textbf{Question 12. } \text{If }f:W\rightarrow W\text{ is defined as }f(x)=x-1,\text{ if }x\text{ is odd and} \ \
\displaystyle f(x)=x+1,\text{ if }x\text{ is even. Show that }f\text{ is invertible. } \text{Find the inverse of }f\text{, where}
\displaystyle W \text{ is the set of all whole numbers} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f : W \to W \text{ is defined as}
\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 function Let } x_1, x_2 \in W \text{ be any two numbers such that } f(x_1) = f(x_2)
\displaystyle \textbf{Case I: When } x_1 \text{ and } x_2 \text{ are odd}
\displaystyle \text{Then, } f(x_1) = f(x_2) \Rightarrow x_1 - 1 = x_2 - 1 \Rightarrow x_1 = x_2
\displaystyle \text{Case II: When } x_1 \text{ and } x_2 \text{ are even}
\displaystyle \text{Then, } f(x_1) = f(x_2) \Rightarrow x_1 + 1 = x_2 + 1 \Rightarrow x_1 = x_2
\displaystyle \text{Thus, in both cases, } f(x_1) = f(x_2) \Rightarrow x_1 = x_2
\displaystyle \text{Case III: When } x_1 \text{ is odd and } x_2 \text{ is even}
\displaystyle \text{Then, } x_1 \neq x_2
\displaystyle \text{Also, } f(x_1) \text{ is even and } f(x_2) \text{ is odd}
\displaystyle \text{So, } f(x_1) \neq f(x_2)
\displaystyle \text{Thus, } x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)
\displaystyle \text{Hence, from cases I, II and III, we can observe that } f(x) \text{ is a one-one function.}
\displaystyle \text{Onto function Clearly, any odd number } 2y+1 \text{ in the codomain } W \text{ is the image of } 2y
\displaystyle \text{Also, any even number } 2y \text{ in the codomain } W \text{ is the image of } 2y+1
\displaystyle \text{Thus, every element in } W \text{ (codomain) has a pre-image in } W \text{ (domain)}
\displaystyle \text{So, } f \text{ is onto.}
\displaystyle \text{Therefore, } f \text{ is bijective and so it is invertible.}
\displaystyle \text{Let } f(x) = y
\displaystyle \Rightarrow x - 1 = y, \text{ if } x \text{ is odd}
\displaystyle \text{and } x + 1 = y, \text{ if } x \text{ is even}
\displaystyle \therefore f^{-1}(y) = x = \begin{cases} y+1, & \text{if } y \text{ is even} \\ y-1, & \text{if } y \text{ is odd} \end{cases}
\displaystyle \text{or } f^{-1}(x) = \begin{cases} x+1, & \text{if } x \text{ is even} \\ x-1, & \text{if } x \text{ is odd} \end{cases}
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\displaystyle \textbf{Question 13. } \text{If }R\text{ is a relation defined on the set of natural numbers } N\text{ as follows }  
\displaystyle R=\{(x,y):x\in N,\ y\in N\text{ and }2x+y=24\}\text{, then find the domain and range of the } \ \
\displaystyle \text{relation }R\text{. Also, find } \text{whether }R\text{ is an } \text{equivalence relation or not.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } R = \{(x, y) : x \in N, \; y \in N \text{ and } 2x + y = 24\}
\displaystyle \text{Now, } x = 1 \Rightarrow y = 22; \qquad [\because \; y = 24 - 2x]
\displaystyle x = 2 \Rightarrow y = 20;
\displaystyle x = 3 \Rightarrow y = 18; \; x = 4 \Rightarrow y = 16;
\displaystyle x = 5 \Rightarrow y = 14; \; x = 6 \Rightarrow y = 12;
\displaystyle x = 7 \Rightarrow y = 10; \; x = 8 \Rightarrow y = 8;
\displaystyle x = 9 \Rightarrow y = 6; \; x = 10 \Rightarrow y = 4
\displaystyle \text{and } x = 11 \Rightarrow y = 2
\displaystyle \text{So, domain of } R = \{1, 2, 3, \ldots, 11\}
\displaystyle \text{and range of } R = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22\}
\displaystyle \text{and } R = \{(1,22), (2,20), (3,18), (4,16), (5,14),
\displaystyle (6,12), (7,10), (8,8), (9,6), (10,4), (11,2)\}
\displaystyle \text{Reflexive Since, for } 1 \in \text{ domain of } R, \; (1,1) \notin R \text{. So, } R
\displaystyle \text{is not reflexive.}
\displaystyle \text{Symmetric We observe that } (1,22) \in R \text{ but }
\displaystyle (22,1) \notin R \text{. So, } R \text{ is not symmetric.}
\displaystyle \text{Transitive We observe that } (7,10) \in R \text{ and }
\displaystyle (10,4) \in R \text{ but } (7,4) \notin R \text{.}
\displaystyle \text{So, } R \text{ is not transitive.}
\displaystyle \text{Thus, } R \text{ is neither reflexive nor symmetric nor transitive.}
\displaystyle \text{Hence, } R \text{ is not an equivalence relation.}
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\displaystyle \textbf{Question 14. } \text{If }A=R-\{3\}\text{ and }B=R-\{1\}\text{. Consider the function} \ \
\displaystyle f:A\rightarrow B\text{ defined by }f(x)=\frac{x-2}{x-3},\text{ for all }x\in A. \text{ Then, show that }f\text{ is bijective. Find }
\displaystyle f^{-1}(x)\text{.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, a function } f : A \to B, \text{ where } A = R - \{3\} \text{ and }
\displaystyle B = R - \{1\}, \text{ defined by } f(x) = \frac{x-2}{x-3}
\displaystyle \text{One-one function}
\displaystyle \text{Let } x_1, x_2 \in A \text{ such that } f(x_1) = f(x_2)
\displaystyle \text{Then, } \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_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6
\displaystyle \Rightarrow -3(x_1-x_2) + 2(x_1-x_2) = 0
\displaystyle \Rightarrow -(x_1-x_2) = 0
\displaystyle \Rightarrow x_1 - x_2 = 0
\displaystyle \Rightarrow x_1 = x_2
\displaystyle \text{Thus, } f(x_1) = f(x_2)
\displaystyle \Rightarrow x_1 = x_2, \; \forall x_1, x_2 \in A
\displaystyle \text{So, } f(x) \text{ is a one-one function.}
\displaystyle \text{Onto function Let } y \in B = R - \{1\} \text{ be any arbitrary element.}
\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}
\displaystyle \text{or } x = \frac{3y-2}{y-1} \qquad (i)
\displaystyle \text{Clearly, } x = \frac{3y-2}{y-1} \text{ is a real number for all } y \neq 1.
\displaystyle \text{Also, } \frac{3y-2}{y-1} \neq 3
\displaystyle \left[\because \; \frac{3y-2}{y-1} = 3 \Rightarrow 3y - 2 = 3y - 3 \Rightarrow 2 = 3 \text{ which is absurd.}\right]
\displaystyle \text{Thus, for each } y \in B, \text{ there exists } x = \frac{3y-2}{y-1} \in A
\displaystyle \text{such that } f(x) = f\left(\frac{3y-2}{y-1}\right)
\displaystyle = \frac{\left(\frac{3y-2}{y-1}\right)-2}{\left(\frac{3y-2}{y-1}\right)-3} = \frac{3y-2-2y+2}{3y-2-3y+3} = y
\displaystyle \text{Hence, } f(x) \text{ is an onto function.}
\displaystyle \text{Therefore, } f(x) \text{ is a bijective function.}
\displaystyle \text{From Eq. (i), we get}
\displaystyle f^{-1}(y) = \frac{3y-2}{y-1}
\displaystyle \text{or } f^{-1}(x) = \frac{3x-2}{x-1}
\displaystyle \text{which is the inverse function of } f(x).
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\displaystyle \textbf{Question 15. } \text{If }A=\{1,2,3,\ldots,9\}\text{ and }R\text{ is the relation in }A\times A \text{ defined by }
\displaystyle (a,b)\ R\ (c,d),\text{ if }a+d=b+c \text{ for }(a,b),(c,d)\text{ in }A\times A. \text{ Prove that }R
\displaystyle \text{ is an equivalence relation. Also, obtain}   \text{ the equivalence } \text{class }[(2,5)]. \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given a relation } R \text{ in } A \times A, \text{ where } A = \{1, 2, 3, \ldots, 9\},
\displaystyle \text{defined as } (a, b) \; R \; (c, d), \text{ if } a + d = b + c.
\displaystyle \text{Reflexive Let } (a, b) \text{ be any arbitrary element of } A \times A.
\displaystyle \text{i.e. } (a, b) \in A \times A, \text{ where } a, b \in A.
\displaystyle \text{Now, as } a + b = b + a \qquad [\because \text{ addition is commutative}]
\displaystyle \therefore (a, b) \; R \; (a, b)
\displaystyle \text{So, } R \text{ is reflexive.}
\displaystyle \text{Symmetric Let } (a, b), (c, d) \in A \times A \text{ such that }
\displaystyle (a, b) \; R \; (c, d).
\displaystyle \text{Then, } a + d = b + c
\displaystyle \Rightarrow b + c = a + d
\displaystyle \Rightarrow c + b = d + a
\displaystyle \Rightarrow (c, d) \; R \; (a, b)
\displaystyle \text{So, } R \text{ is symmetric.}
\displaystyle \text{Transitive Let } (a, b), (c, d), (e, f) \in A \times A \text{ such that }
\displaystyle (a, b) \; R \; (c, d) \text{ and } (c, d) \; R \; (e, f).
\displaystyle \text{Then, } a + d = b + c \text{ and } c + f = d + e
\displaystyle \text{On adding the above equations, we get }
\displaystyle a + d + c + f = b + c + d + e
\displaystyle \Rightarrow a + f = b + e
\displaystyle \Rightarrow (a, b) \; R \; (e, f)
\displaystyle \text{So, } R \text{ is transitive.}
\displaystyle \text{Thus, } R \text{ is reflexive, symmetric and transitive. Hence, }
\displaystyle R \text{ is an equivalence relation.}
\displaystyle \text{Now, for } (2, 5), \text{ we will find } (c, d) \in A \times A \text{ such that }
\displaystyle 2 + d = 5 + c \text{ or } d - c = 3
\displaystyle \text{Clearly, } (2, 5) \; R \; (1, 4) \text{ as } 4 - 1 = 3
\displaystyle (2, 5) \; R \; (2, 5) \text{ as } 5 - 2 = 3
\displaystyle (2, 5) \; R \; (3, 6) \text{ as } 6 - 3 = 3
\displaystyle (2, 5) \; R \; (4, 7) \text{ as } 7 - 4 = 3
\displaystyle (2, 5) \; R \; (5, 8) \text{ as } 8 - 5 = 3
\displaystyle \text{and } (2, 5) \; R \; (6, 9) \text{ as } 9 - 6 = 3
\displaystyle \text{Hence, equivalence class } [(2, 5)]
\displaystyle = \{(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)\}
\\
\displaystyle \textbf{Question 16. } \text{Show that the function }f\text{ in }A=R-\left\{\frac{2}{3}\right\}\text{ defined as} \ \
\displaystyle f(x)=\frac{4x+3}{6x-4}  \text{ is one-one and onto. Hence, find }f^{-1}\text{.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f(x) = \frac{4x+3}{6x-4}
\displaystyle \text{where } x \in A = R - \left\{\frac{2}{3}\right\}
\displaystyle \text{One-one function Let } x_1, x_2 \in A = R - \left\{\frac{2}{3}\right\} \text{ such that }
\displaystyle f(x_1) = f(x_2).
\displaystyle \text{Then, } \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_1x_2 - 16x_1 + 18x_2 - 12 = 24x_1x_2 - 16x_2 + 18x_1 - 12
\displaystyle \Rightarrow -34x_1 = -34x_2
\displaystyle \Rightarrow x_1 = x_2
\displaystyle \text{So, } f \text{ is one-one function.}
\displaystyle \text{Onto function Let } y \text{ be an arbitrary element of } A \text{ (codomain).}
\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)}{4-6y}
\displaystyle \Rightarrow x = \frac{4y+3}{6y-4}
\displaystyle \text{Clearly, } x = \frac{4y+3}{6y-4} \text{ is a real number for all } y \neq \frac{4}{6} = \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 \Rightarrow 9 = -8, \text{ which is absurd}\right]
\displaystyle \text{Thus, for each } y \in A \text{ (codomain), there exists }
\displaystyle x = \frac{4y+3}{6y-4} \in A \text{ (domain) such that }
\displaystyle f\left(\frac{4y+3}{6y-4}\right) = \frac{4\left(\frac{4y+3}{6y-4}\right)+3}{6\left(\frac{4y+3}{6y-4}\right)-4}
\displaystyle = \frac{\frac{16y+12}{6y-4} + 3}{\frac{24y+18}{6y-4} - 4}
\displaystyle = \frac{\frac{16y+12+18y-12}{6y-4}}{\frac{24y+18-24y+16}{6y-4}}
\displaystyle = \frac{34y}{34} = y
\displaystyle \text{Hence, } f \text{ is onto function.}
\displaystyle \text{Since, } f \text{ is bijective, so its inverse exists.}
\displaystyle \therefore x = f^{-1}(y) = \frac{3+4y}{6y-4}
\displaystyle \text{or } f^{-1}(x) = \frac{3+4x}{6x-4}, \; x \neq \frac{2}{3}
\\
\displaystyle \textbf{Question 17. } \text{Consider }f:R^{+}\rightarrow [4,\infty )\text{ given by }f(x)=x^{2}+4. \text{ Show that }
\displaystyle f\text{ is invertible with the inverse }f^{-1}\text{ of }f  \text{ given by }f^{-1}(y)=\sqrt{y-4},\text{ where }
\displaystyle R^{+}\text{ is the set of all } \text{non-negative real numbers. } \ \
\displaystyle \text{Answer:}
\displaystyle \text{Here, function } f : R^{+} \to [4, \infty) \text{ is given by } f(x) = x^2 + 4.
\displaystyle \text{One-one function Let } x, y \in R^{+} \text{ such that } f(x) = f(y)
\displaystyle \Rightarrow x^2 + 4 = y^2 + 4
\displaystyle \Rightarrow x^2 = y^2
\displaystyle \Rightarrow x = y \qquad [\because \text{ we take only positive sign as } x, y \in R^{+}]
\displaystyle \text{Therefore, } f \text{ is a one-one function.}
\displaystyle \text{Onto function Let } y \in [4, \infty), \text{ let } y = x^2 + 4
\displaystyle \Rightarrow x^2 = y - 4 \geq 0 \qquad [\because y \geq 4]
\displaystyle \Rightarrow x = \sqrt{y-4} \geq 0 \qquad [\text{we take only positive sign, as } x \in R^{+}]
\displaystyle \text{Therefore, for any } y \in [4, \infty), \text{ there exists } x = \sqrt{y-4} \in R^{+}
\displaystyle \text{such that } f(x) = f(\sqrt{y-4}) = (\sqrt{y-4})^2 + 4 = y - 4 + 4 = y
\displaystyle \text{Therefore, } f \text{ is onto function.}
\displaystyle \text{Since, } f \text{ is one-one and onto and therefore } f^{-1} \text{ exists.}
\displaystyle \therefore x = f^{-1}(y) = \sqrt{y-4}
\displaystyle \text{or } f^{-1}(x) = \sqrt{x-4}, \; x \in [4, \infty)
\displaystyle \text{Alternate Method}
\displaystyle \text{Let us define } g : [4, \infty) \to R^{+} \text{ by } g(y) = \sqrt{y-4}
\displaystyle \text{Now, } g \circ f(x) = g(f(x)) = g(x^2 + 4)
\displaystyle = \sqrt{(x^2 + 4) - 4} = \sqrt{x^2} = x
\displaystyle \text{and } f \circ g(y) = f(g(y)) = f(\sqrt{y-4})
\displaystyle = (\sqrt{y-4})^2 + 4 = (y-4) + 4 = y
\displaystyle \text{Thus, } g \circ f = I_{R^{+}} \text{ and } f \circ g = I_{[4,\infty)}
\displaystyle \Rightarrow f \text{ is invertible and its inverse function is } g
\displaystyle \therefore f^{-1}(y) = g(y) = \sqrt{y-4} \text{ or } f^{-1}(x) = \sqrt{x-4}
\\
\displaystyle \textbf{Question 18. } \text{Show that }f:N\rightarrow N\text{, given by }  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).} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given function is } 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 function}
\displaystyle \text{From the given function, we observe that}
\displaystyle \text{Case I: When } x \text{ is odd}
\displaystyle \text{Let } f(x_1) = f(x_2) \Rightarrow x_1 + 1 = x_2 + 1 \Rightarrow x_1 = x_2
\displaystyle \therefore f(x_1) = f(x_2) \Rightarrow x_1 = x_2, \; \forall x_1, x_2 \in N
\displaystyle \text{So, } f(x) \text{ is one-one.}
\displaystyle \text{Case II: When } x \text{ is even}
\displaystyle \text{Let } f(x_1) = f(x_2) \Rightarrow x_1 - 1 = x_2 - 1 \Rightarrow x_1 = x_2
\displaystyle \therefore f(x_1) = f(x_2) \Rightarrow x_1 = x_2, \; \forall x_1, x_2 \in N
\displaystyle \text{So, } f(x) \text{ is one-one.}
\displaystyle \text{Hence, from Case I and Case II, } f(x_1) = f(x_2) \Rightarrow x_1 = x_2, \; \forall x_1, x_2 \in N
\displaystyle \text{Therefore, } f(x) \text{ is one-one.}
\displaystyle \text{Onto function Let } y \in N \text{ (codomain) be any arbitrary number.}
\displaystyle \text{If } y \text{ is odd, then there exists an even number } y+1 \in N
\displaystyle \text{such that } f(y+1) = (y+1) - 1 = y
\displaystyle \text{If } y \text{ is even, then there exists an odd number } y-1 \in N
\displaystyle \text{such that } f(y-1) = (y-1) + 1 = y
\displaystyle \text{Thus, every element in } N \text{ (codomain) has a pre-image in } N \text{ (domain)}
\displaystyle \text{Therefore, } f(x) \text{ is an onto function.}
\displaystyle \text{Hence, the function } f(x) \text{ is bijective.}
\\
\displaystyle \textbf{Question 19. } \text{If }f:N\rightarrow N\text{ is defined by} 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}\text{ for all } \\ n\in N. \text{Find whether the function }f\text{ is bijective.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{The given function is } 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{One-one function Here, } f(1) = \frac{1+1}{2} = \frac{2}{2} = 1
\displaystyle \text{and } f(2) = \frac{2}{2} = 1
\displaystyle \therefore f(n) \text{ is not a one-one function because two distinct values in } N
\displaystyle \text{have the same image.}
\displaystyle \textbf{Onto function If } n \text{ is an odd natural number, then } 2n-1 \text{ is also a natural number.}
\displaystyle \text{Now, } f(2n-1) = \frac{(2n-1)+1}{2} = \frac{2n}{2} = n
\displaystyle \text{Again, if } n \text{ is an even natural number, then } 2n \text{ is also a natural number.}
\displaystyle \text{Then, } f(2n) = \frac{2n}{2} = n
\displaystyle \text{From above, for each } n \in N \text{ there exists its pre-image in } N
\displaystyle \therefore \text{Range = Codomain}
\displaystyle \text{Therefore, } f \text{ is onto.}
\displaystyle \text{Hence, } f(x) \text{ is not one-one but it is onto. So, it is not bijective.}
\\
\displaystyle \textbf{Question 20. } \text{Consider }f:R^{+}\rightarrow [-5,\infty)\text{ given by } f(x)=9x^{2}+6x-5.
\displaystyle \text{ Show that }f\text{ is invertible with }  f^{-1}(y)=\left(\frac{\sqrt{y+6}-1}{3}\right).  \text{ Hence, find} \ \
\displaystyle \text{(i) }f^{-1}(10) \ \           \displaystyle \text{(ii) }y,\text{ if }f^{-1}(y)=\frac{4}{3}, \ \
\displaystyle \text{where }R^{+}\text{ is the set of all } \text{non-negative real numbers.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Here, function } f : R^{+} \to [-5, \infty) \text{ is given as}
\displaystyle f(x) = 9x^2 + 6x - 5
\displaystyle \text{Let } y \text{ be any arbitrary element of } [-5, \infty)
\displaystyle \text{Let } y = 9x^2 + 6x - 5
\displaystyle \Rightarrow y = (3x+1)^2 - 6
\displaystyle \Rightarrow (3x+1)^2 = y + 6
\displaystyle \Rightarrow 3x + 1 = \sqrt{y+6} \qquad [\because y \geq -5 \Rightarrow y+6 \geq 0]
\displaystyle \Rightarrow x = \frac{\sqrt{y+6} - 1}{3}
\displaystyle \text{Therefore, } f \text{ is onto, thereby range } f = [-5, \infty)
\displaystyle \text{Let us define } g : [-5, \infty) \to R^{+}
\displaystyle g(y) = \frac{\sqrt{y+6} - 1}{3}
\displaystyle \text{Now, } g \circ f(x) = g(f(x)) = g(9x^2 + 6x - 5)
\displaystyle = \frac{\sqrt{(3x+1)^2 - 6 + 6} - 1}{3} = \frac{3x+1 - 1}{3} = x
\displaystyle \text{and } f \circ g(y) = f(g(y)) = f\left(\frac{\sqrt{y+6} - 1}{3}\right)
\displaystyle = 9\left(\frac{\sqrt{y+6} - 1}{3}\right)^2 + 6\left(\frac{\sqrt{y+6} - 1}{3}\right) - 5
\displaystyle = (\sqrt{y+6} - 1)^2 + 2(\sqrt{y+6} - 1) - 5
\displaystyle = (y+6) - 2\sqrt{y+6} + 1 + 2\sqrt{y+6} - 2 - 5
\displaystyle = y
\displaystyle \text{Therefore, } g \circ f = I_{R^{+}} \text{ and } f \circ g = I_{[-5,\infty)}
\displaystyle \text{Hence, } f \text{ is invertible and the inverse of } f \text{ is given by}
\displaystyle f^{-1}(y) = g(y) = \frac{\sqrt{y+6} - 1}{3}
\displaystyle \text{(i) } f^{-1}(10) = \frac{\sqrt{10+6} - 1}{3} = \frac{4 - 1}{3} = 1
\displaystyle \text{(ii) If } f^{-1}(y) = \frac{4}{3}
\displaystyle \Rightarrow y = 9\left(\frac{4}{3}\right)^2 + 6\left(\frac{4}{3}\right) - 5
\displaystyle = 16 + 8 - 5 = 19
\\
\displaystyle \textbf{Question 21. } \text{Consider }f:R-\left\{\frac{4}{3}\right\}\rightarrow R-\left\{\frac{4}{3}\right\}\text{ given by } f(x)=\frac{4x+3}{3x+4}.  \text{ Show that }
\displaystyle f\text{ is bijective. Find the inverse }  \text{of }f\text{ and hence find } f^{-1}(0) \text{ and } x  \text{ such that }  f^{-1}(x)=2. \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f : R - \left\{-\frac{4}{3}\right\} \to R - \left\{\frac{4}{3}\right\}
\displaystyle \text{defined by } f(x) = \frac{4x+3}{3x+4}
\displaystyle \text{Let } x_1, x_2 \in R - \left\{-\frac{4}{3}\right\} \text{ such that } f(x_1) = f(x_2)
\displaystyle \Rightarrow \frac{4x_1+3}{3x_1+4} = \frac{4x_2+3}{3x_2+4}
\displaystyle \Rightarrow (4x_1+3)(3x_2+4) = (3x_1+4)(4x_2+3)
\displaystyle \Rightarrow 12x_1x_2 + 16x_1 + 9x_2 + 12 = 12x_1x_2 + 9x_1 + 16x_2 + 12
\displaystyle \Rightarrow 7x_1 = 7x_2
\displaystyle \Rightarrow x_1 = x_2
\displaystyle \therefore f \text{ is one-one.}
\displaystyle \text{Let } y \in R - \left\{\frac{4}{3}\right\}, \text{ then } y \neq \frac{4}{3}
\displaystyle \text{The function } f \text{ is onto if there exists } x \in R - \left\{-\frac{4}{3}\right\}
\displaystyle \text{such that } f(x) = y
\displaystyle \text{Now, } f(x) = y \Rightarrow \frac{4x+3}{3x+4} = y
\displaystyle \Rightarrow 4x + 3 = y(3x+4)
\displaystyle \Rightarrow 4x + 3 = 3xy + 4y \Rightarrow 4x - 3xy = 4y - 3
\displaystyle \Rightarrow x(4 - 3y) = 4y - 3
\displaystyle \Rightarrow x = \frac{4y-3}{4-3y} \in R - \left\{-\frac{4}{3}\right\}, \; y \neq \frac{4}{3}
\displaystyle \text{Thus, for any } y \in R - \left\{\frac{4}{3}\right\}, \text{ there exists } x = \frac{4y-3}{4-3y}
\displaystyle \text{Hence, } f \text{ is onto.}
\displaystyle \therefore f^{-1}(y) = \frac{4y-3}{4-3y}
\displaystyle \text{or } f^{-1}(x) = \frac{4x-3}{4-3x}
\displaystyle \text{Now, } f^{-1}(0) = \frac{0-3}{4-0} = -\frac{3}{4}
\displaystyle \text{and also if } f^{-1}(x) = 2
\displaystyle \Rightarrow \frac{4x-3}{4-3x} = 2
\displaystyle \Rightarrow 4x - 3 = 2(4 - 3x)
\displaystyle \Rightarrow 4x - 3 = 8 - 6x
\displaystyle \Rightarrow 10x = 11 \Rightarrow x = \frac{11}{10}
\\
\displaystyle \textbf{Question 22. } \text{Let }f:N\rightarrow N\text{ be a function defined as }f(x)=9x^{2}+6x-5.
\displaystyle \text{ Show that }f:N\rightarrow S,\text{ where }S\text{ is } \text{the range of }f\text{ is invertible. } \text{Find the inverse of }
\displaystyle f\text{ and } \text{hence find }f^{-1}(43)\text{ and }f^{-1}(163). \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have a mapping } f : N \to N \text{ given by } f(x) = 9x^2 + 6x - 5
\displaystyle \textbf{One-one function Let } x_1, x_2 \in N \text{ such that } f(x_1) = f(x_2)
\displaystyle \Rightarrow 9x_1^2 + 6x_1 - 5 = 9x_2^2 + 6x_2 - 5
\displaystyle \Rightarrow 9x_1^2 + 6x_1 = 9x_2^2 + 6x_2
\displaystyle \Rightarrow 9(x_1^2 - x_2^2) + 6(x_1 - x_2) = 0
\displaystyle \Rightarrow 3(x_1 - x_2)(x_1 + x_2) + 2(x_1 - x_2) = 0
\displaystyle \Rightarrow (x_1 - x_2)(3x_1 + 3x_2 + 2) = 0
\displaystyle \Rightarrow x_1 = x_2 \quad [\because 3x_1 + 3x_2 + 2 \neq 0, \; x_1, x_2 \in N]
\displaystyle \text{So, } f \text{ is one-one function.}
\displaystyle \textbf{Onto function Obviously, } f : N \to S \text{ is an onto function, where } S
\displaystyle \text{is the range of } f
\displaystyle \text{Thus, } f : N \to S \text{ is one-one and onto function.}
\displaystyle \therefore f \text{ is invertible function, so its inverse exists.}
\displaystyle \text{Let } f(x) = y, \text{ then } y = 9x^2 + 6x - 5
\displaystyle \Rightarrow y = (3x)^2 + 2 \cdot 3x \cdot 1 + 1 - 6
\displaystyle \Rightarrow y = (3x+1)^2 - 6
\displaystyle \Rightarrow (3x+1)^2 = y + 6
\displaystyle \Rightarrow 3x + 1 = \sqrt{y+6} \quad [\text{taking positive root as } x \in N]
\displaystyle \Rightarrow x = \frac{\sqrt{y+6} - 1}{3}
\displaystyle \therefore f^{-1}(y) = \frac{\sqrt{y+6} - 1}{3}
\displaystyle \text{or } f^{-1}(x) = \frac{\sqrt{x+6} - 1}{3}, \; x \in S
\displaystyle \text{Now, } f^{-1}(43) = \frac{\sqrt{43+6} - 1}{3} = \frac{\sqrt{49} - 1}{3} = \frac{7-1}{3} = 2
\displaystyle \text{and } f^{-1}(163) = \frac{\sqrt{163+6} - 1}{3} = \frac{\sqrt{169} - 1}{3} = \frac{13-1}{3} = 4
\\
\displaystyle \textbf{Question 23. } \text{If }N\text{ denotes the set of all natural numbers and }R\text{ be the relation on} \ \
\displaystyle \text{ }N\times N\text{ defined by }(a,b)\ R\ (c,d),\text{ if } ad(b+c)=bc(a+d).\text{ Show that }R\text{ is an equivalence} \ \
\displaystyle \text{relation.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{ We have, a relation } R \text{ on } N \times N \text{ defined by}
\displaystyle (a, b) \; R \; (c, d), \text{ if } ad(b+c) = bc(a+d)
\displaystyle \textbf{Reflexive Let } (a, b) \in N \times N \text{ be any arbitrary element.}
\displaystyle \text{We have to show } (a, b) \; R \; (a, b), \text{ i.e. to show}
\displaystyle ab(b+a) = ba(a+b) \text{ which is trivially true, as natural numbers are}
\displaystyle \text{commutative under usual multiplication and addition.}
\displaystyle \text{Since, } (a, b) \in N \times N \text{ was arbitrary, therefore } R \text{ is reflexive.}
\displaystyle \textbf{Symmetric Let } (a, b), (c, d) \in N \times N \text{ such that}
\displaystyle (a, b) \; R \; (c, d), \text{ i.e. } ad(b+c) = bc(a+d) \qquad \ldots (i)
\displaystyle \text{To show, } (c, d) \; R \; (a, b)
\displaystyle \text{i.e. to show } cb(d+a) = da(c+b)
\displaystyle \text{From Eq. (i), we have}
\displaystyle ad(b+c) = bc(a+d) \Rightarrow da(c+b) = cb(d+a)
\displaystyle \qquad [\because \text{ natural numbers are commutative under usual addition and multiplication}]
\displaystyle \Rightarrow cb(d+a) = da(c+b) \Rightarrow (c, d) \; R \; (a, b)
\displaystyle \text{Thus, } R \text{ is symmetric.}
\displaystyle \textbf{Transitive Let } (a, b), (c, d) \text{ and } (e, f) \in N \times N
\displaystyle \text{such that } (a, b) \; R \; (c, d) \text{ and } (c, d) \; R \; (e, f).
\displaystyle \text{Now, } (a, b) \; R \; (c, d) \Rightarrow ad(b+c) = bc(a+d)
\displaystyle \Rightarrow \frac{b+c}{bc} = \frac{a+d}{ad}
\displaystyle \Rightarrow \frac{1}{b} + \frac{1}{c} = \frac{1}{a} + \frac{1}{d} \qquad \ldots (ii)
\displaystyle \text{and } (c, d) \; R \; (e, f) \Rightarrow cf(d+e) = de(c+f)
\displaystyle \Rightarrow \frac{d+e}{de} = \frac{c+f}{cf}
\displaystyle \Rightarrow \frac{1}{d} + \frac{1}{e} = \frac{1}{c} + \frac{1}{f} \qquad \ldots (iii)
\displaystyle \text{On adding Eqs. (ii) and (iii), we get}
\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 \Rightarrow R \text{ is transitive.}
\displaystyle \text{Thus, } R \text{ is reflexive, symmetric and transitive, hence } R
\displaystyle \text{is an equivalence relation.}

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