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Match $\text{List I}$ with $\text{List II}$

Let $R_1=\{(1,1), (2,2), (3,3)\}$ and $R_2=\{(1,1), (1,2), (1,3), (1,4)\}$

$\begin{array}{llll} & \text{List I} & & \text{List II} \\ (A) & R_1 \cup R_2 & (I) & \{(1,1), (1,2), (1,3), (1,4), (2,2), (3,3) \} \\ (B) & R_1 – R_2 & (II) & \{(1,1)\} \\ (C) & R_1 \cap R_2 & (III) & \{(1,2), (1,3), (1,4) \} \\ (D) & R_2 – R_1 & (IV) & \{(2,2), (3,3)\} \end{array}$

Choose the correct answer from the options given below:

  1. $A-I, B-II, C-IV, D-III$
  2. $A-I, B-IV, C-III, D-II$
  3. $A-I, B-III, C-II, D-IV$
  4. $A-I, B-IV, C-II, D-III$
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2 Answers

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Given, 

$R_{1}$={(1,2),(2,2),(3,3)} and $R_{2}$ ={(1,1),(1,2),(1,3),(1,4)}

  1. $R_{1}$ U $R_{2}$ :- Presents in $R_{1}$ or $R_{2}$ or both = {(1,1),(1,2),(1,3),(1,4),(2,2),(3,3)}
  2. $R_{1}$ – $R_{2}$ :- Presents in $R_{1}$ but not in $R_{2}$ = {(2,2),(3,3)}
  3. $R_{1}$ ∩ $R_{2}$ :- Presents in both $R_{1}$ and $R_{2}$ = {(1,1)}
  4. $R_{2}$ – $R_{1}$ :- Presents in $R_{2}$ but not in $R_{1}$ = {(1,2),(1,3),(1,4)} 

Correct matching is – A−I,B−IV,C−II,D−III, Option (D) is answer.

Answer:

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