Sets, Relations and FunctionsMCQPYQ June 19Question 1942 of 217
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A={1,2,3,4,…,10}\displaystyle A = \{1, 2, 3, 4, \dots, 10\} a relation on A\displaystyle A, R={(x,y)/x+y=10,x∈A,y∈A}\displaystyle R = \{(x, y) / x + y = 10, x \in A, y \in A\}, then domain of R−1\displaystyle R^{-1} is

Options

A{1,2,3,4,5}\displaystyle \{1, 2, 3, 4, 5\}
B{0,3,5,7,9}\displaystyle \{0, 3, 5, 7, 9\}
C{1,2,4,5,6,7}\displaystyle \{1, 2, 4, 5, 6, 7\}
DNone of these
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Correct Answer

✅ Option d — None of these

All Options:

  • A{1,2,3,4,5}\displaystyle \{1, 2, 3, 4, 5\}
  • B{0,3,5,7,9}\displaystyle \{0, 3, 5, 7, 9\}
  • C{1,2,4,5,6,7}\displaystyle \{1, 2, 4, 5, 6, 7\}
  • DNone of these

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Detailed Solution & Explanation

We are given the set A={1,2,3,4,…,10}\displaystyle A = \{1, 2, 3, 4, \dots, 10\} and a relation R\displaystyle R defined as:
R={(x,y)∣x+y=10,x∈A,y∈A}R = \{(x, y) \mid x + y = 10, x \in A, y \in A\}
Let's list the ordered pairs of R\displaystyle R:
- For x=1\displaystyle x = 1, y=9∈A  ⟹  (1,9)∈R\displaystyle y = 9 \in A \implies (1, 9) \in R
- For x=2\displaystyle x = 2, y=8∈A  ⟹  (2,8)∈R\displaystyle y = 8 \in A \implies (2, 8) \in R
- For x=3\displaystyle x = 3, y=7∈A  ⟹  (3,7)∈R\displaystyle y = 7 \in A \implies (3, 7) \in R
- For x=4\displaystyle x = 4, y=6∈A  ⟹  (4,6)∈R\displaystyle y = 6 \in A \implies (4, 6) \in R
- For x=5\displaystyle x = 5, y=5∈A  ⟹  (5,5)∈R\displaystyle y = 5 \in A \implies (5, 5) \in R
- For x=6\displaystyle x = 6, y=4∈A  ⟹  (6,4)∈R\displaystyle y = 4 \in A \implies (6, 4) \in R
- For x=7\displaystyle x = 7, y=3∈A  ⟹  (7,3)∈R\displaystyle y = 3 \in A \implies (7, 3) \in R
- For x=8\displaystyle x = 8, y=2∈A  ⟹  (8,2)∈R\displaystyle y = 2 \in A \implies (8, 2) \in R
- For x=9\displaystyle x = 9, y=1∈A  ⟹  (9,1)∈R\displaystyle y = 1 \in A \implies (9, 1) \in R
- For x=10\displaystyle x = 10, y=0∉A\displaystyle y = 0 \notin A (since 0\displaystyle 0 is not in A\displaystyle A). So (10,0)∉R\displaystyle (10, 0) \notin R.
Thus:
R={(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1)}R = \{(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)\}
The inverse relation R−1\displaystyle R^{-1} is obtained by swapping the elements of each ordered pair in R\displaystyle R:
R−1={(9,1),(8,2),(7,3),(6,4),(5,5),(4,6),(3,7),(2,8),(1,9)}R^{-1} = \{(9, 1), (8, 2), (7, 3), (6, 4), (5, 5), (4, 6), (3, 7), (2, 8), (1, 9)\}
The domain of R−1\displaystyle R^{-1} is the set of all first components of the ordered pairs in R−1\displaystyle R^{-1}:
Domain(R−1)={1,2,3,4,5,6,7,8,9}\text{Domain}(R^{-1}) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}
*Note: The exact mathematical domain of R−1\displaystyle R^{-1} is {1,2,3,4,5,6,7,8,9}\displaystyle \{1, 2, 3, 4, 5, 6, 7, 8, 9\}, which corresponds to Option D (None of these). The textbook key incorrectly lists Option A ({1,2,3,4,5}\displaystyle \{1, 2, 3, 4, 5\}) as the correct choice.*
Hence, **Option D** is the correct answer.

About This Chapter: Sets, Relations and Functions

Paper

Paper 3: Quantitative Aptitude

Weightage

3-5 Marks

Key Topics

Sets, Relations, Functions

This chapter covers Sets, Relations, Functions and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 3-5 Marks weightage. Focus on understanding core concepts rather than memorizing.

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