Sets, Relations and FunctionsMCQMTP Sep 24 Series IIQuestion 2007 of 222
All Questions

X={x,y,z,w};Y={1,2,3,4};H={(x,1);(y,2);(y,3);(z,4);(x,4)}\displaystyle X=\{x,y,z,w\}; Y=\{1,2,3,4\}; H=\{(x,1);(y,2);(y,3);(z,4);(x,4)\}

Select Your Answer (Keyboard: A, B, C, D)
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — H is not a function from x to y

All Options:

  • AH is a function from x to y
  • BH is not a function from x to y
  • CH is a relation from y to x
  • DNone of these

Detailed Solution & Explanation

Let us check if the relation H={(x,1),(y,2),(y,3),(z,4),(x,4)}\displaystyle H = \{(x,1), (y,2), (y,3), (z,4), (x,4)\} is a function from set X={x,y,z,w}\displaystyle X = \{x,y,z,w\} to set Y={1,2,3,4}\displaystyle Y = \{1,2,3,4\}:

For a relation to be a function from X\displaystyle X to Y\displaystyle Y, two main conditions must be satisfied:
1. **Uniqueness:** Each element in X\displaystyle X must be mapped to **exactly one** element in Y\displaystyle Y. That is, if (a,b)∈H\displaystyle (a, b) \in H and (a,c)∈H\displaystyle (a, c) \in H, then we must have b=c\displaystyle b = c.
2. **Existence:** Every element of the domain X\displaystyle X must have an image in the codomain Y\displaystyle Y.

Let us evaluate these conditions for H\displaystyle H:
- The element x∈X\displaystyle x \in X is mapped to both 1\displaystyle 1 and 4\displaystyle 4 because (x,1)∈H\displaystyle (x,1) \in H and (x,4)∈H\displaystyle (x,4) \in H. Since 1≠4\displaystyle 1 ≠\displaystyle \neq 4, the uniqueness condition is violated.
- The element y∈X\displaystyle y \in X is mapped to both 2\displaystyle 2 and 3\displaystyle 3 because (y,2)∈H\displaystyle (y,2) \in H and (y,3)∈H\displaystyle (y,3) \in H. Since 2≠3\displaystyle 2 ≠\displaystyle \neq 3, the uniqueness condition is again violated.
- The element w∈X\displaystyle w \in X does not appear as the first coordinate in any ordered pair of H\displaystyle H, which means it has no image in Y\displaystyle Y. This violates the existence condition.

Since these conditions are violated, H\displaystyle H is **not a function** from X\displaystyle X to Y\displaystyle Y.

Hence, **Option B** 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.

Related Comparison Tables

More Questions from Sets, Relations and Functions

Ready to Master Sets, Relations and Functions?

Practice all 222 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free