Detailed Solution

Topic: Relations and Functions Mapping

(I) & (II) BIJECTIVITY OF FUNCTION f

1. One-One (Injective) Check: Since every student has a unique roll number, no two different students map to the same roll number. Thus, \( f(x_1) = f(x_2) \) ⇒ \( x_1 = x_2 \). The function is One-One.

2. Onto (Surjective) Check: The Co-domain is the set of all natural numbers \( N \). The Range is the set of assigned roll numbers \( S = \{1, 2, \dots, 30\} \). Since Range \( \neq \) Co-domain, the function is NOT Onto.

Conclusion: Since \( f \) is not onto, it is NOT a bijective function.

(III)(A) ANALYSIS OF RELATION R = {(x, y) : y = 3x}

Set of Roll Numbers \( S = \{1, 2, \dots, 30\} \). We find pairs \( (x, y) \in S \times S \) such that \( y = 3x \).

  • For \( x=1, y=3 \). (1, 3) \(\in R\).
  • For \( x=2, y=6 \). (2, 6) \(\in R\).
  • …
  • For \( x=10, y=30 \). (10, 30) \(\in R\).
  • For \( x=11, y=33 \). \( 33 \notin S \), so no more pairs.

Elements of R: \( \{(1, 3), (2, 6), (3, 9), (4, 12), (5, 15), (6, 18), (7, 21), (8, 24), (9, 27), (10, 30)\} \).

Properties:

  • Reflexive: No, because \( (1, 1) \notin R \) since \( 1 \neq 3(1) \).
  • Symmetric: No, because \( (1, 3) \in R \) but \( (3, 1) \notin R \) since \( 1 \neq 3(3) \).
  • Transitive: No, because \( (1, 3) \in R \) and \( (3, 9) \in R \), but \( (1, 9) \notin R \) since \( 9 \neq 3(1) \).

(III)(B) ANALYSIS OF RELATION R = {(x, y) : y = x³}

We find pairs \( (x, y) \in S \times S \) such that \( y = x^3 \).

  • For \( x=1, y=1^3=1 \). \( (1, 1) \in R \).
  • For \( x=2, y=2^3=8 \). \( (2, 8) \in R \).
  • For \( x=3, y=3^3=27 \). \( (3, 27) \in R \).
  • For \( x=4, y=4^3=64 \). \( 64 \notin S \).

Elements of R: \( \{(1, 1), (2, 8), (3, 27)\} \).

Is R a function?

For a relation on set \( S \) to be a function, every element in the domain \( S \) must have a unique image. Here, elements like 4, 5, …, 30 in \( S \) have no image in \( R \). Therefore, \( R \) is NOT a function.

⚠️ Silly Mistake Audit: The Co-domain Trap

A common mistake is assuming a function is onto just because it is one-one. Always check the Co-domain defined in the problem. In Part (i), the Co-domain is all Natural Numbers (\( N \)), which is much larger than the set of actual roll numbers used (the Range). This difference is why it’s not onto, and thus not bijective.

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