Class 12 Mathematics Chapter 1 Extra Questions (HOTS) – Relations and Functions

Class 12 Mathematics Chapter 1 Extra Questions (HOTS Level) – Relations and Functions

1. Q1 (General/abstract proof). Let R be a relation on set A defined by aRb iff a and b leave the same remainder when divided by a fixed positive integer m. Prove that R is an equivalence relation for any m, and describe the equivalence classes.
   Answer: Reflexive, symmetric, transitive all follow from properties of equality of remainders, so R is an equivalence relation for every m ≥ 1. The equivalence classes are the m residue classes mod m — this generalises Solutions Q9(i) (the m=4 case) to any modulus.
2. Q2 (Reverse-engineering). A relation R on {1,2,3,4,5} is known to be reflexive and symmetric, contains (1,2) and (2,3), and is NOT transitive. List one such relation R with the minimum possible number of ordered pairs.
   Answer: {(1,1),(2,2),(3,3),(4,4),(5,5),(1,2),(2,1),(2,3),(3,2)} — 9 pairs, already fails transitivity since (1,3) is absent.
3. Q3 (Assertion-Reason). Assertion (A): f:R→R, f(x)=x² is not invertible. Reason (R): A function is invertible iff it is bijective. Options: (a) Both true, R explains A (b) Both true, R does not explain A (c) A true, R false (d) A false, R true.
   Answer: (a). f(x)=x² is neither one-one nor onto, so it is not invertible exactly by the criterion in R.
4. Q4 (Fresh-number application). If f:R→R is defined by f(x)=5x−3, find f⁻¹(x) and verify f(f⁻¹(7))=7.
   Answer: f⁻¹(x)=(x+3)/5. f⁻¹(7)=2; f(2)=7. Verified.
5. Q5 (Multi-part composition, fresh numbers). Let f(x)=2x+1, g(x)=x²−4. Find (i) fog(3) (ii) gof(3) (iii) is fog=gof?
   Answer: (i) g(3)=5, f(5)=11. (ii) f(3)=7, g(7)=45. (iii) fog≠gof since 11≠45.
6. Q6 (General proof, harder than the Solutions-post version). If f:A→B and g:B→C are both bijections, prove gof:A→C is a bijection and (gof)⁻¹=f⁻¹og⁻¹.
   Answer: One-one and onto both follow directly by chaining the bijectivity of f and g; the inverse composition (f⁻¹og⁻¹)o(gof)=1_A confirms the inverse formula.
7. Q7 (Word problem, fresh numbers, binary operations). A binary operation * is defined on the integers by a*b=a+b−ab. Check commutativity, associativity, and find the identity.
   Answer: Commutative (symmetric formula), associative ((a*b)*c=a+b+c−ab−ac−bc+abc=a*(b*c)), identity e=0 (a*0=a).
8. Q8 (Reverse-engineering, equivalence classes). A relation R on Z is defined as aRb iff a²=b². Show R is an equivalence relation and find the equivalence class of 5.
   Answer: Reflexive/symmetric/transitive all hold since equality of squares is an equality relation. Equivalence class of 5 is {−5,5}.

Related

• Solutions
• Extra Questions (HOTS)
• Revision Notes
• Class 12 Mathematics Formulas Handbook
• Class 12 Mathematics Book

Frequently Asked Questions

What is the difference between reflexive, symmetric, and transitive properties when checking if a relation is an equivalence relation?
A relation is reflexive if every element is related to itself, symmetric if a relation between two elements works in both directions, and transitive if a relation carrying through one element to another also carries to a third; satisfying all three makes it an equivalence relation.

Last Updated: September 23, 2026

How can you determine whether a given function has an inverse?
A function has an inverse if and only if it is bijective, meaning it is one-one, so no two inputs give the same output, and onto, so every element of the range is covered, allowing the mapping to be reversed uniquely.

Written by Satish

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