ilovepdf_merged (6).pdf). Content covers sections 23.1–23.x.This lesson deepens relations and functions: types of relations (reflexive, symmetric, transitive, equivalence), types of functions (one–one, onto, bijective), composition, and invertible functions.
One–one (injective): f(x₁)=f(x₂) ⇒ x₁=x₂. Onto (surjective): every y in codomain is hit. Bijective: both.
Composition: (g∘f)(x) = g(f(x)). Composition is associative when defined; not commutative in general.
f has an inverse iff f is bijective. Then f⁻¹∘f = id and f∘f⁻¹ = id on appropriate sets.
Most exam-important points from this chapter:
Start every answer with the key definition or standard form from Relations and Functions-II.
Copy the formula, then substitute values; never jump to the number alone.
Check domain, quadrant, non-zero denominators, and applicability of the theorem.
Memorise and apply: Let A = {1, 2, 3} be a set. Then
Memorise and apply: R = {(1, 1), (2, 2), (3, 3), (1, 3), (2, 1)} is a reflexive relation on A.
English questions from NIOS Mathematics (311) public / sample papers (Sample 2024, Apr 2024, Oct 2024, Apr 2025; board-style where scanned papers had no extractable text). Mapped exclusively to this chapter — no cross-chapter duplicates. Use Model Answer for the marking key and Explanation for working.
5 question(s) · Sources: Apr 2024, Sample QP 2024
PYQ1. If f(x)=x² and g(x)=3, find (f∘g)(x).
Model Answer
1) 9 (constant function)
Explanation
1) (f∘g)(x)=f(g(x))=f(3)=3²=9 for all x in the domain of g.
PYQ2. If f(x)=x+3 for x∈ℝ, then f⁻¹(x) is:
Model Answer
1) x − 3
Explanation
1) y=x+3 ⇒ x=y−3 ⇒ f⁻¹(x)=x−3.
PYQ3. Is f : ℤ → ℤ given by f(x)=x+3 a bijection? Justify briefly.
Model Answer
1) Yes — bijective
Explanation
1) One–one: x₁+3=x₂+3
2) x₁=x₂. Onto: for any n∈ℤ, f(n−3)=n. Hence bijective.
PYQ4. If a binary operation * on ℤ is defined by a*b = 3a − b, find (2*3)*4.
Model Answer
1) 5
Explanation
1) 2*3 = 3·2 − 3 =
3. Then (2*3)*4 = 3*4 = 3·3 − 4 = 5.
PYQ5. Define an onto (surjective) function f : A → B.
Model Answer
1) f is onto if for every y∈B there exists x∈A with f(x)=y (range = codomain).
Explanation
1) Surjectivity means every element of the codomain is the image of at least one domain element.
10 English exam-style problems for this chapter (NIOS Mathematics 311). Each question states the formula, gives full step-by-step working with the final answer, plus formal/easy explanations. English only.
Is the relation R={(1,1),(2,2),(1,2)} on A={1,2} reflexive? Justify.
1) For reflexivity we need (2,2) and (1,1).
2) Both are present ⇒ R is reflexive on A.
Answer: Yes, reflexive
R contains every (a,a) for a in A, so R is reflexive.
Both diagonal pairs are in R, so reflexive.
Missing any (a,a) kills reflexivity.
Check every element of A, not only those appearing in pairs.
Is R={(1,2),(2,1),(2,2)} on {1,2} symmetric?
1) (1,2)∈R and (2,1)∈R; (2,2) is fine.
2) No counter-example ⇒ symmetric.
Answer: Yes, symmetric
Whenever (a,b) is in R, (b,a) is also in R.
Pairs come in both directions.
Symmetric does not require reflexive.
One missing reverse pair is enough to fail.
On ℤ, let aRb iff a−b is even. Is R an equivalence relation?
1) Reflexive: a−a=0 even.
2) Symmetric: if a−b even then b−a=−(a−b) even.
3) Transitive: sum of two evens is even ⇒ a−c even.
4) Hence equivalence.
Answer: Yes — equivalence relation
R is reflexive, symmetric and transitive, hence an equivalence relation (parity classes).
Same parity relation: even difference means same even/odd class.
Equivalence classes: even integers and odd integers.
Transitivity uses (a−b)+(b−c)=a−c.
Is f:ℝ→ℝ, f(x)=2x+3 one–one?
1) 2x₁+3=2x₂+3 ⇒ 2x₁=2x₂ ⇒ x₁=x₂.
2) Hence injective (one–one).
Answer: Yes, one–one (injective)
Equal images force equal preimages, so f is injective.
Straight line with non-zero slope is one–one.
Also onto ℝ, hence bijective.
Horizontal line test fails for non-injective functions.
If f(x)=x+1 and g(x)=x², find (g∘f)(3) and (f∘g)(3).
1) (g∘f)(3)=g(4)=16.
2) (f∘g)(3)=f(9)=10.
Answer: (g∘f)(3)=16; (f∘g)(3)=10
Composition is not commutative; the two values differ.
Do f first then g for g∘f; reverse for f∘g.
Always apply the right-hand function first in g∘f.
Bracket carefully.
Find the inverse of f(x)=2x+3, f:ℝ→ℝ.
1) y=2x+3 ⇒ x=(y−3)/2.
2) So f⁻¹(y)=(y−3)/2, or f⁻¹(x)=(x−3)/2.
Answer: f⁻¹(x) = (x − 3)/2
Solving y=f(x) for x yields the inverse function.
Undo: subtract 3, divide by 2.
Check f(f⁻¹(x))=x and f⁻¹(f(x))=x.
Domain/codomain both ℝ here.
Is f(x) = 2x − 3 one-one on ℝ? Justify.
1) Assume 2x₁ − 3 = 2x₂ − 3 ⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂.
2) Hence f is one-one (injective).
Answer: Yes — injective
Linear function with non-zero slope is injective on ℝ.
Different x give different 2x−3.
Horizontal line test: each y once.
Check each algebraic step carefully.
Is f: ℝ→ℝ, f(x)=x² onto? Explain.
1) For y = −1 there is no real x with x² = −1.
2) Hence f is not onto ℝ.
Answer: No — not onto ℝ
Range is [0,∞), not all of ℝ.
Negative numbers are missed.
Onto [0,∞) would be true.
Check each algebraic step carefully.
If f(x)=x+2 and g(x)=3x, find (f∘g)(x).
1) g first: 3x; then f: 3x+2.
2) (f∘g)(x)=3x+2.
Answer: 3x + 2
Composition applies the right-hand function first.
Triple then add two.
(g∘f)(x)=3(x+2)=3x+6 differs.
Check each algebraic step carefully.
Find the inverse of f(x)=4x+1 (as a function ℝ→ℝ).
1) y = 4x + 1 ⇒ x = (y − 1)/4.
2) f⁻¹(x) = (x − 1)/4.
Answer: f⁻¹(x) = (x − 1)/4
Swap and solve; rename the variable.
Undo: subtract 1, divide by 4.
f is bijective on ℝ.
Check each algebraic step carefully.