📘 CLASS XII · MATHEMATICS
Chapter 1 — Relations & Functions
Most Important Questions · CBSE 2026-27
Gulshan Singh | IIT-JEE & NEET Faculty
MCQ · Reflexive Relations
Let A = {a, b}, then the number of reflexive relations defined on A is :
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Answer : (C) 4
Number of reflexive relations on a set with n elements = 2n(n–1).
Here n = 2 ⇒ 22(1) = 22 = 4.
Assertion–Reason
Consider the function f : R → R, defined as f(x) = x³.
Assertion (A) : f(x) is a one-one function.
Reason (R) : f(x) is a one-one function, if co-domain = range.
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Answer : (B) Both A and R are true, but R is not the correct explanation of A.
f(x) = x³ is one-one on R (strictly increasing).
Reason is a true statement but it is not the reason why x³ is one-one.
x³ is one-one because if x₁³ = x₂³ ⇒ x₁ = x₂.
Equivalence Relation · Proof
Let A be the set of all positive integers and a relation R on A × A is defined by
(a, b) R (c, d) ⇔ ad = bc, for all (a, b), (c, d) ∈ A × A.
Show that R is an equivalence relation on A × A.
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Reflexive :
For (a,b) ∈ A × A, ab = ba ⇒ (a,b) R (a,b).
So, R is reflexive.
Symmetric :
Let (a,b) R (c,d) ⇒ ad = bc ⇒ cb = da ⇒ (c,d) R (a,b).
So, R is symmetric.
Transitive :
Let (a,b) R (c,d) and (c,d) R (e,f)
⇒ ad = bc and cf = de
⇒ (ad)(cf) = (bc)(de) ⇒ af = be
⇒ (a,b) R (e,f). So, R is transitive.
Hence R is an equivalence relation.
Parabola Swing · Functions
The swing follows the path of a parabola given by x² = y.
(i) Let f: N → R is defined by f(x) = x². What will be the range ?
(ii) Let f: N → N is defined by f(x) = x². Check if the function is injective or not.
(iii) (a) Let f: {1, 2, 3, 4 …} → {1, 4, 9, 16 …} be defined by f(x) = x². Prove that the function is bijective.
OR
(iii) (b) Let f: R → R is defined by f(x) = x². Show that f is neither injective nor surjective.
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(i) Range :
Rf = {1, 4, 9, 16, …} i.e. set of perfect squares of natural numbers.
(ii) Injective :
Let x₁, x₂ ∈ N and f(x₁) = f(x₂)
⇒ x₁² = x₂² ⇒ x₁ = ± x₂ ⇒ x₁ = x₂ (as x₁, x₂ ∈ N)
∴ f is injective.
(iii) (a) Bijective :
Co-domain = Range = {1, 4, 9, 16, …}.
Since f is one-one and onto, so f is bijective.
OR (iii) (b) Neither injective nor surjective :
f(–1) = f(1) = 1 but –1 ≠ 1 ⇒ not injective.
Co-domain = R, but Range = [0, ∞).
Since Co-domain ≠ Range, f is not surjective.