MCQ · Domain
Domain of f(x) = cos⁻¹ x + sin x is :
📖 Show Answer
Answer : (C) [–1, 1]
cos⁻¹ x is defined only when x ∈ [–1, 1].
sin x is defined for all real x.
∴ Domain = [–1, 1] ∩ R = [–1, 1].
MCQ · Function Type
If f: N → W is defined as
0, if n is odd }
then f is :
(C) a bijection (D) neither surjective nor injective
📖 Show Answer
Answer : (D) neither surjective nor injective
Not injective : f(1) = f(3) = 0 but 1 ≠ 3.
Not surjective : 1 ∈ W (codomain) but no n ∈ N gives f(n) = 1.
∴ neither injective nor surjective.
MCQ · Relation Properties
If R be a relation defined as aRb iff |a – b| > 0, a, b ∈ R then R is :
(C) transitive(D) symmetric and transitive
📖 Show Answer
Answer : (B) symmetric
Not reflexive : |a – a| = 0 not > 0.
Symmetric : |a – b| > 0 ⇒ |b – a| > 0.
Not transitive : e.g. (1,2) and (2,3) but (1,3) ⇒ |1–3| = 2 > 0 but transitive requires all pairs, counterexample fails.
∴ only symmetric.
MCQ · One–one & Onto
For real x, let f(x) = x³ + 5x + 1. Then :
(B) f is onto on R but not one-one
(C) f is one-one and onto on R
(D) f is neither one-one nor onto on R
📖 Show Answer
Answer : (C) f is one-one and onto on R
f'(x) = 3x² + 5 > 0 for all x ∈ R, so f is strictly increasing ⇒ one-one.
Since it’s a cubic polynomial with real coefficients, range = R ⇒ onto.
∴ bijective.
Assertion–Reason
Assertion (A) : Let A = {x ∈ R : –1 ≤ x ≤ 1}. If f : A → A be defined as f(x) = x², then f is not an onto function.
Reason (R) : If y = –1 ∈ A, then x = ±√(–1) ∉ A.
📖 Show Answer
Answer : (A) Both A and R are true and R is the correct explanation of A.
f(x) = x² on [–1,1] has range [0,1], not the whole codomain [–1,1].
–1 has no pre-image because x² = –1 has no real solution.
∴ Assertion is true and Reason correctly explains it.
Assertion–Reason · Domain
Assertion (A) : Let f(x) = ex and g(x) = log x. Then (f + g)x = ex + log x where domain of (f + g) is R.
Reason (R) : Dom(f + g) = Dom(f) ∩ Dom(g).
📖 Show Answer
Answer : (D) Assertion (A) is false, but Reason (R) is true.
Dom(f) = R, Dom(g) = (0, ∞).
Dom(f+g) = R ∩ (0, ∞) = (0, ∞), not R.
∴ Assertion false, Reason true.
Assertion–Reason · Bijective
Assertion (A) : Let Z be the set of integers. A function f : Z → Z defined as f(x) = 3x – 5, ∀ x ∈ Z is a bijective.
Reason (R) : A function is a bijective if it is both surjective and injective.
📖 Show Answer
Answer : (D) Assertion (A) is false, but Reason (R) is true.
f(x) = 3x – 5 is injective on Z but not surjective (e.g. 1 has no pre-image).
So Assertion false. Reason is the correct definition of bijective.
Bijection · Logarithm
If f: R+ → R is defined as f(x) = loga x (a > 0 and a ≠ 1), prove that f is a bijection.
(R+ is the set of all positive real numbers.)
📖 Show Answer
One–one :
Let x₁, x₂ ∈ R+ such that f(x₁) = f(x₂)
⇒ loga x₁ = loga x₂ ⇒ x₁ = x₂.
∴ f is one–one.
Onto :
Let f(x) = y ⇒ loga x = y ⇒ x = ay.
For every y ∈ R, there exists x = ay ∈ R+.
∴ f is onto.
Hence f is a bijection.
Relation · Domain & Range
Let A = {1, 2, 3} and B = {4, 5, 6}. A relation R from A to B is defined as
R = {(x, y) : x + y = 6, x ∈ A, y ∈ B}.
(i) Write all elements of R.
(ii) Is R a function ? Justify.
(iii) Determine domain and range of R.
📖 Show Answer
(i) Elements of R :
R = {(1,5), (2,4)}
(ii) Is R a function ?
R is not a function as 3 ∈ A does not have an image in co-domain B.
(iii) Domain & Range :
Domain of R = {1, 2}, Range of R = {4, 5}.
Relation · Reflexive, Symmetric, Transitive
Let R be a relation defined over N, where N is set of natural numbers, defined as
“mRn if and only if m is a multiple of n, m, n ∈ N.”
Find whether R is reflexive, symmetric and transitive or not.
📖 Show Answer
Reflexive :
Every natural number is a multiple of itself. So xRx. ∴ R is reflexive.
Symmetric :
8 is a multiple of 2 ⇒ 8R2, but 2 is not a multiple of 8.
∴ R is not symmetric.
Transitive :
Let xRy and yRz ⇒ x = my, y = nz ⇒ x = (mn)z ⇒ xRz.
∴ R is transitive.
Relation · Irrational Number
Let R be a relation on set of real numbers R defined as
{(x, y) : x – y + √3 is an irrational number, x, y ∈ R}
Verify R for reflexivity, symmetry and transitivity.
📖 Show Answer
Reflexive :
x – x + √3 = √3 is irrational ⇒ (x,x) ∈ R. ∴ R is reflexive.
Symmetric :
(√3, 2) ∈ R because √3 – 2 + √3 = 2(√3 – 1) irrational.
But (2, √3) ∉ R because 2 – √3 + √3 = 2 rational.
∴ R is not symmetric.
Transitive :
Counterexample: (–√3, √3) ∈ R and (√3, 2) ∈ R but (–√3, 2) ∉ R.
∴ R is not transitive.
Bijection · Piecewise Function
Show that the function f : N → N, where N is a set of natural numbers, given by
n + 1, if n is odd }
is a bijection.
📖 Show Answer
One–one :
Let f(x) = f(y). If both odd or both even, x = y.
If x odd, y even then x+1 = y–1 ⇒ x–y = –2, not possible.
∴ f is one–one.
Onto :
For every y ∈ N, if y is odd, x = y+1 (even); if y is even, x = y–1 (odd).
Thus every y has a pre-image. So range = N = co-domain.
∴ f is onto.
Hence f is bijective.
One–one & Onto · Cubic
Show that the function f : R → R defined by f(x) = 4x³ – 5, ∀ x ∈ R is one–one and onto.
📖 Show Answer
One–one :
Let f(x₁) = f(x₂) ⇒ 4x₁³ – 5 = 4x₂³ – 5 ⇒ x₁³ = x₂³ ⇒ x₁ = x₂.
∴ f is one–one.
Onto :
For any y ∈ R, choose x = ∛((y+5)/4) ∈ R.
Then f(x) = y. So range = R = co-domain.
∴ f is onto.
Hence f is both one–one and onto.
Relation · Domain, Range, Equivalence
A student wants to pair up natural numbers in such a way that they satisfy the equation 2x + y = 41, x, y ∈ N.
Find the domain and range of the relation. Check if the relation thus formed is reflexive, symmetric and transitive. Hence, state whether it is an equivalence relation or not.
📖 Show Answer
Relation R :
R = {(1,39), (2,37), …, (20,1)}
Domain = {1, 2, 3, …, 20}
Range = {1, 3, 5, …, 39} (odd numbers)
Reflexive :
(1,1) ∉ R ⇒ not reflexive.
Symmetric :
(1,39) ∈ R but (39,1) ∉ R ⇒ not symmetric.
Transitive :
(11,19) ∈ R and (19,3) ∈ R but (11,3) ∉ R ⇒ not transitive.
Hence R is not an equivalence relation.
Equivalence Relation · Square Product
Let R be a relation defined on a set N of natural numbers such that
R = {(x, y) : xy is a square of a natural number, x, y ∈ N}.
Determine if the relation R is an equivalence relation.
📖 Show Answer
Reflexive :
x·x = x² is a square ⇒ (x,x) ∈ R. ∴ R is reflexive.
Symmetric :
If xy is a square, then yx is also a square ⇒ (y,x) ∈ R.
∴ R is symmetric.
Transitive :
Let xy = a² and yz = b² ⇒ x = a²/y, z = b²/y.
xz = (ab/y)² ⇒ (x,z) ∈ R.
∴ R is transitive.
Hence R is an equivalence relation.
Relations · Reflexive, Symmetric, Transitive
A classroom teacher writes five relations on A = {1, 2, 3}:
R₂ = {(1,2), (1,3), (3,2)}
R₃ = {(1,2), (2,1), (1,1)}
R₄ = {(1,1), (1,2), (3,3), (2,2)}
R₅ = {(1,1), (1,2), (3,3), (2,2), (2,1), (2,3), (3,2)}
(i) Identify the relation which is reflexive, transitive but not symmetric.
(ii) Identify the relation which is reflexive and symmetric but not transitive.
(iii) (a) Identify the relations which are symmetric but neither reflexive nor transitive.
OR
(iii) (b) What pairs should be added to R₂ to make it an equivalence relation ?
📖 Show Answer
(i) R₄
R₄ is reflexive, transitive but not symmetric (1,2) ∈ R₄ but (2,1) ∉ R₄.
(ii) R₅
R₅ is reflexive and symmetric but not transitive (e.g. (1,2) & (2,3) but (1,3) ∉ R₅).
(iii) (a) R₁ and R₃
Both are symmetric but neither reflexive nor transitive.
OR (iii) (b)
Pairs to be added to R₂ : (1,1), (2,2), (3,3), (2,1), (3,1), (2,3)
Debate Competition · Relations & Functions
Speakers S = {S₁, S₂, S₃, S₄} and Judges J = {J₁, J₂, J₃}.
R = {(x, y) : speaker x is judged by judge y, x ∈ S, y ∈ J}.
(i) How many relations can be there from S to J ?
(ii) A student identifies a function f = {(S₁, J₁), (S₂, J₂), (S₃, J₂), (S₄, J₃)}. Check if it is bijective.
(iii) (a) How many one-one functions can be there from set S to set J ?
OR
(iii) (b) Another student considers R₁ = {(S₁, S₂), (S₂, S₄)} in set S. Write minimum ordered pairs to be included in R₁ so that R₁ is reflexive but not symmetric.
📖 Show Answer
(i) Number of relations = 24×3 = 212 = 4096
(ii) S₂ and S₃ both map to J₂ ⇒ not one–one. Hence not bijective.
(iii) (a) n(S) = 4, n(J) = 3. Since n(S) > n(J), no one–one function exists. Number = 0.
OR (iii) (b)
To make R₁ reflexive and not symmetric, add : (S₁,S₁), (S₂,S₂), (S₃,S₃), (S₄,S₄).
Roll Numbers · Bijection & Relation
Let A be the set of 30 students of class XII. Let f : A → N, f(x) = Roll Number of student x.
(i) Is f a bijective function ?
(ii) Give reasons to support your answer to (i).
(iii) (a) Let R = {(x,y) : x,y are Roll Numbers such that y = 3x}. List elements of R. Is R reflexive, symmetric and transitive ? Justify.
OR
(iii) (b) Let R = {(x,y) : x,y are Roll Numbers such that y = x³}. List elements of R. Is R a function ? Justify.
📖 Show Answer
(i) Yes, f is bijective (assuming roll numbers are 1 to 30, each student has a unique roll number).
(ii) One–one: different students have different roll numbers. Onto: every roll number from 1 to 30 is assigned to a student.
(iii) (a) R = {(1,3), (2,6), (3,9), …, (10,30)}. Not reflexive (1,1) ∉ R. Not symmetric (1,3) ∈ R but (3,1) ∉ R. Not transitive (1,3) and (3,9) but (1,9) ∉ R.
OR (iii) (b) R = {(1,1), (2,8), (3,27)}. Yes, R is a function because each x has exactly one y.