MCQ · Equivalence Classes
A relation R defined on set A = {x : x ∈ Z and 0 ≤ x ≤ 10} as R = {(x, y) : x = y} is given to be an equivalence relation. The number of equivalence classes is :
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Answer : (D) 11
A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
Since R is the equality relation, each element forms its own equivalence class.
∴ 11 equivalence classes.
MCQ · Equivalence Classes Properties
Which of the following statements is not true about equivalence classes Aᵢ (i = 1, 2, … n) formed by an equivalence relation R defined on a set A ?
(B) Aᵢ ∩ Aⱼ ≠ φ, i ≠ j
(C) x ∈ Aᵢ and x ∈ Aⱼ ⇒ Aᵢ = Aⱼ
(D) All elements of Aᵢ are related to each other
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Answer : (B) Aᵢ ∩ Aⱼ ≠ φ, i ≠ j
Equivalence classes are either disjoint or identical. So Aᵢ ∩ Aⱼ = φ for i ≠ j.
∴ (B) is false.
Assertion–Reason · Prime & Composite
Assertion (A) : The relation R = {(x, y) : (x + y) is a prime number and x, y ∈ N} is not a reflexive relation.
Reason (R) : The number ‘2n’ is composite for all natural numbers n.
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Answer : (C) Assertion (A) is true, but Reason (R) is false.
A is true because x+x = 2x is not always prime (e.g. x=1 gives 2 which is prime, but x=2 gives 4 which is not).
R is false because for n=1, 2n = 2 which is prime, not composite.
MCQ · Human Relation
A relation R defined on a set of human beings as
R = {(x, y) : x is 5 cm shorter than y} is :
(B) reflexive and transitive
(C) symmetric and transitive
(D) neither transitive, nor symmetric, nor reflexive
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Answer : (D) neither transitive, nor symmetric, nor reflexive
Not reflexive: x is not 5 cm shorter than itself.
Not symmetric: if x is 5 cm shorter than y, y is not 5 cm shorter than x.
Not transitive: x shorter than y and y shorter than z ⇒ x shorter than z by 10 cm, not 5 cm.
MCQ · Quadratic Function
A function f : R → R defined as f(x) = x² – 4x + 5 is :
(B) surjective but not injective
(C) both injective and surjective
(D) neither injective nor surjective
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Answer : (D) neither injective nor surjective
f(–1) = f(5) = 10 ⇒ not injective.
f(x) = (x–2)² + 1 ≥ 1 ⇒ range = [1, ∞) ≠ R ⇒ not surjective.
MCQ · Cubic on Integers
Let Z denote the set of integers, then function f : Z → Z defined as f(x) = x³ – 1 is :
(B) one-one but not onto
(C) onto but not one-one
(D) neither one-one nor onto
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Answer : (B) one-one but not onto
One–one: x₁³ = x₂³ ⇒ x₁ = x₂.
Not onto: e.g. 0 has no pre-image because x³–1 = 0 ⇒ x³ = 1 ⇒ x = 1, but 0 is in co-domain Z and 1 maps to 0, so actually 0 has pre-image 1.
Wait — check: f(1) = 0, so 0 has pre-image. But what about 2? x³–1 = 2 ⇒ x³ = 3 ⇒ no integer solution. So not onto.
MCQ · Quadratic on Non-negative Reals
Let f: R+ → [–5, ∞) be defined as f(x) = 9x² + 6x – 5, where R+ is the set of all non-negative real numbers. Then, f is :
(B) onto
(C) bijective
(D) neither one-one nor onto
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Answer : (C) bijective
f(x) = 9x² + 6x – 5 = (3x+1)² – 6.
For x ≥ 0, f is strictly increasing ⇒ one–one.
Range = [–5, ∞) = co-domain ⇒ onto.
∴ bijective.
MCQ · x² + 1 on Non-negative Reals
Let R+ denote the set of all non-negative real numbers. Then the function f: R+ → R+ defined as f(x) = x² + 1 is :
(B) onto but not one-one
(C) both one-one and onto
(D) neither one-one nor onto
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Answer : (A) one-one but not onto
One–one: x₁²+1 = x₂²+1 ⇒ x₁ = x₂ (as x₁, x₂ ≥ 0).
Not onto: range = [1, ∞) ≠ R+ = [0, ∞).
∴ one–one but not onto.
MCQ · Linear on Non-negative Reals
A function f : R+ → R (where R+ is the set of all non-negative real numbers) defined by f(x) = 4x + 3 is :
(B) onto but not one-one
(C) both one-one and onto
(D) neither one-one nor onto
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Answer : (A) one-one but not onto
One–one: 4x₁+3 = 4x₂+3 ⇒ x₁ = x₂.
Not onto: range = [3, ∞) ≠ R.
∴ one–one but not onto.
Relation · |x² – y²| < 8
A relation R on set A = {1, 2, 3, 4, 5} is defined as R = {(x, y) : |x² – y²| < 8}.
Check whether the relation R is reflexive, symmetric and transitive.
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Reflexive :
|x² – x²| = 0 < 8 ⇒ (x,x) ∈ R. ∴ reflexive.
Symmetric :
|x² – y²| < 8 ⇒ |y² – x²| < 8 ⇒ (y,x) ∈ R. ∴ symmetric.
Transitive :
(2,3) ∈ R and (3,4) ∈ R but (2,4) ∉ R because |4–16| = 12 not < 8. ∴ not transitive.
Equivalence Relation · a – c = b – d
A relation R is defined on N × N (where N is the set of natural numbers) as :
(a,b) R (c,d) ⇔ a – c = b – d
Show that R is an equivalence relation.
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Reflexive :
a – a = b – b ⇒ (a,b) R (a,b). ∴ reflexive.
Symmetric :
(a,b) R (c,d) ⇒ a – c = b – d ⇒ c – a = d – b ⇒ (c,d) R (a,b). ∴ symmetric.
Transitive :
(a,b) R (c,d) and (c,d) R (e,f) ⇒ a–c = b–d and c–e = d–f
⇒ a–e = b–f ⇒ (a,b) R (e,f). ∴ transitive.
Hence R is an equivalence relation.
Equivalence Relation · x+y divisible by 2
A relation R on set A = {–4, –3, –2, –1, 0, 1, 2, 3, 4} be defined as
R = {(x, y) : x + y is an integer divisible by 2}.
Show that R is an equivalence relation. Also, write the equivalence class [2].
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Reflexive :
x+x = 2x is divisible by 2 ⇒ (x,x) ∈ R. ∴ reflexive.
Symmetric :
If x+y is even, then y+x is even ⇒ (y,x) ∈ R. ∴ symmetric.
Transitive :
x+y even and y+z even ⇒ x and y same parity, y and z same parity ⇒ x and z same parity ⇒ x+z even. ∴ transitive.
[2] :
[2] = {–4, –2, 0, 2, 4} (all even integers in A).
Railway Museum · Parallel & Perpendicular
R = {(l₁, l₂) : l₁ is parallel to l₂}.
(i) Find whether R is symmetric or not.
(ii) Find whether R is transitive or not.
(iii) If one line is y = 3x + 2, find the set of rail lines in R related to it.
OR
Let S = {(l₁, l₂) : l₁ is perpendicular to l₂}. Check whether S is symmetric and transitive.
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(i) R is symmetric: if l₁ ∥ l₂ then l₂ ∥ l₁.
(ii) R is transitive: if l₁ ∥ l₂ and l₂ ∥ l₃ then l₁ ∥ l₃.
(iii) Set of lines = {y = 3x + c : c ∈ R}.
OR
S is symmetric: if l₁ ⟂ l₂ then l₂ ⟂ l₁.
S is not transitive: if l₁ ⟂ l₂ and l₂ ⟂ l₃ then l₁ ∥ l₃, not perpendicular.
Function · 2x/(1+x²)
Show that a function f : R → R defined by f(x) = 2x/(1+x²) is neither one–one nor onto. Further, find set A so that the given function f : R → A becomes an onto function.
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Not one–one :
f(1) = f(–1) = 1 but 1 ≠ –1. ∴ not one–one.
Not onto :
Range of f is [–1, 1] (since |2x/(1+x²)| ≤ 1).
Co-domain is R, so range ≠ co-domain. ∴ not onto.
For onto :
Choose A = [–1, 1]. Then f : R → [–1, 1] is onto.
Linear Function · ax+b
A function f is defined from R → R as f(x) = ax + b, such that f(1) = 1 and f(2) = 3. Find function f(x). Hence, check whether function f(x) is one–one and onto or not.
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Find f(x) :
f(1) = a+b = 1, f(2) = 2a+b = 3
Solving: a = 2, b = –1 ⇒ f(x) = 2x – 1.
One–one :
2x₁–1 = 2x₂–1 ⇒ x₁ = x₂. ∴ one–one.
Onto :
For any y ∈ R, x = (y+1)/2 ∈ R gives f(x) = y. ∴ onto.
Quadratic · x²+x+1
Show that a function f : R → R defined as f(x) = x² + x + 1 is neither one–one nor onto. Also, find all the values of x for which f(x) = 3.
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Not one–one :
f(0) = f(–1) = 1 but 0 ≠ –1. ∴ not one–one.
Not onto :
f(x) = (x+½)² + ¾ ≥ ¾ ⇒ range = [¾, ∞) ≠ R. ∴ not onto.
f(x) = 3 :
x²+x+1 = 3 ⇒ x²+x–2 = 0 ⇒ (x–1)(x+2) = 0 ⇒ x = 1, –2.
Equivalence Relation · a/c = b/d
A relation R is defined on N × N (where N is the set of natural numbers) as
(a, b) R (c, d) ⇔ a/c = b/d.
Show that R is an equivalence relation.
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Reflexive :
a/a = b/b ⇒ (a,b) R (a,b). ∴ reflexive.
Symmetric :
a/c = b/d ⇒ c/a = d/b ⇒ (c,d) R (a,b). ∴ symmetric.
Transitive :
a/c = b/d and c/e = d/f ⇒ (a/c)(c/e) = (b/d)(d/f) ⇒ a/e = b/f ⇒ (a,b) R (e,f). ∴ transitive.
Hence R is an equivalence relation.
Equivalence Relation · Divisible by 5
A relation R on set A = {x : –10 ≤ x ≤ 10, x ∈ Z} is defined as
R = {(x, y) : (x – y) is divisible by 5}.
Show that R is an equivalence relation. Also, write the equivalence class [5].
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Reflexive :
x–x = 0 divisible by 5 ⇒ (x,x) ∈ R. ∴ reflexive.
Symmetric :
If x–y divisible by 5, then y–x divisible by 5 ⇒ (y,x) ∈ R. ∴ symmetric.
Transitive :
x–y and y–z divisible by 5 ⇒ x–z = (x–y)+(y–z) divisible by 5 ⇒ (x,z) ∈ R. ∴ transitive.
[5] :
[5] = {–10, –5, 0, 5, 10}.
MCQ · Relation in N
Let R be a relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Then
(B) (6, 8) ∈ R
(C) (3, 8) ∈ R
(D) (2, 4) ∈ R
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Answer : (B) (6, 8) ∈ R
a = b – 2 and b > 6. For (6,8): a=6, b=8 ⇒ 6 = 8–2 and 8 > 6. ✓