The relation R on the set A = {1, 2, 3} given by R = {(1, 2), (2, 1)} is:
Relations and Functions — Important Questions
SUMMARY: This chapter explores the fundamental concepts of relations and functions, including their types and properties.
KEY TOPICS: types of relations, equivalence relation, types of functions, one-one and onto functions, composition of functions, inverse of a function, binary operations, domain and range
If f: R → R, f(x) = x² + 1, then f is:
Check answerHide answer
Let f(x) = 2x + 3 and g(x) = x − 1. Then (fog)(2) equals:
Check answerHide answer
The number of equivalence relations on the set {1, 2, 3} that contain (1, 2) is:
Check answerHide answer
For a one-one function f: A → B, n(A) = 5 and n(B) = 7, the number of one-one functions from A to B is:
Check answerHide answer
Define an equivalence relation and give one example.
View sample solutionHide solution
Show that the function f: N → N defined by f(n) = 2n is one-one but not onto.
View sample solutionHide solution
For f(x) = x + 1 and g(x) = x², find (fog)(x) and (gof)(x).
View sample solutionHide solution
State the necessary and sufficient condition for a function f: A → B to be invertible.
View sample solutionHide solution
On Z define R: aRb ⇔ a − b is divisible by 3. Show R is reflexive.
View sample solutionHide solution
Show that the relation R on N × N defined by (a, b) R (c, d) ⇔ a + d = b + c is an equivalence relation.
View sample solutionHide solution
Show that f: R − {−4/3} → R − {4/3} given by f(x) = (4x + 3) / (3x + 4) is a bijection.
View sample solutionHide solution
Show that the relation R on Z given by aRb ⇔ a − b is divisible by 5 is an equivalence relation. Find the equivalence class of 0.
View sample solutionHide solution
Let f: R → R, f(x) = x³ + 4. Show that f is bijective and find f⁻¹.
View sample solutionHide solution
Let A = {1, 2, 3, ..., 9} and R be the relation in A × A defined by (a, b) R (c, d) iff a + d = b + c. Show that R is an equivalence relation. Also find equivalence class of (2, 5).
View sample solutionHide solution
Differentiate between reflexive symmetric and transitive relations in tabular form with one example each.
Assertion (A): The identity relation is an equivalence relation.
Reason (R): It is reflexive, symmetric and transitive — every element is related only to itself.
Show explanationHide explanation
Assertion (A): The function f: R → R defined by f(x) = 2x + 3 is bijective.
Reason (R): A linear function with non-zero slope is one-one and onto.
Show explanationHide explanation
Assertion (A): Composition of functions is commutative for all f and g.
Reason (R): Composition is associative: (f o g) o h = f o (g o h).
Show explanationHide explanation
Assertion (A): A function has an inverse if and only if it is bijective.
Reason (R): Bijection guarantees a unique correspondence between domain and range.
Show explanationHide explanation
Assertion (A): The relation (a, b) R (c, d) ⇔ a + d = b + c on N × N is an equivalence relation.
Reason (R): It is reflexive, symmetric and transitive on the Cartesian product.
Show explanationHide explanation
Statement 1: Every reflexive relation is symmetric.
Statement 2: Every symmetric relation is reflexive.
Show answerHide answer
Statement 1: The function sin x is many-one on R.
Statement 2: The function sin x restricted to [−π/2, π/2] is one-one.
Show answerHide answer
Statement 1: Composition of functions is associative.
Statement 2: Composition of functions is commutative.
Show answerHide answer
Statement 1: The inverse of the identity function is itself.
Statement 2: The inverse of a bijective function is also bijective.
Show answerHide answer
Statement 1: A set with n elements has 2^(n²) total relations.
Statement 2: The number of reflexive relations on such a set is 2^(n² − n).
Show answerHide answer
-
Which of the following properties is satisfied by R?AReflexiveBSymmetricCTransitiveDAll of these
-
The equivalence class of 0, denoted [0], is best described as which of the listed sets?A{0, 5, 10}B{1, 6, 11}C{2, 7, 12}DAll sets above
-
Explain why R is an equivalence relation and describe the partition of Z it creates.
Show answersHide answers
-
Since each student has a unique roll number and every roll number is used the function f is:AOne-one but not ontoBOnto but not one-oneCBoth one-one and ontoDNeither
-
Because f is bijective:AInverse existsBInverse does not existCInverse is many-valuedDCannot decide
-
Explain in words why a bijection guarantees the existence of an inverse function.
Show answersHide answers
-
The composite (g ∘ f)(x) is:A2x + 5B2x + 10C2(x + 5)Dx + 10
-
The composite (f ∘ g)(x) is:A2x + 5B2x + 10Cx + 5D4x + 5
-
Compute (g ∘ f)(3) and (f ∘ g)(3) and comment on whether composition is commutative.
Show answersHide answers
Study the table of relations on set A = {1, 2, 3} and answer:
| Relation | Reflexive | Symmetric | Transitive |
|---|---|---|---|
| R1 = {(1,1),(2,2),(3,3)} | Yes | Yes | Yes |
| R2 = {(1,2),(2,1)} | No | Yes | No |
| R3 = {(1,2),(2,3),(1,3)} | No | No | Yes |
| R4 = {(1,1),(2,2),(1,2),(2,1)} | No | Yes | Yes |
-
Which relation is an equivalence relation on A?AR1BR2CR3DR4
-
Which relation is symmetric and transitive but not reflexive?AR2BR3CR4D
-
Explain why R4 fails to be an equivalence relation despite being symmetric and transitive.
Show answersHide answers
Study the function values and answer:
| x | f(x) = x² | g(x) = x + 1 | (f ∘ g)(x) |
|---|---|---|---|
| 0 | 0 | 1 | 1 |
| 1 | 1 | 2 | 4 |
| 2 | 4 | 3 | 9 |
| −1 | 1 | 0 | 0 |
-
The composite (f ∘ g)(x) equals:A(x + 1)²Bx² + 1Cx²(x + 1)Dx + 1
-
Considering f: R → R defined by f(x) = x² the function is:AOne-one ontoBOne-one not ontoCMany-one ontoDMany-one not onto
-
Explain whether f(x) = x², considered as a function from R to R, is one-one or onto.
Show answersHide answers
For each relation R defined on A = {1, 2, 3} below, identify whether it is reflexive, symmetric or transitive.
| Relation R | Pairs |
|---|---|
| R1 | {(1,1),(2,2),(3,3),(1,2)} |
| R2 | {(1,2),(2,1)} |
| R3 | {(1,1),(2,2),(3,3),(1,2),(2,1)} |
| R4 | {(1,2),(2,3)} |
Study the graphs of f(x) = x² and g(x) = x³ on R and answer:
-
As a function from R to R, f(x) = x² is:AOne-one and ontoBOne-one but not ontoCMany-one but ontoDMany-one and not onto
-
As a function from R to R, g(x) = x³ is:AOne-one and ontoBOne-one but not ontoCMany-one but ontoDMany-one and not onto
-
Use the horizontal-line test on each graph to justify your answers.
Show answersHide answers
Make a full Mathematics paper on Relations and Functions.
Pick the question mix, set the marks, hit generate. You get a ready-to-print paper with an answer key.
Generate your paper — free