Sample QuestionsRelations and Functions questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
Let $A=\{1,2,3\}, B=\{4,5,6,7\}$ and let $f=\{(1,4),(2,5)$, $(3,6)\}$ be a function from $A$ to $B$. Based on the given information, $f$ is best defined as
View full solution →A relation $R$ in set $A=\{1,2,3\}$ is defined as $R=\{(1,1),(1,2),(2,2),(3,3)\}$. Which of the following ordered pair in $R$ shall be removed to make it an equivalence relation in $A$ ?
- A
$(1,1)$
- B
$(1,2)$
- C
$(2,2)$
- D
$(3,3)$
View full solution →The function $f: N \rightarrow N$ is defined by $f(n)=\left\{\begin{array}{ll}\frac{n+1}{2}, & \text { if } n \text { is odd } \\ \frac{n}{2}, & \text { if } n \text { is even }\end{array}\right.$
The function $f$ is
View full solution →The number of equivalence relations in the set $\{1,2,3\}$ containing the elements $(1,2)$ and $(2,1)$ is
View full solution →The function $f: R \rightarrow R$ defined by $f(x)=4+3 \cos x$ is
View full solution →Assertion (A): The relation $R=\{(x, y):(x+y)$ is a prime number and $x, y \in N\}$ is not a reflexive relation.
Reason (R) : The number ' $2 n$ ' is composite for all natural numbers $n$.
View full solution →Assertion $( A )$ : The relation $f:\{1,2,3,4\} \rightarrow\{x, y$, $z, p\}$ defined by $f=\{(1, x),(2, y),(3, z)\}$ is a bijective function.
Reason $( R )$ : The function $f:\{1,2,3\} \rightarrow\{x, y, z, p\}$ such that $f=\{(1, x),(2, y),(3, z)\}$ is one-one.
- A
Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of (A).
- B
Both $(A)$ and $(R)$ are true but (R) is not the correct explanation of (A).
- C
(A) is true but $( R )$ is false.
- D
(A) is false but ( $R$ ) is true.
View full solution →Assertion (A) : Let $f:(e, \infty) \rightarrow R$ defined by $f(x)=\log (\log (\log x))$ is bijective.
Reason (R) : A function $f$ will be bijective if $f$ is both one-one and onto.
- ✓
Both (A) and (R) are true and (R) is the correct explanation of (A).
- B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
- C
(A) is true but (R) is false.
- D
(A) is false but (R) is true.
Answer: A.
View full solution →Assertion (A) : Relation $R$ defined in the set $A$ as $R\{(x, y): y-x$ is an integer, $x, y \in R\}$ is an equivalence relation.
Reason (R) : Relation $R$ defined in the set $B$ as $R\{(x, y): x=\alpha y$ for some rational number $\alpha$, $x, y \in R\}$ is an equivalence relation.
- A
Both (A) and (R) are true and (R) is the correct explanation of (A).
- B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
- ✓
(A) is true but (R) is false.
- D
(A) is false but (R) is true.
Answer: C.
View full solution →Assertion (A) : If $f: R \rightarrow R$ defined by $f(x)=7 x-[7 x]$, where [.] denotes greatest integer $\leq x \forall x \in R$, then $f$ is not one-one function.
Reason (R) : Fractional part functions are always many-one.
- ✓
Both (A) and (R) are true and (R) is the correct explanation of (A).
- B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
- C
(A) is true but (R) is false.
- D
(A) is false but (R) is true.
Answer: A.
View full solution →Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
Find the number of all onto functions from the set {1, 2, 3, ...., n} to itself.
Show that the function f : R $\rightarrow$ R given by f(x) = x3 is injective.
View full solution →Let $\;f{\text{ }}:{\text{ }}R{\text{ }} \to {\text{ }}R$ be defined as f (x) = 3x. Choose the correct answer.
View full solution →Given a non empty set X, consider P (X) which is the set of all subsets of X.
Define the relation R in P (X) as follows:
For subsets A, B in P (X), ARB if and only if A $\subset$ B. Is R an equivalence relation on P (X)? Justify your answer.
View full solution →Show that the function f : R $\rightarrow$ {x $\in$ R : -1 < x < 1} defined by $f(x) = \frac{x}{{1 + |x|}}$, x $\in$ R is one-one and onto function.
View full solution →Let A and B be sets. Show that f : A $\times $ B $\rightarrow$ B $\times $ A such that f(a, b) = (b, a) is a bijective function.
View full solution →State whether the function is one-one, onto or bijective. Justify your answer. f: R $\rightarrow$ R defined by f(x) = 1+ x2
View full solution →State whether the function is one-one, onto or bijective. Justify your answer. f: R $\rightarrow$ R defined by f(x) = 3 - 4x.
View full solution →Let A = {-1, 0, 1, 2}, B = {-4, -2, 0, 2} and f, g : A $\rightarrow$ B be the functions defined by f(x) = x2 - x, x $\in$ A and $g(x) = 2\left| {x - \frac{1}{2}} \right| - 1,x \in A$. Are f and g equal? Justify your answer.
(Hint: One may note that two functions f : A $\rightarrow$ B and g : A $\rightarrow$ B such that f(a) = g(a) $\forall$ a $\in$ A, are called equal functions).
View full solution →If f: N $\to$ N is defined by f(n) = $\left\{ \begin{array} { l } { \frac { n + 1 } { 2 } , \text { if } n \text { is odd } } \\ { \frac { n } { 2 } , \text { if } n \text { is even } } \end{array} \right.$for all n $ \in$ N. State whether the function f is bijective. Justify your answer.
View full solution →Let A = R - {3} and B = R - {1}. Consider the function f : A $\rightarrow$ B defined by $f(x) = \left( {\frac{{x - 2}}{{x - 3}}} \right)$ Is f one-one and onto? Justify your answer.
View full solution →Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a - b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation.
If A = { 1, 2, 3}, B = { 4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
Consider the mapping f: A → B is defined by f(x) = x - 1 such that f is a bijection.
Based on the above information, answer the following questions.
- Domain of f is:
- R - {2}
- R
- R - {1, 2}
- R - {0}
- Range of f is:
- R
- R - {2}
- R - {0}
- R - {1, 2}
- If g: R - {2} → R - {1} is defined by g(x) = 2f(x) - 1, then g(x) in terms of x is:
- $\frac{\text{x}+2}{\text{x}}$
- $\frac{\text{x}+1}{\text{x}-2}$
- $\frac{\text{x}-2}{\text{x}}$
- $\frac{\text{x}}{\text{x}-2}$
- The function g defined above, is:
- One-one
- Many-one
- into
- None of these
- A function f(x) is said to be one-one iff.
- f(x1) = f(x2) ⇒ -x1 = x2
- f(-x1) = f(-x2) ⇒ -x1 = x2
- f(x1) = f(x2) ⇒ x1 = x2
- None of these
View full solution →A relation R on a set A is said to be an equivalence relation on A iff it is:
- Reflexive i.e., $(\text{a, a})\in\ \text{R} \ \forall \ \text{a}\in\text{A}.$
- Symmetric i.e., $(\text{a, b})\in\ \text{R} \Rightarrow \text{(b, a) } \in\text{R}\ \forall \ \text{a, b}\in\text{A}.$
- Transitive i.e., $(\text{a, b})\in\ \text{R} \ \text{and}\ \text{(b, c) } \in\text{R}\Rightarrow\text{(a, c)}\in\text{R}\ \forall \ \text{a, b, c}\in\text{A}.$
Based on the above information, answer the following questions.
- If the relation R = {(1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} defined on the set A = {1, 2, 3}, then R is:
- Reflexive
- Symmetric
- Transitive
- Equivalence
- If the relation R = {(1, 2), (2, 1), (1, 3), (3, 1)} defined on the set A = {1, 2, 3}, then R is:
- Reflexive
- Symmetric
- Transitive
- Equivalence
- If the relation R on the set N of all natural numbers defined as R = {(x, y): y = x + 5 and x < 4}, then R is:
- Reflexive
- Symmetric
- Transitive
- Equivalence
- If the relation R on the set A = {1, 2, 3, ........., 13, 14} defined as R = {(x, y): 3x - y = O}, then R is:
- Reflexive
- Symmetric
- Transitive
- Equivalence
- If the relation R on the set A = {I, 2, 3} defined as R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}, then R is:
- Reflexive only
- Symmetric only
- Transitive only
- Equivalence
View full solution →