MCQ 2511 Mark
Let $R$ be a relation in the set $N$ given by $R=\{(a, b): a=b-2, b>6\}$. Then
- A
$(8,7) \in R$
- ✓
$(6,8) \in R$
- C
$(3,8) \in R$
- D
$(2,4) \in R$
AnswerCorrect option: B. $(6,8) \in R$
(b) : Given, $R=\{(a, b): a=b-2, b>6\}$
Since, $b>6$, so $(2,4) \notin R$
Also, $(3,8) \notin R$ as $3 \neq 8-2$
and $(8,7) \notin R$ as $8 \neq 7-2$
Now, for $(6,8)$, we have
$8>6$ and $6=8-2$, which is true
$\therefore \quad(6,8) \in R$
View full question & answer→MCQ 2521 Mark
Which of the following functions is not one-one?
- A
$\frac{3-x}{3+x}$
- B
$\sqrt{x}$
- ✓
$x^2+1$
- D
AnswerCorrect option: C. $x^2+1$
(c) : Since $f(x)=f(-x)=x^2+1$ for all $x \in R$, therefore, $f$ is not one-one.
View full question & answer→MCQ 2531 Mark
Let $A=\{1,2,3\}$ and $B=\{1,2,4\}$, then $f=\{(1,1),(1,2),(2,1),(3,4)\}$ is a
- A
one-one function from $A$ to $B$
- B
bijection from $A$ to $B$
- C
surjection from $A$ to $B$
- ✓
Answer(d) : Here, $f$ is not a function from $A$ to $B$ as $f(1)$ is not unique.
View full question & answer→MCQ 2541 Mark
Let $f: R \rightarrow R$ be defined by $f(x)=\frac{1}{x}$ $\forall x \in R$. Then $f$ is
AnswerCorrect option: D. $f$ is not defined
(d) : Since, $\frac{1}{x}$ is not defined for $x=0$
$\therefore \quad f: R \rightarrow R$ can not be defined.
View full question & answer→MCQ 2551 Mark
Let $A=\{1,2,3, \ldots, n\}$ and $B=\{a, b\}$. Then the number of surjections from $A$ into $B$ is
- A
${ }^n P_2$
- ✓
$2^n-2$
- C
$2^n-1$
- D
AnswerCorrect option: B. $2^n-2$
(b) : If $f: A \rightarrow B$ is a function, then $f(1)$ can be chosen in two ways, $f(2)$ can be chosen in two ways, ..., $f(n)$ can be chosen in two ways.
Hence, $f$ can be chosen in $2 \times 2 \times \ldots \times 2=2^n$ ways
In total there are $2^n$ functions possible. Out of these two functions $f_1$ and $f_2$, defined as $f_1(i)=a \forall i=1,2, \ldots, n$ and $f_2(i)=b \forall i=1,2, \ldots, n$ are not surjective as range of $f_1$ is $\{a\} \neq B$ and $f_2$ is $\{b\} \neq B$.
Hence, the number of surjections from $A$ to $B$ is $2^n-2$.
View full question & answer→MCQ 2561 Mark
If the set $A$ contains 5 elements and the set $B$ contains 6 elements, then the number of one-one and onto mappings from $A$ to $B$ is
Answer(c) : As $A$ contains 5 elements.
$\therefore \quad$ For any one-one onto mapping $f: A \rightarrow B, f(A)$ also contains 5 elements but $B$ contains 6 elements.
$\therefore f(A) \neq B$.
So, no one-one mapping from $A$ to $B$ can be onto.
View full question & answer→MCQ 2571 Mark
Let $A=\{1,2,3\}$ and consider the relation $R=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$. Then $R$ is
- ✓
reflexive but not symmetric
- B
reflexive but not transitive
- C
- D
neither symmetric nor transitive
AnswerCorrect option: A. reflexive but not symmetric
(a) : $(1,1),(2,2),(3,3) \in R$
$\therefore \quad R$ is reflexive but it is not symmetric.
Also, $R$ is transitive.
View full question & answer→MCQ 2581 Mark
Let us define a relation $R$ in $R$ as $a R b$ if $a \geq b$. Then $R$ is
- A
- ✓
reflexive, transitive but not symmetric
- C
symmetric, transitive but not reflexive
- D
neither transitive nor reflexive but symmetric
AnswerCorrect option: B. reflexive, transitive but not symmetric
(b) : Given $a R b, a \geq b$
(i) Now $a \geq a$ is true for all real number
$\therefore \quad R$ is reflexive.
(ii) Let $(a, b) \in R, a \geq b$
Now, $a \geq b$ but does not imply $b \geq a$.
$\therefore \quad(b, a) \notin R \therefore R$ is not symmetric.
(iii) Let $(a, b) \in R$ and $(b, c) \in R \Rightarrow a \geq b$ and $b \geq c$
$\therefore a \geq c \Rightarrow(a, c) \in R \therefore R$ is transitive.
View full question & answer→MCQ 2591 Mark
If a relation $R$ on the set $\{1,2,3\}$ be defined by $R=\{(1,2)\}$, then $R$ is
View full question & answer→MCQ 2601 Mark
The maximum number of equivalence relations on the set $A=\{1,2,3\}$ are
AnswerThe smallest equivalence relation is the identity relation $R_1=\{(1,1),(2,2),(3,3)\}$
Then, two ordered pairs of two distinct elements can be added to give three more equivalence relations.
$R_2=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$
Similarly $R_3$ and $R_4$.
Finally the largest equivalence relation, that is the universal relation.
$R_5=\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1),(2,3),$
$(3,2)\}$
View full question & answer→MCQ 2611 Mark
Consider the non-empty set consisting of children in a family and a relation $R$ defined as $a R b$ if $a$ is brother of $b$. Then $R$ is
- A
symmetric but not transitive
- ✓
transitive but not symmetric
- C
neither symmetric nor transitive
- D
both symmetric and transitive
AnswerCorrect option: B. transitive but not symmetric
(b) : Given $a R b \Rightarrow a$ is brother of $b$.
But $b \not R a \quad[\because b$ may or may not be brother of $a]$
$\therefore \quad R$ is not symmetric.
Let $a R b$ and $b R c$
$\Rightarrow \quad a$ is brother of $b$ and $b$ is brother of $c$.
$\therefore \quad a$ is brother of $c \Rightarrow(a, c) \in R . \therefore R$ is transitive.
View full question & answer→MCQ 2621 Mark
Let $T$ be the set of all triangles in the Euclidean plane, and let a relation $R$ on $T$ be defined as $a R b$ if $a$ is congruent to $b \forall a, b \in T$. Then $R$ is
- A
reflexive but not transitive
- B
transitive but not symmetric
- ✓
- D
Answer(c) : (i) We know that every triangle is congruent to itself.
$\therefore \quad\left(T_1, T_1\right) \in R$ for all $T_1 \in T$. Thus, $R$ is reflexive.
(ii) Let $\left(T_1, T_2\right) \in R$
$\Rightarrow \quad T_1$ is congruent to $T_2$.
$\Rightarrow \quad T_2$ is congruent to $T_1$.
$\therefore \quad\left(T_2, T_1\right) \in R$
Thus, $R$ is symmetric.
(iii) Let $\left(T_1, T_2\right) \in R$ and $\left(T_2, T_3\right) \in R$.
$\Rightarrow \quad T_1$ is congruent to $T_2$ and $T_2$ is congruent to $T_3$.
$\therefore \quad T_1$ is congruent to $T_3$
$\Rightarrow \quad\left(T_1, T_3\right) \in R$.
Thus, $R$ is transitive.
$\therefore \quad R$ is an equivalence relation.
View full question & answer→MCQ 2631 Mark
The number of bijective functions from set $A$ to itself when $A$ contains 106 elements is
Answer(c) : The total number of bijections from a set containing $n$ elements to itself is $n$ ! Hence, required number $=(106) !$
View full question & answer→MCQ 2641 Mark
Set $A$ has three elements and set $B$ has four elements. The number of injections that can be defined from $A$ to $B$ is
Answer(c) : Since $3<4$, injective functions from $A$ to $B$ are defined and the total number of such functions is
$
{ }^4 P_3=\frac{4 !}{(4-3) !}=4 \times 3 \times 2 \times 1=24 .
$
View full question & answer→MCQ 2651 Mark
Let $f: N \rightarrow N$, where $f(x)=x-(-1)^x$, then $f$ is
Answer(c) : $f(x)=\left\{\begin{array}{ll}x-1, & x \text { is even } \\ x+1, & x \text { is odd }\end{array}\right.$, which is clearly one-one and onto.
View full question & answer→MCQ 2661 Mark
The function $f: R \rightarrow R$ defined by $f(x)=6^x+6^{|x|}$ is
View full question & answer→MCQ 2671 Mark
Which of the following statements is false?
- A
$f: A \rightarrow B$ is one-one iff $x_1 \neq x_2$ in $A$ $\Rightarrow f\left(x_1\right) \neq f\left(x_2\right)$ in $B$
- B
$f: A \rightarrow B$ is onto iff for each $y$ in $B$, there is some $x$ in $A$ s.t. $f(x)=y$
- C
$f: A \rightarrow B$ is invertible iff $f$ is both one-one and onto.
- ✓
A real valued function $f$ (of a real variable) is invertible iff $f$ is only one-one.
AnswerCorrect option: D. A real valued function $f$ (of a real variable) is invertible iff $f$ is only one-one.
View full question & answer→MCQ 2681 Mark
Let $f: R \rightarrow R$ be defined by $f(x)=x+|x|$. Then $f(x)$ is
Answer(d) : Given, $f(x)=x+|x|$
Now, $f(-2)=-2+|-2|=-2+2=0$
and $f(-3)=-3+|-3|=-3+3=0$
Hence, $f$ is not one-one
Also, $f(x)=\left\{\begin{array}{ll}x+x & \text { if } x \geq 0 \\ x-x & \text { if } x<0\end{array} \Rightarrow f(x)=\left\{\begin{array}{ll}2 x, & x \geq 0 \\ 0, & x<0\end{array}\right.\right.$
Thus, $f(x)=2 x \geq 0$ for all $x \geq 0$ and $f(x)=0$ for $x<0$. This means that $f(x)$ cannot be negative for any $x \in R$. So, $f$ is not onto. Note that $R_f=[0, \infty)$, which is a proper subset of $R$.
View full question & answer→MCQ 2691 Mark
Let $R$ be a relation on the set $N$ be defined by $\{(x, y): x, y \in N, 2 x+y=41\}$. Then, $R$ is
Answer(d) $: R=\{(x, y): x, y \in N, 2 x+y=41\}$
Reflexive : $(1,1) \notin R$ as $2 \cdot 1+1=3 \neq 41$. So, $R$ is not reflexive.
Symmetric : $(1,39) \in R$ but $(39,1) \notin R$. So $R$ is not symmetric.
Transitive : $(20,1) \in R$ and $(1,39) \in R$. But $(20,39) \notin R$, so $R$ is not transitive.
View full question & answer→MCQ 2701 Mark
If $A$ and $B$ are finite sets containing respectively $m$ and $n$ elements, then find the number of relation that can be defined from $A$ to $B$.
- A
$m n$
- ✓
$2^{m n}$
- C
$m+n$
- D
$3^{m n}$
AnswerCorrect option: B. $2^{m n}$
(b) : Let $R$ be a relation from $A$ to $B$, then $R \subset A \times B$. This means that the number of relations from $A$ to $B$ is equal to the number of subsets of $A \times B$.
Now, $O(A)=m$ and $O(B)=n$
$\Rightarrow O(A \times B)=m n$
$\therefore \quad$ Number of subsets of $A \times B=2^{m n}$
$\left(\because O(\right.$ Power set of $\left.A)=2^{O(A)}\right)$
$\therefore \quad$ Number of relations from $A$ to $B=2^{m n}$
View full question & answer→MCQ 2711 Mark
For the set $A=\{1,2,3\}$, define a relation $R$ on the set $A$ as follows :
$R=\{(1,1),(2,2),(3,3),(1,3)\}$
How many ordered pairs to be added to $R$ to make it the smallest equivalence relation?
Answer(a): Here, $A=\{1,2,3\}$ and the relation $R=\{(1,1),(2,2)(3,3)(1,3)\}$.
Clearly, $R$ is reflexive but not symmetric as $(1,3) \in R$ but $(3,1) \notin R$.
We shall include $(3,1)$ to the above relation to make it smallest equivalence relation
$
R^{\prime}=\{(1,1),(2,2),(3,3)(1,3),(3,1)\} \text {. }
$
$R^{\prime}$ is certainly transitive as transitivity is not contradicted.
View full question & answer→MCQ 2721 Mark
Let $A=\{a, b, c\}$ and let $R=\{(a, a),(a, b)$, $(b, a)\}$. Then, $R$ is
- A
reflexive and symmetric but not transitive
- B
reflexive and transitive but not symmetric
- ✓
symmetric and transitive but not reflexive
- D
AnswerCorrect option: C. symmetric and transitive but not reflexive
(c) : $R$ is symmetric and transitive but not reflexive.
View full question & answer→MCQ 2731 Mark
Let $S$ be the set of all real numbers and let $R$ be a relation on $S$ defined by $a R b \Leftrightarrow a^2+b^2=1$. Then, $R$ is
- ✓
symmetric but neither reflexive nor transitive
- B
reflexive but neither symmetric nor transitive
- C
transitive but neither reflexive nor symmetric
- D
AnswerCorrect option: A. symmetric but neither reflexive nor transitive
(a) : $\left(1^2+1^2\right) \neq 1$. So, $R$ is not reflexive.
Now, $a R b a^2+b^2=1 \Rightarrow b^2+a^2=1 \Rightarrow b R a$.
Hence, $R$ is symmetric.
Also, 1 R 0 and $0 R 1$. But, 1 is not related to 1 .
Hence, $R$ is not transitive.
View full question & answer→