MCQ 2011 Mark
Let $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}=\text{B}$ and $\text{C}=\{\text{x}\in\text{R}:\text{x}\geq0\}$ and let $\text{S}=\{(\text{x, y})\in\text{A}\times\text{B}:\text{x}^2+\text{y}^2=1\}$ and $\text{S}_0=\{(\text{x, y})\in\text{A}\times\text{C}:\text{x}^2+\text{y}^2=1\}.$ Then,
- ✓
$S$ defines a function from $A$ to $B.$
- B
$S^o$ defines a function from $A$ to $C.$
- C
$S^o$ defines a function from $A$ to $B.$
- D
$S$ defines a function from $A$ to $C.$
AnswerCorrect option: A. $S$ defines a function from $A$ to $B.$
Given that $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}=\text{B}$ and $\text{C}=\{\text{x}\in\text{R}:\text{x}\geq0\}$ and $\text{S}=\{(\text{x, y})\in\text{A}\times\text{B}:\text{x}^2+\text{y}^2=1\}$ and $\text{S}_0=\{(\text{x, y})\in\text{A}\times\text{C}:\text{x}^2+\text{y}^2=1\}$
$\text{x}^2+\text{y}^2=1$
$\Rightarrow\ \text{y}^2=1-\text{x}^2$
$\Rightarrow\ \text{y}=\sqrt{1-\text{x}^2}$
$\text{y}\in\text{B}$
Hence$, S$ defines a function from $A$ to $B.$
View full question & answer→MCQ 2021 Mark
Let $\text{f}:\text{R}-\Big\{\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be defined by $\text{f(x)}=\frac{3\text{x}+2}{5\text{x}-3}.$ Then,
AnswerCorrect option: A. $f^{-1}(x)=f(x)$
Given function is $\text{f}:\text{R}-\Big\{\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be defined by $\text{f(x)}=\frac{3\text{x}+2}{5\text{x}-3}$
fo $f(x) = f(f(x))$
$=\text{f}\Big(\frac{3\text{x}+2}{5\text{x}-3}\Big)$
$=\frac{3\big(\frac{3\text{x}+2}{5\text{x}-3}\big)+2}{5\big(\frac{3\text{x}+2}{5\text{x}-3}\big)-3}$
After solving you will get
$f(f(x)) = x$
Also, $f^{-1}(x) = f(x)$ you can check.
View full question & answer→MCQ 2031 Mark
Choose the correct answer from the given four options.
Let $f : [0, 1] \rightarrow [0, 1]$ be defined by $\text{f}(\text{x})=\begin{cases}\text{x, if x is rational}1-\text{x, if x is irrational}\end{cases}\}$ Then $\text{(fof)x}$ is:
AnswerWe are given that$, f : [0, 1] \rightarrow [0, 1]$ be defined by $\text{f}(\text{x})=\begin{cases}\text{x, if x is rational}1-\text{x, if x is irrational}\end{cases}\}$
Now, $(\text{fof})\text{x} = \text{f}(\text{f(x)})$
$=\text{x}$
View full question & answer→MCQ 2041 Mark
If $f(x)=\left(a x^2-b\right)^3$, then the function $g$ such that $f\{g(x)\}=g\{f(x)\}$ is given by:
- A
$\text{g(x)}=\Big(\frac{\text{b}-\text{x}^\frac{1}{3}}{\text{a}}\Big)^\frac{1}{2}$
- B
$\text{g(x)}=\frac{1}{(\text{ax}^2+\text{b})^3}$
- C
$\text{g(x)}=(\text{ax}^2+\text{b})^\frac{1}{3}$
- ✓
$\text{g(x)}=\Big(\frac{\text{x}^\frac{1}{3}+\text{b}}{\text{a}}\Big)^\frac{1}{2}$
AnswerCorrect option: D. $\text{g(x)}=\Big(\frac{\text{x}^\frac{1}{3}+\text{b}}{\text{a}}\Big)^\frac{1}{2}$
View full question & answer→MCQ 2051 Mark
Find the identity element in the set $I^{+}$ of all positive integers defined by $a \times b=a+b$ for $a \ | | \ a, b \in I^{+}$.
View full question & answer→MCQ 2061 Mark
The function $f : R \rightarrow R$ defined by $f(x) = 3 – 4x$ is:
- ✓
- B
- C
None one$-$one.
- D
None one$-$onto.
View full question & answer→MCQ 2071 Mark
If $f: A \rightarrow B$ given by $3^{f(x)}+2^{-x}=4$ is a bijection, then
- A
$\text{A}=\{\text{x}\in\text{R}:-1<\text{x}<\infty\},$
$\text{B}=\{\text{x}\in\text{R}:2<\text{x}<4\}$
- B
$\text{A}=\{\text{x}\in\text{R}:-3<\text{x}<\infty\},$
$\text{B}=\{\text{x}\in\text{R}:2<\text{x}<4\}$
- C
$\text{A}=\{\text{x}\in\text{R}:-2<\text{x}<\infty\},$
$\text{B}=\{\text{x}\in\text{R}:2<\text{x}<4\}$
- ✓
Answer$\text{f}:\text{A}\rightarrow\text{B}$
$3^\text{f(x)}+2^{-\text{x}}=4$
$\Rightarrow\ 3^{\text{f(x)}}=4-2^{-\text{x}}$
Taking $\log$ on both the sides,
$\text{f(x)}\log3=\log(4-2^{-\text{x}})$
$\Rightarrow\ \text{f(x)}=\frac{\log(4-2^{-\text{x}})}{\log3}$
Logaritmic function will only be defined if $4-2^{-\text{x}}>0$
$\Rightarrow\ 4>2^{-\text{x}}$
$\Rightarrow\ 2^2>2^{-\text{x}}$
$\Rightarrow\ 2>-\text{x}$
$\Rightarrow-2<\text{x}$
$\Rightarrow\ \text{x}\in(-2,\infty)$
That means $\text{A}=\{\text{x}\in\text{R}:-2<\text{x}<\infty\}$
As we know that, $\text{f(x)}=\frac{\log(4-2^{-\text{x}})}{\log3}$
We take $\text{x}=0\in(-2,\infty)$
$\Rightarrow f(x) = 1$ which does not belong to any of the options.
View full question & answer→MCQ 2081 Mark
If $\text{g(f(x))}=|\sin\text{x}|$ and $\text{f(g(x))}=(\sin\sqrt{\text{x}})^2,$ then
- ✓
$\text{f(x)}=\sin^2\text{x},\ \text{g(x)}=\sqrt{\text{x}}$
- B
$\text{f(x)}=\sin\text{x},\ \text{g(x)}=|\text{x}|$
- C
$\text{f(x)}=\text{x}^2,\ \text{g(x)}=\sin\sqrt{\text{x}}$
- D
$f$ and $g$ can not be determined.
AnswerCorrect option: A. $\text{f(x)}=\sin^2\text{x},\ \text{g(x)}=\sqrt{\text{x}}$
If we solve it by the trial$-$and$-$error method, we can see that $(a)$ satisfies the given condition.
From $(a):$
$\text{f(x)}=\sin^2\text{x}$ and $\text{g(x)}=\sqrt{\text{x}}$
$\Rightarrow\ \text{f(g(x))}=\text{f}(\sqrt{\text{x}})=\sin^2\sqrt{\text{x}}$
$=(\sin\sqrt{\text{x}})^2$
View full question & answer→MCQ 2091 Mark
Consider the function $f$ in $\text{A}=\text{R}-\{\frac{2}{3}\}$ defiend as $\text{f(x)}=\frac{4\text{x}+3}{6\text{x}-4}.$ Find $f^{-1}$
- ✓
$\frac{3+4\text{x}}{6\text{x}-4}$
- B
$\frac{6\text{x}-4}{3+4\text{x}}$
- C
$\frac{3-4\text{x}}{6\text{x}-4}$
- D
$\frac{9+2\text{x}}{6\text{x}-4}$
AnswerCorrect option: A. $\frac{3+4\text{x}}{6\text{x}-4}$
View full question & answer→MCQ 2101 Mark
Mark the correct alternative in the following question for the binary operation $*$ on $Z$ defined by $a^ * b = a + b + 1,$ the identity element is:
AnswerWe have,
$a^ * b = a + b + 1$
Let e be the identity element of $*.$ Then,
$a^ * e = a = e^ * a$
$a + e + 1 = a$
$e = a - a - 1$
$e = -1$
View full question & answer→MCQ 2111 Mark
Find which of the binary operations are commutative and which are associative. Consider a binary operation * on N defined as $a^* b=a^3+b^3$. Choose the correct answer.
- A
Is * both associative and commutative?
- ✓
Is * commutative but not associative?
- C
Is * commutative but not associative?
- D
Is * neither commutative nor associative?
AnswerCorrect option: B. Is * commutative but not associative?
$a^* b=a^3+b^3=b^3+a^3=b * a$
$\therefore$ The operation is commutative.
Again, $(a * b) * c=a *\left(a^3+b^3\right)=a^3\left(a^3+b^3\right)^3$
And $(a * b) * c=\left(a^3+b^3\right) * c=\left(a^3+b^3\right)^3+c^3 \neq a *(b * c)$
$\therefore$ The operation * is not associative.
Therefore, option (B) is correct.
b. Is * commutative but not associative?
View full question & answer→MCQ 2121 Mark
Let $A = \{2, 3, 4, 5, ..., 17, 18\}.$ Let $'\simeq\ '$ be the equivalence relation on $A \times A,$ cartesian product of $A$ with itself, defined by $(\text{a, b})\simeq(\text{c, d)}$ if $ad = bc.$ Then, the number of ordered pairs of the equivalence class of $(3, 2)$ is:
AnswerThe ordered pairs of the equivalence class of $(3, 2)$ are $\{(3, 2), (6, 4), (9, 6), (12, 8), (15, 10), (18, 12)\}.$
We observe that these are $6$ pairs.
View full question & answer→MCQ 2131 Mark
The number of bijective functions from set $A$ to itself when $A$ contains $106$ elements is:
- A
$106$
- B
$(106)^2$
- ✓
$106!$
- D
$2^{106}$
AnswerCorrect option: C. $106!$
View full question & answer→MCQ 2141 Mark
Let $\times$ be the binary operation on $N$ given by $a \times b = \text{HCF} (a, b)$ where$, a, b \in N.$ Find the value of $22 \times 4.$
View full question & answer→MCQ 2151 Mark
A relation $R$ is defined from $\{2, 3, 4, 5\}$ to $\{3, 6, 7, 10\}$ by $: xRy \Leftrightarrow x$ is relatively prime to $y.$ Then, domain of $R$ is:
- A
$\{2, 3, 5\}$
- B
$\{3, 5\}$
- C
$\{2, 3, 4\}$
- ✓
$\{2, 3, 4, 5\}$
AnswerCorrect option: D. $\{2, 3, 4, 5\}$
Given that relation $R$ is defined from $\{2, 3, 4, 5\}$ to $\{3, 6, 7, 10\}$ by $:\ xRy \Leftrightarrow x$ is relatively prime to $y. R$ can be written as,
$\{(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)\}$
Here we can see that domain means $x$ element which is $2\leq\text{x}\leq5.$
Hence$, \{2, 3, 4, 5\}$
View full question & answer→MCQ 2161 Mark
Let $A = \{1, 2, 3\}. $Then, the number of relations containing $(1, 2)$ and $(1, 3)$ which are reflexive and symmetric but not transitive is:
AnswerGiven that $A = \{1, 2, 3\}$
To find the number of relations containing $(1, 2)$ and $(1, 3)$
then $R$ can be written as $\{(1, 2), (1, 3), (1, 1), (2, 2), (3, 3), (2, 1), (3, 1)\}$
Here, we can see that
$(3, 1)$ and $(1, 2)$
$ \Rightarrow (3, 2)$ which is not belongs to $R.$
The number of relations containing $(1, 2)$ and $(1, 3)$
Which are reflexive and symmetric but not transitive is $1.$
View full question & answer→MCQ 2171 Mark
Let $\text{f}:(2,\infty)\rightarrow\ \text{X}$ be defined by $f(x)=4 x-x^2$. Then, $f$ is invertible if $X=$
- A
$(2,\infty)$
- B
$(-\infty,2)$
- ✓
$(-\infty,4)$
- D
$(4,\infty)$
AnswerCorrect option: C. $(-\infty,4)$
Since $f$ is invertible, range of $f = co-$domain of $f = X$
So, we need to find the range of f to find $X.$
For finding the range, let
$f(x) = y$
$\Rightarrow 4 x-x^2=y$
$\Rightarrow x^2-4 x=-y$
$\Rightarrow x^2-4 x+4=4-y$
$\Rightarrow(x-2)^2=4-y$
$\Rightarrow\ \text{x}-2=\pm\sqrt{4-\text{y}}$
$\Rightarrow\ \text{x}=2\pm\sqrt{4-\text{y}}$
This is defined only when
$4-\text{y}\geq0$
$\Rightarrow\ \text{y}\leq4$
$X =$ Range of $\text{f}=(-\infty,4)$
View full question & answer→MCQ 2181 Mark
Consider the binary operation $*$ defined on $Q − \{1\}$ by the rule $a^ * b = a + b − ab$ for all $a, b \in Q − \{1\}.$ The identity element in $Q − \{1\}$ is:
- ✓
$0$
- B
$1$
- C
$\frac{1}2$
- D
$-1$
AnswerLet e be the identity element in $Q - \{1\}$ with respect to $*$ such that
$a^ * e = a = e^ * a, \forall\text{ a}\in\text{Q}-\{-1\}$
$a^ * e = a$ and $e^ * a = a, \forall\text{ a}\in\text{Q}-\{-1\}$
Then,
$a + e - ae = a$ and $e + a - ea = a, \forall\text{ a}\in\text{Q}-\{-1\}$
$e(1 - a) = 0, \forall\text{ a}\in\text{Q}-\{-1\}$
$\text{e}=0\in\text{Q}-\{-1\}$
$[\because\text{ a}\neq1]$
Thus$, 0$ is the identity element in $Q - \{1\}$ with respect to $*.$
View full question & answer→MCQ 2191 Mark
The function $f: R \rightarrow R$ defined as $f(x)=x^3$ is
AnswerLet $x_1, x_2 \in R$ be such that $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow x_1^3=x_2^3 \Rightarrow x_1=x_2 \Rightarrow f$ is one-one.
Let $f(x)=x^3=y$ for some arbitrary element $y \in R \Rightarrow x=y^{1 / 3}$
$\Rightarrow f\left(y^{1 / 3}\right)=y$
Every image $y \in R$ has a unique pre-image in $R$.
$\Rightarrow f$ is onto
$\therefore \quad f$ is one-one and onto.
View full question & answer→MCQ 2201 Mark
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
AnswerAs every pre-image $x \in A$, has a unique image $y \in B$.
$\Rightarrow \quad f$ is injective function.
View full question & answer→MCQ 2211 Mark
Let the relation $R$ in the set $A=\{x \in Z: 0 \leq x \leq 12\}$, given by $R=\{(a, b):|a-b|$ is a multiple of 4. $\}$ Then [1], the equivalence class containing 1 , is
- A
$\{1,5,9\}$
- B
$\{0,1,2,5\}$
- C
$\phi$
- D
$A$
AnswerWe have, $R=\{(a, b):|a-b|$ is a multiple of 4$\}$
$\therefore \quad$ The set of elements related to 1 is $\{1,5,9\}$.
So, equivalence class for $[1]$ is $\{1,5,9\}$
View full question & answer→MCQ 2221 Mark
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
$(3,3)$
AnswerWe have, $(1,2) \in R$ but $(2,1) \notin R$
So, $(1,2)$ should be removed from $R$ to make it an equivalence relation.
View full question & answer→MCQ 2231 Mark
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
AnswerGiven, $f(x)=\left\{\begin{array}{cl}\frac{n+1}{2}, & \text { if } n \text { is odd } \\ \frac{n}{2}, & \text { if } n \text { is even }\end{array}\right.$
Now, $f(1)=\frac{1+1}{2}=1, f(2)=\frac{2}{2}=1$
$\Rightarrow f(1)=f(2)$ but $1 \neq 2 \therefore f$ is not one-one.
But $f$ is onto $(\because$ range of $f$ is $N$.)
View full question & answer→MCQ 2241 Mark
Let $f: R \rightarrow R$ be defined by $f(x)=1 / x$, for all $x \in R$, Then, $f$ is
AnswerGiven $f(x)=\frac{1}{x}$, for all $x \in R$ At $x=0 \in R, f(x)$ is not defined.
View full question & answer→MCQ 2251 Mark
The number of functions defined from $\{1,2,3,4,5\} \rightarrow\{a, b\}$ which are one-one is
Answer$\because f: X \rightarrow Y$ is one-one, if different element of $X$ have different image in $Y$ under $f$. But here, no such situation is possible.
View full question & answer→MCQ 2261 Mark
The function $f: R \rightarrow R$ defined by $f(x)=4+3 \cos x$ is
AnswerWe have, $f(x)=4+3 \cos x, \forall x \in R$
At $x=\frac{\pi}{2}, f\left(\frac{\pi}{2}\right)=4+3 \cos \frac{\pi}{2}=4 \Rightarrow f\left(-\frac{\pi}{2}\right)=4+3 \cos \left(-\frac{\pi}{2}\right)=4$
Since, $f\left(\frac{\pi}{2}\right)=f\left(-\frac{\pi}{2}\right)$, But $\frac{\pi}{2} \neq-\frac{\pi}{2}$
Therefore, $f$ is not one-one.
As $-1 \leq \cos x \leq 1, \forall x \in R \Rightarrow 1 \leq 4+3 \cos x \leq 7, \forall x \in R$
$\Rightarrow f(x) \in[1,7]$, where $[1,7]$ is subset of $R . \therefore f$ is not onto.
View full question & answer→MCQ 2271 Mark
The number of equivalence relations in the set $\{1,2,3\}$ containing the elements $(1,2)$ and $(2,1)$ is
Answer5. (c) : Equivalence relations in the set $\{1,2,3\}$ containing the elements $(1,2)$ and $(2,1)$ are
$R_1=\{(1,2),(2,1),(1,1),(2,2),(3,3)\}$
$R_2=\{(1,2),(2,1),(1,3),(3,1)(2,3),(3,2),(1,1),(2,2),(3,3)\}$
$\therefore \quad$ Number of equivalence relations is 2 .
View full question & answer→MCQ 2281 Mark
A relation $R$ is defined on $N$. Which of the following is the reflexive relation?
- A
$R=\{(x, y): x>y, x, y \in N\}$
- B
$R=\{(x, y): x+y=10, x, y \in N\}$
- C
$R=\{(x, y): x y$ is the square number, $x, y \in N\}$
- D
$R=\{(x, y): x+4 y=10, x, y \in N\}$
AnswerConsider, $R=\{(x, y)$ : xy is the square number, $x, y \in N\}$
As, $x x=x^2$, which is the square of natural number $x$.
$\Rightarrow \quad(x, x) \in R$. So, $R$ is reflexive.
View full question & answer→MCQ 2291 Mark
A relation $R$ is defined on $Z$ as $a R b$ if and only if $a^2-7 a b+6 b^2=0$. Then, $R$ is
- A
- B
symmetric but not reflexive
- C
transitive but not reflexive
- D
reflexive but not symmetric
AnswerGiven, $a R b, a, b \in Z$
Reflexive: For $a \in Z$, we have
$a^2-7 a \cdot a+6 a^2=a^2-7 a^2+6 a^2=0 \Rightarrow(a, a) \in R$
$\therefore \quad$ Relation is reflexive.
Symmetric: Since, $(6,1) \in R$
As, $6^2-7 \times 6 \times 1+6 \times 1^2=36-42+6=0$
But $(1,6) \notin R . \therefore$ Relation is not symmetric.
View full question & answer→MCQ 2301 Mark
A function $f: R \rightarrow R$ defined as $f(x)=x^2-4 x+5$ is:
- A
injective but not surjective
- B
surjective but not injective
- C
both injective and surjective
- ✓
neither injective nor surjective
AnswerCorrect option: D. neither injective nor surjective
Given, $f(x)=x^2-4 x+5$
Here $f(0)=f(4)=5$
Hence, $f(x)$ is not one$-$one.
To check whether the function is onto or not,
we have to find range of function.
$\text { Let } y=x^2-4 x+5$
$\Rightarrow x^2-4 x+5-y=0$
$\therefore D=(4)^2-4(1)(5-y) \geq 0 \forall x \in R$
$\Rightarrow 16-20+4 y \geq 0$
$\Rightarrow 4 y-4 \geq 0$
$\Rightarrow 4(y-1) \geq 0$
$\Rightarrow y \geq 1$
Hence, range $=(1, \infty)$
Here, Co$-$domain $\neq$ Range
So, $f(x)$ is not onto.
View full question & answer→MCQ 2311 Mark
Let $f: R_{+} \rightarrow[-5, \infty)$ be defined as $f(x)=9 x^2+6 x-5$, where $R_{+}$is the set of all non-negative real numbers. Then, $f$ is :
AnswerCorrect option: A. one$-$one
$\text {Let } f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow 9 x_1^2+6 x_1-5=9 x_2^2+6 x_2-5$
$\Rightarrow 9\left(x_1^2-x_2^2\right)+6\left(x_1-x_2\right)=0$
$\Rightarrow \quad\left(x_1-x_2\right)\left[9\left(x_1+x_2\right)+6\right]=0$
$\Rightarrow x_1-x_2=0 \text { as } 9\left(x_1+x_2\right)+6>0$
$\Rightarrow x_1=x_2 \quad\left(\because x_1, x_2 \text { are non-negative real numbers }\right)$Thus, $f$ is one$-$one.
Let $y \in[-5, \infty)$ be such that $f(x)=y$
Now, $f(x)=9 x^2+6 x-5=9 x^2+6 x+1-6=(3 x+1)^2-6$
$\Rightarrow y+6=(3 x+1)^2$
$\Rightarrow 3 x+1=\sqrt{y+6} \Rightarrow x=\frac{-1+\sqrt{y+6}}{3}$
$\therefore f\left(\frac{-1+\sqrt{y+6}}{3}\right)=y$
$\therefore f(x)$ is onto.
So, the given function is bijective.
View full question & answer→MCQ 2321 Mark
Let $R$ be a relation in the set $N$ given by $R=\{(a, b): a=b-2, b>6\}$. Then
- A
$(8,7) \in R$
- B
$(6,8) \in R$
- C
$(3,8) \in R$
- D
$(2,4) \in R$
AnswerGiven, $R=\{(a, b): a=b-2, b>6\}$
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 2331 Mark
Let $A=\{3,5\}$. Then number of reflexive relations on $A$ is
AnswerTotal number of reflexive relations on a set having $n$ number of elements $=2^{n^2-n}$
Here, $n=2$
$\therefore \quad$ Required number of reflexive relations $=2^{2^2-2}$ $=2^{4-2}=2^2=4$
View full question & answer→MCQ 2341 Mark
The relation $R$ in the set $\{1,2,3\}$ given by $R=\{(1,2)$, $(2,1),(1,1)\}$ is
- A
symmetric and transitive, but not reflexive
- B
reflexive and symmetric, but not transitive
- C
symmetric, but neither reflexive nor transitive
- D
AnswerGiven, $R=\{(1,2),(2,1),(1,1)\}$ is a relation on set $\{1,2,3\}$
Reflexive : Clearly $(2,2),(3,3) \notin R$
$\therefore \quad R$ is not a reflexive relation.
Symmetric: Now, $(1,2) \in R$ and $(2,1) \in R \therefore R$ is symmetric.
Transitive: Now, $(2,1) \in R$ and $(1,2) \in R$ but $(2,2) \notin R$
$\therefore \quad R$ is not transitive relation.
$R$ is symmetric, but neither reflexive nor transitive.
View full question & answer→MCQ 2351 Mark
Let $A=\{1,3,5\}$. Then the number of equivalence relations in $A$ containing $(1,3)$ is
AnswerEquivalence relations in the set containing the element $(1,3)$ are
$R_1=\{(1,1),(3,3),(1,3),(3,1),(5,5)\}$
$R_2=\{(1,1),(3,3),(5,5),(1,5),(5,1),(3,5),(5,3),(1,3),(3,1)\}$
$\therefore $ There are $2$ possible equivalence relations.
View full question & answer→MCQ 2361 Mark
Let $R$ be the relation defined on $N \times N$ by the rule $(a, b) R(c, d) \Leftrightarrow a+d=b+c$, then $R$ is
AnswerHere, $(a, b) R(a, b)$ for all $(a, b) \in N \times N$
$\Rightarrow R$ is reflexive.
$(\because a+b=b+a)$
Let $(a, b) R(c, d)$
$\Rightarrow a+d=b+c$
$\Rightarrow d+a=c+b$
$\Rightarrow c+b=d+a$
$\Rightarrow (c, d) R(a, b)$
$\therefore R$ is symmetric.
Let $(a, b) R(c, d)$ and $(c, d) R(e, f)$
$\Rightarrow a+d=b+c$ and $c+f=d+e$
$\Rightarrow(a+d)+(c+f)=(b+c)+(d+e)$
$\Rightarrow a+f=b+e$
$\Rightarrow(a, b) R(e, f)$
$\therefore R$ is transitive
View full question & answer→MCQ 2371 Mark
The diagram given below shows that
- A
$f$ is a function from $A$ to $B$
- B
$f$ is a one-one function from $A$ to $B$
- C
$f$ is an onto function from $A$ to $B$
- ✓
$f$ is not a function
AnswerCorrect option: D. $f$ is not a function
(d) : As $f(a)$ is not unique, thus $f$ is not a function.
View full question & answer→MCQ 2381 Mark
Consider the following statements on a set $A=\{1,2,3\}$ :
(i) $\quad R=\{(1,1),(2,2)\}$ is a reflexive relation on $A$.
(ii) $R=\{(3,3)\}$ is symmetric and transitive but not a reflexive relation on $A$.
Which of the statements given above is/are correct?
Answer(b) : (i) is not correct as $(3,3) \notin R$ and hence $R$ is not reflexive.
(ii) is correct as the relation $R=\{(3,3)\}$ is symmetric and transitive but not reflexive as $(1,1) \notin R,(2,2) \notin R$.
View full question & answer→MCQ 2391 Mark
The signum function, $f: R \rightarrow R$ is given by $f(x)=\left\{\begin{array}{ll}1, & x>0 \\ 0, & x=0 \\ -1, & x<0\end{array}\right.$ is
AnswerWe have, $f(1)=f(2)$
$=f(3)=1, f(0)=0$
$f(-1)=f(-2)=f(-3)=-1$
Hence, function $f$ is not one$-$one.
View full question & answer→MCQ 2401 Mark
Let $f: R \rightarrow R$ be defined by the smallest integer function $f(x)=[x]$, then $f$ is
Answer(d) : Let $f: A \rightarrow B$ such that $f(x)=[x]$.
We have, $[1.4]=[1.6]=2$
Here, two elements in $A, 1.4$ and 1.6 have the same image i.e., 2 in $B$.
Thus, $f(x)=[x]$ is a not one-one function.
Here, codomain is not equal to range of function, so, function is not onto. View full question & answer→MCQ 2411 Mark
The mapping $f: N \rightarrow N$ given by $f(n)=1+n^2$, $n \in N$ where $N$ is the set of natural numbers, is
Answer(c) : Since, $f(n)=1+n^2$
For one-one, $1+n_1^2=1+n_2^2, n_1, n_2 \in N$
$
\Rightarrow n_1^2-n_2^2=0 \Rightarrow n_1=n_2 \quad\left(\because n_1+n_2 \neq 0\right)
$
$\therefore f(n)$ is one-one.
Clearly, $f(n)$ is not onto.
View full question & answer→MCQ 2421 Mark
Let $X=\{0,1,2,3\}$ and $Y=\{-1,0,1,4,9\}$ and a function $f: X \rightarrow Y$ defined by $y=x^2$, is
Answer(b) : $y(0)=0, y(1)=1, y(2)=4, y(3)=9$. No two different values of $x$ (where $x \in X$ ) gives same image. Also -1 is element of set $Y$, which does not have its pre-image in set $X$. So, function is one-one into.
View full question & answer→MCQ 2431 Mark
Let $X=\{-1,0,1\}, Y=\{0,2\}$ and afunction $f: X \rightarrow Y$ defined by $y=2 x^4$, is
Answer(b): We have, $y=2 x^4$
$
\therefore \quad y(-1)=y(1)=2, y(0)=0
$
Here, we see that for two different values of $x$, we will get a same image so, function is not one-one and no element of $y$ is left, which do not have pre-image. So, function is onto.
View full question & answer→MCQ 2441 Mark
Let $R$ be an equivalence relation on a finite set $A$ having $n$ elements. Then, the number of ordered pairs in $R$ is
AnswerCorrect option: B. Greater than or equal to $n$
(b) : As $R$ is an equivalence relation on set $A$.
Hence, $R$ has atleast $n$ ordered pairs.
View full question & answer→MCQ 2451 Mark
How many reflexive relations are possible in a set $A$ whose $n(A)=3$ ?
Answer(a) : Number of reflexive relations on a set having $n$ elements $=2^{n(n-1)}$
So, required number of reflexive relations $=2^{3(3-1)}=2^6$
View full question & answer→MCQ 2461 Mark
Which one of the following relations on $R$ is an equivalence relation?
- ✓
$a R_1 b \Leftrightarrow|a|=|b|$
- B
$a R_2 b \Leftrightarrow a \geq b$
- C
$a R_3 b \Leftrightarrow a$ divides $b$
- D
$a R_4 b \Leftrightarrow a
AnswerCorrect option: A. $a R_1 b \Leftrightarrow|a|=|b|$
(a) : (i) Reflexive : $a \in R, a R_1 a \Rightarrow|a|=|a|$
(ii) Symmetric : $a, b \in R$
$
a R_1 b \Rightarrow|a|=|b| \Rightarrow|b|=|a| \Rightarrow b R_1 a
$
(iii) Transitive $: a, b, c \in R$
$
a R_1 b \Rightarrow|a|=|b|, b R_1 c \Rightarrow|b|=|c| \text {. So, }|a|=|c| \Rightarrow a R_1 c
$
$\Rightarrow R_1$ is an equivalence relation on $R$.
View full question & answer→MCQ 2471 Mark
Let $A=\{1,2,3,4\}$ and $R$ be a relation in $A$ given by $R=\{(1,1),(2,2),(3,3),(4,4),(1,2)$, $(2,1),(3,1)\}$. Then, $R$ is
Answer(a) : Reflexive: $(1,1),(2,2),(3,3),(4,4) \in R$;
$\therefore \quad R$ is reflexive.
View full question & answer→MCQ 2481 Mark
If $R$ and $R^{\prime}$ are symmetric relations (not disjoint) on a set $A$, then the relation $R \cap R^{\prime}$ is
Answer(b) : Given $R$ and $R^{\prime}$ are not disjoint, so there is atleast one ordered pair, say, $(a, b) \in R \cap R^{\prime}$.
$
\Rightarrow \quad(a, b) \in R \text { and }(a, b) \in R^{\prime}
$
As $R$ and $R^{\prime}$ are symmetric relations, we get $(b, a) \in R$ and $(b, a) \in R^{\prime} \Rightarrow(b, a) \in R \cap R^{\prime}$
Hence, $R \cap R^{\prime}$ is symmetric.
View full question & answer→MCQ 2491 Mark
Let $R$ be the relation on the set of all real numbers defined by $a R b$ iff $|a-b| \leq 1$. Then, $R$ is
Answer(a) : Reflexive : $|a-a|=0<1 \quad \therefore a R a \forall a \in R$
$\therefore \quad R$ is reflexive.
Symmetric : $a R b \Rightarrow|a-b| \leq 1 \Rightarrow|b-a| \leq 1 \Rightarrow b R a$
$\therefore \quad R$ is symmetric.
Transitive : $1 R 2$ and $2 R 3$ but $1 \not R 3[\because|1-3|=2>1]$
$\therefore \quad R$ is not transitive.
View full question & answer→MCQ 2501 Mark
Let $L$ denote the set of all straight lines in a plane. Let a relation $R$ be defined by $\alpha R \beta \Leftrightarrow \alpha \perp \beta, \alpha, \beta \in L$. Then, $R$ is
Answer(b) : Given, $\alpha R \beta \Leftrightarrow \alpha \perp \beta \Leftrightarrow \beta \perp \alpha \Rightarrow \beta R \alpha$ Hence, $R$ is symmetric.
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