MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 12

Let

$A = \left\{ {1,2,3} \right\}$ and

$R = \left\{ {\left( {1,1} \right),\left( {1,3} \right),\left( {3,1} \right),\left( {2,2} \right),\left( {2,1} \right),\left( {3,3} \right)} \right\}$ , then the relation

$R$ and

$A$ is

0%
reflexive
0%
symmetric
0%
transitive
0%
equivalence

Explanation

$A=\{ 1,2,3\} \\ R=\{ (1,1),(1,3),(3,1),(2,2),(2,1),(3,3)\}$

The relation

$R$ on

$A$ is reflexive.

Since

$(1,1),(2,2),(3,3)$ exist.

Also

$(1,3),(3,1)$ exist but

$(1,2),(2,1)$ does not . Hence it is not symmetric.

$A$ and

$B$ are independent event such that

$P(A \cap B')=\dfrac {3}{25}$ and

$P(A' \cap B)=\dfrac {8}{25}$ , then

$P(A)=$

0%
$1/5$
0%
$3/8$
0%
$2/5$
0%
$4/5$

Explanation

$A$ &

$B$ are independent

$P(A\cap B)=\dfrac {3}{25}$ and

$P(A'\cap B)=\dfrac {8}{25}\quad P(A)=(?)$

$\rightarrow \ P(A)+P(B)=1---(i)$ (

$A$ &

$B$ are independent )

$\rightarrow \ P(A\cap B')=P(A)-P(A\cap B)$

$P(A)-P(A\cap B)=\dfrac {3}{25}----(ii)$

$\rightarrow \ P(A' \cap B)=P(B)-P(A\cap B)$

$P(B)-P(A\cap B)=\dfrac {8}{25}-----(iii)$

$\rightarrow \$ solving equation

$(ii)$ and

$(iii)$

$\dfrac {\,\,\, P\left( A \right) -P\left( A\cap B \right) =\dfrac { 3 }{ 25 } \\\,\,\, P\left( B \right) -P\left( A\cap B \right) =\dfrac { 8 }{ 25 } \\ -\quad \,\,\,\,\,\,\,+\quad \quad\quad\quad- }{ P\left( A \right) -P\left( B \right) =\dfrac { 3 }{ 25 } -\dfrac { 8 }{ 25 } \\ P\left( A \right) -P\left( B \right) =\dfrac { 3-8 }{ 25 } }$

$=\dfrac {-5}{25}$

$\rightarrow \ P(A)-P(B)=\dfrac {-1}{5}$

$P(A)-(1-P(A))=\dfrac {-1}{5}$

$P(A)-I+P(A)=\dfrac {-1}{5}$

$\therefore \ 2P(A)=\dfrac {-1}{5}+1$

$\therefore \ 2P(A)=\dfrac {4}{5}$

$\therefore \ P(A)=\dfrac {4}{5\times 2}$

$\therefore \ P(A)=\dfrac {2}{5}$

Let

$S=\{x \in R : x\geq 0$ and

$2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0\}$ . Then S?

0%
Contains exactly one element
0%
Contains exactly two elements
0%
Contains exactly four elements
0%
Is an empty set

The domain of

$\dfrac{1}{\sqrt{x-x^{2}}}+\sqrt{3x-1-2x^{2}}$ is

0%
$(\dfrac{1}{2},11)$
0%
$(\dfrac{1}{2},31)$
0%
$(\dfrac{1}{2},17)$
0%
$(\dfrac{1}{2},41)$

Explanation

$+(x) = \dfrac {1}{\sqrt {x (1 - x)}} + \sqrt {3x - 1 - 2x^{2}}$

Domain of

$f(x)$ : -

$x (1 - x) > 0$ and

$3x - 1 - 2x^{2} \geq 0$

$2x^{2} - 3x + 1 \leq 0$

$2x (x - 1) - 1 (x - 1) \leq 0$

$(2x - 1) (x - 1)\leq 0$

Thus taking the intersection of above

$2$ .

$x \leftarrow [\dfrac {1}{2}, 1)$ .

1349793_1241815_ans_95d98ed54cd643f6bcf8cb03f3038873.png

The set of values of '

$b$ ' for which the origin and the point

$(1,1)$ lie on the same side of the straight line

${ a }^{ 2 } x + ab y + 1 = 0 \;\;\forall \;\; a\in R,\; b > 0$ are:

0%
$b\in (2,4)$
0%
$b\in (0,2)\quad$
0%
$b\in (0,4)\quad$
0%
$b\in (2,6)$

Total number of equivalence relation defined in the set

$s=\{a, b, c\}$ is

0%
$5$
0%
$31$
0%
$2^3$
0%
$3^3$

Explanation

By pascal's triangle

$1\\ 1\quad \quad 1\\ 1\quad \quad 3\quad \quad 1\\ 1\quad \quad 7\quad \quad 6\quad \quad 1\\ 1\quad \quad 15\quad 25\quad 10\quad \quad 1\\ 1\quad \quad 31\quad 90\quad 65\quad \quad 15\quad \quad 1$

Here no. of elements is

$3$ .

$\therefore$ No. of equivalent relations

$=1+3+1=5$ .

Let A be a finite set and n(A) = 5, then the number of relations which are not symmetric is _____________________.

0%
$2^{25}$
0%
$2^{15}$
0%
$2^{25} - 2^{10}$
0%
$2^{25} - 2^{15}$

$x\ \epsilon \ R$ , the number of solutions of

$\sqrt{2x+1}-\sqrt{2x-1}-1$ is

0%
$1$
0%
$2$
0%
$4$
0%
$infinite$

Explanation

We have,

$x\in R$

Then,

$\sqrt{2x+1}-\sqrt{2x-1}-1$

Now,

Put $x=1,2,3,4.........$

So,

$\sqrt{2\left( 1 \right)+1}-\sqrt{2\left( 1 \right)-1}-1=\sqrt{3}-2$

$\sqrt{2\left( 2 \right)+1}-\sqrt{2\left( 2 \right)-1}-1=\sqrt{5}-\sqrt{3}-1$

And so on….

Then,

Infinite number of solution.

Hence, this is the answer.

$\sqrt { \log _ { 4 } \left( \log _ { 2 } \left( \log _ { 2 } \left( x ^ { 2 } - 2 x + a \right) \right) \right. } )$ is defined

$\forall x \in R$ , then

0%
$[ 10 , \infty )$
0%
$[ 9 , \infty )$
0%
$[ 2 , \infty )$
0%
$[ 5 , \infty )$

The domain of log

$\left( { tan }^{ -1 }x \right)$ is :-

0%
R
0%
${ R }^{ + }$
0%
$\left( 0,\infty \right)$
0%
$\left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right)$

Let

$T$ be the set of all triangle in the Euclidean plane, and let a relation

$R$ on

$T$ be defined as

$aRb$ if a is congruent to

$b\ \forall \ a, b\ \epsilon \ T$ . Then

$R$ is

0%
reflective but not transitive
0%
transitive but not symmetric
0%
equivalence
0%
none of these

Let S be the set of all real nos, then the relation

$R = \{ ( a , b ) : | a - b |$ is a multiple of 4

$\}$ then the relation.

0%
Reflexive but not symmetric
0%
Symmetric but not reflexive
0%
Reflexive only
0%
Equivalence

Which of the following is true for a relation :

$A\longrightarrow A$

0%
If R is reflexive then domain of R is A
0%
If R is symmetric then domain of R is A
0%
If R be an increasing function, then R is transitive
0%
If R is one one function then R is symmetric

Let

$A\equiv \left\{1,2,3,4\right\},\ B\equiv \left\{a,b,c\right\}$ , then number of function from

$A\rightarrow B$ , which are not onto is

0%
$9$
0%
$24$
0%
$45$
0%
$6$

Explanation

$A \equiv\{1,2,3,4\} \\$

$B \equiv\{a, b, c\}$

Total function

$(A \rightarrow B)$

$\text { Total onto function }(A \rightarrow B)$

$=4 c_{3} \times 3 ! \times 3$

$=\frac{4 !}{3 ! 1 !} \times 3 ! \times 3$

$=41 \times 3=72$

Hence, no. of

$f^{n}$ which are not onto

$=81-72 \\$

$=9$

If domain of

$y = f\left( x \right)$ is

$\left[ { - 3,\,\,2} \right]$ then domain of

$y = f\left( {\left| {\left[ x \right]} \right|} \right)$ is

0%
$[-3, 2]$
0%
$[-2, 3)$
0%
$[-3, 3]$
0%
$[-2, 3]$

Let

$R = \{ ( 3,3 ) , ( 6,6 ) , ( 9,9 ) , ( 6,12 ) , ( 3,9 ) , ( 3,12 ) , ( 3,6 ) \}$ be a relation on the set

$A=\{ 3,6,9,12\} .$ Then the relation

$R ^ { - 1 }$ is

0%
reflexive and transitive.
0%
not symmetric
0%
transitive
0%
all the above

Explanation

$\begin{array}{l} We\, have \\ R=\left\{ { \left( { 3,3 } \right) ,\left( { 6,6 } \right) \left( { 9,9 } \right) \left( { 12,12 } \right) ,\, \left( { 6,12 } \right) ,\left( { 3,12 } \right) ,\, \left( { 3,6 } \right) } \right\} \\ and, \\ A=\left\{ { 3,6,9,12 } \right\} \\ Then \\ The\, relation\, is\, reflexive\, and\, transitive\, only.\, because \\ \left\{ { \left( { 3,3 } \right) ,\left( { 6,6 } \right) \left( { 9,9 } \right) \left( { 12,12 } \right) } \right\} \, \, it\, \, is\, { { Re } }flexive \\ and,\, \left( { 6,12 } \right) ,\left( { 3,12 } \right) ,\, \left( { 3,6 } \right) \left[ { it\, is\, transitive. } \right] \\ \end{array}$

Hence, option

$A$ is the correct answer.

Consider the following relations :-

$R=\{ (x,y):x,y$ are real numbers and

$x =w y$ for some rational number

$w \}$ :

$S = \{ \left( \dfrac { m } { n } , \dfrac { p } { q } \right) : m , n , p$ and

$q$ are integers such that

${ n },{ q } \neq 0$ and

${ qm }={ pn }\}.$ Then :

0%
$R$ is an equivalence relation but $S$ is not an equivalence relation
0%
Neither $R$ nor $S$ is an equivalence relation
0%
$S$ is an equivalence relation but $R$ is not an equivalence relation
0%
$R$ and $S$ both are equivalence relations

Explanation

$x=w y$ for a function to be a reflexive function

$(a, a) \in R \quad \Rightarrow \quad a=\omega a$

$w=1$ (only)

$\therefore R$ is not a reflexive function.

$\begin{aligned} S:\left\{\left(\frac{m}{n}, \frac{p}{q}\right)\right.& m q=n p \\ \Rightarrow \frac{q}{n} &=\frac{p}{m} \Rightarrow \frac{m}{n}=\frac{p}{q} \end{aligned}$

$\left(\frac{a}{b}, \frac{a}{b}\right) \in S$ , then

$S$ will be reflexive function If

$S \in\left(\frac{a}{b}, \frac{c}{d}\right)$ and

$\in\left(\frac{c}{d}, \frac{a}{b}\right) \Rightarrow S$ is symmetric for a function to be transitive,

$\begin{aligned} S \in\left(\frac{1}{2}, \frac{2}{4}\right), & S \in\left(\frac{2}{4}, \frac{4}{8}\right) \\ s \in(a, b), S \in(b, c) & \Rightarrow S \in(a, c) \\ \therefore \frac{1}{2}=\frac{4}{8}=\frac{1}{2} \quad & \Rightarrow S \in\left(\frac{1}{2}, \frac{4}{8}\right) \\ & \therefore \text { S is transitive } \end{aligned}$

$\begin{array}{l} \text { Hence, } S \text { is an equivalence relation. } \\ \Rightarrow \text { (c) is the correct option } \end{array}$

$A=\left\{1,2,3\right\}$ , the number of symmetric relation in

$A$ is

0%
$64$
0%
$8$
0%
$324$
0%
$328$

Let

$S = \{ 1,2,3 , \ldots , 100 \} .$ The number of non-empty subsets

$A$ of

$S$ such that the product of elements in

$A$ is even is :-

0%
$2 ^ { 50 } \left( 2 ^ { 50 } - 1 \right)$
0%
$2 ^ { 100 } - 1$
0%
$2 ^ { 50 } - 1$
0%
$2 ^ { 50 } + 1$

Explanation

$\mathrm { S } = \{ 1,2,3 - \ldots - 100 \}$

= Total non empty subsets-subsets with product of element is odd

$= 2 ^ { 100 } - 1 - 1 \left[ \left( 2 ^ { 50 } - 1 \right) \right]$

$= 2 ^ { 100 } - 2 ^ { 50 }$

$= 2 ^ { 50 } \left( 2 ^ { 50 } - 1 \right)$

$A=\left\{ 2,3,5 \right\} ,B=\left\{ 4,6,8 \right\}$ , then the relation from

$A$ into

$B$ is

0%
$\left\{(2,4),(3,5),(5,6)\right\}$
0%
$\left\{(2,4),(5,8),(6,5)\right\}$
0%
$\left\{(2,4),(3,6),(5,6)\right\}$
0%
$\left\{(2,5),(3,6)\right\}$

If A= {a, b} then possible number of relation on the set A.

0%
$2$
0%
$4$
0%
$16$
0%
none of these

If n(A)=4, n(B)=3,

$n(A\times B\times C)=24,then\quad n(C)$ is equal to

0%
288
0%
1
0%
2
0%
12

Explanation

$n\left(A\right)=4\,\,n\left(B\right)=3\,\,n\left(A\times B\times C\right)=24$

We have

$n\left(A\times B\times C\right)=n\left(A\right)\times n\left(B\right)\times n\left(C\right)$

$\Rightarrow 24=4\times 3\times n\left(C\right)$

$\therefore n\left(C\right)=\dfrac{24}{12}=2$

If a set

$A$ has

$n$ elements then the number of relations defined on

$A$ is

0%
$2^{ \left( { n }^{ 2 } \right) }$
0%
$2^{ \left( { n }^{ 2 } \right) }-1$
0%
$2^{ n }$
0%
None of these.

$R$ is a relation on the set

$A=\left\{ 1,2,3,4,5,6,7,8,9 \right\}$ given by

$xRy\Leftrightarrow y=3x$ , then

$R=$

0%
$\left\{ \left( 3,1 \right) ,\left( 6,2 \right) ,\left( 8,2 \right) ,\left( 9,3 \right) \right\}$
0%
$\left\{ \left( 3,1 \right) ,\left( 6,2 \right) ,\left( 9,3 \right) \right\}$
0%
$\left\{ \left( 3,1 \right) ,\left( 2,6 \right) ,\left( 3,9 \right) \right\}$
0%
none of these

Explanation

Given,

$R$ is a relation on the set

$A=\left\{ 1,2,3,4,5,6,7,8,9 \right\}$ given by

$xRy\Leftrightarrow y=3x$ , then

$R=\{(1,3),(6,2),(9,3)\}$ .

Let A and B be two sets such that

$A\times B=\left\{ \left( a,1 \right) ,\left( b,3 \right) ,\left( a,3 \right) ,\left( b,1 \right) ,\left( a,2 \right) ,\left( b,2 \right) \right\} ,$ then

0%
$A=\left\{ 1,2,3 \right\}$ and $B=\left\{ a,b \right\}$
0%
$A=\left\{ a,b \right\}$ and $B=\left\{ 1,2,3 \right\}$
0%
$A=\left\{ 1,2,3 \right\}$ and $B\subset \left\{ a,b \right\}$
0%
$A\subset \left\{ a,b \right\}$ and $B\subset \left\{ 1,2,3 \right\}$

Explanation

$A$ is the first element in the cartesian product

$A\times B=\left\{a\,,b\,,a\,,b\,,a\,,b\right\}$

and

$B$ is the second element in the cartesian product

$A\times B=\left\{1,\,3\,,3\,,1\,,2\,,2\right\}$

$\therefore$ elements of

$A=\left\{a,b\right\}$ and

$B=\left\{1\,,2\,,3\right\}$

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

0%
{2,3,5}
0%
{3,5}
0%
{2,5}
0%
{2,3,4,5}

Explanation

Given:

From

$\left\{2,3,4,5\right\}$ to

$\left\{3,6,7,10\right\}$

$x$ is related to

$y$ iff

$x$ is relatively prime to

$y$

$2$ is relatively prime to

$3,7$

$3$ is relatively prime to

$7,10$

$4$ is relatively prime to

$3,7$

$5$ is relatively prime to

$3,6,7$

So, domain of

$R$ is

$\left\{2,3,4,5\right\}$

If the cardinality of a set

$A$ is

$4$ and that of a set

$B$ is

$3$ , then what is the cardinality of the set

$A\Delta B$ .

0%
$1$
0%
$5$
0%
$7$
0%
$Cannot\ be\ determined$

Explanation

$n(A\Delta B)=n(A)+n(B)-n(A\cap B)$

Here we don`t know the $n(A\cap B)$

So the answer cannot be Determined.

Let

$A$ be set of first ten natural numbers and

$R$ be a relation on A, defined by

$(x,y)\in R\Rightarrow x+2y=10$ , then domain of

$R$ is

0%
{1, 2, 3, ......,10}
0%
{2, 4, 6, 8}
0%
{1, 2, 3, 4}
0%
{2, 4, 6, 8, 10}

What type of a relation is "Less than" in the set of real numbers?

0%
only symmertric
0%
only transitive
0%
only reflexive
0%
equivalence relation

Explanation

$a\in\,R,\,a$ is not less than

$a$ .

$\Rightarrow\,$ The relation “less than” is not reflexive.

$a,b\in R,$ and

$a$ is less than

$b$ then

$b$ is not less than

$a$ .

$\Rightarrow\,$ The relation “less than” is not symmetric.

$a,b,c\in\,R,$ and

$a$ is less than

$b$ and

$b$ is less than

$c$ then

$a$ is less than

$c$ .

$\Rightarrow\,$ The relation “less than” is transitive.

$\Rightarrow\,$ The relation “less than” is not an equivalence relation.

Hence option

$(b)$ is correct.

For real numbers x and y, define xRy if

$x-y+\sqrt{2}$ is an irrational number. Then the relation R is?

0%
Not Reflexive
0%
Not Symmetric
0%
Not Transitive
0%
None of these

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 12 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 12 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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