MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 11

$x*y=\sqrt {\dfrac {(x+y)(y^{2}-12x)}{(x-2)(y-7)}}$ then what will be the value of

$5*9$ ?

0%
$7$
0%
$8$
0%
$10$
0%
$9$

Explanation

$x\ast y=\sqrt{\dfrac{(x+y)(y^{2}-12x)}{(x-2)(y-7)}}$

$5\ast 9=\sqrt{\dfrac{(5+9)(81-60)}{(3)(2)}}=\sqrt{\dfrac{(14)(21)}{(3)(2)}}=\sqrt{(7)(7)}$

$=\sqrt{49}$

$=7$

Let

$R$ be a reflexive relation in a finite set having

$n$ elements and let there be

$m$ ordered pairs in

$R$ . Then,

0%
$m \ge n$
0%
$m \le n$
0%
$m = n$
0%
None of these

Explanation

$\underset{(n\quad values\quad possible)}{\underset{\downarrow}{a R a}}$

$\underset{n\quad elements}{\underset{\downarrow}{a\in A}}$

Given m ordered pairs. [We know n values possible that is at least n values cut of m]

$\therefore m\geq n$ .

Find the domain of the function defined as

$f(x)=\dfrac{1}{1-x^2}$ .

0%
$x\epsilon R -\{1,-1\}$
0%
$x\epsilon R -\{0\}$
0%
$x\epsilon R -\{-1\}$
0%
$x\epsilon R -\{1\}$

Explanation

$f\left(x\right)=\dfrac{1}{1-x^{2}}$

For existance

$1-x^{2}\neq 0$

$\Rightarrow x\neq+1,-1$

$\Rightarrow x\in R-\left\{1,-1\right\}$

Let S be a non-empty set. Let R be a relation on P(S), defined as ARB

$\Leftrightarrow A\cap B\neq \phi$ . The relation R is?

0%
Reflexive
0%
Symmetric
0%
Transitive
0%
None of these

$f:\,R\, \to R\,$ is an even function which is differentiable on R and

${f^N}\left( \pi \right) = 1\,the\,{f^N}\left( { - \pi } \right)\,{\rm{is}}$

0%
-1
0%
0
0%
1
0%
2

Let

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

$B = \left\{ {4,\,5} \right\}$ . Consider a relation defined from set A to set B, then R is equal to

0%
A
0%
B
0%
$A \times B$
0%
$B \times A$

Explanation

A relation from set

$A$ to set

$B$ is called

$A\times B$ and have following elements

$A\times B=\left\{(a, 4),\ (a, 5),\ (b, 4),\ (b, 5),\ (c, 4),\ (c, 5)\right\}$

Total

$6$ elements

$C$ is correct

$A = \{2, 3, 5\}$ and

$B = \{5, 7\}$ , find the set with highest number of elements:

0%
$A \times B$
0%
$B \times A$
0%
$A \times A$
0%
$B \times B$

Explanation

$A=\left \{ 2,3,5 \right \}$

$B=\left \{ 5,7 \right \}$

$A\times B=\left \{ (2,5),(2,7),(3,5),(3,7),(5,5),(5,7) \right \}$

$B\times A=\left \{ (5,2),(5,3),(5,5),(7,2),(7,3),(7,5) \right \}$

$A\times A=\left \{ (2,2),(2,3),(2,5),(3,2),(3,3),(3,5),(5,2),(5,3),(5,5) \right \}$

$B\times B=\left \{ (5,5),(5,7),(7,5),(7,7) \right \}$

$\therefore A\times A$ has the highest number of elements

For two sets

$A$ and

$B$ ,

$A\times B=B\times A$ .

0%
True
0%
False

Explanation

The given statement is false.

Example:-

Let us consider

$A=\{1,2\}$ and

$B=\{3,4\}$ .

Now

$A\times B=\{(1,3),(1,4),(2,3),(2,4)\}$ ......(1).

And

$B\times A=\{(3,1),(3,2),(4,1),(4,2)\}$ ..........(2).

Form (1) and (2) it's evident that

$A\times B\ne B\times A$ .

$f:R \to R$ is defined by

$f\left( x \right) = {x^2} - 3x + 2$ and

$f\left( {{x^2} - 3x - 2} \right) = a{x^4} + b{x^3} + c{x^2} + dx + e$ then

$a + b + c + d + e =$

0%
$1$
0%
$2$
0%
$30$
0%
$20$

Let

$A = \left\{ {x,\,y,\,z} \right\}$ and

$B = \left\{ {1,\,2} \right\}$ . The number relations from A to B is

0%
64
0%
32
0%
16
0%
28

Explanation

$\Rightarrow$

$A=\{X,\,Y,\,Z\}$ and

$B=\{1,\,2\}$ [ Given ]

$\Rightarrow$ Number of relation from

$A$ to

$B$

$=2^{Number\,of\,elements\,in\,A\times B}$

$=2^{n(A)\times n(B)}$

$\Rightarrow$ Number of elements in set

$A=3$

$\Rightarrow$ Number of elements in set

$B=2$

$\Rightarrow$ Number of relation from

$A$ to

$B$

$=2^{n(A)\times n(B)}$

$=2^{3\times 2}$

$=2^6$

$=2\times 2\times 2\times 2\times 2\times 2$

$=64$

Let R be a relation from N to N defined by

$R = \left\{ {\left( {a,\,b} \right):a,\,b\, \in \,N\,\,and\,\,a = {b^2}} \right\}$

0%
$\left( {a,a} \right) \in R\,$ , for all $\,a \in N$
0%
$\left( {a,b} \right) \in R$ , implies $\left( {b,a} \right) \in R$
0%
$\left( {a,b} \right) \in R,\,\left( {b,c} \right) \in \,R$ implies $\left( {a,c} \right) \in R$
0%
None of these

$A = \left \{1, 2, 3, 4\right \}$ and

$I_{A}$ be the identify relation on

$A$ , then

0%
$(1, 2) \epsilon I_{A}$
0%
$(2, 2) \epsilon I_{A}$
0%
$(2, 1) \epsilon I_{A}$
0%
$(3, 4) \epsilon I_{A}$

Let

$R = \left \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 2)\right \}$ be a relation on the set

$A = \left \{1, 2, 3, 4\right \}$ . The relation

$R$ is

0%
Reflexive
0%
Transitive
0%
Not symmetric
0%
none of these

Let

$A$ and

$B$ be two sets containing four and two elements respectively.Then the number of subset of the set

$A \times B$ , each having at least three elements is

0%
$256$
0%
$275$
0%
$510$
0%
$219$

Explanation

$A$ and

$B$ containing

$4$ and

$2$ respectively

set A has 4 elements

set B has 2 elements

total number of element in

$(A\times B)=4\times 2=8$

total number of subsets of

$(A\times B)={2}^{8}=256$

Number of subsets having 0 element

$={ _{ }^{ 8 }{ C } }_{ 0 }=1$

Number of subsets having 1 element

$={ _{ }^{ 8 }{ C } }_{ 1 }=8$

number of subsets having 2 element

$={ _{ }^{ 8 }{ C } }_{ 2 }=\cfrac{8\times 7}{2}=28$

Number of subsets having atleast 3 elements

$=256-28-8-1=219$

If x is real, then

$\dfrac{x^2+2x+c}{x^2+4x+3c}$ can take all real values if?

0%
$0 < c < 2$
0%
$0 < c < 1$
0%
$-1 < c < 1$
0%
None of the above

Explanation

Let us assume that

$\dfrac{x^2+2x+c}{x^2+4x+3c}=y$

$\Rightarrow yx^2+4xy+3yc=x^2+2x+c$

$\Rightarrow x^2(y-1)+2x(2y-1)+3yc-c=0$
For x to be real, Discrimination of this equation should be

$\geq 0$

$\therefore 4(2y-1)^2-4\times (y-1)\times (3yc-c)\geq 0$

$4(4y^2-4y+1)-4(3cy^2-4cy+c)\geq 0$
Dividing both sides by

$4$ , we get

$(4-3c)y^2+4y(c-1)+1-c\geq 0$
For this equation to hold true, coefficient of

$y^2$ should be greater than

$0$ and D

$< 0$

$\Rightarrow 4-3c > 0$ i.e.

$c < \dfrac{4}{3}$ ..

$(1)$
Also,

$16(c^2-2c+1)-4\times (1-c)\times (4-3c) < 0$

$\Rightarrow 16c^2-32c+16-4(4-7a+3a^2) < 0$

$\Rightarrow 16c^2-32c+16-12c^2+28c-16 < 0$

$\Rightarrow 4c(c-1) < 0$

$\therefore 0 < c < 1$ ..

$(2)$
From

$(1)$ and

$(2)$ , we can say that

$0 < c < 1$ is the required condition.

$a\ne R$ and the equation

$-3(x-[x])^2+2(x-[x])+a^2=0$ ( where

$[x]$ denotes the greatest integer

$\le x$ ) has no integral solution, then all possible values of a lie in the interval :

0%
$(-2,-1)$
0%
$(-\infty ,-2)\cup(2,\infty)$
0%
$(-1 ,0)\cup(0,1)$
0%
$(1,2)$

Explanation

REF.Image

$x-[x]=\left \{ x \right \}$ , let

$\left \{ x \right \}=y$

As x is non-integral,

$0<y<1$

$(y\neq 0)$

equation becomes,

$-3y^{2}+2y+a^{2}=0$

$\Rightarrow a^{2}=3y^{2}-2y=y(3y-2)$

Drawing graph of

$3y^{2}-2y=0$ (parabola)

Minima

$\Rightarrow \frac{d}{dy}(3y^{2}-2y)=0\Rightarrow 6y-2=0\Rightarrow y=1/3$

Taking positive solutions

$(a^{2}\geq 0)$

we get

$0 < a^{2} < 1$

$\Rightarrow a\epsilon (-1,0)\cup (0,1)$

Option (C)

1057641_1180978_ans_34a47b492bcf45449fcbfb26bec67ee3.jpg

${x\epsilon R:\frac{14x}{x+1}-\frac{9x-30}{x-4}<0}$ is equal to

0%
(-1,4)
0%
(1,4) $\cup$ (5,7)
0%
(1,7)
0%
(-1,1) $\cup$ (4,6)

Explanation

$x\epsilon R :\frac{14x}{x+1}-\frac{9x-30}{x-4}< 0$

$x+1\neq 0$

$x-4 \neq 0$

$\Rightarrow x \neq -1$

$x \neq 4$

$14(x-4)-(x+1)(7x-30)< 0$

$14x^{2}-56x-(9x^{2}-30x+9x-30)< 0$

$14x^{2}-56x-9x^{2}+30x-9x+30< 0$

$5x^{2}-35x+30< 0$

$x^{2}-7x+6< 0$

$x^{2}-6x+6< 0$

$x(x-6)-1(x-6)< 0$

$(x-1)(x-6)< 0$

$\Rightarrow x> 1$ &

$x< 6$

$\Rightarrow x \epsilon ; (1,4)\cup (4,6)$

1170962_1200512_ans_522c5672e17c4ba0a46428c4c111ac10.jpeg

$\left| { z }_{ 1 }-a \right| <a,\left| { z }_{ 2 }-a \right| <b,\left| { z }_{ 3 }-a \right| <c$ ,

$(a,b,c\in R)$ then

$\left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right|$ is

0%
less than $(a+b+c)$
0%
more than $(a+b+c)$
0%
less than $2(a+b+c)$
0%
more than 2 $(a+b+c)$

Explanation

Given

$,$

$\left| {{z_1} - a} \right| < a$

$\left| {{z_2} - b} \right| < b$

$\left| {{z_3} - c} \right| < c$

Then

$,$

$\left| {{z_1} - a} \right| + \left| {{z_2} - b} \right| + \left| {{z_3} - c} \right| < a + b + c$

$\left| {{z_1} + {z_2} + {z_3}} \right| < \left| {{z_1}} \right| + \left| {{z_2}} \right| + \left| {{z_3}} \right|$

and

$,$

$\left| {\left( {{z_1} - a} \right) + \left( {{z_2} - b} \right) + \left( {{z_3} - c} \right) + a + b + c} \right|$

$< \left| {{z_1} - a} \right| + \left| {{z_2} - b} \right| + \left| {{z_3} - c} \right| + \left| {a + b + c} \right|$

$\left| {{z_1} + {z_2} + {z_3}} \right|$

$< a + b + c + a + b + c$

$\left| {{z_1} + {z_2} + {z_3}} \right|$

$<2 \left( {a + b + c} \right)$

$\left| {{z_1} + {z_2} + {z_3}} \right|$ less then

$2\left( {a + b + c} \right)$

Hence

$,$ Option

$(C)$ is correct

$.$

The area of the region

$R = \{(x, y) : |x| \le |y| \, \text{and} \, x^2 + y^2 \le 1 \}$ is

0%
$\dfrac{3 \pi}{8}$ sq. units
0%
$\dfrac{5 \pi}{8}$ sq. units
0%
$\dfrac{\pi}{2}$ sq. units
0%
$\dfrac{\pi}{8}$ sq. units

$f:R\rightarrow R,g:R\rightarrow R$ are defined by

$f(x)=5x-3,g(x)={ x }^{ 2 }+3,$ then

$(go{ f }^{ -1 })(3)=\\$

0%
$\frac { 25 }{ 3 }$
0%
$\frac { 111}{ 25 }$
0%
$\frac { 9 }{ 25 }$
0%
$\frac { 25 }{ 111 }$

Let A and B be sets containing 2 and 4 elements respecetively. The number of subsets

$A \times B$ having 3 or more elements is

0%
$219$
0%
$211$
0%
$256$
0%
$220$

Explanation

Let

$A=\left\{x,y\right\}$

$B=\left\{a,bc,d\right\}$

$A\times B$ has

$2\times 4=8$ elements

Total substance of

$A\times B={2}^{8}=256$

$\therefore\,$ Total number of subsets of

$A\times B$ having

$3$ or more elements

$=256-\left(1\,null \,set+8\,single\,ton\,set-^{8}C_{2}\,having\,2\,elements\right)$

$=256-1-8-\dfrac{8!}{6!2!}$

$=256-1-8-\dfrac{8\times 7\times 6!}{6!2!}$

$=256-1-8-28=219$

A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by :

$(x,y)\in\;R\; \rightarrow x$ is relatively prime to y. Then, domain of R is

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

Explanation

Given set

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

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

is defined as

$\left( x,y \right) \in R\Rightarrow x$ is relatively prime to

$y$

$(2,3)\in R$

Similarly,

$2$ is relatively prime to

$7$ so that

$(2,7)\in R$

So, we get

$R=\left\{ \left( 2,3 \right) ,\left( 2,7 \right) ,\left( 3,7 \right) ,\left( 3,10 \right) ,\left( 4,3 \right) ,\left( 4,7 \right) ,\left( 5,3 \right) ,\left( 5,6 \right) ,\left( 5,7 \right) , \right\}$

Thus domain

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

$R$ is the relation from set

$A$ to a set

$B$ and

$S$ is the relation from

$B$ to a set

$C$ , then the relation

$SoR$

0%
If from $A$ to $C$
0%
If from $C$ to $A$
0%
Does not exist
0%
None of these

The domain of

$f(x)=\frac{1}{\sqrt{(x-1)(x-2)(x-3)}}$ is

0%
$(-\infty ,1)\cup (3,\infty )$
0%
$(1,2)\cup (3,\infty )$
0%
$(-\infty ,2)$
0%
R

Explanation

${\textbf{Step -1: Put x=0 and change the sign as negative multiply by negative is positive}}$ ${\textbf{and positive multiply by negative is negative }}$

$x = 0 \Rightarrow - \times - \times - = -$

${\textbf{Step -2:Put x=1.5 and change the sign as positive multiply by negative is negative}}$ ${\textbf{and negative multiply by negative is positive }}$

$x = 1.5 \Rightarrow + \times - \times - = +$

${\textbf{Step -3:Put x=2.5 and change the sign as positive multiply by positive is positive}}$ ${\textbf{and positive multiply by negative is negative}}$

$x = 2.5 \Rightarrow + \times + \times - = -$

${\textbf{Step -4: Put x=4 and change the sign as positive multiply by positive is always positive }}$

$x = 4 \Rightarrow + \times + \times + = +$

${\text{x is positive between 1 and 2 and between 3 and infinity so }} x \in (1,2) \cup (3,\infty )$

$\textbf{Hence option B is correct}$

2165197_1242027_ans_f45f3de2e38348f4b12a993c3658da70.jpg

Let

$A=\left\{ 2,4,6,8 \right\}$ . A relation

$R$ on

$A$ is defined by

$R={(2,4),(4,2),(4,6),(6,4)}$ . Then

$R$ is:

0%
$anti-symmetric$
0%
$reflexive$
0%
$symmetric$
0%
$transitive$

For A= { 1, 2, 3} thel relation R = {(1,1), (2,2)}, (3,3)} is

0%
reflective only
0%
symmetric only
0%
transitive only
0%
equivalence

The relation

$R$ on the set

$Z$ of all integer numbers defined by

$(x,y)\ \epsilon \ R\ \Leftrightarrow x-y$ is divisible by

$n$ is

0%
Equivalence
0%
Symmetric only
0%
Reflexive only
0%
Transitive only

Explanation

$\begin{array}{l} R=\left\{ { \left( { x,y } \right) :x,y\in z,\, \, x-y\, \, is\, \, divisible\, \, by\, \, n } \right\} \\ For\, \, { { Reflexive } }, \\ x\in z \\ So\Rightarrow \left( { x-x } \right) is\, \, divisible\, \, by\, \, n \\ \Rightarrow \left( { x,x } \right) \in z \end{array}$

So, Relation is Reflexive

$\begin{array}{l} For\, \, Symmetric\Rightarrow Let\left( { x,y } \right) \in R \\ \Rightarrow \left( { x-y } \right) is\, \, divisible\, \, by\, \, n. \\ \Rightarrow \frac { { x-y } }{ n } =c,{ { Remainder } }\, \, is\, \, 0. \\ \Rightarrow \frac { { y-x } }{ n } =-c,{ { Remainder\quad is\quad also\quad 0 } }{ { . } } \\ \Rightarrow \left( { y-x } \right) is\, \, divisible\, \, by\, \, n \\ \left( { y,x } \right) \in R\, \, So,R\, \, is\, \, Symmetric. \\ For\, \, Transitive\Rightarrow Let\left( { x,y } \right) \in R\& \left( { y,z } \right) \in R \\ \Rightarrow \left( { x-y } \right) is\, \, divisible\, \, by\, \, n\Rightarrow \left( { y-q } \right) is\, \, divisible\, \, by\, \, n \end{array}$

add both these equation (i) & (ii)

$\begin{array}{l} \left( { x-y } \right) =xc,c\in z-v,\left( { y-q } \right) =na,\left( { a\in z } \right) \\ \Rightarrow \left( { x-y+y-q } \right) =nc+na,c,a\in z \\ \Rightarrow \left( { x-q } \right) =n\left( { a+c } \right) ,c,a\in z \\ \left( { x-q } \right) is\, \, divisible\, \, by\, \, n \\ \Rightarrow \left( { x,q } \right) \in R \end{array}$

So, R is Transitive

So, the R is Reflexive, Symmetric and Transitive
Then it is Equivalence Relation.

The relation

$“$ is congruent to

$"$ on the set of all triangles in a plane is

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

Explanation

$\begin{array}{l} Now,T=set\, \, of\, \, all\, \, triangles & \\ R:T\to T & \\ R=\left\{ { \left( { { T_{ 1 } },{ T_{ 2 } } } \right) :{ T_{ 1 } }\cong { T_{ 2 } } } \right\} & \\ \\ \left( i \right) For\, \, { { reflexive } } & \\ { T_{ 1 } }\cong { T_{ 1 } } & \left[ { all\, \, triangles\cong to\, \, itself } \right] \\ So,\left( { { T_{ 1 } },{ T_{ 1 } } } \right) \in R & \end{array}$

So it is a reflexive relation.

$\begin{array}{l} \left( { ii } \right) For\, \, symmetric \\ Let\left( { { T_{ 1 } },{ T_{ 2 } } } \right) \in R\Rightarrow { T_{ 1 } }\cong { T_{ 2 } } \\ \Rightarrow { T_{ 2 } }\cong { T_{ 1 } } \\ So,\left( { { T_{ 2 } },{ T_{ 1 } } } \right) \in R \end{array}$

Thus, it is a symmetric relation.

$\begin{array}{l} \left( { iii } \right) For\, \, transitive \\ Let\left( { { T_{ 1 } },{ T_{ 2 } } } \right) \in R\Rightarrow { T_{ 1 } }\cong { T_{ 2 } }\to \left( i \right) \\ \left( { { T_{ 2 } },{ T_{ 3 } } } \right) \in R\Rightarrow { T_{ 2 } }\cong { T_{ 3 } }\to \left( { ii } \right) \\ From\, \, equation\, \left( i \right) \& \left( { ii } \right) ,\, we\, \, get \\ { T_{ 1 } }\cong { T_{ 2 } }\cong { T_{ 3 } } \\ { T_{ 1 } }\cong { T_{ 3 } }\Rightarrow \left( { { T_{ 1 } },{ T_{ 3 } } } \right) \in R \end{array}$

Thus, it is a transitive.

Hence, given relation is an equivalence.

The domain of

$\dfrac { 10 ^ { x } + 10 ^ { - x } } { 10 ^ { x } - 10 ^ { - x } }$ is

0%
$R$
0%
$R - \{ 0 \}$
0%
$R - \{ 1 \}$
0%
$R ^ { + }$

Explanation

For domain only the denominator must not be equal to ZERO

It means

$10^x-10^{-x} \neq 0\\or\> x\neq 0$

So domain is R-{0}

Numerator is fine with any value of x.

The number of equivalence relations on a five element set is

0%
$32$
0%
$42$
0%
$50$
0%
$52$

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

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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