MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 10

Let

$n$ be a fixed positive integer. Define a relation

$R$ in the set

$Z$ of integers by

$aRb$ if and only if

$\dfrac {n}{a - b}$ . The relation

$R$ is

0%
Reflexive
0%
Symmetric
0%
Transitive
0%
An equivalence relation

Explanation

Given

$aRb$ such that

$R:\dfrac{n}{a-b}\epsilon Z$

$(A) aRa:$

$\dfrac{n}{a-a}=\infty$ and

$\infty$ does not belongs to

$Z$

So not a reflexive relation and hence (A) and (D) options are ruled out.

$(B)aRb\epsilon Z$ we need to check whether

$bRa\epsilon Z$ or not.

$\dfrac{n}{a-b}\epsilon Z$ Now,

$\dfrac{n}{b-a}\rightarrow -\dfrac{n}{a-b}$

$\dfrac{n}{a-b}$ is integer so negative of it also integer

$\therefore bRa\epsilon Z$

Hence relation

$R$ is Symmetric.

$(C)$ Now

$aRb\epsilon Z$ and

$bRc\epsilon Z$ but we can't say anything about

$aRc\epsilon Z.$ Hence not a transitive relation.

If two sets

$A$ and

$B$ are having

$39$ elements in common, then the number of elements common to each of the sets

$A\times B$ and

$B\times A$ are

0%
${ 2 }^{ 39 }$
0%
${ 39 }^{ 2 }$
0%
$78$
0%
$351$

Explanation

If set

$A$ and set

$B$ have

$39$ common elements, then the number of common elements in set

$A\times B$ and set

$B\times A\,=39^2$

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

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

$\left( { gof }^{ -1 } \right) \left( 3 \right) =$

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

Explanation

$f\left( x \right) =5x-3$

${ f }^{ -1 }\left( x \right) =\frac { x+3 }{ 5 }$

$g\left( x \right) ={ x }^{ 2 }+3$

${ gof }^{ -1 }\left( 3 \right) =g\left( { f }^{ -1 }\left( 3 \right) \right) =g\left( \frac { 6 }{ 5 } \right) =\frac { 111 }{ 25 }$

$\int\dfrac{2\cos x-\sin x+\lambda}{\cos x-\sin x-2}dx=A In\left|\cos x+\sin x-2\right|+Bx+C$ . Then the ordered triplet

$\left(A,B,\lambda\right)$ , is

0%
$\left(\dfrac{1}{2},\dfrac{3}{2},-1\right)$
0%
$\left(\dfrac{3}{2},\dfrac{1}{2},-1\right)$
0%
$\left(\dfrac{1}{2},-1, \dfrac{3}{2}\right)$
0%
$\left(\dfrac{3}{2},-1, \dfrac{1}{2}\right)$

Let X be the set of all citizens of India. Elements x, y in X are said to be related if the difference of their age is 5 years. Which one of the following is correct ?

0%
The relation is an equivalence relation on X.
0%
The relation is symmetric but neither reflexive nor transitive.
0%
The relation is reflexive but neither symmetric nor transitive.
0%
None of the above

Explanation

given that,

X = {All citizens of India}

and R=

$\{ (x,y):x,y\epsilon X,\left| x-y \right| =5\}$

i) For reflexive

$\left| x-x \right| =0\neq 5$

$\therefore (x,x)\in R$

So,

$R$ is not reflexive

ii) For symmetric

Let

$xRy \Rightarrow \left| x-y \right|=5 \Rightarrow \left| y-x \right| =5$

$\Rightarrow yRx$

So,

$R$ is symmetric

iii) For transitive

Let

$x,y,z \in X$ such that

$xRy \Rightarrow \left| x-y \right|=5$

and

$yRz \Rightarrow \left| y-z \right|=5$

But

$\left| x-z \right| \neq 5$

So,

$R$ is not transitive

Hence, the relation is symmetric but neither reflexive nor transitive

Find the correct co-related number.

$5:36::6:?$

0%
$48$
0%
$50$
0%
$49$
0%
$56$

Explanation

From the given co- relation we can write as

${ \Rightarrow (5+1) }^{ 2 }=36$

${ \therefore (6+1) }^{ 2 }=49$

$C$ is correct answer

Let

$S$ be a relation on

$\mathbb{R}^{+}$ defined by

$xSy\Leftrightarrow { x }^{ 2 }-{ y }^{ 2 }=2\left( y-x \right)$ , then

$S$ is

0%
Only reflecxive
0%
Only symmetric
0%
Only Trasitive
0%
Equivalence

Explanation

Let

$S$ be the relation on

$\mathbb{R}^{+}$ defined by

$xSy\Leftrightarrow x^2-y^2=2(y-x)$ .

Now,

$xSx$ holds as

$x^2-x^2=2(x-x)$ as

$0=0$ . So,

$S$ is reflexive.

Also,

$xSy\Leftrightarrow ySx$ holds as

$x^2-y^2=2(y-x)\Leftrightarrow y^2-x^2=2(x-y)$ . So,

$S$ is symmetric.

Let,

$xSy,\ ySz$ holds.

Then,

$x^2-y^2=2(y-x)$ .........(1) and

$y^2-z^2=2(z-y)$ .......(2).

Now, adding (1) and (2) we get,

$x^2-z^2=2(z-x)\Rightarrow xSz$ holds. So,

$S$ is also transitive.

Moreover

$S$ is an equivalence relation.

${C}_{r}$ stands for

${ _{ }^{ n }{ C } }_{ r }$ then

$\left( { C }_{ 0 }+{ C }_{ 1 } \right) +\left( { C }_{ 1 }+{ C }_{ 2 } \right) +....\left( { C }_{ n-1 }+{ C }_{ n } \right) \quad$ is equal to

0%
${ 2 }^{ n }-1$
0%
${ 2 }^{ n+1 }+1$
0%
${ 2 }^{ n+1 }-1$
0%
${ 2 }^{ n+1 }-2$

Explanation

$(1+x)^{n}=^{}C_0+^{}C_{1}x+^{}C_2x^2+C_{3}x^{3}...........+^{}C_{n}x^{n}$

Put x =1

Using the relation of sum of coefficent

$(C_0+C_1+$

$C_2+C_3....C_n)=2^n$

According to the question ,

$(C_0+C_1)+$

$(C_1+C_2)+$

$(C_2+C_3)+$

$(C_{n-1}+C_n)\rightarrow$

on simplifying

$(C_0+C_1+$

$C_2+C_3....C_n)+$

$(C_1+C_2+C_3+....$

$C_{n-1})\rightarrow$

Adding and subtracting

$C_0and C_n$ to above equation

$(C_0+C_1+$

$C_2+C_3....C_n)+$

$(C_0+C_1+C_2+C_3+....$

$C_{n})-C_0-C_n\rightarrow$

$=2^n+2^n-1-1=2^{n+1}-2$

For real values of x ,the range of

$\frac {x^2+2x+1}{X^2+2x-1}$ is

0%
$(-\infty,0) \cup (1,\infty)$
0%
$\begin{bmatrix}\frac{1}{2},2\end{bmatrix}$
0%
$\begin{bmatrix} -\infty,\frac {-2}{9} \end{bmatrix}\cup (1,\infty)$
0%
$(-\infty,-6)\cup(-2,\infty)$

Explanation

Let

$y=\frac{x^2+2x+1} {x^2+2x-1}$

$\Rightarrow yx^2+2xy-y=x^2+2x+1$

$\Rightarrow (y-1)x^2+2x(y-1)-(y+1)=0$

$B^2-4AC\ge 0$ &

$y\neq 1,y\neq 0$

Let

$R$ be a relation defined as

$aRb$ if

$1 + ab > 0$ , then the relation

$R$ is

0%
reflexive and symmetric
0%
symmetric but not reflexive
0%
transitive
0%
equivalence

Explanation

Given relation is

$aRb$ is

$1+ab>0$ ,

Considering both

$a$ and

$b$ are real numbers,

We know that

$ab=ba$ ,

$\Longrightarrow aRb=1+ab>0=1+ba>0=bRa$ ,

$\therefore$

$R$ is a symmetric relation.

Now,

$aRa=1+{ a }^{ 2 }$ as

$a^2$ is always a positive real number

$\therefore 1+a^2 >0$

$\therefore$

$R$ is a reflexive relation.

Now consider

$aRb$ which implies

$1+ab>0$ and also

$bRc$ which implies

$1+bc>0$

If we take

$a=0.5,$

$b=-0.5$ and

$c=-4$ , then

$1+(0.5)(-0.5) = 0.75>0$ and

$1+(-0.5)(-4) = 3>$

$\Rightarrow$ Both

$aRb$ and

$bRc$ are satisfied

But,

$aRc=1+(0.5)(-4) =-2<0$

$\therefore$

$aRc$ is not a relation

Hence

$R$ is not a equivalence relation, but is a reflexive and symmetric relation.

Let N denote the set of all natural numbers. Define two binary relations on N as

$R_1=\{(x, y)\epsilon N\times N : 2x+y=10\}$ and

$R_2=\{(x, y)\epsilon N\times N:x+2y=10\}$ . Then.

0%
Both $R_1$ and $R_2$ are transitive relations
0%
Both $R_1$ and $R_2$ are symmetric relations
0%
Range of $R_2$ is $\{1, 2, 3, 4\}$
0%
Range of $R_1$ is $\{2, 4, 8\}$

Explanation

Define two binary relations on N as

$R_1=\{(x, y)\epsilon N\times N : 2x+y=10\}$ and

$R_2=\{(x, y)\epsilon N\times N:x+2y=10\}$

From

$R_{1}$ ,

$2x+y=10$ and

$x, y\in N$

So, possible values for

$x$ and

$y$ are:

$x=1, y=8$ i.e

$(1,8)$

$x=2, y=6$ i.e

$(2,6)$

$x=3, y=4$ i.e

$(3,4)$

$x=4, y=2$ i.e

$(4,2)$

$R_{1}=\{(1,8),(2,6),(3,4),(4,2)\}$

Therefore, Range of

$R_{1}$ is

$\{2,4,6,8\}$

$R_{1}$ is not symmetric

Also,

$R_{1}$ is not transitive because

$(3,4),(4,2)\in R_{1}$ but

$(3,2) \notin R_{1}$

Thus, options A,B and D are incorrect.

From

$R_{2}$ ,

$x+2y=10$ and

$x, y\in N$

So, possible values for

$x$ and

$y$ are:

$x=8, y=1$ i.e

$(8,1)$

$x=6, y=2$ i.e

$(6,2)$

$x=4, y=3$ i.e

$(4,3)$

$x=2, y=4$ i.e

$(2,4)$

$R_{2}=\{(8,1),(6,2),(4,3),(2,4)\}$

Therefore, Range of

$R_{2}$ is

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

$R_{2}$ is not symmetric

Hence, option C is correct.

Let

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

$B=\left\{ 1,2 \right\}$ . Consider a relation

$R$ defined from set

$A$ to set

$B$ . Then

$R$ is equal to set

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

Explanation

The relation defined from a set

$A$ to set

$B$ is defined as a subset of

$A\times B$ .

$R:A\rightarrow B=\{(x,y)|x\in A, y\in B , (x,y)\in A\times B\}$

Here,

$A\times B=\{(a,1),(a,2),(b,1),(b,2),(c,1),(c,2)\}$

Relation from A to B will be a subset of this

$A\times B$ .

The maximum number of equivalence relation on the set

$A=\{1,2,3,4\}$ are

0%
$15$
0%
$13$
0%
$20$
0%
$5$

The number of reflexive relations of a set with three elements is equal to

0%
$2^{12}$
0%
$2^9$
0%
$2^6$
0%
$2^3$

The relation P defined from R to R as a P b

$\Leftrightarrow$ 1 + ab > 0, for all a, b

$\epsilon$ R is

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

Explanation

$Pb\Leftrightarrow 1+ab>0$

i)Reflexive:

$a\times a$ for any real value except 0 is positive.Hence

$0\times 0+1>1\Rightarrow a\times a+1>0$

ii)Symmetric: if

$a\times b+1>0$ then

$b\times a+1$ will also be

$>0$

iii)Not Transitive: a=-2,b=0\Rightarrow -2\times 0+1>0$$

$b=0,c=4\Rightarrow 0\times 4+1>0$

but a is not related to c

$\because a=-2,c=4\Rightarrow -2\times 4+1<0$

If R is a relation on a finite set having n elements, then the number of relations on A is :

0%
$2^n$
0%
$2^{n^2}$
0%
$n^2$
0%
$n^n$

Explanation

Here, the required number of relations on A can be found out by the formula

$2^{n^2}$ .

Hence, option B is correct.

Let

$\rho$ be a relation defined on

$N$ , the set of natural numbers, as

$\rho =\left\{ \left( x,y \right) \in N\times N:2x+y=41 \right\}$ then

0%
$\rho$ is an equivalence relation
0%
$\rho$ is only reflexive relation
0%
$\rho$ is only symmetric relation
0%
$\rho$ is not transitive

Explanation

$\rho =\left\{ \left( x,y \right) \in N\times N:2x+y=41 \right\}$
for reflexive relation

$xRx\Rightarrow 2x+x=41\Rightarrow x=\cfrac { 41 }{ 3 } \notin N$
for symmetric

$\Rightarrow xRy\Rightarrow 2x+y=41\neq yRx\quad$ (Not symmetric)
for transitive

$xRy\Rightarrow 2x+y=41$ and

$\quad yRz\Rightarrow 2y+z=41,x$ (Not transitive)

$A=\left\{2,3,5\right\}, B=\left\{2,5,6\right\}$ , then

$\left( A-B \right) \times \left( A\cap B \right)$ is

0%
$\left\{(3,2),(3,3),(3,5)\right\}$
0%
$\left\{(3,2),(3,5),(3,6)\right\}$
0%
$\left\{(3,2),(3,5)\right\}$
0%
$none\ of\ these$

Explanation

Ans.

$(c)$ .

$A-B=\{ 3\} ,A\cap B=(2,5)$ .

$\therefore \left( A-B \right) \times \left( A\cap B \right) =\{ (3,2),(3,5)\}$ .

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

$B=\left\{3,8\right\},$ then

$\left( A\cup B \right) \times \left( A\cap B \right)$ is

0%
$\left\{(3,1),(3,2),(3,3),(3,8)\right\}$
0%
$\left\{(1,3),(2,3),(3,3),(8,3)\right\}$
0%
$\left\{(1,2),(2,2),(3,3),(8,8)\right\}$
0%
$\left\{(8,3),(8,2),(8,1),(8,8)\right\}$

Explanation

$\left( A\cup B \right) =\{ 1,2,3,8\}$

$\left( A\cap B \right) =\{ 3\}$

$\therefore \left( A\cup B \right) \times \left( A\cap B \right)$

$=\left\{ (1,3),(2,3),(3,3),(8,3) \right\}$

Ans.

$(b)$ .

Let

$R = gS - 4$ . When

$S = 8, R = 16$ . When

$S = 10, R$ is equal to

0%
$11$
0%
$14$
0%
$20$
0%
$21$
0%
None of these

Explanation

$16 = 8g - 4; \therefore g = 2\dfrac {1}{2}; \therefore R = \left (2\dfrac {1}{2}\right ) 10 - 4 = 21$ .

Let n be a fixed positive integer. Define a relation R on I (the set of all integers) as follows:
aRB iff n/ (a -b), that is iff a - b is divisible by n. Then R is an equivalence relation on I.

0%
True
0%
False

$\alpha$ ,

$\beta$ be a straight lines in a plane, then

$R_1$ and

$R_2$ is reflexive, symmetric and transitive

$\alpha , R_1, \beta$ if

$\alpha \perp \beta$ and

$\alpha R_2 \beta$ if

$\alpha || \beta$ .

0%
True
0%
False

Explanation

$R_1 : \alpha \bot \beta \ \Rightarrow$ symmetric

$\to \ \beta \bot \alpha$

$\alpha$ is not

$\bot \alpha$

$\Rightarrow \$ Not reflection

$\to$ if

$\alpha \bot \beta .\ \beta \bot r$

$\alpha$ may or may not be

$\bot \ r$

So,

$R-1$ is not transitive

$R_2 : \alpha \parallel \beta$

$\alpha \parallel \alpha$ (reflection)

$\alpha \parallel \beta \to \beta \parallel \alpha$ (symmetric)

$\alpha \parallel \beta .\ \beta \parallel \alpha$

Then

$\alpha \parallel r$ (transitive)

1438100_998508_ans_faf803f208cd4fd1bda4c22a849973f5.png

If the line

$x=\alpha$ divides the area of the region

$R=\left\{(x,y)\in \mathbb{R}^2:x^3\le y\le x,0\le x\le 1\right\}$ into two equal parts, then

0%
$0\lt \alpha\le \dfrac{1}{2}$
0%
$\dfrac{1}{2}\lt \alpha\lt 1$
0%
$2\alpha^4-4\alpha^2+1=0$
0%
$\alpha^4+4\alpha^2-1=0$

Explanation

So,

$\int _{ 0 }^{ \alpha }{ (x-{ x }^{ 3 }) } dx=\int _{ \alpha }^{ 1 }{ (x-{ x }^{ 3 }) } dx\\ { [\cfrac { { x }^{ 2 } }{ 2 } -\cfrac { { x }^{ 4 } }{ 4 } ] }_{ 0 }^{ \alpha }={ [\cfrac { { x }^{ 2 } }{ 2 } -\cfrac { { x }^{ 4 } }{ 4 } ] }_{ \alpha }^{ 1 }\\ \cfrac { { \alpha }^{ 2 } }{ 2 } -\cfrac { { \alpha }^{ 4 } }{ 4 } =\cfrac { { 1 }^{ 2 } }{ 2 } -\cfrac { { 1 }^{ 4 } }{ 4 } -\cfrac { { \alpha }^{ 2 } }{ 2 } +\cfrac { { \alpha }^{ 4 } }{ 4 } \\ \cfrac { { 2\alpha }^{ 2 } }{ 2 } -\cfrac { 2{ \alpha }^{ 4 } }{ 4 } =\cfrac { { 1 } }{ 4 } ={ \alpha }^{ 2 }-\cfrac { { \alpha }^{ 4 } }{ 2 } \\ \cfrac { { 1 } }{ 2 } ={ 2\alpha }^{ 2 }-{ \alpha }^{ 4 }\\ { 2\alpha }^{ 4 }-4{ \alpha }^{ 2 }+1=0$

Thus,

${ 2\alpha }^{ 4 }-4{ \alpha }^{ 2 }+1=0$

1044103_1003631_ans_7a34d11d2f4941df87a39149762355f7.PNG

$aN = (ax/x \ \epsilon \ N)$ and

$bN \cap cN = dN,$ , where

$b, c \ \epsilon N$ are relatively prime, then

0%
d = bc
0%
c = bd
0%
b = cd
0%
none

Let

$f\left( x \right) :\begin{cases} x,\quad x\quad is\quad rational \\ 0,\quad x\quad is\quad irrational \end{cases}$

and

$g\left( x \right) :\begin{cases} 0,\quad x\quad is\quad rational \\ x,\quad x\quad is\quad irrational \end{cases}$

$f:R\rightarrow R$ and

$g:R\rightarrow R$ , then

$(f-g)$ is

0%
one-one and into
0%
neither one-one nor onto
0%
many-one and onto
0%
one-one and onto

Explanation

When x is rational

$f(x)=x,g(x)=0$

$\therefore (f-g)=x-0=x$ .So for each rational no. in

$(f-g)$ there is the same rational number in co domain

$f(x)=0,g(x)=x$ [when x is irrational]

$\therefore (f-g)=0-x=-x$

So for each irrational number x in domain, we have

$(-x)$ in co domain.

Conversely for each negative irrational number, there is an element in domain.

But for positive irrational no. there is no corresponding element.

Thus

$(f-g)$ is one -one and into.

$f:R \to R$ is a function satisfying

$f(x + y) = f(xy)$ for all

$x, y \in R$ and

$f\left(\dfrac{3}{4}\right) = \left( \dfrac{3}{4} \right)$ , then

$f\left( \dfrac{9}{16} \right)$ =

0%
$\dfrac{3}{4}$
0%
$\dfrac{9}{16}$
0%
$\dfrac{\sqrt{3}}{2}$
0%
$0$

Explanation

$f(x+y)=f(xy)\quad -(i)\\ f(\cfrac { 3 }{ 4 } )=(\cfrac { 3 }{ 4 } )\\ f(\cfrac { 9 }{ 16 } )=f(\cfrac { 3 }{ 4 } \times \cfrac { 3 }{ 4 } )\\ from(i)\\ f(\cfrac { 3 }{ 4 } \times \cfrac { 3 }{ 4 } )=f(\cfrac { 3 }{ 4 } +\cfrac { 3 }{ 4 } )\\ f(\cfrac { 3 }{ 4 } \times 2)=f(\cfrac { 3 }{ 2 } )\\ \because f(\cfrac { 3 }{ 4 } )=\cfrac { 3 }{ 4 } \quad constant\quad func.\\ \therefore f(\cfrac { 3 }{ 2 } )=\cfrac { 3 }{ 4 }$

$R$ is the largest equivalence relation on a set

$A$ and

$S$ is any relation on

$A$ , then

0%
$R\subset S$
0%
$S\subset R$
0%
$R=S$
0%
none of these

Explanation

$S$ is a subset of

$R$

Since

$R$ is the largest equivalence relation on set

$A$ ,

$R$ is a subset of

$A\times A$

Since

$S$ is any relation

$A$

$S$ is contained in

$R$

and

$R$ is a subset of

$a\times A$

So,

$S\subset R$

Let

$s = \{(x, y)| \sin y = \sin x; x, y \in R \}$ , then

$s$ is

0%
Not transitive
0%
equivalence
0%
transitive but not reflexive
0%
partial order relation

Find the domain of the function defined as

$f(x)=\dfrac{x+1}{2x+3}$ .

0%
$x\epsilon$ R $-$ { $\dfrac{3}{2}$ }
0%
$x\epsilon$ R $-$ { $\dfrac{-3}{2}$ }
0%
$x\epsilon$ R
0%
- $x\epsilon$ R $-$ { $\dfrac{2}{3}$ }

Explanation

Given:

$f\left ( x \right ) = \frac{x + 1}{2x + 3}$

$2x + 3 = 0$

$x = -\frac{3}{2}$

The domain is $R - \left \{ -\frac{3}{2} \right \}$ .

Hence, the domain of the function is $R - \left \{ -\frac{3}{2} \right \}$ .

Let R be a relation from

$A=\left\{ 1,2,3,4 \right\} to B=\left\{ 1,3,5 \right\}$ such that

$R=[(a,b):a<b,where\ a\epsilon A\ and\ b\epsilon B]$ . What is

$R$ equal to?

0%
$(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)$
0%
$(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)$
0%
$(3,3),(3,5),(5,3),(5,5)$
0%
$(3,3),(3,4),(4,5)$

Explanation

$A=\{ 1,2,3,4\} \quad ,\quad B=\{ 1,3,5\} \\ R=[(a,b):a<b,where\quad a\epsilon A\quad \& \quad b\epsilon B]\\ R=[(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)]$

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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