Relations Test 26

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Relations Test 26

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Weekly Quiz Competition

Question 1
1 / -0

The number of ordered pairs $\left( m,n \right)$ , where $m, n \in \left\{ 1,2,3,\dots ,50 \right\}$ , such that ${ 6 }^{ m }+{ 9 }^{ n }$ is a multiple of $5$ is

A
$1250$

B
$2500$

C
$625$

D
$500$

Solution

For a number to be divisible by $5$ the last digit shall be a multiple of $5$
Any power of 6 has its last digit(or unit's digit) equal to 6.
The power of $9$ of the form
$4p+1$ has it's last digit as 9
$4p+2$ has it's last digit as 1
$4p+3$ has it's last digit as 9
$4p$ has it's last digit as 1
Thus $6^{m}+9^{n}$ will be a multiple of $5$ only if $m$ is any number from the given set and $n$ is a number of the form $4p+1$ or $4p+3$ .
Therefore the number of ways to select $m$ is $\binom{50}{1}$ and the number of ways to select $n$ from the given set is $\binom{25}{1}$
hence the total number of ordered pairs $(m,n)$ is $\binom{50}{1} \times \binom{25}{1}=50 \times 25=1250$
Question 2
1 / -0

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

A
Reflexive

B
Symmetric

C
Transitive

D
An equivalence relation

Solution

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, $\dfrac{n}{b-a}\rightarrow -\dfrac{n}{a-b}$
As $\dfrac{n}{a-b}$ 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.
Question 3
1 / -0

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

A
${ 2 }^{ 39 }$

B
${ 39 }^{ 2 }$

C
$78$

D
$351$

Solution

If set $A$ and set $B$ have $39$ common elements, then the number of common elements in set $A\times B$ and set $B\times A\,=39^2$
Question 4
1 / -0

Let $S$ be a relation on $\mathbb{R}^{+}$ defined by $xSy\Leftrightarrow { x }^{ 2 }-{ y }^{ 2 }=2\left( y-x \right)$ , then $S$ is

A
Only reflecxive

B
Only symmetric

C
Only Trasitive

D
Equivalence

Solution

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.
Question 5
1 / -0

If ${C}_{r}$ 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

A
${ 2 }^{ n }-1$

B
${ 2 }^{ n+1 }+1$

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

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

Solution

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

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

$(C_0+C_1+$ $C_2+C_3....C_n)+$ $(C_0+C_1+C_2+C_3+....$ $C_{n})-C_0-C_n\rightarrow$

$=2^n+2^n-1-1=2^{n+1}-2$
Question 6
1 / -0

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

A
$\rho$ is an equivalence relation

B
$\rho$ is only reflexive relation

C
$\rho$ is only symmetric relation

D
$\rho$ is not transitive

Solution

$\rho =\left\{ \left( x,y \right) \in N\times N:2x+y=41 \right\}$
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)
Question 7
1 / -0

For real values of x ,the range of $\frac {x^2+2x+1}{X^2+2x-1}$ is

A
$(-\infty,0) \cup (1,\infty)$

B
$\begin{bmatrix}\frac{1}{2},2\end{bmatrix}$

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

D
$(-\infty,-6)\cup(-2,\infty)$

Solution

Let $y=\frac{x^2+2x+1} {x^2+2x-1}$
$\Rightarrow yx^2+2xy-y=x^2+2x+1$
$\Rightarrow (y-1)x^2+2x(y-1)-(y+1)=0$
$B^2-4AC\ge 0$ & $y\neq 1,y\neq 0$
Question 8
1 / -0

Let $R$ be a relation defined as $aRb$ if $1 + ab > 0$ , then the relation $R$ is

A
reflexive and symmetric

B
symmetric but not reflexive

C
transitive

D
equivalence

Solution

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.
Question 9
1 / -0

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

A
$A$

B
$B$

C
$A\times B$

D
$B\times A$

Solution

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 $A\times B$ .
Question 10
1 / -0

If $f:R\rightarrow R,g\quad :R\rightarrow R$ are defined by $f\left( x \right) =5x-3,g(x)={ x }^{ 2 }+3,$ then $\left( { gof }^{ -1 } \right) \left( 3 \right) =$

A
$\frac { 25 }{ 3 }$

B
$\frac { 111 }{ 25 }$

C
$\frac { 9 }{ 25 }$

D
$\frac { 25 }{ 111 }$

Solution

$f\left( x \right) =5x-3$
${ f }^{ -1 }\left( x \right) =\frac { x+3 }{ 5 }$
$g\left( x \right) ={ x }^{ 2 }+3$
${ gof }^{ -1 }\left( 3 \right) =g\left( { f }^{ -1 }\left( 3 \right) \right) =g\left( \frac { 6 }{ 5 } \right) =\frac { 111 }{ 25 }$

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Re-Attempt Weekly Quiz Competition

Relations Test 26

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Relations Test 26

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SHARING IS CARING

Question 1

125012501250

250025002500

625625625

500500500

Solution

Question 2

Reflexive

Symmetric

Transitive

An equivalence relation

Solution

Question 3

239{ 2 }^{ 39 }239

392{ 39 }^{ 2 }392

787878

351351351

Solution

Question 4

Only reflecxive

Only symmetric

Only Trasitive

Equivalence

Solution

Question 5

2n−1{ 2 }^{ n }-12n−1

2n+1+1{ 2 }^{ n+1 }+12n+1+1

2n+1−1{ 2 }^{ n+1 }-12n+1−1

2n+1−2{ 2 }^{ n+1 }-22n+1−2

Solution

Question 6

ρ\rhoρ is an equivalence relation

ρ\rhoρ is only reflexive relation

ρ\rhoρ is only symmetric relation

ρ\rhoρ is not transitive

Solution

Question 7

(−∞,0)∪(1,∞)(-\infty,0) \cup (1,\infty)(−∞,0)∪(1,∞)

[12,2]\begin{bmatrix}\frac{1}{2},2\end{bmatrix}[21​,2​]

[−∞,−29]∪(1,∞)\begin{bmatrix} -\infty,\frac {-2}{9} \end{bmatrix}\cup (1,\infty)[−∞,9−2​​]∪(1,∞)

(−∞,−6)∪(−2,∞)(-\infty,-6)\cup(-2,\infty)(−∞,−6)∪(−2,∞)

Solution

Question 8

reflexive and symmetric

symmetric but not reflexive

transitive

equivalence

Solution

Question 9

AAA

BBB

A×BA\times BA×B

B×AB\times AB×A

Solution

Question 10

253\frac { 25 }{ 3 }325​

11125\frac { 111 }{ 25 } 25111​

925\frac { 9 }{ 25 } 259​

25111\frac { 25 }{ 111 } 11125​

Solution

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$1250$

$2500$

$625$

$500$

${ 2 }^{ 39 }$

${ 39 }^{ 2 }$

$78$

$351$

${ 2 }^{ n }-1$

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

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

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

$\rho$ is an equivalence relation

$\rho$ is only reflexive relation

$\rho$ is only symmetric relation

$\rho$ is not transitive

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

$\begin{bmatrix}\frac{1}{2},2\end{bmatrix}$

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

$(-\infty,-6)\cup(-2,\infty)$

$A$

$B$

$A\times B$

$B\times A$

$\frac { 25 }{ 3 }$

$\frac { 111 }{ 25 }$

$\frac { 9 }{ 25 }$

$\frac { 25 }{ 111 }$