Relations Test 25

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Relations Test 25

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Weekly Quiz Competition

Question 1
1 / -0

If $A$ is a finite set having $n$ elements, then the number of relations which can be defined in $A$ is

A
${ 2 }^{ n }$

B
${ n }^{ 2 }$

C
${ 2 }^{ { n }^{ 2 } }$

D
${ n }^{ n }$

Solution

A relation is simply a subset of cartesian product $A\times A$ .
If ${A\times A}= [(a_1,a_1), (a_1,a_2),.....(a_1,a_n),$
$(a_2,a_1),(a_2,a_2),........(a_2,a_n)$
$......$
$(a_n,a_1), (a_n,a_2).........(a_n,a_n)]$
We can select first element of ordered pair in $n$ ways and second element in $n$ ways.
So, clearly this set of ordered pairs contain $n^2$ pairs.
Now, each of these $n^{2}$ ordered pairs can be present in the relation or can't be. So, there are $2$ possibilities for each of the $n^{2}$ ordered pairs.
Thus, the total no. of relations is $2^{n^{2}}$ .
Hence, option C is correct.
Question 2
1 / -0

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 ?

A
The relation is an equivalence relation on X.

B
The relation is symmetric but neither reflexive nor transitive.

C
The relation is reflexive but neither symmetric nor transitive.

D
None of the above

Solution

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
Question 3
1 / -0

Let R be a relation from A = {1, 2, 3, 4} to B = {1, 3, 5} such that
R = [(a, b) : a < b, where a $\varepsilon$ A and b $\varepsilon$ B].
What is RoR $^{-1}$ equal to?

A
${(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}$

B
${(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}$

C
${(3, 3), (3, 5), (5, 3), (5, 5)}$

D
${(3, 3), (3, 4), (4, 5)}$

Solution

$R$ gives the set $\{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\}$
$R^{-1}$ gives the set $\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4),(3,3)\}$
To compute $RoR^{-1}$ , we pick 1 element from $R^{-1}$ and its corresponding relation from $R$ .
eg: $(3,1)\in R^{-1}$ and $(1,3)\in R\implies (3,3)\in RoR^{-1}$
Similarly, computing for all such pairs, we have
$RoR^{-1}=\{(3,3),(3,5),(5,3),(5,5)\}$
Question 4
1 / -0

Find the correct co-related number.
$5:36::6:?$

A
$48$

B
$50$

C
$49$

D
$56$

Solution

From the given co- relation we can write as
${ \Rightarrow (5+1) }^{ 2 }=36$
${ \therefore (6+1) }^{ 2 }=49$
So $C$ is correct answer
Question 5
1 / -0

Let $A = \{ 1,2,3,4 \}$ and $R$ be a relation in $A$ given by $R = \{ (1,1) , (2,2) (3,3) , (4,4) , (1,2) , (3,1) , (1,3) \}$ then $R$ is :

A
Reflexive and transitive only

B
Transitive and symmetric only

C
An equivalence relation

D
None of the above

Solution

$A = \{ 1, 2, 3, 4\}$
$R = \{ (1,1) , (2,2),(3,3),(4,4),(1,2),(3,1)(1,3) \}$
clearly $(1,1) \epsilon R, (1,2) \epsilon (2,2) \epsilon R , (3,3) \epsilon R, (4,4) \epsilon R$
$\Rightarrow R$ is Reflective.
also $(1,1) \epsilon R, (1,2) \epsilon R \Rightarrow (1,2) R$
and $(1,2) \epsilon R, (1,3) \epsilon R \Rightarrow (1,3) \epsilon R \Rightarrow R$ is transitive
But for $(1,2) \epsilon R (2,1) \notin R, \Rightarrow R$ is not symmetric
$\Rightarrow R$ is reflective and transitive only.
Question 6
1 / -0

How many ordered pairs of (m, n) integers satisfy $\displaystyle\frac{m}{12}=\frac{12}{n}$ ?

A
$30$

B
$15$

C
$12$

D
$10$

Solution

$m \times n = 144$
144 can be written in multiplication of two integers in the following ways:

$1 \times 144 = 2 \times 72 = 3 \times 48 = 4 \times 36 = 6 \times 24 = 8 \times 18 = 9 \times 16 = 12 \times 12$

For all the factors except $(12, 12)$ there would be 14 ordered pairs.
Adding one more, yields 15 possibilities
Question 7
1 / -0

If ${ y }^{ 2 }={ x }^{ 2 }-x+1$ and $\quad { I }_{ n }=\int { \cfrac { { x }^{ n } }{ y } } dx$ and $A{ I }_{ 3 }+B{ I }_{ 2 }+C{ I }_{ 1 }={ x }^{ 2 }y$ then ordered triplet $A,B,C$ is

A
$\quad \left( \cfrac { 1 }{ 2 } ,-\cfrac { 1 }{ 2 } ,1 \right)$

B
$\left( 3,1,0 \right)$

C
$\left( 1,-1,2 \right)$

D
$\left( 3,-\cfrac { 5 }{ 2 } ,2 \right)$

Solution

We have given

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

${I_n} =\displaystyle \int {\dfrac{{{x^n}}}{y}dx}$
And,

$A{I_3} + B{I_2} + C{I_1} = {x^2}y$

Order triplet $A,\, B,\,C$ $= \left( {\dfrac{1}{2},\,\dfrac{{ - 1}}{2},1} \right)$

Hence, the option $(A)$ is correct.
Question 8
1 / -0

The number of ordered pairs $(x, y)$ of real numbers that satisfy the simultaneous equations.
$x + y^{2} = x^{2} + y = 12$ is.

A
$0$

B
$1$

C
$2$

D
$4$

Solution

$x + y^{2} = x^{2} + y = 12$
$x + y^{2} = x^{2} + y$
$x - y = x^{2} - y^{2} \Rightarrow x = y, x + y = 1$
when $x = y, x^{2} + x = 12$
$x^{2} + x - 12 = 0$
$x^{2} + 4x - 3x - 12 = 0$
$(x + 4)(x - 3) = 0$
$x = -4, 3$
$(3, 3),(-4, -4)$
When $y = 1 - x$
$x + (1 - x)^{2} = 12$
$x^{2} - x - 11 = 0$
$x = \dfrac {1\pm \sqrt {1 + 44}}{2}$ for two value of $x$ there are two value of $y$ .
So four pair.
Question 9
1 / -0

$R$ is a relation on $N$ given by $R = \left \{(x, y)|4x + 3y = 20\right \}$ . Which of the following doesnot belong to $R$ ?

A
$(-4, 12)$

B
$(5, 0)$

C
$(3, 4)$

D
$(2, 4)$

Solution

$R$ is given by $\{(x,y)|4x+3y=20\}$
$4 \times (-4)+3 \times 12=-16+36=20$
$4 \times 5+3 \times 0=20+0=20$
$4 \times 3+3 \times 4=12+12=24\neq 20$
$4 \times 2+3 \times 4=8+12=20$
So, among the given options $C$ does not satisfy the given condition.
All except $C$ , does not belong to the set $R.$
Hence, option C is correct.
Question 10
1 / -0

If $n (A) = 5$ and $n (B) = 7 ,$ then the number of relations on $A \times B$ is :

A
$2^{35}$

B
$2^{49}$

C
$2^{25}$

D
$2^{70}$

E
$2^{35 \times35 }$

Solution

Given, $n (A) = 5$ and $n (B) = 7$
Therefore, number of relations on $A \times B = 2^{[ n(A) \times n(B) ]}$
$= 2^{(5 \times 7)} = 2^{(35)}$

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Re-Attempt Weekly Quiz Competition

Relations Test 25

Result

Relations Test 25

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Question 1

2n{ 2 }^{ n }2n

n2{ n }^{ 2 }n2

2n2{ 2 }^{ { n }^{ 2 } }2n2

nn{ n }^{ n }nn

Solution

Question 2

The relation is an equivalence relation on X.

The relation is symmetric but neither reflexive nor transitive.

The relation is reflexive but neither symmetric nor transitive.

None of the above

Solution

Question 3

(1,3),(1,5),(2,3),(2,5),(3,5),(4,5){(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)

(3,1),(5,1),(3,2),(5,2),(5,3),(5,4){(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)

(3,3),(3,5),(5,3),(5,5){(3, 3), (3, 5), (5, 3), (5, 5)}(3,3),(3,5),(5,3),(5,5)

(3,3),(3,4),(4,5){(3, 3), (3, 4), (4, 5)}(3,3),(3,4),(4,5)

Solution

Question 4

484848

505050

494949

565656

Solution

Question 5

Reflexive and transitive only

Transitive and symmetric only

An equivalence relation

None of the above

Solution

Question 6

303030

151515

121212

101010

Solution

Question 7

(12,−12,1)\quad \left( \cfrac { 1 }{ 2 } ,-\cfrac { 1 }{ 2 } ,1 \right) (21​,−21​,1)

(3,1,0)\left( 3,1,0 \right) (3,1,0)

(1,−1,2)\left( 1,-1,2 \right) (1,−1,2)

(3,−52,2)\left( 3,-\cfrac { 5 }{ 2 } ,2 \right) (3,−25​,2)

Solution

Question 8

000

111

222

444

Solution

Question 9

(−4,12)(-4, 12)(−4,12)

(5,0)(5, 0)(5,0)

(3,4)(3, 4)(3,4)

(2,4)(2, 4)(2,4)

Solution

Question 10

235 2^{35} 235

249 2^{49} 249

225 2^{25} 225

270 2^{70} 270

235×35 2^{35 \times35 } 235×35

Solution

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${ 2 }^{ n }$

${ n }^{ 2 }$

${ 2 }^{ { n }^{ 2 } }$

${ n }^{ n }$

${(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}$

${(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}$

${(3, 3), (3, 5), (5, 3), (5, 5)}$

${(3, 3), (3, 4), (4, 5)}$

$48$

$50$

$49$

$56$

$30$

$15$

$12$

$10$

$\quad \left( \cfrac { 1 }{ 2 } ,-\cfrac { 1 }{ 2 } ,1 \right)$

$\left( 3,1,0 \right)$

$\left( 1,-1,2 \right)$

$\left( 3,-\cfrac { 5 }{ 2 } ,2 \right)$

$0$

$1$

$2$

$4$

$(-4, 12)$

$(5, 0)$

$(3, 4)$

$(2, 4)$

$2^{35}$

$2^{49}$

$2^{25}$

$2^{70}$

$2^{35 \times35 }$