Relations Test 10

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Relations Test 10

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Weekly Quiz Competition

Question 1
1 / -0

Which one of the following is an elementary symmetric function of $$x_{1},x_{2},x_{3},x_{4}$$.

A
$$x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}$$

B
$$x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}$$

C
$$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$$

D
$$x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}$$

Solution

Numbers of elements of is $$\{x_{1},x_{2},x_{3},x_{4}\}$$.
$$\therefore n(\mathrm{A})=4$$
$$\therefore$$ Number of elementary functions is given by$$^{4}\textrm{c}_{2}=6$$
Hence, option 'D' is correct.
Question 2
1 / -0

The solution of $$8x\equiv 6(mod \ 14) $$ is

A
$$\{8, 6\}$$

B
$$\{6,14\}$$

C
$$\{6,13\}$$

D
$$\{8,14,6\}$$

Solution

Since,$$8\mathrm{x}\equiv 6(mod \ 14)$$
i.e., $$8\mathrm{x}-6=14\mathrm{k}$$ for $$\mathrm{k}\in \mathrm{I}$$.
$$ \Rightarrow 8x = 14k+6$$
$$ \Rightarrow 4x = 7k+3$$
The values $$6$$ and $$13$$ satisfy this equation (when $$k =3$$ and $$k=7$$),
while $$8, 14$$ and $$16$$ do not.
Question 3
1 / -0

If relation R$$=\left \{ (x, x+2) : x \in N, 1 \leq x <4 \right \}$$ then R is

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

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

C
$$\{(1, 3), (2, 4), (3, 5), (4, 6)\}$$

D
none of these

Solution

We have $$x\in N, 1\le x<4$$
$$\Rightarrow x =1,2,3$$
Hence $$R =\{(1,3),(2,4),(3,5)\}$$
Question 4
1 / -0

The minimum number of elements that must be added to the relation $$R=\{(1,2,),(2,3)\} $$ on the set of natural numbers so that it is an equivalence is

A
$$4$$

B
$$7$$

C
$$6$$

D
$$5$$

Solution

for equivalence relation R
$$(a,a)\in R$$
if $$(a,b)\in R$$
then $$(b,a)\in R$$
if $$(a,b)\in R$$
$$(b,c)\in R$$
then $$(a,c)\in R$$
$$R={(1,2), (2,3), (1,1), (2,2), (3,3), (2,1), (3,2), (1,3), (3,1)}$$
No of added ordered pairs $$=9-2=7$$
Question 5
1 / -0

If $$R$$ is the relation 'less than' from $$A=\{1, 2, 3, 4, 5\}$$ to $$B=\{1, 4\}$$, the set of ordered pairs corresponding to $$R$$, then the inverse of $$R$$ is

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

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

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

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

Solution

$$\because R$$ is the relation 'less than' from $$A$$ to $$B$$.

So $$R=\{(1,4), (2,4), (3,4)\}$$

$$R^{-1}=\{(4,1), (4,2), (4, 3)\}$$
Question 6
1 / -0

Let $$R = \{(2,3),(3, 4)\}$$ be relation defined on the set of natural numbers. The minimum number of ordered pairs required to be added in $$R$$ so that enlarged relation becomes an equivalence relation is

A
$$3$$

B
$$5$$

C
$$7$$

D
$$9$$

Solution

We know that a relation is an equivalence relation if it is reflexive, symmetric and transitive.
For reflexive relation $$(a, a)\in R$$
For symmetric: If $$(a,b)\in R,$$ then $$(b,a)\in R$$
For transitive relation:
Let $$(a,b)\in R$$ and $$(b,c)\in R,$$ then $$(a,c)\in R$$
$$\therefore $$ Required $$R= \{(2,3), (3,4), (2,2), (3,3), (2,4), (4,2), (3,2), (4,3), (4,4)\}$$
So minimum number of ordered pairs to be added $$=7$$.
Question 7
1 / -0

If $$A =\{1,2,3\}$$, $$B=\{1,4,6,9\}$$ and $$R$$ is a relation from $$A$$ to $$B$$ defined by '$$x$$ is greater than $$y$$'. The range of $$R$$ is

A
$$\{1, 4, 6, 9\}$$

B
$$\{4, 6, 9\}$$

C
$$\{1\}$$

D
$$\{4, 6\}$$

Solution

$$R=\left \{ ( x,y \right ): x\in A, y\in B$$ and $$x> y\}$$
$$R= \left \{ \left ( 2,1 \right ),\left ( 3,1 \right ) \right \}$$
Therefore, $$R=\left \{ 1 \right \}$$
Question 8
1 / -0

If $$ X =\{1, 2,3,4,5\} $$ and $$Y =\{1,3,5,7,9\}$$, determine which of the following sets represent a relation and also a mapping?

A
$$R_{1}= \{(x,y)$$:$$ y=x+2, x \in Y,y \in Y\}$$

B
$$R_{2}=\{(1,1), (1,3), (3,5), (4,7), (5,9)\}$$

C
$$R_{3}=\{(1,1), (2,3), (3,5), (3,7), (5,7)\}$$

D
$$R_{4}=\{(1,3), (2,5), (4,7), (5,9), (3,1)\}$$

Solution

Here we will check all options one by one:
Option A:
$$R_1 = \{(x,y): y= x+2, x\in X, y\in Y\}$$
$$\implies \mathrm{R}_{1}=\{(1,3),(2, 4),(3,5),(4,6),(5,7)\}$$
Since $$4$$ amd $$6$$ are the images of $$2$$ and $$4$$ respectively but $$4$$ and $$6$$ do not belong to $$Y$$
$$\therefore (2, 4),(4,6)\not\in X \times Y$$
Hence $$R_{1}$$ is not a relation as well as not a mapping.

Option B:
$$R_{2}$$: It is a relation but not a mapping because the element $$1$$ has two different images.

Option C:
$$R_{2}$$: It is a relation but not a mapping because the element $$3$$ has two different images.

Option D:
$$R_{4}$$: It is both a relation and a mapping because every element in $$X$$ is mapped to the elements in $$Y$$. Also, every element of $$X$$ has a one and only one image in $$Y$$ and every element in $$Y$$ has its pre-image in $$X$$. Hence, it is also one-one and onto mapping and hence it is a bijection.
Question 9
1 / -0

If $$A$$ is the set of even natural numbers less than $$8$$ and $$B$$ is the set of prime numbers less than $$7$$, then the number of relations from $$A$$ to $$B$$ is

A
$$2^{9}$$

B
$$9^{2}$$

C
$$3^{2}$$

D
$$2^{9}-1$$

Solution

$$A = \{2, 4, 6\}, B = \{2,3,5\}$$
$$n (A\times B) = 3\times 3 = 9$$
Number of relations from $$A$$ to $$B $$ is $$2^9$$
Question 10
1 / -0

If A $$=$${$$x : x^{2}-3x+2= 0$$}, and $$R$$ is a universal relation on $$A$$, then $$R$$ is

A
$$\{(1,1),(2, 2)\}$$

B
$$\{(1,1)\}$$

C
$$\phi $$

D
$$\{(1,1),(1, 2)(2,1),(2,2)\}$$

Solution

Consider, $$x^2-3x+2=0$$
$$\implies x^2-2x-x+2=0$$
$$\implies (x-2)(x-1)=0$$
$$\implies x=1,2$$
$$\therefore A=\{1,2\}$$
Also R is universal relation on set A, then every element of set A is related every other element of A
So $$R= \{(1,1), (1,2), (2,1), (2,2)\}$$.

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Relations Test 10

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Relations Test 10

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SHARING IS CARING

Question 1

$$x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}$$

$$x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}$$

$$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$$

$$x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}$$

Solution

Question 2

$$\{8, 6\}$$

$$\{6,14\}$$

$$\{6,13\}$$

$$\{8,14,6\}$$

Solution

Question 3

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

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

$$\{(1, 3), (2, 4), (3, 5), (4, 6)\}$$

none of these

Solution

Question 4

$$4$$

$$7$$

$$6$$

$$5$$

Solution

Question 5

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

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

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

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

Solution

Question 6

$$3$$

$$5$$

$$7$$

$$9$$

Solution

Question 7

$$\{1, 4, 6, 9\}$$

$$\{4, 6, 9\}$$

$$\{1\}$$

$$\{4, 6\}$$

Solution

Question 8

$$R_{1}= \{(x,y)$$:$$ y=x+2, x \in Y,y \in Y\}$$

$$R_{2}=\{(1,1), (1,3), (3,5), (4,7), (5,9)\}$$

$$R_{3}=\{(1,1), (2,3), (3,5), (3,7), (5,7)\}$$

$$R_{4}=\{(1,3), (2,5), (4,7), (5,9), (3,1)\}$$

Solution

Question 9

$$2^{9}$$

$$9^{2}$$

$$3^{2}$$

$$2^{9}-1$$

Solution

Question 10

$$\{(1,1),(2, 2)\}$$

$$\{(1,1)\}$$

$$\phi $$

$$\{(1,1),(1, 2)(2,1),(2,2)\}$$

Solution

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