Relations Test 1

Result

Relations Test 1

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

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.

A
Both $$R_1$$ and $$R_2$$ are transitive relations

B
Both $$R_1$$ and $$R_2$$ are symmetric relations

C
Range of $$R_2$$ is $$\{1, 2, 3, 4\}$$

D
Range of $$R_1$$ is $$\{2, 4, 8\}$$

Solution

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.
Question 2
1 / -0

Consider the following two binary relations on the set $$A = \left \{a, b, c\right \} : R_{1} = \left \{(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)\right \}$$
and $$R_{2} = \left \{(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)\right \}$$ Then

A
$$R_{2}$$ is symmetric but it is not transitive

B
Both $$R_{1}$$ and $$R_{2}$$ are transitive

C
Both $$R_{1}$$ and $$R_{2}$$ are not symmetric

D
$$R_{1}$$ is not symmetric but it is transitive

Solution

$$R_2$$ is symmetric as for any $$(a_1, a_2) \in R_2$$, we have $$(a_2, a_1) \in R_2$$.
But $$R_1$$ is not symmetric as $$(b,c) \in R_1$$ but $$(c,b) \notin R_1$$

For checking transitivity, we observe for $$R_2$$ that $$(b,a) \in R_2$$, $$(a,c) \in R_2$$ but $$(b,c) \notin R_2$$.
Similarly, for $$R_1$$, $$(b,c) \in R_1$$, $$(c,a) \in R_1$$ but $$(b,a) \notin R_1$$. So neither $$R_1$$ nor $$R_2$$ is transitive.
So, the correct answer is option A.
Question 3
1 / -0

Let $$S = \{ 1,2,3 , \ldots , 100 \} .$$ The number of non-empty subsets $$A$$ of $$S$$ such that the product of elements in $$A$$ is even is :-

A
$$2 ^ { 50 } \left( 2 ^ { 50 } - 1 \right)$$

B
$$2 ^ { 100 } - 1$$

C
$$2 ^ { 50 } - 1$$

D
$$2 ^ { 50 } + 1$$

Solution

$$\mathrm { S } = \{ 1,2,3 - \ldots - 100 \}$$

= Total non empty subsets-subsets with product of element is odd

$$= 2 ^ { 100 } - 1 - 1 \left[ \left( 2 ^ { 50 } - 1 \right) \right]$$

$$= 2 ^ { 100 } - 2 ^ { 50 }$$

$$= 2 ^ { 50 } \left( 2 ^ { 50 } - 1 \right)$$
Question 4
1 / -0

$$A$$ and $$B$$ are two sets having $$3$$ and $$4$$ elements respectively and having $$2$$ elements in common. The number of relations which can be defined from $$A$$ to $$B$$ is:

A
$$2^{5}$$

B
$$2^{10}-1$$

C
$$2^{12}-1$$

D
$$2^{12}$$

Solution

Given:
$$n(A) = 3 = m$$
$$n(B) = 4 = n$$

Number of relation
$$= { 2 }^{ mn }$$
$$= { 2 }^{ 3\times 4 }$$
$$= { 2 }^{ 12 }$$.
Question 5
1 / -0

Let $$R$$ be a relation from a set $$A$$ to a set $$B$$, then:

A
$$R=A\cup B$$

B
$$R=A\cap B$$

C
$$R\subseteq A\times B$$

D
$$R\subseteq B\times A$$

Solution

$$R : $$$$A\rightarrow B$$
then $$R$$ is a subset $$A\times B$$
$$\therefore$$ $$R\quad \subseteq \quad A\times B$$
Question 6
1 / -0

Let $$x$$ be a real number $$\left [ x \right ]$$ denotes the greatest integer function, and $$\left \{ x \right \}$$ denotes the fractional part and $$(x)$$ denotes the least integer function,then solve the following.
$$\left [ 2x \right ]-2x=\left [ x+1 \right ]$$

A
$$\left \{ -1,\displaystyle-\frac{1}{2} \right \}$$

B
$$\left \{ -1,\displaystyle \frac{1}{2} \right \}$$

C
$$\left \{ 1,\displaystyle-\frac{1}{2} \right \}$$

D
$$\left \{ 1,\displaystyle \frac{1}{2} \right \}$$

Solution

$$\left [ 2x \right ]-2x=\left [ x+1 \right ]$$

$$\Rightarrow \left \{2x\right\}=\left [ x \right ]+1$$ (i)

But range of fractional part is $$[0,1)$$

$$\Rightarrow 0\leq [x]+1 < 1$$

$$\Rightarrow -1 \leq [x] < 0$$

$$ \Rightarrow -1 \leq x < 0$$

$$\Rightarrow [x] =-1$$

Thus (i) becomes $$\{2x\} =0 $$ it means that fraction part is zero for that,

$$\therefore x = -\cfrac{1}{2}, -1$$ in $$[-1,0)$$
Question 7
1 / -0

If $$A = \{x, y\}$$ and $$B = \{3, 4, 5, 7, 9\}$$ and $$C = \{4, 5, 6, 7\}$$, find $$\displaystyle A\times (B\cap C)$$

A
$$\displaystyle \left \{ (x,4),(x,5),(x,7),(y,4),(y,5),(y,7) \right \}$$

B
$$\displaystyle \left \{ (x,3),(x,5),(x,7),(y,4),(y,5),(y,7) \right \}$$

C
$$\displaystyle \left \{ (x,4),(x,5),(x,7),(y,4),(y,5),(y,9) \right \}$$

D
none of the above

Solution

$$\displaystyle B\cap C=\left \{ 3,4,5,7,9 \right \}\cap \left \{ 4,5,6,7 \right \}=\left \{ 4,5,7 \right \}$$
$$\displaystyle \therefore A\times (B\cap C)=\left \{ x,y \right \}\times \left \{ 4,5,7 \right \}$$
=$$\displaystyle \left \{ (x,4),(x,5),(x,7),(y,4),(y,5),(y,7) \right \}$$
Question 8
1 / -0

$$\displaystyle x^{2} = xy$$ is a relation (defined on set R) which is

A
Symmetric

B
Reflexive

C
Transitive

D
None of these

Solution

Given , R:$$x^{2}=xy$$
Symmetric: If $$ (x,x) \in $$ R
$$x^{2}=x \times x$$
$$x^{2}=x^{2},(x,x) \in$$ R
So the relation is symmetric
Question 9
1 / -0

Which one of the following relations on R (set of real numbers) is an equivalence relation

A
$$\displaystyle aR_{1}b\Leftrightarrow \left | a \right |= \left | b \right |$$

B
$$\displaystyle aR_{2}b\Leftrightarrow a\geq b $$

C
$$\displaystyle aR_{3}b\Leftrightarrow a$$ divides $$b $$

D
$$\displaystyle aR_{4}b\Leftrightarrow a< b $$

Solution

$$a R_1 b \iff |a| = |b| $$

$$a=b$$

$$R_1 $$is a reflexive symmetric and transitive.( as we know $$a=b$$)

$$\therefore (a,a) $$ is present.

$$\therefore (a,b) (b,a)$$ will also be present.

So it is an equivalence relation .
Question 10
1 / -0

If $$\displaystyle A\prime $$ is symmetric to A and $$\displaystyle B\prime $$ is symmetric to B with respect to a line of symmetry, then

A
$$\displaystyle AB>A'B'$$

B
$$\displaystyle AB=A'B'$$

C
$$\displaystyle AB\neq A'B'$$

D
None of these

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Relations Test 1

Result

Relations Test 1

Score

-

Rank

-

SHARING IS CARING

Question 1

Both $$R_1$$ and $$R_2$$ are transitive relations

Both $$R_1$$ and $$R_2$$ are symmetric relations

Range of $$R_2$$ is $$\{1, 2, 3, 4\}$$

Range of $$R_1$$ is $$\{2, 4, 8\}$$

Solution

Question 2

$$R_{2}$$ is symmetric but it is not transitive

Both $$R_{1}$$ and $$R_{2}$$ are transitive

Both $$R_{1}$$ and $$R_{2}$$ are not symmetric

$$R_{1}$$ is not symmetric but it is transitive

Solution

Question 3

$$2 ^ { 50 } \left( 2 ^ { 50 } - 1 \right)$$

$$2 ^ { 100 } - 1$$

$$2 ^ { 50 } - 1$$

$$2 ^ { 50 } + 1$$

Solution

Question 4

$$2^{5}$$

$$2^{10}-1$$

$$2^{12}-1$$

$$2^{12}$$

Solution

Question 5

$$R=A\cup B$$

$$R=A\cap B$$

$$R\subseteq A\times B$$

$$R\subseteq B\times A$$

Solution

Question 6

$$\left \{ -1,\displaystyle-\frac{1}{2} \right \}$$

$$\left \{ -1,\displaystyle \frac{1}{2} \right \}$$

$$\left \{ 1,\displaystyle-\frac{1}{2} \right \}$$

$$\left \{ 1,\displaystyle \frac{1}{2} \right \}$$

Solution

Question 7

$$\displaystyle \left \{ (x,4),(x,5),(x,7),(y,4),(y,5),(y,7) \right \}$$

$$\displaystyle \left \{ (x,3),(x,5),(x,7),(y,4),(y,5),(y,7) \right \}$$

$$\displaystyle \left \{ (x,4),(x,5),(x,7),(y,4),(y,5),(y,9) \right \}$$

none of the above

Solution

Question 8

Symmetric

Reflexive

Transitive

None of these

Solution

Question 9

$$\displaystyle aR_{1}b\Leftrightarrow \left | a \right |= \left | b \right |$$

$$\displaystyle aR_{2}b\Leftrightarrow a\geq b $$

$$\displaystyle aR_{3}b\Leftrightarrow a$$ divides $$b $$

$$\displaystyle aR_{4}b\Leftrightarrow a< b $$

Solution

Question 10

$$\displaystyle AB>A'B'$$

$$\displaystyle AB=A'B'$$

$$\displaystyle AB\neq A'B'$$

None of these

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.