Relations Test 30

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Relations Test 30

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Weekly Quiz Competition

Question 1
1 / -0

The domain of $$\dfrac { 10 ^ { x } + 10 ^ { - x } } { 10 ^ { x } - 10 ^ { - x } }$$ is

A
$$R$$

B
$$R - \{ 0 \}$$

C
$$R - \{ 1 \}$$

D
$$R ^ { + }$$

Solution

For domain only the denominator must not be equal to ZERO
It means $$10^x-10^{-x} \neq 0\\or\> x\neq 0$$
So domain is R-{0}
Numerator is fine with any value of x.
Question 2
1 / -0

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

A
$$\frac { 25 }{ 3 } $$

B
$$\frac { 111}{ 25 } $$

C
$$\frac { 9 }{ 25 } $$

D
$$\frac { 25 }{ 111 } $$
Question 3
1 / -0

If $$R$$ is the relation from set $$A$$ to a set $$B$$ and $$S$$ is the relation from $$B$$ to a set $$C$$, then the relation $$SoR$$

A
If from $$A$$ to $$C$$

B
If from $$C$$ to $$A$$

C
Does not exist

D
None of these

Solution
Question 4
1 / -0

A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by :$$(x,y)\in\;R\; \rightarrow x$$ is relatively prime to y. Then, domain of R is

A
{2, 3, 5}

B
{3, 5}

C
{2, 3, 4}

D
{2, 3, 4,5}

Solution

Given set $$A=\left\{ 2,3,4,5 \right\} $$ to $$B=\left\{ 3,6,7,10 \right\} $$
is defined as $$\left( x,y \right) \in R\Rightarrow x$$ is relatively prime to $$y$$
So $$(2,3)\in R$$
Similarly, $$2$$ is relatively prime to $$7$$ so that $$(2,7)\in R$$
So, we get $$R=\left\{ \left( 2,3 \right) ,\left( 2,7 \right) ,\left( 3,7 \right) ,\left( 3,10 \right) ,\left( 4,3 \right) ,\left( 4,7 \right) ,\left( 5,3 \right) ,\left( 5,6 \right) ,\left( 5,7 \right) , \right\} $$
Thus domain $$\left\{ 2,3,4,5 \right\} $$
Question 5
1 / -0

If $$A$$ and $$B$$ are independent event such that $$P(A \cap B')=\dfrac {3}{25}$$ and $$P(A' \cap B)=\dfrac {8}{25}$$, then $$P(A)=$$

A
$$1/5$$

B
$$3/8$$

C
$$2/5$$

D
$$4/5$$

Solution

$$A$$ & $$B$$ are independent
$$P(A\cap B)=\dfrac {3}{25}$$ and $$P(A'\cap B)=\dfrac {8}{25}\quad P(A)=(?)$$
$$\rightarrow \ P(A)+P(B)=1---(i)$$ ($$A$$& $$B$$ are independent )
$$\rightarrow \ P(A\cap B')=P(A)-P(A\cap B)$$
$$P(A)-P(A\cap B)=\dfrac {3}{25}----(ii)$$
$$\rightarrow \ P(A' \cap B)=P(B)-P(A\cap B)$$
$$P(B)-P(A\cap B)=\dfrac {8}{25}-----(iii)$$
$$\rightarrow \ $$ solving equation $$(ii)$$ and $$(iii)$$
$$\dfrac {\,\,\, P\left( A \right) -P\left( A\cap B \right) =\dfrac { 3 }{ 25 } \\\,\,\, P\left( B \right) -P\left( A\cap B \right) =\dfrac { 8 }{ 25 } \\ -\quad \,\,\,\,\,\,\,+\quad \quad\quad\quad- }{ P\left( A \right) -P\left( B \right) =\dfrac { 3 }{ 25 } -\dfrac { 8 }{ 25 } \\ P\left( A \right) -P\left( B \right) =\dfrac { 3-8 }{ 25 } } $$
$$=\dfrac {-5}{25}$$
$$\rightarrow \ P(A)-P(B)=\dfrac {-1}{5}$$
$$P(A)-(1-P(A))=\dfrac {-1}{5}$$
$$P(A)-I+P(A)=\dfrac {-1}{5}$$
$$\therefore \ 2P(A)=\dfrac {-1}{5}+1$$
$$\therefore \ 2P(A)=\dfrac {4}{5}$$
$$\therefore \ P(A)=\dfrac {4}{5\times 2}$$
$$\therefore \ P(A)=\dfrac {2}{5}$$
Question 6
1 / -0

The relation $$R$$ on the set $$Z$$ of all integer numbers defined by $$(x,y)\ \epsilon \ R\ \Leftrightarrow x-y$$ is divisible by $$n$$ is

A
Equivalence

B
Symmetric only

C
Reflexive only

D
Transitive only

Solution

$$\begin{array}{l} R=\left\{ { \left( { x,y } \right) :x,y\in z,\, \, x-y\, \, is\, \, divisible\, \, by\, \, n } \right\} \\ For\, \, { { Reflexive } }, \\ x\in z \\ So\Rightarrow \left( { x-x } \right) is\, \, divisible\, \, by\, \, n \\ \Rightarrow \left( { x,x } \right) \in z \end{array}$$
So, Relation is Reflexive
$$\begin{array}{l} For\, \, Symmetric\Rightarrow Let\left( { x,y } \right) \in R \\ \Rightarrow \left( { x-y } \right) is\, \, divisible\, \, by\, \, n. \\ \Rightarrow \frac { { x-y } }{ n } =c,{ { Remainder } }\, \, is\, \, 0. \\ \Rightarrow \frac { { y-x } }{ n } =-c,{ { Remainder\quad is\quad also\quad 0 } }{ { . } } \\ \Rightarrow \left( { y-x } \right) is\, \, divisible\, \, by\, \, n \\ \left( { y,x } \right) \in R\, \, So,R\, \, is\, \, Symmetric. \\ For\, \, Transitive\Rightarrow Let\left( { x,y } \right) \in R\& \left( { y,z } \right) \in R \\ \Rightarrow \left( { x-y } \right) is\, \, divisible\, \, by\, \, n\Rightarrow \left( { y-q } \right) is\, \, divisible\, \, by\, \, n \end{array}$$
add both these equation (i) & (ii)
$$\begin{array}{l} \left( { x-y } \right) =xc,c\in z-v,\left( { y-q } \right) =na,\left( { a\in z } \right) \\ \Rightarrow \left( { x-y+y-q } \right) =nc+na,c,a\in z \\ \Rightarrow \left( { x-q } \right) =n\left( { a+c } \right) ,c,a\in z \\ \left( { x-q } \right) is\, \, divisible\, \, by\, \, n \\ \Rightarrow \left( { x,q } \right) \in R \end{array}$$
So, R is Transitive
So, the R is Reflexive, Symmetric and Transitive
Then it is Equivalence Relation.
Question 7
1 / -0

The relation $$“$$is congruent to$$"$$ on the set of all triangles in a plane is

A
reflexive only

B
symmetric only

C
transitive only

D
equivalence

Solution

$$\begin{array}{l} Now,T=set\, \, of\, \, all\, \, triangles & \\ R:T\to T & \\ R=\left\{ { \left( { { T_{ 1 } },{ T_{ 2 } } } \right) :{ T_{ 1 } }\cong { T_{ 2 } } } \right\} & \\ \\ \left( i \right) For\, \, { { reflexive } } & \\ { T_{ 1 } }\cong { T_{ 1 } } & \left[ { all\, \, triangles\cong to\, \, itself } \right] \\ So,\left( { { T_{ 1 } },{ T_{ 1 } } } \right) \in R & \end{array}$$
So it is a reflexive relation.

$$\begin{array}{l} \left( { ii } \right) For\, \, symmetric \\ Let\left( { { T_{ 1 } },{ T_{ 2 } } } \right) \in R\Rightarrow { T_{ 1 } }\cong { T_{ 2 } } \\ \Rightarrow { T_{ 2 } }\cong { T_{ 1 } } \\ So,\left( { { T_{ 2 } },{ T_{ 1 } } } \right) \in R \end{array}$$
Thus, it is a symmetric relation.

$$\begin{array}{l} \left( { iii } \right) For\, \, transitive \\ Let\left( { { T_{ 1 } },{ T_{ 2 } } } \right) \in R\Rightarrow { T_{ 1 } }\cong { T_{ 2 } }\to \left( i \right) \\ \left( { { T_{ 2 } },{ T_{ 3 } } } \right) \in R\Rightarrow { T_{ 2 } }\cong { T_{ 3 } }\to \left( { ii } \right) \\ From\, \, equation\, \left( i \right) \& \left( { ii } \right) ,\, we\, \, get \\ { T_{ 1 } }\cong { T_{ 2 } }\cong { T_{ 3 } } \\ { T_{ 1 } }\cong { T_{ 3 } }\Rightarrow \left( { { T_{ 1 } },{ T_{ 3 } } } \right) \in R \end{array}$$
Thus, it is a transitive.

Hence, given relation is an equivalence.
Question 8
1 / -0

The number of equivalence relations on a five element set is

A
$$32$$

B
$$42$$

C
$$50$$

D
$$52$$

Solution
Question 9
1 / -0

Let $$A = \left\{ {1,2,3} \right\}$$ and $$R = \left\{ {\left( {1,1} \right),\left( {1,3} \right),\left( {3,1} \right),\left( {2,2} \right),\left( {2,1} \right),\left( {3,3} \right)} \right\}$$, then the relation $$R$$ and $$A$$ is

A
reflexive

B
symmetric

C
transitive

D
equivalence

Solution

$$A=\{ 1,2,3\} \\ R=\{ (1,1),(1,3),(3,1),(2,2),(2,1),(3,3)\} $$
The relation $$R$$ on $$A$$ is reflexive.
Since $$(1,1),(2,2),(3,3)$$exist.
Also $$(1,3),(3,1)$$ exist but $$(1,2),(2,1)$$ does not . Hence it is not symmetric.
Question 10
1 / -0

Let $$S=\{x \in R : x\geq 0$$ and $$2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0\}$$. Then S?

A
Contains exactly one element

B
Contains exactly two elements

C
Contains exactly four elements

D
Is an empty set

Solution

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Re-Attempt Weekly Quiz Competition

Relations Test 30

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Relations Test 30

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SHARING IS CARING

Question 1

$$R$$

$$R - \{ 0 \}$$

$$R - \{ 1 \}$$

$$R ^ { + }$$

Solution

Question 2

$$\frac { 25 }{ 3 } $$

$$\frac { 111}{ 25 } $$

$$\frac { 9 }{ 25 } $$

$$\frac { 25 }{ 111 } $$

Question 3

If from $$A$$ to $$C$$

If from $$C$$ to $$A$$

Does not exist

None of these

Solution

Question 4

{2, 3, 5}

{3, 5}

{2, 3, 4}

{2, 3, 4,5}

Solution

Question 5

$$1/5$$

$$3/8$$

$$2/5$$

$$4/5$$

Solution

Question 6

Equivalence

Symmetric only

Reflexive only

Transitive only

Solution

Question 7

reflexive only

symmetric only

transitive only

equivalence

Solution

Question 8

$$32$$

$$42$$

$$50$$

$$52$$

Solution

Question 9

reflexive

symmetric

transitive

equivalence

Solution

Question 10

Contains exactly one element

Contains exactly two elements

Contains exactly four elements

Is an empty set

Solution

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