If $$A = \left \{1, 2, 3, 4, 5\right \}$$ then

The number of relation on $$A$$ is $$2^{25}$$

The number of relation on $$A$$ is $$2^{20}$$

The number of reflexive relation on $$A$$ is $$20^{20}$$

The number of reflexive relation on $$A$$ is $$2^{15}$$

If $$g(f(x))= |sinx|$$ and $$g(f(x))= sin^2 \sqrt{x},$$ then

$$f(x)= \sqrt{x}, \ g(x)= sin^2x$$

$$f(x) = sin \sqrt{x}, \ g(x)= x^2$$

$$f(x)= \sqrt{sinx}, \ g(x)= x^2$$

If $$f(x) = -{ \frac { x\left| x \right| }{ 1+x^{ 2 } } }$$ then $$f^{-1} (x) $$ equals

$$(Sgn x) \sqrt { \frac { x\left| x \right| }{ 1-\left| x \right| } } $$

$$\sqrt { \frac { x\left| x \right| }{ 1-\left| x \right| } } $$

MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 16

Let

$R:N \to N$ be defined by

$R = \{ (a,b):a,b \in N$ and

$a = {b^2}\}$ then, which of the following is true?

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$(a,a) \in R,\forall a \in N$
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$(a,b) \in R, \Rightarrow (b,a) \in R$
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$(a,b) \in R,(b,c) \in R \Rightarrow (a,c) \in R$
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None of the above

Let

$A$ be set of first ten natural numbers and

$R$ be a relation on

$A$ defined by (x , y)

$\in$

$R$

$\Rightarrow$ x + 2y = 10 , then domain of

$R$ is

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$\left \{ 1 , 2 , 3 ,............, 10 \right \}$
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$\left \{ 2 , 4 , 6 , 8 \right \}$
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$\left \{ 1 , 2 , 3 , 4 \right \}$
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$\left \{ 2 , 4 , 6 , 8 , 10 \right \}$

Explanation

$\begin{aligned} A &=\{1,2,3, \cdots \cdots, 10\} \\ R &=\{(x, y): x+2 y=10, x, y \in A\} \\ & \therefore y=\dfrac{10-x}{2} \\ R &=\{(2,4),(4,3),(6,2),(8,1)\} \\ \therefore & \text { Domain of } R \text { is }\{2,4,6,8\} \end{aligned}$

Correct option is (B).

$h=\left\{ ((x,y),(x-y, x+y))/ x,y \epsilon N \right\}$ is a relation on NxN, then domain of h

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$$\left\{ (x,y):\quad x
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$\left\{ (x,y):\quad x\le y,x,y\epsilon N \right\}$
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$\left\{ (x,y):\quad x>y,x,y\epsilon N \right\}$
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$\left\{ (x,y):\quad x \ge y,x,y\epsilon N \right\}$

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

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${(1, 1), (2, 2)}$
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${(1, 1)}$
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$\{ \phi \}$
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None of these.

Let

$R_1, R_2$ are relation defined on

$Z$ such that

$aR_1b \Longleftrightarrow (a - b)$ is divisible by

$3$ and

$a \,R_2 b \Longleftrightarrow (a - b)$ is divisible by

$4$ . Then which of the two relation

$(R_1 \cup R_2), (R_1 \cap R_2)$ is an equivalence relation?

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$(R_1 \cup R_2)$ Only
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$(R_1 \cap R_2)$ Only
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Both $(R_1 \cup R_2), (R_1 \cap R_2)$
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Neither $(R_1 \cup R_2)$ nor $(R_1 \cap R_2)$

Explanation

Givn relations defined on z are,

$R_1=aR_1b\leftrightarrow (a-b)$ is divisible by 3.

and

$R_2=aR_2b\leftrightarrow (a-b)$ is divisible by 4.

Consider,

$aR_1b\leftrightarrow a-b$ is divisible by 3.

Reflexive:

$aR_1a\leftrightarrow a-a$ is divisible by 3

True

$\therefore$ the given relation

$R_1$ is a reflexive relation.

Symmetric:

$aR_1b\leftrightarrow a-b$ is divisible by 3.

$\Rightarrow a-b=3k$ ; where

$k$ is an integer.

then

$bR_1a\leftrightarrow b-a$ is divisible by 3.

because

$b-a=-(a-b)=-3k$ which is divisible by 3.

$\therefore$ the given relation

$R$ is a symmetric relation.

Transitive:

$aR_1b\leftrightarrow a-b$ is divisible by 3.

$\Rightarrow a-b=3k; k$ is an integer.

$b R_1c\leftrightarrow b-c$ is divisible by 3.

$\Rightarrow b-c=3p;p$ is integer.

Then

$aR_1c\leftrightarrow a-c$ is divisible by 3.

Because

$(a-b)+(b-c)=a-c$

$\Rightarrow 3k+3p=3(k+p)$ which is divisible by 3.

$\therefore$ The given relation

$R_1$ is transitive relation.

$\therefore R_1$ satisfies the reflexive, symmetric and transitive relation properties.

$\therefore R_1$ is an equivalence relation.

Similarly

$R_2$ is also an equivalence relation.

And as we know that,
"If

$R_1$ and

$R_2$ are equivalence relations then

$R_1\cap R_2$ is also an equivalence relation."

$\therefore R_1\cap R_2$ is an equivalence relation.

Let

$R_1$ be a relation defined by

$R_1 =$ {

$(a, b)| a >b , a, b \epsilon R$ }. Then

$R_1$ is

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an equivalence relation on R
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reflexive, transitive but not symmetric
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symmetric, transitive but nor reflective
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neither transitive nor reflective but symmetric

If a relation { R } on the set { 1,2,3 } be definedby R = { ( 1,2 ) } then { R } is

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reflexive
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transitive
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Symmetric
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None of these

The relation R defined in N as

$aRb \Leftrightarrow b$ is divisible by a is

0%
Reflexive but not symmetric
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Symmetric but not transitive
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Symmetric and transitive
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Equivalence relation

$(x^{2}-2)+(y+3)i=7+4i$ then x and y are

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$\pm 3 and 4$
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$\pm \sqrt{5} and 4$
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$\pm 3 and 1$
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None of these

Let T be the set of all triangles in the Euclidean plane, and let a relation R' on T be defined as aRb if a is congruent to b for all a, b

$\in$ T. Then, R is?

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Reflexive but not symmetric
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Transitive but not symmetric
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Equivalence
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None of these

Let

$R = \left \{(2, 3), (3, 3), (2, 2), (5, 5), (2, 4), (4, 4), (4, 3)\right \}$ be a relation on the set

$\left \{2, 3, 4, 5\right \}$ , then

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$R$ is reflexive and symmetric but not transitive
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$R$ is reflexive and transitive but not symmetric
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$R$ is symmetric and transitive but not reflexive
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$R$ is an equivalence relation

$A = \left \{1, 2, 3, 4, 5\right \}$ then

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The number of relation on $A$ is $2^{25}$
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The number of relation on $A$ is $2^{20}$
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The number of reflexive relation on $A$ is $20^{20}$
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The number of reflexive relation on $A$ is $2^{15}$

$A=\left\{1,2,4 \right\}$ ,

$B=\left\{1,4,6,9 \right\}$ and

$R$ is a relation from

$A$ to

$B$ defined by

$x$ is greater than

$y$ . The rangle of

$R$ is :

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$\left\{1,4,6,9 \right\}$
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$\left\{4,6,9 \right\}$
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$\left\{ 1 \right\}$
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none of these

$g(f(x))= |sinx|$ and

$g(f(x))= sin^2 \sqrt{x},$ then

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$f(x)= \sqrt{x}, \ g(x)= sin^2x$
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$f(x) = sin \sqrt{x}, \ g(x)= x^2$
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$f(x)= \sqrt{sinx}, \ g(x)= x^2$
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$f(x)= |x|, \ g(x)= sin^2 x$

$A=\{ 1,2,3\}$ then the number of equivalance relation containing the element {1,2} is :

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1
0%
2
0%
3
0%
14

Let S be the set of all straight lines in a plane. Let R be a relation on S defined by

$a$ R

$b\Leftrightarrow a||b$ . Then, R is?

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Reflexive and symmetric but not transitive
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Reflexive and transitive but not symmetric
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Symmetric and transitive but not reflexive
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An equivalence relation

$f(x) = -{ \frac { x\left| x \right| }{ 1+x^{ 2 } } }$ then

$f^{-1} (x)$ equals

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$\sqrt { \frac { x\left| x \right| }{ 1-\left| x \right| } }$
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$(Sgn x) \sqrt { \frac { x\left| x \right| }{ 1-\left| x \right| } }$
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$- \sqrt { \frac {x} {1-x}}$
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None of these

Explanation

$\textbf{Step-1: Apply the concept of function & simplify}$

$\text{Let y = f(x)}$

$= \dfrac{-x^2}{1+x^2},x\geq0$

$= \dfrac{x^2}{1+x^2}, x<0$

$f^{-1}(y)=$

$\sqrt{\dfrac{-y}{1+y}},y<0$

$= -\sqrt{\dfrac{y}{1-y}},y>0$

$f'^{-1}(x) = \sqrt{\dfrac{-x}{1+x}}, x<0$

$=\sqrt{\dfrac{-x}{1-x}},x>0$

$\text{= (sgn x)} \sqrt{\dfrac{|x|}{1-|x|}}$

$\textbf{Hence, option B}$

From the following relations defined on set Z of integers, which of the relation is not equivalence relation

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$aR_{1} b \Leftrightarrow ( a + b)$ is an even integer
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$aR_{1} b \Leftrightarrow ( a - b)$ is an even integer
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$aR_{3} b \Leftrightarrow ( a + b) a < b$
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$aR_{4} b \Leftrightarrow ( a + b) a = b$

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 16 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 16 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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