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CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 16 - MCQExams.com
CBSE
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).
If $$h=\left\{ ((x,y),(x-y, x+y))/ x,y \epsilon N \right\} $$ 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\} $$
If $$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$$ is
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an equivalence relation on R
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reflexive, transitive but not symmetric
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symmetric, transitive but nor reflective
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neither transitive nor reflective but symmetric
If a relation { R } on the set { 1,2,3 } be defined
by R = { ( 1,2 ) } then { R } is
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reflexive
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transitive
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Symmetric
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None of these
The relation R defined in N as $$aRb \Leftrightarrow b$$ is divisible by a is
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Reflexive but not symmetric
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Symmetric but not transitive
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Symmetric and transitive
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Equivalence relation
If $$(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$$ is an equivalence relation
If $$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}$$
If $$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\} $$
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none of these
If $$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$$
If $$A=\{ 1,2,3\} $$ then the number of equivalance relation containing the element {1,2} is :
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1
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2
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3
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14
Let S be the set of all straight lines in a plane. Let R be a relation on S defined by $$a$$ R $$b\Leftrightarrow a||b$$. Then, R is?
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Reflexive and symmetric but not transitive
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Reflexive and transitive but not symmetric
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Symmetric and transitive but not reflexive
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An equivalence relation
If $$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 $$
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