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CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 7 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Relations
Quiz 7
Let $$X=\left\{ 1,2,3,4 \right\} $$ and $$Y=\left\{ 1,2,3,4 \right\} $$. Which of the following is a relation from $$X$$ to $$Y$$.
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$${R}_{1}=\left\{ (x,y)| y=2+x, x\in X, y\in Y \right\} $$
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$${R}_{2}=\left\{ (1,1),(2,1),(3,3),(4,3),(5,5) \right\} $$
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$${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\} $$
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$${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\} $$
Explanation
Given $$X=\left\{ 1,2,3,4 \right\} $$ and $$Y=\left\{ 1,2,3,4 \right\} $$
$$X\times Y=\{ (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)$$
$$(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)\} $$
Now, $$R_1=\{(1,3),(2,2)\}$$
Since, $$R_1\subset X\times Y$$
Hence, $$R_1$$ is a relation from $$X$$ to $$Y$$
$${R}_{2}=\left\{ (1,1),(2,1),(3,3),(4,3),(5,5) \right\} $$
Since, $$(5,5) \in R_2$$ but $$\notin X\times Y$$
So, $$R_2$$ is not a relation from $$X$$ to$$Y$$
$${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\} $$
Since, $$(3,5),(3,7),(5,7) \in R_3$$ but $$\notin X\times Y$$
So, $$R_3$$ is not a relation from $$X$$ to $$Y$$
$${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\} $$
Since, $$(2,5),(7,9) \in R_4$$ but $$\notin X\times Y$$
So, $$R_4$$ is not a relation from $$X$$ to $$Y$$
Let $$L$$ be the set of all straight lines in the Euclidean plane. Two lines $${l}_{1}$$ and $${l}_{2}$$ are said to be related by the relation $$R$$ if $${l}_{1}$$ is parallel to $${l}_{2}$$. Then the relation $$R$$ is-
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Reflexive
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Symmetric
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Transitive
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Equivalence
Explanation
$$l_1 R l_1 \Rightarrow l_1$$ is parallel to itself. Reflexive
$$l_1 Rl_2\Rightarrow l_2Rl_1$$ which is true. Symmetric
$$l_1Rl_2$$ and $$l_2Rl_3 \Rightarrow l_1Rl_3$$ which is true. Transitive. Hence all options are correct.
Let $$g\left( x \right)=1+x-\left[ x \right] $$ and $$f\left( x \right)=\begin{cases} \begin{matrix} -1 & x<0 \end{matrix} \\ \begin{matrix} 0 & x=0 \end{matrix} \\ \begin{matrix} 1 & x>0 \end{matrix} \end{cases}$$. Then for all $$x,f\left[ g\left( x \right) \right] $$ is equal to
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$$x$$
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$$1$$
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$$f(x)$$
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$$g(x)$$
Explanation
We have,
$$g\left( x \right)=\begin{cases} 1+n-n=1,x=n\in Z \\ 1+n+k-n=1+k,x=n+k \end{cases}$$
where $$n\in Z,0<k<1$$
Now $$f\left[ g\left( x \right) \right] =\begin{cases} \begin{matrix} -1 & g\left( x \right)<0 \end{matrix} \\ \begin{matrix} 0 & g\left( x \right)=0 \end{matrix} \\ \begin{matrix} 1 & g\left( x \right)>1 \end{matrix} \end{cases}$$
Clearly, $$g(x)>0$$ for all $$x$$. So $$f\left[ g\left( x \right) \right] =1$$, for all $$x$$.
If $$A = \left\{1,2,3\right\}$$, $$B = \left\{1,4,6,9\right\}$$ and $$R$$ is relation from $$A$$ to $$B$$ defined by $$'x'$$ is greater than $$'y'$$. Then range 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
Explanation
Here $$R$$ is a relation $$A$$ to $$B$$ defined by $$x$$ is greater than $$y$$.
$$\therefore\quad R = \left\{(2,1); (3,1)\right\}$$. Hence, range of $$R = \left\{1\right\}$$
from the given statement $$N$$ denotes the natural number and $$W$$ denotes the whole number, so which statement in the following is correct
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N=W
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N $$\subset$$ W
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W $$\subset$$ N
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N $$\cong$$ W
Explanation
Natural numbers are naturally occurring counts of an object $$1, 2, 3, 4,… $$forming an infinite list in which the first number is $$1$$ and every next number is equal to the preceding number $$+1.$$
Whole number include the set of all Natural numbers and also $$0.$$
Let $$N$$ and $$W$$ elements of natural and whole numbers.
$$N=\{1,2,3,4,....\infty\}$$
$$W=\{0,1,2,3,4,...\infty\}$$
So, here we can say $$N$$ is subset of $$W.$$
$$\therefore$$ $$N\subset W$$ is correct statement.
Which one of the following relations on $$R$$ is equivalence relation
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$$x \space R_1 \space y \Leftrightarrow |x| = |y|$$
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$$x \space R_2 \space y \Leftrightarrow x \ge y$$
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$$x \space R_3 \space y \Leftrightarrow x | y$$
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$$x \space R_4 \space y \Leftrightarrow x < y$$
Explanation
Let's take option A,
$$x \space R_1 \space y \Leftrightarrow |x| = |y|$$
Now, $$|x|=|x|$$
$$\Rightarrow xR_{1}x$$
Hence, $$R_1$$ is reflexive.
Now, let $$xR_1 y$$
$$\Rightarrow |x|=|y|$$
or $$|y|=|x|$$
$$\Rightarrow yR_{1}x$$
Hence, $$R_1$$ is symmetric
Now, let $$xR_1 y, yR_1 z$$
$$\Rightarrow |x|=|y|, |y|=|z|$$
$$\Rightarrow |x|=|z|$$
$$\Rightarrow xR_{z}x$$
Hence, $$R_1$$ is transitive.
Hence, $$R_1$$ is an equivalence relation.
Clearly, $$R_2$$ is not an equivalence relation as $$2 \ge 3$$ does not implies $$3\ge 2$$
Also, $$R_3$$ is not an equivalence relation as $$x$$ divides $$y $$, does not implies $$y$$ divides $$x$$
Also $$R_4$$ is not an equivalence relation because $$2 <3$$ but $$3 >2$$
Given the relation $$R = \left\{(1,2), (2,3)\right\}$$ on the set $$A = \left\{1,2,3\right\}$$, the minimum number of ordered pairs which when added to $$R$$ make it an equivalence relation is
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$$5$$
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$$6$$
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$$7$$
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$$8$$
Explanation
$$R$$ is reflexive if it contains $$(1,1) (2,2) (3,3)$$
$$\because (1,2) \in R,\space (2,3) \in R$$
$$\therefore R$$ is symmetric if $$(2,1), (3,2) \in R$$
Now, $$R = \left\{(1,1), (2,2), (3,3), (2,1), (3,2), (2,3), (1,2)\right\}$$
$$R$$ will be transitive if $$(3,1); (1,3)\in R$$. Thus, $$R$$ becomes and
equivalence relation by adding $$(1,1) (2,2) (3,3) (2,1) (3,2) (1,3) (1,2)$$. Hence,
the total number of ordered pairs is $$7$$.
$$A$$ 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
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$$2^5$$
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$$2^{10} - 1$$
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$$2^{12} - 1$$
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$$none\ of\ these$$
Explanation
`if $$A$$ and $$B$$ are two sets which contains 3 and 4 elements respectively.
Then no of elements in relation from $$A$$ to $$B$$ is $$2^(3\times 4)$$
means $$2^{12}$$
If $$A = \left\{2,3\right\}$$ and $$B = \left\{1,2\right\}$$, then $$A \times B$$ is equal to
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$$\left\{(2,1), (2,2), (3,1), (3,2)\right\}$$
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$$\left\{(1,2), (1,3), (2,2), (2,3)\right\}$$
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$$\left\{(2,1), (3,2)\right\}$$
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$$\left\{(1,2), (2,3)\right\}$$
Explanation
If $$A$$ and $$B$$ are any two non-empty sets.
then $$A\times B$$$$=\left\{(x,y):x\in A and y\in B\right\}$$
As $$A = \left\{2,3\right\}$$ and $$B = \left\{1,2\right\}$$
$$A \times B$$$$=\left\{(2,1), (2,2), (3,1), (3,2)\right\}$$
Hence, option A.
Let $$X = \left\{1,2,3,4\right\}$$ and $$Y = \left\{1,3,5,7,9\right\}$$. Which of the following is relations from $$X$$ to $$Y$$
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$$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$$
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$$R_2 = \left\{(1,1),(2,1),(3,3),(4,3),(5,5)\right\}$$
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$$R_3 = \left\{(1,1),(1,3),(3,5),(3,7),(5,7)\right\}$$
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$$R_4 = \left\{(1,3),(2,5), (2,4), (7,9)\right\}$$
Explanation
We have
$$X = \left\{1,2,3,4\right\}$$ and $$Y = \left\{1,3,5,7,9\right\}$$.
$$X \times Y=\{(1,1),(1,3),(1,5),(1,7),(1,9), (2,1),(2,3),(2,5),$$
$$(2,7),(2,9), (3,1),(3,3),(3,5),(3,7),(3,9),(4,1),(4,3),(4,5),(4,7),(4,9)\}$$
Let's take option A,
$$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$$
$$R_{1}= \{(1,3),(2,5),(3,7), (4,9)\}$$
Since, $$R_1 \subseteq X\times Y$$
So, $$R_1$$ is a relation from $$X$$ to $$Y$$
Clearly, $$R_2$$ is not a relation from $$X$$ to $$Y$$ as $$(5,5) \notin X\times Y$$
Similarly, $$R_3$$ is not a relation from $$X$$ to $$Y$$ as $$(5,7) \notin X\times Y$$
In the same manner, $$R_4$$ is also not a relation from $$X$$ to $$Y$$ as $$(7,9) \notin X\times Y$$
If $$R$$ is a relation from a finite set $$A$$ having $$m$$ elements to a finite set $$B$$ having $$n$$ elements, then the number of relations from $$A$$ to $$B$$ is:
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$$2^{mn}$$
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$$2^{mn} - 1$$
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$$2mn$$
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$$m^n$$
Explanation
$$R$$ is a relation from a finite set $$A$$ having $$m$$ elements to a finite set $$B$$ having $$n$$ elements.
Number of elements in $$A\times B$$ $$=mn$$
No of relations $$=$$ number of subsets of $$A\times B$$ $$=2^{mn}$$
Hence, option A is correct.
The relation $$R$$ is defined in $$A = \left\{1,2,3\right\}$$ by $$a\ R$$ $$b$$ if $$|a^2 - b^2| \le 5$$. Which of the following is false?
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$$R = \left\{(1,1), (2,2), (3,3), (2,1), (1,2), (2,3), (3,2)\right\}$$
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$$R^{-1} = R$$
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Domain of $$R = \left\{1,2,3\right\}$$
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Range of $$R = \left\{5\right\}$$
Explanation
$$R=\left\{ (1,1),(2,2),(3,3),(2,1),(1,2),(2,3),(3,2) \right\} $$
Clearly $$R^{-1} = R$$
Domain $$=\{1,2,3\}$$ Range $$=\{1,2,3\}$$
Hence option 'D' is correct choice.
If $$R = \{(x, y):3x + 2y = 15 \text{ and }x\,, y\,\in \, N\}$$, the range of the relation R is .........
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$$\{1, 2\}$$
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$$\{1, 2,..., 5\}$$
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$$\{1, 2,..., 7\}$$
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$$\{3, 6\}$$
Explanation
Since x and y belong to natural numbers we need to find positive integral solutions for $$3x+2y=15$$
$$2y=15-3x$$
$$\Rightarrow y =\dfrac{15-3x}{2}$$
For $$x=0,2,4$$ $$y$$ will become fractional which is not possible
For $$x\geq 5$$ y becomes 0 or negative which again is not possible.
Hence for $$x=1,3$$ y= 6,3$$ respectively.
Hence the range is $$\{3,6\}$$
Let $$n(A) = n$$. Then the number of all relations on $$A$$ is
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$$\displaystyle 2^{n}$$
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$$\displaystyle 2^{\left ( n \right )!}$$
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$$\displaystyle 2^{n^{2}}$$
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none
Explanation
For any set $$A$$ such that $$n(A)=n$$
then number of all relations on $$A$$ is $${ 2 }^{ n^{2} }$$
As the total number of Relations that can be defined from a set $$A$$ to $$B$$ is the number of possible subsets of $$ A \times B$$. If $$n(A) = p$$ and $$n(B) = q$$ then $$n(A \times B) = pq$$ and the number of subsets of $$A \times B$$ = $$2^{pq}$$.
If $$R$$ is an equivalence relation in a set $$A$$, then $$R^{-1}$$ is
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Reflexive but not symmetric
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Symmetric but not transitive
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An equivalence relation
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None of these
Explanation
Let R be a relation on A i.e. $$R\subseteq A\times A$$
$$R=\{(a,b)|a,b \in A \}$$
Also, given $$R$$ is equivalence relation,
Now, let $$R^{-1}=\{(b,a) | (a,b) \in R\}$$
We will check whether $$R^{-1}$$ is reflexive, symmetric, transitive or an equivalence relation.
Reflexive:
Since, $$R$$ is reflexive
$$\Rightarrow (a,a)\in R$$
$$\Rightarrow (a,a) \in R^{-1}$$ (by def of $$R^{-1}$$)
Hence, $$R^{-1}$$ is reflexive.
Symmetric: Let $$(b,a) \in R^{-1}$$
$$\Rightarrow (a,b) \in R$$
(by def of $$R^{-1}$$)
$$\Rightarrow (b,a) \in R$$ (Since, R is symmetric)
$$\Rightarrow (a,b) \in R^{-1}$$
(by def of $$R^{-1}$$)
Hence, $$R^{-1}$$ is symmetric.
Transitive : Let $$(b,a), (a,c) \in R^{-1}$$
$$\Rightarrow (a,b), (c,a) \in R$$
(by def of $$R^{-1}$$)
or $$(c,a)(a,b) \in R$$
$$\Rightarrow (c,b) \in R$$ (since, R is transitive.)
$$\Rightarrow (b,c) \in R^{-1}$$
Hence, $$R^{-1}$$ is transitive.
Hence, $$R^{-1}$$ is an equivalence relation.
If $$\displaystyle A=\left\{ 2,4,5 \right\} ,B=\left\{ 7,8,9 \right\} $$ then $$\displaystyle n\left( A \times B \right) $$ is equal to
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$$6$$
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$$9$$
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$$3$$
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$$0$$
Explanation
$$\displaystyle A=\left\{ 2,4,5 \right\} ,B=\left\{ 7,8,9 \right\} $$
$$\Rightarrow n(A)=3$$ and $$n(B)=3$$
$$\therefore n(A\times B)=n(A)n(B)=9$$
Hence, option B.
If $$(3p+q,p-q)=(p-q,3p+q)$$, then:
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$$p=q=0$$
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$$p=q$$
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$$p=2q$$
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$$p+q=0$$
Explanation
As the ordered pairs are equal,
$$ 3p + q = p - q $$
$$ => 2p = -2q $$
$$ => 2p + 2q = 0 $$
$$ => p + q = 0 $$
Let $$A = \left\{p,q,r\right\}$$. Which of the following is an equivalence relation in $$A$$?
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$$R_1 = \left\{(p,q), (q,r), (p,r), (p,p)\right\}$$
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$$R_2 = \left\{(r,q), (r,p), (r,r), (q,q)\right\}$$
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$$R_3 = \left\{(p,p), (q,q), (r,r), (p,q)\right\}$$
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None of these
Explanation
Given $$A=\{p,q,r\}$$
$$R_1 = \left\{(p,q), (q,r), (p,r), (p,p)\right\}$$
$$R_1$$ is not reflexive as it does not have $$(q,q),(r,r)$$ . So, $$R_1$$ is not an equivalence relation on A.
$$R_2 = \left\{(r,q), (r,p), (r,r), (q,q)\right\}$$
$$R_2$$ is not reflexive as it does not have $$(p,p)$$ . So, $$R_2$$ is not an equivalence relation on A.
$$R_3 = \left\{(p,p), (q,q), (r,r), (p,q)\right\}$$
$$R_3$$ is reflexive but not symmetric as $$(p,q)$$ is there ,but not $$(q,p)$$
If $$A = \{1, 2 \}$$ and $$B = \{3, 4\}$$ then find $$A \times B$$
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$$A \times B = \{(1,3),(1,2),(2,3),(2,4)\}$$
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$$A \times B = \{(1,3),(1,4),(2,3),(2,4)\}$$
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$$A \times B = \{(1,3),(1,4),(2,1),(2,4)\}$$
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$$A \times B = \{(1,3),(1,4),(2,3),(2,1)\}$$
If $$\displaystyle R_{n}=\left \{ x:\frac{-1}{n}< x< \frac{1}{n} \right \}$$ then $$\displaystyle R_{5}\cup R_{15}=$$ ________
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$$\displaystyle R_{5}$$
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$$\displaystyle R_{15}$$
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$$\displaystyle R_{3}$$
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$$\displaystyle R_{20}$$
Explanation
$$ R_5 = $$ { $$ x:-\frac {1}{5} < x < \frac {1}{5} $$ }
And
$$ R_{15} = $$ {$$ x:-\frac {1}{15} < x < \frac {1}{15} $$ }
But, $$ \frac {1}{15} < x < \frac {1}{15} $$ is within the limits of $$ \frac {1}{5} < x < \frac {1}{5} $$
Hence, $$ R_5 \cup R_{15} = \frac {1}{5} < x < \frac {1}{5} = R_5 $$
a R b if "a and b are animals in different zoological parks" then R is
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only reflexive
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only symmetric
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only transitive
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equivalence
Explanation
A relation is said to be symmetric if an element a is related to b then b is also related to a.
Since the relation defines about the animals in different parks, it will be symmetric as the animals are same.
R is a relation on set A then $$\displaystyle \left [ \left ( R^{-1} \right )^{-1} \right ]^{-1}$$ is_______
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R
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$$\displaystyle R^{-1}$$
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$$\displaystyle A\times A$$
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None of these
Explanation
We know that Inverse of inverse of a Relation is the original Relation itself.
So, $$ \left[ \left( { R }^{ -1 } \right) ^{ -1 } \right] ^{ -1 } = \left[ R \right] ^{ -1 } $$
A relation which satisfies reflexive symmetric and transitive is ________ relation
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an identity
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a constant
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an equivalence
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None of these
Explanation
A relation which satisfies reflexive symmetric and transitive is _called an equivalence relation
The domain of the function f(x) = $$\displaystyle \frac{\left | x \right |-2}{\left | x \right |-3}$$ is ________
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R
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R - {2, 3}
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R - {2, -2}
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R - {-3, 3}
Explanation
The function has a real value only when denominator of the fraction is not equal to zero.
So, $$\left| x \right| - 3 \neq = 0 $$
$$ => \left| x \right| \neq = 3 $$
$$ => x \neq = 3 ; -3 $$
So, the domain of the function is $$ R - $$ { $$ -3,3 $$ }
If $$\displaystyle R=R^{-1}$$ then the relation R is ________
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reflexive
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symmetric
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anti-symmetric
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transitive
Explanation
When a relation is equal to its inverse, then the relation is symmetric.
Write the properties that the relation "is greter that" satisfies in the set of all positive integers
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Reflexive
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Symmetric
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Antisymmetric
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Transitive
Explanation
If in a relation, an element $$ a $$ is related to $$ b $$, and $$ b $$ is related to a third element $$ c $$, the relation $$ R $$ is said to be transitive if $$ a $$ is also related to $$ c $$
Clearly, when $$ a > b $$ and $$ b > c $$ then $$ a $$ $$ > c $$
Hence, the relation is transitive.
If $$\displaystyle n\left ( A\times B \right )=36$$ then n(A) can possibly be____
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$$7$$
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$$8$$
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$$9$$
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$$10$$
Explanation
$$n(A\times B)=n(A)\times n(B)$$
Hence $$n(A)$$ must be a factor of $$36$$. only possible answer is $$B:9$$
The domain of the relation R = $$\displaystyle \left \{ \left ( x,y \right ):x,y\epsilon N \ and\ x+y\leq 3 \right \}$$ is____
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{1,2,3}
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{1,2}
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{...-1,0,1,2,3}
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None of these
Explanation
The only natrual numbers, less than equal to $$ 3 $$ are $$ 1, 2, 3 $$
From these, only the sum of $$ 1 ,2 $$ gives $$ 3 $$
So the domain of he relation is {$$ 1,2$$}
The relation 'is a sister of' in the set of human beings is____
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only transitive
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only symmetric
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equivalent
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None of these
Explanation
A relation is said to be transitive when an element $$ a $$ is related to $$ b $$ and element $$ b $$ is related to $$ c $$, then $$ c $$ is also related to $$ a $$
We know that if $$ a $$ is the sister of $$ b $$ and $$ b $$ is sister of $$ c $$, then $$ c $$ is also a sister of $$ a $$
Thus, the relation " is a sister of " is a transitive relation.
The relation 'is a factor of' on the set of natural numbers is not___________
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reflective
0%
symmetric
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anti symmetric
0%
transitive
Explanation
A relation is said to be symmetric if an element a is related to b then b is also related to a.
Suppose factor of a number say $$ 10 = 1,2,5,10 $$
We can notice that factors of $$ 10 $$ are all these numbers, but the reverse is not true for each factor. That is $$ 10 $$ is not a factor of $$ 1 $$ or $$ 2 $$ or $$ 5 $$.
So, this relation is not symmetric.
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