Differentiation (Part-38) | S N De

 

In the  previous article , we have discussed about Exercise-2A of the chapter  RELATION AND MAPPING (OR FUNCTION): Part-1  as a part of   S.N.Dey Math solution series. 

In this article, we will discuss  about all solutions of Short and Very Short Answer Type Questions  (Exercise-2B ). So, let us start.

Very Short Answer Type Questions  (Exercise-2B )

1. Define a relation from the set $\,A\,$ to the set $\,B\,$.  

What do you mean by the domain, range and co-domain of a relation?

Sol. If $\, \mathbb R\,$ be a relation from a set $\,A\,$ to a set $\,B\,$,then the domain  of the relation $\, \mathbb R\,$ is the set of all first elements (or co-ordinates) of the ordered pairs which belong to $\, \mathbb R\,$. Thus, Dom. ($\, \mathbb R\,$) [i.e., domain of $\, \mathbb R] = \{x: (x, y) =  \mathbb R\}$

The range of the relation $\, \mathbb R\,$ is the set of all second elements

(or co-ordinates) of the ordered pairs which belong to $\, \mathbb R\,$.  

Thus, Range ($\, \mathbb R\,$) [ie., range of $\, \mathbb R] = \{y: (x, y) = \mathbb R\,\}$.

 From the above definitions it is evident that the domain of a relation from $\,A\,$ to $\,B\,$ is a subset of $\,A\,$ while its range is a subset  of $\,B\,$.

The set $\,B\,$ is called the co-domain of relation $\, \mathbb R\,$.

2. Find all the relations from the set $\,A=(1, 2)\,$ to the set $\,B = {3}.\,$

Sol. The relations from the set $\,A=(1, 2)\,$ to the set $\,B = {3}\,$ is given by : $\,\, \phi, \{(1,3)\},\,\{(2,3)\},\,\{(1,3),(2,3)\}$

3. If $\,\mathbb R\,$ is the relation "is greater than" from $\,A = \{1, 2, 3, 4, 5\}\,$ to $\,\,B = \{1, 3, 4\}\,$, write $\,\mathbb R\,$ as the set of ordered pairs. Also find $\,\mathbb R^{-1}\,$.

Sol. We have,   $\,\mathbb R\,$ is the relation "is greater than" from $\,A = \{1, 2, 3, 4, 5\}\,$ to $\,\,B = \{1, 3, 4\}\,$, which means if $\,\mathbb R\,$ be the relation from $\,A\,$ to $\,B\,\,\,\,$defined by , $\,\,(x,y) \in \mathbb R \Rightarrow x >y.$ Under the defined relation $\,\mathbb R\,$ from $\,A\,$ to $\,B\,\,\,\,$, we can say that as a set of ordered pairs the relation $\,\mathbb R\,$ is given by : $\,\, \mathbb R=\{(2, 1), (3, 1), (4, 1), (4, 3), (5, 1),\\ (5, 3), (5,4)\}$.

By definition , the inverse relation of $\,\mathbb R\,$, i.e.,  $\,\mathbb R^{-1}=\{(y,x): (x,y) \in \mathbb R\}\\=\{(1, 2), (1, 3), (1, 4), (3, 4), (1, 5),\\ (3, 5),(4,5)\}\,$

4. Let $\,\,S = \{a, b, c, d, e\}\,$ and $\,\mathbb R\,\,$ be a relation on $\,S\,$ defined by,

$\,\mathbb R=\{(b, a), (b, d), (d, b), (d, d), (e, b)\},\,\,$  

find the (i) domain (ii) range and (iii) inverse of $\,\mathbb R\,\,$. 

Sol. By definition, we have 

Dom. ($ \mathbb R\,) = \{x: (x, y) = \mathbb R\} = \{b, d, e\}\,$

Range $\,\,(\mathbb R) = \{y: (x, y) =\mathbb R\} = \{a,b,d\}\,\,$ and  

By definition , the inverse relation of $\,\mathbb R\,$, i.e.,  $\,\mathbb R^{-1}=\{(y,x): (x,y) \in \mathbb R\}\\=\{(a, b), (d, b), (b, d), (d, d), (b, e)\}\,$

5. A relation $\,\mathbb R\,\,$ is defined on the set $\,\,A = \{2, 3, 4, 6\}\,\,$ as follows:

$\,\,(x, y) = \mathbb R \Rightarrow x\,\,$ and $\,y\,$ are relatively prime.   Write $\mathbb R$ as a set of ordered pairs. Also find Dom. ($\mathbb R$) and Range ($\mathbb R$).

Sol. $\,\,\mathbb R= \{(2, 3), (3, 2), (3, 4), (4, 3)\} \rightarrow(1)$.  

By definition, we get from (1),  

 Dom. ($\, \mathbb R\,$) [i.e., domain of $\, \mathbb R] = \{x: (x, y) =  \mathbb R\}\\ \implies  \text{Dom.} (\mathbb R) = (2, 3, 4)\,\,\,$   

and Range $(\mathbb R) = \{2, 3, 4\}$

# 6. Find $\,\,\mathbb R^{-1}$ in each of the following cases: $(i)\,\,\mathbb R= \{(1, 2), (1, 3), (2, 4), (2, 3), (3, 2), (4,3)\} \\ (ii) \,\,\mathbb R=\{(x, y): x \in \mathbb N, y \in \mathbb N \wedge 2x+y=10\}$

Sol. (i) By definition , the inverse relation of $\,\mathbb R\,$, i.e.,  

$ \,\mathbb R^{-1}=\{(y,x): (x,y) \in \mathbb R\}\\=\{(2, 1), (3, 1), (4, 2), (3, 2), (2, 3), (3,4)\}\,$  

(ii) In the second case, $\,\,\mathbb R $ can be redefined as $ \,\,\mathbb R=\{(x, y): x \in \mathbb N, y \in  \mathbb N \wedge 2x+y=10\}\\=\{(1,8),(2,6),(3,4),(4,2)\}$

and hence  

$ \,\mathbb R^{-1}=\{(y,x): (x,y) \in \mathbb R\}\\=\{(8, 1), (6, 2), (4, 3), (2,4)\}\,$  

 7. Let $\,\,A = \{2, 3, 4, 5\}\,\,$ and $\,B=\{8, 9, 10, 11\}\,\,$ and let $\,\mathbb R \,$ be a relation from $\,A\,$ to $\,B\,$ defined by, $\,x \,\,\mathbb R\,\, y\, \implies $" x divides y". Find $\,\mathbb R \,$ as a set of ordered pairs and also find its domain and range.

Sol. We have,   $\,\mathbb R\,$ is the relation "x divides y" from $\,A = \{2, 3, 4, 5\}\,$ to $\,\,B = \{8, 9, 10, 11\}\,$, which means if $\,\mathbb R\,$ be the relation from $\,A\,$ to $\,B\,\,\,\,$defined by , $\,\,(x,y) \in \mathbb R \Rightarrow x \,\,\text{divides}\,\,y.$

 Under the defined relation $\,\mathbb R\,$ from $\,A\,$ to $\,B\,\,\,\,$, we can say that as a set of ordered pairs the relation $\,\mathbb R\,$ is given by : $\,\, \mathbb R=\{(2, 8), (2, 10), (3, 9), (4, 8), (5, 10)\}$.

Dom.$\,(\mathbb R) = \{2, 3, 4, 5\}\,\,$ and  

Range $\,( \mathbb R)=\{8, 9, 10\}$

Short Answer Type Questions  (Exercise-2B )

1. Write the following relations as the sets of ordered pairs: $\,(i)\,$ A relation $\, \mathbb R\,$ on the set $\,A\,$ of first six natural numbers defined by,   

$\,(x, y) \in \mathbb R \implies x$ is relatively prime to $\,y\,$

(ii) A relation $\,\mathbb R\,$ defined on the set of natural numbers N by, $\,\,(x, y) \in \,\mathbb R,\,\,\,\, 2x + y = 10\,\,$ for all $\,\,x, y \in \mathbb N$       

(iii) A relation $\,\mathbb R\,$ defined on the set $\,\,A = \{2, 3, 4, 5, 6\}\,$ by, $a \,\mathbb R\, b \implies  |a-b|\,\,$ is divisible by $\,3\,$ 

Sol. (i) $\,\,A=\{1,2,3,4,5,6\} $ 

So, by the given problem we have,    

$\,\mathbb R = \{(x,y): x,y \in A , \text{where, x is relatively  prime to y}\}\\=\{(2, 3), (3, 2), (3, 4), (4, 3), (2, 5), (5, 2), \\(3, 5), (5, 3), (4, 5), (5, 4), (5, 6), (6,5)\}$

(ii)$\,\,2x+y=10 \\ \implies y=(10-2x) \wedge x,y \in \mathbb N $

For respective values of $x=1,2,3,4\,$ we get, $y=8,6,4,2.$ 

Hence, $\,\,\mathbb R=\{ (x,y): x,y \in \mathbb N \wedge  2x + y = 10\}\\=\{(1, 8), (2, 6), (3, 4), (4, 2)\}$

(iii) $\mathbb R=\{(a,b): |a-b| \,\text{is divisible by}\, \,3\, \wedge a,b \in A\} \\ \implies R=\{(2, 2), (3, 3), (4, 4), (5, 5), (6, 6), \\(2, 5),(5,2), (3, 6), (6, 3)\}$

2. A relation $\,\mathbb R\,$ on the set of natural numbers $\,\mathbb N\,$is defined as follows: $\,\mathbb R\,=\{(x, y); x+5y = 20,\,\, x, y \in \mathbb N\};$ find the domain and range of $\mathbb R$. 

Sol. $x+5y = 20 ,\, x,y \in \mathbb N \\  \implies y=20-5y. $

Now, $\,\mathbb R\,=\{(x, y); x+5y = 20,\,\, x, y \in \mathbb N\}; \\ \implies R=\{(15, 1), (10, 2), (5, 3)\} \\ \text{Dom.} (R) = \{15, 10, 5\} \\ \text{and Range} (R) =\{1, 2, 3\}$

3. Let $\,\,A = \{1, 2, 3, 4, 5, 6, 7, 8\}\,\,$ and a relation $\,\mathbb R\,$ on $\,A\,$ is 

given by, $\,\mathbb R=\{(x, y):x \in A, \,\,\, y \in A \,\text{and}\, 2x + y = 12\}$.

Find $\,\mathbb R\,$ and $\,\mathbb R^{-1}$ as sets of ordered pairs. Also find domains and ranges of $\,\mathbb R\,$ and $\,\mathbb R^{-1}$.

Sol. $\, 2x + y = 12, \,\, x,y \in A \\ \implies y=12-2x,\,\, x,y \in A\,$  

Now, for $\,\,x=2,3,4,5\,\,$ respective values of $y=8,6,4,2.$  

Now, $\,\,\mathbb R=\{(x,y): x \in A ,\, y \in A  \wedge 2x+y=12\}\\=\{(2, 8), (3, 6), (4, 4), (5, 2)\} $   

Hence , $\mathbb R^{-1}= \{(y,x): (x,y) \in \mathbb R\}\\=\{(8, 2), (6, 3), (4, 4), (2, 5)\}$

Dom. $( \mathbb R) =\{x: (x,y) \in \mathbb R\}= \{2, 3, 4, 5\}$  

Range $(\mathbb R)=\{y:(x,y) \in  \mathbb R\}=\{8,6,4,2\}$

Now, Dom $(\mathbb R^{-1})=\text{Range} (\mathbb R) \\=\{8,6,4,2\}$

And Range $(\mathbb R^{-1})=\text{Dom}(\mathbb R)= \{2, 3, 4, 5\}$ 

4. Find the domain and range of each of the following relations:

(i) $R_1=\{(a,\frac1a): 0 <a<5 \wedge a\,\, \text{is an integer}\}$

Sol. $R_1=\{(a,\frac1a): 0 <a<5 \wedge a\ \in I,\,\,\text{I being set of integers}\}\\ \Rightarrow R_1=\{(1,1),(2,\frac12),(3,\frac13),(4,\frac14)\}$  

Now, Domain of $R_1=\{a: (a,\frac1a) \in R_1\}=\{1,2,3,4\}$  

Range of $(R_1)=\{\frac1a:(a,\frac1a )\in R_1\}\\=\{1,\frac12,\frac13,\frac14\}$

(ii) $R_2=\{(x,y): x,y \in I \wedge xy=4\}\\ \implies R_2=\{(1,4),(2,2),(4,1),(-1,-4),\\(-2,-2),(-4,-1) \}$ 

Dom. $(R_2)=\{x: (x,y) \in R_2\}\\=\{1,2,4,-1,-2,-4\}$

Range $(R_2)=\{y: (x,y) \in R_2\}\ \\=\{1,2,4,-1,-2,-4\}$

(iii) $R_3=\{(x,y): x,y \in \mathbb N \wedge 2x+y=41 \}$  

Sol. $\,\,2x+y=41,\,\,x,y \in \mathbb N \\ \implies y=41-2x.$  

Now, for $\,\,x=1,2,3,4, \cdots,20, \,$ we get the respective values of $\,\,y=39,37,35,33,\cdots ,1.$  

So, $R_3=\{(x,y): x,y \in \mathbb N \wedge 2x+y=41 \}\\ \implies R_3=\{(1,39),(2,37), \\(3,35),(4,33),\cdots ,(20,1)\}$  

Dom. $(R_3)=\{x: (x,y) \in R_3\}\\=\{1,2,3,4,...,20\}$

Range of $(R_3)=\{y: (x,y) \in R_3\}\\=\{39,37,35,33,...,1\}$

(iv) $R_4=\{(x,y): x,y \in I \wedge x^2+y^2=25\}$

 

Sol. $\,\, x^2+y^2=25 \\ \implies y =\pm \sqrt{25-x^2},\,\,x,y \in I.$  

For $\,\,x=0, \pm3,\pm 4, \pm 5$ , the respective values of $\,y=\pm5, \pm 4,\pm3,0.$ 

Now,$R_4=\{(x,y): x,y \in I \wedge x^2+y^2=25\} \\ \implies R_4=\{(0,5),(0,-5),(3,4),(3,-4),\\(-3,4),(4,3),(-4,3),(4,-3),(5,0),(-5,0)\}$

Dom. $(R_4)=\{x: (x,y) \in R_3\}\\=\{0,3,-3,4,-4,5,-5\}$  

Range of $(R_4)=\{y: (x,y) \in R_3\}\\=\{5,-5,4,-4,3,-3,0\}$

4(v) $R_5=\{(x-5,2x-7): \text{x is an odd natural no.} <10\}$ 

Sol. The odd natural numbers which are less than $\,10\,$ are given by $\,1,3,5,7,9.$  

Now, for $x=1, (x-5,2x-7)=(-4,-5),.....$ and so on.  

Hence from the given problem, we get  

 $R_5=\{(x-5,2x-7): \text{x is an odd natural no.} <10\}\\=\{(-4,-5),(-2,-1),(0,3),(2,7),(4,11)\}$  

Now, dom. $(R_5)=\{x: (x,y) \in R_5\}\\=\{-4,-2,0,2,4\}$ and  

Range of $(R_5)=\{y: (x,y) \in R_5\}\\=\{-5,-7,3,7,11\}$

4(vi) $\,R_6=\{(x,x^2-31): \text{x is prime no.} <12\}$

Sol. The prime numbers less than 12 are given by: $2,3,5,7,11.$  

Now, for $x=2, \,\, x^2-31=-27; \\ x=3, \,\,x^2-31=-22, \\ x=5,\,\,x^2-31=-6 \\  x=7,\,\,x^2-31=18\\  x=11,\,\,x^2-31=90\\$  

So, by the given problem, we have  

$\,R_6=\{(x,x^2-31): \text{x is prime no.} <12\}\\=\{(2,-27),(3,-22),(5,-6),(7,18),(11,90)\}$  

Now, dom. $(R_6)=\{x: (x,y) \in R_6\}\\=\{2,3,5,7,11\}$             

 and  Range of $(R_6)=\{y: (x,y) \in R_6\}\\=\{-27,-22,-6,18,90\}$

4(vii) $R=\{(x,y): \text{x is an integer} \wedge |x|<3\\  \wedge y=|x-3|\}$

Sol. Since $\,x\,$ is an integer and $|x|<3 \\ \Rightarrow x=-2,-1,0,1,2$  

and hence corresponding values of $\,y=|x-3|\,$ is $\,5,4,3,2,1.$

Hence, $\,R=\{(x,y): \text{x is an integer} \\ \wedge |x|<3 \wedge y=|x-3|\} \\ \Rightarrow R=\{(-2,5),(-1,4),(0,3),(1,2),(2,1)\}$

Now, dom. $(R)=\{x: (x,y) \in R\}\\=\{-2,-1,0,1,2\}$             

 and  Range of $(R)=\{y: (x,y) \in R\}\\=\{5,4,3,2,1\}$

4(viii) $\,\,S=\{(x,y): x,y \in \mathbb N \wedge x+3y=12\}$

Sol. $\,\,x+3y=12 \\ \implies x=12-3y,\,x,y \in \mathbb N.$   

Now, for $\,\,y=1,2,3\,$ the corresponding values of $x=9,6,3.$  

Hence, $\,\,S=\{(x,y): x,y \in \mathbb N \wedge x+3y=12\}\\ \implies S=\{(9,1),(6,2),(3,3)\}$ 

Now, dom. $(S)=\{x: (x,y) \in R\}\\=\{9,6,3\}$             

 and  Range of $(S)=\{y: (x,y) \in R\}\\=\{1,2,3\}$

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