Relations and Functions

The $fx$ n is $f(x)=\frac{1}{x}$ , which is a $f:R_{*}$ $\to$ $R_{*}$ and $R_{*}$ is set of all non-zero real numbers

For, $x_1,x_2\in R_{*},f(x_1)=f(x_2)$

$\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}$

$\Rightarrow x_1=x_2$ So, $f$ is one-one

For, $y\in R_{*},$ $x=\frac{1}{f(x)}=\frac{1}{y}$ such that

So, $f(x)=y$

So, every element in the co-domain has a pre-image in $f$

So, $f$ is onto

If $f:N\to R_{*}$ such that $f(x)=\frac{1}{x}$

For, $x_1,x_2\in N,$ $f(x_1)=f(x_2)$

$\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}$

$\Rightarrow x_1=x_2$ So, $f$ is one-one

For, $y\in R_{*}$ and $f(x)=y$ we have $x=\frac{1}{y}\notin N$

Eg., $3\in R_{*}$ so $x=\frac{1}{3}\notin N$

So, $f$ is not onto

The given relation in set N defined by

$R=\left\{\left(a,b\right):a=b-2,b>6\right\}$

For (2,4),        4>6 is not true

For (3,8),     8>6  but  3= 8-2 ⇒3=6 is not true

For (6,8),      8>6 and 6= 8-2 ⇒6=6 is true

And for (8,7), 7>6 but 8= 7-2 ⇒8=5 is not true

Hence, option (C) is correct

The set in $A=\left\{1,2,3,4\right\}$

The relation in this set $A$ is given by

$R=\left\{\left(1,2\right),\left(2,2\right),\left(1,1\right),\left(4,4\right),\left(1,3\right),\left(3,3\right),\left(3,2\right)\right\}$

$R$ is reflexive as $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right)\in R$

As, $\left(1,2\right)\in R$ but $\left(2,1\right)\notin R$

$R$ is not symmetric

For $\left(1,2\right)\in R$ and $\left(2,2\right)\in R;\left(1,2\right)\in R$

And for $\left(1,3\right)\in R$ and $\left(3,2\right)\in R;\left(1,3\right)\in R$

∴ $R$ is transitive

Hence, option (B) is correct

The given relation in the set $L=$ all lines in $XY-$ plane is defined as

$R=\{\left(L_1,L_2\right):L_1$ is parallel to $L_2\}$

Let $L_1\in A$ then as $L_1$ is parallel to $L_1$ ,

$\left(L_1,L_1\right)\in R$

So, $R$ is reflexive

Let $L_1,L_2\in A$ and $\left(L_1,L_1\right)\in R$

Then, $L_1$ is parallel to $L_2$

$L_2$ is parallel to $L_1$

So, $\left(L_2,L_1\right)\in R$

i.e., $R$ is symmetric

Let $L_1,L_2,L_3\in A$ and $\left(L_1,L_2\right)$ and $\left(L_2,L_3\right)\in R$

Then, $L_1?L_2$ and $L_2?L_3$

So, $L_1?L_3$

i.e., $(L_1,L_2)\in R$

So, $R$ is transitive

Hence, $R$ is an equivalence relation

The set of lines related to $y=2x+4$ is given by the equation $y=2x+C$ where $C$ is some constant.

The given relation in set A of points in a plane is

R= $\{\left(P,Q\right):$ distance of point P from origin=distance of point Q from $origin\}$

If O is the point of origin

R= $\left\{\left(P,Q\right):PO=QO\right\}$

Then, for $P\in A$ we have PO=PO

So,  $\left(P,P\right)\in R$

i.e., P is reflexive

for,  $P,Q\in A$ and $\left(P,Q\right)\in R$ we have

PO=QO

QO=PO i.e.,  $\left(Q,P\right)\in R$

i.e., R is symmetric

for $P,Q,S\in A$ and $\left(P,Q\right)\&\left(Q,S\right)\in R$

PO=QO and QO=SO

PO=SO

i.e.,  $\left(P,S\right)\in R$

so, R is transitive

Hence, R is an equivalence relation

For a point $P\ne \left(o,o\right)$ the set of all points related to P i.e., distance from origin to the points are equal is a circle with center at origin (o, o) by the definition of circle

Let A= $\left\{a,b,c\right\}$

(i) R= $\left\{\left(a,b\right),\left(b,a\right)\right\}$ is a relation in set A

So, $\left(a,b\right)\in R$ and $\left(b,a\right)\in R\to$ Symmetric

$\left(a,a\right)\notin R\to$ not reflexive

$\left(a,b\right)\in R,\left(b,a\right)\in R$ but $\left(a,a\right)\notin R$ $\to$ not transitive

(ii) R= $\left\{\left(a,b\right),\left(b,c\right),\left(a,c\right)\right\}$ is a relation in set A

So, $\left(a,a\right)\notin R\to$ not reflexive

$\left(a,b\right)\in R$ but $\left(b,a\right)\notin R\to$ not symmetric

$\left(a,b\right)\in R\&\left(b,c\right)\in R$ and also $\left(a,c\right)\in R\to$ transitive

(iii) R= $\left\{\left(a,a\right),\left(b,b\right),\left(c,c\right),\left(a,b\right),\left(b,a\right),\left(a,c\right),\left(c,a\right)\right\}$

So, $\left(a,a\right),\left(b,b\right),\left(c,c\right)\in R\to$ Reflexive

$\left(a,b\right)\in R\to \left(b,a\right)\in R\to$ Symmetric

$\left(a,c\right)\in R\to \left(c,a\right)\in R$

$\left(b,a\right)\in R$ and $\left(a,c\right)\in R$

But $\left(b,c\right)\notin R\to$ not transitive

(iv) R= $\left\{\left(a,a\right),\left(b,b\right),\left(c,c\right),\left(a,b\right),\left(b,c\right),\left(a,c\right)\right\}$ is s relation in set A

So, $\left(a,a\right),\left(b,b\right),\left(c,c\right)\in R\to$ reflexive

$\left(a,b\right)\&\left(b,c\right)\in R$ so, $\left(a,c\right)\in R\to$ transitive

$\left(a,b\right)\in R$ but $\left(b,a\right)\notin R\to$ not symmetric

(v) R= $\left\{\left(a,a\right),\left(a,b\right),\left(b,a\right)\right\}$

So, $\left(b,b\right)\notin R\to$ not reflexive

$\left(a,b\right)\in R$ and $\left(b,a\right)\in R\to$ symmetric

And $\left(a,b\right)\in R\&\left(b,a\right)\in R$

and also $\left(a,a\right)\in R\to$ transitive