Chapter 1: Relations and Functions

Ans: The function \[\text{f: R }\to \text{  }\!\!\{\!\!\text{ x}\in \text{R:-1  x  1 }\!\!\}\!\!\text{ }\] is defined as \[\text{f}\left( \text{x} \right)\text{=}\dfrac{\text{x}}{\text{1+ }\!\!|\!\!\text{ x }\!\!|\!\!\text{ }}\text{,x}\in \text{R}\].

For the function \[\text{f}\] to be one – one:

\[\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)\], where \[\text{x, y}\in \text{R}\].

\[\Rightarrow \dfrac{\text{x}}{\text{1+ }\!\!|\!\!\text{ x }\!\!|\!\!\text{ }}\text{=}\dfrac{\text{y}}{\text{1+ }\!\!|\!\!\text{ y }\!\!|\!\!\text{ }}\] 

Assuming that \[\text{x}\] is positive and \[\text{y}\] is negative:

\[\dfrac{\text{x}}{\text{1+x}}\text{=}\dfrac{\text{y}}{\text{1+y}}\]

\[\Rightarrow \text{2xy=x-y}\]

Since, \[\text{x  y}\Rightarrow \text{x-y  0}\].

But \[\text{2xy}\] is negative.

Therefore, \[\text{2xy }\ne \text{ x - y}\].

Hence, \[\text{x}\] being positive and \[\text{y}\] being negative is not possible. Similarly \[\text{x}\] being negative and \[\text{y}\] being positive can also be ruled out.

So, \[\text{x}\] and \[\text{y}\] have to be either positive or negative.

Assuming that both \[\text{x}\] and \[\text{y}\] are positive:

\[\text{f(x)=f(y)}\]

\[\Rightarrow \dfrac{\text{x}}{\text{1+x}}\text{=}\dfrac{\text{y}}{\text{1+y}}\]

\[\Rightarrow \text{x+xy=y+xy}\]

\[\Rightarrow \text{x=y}\]

Assuming that both \[\text{x}\] and \[\text{y}\] are negative:

\[\text{f(x)=f(y)}\]

\[\Rightarrow \dfrac{\text{x}}{\text{1+x}}\text{=}\dfrac{\text{y}}{\text{1+y}}\]

\[\Rightarrow \text{x+xy=y+xy}\]

\[\Rightarrow \text{x=y}\]

Therefore, the function \[\text{f}\] is one – one.

For onto:

\[\text{y}\in \text{R}\] such that \[\text{-1  y  1}\].

If \[\text{y}\] is negative, then, there exists \[\text{x = }\dfrac{\text{y}}{\text{1+y}}\in \text{R}\] such that

\[\text{f}\left( \dfrac{\text{y}}{\text{1+y}} \right)\text{=}\dfrac{\left( \dfrac{\text{y}}{\text{1+y}} \right)}{\text{1+}\left| \dfrac{\text{y}}{\text{1+y}} \right|}\]

\[\text{=}\dfrac{\dfrac{\text{y}}{\text{1+y}}}{\text{1+}\left( \dfrac{\text{-y}}{\text{1+y}} \right)}\]

\[\text{=}\dfrac{\text{y}}{\text{1+y-y}}\]

\[\text{=y}\] 

If \[\text{y}\] is positive, then, there exists \[\text{x = }\dfrac{\text{y}}{\text{1-y}}\in \text{R}\] such that

\[\text{f}\left( \dfrac{\text{y}}{\text{1-y}} \right)\text{=}\dfrac{\left( \dfrac{\text{y}}{\text{1-y}} \right)}{\text{1+}\left| \dfrac{\text{y}}{\text{1-y}} \right|}\]

\[\text{=}\dfrac{\dfrac{\text{y}}{\text{1-y}}}{\text{1+}\left( \dfrac{\text{-y}}{\text{1-y}} \right)}\]

\[\text{=}\dfrac{\text{y}}{\text{1-y+y}}\]

\[\text{=y}\]

Therefore, the function \[\text{f}\] is onto.

Hence the given function \[\text{f}\] is both one–one and onto.

Ans: The given function \[\text{f: R }\to \text{ R}\] is given as \[\text{f}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{3}}}\].

For the function \[\text{f}\] to be one – one:

\[\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)\] where \[\text{x, y}\in \text{R}\].

\[\Rightarrow {{\text{x}}^{\text{3}}}\text{=}{{\text{y}}^{\text{3}}}\]           …… (1) 

We need to show that \[\text{x=y}\].

Assuming that \[\text{x}\ne \text{y}\], then,

\[\Rightarrow {{\text{x}}^{\text{3}}}\ne {{\text{y}}^{\text{3}}}\] 

Since this is a contradiction to (1), therefore, \[\text{x=y}\].

Hence, the function \[\text{f}\] is injective.

Ans: We know that every set is a subset of itself, ${ARA}$ for all ${A}\in {P}\left( {X} \right)$

Therefore ${R}$ is reflexive.

Let ${ARB}\Rightarrow {A}\subset {B}$.

This does not mean that ${B}\subset {A}$.

If ${A = }\left\{ {1, 2} \right\}$ and ${B = }\left\{ {1, 2, 3} \right\}$, then it cannot be implied that ${B}$ is related to ${A}$.

Therefore ${R}$ is not symmetric.

If ${ARB}$ and ${BRC}$ , then;

${A}\subset {B}$ and ${B}\subset {C}$

$\Rightarrow {A}\subset {C}$ 

$\Rightarrow {ARC}$ 

Therefore ${R}$ is transitive.

Hence, ${R}$ is not an equivalence relation as it is not symmetric.

Ans: The total number of onto maps from ${ }\!\!\{\!\!{ 1, 2, 3, }...{ , n }\!\!\}\!\!{ }$ to itself will be same as the total number of permutations on ${n}$ symbols ${1, 2, 3, }...{ , n}$.

Since the total number of permutations on ${n}$ symbols ${1, 2, 3, }...{ , n}$ is ${n}$, thus total number of onto maps from ${ }\!\!\{\!\!{ 1, 2, 3, }...{ , n }\!\!\}\!\!{ }$ to itself are ${n}$.

Ans: Let ${A=}\left\{ {-1, 0, 1, 2} \right\}{,B=}\left\{ {-4, -2, 0, 2} \right\}$ and ${f,g:A }\to { B}$ are defined by ${f}\left( {x} \right){=}{{{x}}^{{2}}}{-x,}\,{x}\in {A}$ and ${g}\left( {x} \right){=2}\left| {x-}\dfrac{{1}}{{2}} \right|{-1,x}\in {A}$.

${f}\left( {-1} \right){=}{{\left( {-1} \right)}^{{2}}}{-}\left( {-1} \right)$

${=1+1}$

${=2}$

And,

${g}\left( {-1} \right){=2}\left| \left( {-1} \right){-}\dfrac{{1}}{{2}} \right|{-1}$

${=2}\left( \dfrac{{3}}{{2}} \right){-1}$

${=3-1}$

${=2}$

$\Rightarrow {f}\left( {-1} \right){=g}\left( {-1} \right)$

${f}\left( {0} \right){=}{{\left( {0} \right)}^{{2}}}{-}\left( {0} \right)$

${=0}$

And,

${g}\left( {0} \right){=2}\left| \left( {0} \right){-}\dfrac{{1}}{{2}} \right|{-1}$

${=1-1}$

${=0}$

$\Rightarrow {f}\left( {0} \right){=g}\left( {0} \right)$

${f}\left( {1} \right){=}{{\left( {1} \right)}^{{2}}}{-}\left( {1} \right)$

${=1-1}$

${=0}$

And,

${g}\left( {1} \right){=2}\left| \left( {1} \right){-}\dfrac{{1}}{{2}} \right|{-1}$

${=2}\left( \dfrac{1}{{2}} \right){-1}$

${=1-1}$

${=0}$

$\Rightarrow {f}\left( {1} \right){=g}\left( {1} \right)$

${f}\left( {2} \right){=}{{\left( {2} \right)}^{{2}}}{-}\left( {2} \right)$

${=4-2}$

${=2}$

And,

${g}\left( {2} \right){=2}\left| \left( {2} \right){-}\dfrac{{1}}{{2}} \right|{-1}$

${=2}\left( \dfrac{{3}}{{2}} \right){-1}$

${=3-1}$

${=2}$

$\Rightarrow {f}\left( {2} \right){=g}\left( {2} \right)$

Therefore, ${f}\left( {a} \right){=g}\left( {a} \right)\forall {a}\in {A}$. Hence the functions ${f}$ and ${g}$ are equal.

Ans: We are given a set \[\text{A=}\left\{ \text{1, 2, 3} \right\}\].

Let us take the relation \[\text{R}\], containing \[\left( \text{1, 2} \right)\] and \[\left( \text{1, 3} \right)\], as \[\text{R=}\left\{ \left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{1, 2} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{2, 1} \right)\text{, }\left( \text{3, 1} \right) \right\}\].

As we can see that \[\left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\in \text{R}\], therefore relation \[\text{R}\] is reflexive.

Since \[\left( \text{1, 2} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{2, 1} \right)\in \text{R}\], the relation \[\text{R}\] is symmetric.

The relation Relation \[\text{R}\] is not transitive because \[\left( \text{1, 2} \right)\text{,}\,\left( \text{3, 1} \right)\in \text{R}\], but \[\left( \text{3, 2} \right)\notin \text{R}\].

The relation Relation \[\text{R}\] will become transitive on adding and two pairs \[\left( \text{3, 2} \right)\text{, }\left( \text{2, 3} \right)\].

Therefore the total number of desired relations is one.

The correct answer is option \[(A)\] \[\text{1}\].

Ans: We are given a set \[\text{A=}\left\{ \text{1, 2, 3} \right\}\].

Let us take the relation \[\text{R}\], containing \[\left( \text{1, 2} \right)\] as \[\text{R=}\left\{ \left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{1, 2} \right)\text{, }\left( \text{2, 1} \right) \right\}\].

Now the pairs left are \[\left( \text{2, 3} \right)\text{, }\left( \text{3, 2} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{3, 1} \right)\] 

In order to add one pair, say \[\left( \text{2, 3} \right)\], we must add \[\left( \text{3, 2} \right)\] for symmetry. And we are required to add \[\left( \text{1, 3} \right)\text{, }\left( \text{3, 1} \right)\] for transitivity.

So, only the equivalence relation (bigger than \[\text{R}\]) is the universal relation.

Therefore, the total number of equivalence relations containing \[\left( \text{1, 2} \right)\] are two.

Hence, the correct answer is (B) \[\text{2}\].