NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 1 - Relation and Functions

Miscellaneous Exercise

1. Show that function \[\text{f: R }\to \text{  }\!\!\{\!\!\text{ x}\in \text{R:-1  x  1 }\!\!\}\!\!\text{ }\] defined by \[\text{f}\left( \text{x} \right)\text{=}\dfrac{\text{x}}{\text{1+ }\!\!|\!\!\text{ x }\!\!|\!\!\text{ }}\text{,x}\in \text{R}\] is one – one and onto function.

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.

2. Show that the function \[\text{f: R }\to \text{ R}\] given by \[\text{f}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{3}}}\] is injective.

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.

3. Given a non-empty set ${X}$, consider ${P}\left( {X} \right)$ which is the set of all subsets of ${X}$. Define the relation ${R}$ in ${P}\left( {X} \right)$ as follows:

For subsets ${A,B}$ in ${P}\left( {X} \right)$, ${ARB}$ if and only if ${A}\subset {B}$. Is ${R}$ an equivalence relation on ${P}\left( {X} \right)$? Justify your answer.

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.

4. Find the number of all onto functions from the set ${ }\!\!\{\!\!{ 1, 2, 3, }...{, n }\!\!\}\!\!{ }$ to itself.

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 ${n}$.

5. Let ${{A=}\left\{ {-1, 0, 1, 2} \right\}{,B=}\left\{ {-4, -2, 0, 2} \right\}}$ and ${{f,g:A }\to { B}}$ be functions defined by ${{f}\left( {x} \right){=}{{{x}}^{{2}}}{-x,}\,{x}\in {A}}$ and ${{g}\left( {x} \right){=2}\left| {x-}\dfrac{{1}}{{2}} \right|{-1,x}\in {A}}$. Are ${{f}}$ and ${{g}}$ equal? Justify your answer. (Hint: One may note that two function ${{f:A }\to { B}}$ and ${{g:A }\to { B}}$ such that ${{f}\left( {a} \right){=g}\left( {a} \right)\forall {a}\in {A}}$, are called equal functions)

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.

6. Let \[\mathbf{\text{A=}\left\{ \text{1, 2, 3} \right\}}\] Then number of relations containing \[\mathbf{\left( \text{1, 2} \right)}\] and \[\mathbf{\left( \text{1, 3} \right)}\]  which are reflexive and symmetric but not transitive is

(A) \[\mathbf{\text{1}}\]

(B) \[\mathbf{\text{2}}\]

(C) \[\mathbf{\text{3}}\]

(D) \[\mathbf{\text{4}}\]

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 \[(A)\] \[\text{1}\].

7. Let \[\mathbf{\text{A=}\left\{ \text{1, 2, 3} \right\}}\] Then number of equivalence relations containing \[\mathbf{\left( \text{1, 2} \right)}\] is

(A) \[\mathbf{\text{1}}\]

(B) \[\mathbf{\text{2}}\]

(C) \[\mathbf{\text{3}}\]

(D) \[\mathbf{\text{4}}\] 

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) \[\text{2}\].

Conclusion

Miscellaneous Exercise Class 12 Chapter 1 of Maths is important for understanding various concepts thoroughly. Relations and Functions Miscellaneous Exercise Class 12 covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorizing solutions. 

Class 12 Maths Chapter 1: Exercises Breakdown

CBSE Class 12 Maths Chapter 1 Other Study Materials

Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.