# Miscellaneous Exercise Relations and Function Solution - NCERT Class 12

## Chapter 1 Ex.1.ME Question 1

Let $$f:R \to R$$ be defined as $$f(x)=\text{1}0x~+\text{7}$$. Find the function $$g:R \to R$$ such that $$gof = fog = {{\rm{I}}_R}$$.

### Solution

$$f:R \to R$$ is defined as $$f(x)=\text{1}0x~+\text{7}$$

For one-one:

$$f\left( x \right) = f\left( y \right)$$ where $$x,y \in R$$

\begin{align}& \Rightarrow 10x + 7 = 10y + 7\\& \Rightarrow x = y\end{align}

$$\therefore f$$is one-one.

For onto:

$$y \in R,\,\,{\text{Let }}y = 10x + 7$$

$\Rightarrow x = \frac{{y - 7}}{{10}} \in R$

For any $$y \in R,$$ there exists $$x = \frac{{y - 7}}{{10}} \in R$$ such that

$$f\left( x \right) = f\left( {\frac{{y - 7}}{{10}}} \right) = 10\left( {\frac{{y - 7}}{{10}}} \right) + 7 = y - 7 + 7 = y$$

$$\therefore f$$is onto.

Thus, $$f$$ is an invertible function.

Let us define $$g:R \to R$$as$$g\left( y \right) = \frac{{y - 7}}{{10}}$$.

Now,

$$gof\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {10x + 7} \right) = \frac{{\left( {10x + 7} \right) - 7}}{{10}} = \frac{{10x}}{{10}} = 10$$

And,

$$fog\left( y \right) = f\left( {g\left( y \right)} \right) = f\left( {\frac{{y - 7}}{{10}}} \right) = 10\left( {\frac{{y - 7}}{{10}}} \right) + 7 = y - 7 + 7 = y$$

$$\therefore gof = {{\text{I}}_R}$$ and $$fog = {{\text{I}}_R}$$

Hence, the required function $$g:R \to R$$ as$$g\left( y \right) = \frac{{y - 7}}{{10}}$$.

## Chapter 1 Ex.1.ME Question 2

Let $$f:W \to W$$ be defined as$$f(n) = n - 1$$, if is odd and $$f\left( n \right) = n + 1$$, if $$n$$ is even. Show that $$f$$ is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

### Solution

$$f:W \to W$$is defined as $$f\left( n \right) = \left\{ \begin{gathered} n - 1\,,\,\,{\text{If }}n{\text{ is odd}} \hfill \\ n + 1\,,\,\,{\text{If }}n{\text{ is even}} \hfill \\ \end{gathered} \right\}$$

For one-one:

$$f\left( n \right) = f\left( m \right)$$

If $$n$$is odd and$$m$$ is even, then we will have$$n - 1 = m + 1$$.

$$\Rightarrow n - m = 2$$

Similarly, the possibility of $$n$$ being even and $$m$$ being odd can also be ignored under a similar argument.

∴Both $$n$$ and $$m$$ must be either odd or even.

Now, if both $$n$$ and $$m$$ are odd, then we have:

\begin{align} & f\left( n \right)=~f\left( m \right) \\ & \Rightarrow ~n~-\text{ 1 }=~m~-\text{1 } \\ & \Rightarrow ~n~=~m \\ \end{align}

Again, if both $$n$$ and $$m$$ are even, then we have:

\begin{align} & f\left( n \right)=~f\left( m \right) \\ & \Rightarrow ~n~+\text{ 1 }=~m~+\text{1} \\ & \Rightarrow ~n~=~m \\ \end{align}

$$\therefore f$$ is one-one.

For onto:

Any odd number $$2r + 1$$in co-domain N is the image of $$2r$$ in domain N and any even number $$2r$$ in co-domain N is the image of $$2r + 1$$in domain N.

$$\therefore f$$ is onto.

$$f$$is an invertible function.

Let us define $$g:W \to W$$ as $$f\left( m \right) = \left\{ \begin{array}{l}m - 1\,,\,\,{\text{If }}m{\text{ is odd}}\\m + 1,\,\,\,{\text{If }}m{\text{ is even}}\end{array} \right\}$$

When $$r$$ is odd

$$gof\left( n \right) = g\left( {f\left( n \right)} \right) = g\left( {n - 1} \right) = n - 1 + 1 = n$$

When $$r$$ is even

$$gof\left( n \right) = g\left( {f\left( n \right)} \right) = g\left( {n + 1} \right) = n + 1 - 1 = n$$

When $$m$$ is odd

$$fog\left( n \right) = f\left( {g\left( m \right)} \right) = f\left( {m - 1} \right) = m - 1 + 1 = m$$

When $$m$$ is even

$$fog\left( m \right) = f\left( {g\left( m \right)} \right) = f\left( {m + 1} \right) = m + 1 - 1 = m$$

$$\therefore gof = {{\text{I}}_W}$$ and $$fog = {{\text{I}}_W}$$

$$f$$ is invertible and the inverse of $$f$$ is given by $${f^{ - 1}} = g$$, which is the same as $$f$$.

inverse of $$f$$ is $$f$$ itself.

## Chapter 1 Ex.1.ME Question 3

If $$f:R \to R$$ be defined as $$f(x) = {x^2} - 3x + 2$$, find $$f\left( {f\left( x \right)} \right)$$.

### Solution

$$f:R \to R$$ is defined as $$f(x) = {x^2} - 3x + 2$$.

\begin{align}f\left( {f\left( x \right)} \right) &= f\left( {{x^2} - 3x + 2} \right)\\& = {\left( {{x^2} - 3x + 2} \right)^2} - 3\left( {{x^2} - 3x + 2} \right) + 2\\& = \left( {{x^4} + 9{x^2} + 4 - 6{x^3} - 12x + 4{x^2}} \right) + \left( { - 3{x^2} + 9x - 6} \right) + 2\\ &= {x^4} - 6{x^3} + 10{x^2} - 3x\end{align}

## Chapter 1 Ex.1.ME Question 4

Show that function $$f:R \to \left\{ {x \in R: - 1 < x < 1} \right\}$$ be defined by $$f(x) = \frac{x}{{1 + \left| x \right|}}$$, $$x \in R$$ is one-one and onto function.

### Solution

$$f:R\to \left\{ x\in R:-1<x<1 \right\}$$ is defined by $$f(x) = \frac{x}{{1 + \left| x \right|}}$$, $$x \in R$$.

For one-one:

$$f\left( x \right) = f\left( y \right)$$  where $$x,y \in R$$

$$\Rightarrow \frac{x}{{1 + \left| x \right|}} = \frac{y}{{1 + \left| y \right|}}$$

If $$x$$ is positive and $$y$$ is negative,

\begin{align}\frac{x}{{1 + \left| x \right|}} = \frac{y}{{1 + \left| y \right|}}\\ \Rightarrow 2xy = x - y\end{align}

Since, $$x$$ is positive and $$y$$ is negative,

$$x > y \Rightarrow x - y > 0$$

$$2xy$$ is negative.

$$2xy \ne x - y$$

Case of $$x$$ being positive and $$y$$ being negative, can be ruled out.

$$\therefore\; x$$ and $$y$$ have to be either positive or negative.

If $$x$$ and $$y$$ are positive,

\begin{align}&f\left( x \right) = f\left( y \right)\\ &\Rightarrow \frac{x}{{1 + x}} = \frac{y}{{1 + y}}\\& \Rightarrow x - xy = y - xy\\ &\Rightarrow x = y\end{align}

$$\therefore f$$ is one-one.

For onto:

Let $$y \in R$$such that $$- 1 < y < 1$$.

If $$x$$ is negative, then there exists $$x = \frac{y}{{1 + y}} \in R$$ such that

$f\left( x \right) = f\left( {\frac{y}{{1 + y}}} \right) = \frac{{\left( {\frac{y}{{1 + y}}} \right)}}{{1 + \left| {\frac{y}{{1 + y}}} \right|}} = \frac{{\frac{y}{{1 + y}}}}{{1 + \left( {\frac{{ - y}}{{1 + y}}} \right)}} = \frac{y}{{1 + y - y}} = y$

If $$x$$ is positive, then there exists $$x = \frac{y}{{1 - y}} \in R$$such that

$f\left( x \right) = f\left( {\frac{y}{{1 - y}}} \right) = \frac{{\left( {\frac{y}{{1 - y}}} \right)}}{{1 + \left| {\frac{y}{{1 - y}}} \right|}} = \frac{{\frac{y}{{1 - y}}}}{{1 + \left( {\frac{y}{{1 - y}}} \right)}} = \frac{y}{{1 - y + y}} = y$

$$\therefore f$$is onto.

Hence, $$f$$ is one-one and onto.

## Chapter 1 Ex.1.ME Question 5

Show that function $$f:R \to R$$ be defined by$$f(x) = {x^3}$$ is injective.

### Solution

$$f:R \to R$$ is defined by$$f(x)={{x}^{3}}$$

For one-one:

$$f\left( x \right) = f\left( y \right)$$ where$$x,y \in R$$

${x^3} = {y^3}........................\left( 1 \right)$

We need to show that $$x = y$$

Suppose $$x \ne y$$, their cubes will also not be equal.

$\Rightarrow {x^3} \ne {y^3}$

This will be a contradiction to $$\left( 1 \right)$$.

$$\therefore x = y$$. Hence, $$f$$ is injective.

## Chapter 1 Ex.1.ME Question 6

Give examples of two functions $$f:N \to Z$$ and $$g:Z \to Z$$ such that $$gof$$ is injective but $$g$$ is not injective.

(Hint: Consider $$f\left( x \right) = x$$and $$g\left( x \right) = \left| x \right|$$)

### Solution

Define $$f:N \to Z$$as $$f\left( x \right) = x$$and $$g:Z \to Z$$as $$g\left( x \right) = \left| x \right|$$

Let us first show that $$g$$is not injective.

\begin{align}\left( { - 1} \right) = \left| { - 1} \right| = 1\\\left( 1 \right) = \left| 1 \right| = 1\\\therefore \left( { - 1} \right) = g\left( 1 \right),\,\,\,\,\,{\text{but }} - 1 \ne 1\end{align}

$$\therefore g$$is not injective.

$$gof:N \to Z$$is defined as $$gof\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( x \right) = \left| x \right|$$

$$x,y \in N$$such that $$gof\left( x \right) = gof\left( y \right)$$

$$\Rightarrow \left| x \right| = \left| y \right|$$

Since $$x,y \in N$$, both are positive.

\begin{align}\therefore \left| x \right| = \left| y \right|\\ \Rightarrow x = y\end{align}

$$\therefore$$ $$gof$$ is injective.

## Chapter 1 Ex.1.ME Question 7

Given examples of two functions $$f:N \to N$$ and $$g:N \to N$$ such that $$gof$$ is onto but $$f$$ is not onto.

(Hint: Consider $$f\left( x \right) = x + 1$$and $$g\left( x \right) = \left\{ \begin{array}{l}x - 1,\,\,\,{\text{if }}x > 1\\1,\,\,\,\,\,\,\,\,\,\,\,{\text{if }}x = 1\end{array} \right\}$$)

### Solution

Define $$f:N \to Z$$as $$f\left( x \right) = x + 1$$and $$g:Z \to Z$$as $$g\left( x \right) = \left\{ \begin{array}{l}x - 1,\,\,\,{\text{if }}x > 1\\1,\,\,\,\,\,\,\,\,\,\,\,{\text{if }}x = 1\end{array} \right\}$$

Let us first show that $$g$$ is not onto.

Consider element 1 in co-domain $$N$$. This element is not an image of any of the elements in domain $$N$$.

$$\therefore f$$ is not onto.

$$g:N \to N$$ is defined by

$$gof\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {x + 1} \right) = x + 1 - 1 = x \qquad \left[ {x \in N \Rightarrow x + 1 > 1} \right]$$

For $$y \in N$$, there exists $$x = y \in N$$ such that $$gof\left( x \right) = y$$.

$$\therefore gof$$ is onto.

## Chapter 1 Ex.1.ME Question 8

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 you answer.

### Solution

Since 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 cannot be implied to $$B \subset A$$.

If $$A = \left\{ {{\text{1}},{\text{2}}} \right\}$$ and $$B = \left\{ {{\text{1}},{\text{ 2}},{\text{3}}} \right\}$$, 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$$B \subset C$$.

\begin{align} &\Rightarrow A \subset C \hfill \\& \Rightarrow ARC\end{align}

$$\therefore\;R$$ is transitive.

$$R$$ is not an equivalence relation as it is not symmetric.

## Chapter 1 Ex.1.ME Question 9

Given a non-empty set $$X$$, consider the binary operation $$^*$$: $${\text{P}}\left( X \right) \times {\text{P}}\left( X \right) \to {\text{P}}\left( X \right){\text{ }}$$given by $$A*B = A \cap B\,\,\,\forall \,A,B$$ in $$P\left( X \right)$$ is the power set of $$X$$. Show that $$X$$ is the identity element for this operation and $$X$$ is the only invertible element in $$P\left( X \right)$$with respect to the operation $$^*$$.

### Solution

$${\text{P}}\left( X \right) \times {\text{P}}\left( X \right) \to {\text{P}}\left( X \right){\text{ }}$$given by $$A*B = A \cap B\,\,\,\forall \,A,B$$ in $$P\left( X \right)$$

$$A \cap X = A = X \cap A$$ for all $$A \in P\left( X \right)$$

$$\Rightarrow A*X = A = X*A$$ for all $$A \in P\left( X \right)$$

$$X$$ is the identity element for the given binary operation $$^*$$.

An element $$A \in P\left( X \right)$$is invertible if there exists $$B \in P\left( X \right)$$such that

$$A*B = X = B*A$$ [As $$X$$is the identity element]

Or

$$A \cap B = X = B \cap A$$

This case is possible only when $$A = X = B$$.

$$X$$ is the only invertible element in $$P\left( X \right)$$with respect to the given operation $$^*$$.

## Chapter 1 Ex.1.ME Question 10

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

### Solution

Onto functions from the set $$\left\{ {1,2,3, \ldots ,n} \right\}$$ to itself is simply a permutation on $$n$$symbols

$$1,2,3, \ldots ,n$$.

Thus, the total number of onto maps from $$\left\{ {1,2,3, \ldots ,n} \right\}$$ to itself is the same as the total number of permutations on $$n$$ symbols $$1,2,3, \ldots ,n$$, which is$$n!$$.

## Chapter 1 Ex.1.ME Question 11

Let $${\text{S}} = \left\{ {a,b,c} \right\}$$and $$T = \left\{ {{\text{1}},{\text{ 2}},{\text{3}}} \right\}$$. Find $${{F}^{-1}}$$ of the following functions $$F$$ from $$S$$ to $$T$$, if it exists.

(i) $$F = \left\{ {\left( {a,3} \right),\left( {b,2} \right),\left( {c,1} \right)} \right\}$$

(ii) $$F = \left\{ {\left( {a,2} \right),\left( {b,1} \right),\left( {c,1} \right)} \right\}$$

### Solution

$${\text{S}} = \left\{ {a,b,c} \right\},T = \left\{ {{\text{1}},{\text{ 2}},{\text{3}}} \right\}$$

(i) $$F:S \to T$$is defined by $$F = \left\{ {\left( {a,3} \right),\left( {b,2} \right),\left( {c,1} \right)} \right\}$$

$$\Rightarrow F\left( a \right) = 3,\,\,F\left( b \right) = 2,\,\,F\left( c \right) = 1$$

Therefore, $${F^{ - 1}}:T \to S$$is given by $${F^{ - 1}} = \left\{ {\left( {3,a} \right),\left( {2,b} \right),\left( {1,c} \right)} \right\}$$

(ii) $$F:S \to T$$is defined by $$F = \left\{ {\left( {a,2} \right),\left( {b,1} \right),\left( {c,1} \right)} \right\}$$

Since, $$F\left( b \right) = F\left( c \right) = 1$$, $$F$$ is not one-one.

Hence, $$F$$ is not invertible i.e., $${F^{ - 1}}$$ does not exists.

## Chapter 1 Ex.1.ME Question 12

Consider the binary operations$$^*$$: $$R \times R \to R$$and $$o:R \times R \to R$$ defined as $$a*b = \left| {a - b} \right|$$ and $$aob = a,\,\,\forall a,b \in R$$. Show that $$^*$$ is commutative but not associative $$o$$is associative but not commutative. Further, show that$$\forall a,b,c \in R$$, $$a*\left( {boc} \right) = \left( {a*b} \right)o\left( {a*c} \right)$$. [ If it is so, we say that the operation $$^*$$ distributes over the operation$$o$$]. Does $$o$$ distribute over$$^*$$? Justify your answer.

### Solution

It is given that $$^*$$: $$R \times R \to R$$and $$o:R \times R \to R$$ defined as $$a*b = \left| {a - b} \right|$$ and $$aob = a,\,\,\forall a,b \in R$$.

For $$a,b \in R$$, we have $$a*b = \left| {a - b} \right|$$ and $$b*a = \left| {b - a} \right| = \left| { - \left( {a - b} \right)} \right| = \left| {a - b} \right|$$

$$\therefore a*b = b*a$$

$$\therefore$$ The operation $$^*$$is commutative.

\begin{align}\left( {1*2} \right)*3 = \left( {\left| {1 - 2} \right|} \right)*3 = 1*3 = \left| {1 - 3} \right| = 2\\1*\left( {2*3} \right) = 1*\left( {\left| {2 - 3} \right|} \right) = 1*1 = \left| {1 - 1} \right| = 0\end{align}

$$\therefore \left( {1*2} \right)*3 \ne 1*\left( {2*3} \right)$$ where $$1,2,3 \in R$$

$$\therefore$$The operation * is not associative.

Now, consider the operation$$o$$:

It can be observed that $$1o2 = 1$$and $$2o1 = 2$$.

$$\therefore 1o2 \ne 2o1$$ (where$$1,2 \in R$$)

$$\therefore$$The operation $$o$$is not commutative.

Let $$a,~b,~c~\in R$$. Then, we have:

\begin{align}\left( {aob} \right)oc = aoc = a\\ao\left( {boc} \right) = aob = a\\ \Rightarrow \left( {aob} \right)oc = ao\left( {boc} \right)\end{align}

$$\therefore$$The operation $$o$$ is associative.

Now, let $$a,~b,~c~\in ~R$$, then we have:

\begin{align}&a*\left( {boc} \right) = a*b = \left| {a - b} \right|\\&\left( {a*b} \right)o\left( {a*c} \right) = \left( {\left| {a - b} \right|} \right)o\left( {\left| {a - c} \right|} \right) = \left| {a - b} \right|\end{align}

Hence, $$a*\left( {boc} \right) = \left( {a*b} \right)o\left( {a*c} \right)$$

Now,

$$\begin{gathered} 1o\left( {2*3} \right) = 1o\left( {\left| {2 - 3} \right|} \right) = 1o1 = 1 \hfill \\ \left( {1o2} \right)*\left( {1o3} \right) = 1*1 = \left| {1 - 1} \right| = 0 \hfill \\ \end{gathered}$$

$$\therefore 1o\left( {2*3} \right) \ne \left( {1o2} \right)*\left( {1o3} \right)$$ where $$1,2,3 \in R$$

$$\therefore$$The operation $$o$$does not distribute over$$^*$$.

## Chapter 1 Ex.1.ME Question 13

Given a non - empty set $$X$$, let $$^*$$: $${\text{P}}\left( X \right) \times {\text{P}}\left( X \right) \to {\text{P}}\left( X \right)$$be defined as $$A*B = \left( {A - B} \right) \cup \left( {B - A} \right)$$, $$\forall A,B \in P\left( X \right)$$. Show that the empty set $$\Phi$$ is the identity for the operation $$^*$$ and all the elements $$A$$ of $$P\left( X \right)$$ are invertible with $${A^{ - 1}} = A$$.

(Hint: $$\left( {A - \Phi } \right) \cup \left( {\Phi - A} \right) = A$$ and $$\left( {A - A} \right) \cup \left( {A - A} \right) = A*A = \Phi$$).

### Solution

It is given that $$^*$$: $${\text{P}}\left( X \right) \times {\text{P}}\left( X \right) \to {\text{P}}\left( X \right)$$is defined as $$A*B = \left( {A - B} \right) \cup \left( {B - A} \right)$$,$$\forall A,B \in P\left( X \right)$$

$$A \in P\left( X \right)$$ then,

\begin{align}A*\Phi = \left( {A - \Phi } \right) \cup \left( {\Phi - A} \right) = A \cup \Phi = A\\\Phi *A = \left( {\Phi - A} \right) \cup \left( {A - \Phi } \right) = \Phi \cup A = A\\\therefore A*\Phi = A = \Phi *A\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{for all }}A \in P\left( X \right)\end{align}

$$\Phi$$ is the identity for the operation $$^*$$.

Element $$A\in P\left( X \right)$$ will be invertible if there exists $$B \in P\left( X \right)$$such that

$$A*B = \Phi = B*A$$ [As $$\Phi$$is the identity element]

$$A*A = \left( {A - A} \right) \cup \left( {A - A} \right) = \Phi \cup \Phi = \Phi$$ for all $$A \in P\left( X \right)$$.

All the elements $$A$$of $$P\left( X \right)$$are invertible with $${A^{ - 1}} = A$$.

## Chapter 1 Ex.1.ME Question 14

Define a binary operation $$^*$$ on the set $$\left\{ {0,{\text{1}},{\text{2}},{\text{3}},{\text{4}},{\textabc}} \right\}$$ as

$$a + b = \left\{ \begin{array}{l}a + b,\,\,\,\,\,\,\,\,\,\,\,\,{\text{if }}a + b < 6\\a + b - 6\,\,\,\,\,\,{\text{if }}a + b \ge 6\end{array} \right\}$$

Show that zero is the identity for this operation and each element $$a \ne 0$$ of the set is invertible with $$6 - a$$ being the inverse of $$a$$.

### Solution

Let $$X = \left\{ {0,{\text{1}},{\text{2}},{\text{3}},{\text{4}},{\text{5}}} \right\}$$

The operation $$^*$$ is defined as $$a + b = \left\{ \begin{gathered} a + b,\,\,\,\,\,\,\,\,\,\,\,\,{\text{if }}a + b < 6 \hfill \\ a + b - 6\,,\,\,\,\,{\text{if }}a + b \geqslant 6 \hfill \\ \end{gathered} \right\}$$

An element $$e \in X$$ is the identity element for the operation $$^*$$, if$$a*e = a = e*a\,\,\,\,\,\,\forall a \in X$$

For $$a \in X$$,

\begin{align}a*0 = a + 0 = a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {a \in X \Rightarrow a + 0 < 6} \right]\\0*a = 0 + a = a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {a \in X \Rightarrow 0 + a < 6} \right]\end{align}

$$\therefore a*0 = a = 0*a\,\,\,\forall a \in X$$

Thus, 0 is the identity element for the given operation $$^*$$.

An element $$a \in X$$ is invertible if there exists $$b \in X$$ such that $$a*b = 0 = b*a$$.

i.e., $$\left\{ \begin{array}{l}a + b = 0 = b + a,\,\,\,\,\,\,\,\,\,\,\,\,{\text{if }}a + b < 6\\a + b - 6 = 0 = b + a - 6\,\,\,\,\,\,{\text{if }}a + b \ge 6\end{array} \right\}$$

$$\Rightarrow a = - b\,\,{\text{or }}b = 6 - a$$

$$X = \left\{ {0,{\text{1}},{\text{2}},{\text{3}},{\text{4}},{\text{5}}} \right\}$$ and $$a,b \in X$$. Then $$a \ne - b$$.

$$\therefore b = 6 - a$$ is the inverse of $$a$$ for all $$a \in X$$.

Inverse of an element $$a \in X$$, $$a \ne 0$$ is $$6 - a$$ i.e., $$a - 1 = 6 - a$$.

## Chapter 1 Ex.1.ME Question 15

Let$$A = \left\{ { - {\text{1}},0,{\text{1}},{\text{2}}} \right\}$$, $$B = \left\{ { - {\text{4}}, - {\text{2}},0,{\text{2}}} \right\}$$and $$f,g:A \to B$$ be functions defined by $${x^2} - x$$, $$x \in A$$and $$g\left( x \right) = 2\left| {x - \frac{1}{2}} \right| - 1,\,\,x \in A$$. Are $$f$$ and $$g$$ equal?

### Solution

It is given that $$A = \left\{ { - {\text{1}},0,{\text{1}},{\text{2}}} \right\}$$, $$B = \left\{ { - {\text{4}}, - {\text{2}},0,{\text{2}}} \right\}$$

Also, $$f,g:A \to B$$ is defined by $${x^2} - x$$, $$x \in A$$ and $$g\left( x \right) = 2\left| {x - \frac{1}{2}} \right| - 1,\,\,x \in A$$.

\begin{align}&f\left( { - 1} \right) = {\left( { - 1} \right)^2} - \left( { - 1} \right) = 1 + 1 = 2\\&g\left( { - 1} \right) = 2\left| {\left( { - 1} \right) - \frac{1}{2}} \right| - 1 = 2\left( {\frac{3}{2}} \right) - 1 = 3 - 1 = 2\\& \Rightarrow f\left( { - 1} \right) = g\left( { - 1} \right)\end{align}

\begin{align}&f\left( 0 \right) = {\left( 0 \right)^2} - 0 = 0\\&g\left( 0 \right) = 2\left| {0 - \frac{1}{2}} \right| - 1 = 2\left( {\frac{1}{2}} \right) - 1 = 1 - 1 = 0\\& \Rightarrow f\left( 0 \right) = g\left( 0 \right)\end{align}

\begin{align}&f\left( 1 \right) = {\left( 1 \right)^2} - 1 = 0\\&g\left( 1 \right) = 2\left| {1 - \frac{1}{2}} \right| - 1 = 2\left( {\frac{1}{2}} \right) - 1 = 1 - 1 = 0\\& \Rightarrow f\left( 1 \right) = g\left( 1 \right)\end{align}

\begin{align}&f\left( 2 \right) = {\left( 2 \right)^2} - 2 = 2\\&g\left( 2 \right) = 2\left| {2 - \frac{1}{2}} \right| - 1 = 2\left( {\frac{3}{2}} \right) - 1 = 3 - 1 = 2\\& \Rightarrow f\left( 2 \right) = g\left( 2 \right)\end{align}

$$\therefore f\left( a \right) = g\left( a \right)\,\,\,\,\,\,\,\forall a \in A$$

Hence, the functions $$f$$ and $$g$$ are equal.

## Chapter 1 Ex.1.ME Question 16

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

A. $$1$$

B. $$2$$

C. $$3$$

D. $$4$$

### Solution

The given set is $$A = \left\{ {1,2,3} \right\}$$.

The smallest relation containing $$\left( {{\text{1}},{\text{2}}} \right)$$ and $$\left( {{\text{1}},{\text{3}}} \right)$$ which are reflexive and symmetric but not transitive is given by,

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

This is because relation $$R$$ is reflexive as $$\left\{ {\left( {1,1} \right),\left( {2,2} \right),\left( {3,3} \right)} \right\} \in R$$.

Relation $$R$$ is symmetric as $$\left\{ {\left( {1,2} \right),\left( {2,1} \right)} \right\} \in R\,\,\,{\text{and}}\,\,\,\left\{ {\left( {1,3} \right)\left( {3,1} \right)} \right\} \in R$$.

Relation $$R$$ is transitive as $$\left\{ {\left( {3,1} \right),\left( {1,2} \right)} \right\} \in R$$ but $$\left( {3,2} \right) \notin R$$.

Now, if we add any two pairs $$\left( {{\text{3}},{\text{2}}} \right)$$ and $$\left( {{\text{2}},{\text{3}}} \right)$$ (or both) to relation $$R$$, then relation $$R$$ will become transitive.

Hence, the total number of desired relations is one.

## Chapter 1 Ex.1.ME Question 17

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

A. $$1$$

B. $$2$$

C. $$3$$

D. $$4$$

### Solution

The given set is $$A = \left\{ {1,2,3} \right\}$$.

The smallest equivalence relation containing $$\left( {{\text{1}},{\text{2}}} \right)$$ is given by;

$${R_1} = \left\{ {\left( {1,1} \right),\left( {2,2} \right),\left( {3,3} \right),\left( {1,2} \right),\left( {2,1} \right)} \right\}$$

Now, we are left with only four pairs i.e., $$\left( {{\text{2}},{\text{3}}} \right),\left( {{\text{3}},{\text{2}}} \right),\left( {{\text{1}},{\text{3}}} \right){\text{and }}\left( {{\text{3}},{\text{1}}} \right).$$

If we odd any one pair [say$$\left( {{\text{2}},{\text{3}}} \right)$$] to$${R_{\text{1}}}$$, then for symmetry we must add$$\left( {{\text{3}},{\text{2}}} \right)$$. Also, for transitivity we are required to add $$\left( {{\text{1}},{\text{3}}} \right)$$ and $$\left( {{\text{3}},{\text{1}}} \right)$$.

Hence, the only equivalence relation (bigger than $${R_1}$$) is the universal relation.

This shows that the total number of equivalence relations containing $$\left( {{\text{1}},{\text{2}}} \right)$$ is two.

## Chapter 1 Ex.1.ME Question 18

Let $$f:R \to R$$ be the Signum Function defined as$$f\left( x \right) = \left\{ \begin{array}{l}1,\,\,\,x > 0\\0,\,\,\,x = 0\\ - 1,\,\,\,x < 0\end{array} \right\}$$ and $$g:R \to R$$ be the greatest integer function given by $$g\left( x \right) = \left[ x \right]$$, where $$\left[ x \right]$$ is greatest integer less than or equal to $$x$$. Then does $$fog$$ and $$gof$$ coincide in $$\left( {0,{\text{1}}} \right]$$?

### Solution

It is given that $$f:R \to R$$be the Signum Function defined as$$f\left( x \right) = \left\{ \begin{array}{l}1,\,\,\,x > 0\\0,\,\,\,x = 0\\ - 1,\,\,\,x < 0\end{array} \right\}$$

Also$$g:R \to R$$is defined as $$g\left( x \right) = \left[ x \right]$$, where $$\left[ x \right]$$is greatest integer less than or equal to $$x$$.

Now let $$x \in (0,1]$$,

$$\left[ x \right] = 1$$if $$x = 1$$ and $$\left[ x \right] = 0$$ if $$0 < x < 1$$.

\begin{align} \therefore fog\left( x \right) &= f\left( {g\left( x \right)} \right) = f\left( {\left[ x \right]} \right) = \left\{ \begin{gathered} f\left( 1 \right),\,\,\,\,{\text{if }}x = 1 \hfill \\ f\left( 0 \right),\,\,\,\,{\text{if }}x \in \left( {0,1} \right) \hfill \\ \end{gathered} \right\} = \left\{ \begin{gathered} 1,\,\,\,{\text{if }}x = 1 \hfill \\ 0,\,\,\,{\text{if }}x \in \left( {0,1} \right) \hfill \\ \end{gathered} \right\} \hfill \\ gof\left( x \right)& = g\left( {f\left( x \right)} \right) \hfill \\ & = g\left( 1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {x > 0} \right] \hfill \\& = \left[ 1 \right] = 1 \hfill \\ \end{align}

Thus, when $$x \in \left( {0,1} \right)$$, we have $$fog\left( x \right) = 0$$and $$gof\left( x \right) = 1$$.

Hence, $$fog$$ and $$gof$$ does not coincide in $$\left( {0,{\text{1}}} \right]$$.

## Chapter 1 Ex.1.ME Question 19

Number of binary operations on the set $$\{a,b\}$$are

A. $$10$$

B. $$16$$

C. $$20$$

D. $$8$$

### Solution

A binary operation $$^*$$ on $$\{a,~b\}$$is a function from $$\left\{ {a,b} \right\} \times \left\{ {a,b} \right\} \to \left\{ {a,b} \right\}$$

i.e., $$^*$$ is a function from $$\left\{ {\left( {a,a} \right),\left( {a,b} \right),\left( {b,a} \right),\left( {b,b} \right)} \right\} \to \left\{ {a,b} \right\}$$

Hence, the total number of binary operations on the set $$\{a,~b\}$$ is $${2^4} = 16$$.