Q: Let $\quad f: R \rightarrow R \quad$ be defined as $f(x)=10 x+7$. Find the function $g: R \rightarrow R \quad$ such that $g \circ f=f \circ g=\mathrm{I}_{R}$.

$f: R \rightarrow R$ is defined as $f(x)=10 x+7$ For one-one: $f(x)=f(y)$ where $x, y \in R$ $\Rightarrow 10 x+7=10 y+7$ $\Rightarrow x=y$ $\therefore f$ is one-one. For onto: $y \in R$, Let $y=10 x+7$ $\Rightarrow x=\frac{y-7}{10} \in R$ For any $y \in R$, there exists $x=\frac{y-7}{10} \in R$ such that $f(x)=f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=y-7+7=y$ $\therefore f$ is onto. Thus, $f$ is an invertible function. Let us define $g: R \rightarrow R$ as $g(y)=\frac{y-7}{10}$. Now, $g \circ f(x)=g(f(x))=g(10 x+7)=\frac{(10 x+7)-7}{10}=\frac{10 x}{10}=10$ And, $f \circ g(y)=f(g(y))=f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=y-7+7=y$ $\therefore g o f=\mathrm{I}_{R}$ and $f o g=\mathrm{I}_{R}$ Hence, the required function $g: R \rightarrow R$ as $g(y)=\frac{y-7}{10}$.

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

$f: W \rightarrow W$ is defined as $f(n)=\left\{\begin{array}{l}n-1, \text { If } n \text { is odd } \\ n+1, \text { If } n \text { is even }\end{array}\right\}$ For one-one: $f(n)=f(m)$ 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. $\therefore$ Both $n$ and $m$ must be either odd or even. Now, if both $n$ and $m$ are odd, then we have: $f(n)=f(m)$ $\Rightarrow n-1=m-1$ $\Rightarrow n=m$ Again, if both $n$ and $m$ are even, then we have: $f(n)=f(m)$ $\Rightarrow n+1=m+1$ $\Rightarrow n=m$ $\therefore f$ is one-one. For onto: Any odd number $2 r+1$ in co-domain N is the image of $2 r$ in domain N and any even number $2 r$ in co-domain N is the image of $2 r+1$ in domain N . $\therefore f$ is onto. $f$ is an invertible function. Let us define $g: W \rightarrow W$ as $f(m)=\left\{\begin{array}{l}m-1, \text { If } m \text { is odd } \\ m+1, \text { If } m \text { is even }\end{array}\right\}$ When $r$ is odd $\operatorname{gof}(n)=g(f(n))=g(n-1)=n-1+1=n$ When $r$ is even $g \circ f(n)=g(f(n))=g(n+1)=n+1-1=n$ When $m$ is odd fog $(n)=f(g(m))=f(m-1)=m-1+1=m$ When $m$ is even $f o g(m)=f(g(m))=f(m+1)=m+1-1=m$ $\therefore g \circ f=\mathrm{I}_{W}$ and $f \circ g=\mathrm{I}_{W}$ $f$ 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.

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

$f: R \rightarrow\{x \in R:-1 y \Rightarrow x-y>0$ $2 x y$ is negative. $2 x y \neq x-y$ 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, $f(x)=f(y)$ $\Rightarrow \frac{x}{1+x}=\frac{y}{1+y}$ $\Rightarrow x-x y=y-x y$ $\Rightarrow x=y$ $\therefore f$ 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 $f(x)=f\left(\frac{y}{1+y}\right)=\frac{\left(\frac{y}{1+y}\right)}{1+\left|\frac{y}{1+y}\right|}=\frac{\frac{y}{1+y}}{1+\left(\frac{-y}{1+y}\right)}=\frac{y}{1+y-y}=y$ If $x$ is positive, then there exists $x=\frac{y}{1-y} \in R$ such that $f(x)=f\left(\frac{y}{1-y}\right)=\frac{\left(\frac{y}{1-y}\right)}{1+\left|\frac{y}{1-y}\right|}=\frac{\frac{y}{1-y}}{1+\left(\frac{y}{1-y}\right)}=\frac{y}{1-y+y}=y$ $\therefore f$ is onto. Hence, $f$ is one-one and onto.

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

$f: R \rightarrow R$ is defined by $f(x)=x^{3}$ For one-one: $f(x)=f(y) \quad$ where $x, y \in R$ $$ \begin{equation*} x^{3}=y^{3} \tag{1} \end{equation*} $$ We need to show that $x=y$ Suppose $x \neq y$, their cubes will also not be equal. $\Rightarrow x^{3} \neq y^{3}$ This will be a contradiction to $(1)$. $\therefore x=y$. Hence, $f$ is injective.

Q: Give examples of two functions $f: N \rightarrow Z$ and $g: Z \rightarrow Z$ such that $g o f$ is injective but $g$ is not injective. (Hint: Consider $f(x)=x$ and $g(x)=|x|$ )

Define $f: N \rightarrow Z$ as $f(x)=x$ and $g: Z \rightarrow Z$ as $g(x)=|x|$ Let us first show that $g$ is not injective. $(-1)=|-1|=1$ $(1)=|1|=1$ $\therefore(-1)=g(1)$, but $-1 \neq 1$ $\therefore g$ is not injective. gof : $N \rightarrow Z$ is defined as $g \circ f(x)=g(f(x))=g(x)=|x|$ $x, y \in N$ such that $g o f(x)=g o f(y)$ $\Rightarrow|x|=|y|$ Since $x, y \in N$, both are positive. $\therefore|x|=|y|$ $\Rightarrow x=y$ $\therefore$ gof is injective.

Q: Given examples of two functions $f: N \rightarrow N$ and $g: N \rightarrow N$ such that gof is onto but $f$ is not onto. (Hint: Consider $f(x)=x+1$ and $g(x)=\left\{\begin{array}{ll}x-1, & \text { if } x>1 \\ 1, & \text { if } x=1\end{array}\right\}$ )

Define $f: N \rightarrow Z$ as $f(x)=x+1$ and $g: Z \rightarrow Z$ as $\quad g(x)=\left\{\begin{array}{ll}x-1, & \text { if } x>1 \\ 1, & \text { if } x=1\end{array}\right\}$ 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. $g: N \rightarrow N$ is defined by $g \circ f(x)=g(f(x))=g(x+1)=x+1-1=x \quad[x \in N \Rightarrow x+1>1]$ For $y \in N$, there exists $x=y \in N$ such that $\operatorname{gof}(x)=y$. $\therefore$ gof is onto.

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

$\mathrm{P}(X) \times \mathrm{P}(X) \rightarrow \mathrm{P}(X)$ given by $A^{*} B=A \cap B \forall A, B$ in $P(X)$ $A \cap X=A=X \cap A$ for all $A \in P(X)$ $\Rightarrow A^{*} X=A=X^{*} A$ for all $A \in P(X)$ $X$ is the identity element for the given binary operation *. An element $A \in P(X)$ is invertible if there exists $B \in P(X)$ 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 $P(X)$ with respect to the given operation *.

Question 1

Let $\quad f: R \rightarrow R \quad$ be defined as $f(x)=10 x+7$. Find the function $g: R \rightarrow R \quad$ such that $g \circ f=f \circ g=\mathrm{I}_{R}$.

Accepted Answer

$f: R ightarrow R$ is defined as $f(x)=10 x+7$

For one-one:

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

$\Rightarrow 10 x+7=10 y+7$

$\Rightarrow x=y$

$	herefore f$ is one-one.

For onto:

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

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

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

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

$	herefore f$ is onto.

Thus, $f$ is an invertible function.

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

Now,

$g \circ f(x)=g(f(x))=g(10 x+7)=\frac{(10 x+7)-7}{10}=\frac{10 x}{10}=10$

And,

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

$	herefore g o f=\mathrm{I}_{R}$ and $f o g=\mathrm{I}_{R}$

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

Question 2

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

Accepted Answer

$f: W ightarrow W$ is defined as $f(n)=\left\{\begin{array}{l}n-1, 	ext { If } n 	ext { is odd } \ n+1, 	ext { If } n 	ext { is even }\end{array}ight\}$

For one-one:

$f(n)=f(m)$

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.

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

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

$f(n)=f(m)$

$\Rightarrow n-1=m-1$

$\Rightarrow n=m$

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

$f(n)=f(m)$

$\Rightarrow n+1=m+1$

$\Rightarrow n=m$

$	herefore f$ is one-one.

For onto:

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

$	herefore f$ is onto.

$f$ is an invertible function.

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

When $r$ is odd

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

When $r$ is even

$g \circ f(n)=g(f(n))=g(n+1)=n+1-1=n$

When $m$ is odd

fog $(n)=f(g(m))=f(m-1)=m-1+1=m$

When $m$ is even

$f o g(m)=f(g(m))=f(m+1)=m+1-1=m$

$	herefore g \circ f=\mathrm{I}_{W}$ and $f \circ g=\mathrm{I}_{W}$

$f$ 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.

Question 3

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

Accepted Answer

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

$f(f(x))=f\left(x^{2}-3 x+2\right)$

$=\left(x^{2}-3 x+2\right)^{2}-3\left(x^{2}-3 x+2\right)+2$

$=\left(x^{4}+9 x^{2}+4-6 x^{3}-12 x+4 x^{2}\right)+\left(-3 x^{2}+9 x-6\right)+2$

$=x^{4}-6 x^{3}+10 x^{2}-3 x$

Question 4

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

Accepted Answer

$f: R ightarrow\{x \in R:-1y \Rightarrow x-y>0$ $2 x y$ is negative. $2 x y eq x-y$ Case of $x$ being positive and $y$ being negative, can be ruled out. $ herefore x$ and $y$ have to be either positive or negative. If $x$ and $y$ are positive, $f(x)=f(y)$ $\Rightarrow \frac{x}{1+x}=\frac{y}{1+y}$ $\Rightarrow x-x y=y-x y$ $\Rightarrow x=y$ $ herefore f$ is one-one. For onto: Let $y \in R$ such that $-1

Question 5

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

Accepted Answer

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

For one-one:

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

$$

\begin{equation*}

x^{3}=y^{3} 	ag{1}

\end{equation*}

$$

We need to show that $x=y$

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

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

This will be a contradiction to $(1)$.

$	herefore x=y$. Hence, $f$ is injective.

Question 6

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

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

Accepted Answer

Define $f: N ightarrow Z$ as $f(x)=x$ and $g: Z ightarrow Z$ as $g(x)=|x|$

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

$(-1)=|-1|=1$

$(1)=|1|=1$

$	herefore(-1)=g(1)$, but $-1 
eq 1$

$	herefore g$ is not injective.

gof : $N ightarrow Z$ is defined as $g \circ f(x)=g(f(x))=g(x)=|x|$

$x, y \in N$ such that $g o f(x)=g o f(y)$

$\Rightarrow|x|=|y|$

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

$	herefore|x|=|y|$

$\Rightarrow x=y$

$	herefore$ gof is injective.

Question 7

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

(Hint: Consider $f(x)=x+1$ and $g(x)=\left\{\begin{array}{ll}x-1, & 	ext { if } x>1 \ 1, & 	ext { if } x=1\end{array}ight\}$ )

Accepted Answer

Define $f: N ightarrow Z$ as $f(x)=x+1$ and $g: Z ightarrow Z$ as $\quad g(x)=\left\{\begin{array}{ll}x-1, & 	ext { if } x>1 \ 1, & 	ext { if } x=1\end{array}ight\}$

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 $N$.

$	herefore f$ is not onto.

$g: N ightarrow N$ is defined by

$g \circ f(x)=g(f(x))=g(x+1)=x+1-1=x \quad[x \in N \Rightarrow x+1>1]$

For $y \in N$, there exists $x=y \in N$ such that $\operatorname{gof}(x)=y$.

$	herefore$ gof is onto.

Question 8

Given a non-empty set $X$, consider $P(X)$ which is the set of all subsets of $X$.

Define the relation $R$ in $P(X)$ as follows:

For subsets $A, B$ in $P(X), A R B$ if and only if $A \subset B$. Is $R$ an equivalence relation on $P(X)$ ? Justify you answer.

Accepted Answer

Since every set is a subset of itself, $A R A$ for all $A \in P(X)$.

$	herefore R$ is reflexive.

Let $A R B \Rightarrow A \subset B$

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

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

$	herefore R$ is not symmetric.

If $A R B$ and $B R C$, then $A \subset B$ and $B \subset C$.

$\Rightarrow A \subset C$

$\Rightarrow A R C$

$	herefore R$ is transitive.

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

Question 9

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

Accepted Answer

$\mathrm{P}(X) 	imes \mathrm{P}(X) ightarrow \mathrm{P}(X)$ given by $A^{*} B=A \cap B \forall A, B$ in $P(X)$

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

$\Rightarrow A^{*} X=A=X^{*} A$ for all $A \in P(X)$

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

An element $A \in P(X)$ is invertible if there exists $B \in P(X)$ 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 $P(X)$ with respect to the given operation *.

Question 10

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

Accepted Answer

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

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

Question 11

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

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

ii. $\quad F=\{(a, 2),(b, 1),(c, 1)\}$

Solution: $\mathrm{S}=\{a, b, c\}, T=\{1,2,3\}$

i. $\quad F: S \rightarrow T$ is defined by $F=\{(a, 3),(b, 2),(c, 1)\}$

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

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

ii. $\quad F: S \rightarrow T$ is defined by $F=\{(a, 2),(b, 1),(c, 1)\}$

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

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

Accepted Answer

(No solution provided.)

Question 12

Consider the binary operations ${ }^{*}: R 	imes R ightarrow R$ and $o: R 	imes R ightarrow R$ defined as $a^{*} b=|a-b|$ and $a o b=a, \forall a, b \in R$. Show that *is commutative but not associative 0 is associative but not commutative. Further, show that $\forall a, b, c \in R, a^{*}(b o c)=\left(a^{*} bight) o\left(a^{*} cight)$. [ If it is so, we say that the operation * distributes over the operation 0 ]. Does 0 distribute over*? Justify your answer.

Accepted Answer

It is given that ${ }^{*}: R 	imes R ightarrow R$ and $o: R 	imes R ightarrow R$ defined as $a^{*} b=|a-b|$ and $a o b=a, \forall a, b \in R$. For $a, b \in R$, we have $a^{*} b=|a-b|$ and $b^{*} a=|b-a|=|-(a-b)|=|a-b|$

$	herefore a^{*} b=b^{*} a$

$	herefore$ The operation *is commutative.

$(1 * 2) * 3=(|1-2|) * 3=1 * 3=|1-3|=2$

$1 *(2 * 3)=1 *(|2-3|)=1 * 1=|1-1|=0$

$	herefore(1 * 2) * 3 
eq 1 *(2 * 3)$

where $1,2,3 \in R$

$	herefore$ The operation * is not associative.

Now, consider the operation 0 :

It can be observed that $1 o 2=1$ and $2 o 1=2$.

$	herefore 1 o 2 
eq 2 o 1 \quad($ where $1,2 \in R)$

$	herefore$ The operation 0 is not commutative.

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

$(a o b) o c=a o c=a$

$a o(b o c)=a o b=a$

$\Rightarrow(a o b) o c=a o(b o c)$

$	herefore$ The operation 0 is associative.

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

$a^{*}(b o c)=a^{*} b=|a-b|$

$\left(a^{*} bight) o\left(a^{*} cight)=(|a-b|) o(|a-c|)=|a-b|$

Hence, $a^{*}(b o c)=\left(a^{*} bight) o\left(a^{*} cight)$

Now,

$1 o(2 * 3)=1 o(|2-3|)=1 o 1=1$

$(1 o 2) *(1 o 3)=1 * 1=|1-1|=0$

$	herefore 1 o(2 * 3) 
eq(1 o 2) *(1 o 3)$

where $1,2,3 \in R$

$	herefore$ The operation 0 does not distribute over*.

Question 13

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

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

Accepted Answer

It is given that *: $\mathrm{P}(X) 	imes \mathrm{P}(X) ightarrow \mathrm{P}(X)$ is defined as $A^{*} B=(A-B) \cup(B-A), \forall A, B \in P(X) A \in P(X)$ then,

$A^{*} \Phi=(A-\Phi) \cup(\Phi-A)=A \cup \Phi=A$

$\Phi * A=(\Phi-A) \cup(A-\Phi)=\Phi \cup A=A$

$	herefore A^{*} \Phi=A=\Phi^{*} A \quad$ for all $A \in P(X)$

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

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

$A^{*} B=\Phi=B^{*} A$

[As $\Phi$ is the identity element]

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

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

Question 14

Define a binary operation * on the set $\{0,1,2,3,4,5\}$ as

$a+b=\left\{\begin{array}{ll}a+b, & 	ext { if } a+b<6 \ a+b-6 & 	ext { if } a+b \geq 6\end{array}ight\}$

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

Accepted Answer

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

The operation *is defined as $a+b=\left\{\begin{array}{ll}a+b, & 	ext { if } a+b<6 \ a+b-6, & 	ext { if } a+b \geq 6\end{array}ight\}$

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

$a^{*} 0=a+0=a \quad[a \in X \Rightarrow a+0<6]$

$0 * a=0+a=a \quad[a \in X \Rightarrow 0+a<6]$

$	herefore a^{*} 0=a=0 * a \quad \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., $\left\{\begin{array}{l}a+b=0=b+a, \quad 	ext { if } a+b<6 \ a+b-6=0=b+a-6 \quad 	ext { if } a+b \geq 6\end{array}ight\}$

$\Rightarrow a=-b$ or $b=6-a$

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

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

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

Question 15

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

Accepted Answer

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

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

$f(-1)=(-1)^{2}-(-1)=1+1=2$

$g(-1)=2\left|(-1)-\frac{1}{2}ight|-1=2\left(\frac{3}{2}ight)-1=3-1=2$

$\Rightarrow f(-1)=g(-1)$

$f(0)=(0)^{2}-0=0$

$g(0)=2\left|0-\frac{1}{2}ight|-1=2\left(\frac{1}{2}ight)-1=1-1=0$

$\Rightarrow f(0)=g(0)$

$f(1)=(1)^{2}-1=0$

$g(1)=2\left|1-\frac{1}{2}ight|-1=2\left(\frac{1}{2}ight)-1=1-1=0$

$\Rightarrow f(1)=g(1)$

$f(2)=(2)^{2}-2=2$

$g(2)=2\left|2-\frac{1}{2}ight|-1=2\left(\frac{3}{2}ight)-1=3-1=2$

$\Rightarrow f(2)=g(2)$

$	herefore f(a)=g(a) \quad \forall a \in A$

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

Question 16

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

A. 1

B. 2

C. 3

D. 4

Accepted Answer

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

The smallest relation containing $(1,2)$ and $(1,3)$ which are reflexive and symmetric but not transitive is given by,

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

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

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

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

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

Hence, the total number of desired relations is one.

The correct answer is A .

Question 17

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

A. 1

B. 2

C. 3

D. 4

Accepted Answer

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

The smallest equivalence relation containing $(1,2)$ is given by; $R_{1}=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$

Now, we are left with only four pairs i.e., (2,3),(3,2),(1,3)and (3,1).

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

Hence, the only equivalence relation (bigger than $R_{1}$ ) is the universal relation.

This shows that the total number of equivalence relations containing $(1,2)$ is two. The correct answer is B.

Question 18

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

Accepted Answer

It is given that $f: R ightarrow R$ be the Signum Function defined as

$$

f(x)=\left\{\begin{array}{l}

1, x>0 \

0, x=0 \

-1, x<0

\end{array}ight\}

$$

Also $g: R ightarrow R$ is defined as $g(x)=[x]$, where $[x]$ is greatest integer less than or equal to $x$. Now let $x \in(0,1]$,

$[x]=1$ if $x=1$ and $[x]=0$ if $0<x<1$.

$$

\begin{aligned}

& 	herefore f \circ g(x)=f(g(x))=f([x])=\left\{\begin{array}{ll}

f(1), & 	ext { if } x=1 \

f(0), & 	ext { if } x \in(0,1)

\end{array}ight\}=\left\{\begin{array}{ll}

1, & 	ext { if } x=1 \

0, & 	ext { if } x \in(0,1)

\end{array}ight\} \

& g o f(x)=g(f(x)) \

& \quad \begin{aligned}

& =g(1) \

& =[1]=1

\end{aligned}

$$

Thus, when $x \in(0,1)$, we have $f \circ g(x)=0$ and $g \circ f(x)=1$.

Hence, $f o g$ and $g o f$ does not coincide in $(0,1]$.

Question 19

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

A. 10

B. 16

C. 20

D. 8

Accepted Answer

A binary operation * on $\{a, b\}$ is a function from $\{a, b\} 	imes\{a, b\} ightarrow\{a, b\}$

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

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

The correct answer is B.