Solution

It is given that $f: \mathbf{R} \to{x \in \mathbf{R}:-1<x<1}$ is defined as $f(x)=\frac{x}{1+|x|}, x \in \mathbf{R}$.

Suppose $f(x)=f(y)$, where $x, y \in \mathbf{R}$.

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

It can be observed that if $x$ is positive and $y$ is negative, then we have:

$ \frac{x}{1+x}=\frac{y}{1-y} \Rightarrow 2 x y=x-y $

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

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

But, $2 x y$ is negative.

Then, $2 x y \neq x-y$.

Thus, the case of $x$ being positive and $y$ being negative can be ruled out.

Under a similar argument, $x$ being negative and $y$ being positive can also be ruled out $\therefore x$ and $y$ have to be either positive or negative.

When $x$ and $y$ are both positive, we have:

$ f(x)=f(y) \Rightarrow \frac{x}{1+x}=\frac{y}{1+y} \Rightarrow x+x y=y+x y \Rightarrow x=y $

When $x$ and $y$ are both negative, we have:

$ f(x)=f(y) \Rightarrow \frac{x}{1-x}=\frac{y}{1-y} \Rightarrow x-x y=y-y x \Rightarrow x=y $

$\therefore f$ is one-one.

Now, let $y \in \mathbf{R}$ such that $-1<y<1$.

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

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

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

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

$\therefore f$ is onto.

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

Solution

$f: \mathbf{R} \to \mathbf{R}$ is given as $f(x)=x^{3}$.

Suppose $f(x)=f(y)$, where $x, y \in \mathbf{R}$.

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

Now, we need to show that $x=y$.

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

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

However, this will be a contradiction to (1).

$\therefore x=y$

Hence, $f$ is injective.

Solution

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

$\therefore R$ is reflexive.

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

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

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

$\therefore R$ is not symmetric.

Further, if $A R B$ and $B R C$, then $A \subset B$ and $B \subset C$.

$\Rightarrow A \subset C$

$\Rightarrow A R C$

$\therefore R$ is transitive.

Hence, $R$ is not an equivalence relation since it is not symmetric.

Solution

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

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

Solution

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

Also, it is given that $f, g: A \to B$ are defined by $f(x)=x^{2}-x, x \in A$ and

$ g(x)=2|x-\frac{1}{2}|-1, x \in A $

It is observed that:

$ \begin{aligned} & f(-1)=(-1)^{2}-(-1)=1+1=2 \\ & g(-1)=2|(-1)-\frac{1}{2}|-1=2(\frac{3}{2})-1=3-1=2 \\ & \Rightarrow f(-1)=g(-1) \\ & f(0)=(0)^{2}-0=0 \\ & g(0)=2|0-\frac{1}{2}|-1=2(\frac{1}{2})-1=1-1=0 \\ & \Rightarrow f(0)=g(0) \\ & f(1)=(1)^{2}-1=1-1=0 \\ & g(1)=2|1-\frac{1}{2}|-1=2(\frac{1}{2})-1=1-1=0 \\ & \Rightarrow f(1)=g(1) \\ & f(2)=(2)^{2}-2=4-2=2 \\ & g(2)=2|2-\frac{1}{2}|-1=2(\frac{3}{2})-1=3-1=2 \\ & \Rightarrow f(2)=g(2) \\ & \therefore f(a)=g(a) \forall a \in A \end{aligned} $

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

Solution

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

The smallest relation containing $(1,2)$ and $(1,3)$ which is 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 since $(1,2),(2,1) \in R$ and $(1,3),(3,1) \in R$.

But relation $R$ is not transitive as $(3,1),(1,2) \in R$, but $(3,2) \notin R$.

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.

Solution

It is given that $A={1,2,3}$.

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.