Solution

It is given that $f: \mathbf{R} * \to \mathbf{R} *$ is defined by $f(x)=\frac{1}{x}$.

One-one:

$ \begin{aligned} & f(x)=f(y) \\ & \Rightarrow \frac{1}{x}=\frac{1}{y} \\ & \Rightarrow x=y \end{aligned} $

$\therefore f$ is one-one.

Onto:

It is clear that for $y \in \mathbf{R} _{*}$, there exists $x=\frac{1}{y} \in \mathbf{R}$. (Exists as $.y \neq 0)$ such that

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

$\therefore f$ is onto.

Thus, the given function $(f)$ is one-one and onto.

Now, consider function $g: \mathbf{N} \to \mathbf{R} *$ defined by

$ g(x)=\frac{1}{x} $

We have,

$ g(x_1)=g(x_2) \Rightarrow \frac{1}{x_1}=\frac{1}{x_2} \Rightarrow x_1=x_2 $

$\therefore g$ is one-one.

Further, it is clear that $g$ is not onto as for $1.2 \in \mathbf{R} *$ there does not exit any $x$ in $\mathbf{N}$ such

that $g(x)=\frac{1}{1.2}$.

Hence, function $g$ is one-one but not onto.

Solution

(i) $f: \mathbf{N} \to \mathbf{N}$ is given by,

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

It is seen that for $x, y \in \mathbf{N}, f(x)=f(y) \Rightarrow x^{2}=y^{2} \Rightarrow x=y$.

$\therefore f$ is injective.

Now, $2 \in \mathbf{N}$. But, there does not exist any $x$ in $\mathbf{N}$ such that $f(x)=x^{2}=2$.

$\therefore f$ is not surjective.

Hence, function $f$ is injective but not surjective.

(ii) $f: \mathbf{Z} \to \mathbf{Z}$ is given by,

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

It is seen that $f(-1)=f(1)=1$, but $-1 \neq 1$.

$\therefore f$ is not injective.

Now, $-2 \in \mathbf{Z}$. But, there does not exist any element $x \in \mathbf{Z}$ such that $f(x)=x^{2}=-2$.

$\therefore f$ is not surjective.

Hence, function $f$ is neither injective nor surjective.

(iii) $f: \mathbf{R} \to \mathbf{R}$ is given by,

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

It is seen that $f(-1)=f(1)=1$, but $-1 \neq 1$.

$\therefore f$ is not injective.

Now, $-2 \in \mathbf{R}$. But, there does not exist any element $x \in \mathbf{R}$ such that $f(x)=x^{2}=-2$. $\therefore f$ is not surjective.

Hence, function $f$ is neither injective nor surjective.

(iv) $f: \mathbf{N} \to \mathbf{N}$ given by,

$f(x)=x^{3}$

It is seen that for $x, y \in \mathbf{N}, f(x)=f(y) \Rightarrow x^{3}=y^{3} \Rightarrow x=y$.

$\therefore f$ is injective.

Now, $2 \in \mathbf{N}$. But, there does not exist any element $x$ in domain $\mathbf{N}$ such that $f(x)=x^{3}=$

$\therefore f$ is not surjective

Hence, function $f$ is injective but not surjective.

(v) $f: \mathbf{Z} \to \mathbf{Z}$ is given by,

$f(x)=x^{3}$

It is seen that for $x, y \in \mathbf{Z}, f(x)=f(y) \Rightarrow x^{3}=y^{3} \Rightarrow x=y$.

$\therefore f$ is injective.

Now, $2 \in \mathbf{Z}$. But, there does not exist any element $x$ in domain $\mathbf{Z}$ such that $f(x)=x^{3}=2$.

$\therefore f$ is not surjective.

Hence, function $f$ is injective but not surjective.

Solution

$f: \mathbf{R} \to \mathbf{R}$ is given by,

$f(x)=[x]$

It is seen that $f(1.2)=[1.2]=1, f(1.9)=[1.9]=1$.

$\therefore f(1.2)=f(1.9)$, but $1.2 \neq 1.9$.

$\therefore f$ is not one-one.

Now, consider $0.7 \in \mathbf{R}$.

It is known that $f(x)=[x]$ is always an integer. Thus, there does not exist any element $x$ $\in \mathbf{R}$ such that $f(x)=0.7$.

$\therefore f$ is not onto.

Hence, the greatest integer function is neither one-one nor onto.

Solution

$f: \mathbf{R} \to \mathbf{R}$ is given by,

$f(x)=|x|=\begin{cases} x, \text{ if } x \geq 0 \\ -x, \text{ if } x<0 \end{cases} .$

It is seen that $f(-1)=|-1|=1, f(1)=|1|=1$.

$\therefore f(-1)=f(1)$, but $-1 \neq 1$.

$\therefore f$ is not one-one.

Now, consider $-1 \in \mathbf{R}$.

It is known that $f(x)=|x|$ is always non-negative. Thus, there does not exist any element $x$ in domain $\mathbf{R}$ such that $f(x)=|x|=-1$.

$\therefore f$ is not onto.

Hence, the modulus function is neither one-one nor onto.

Solution

$f: \mathbf{R} \to \mathbf{R}$ is given by,

$$ f(x)=\begin{matrix} 1, \text{ if } x>0 \\ 0, \text{ if } x=0 \\ -1, \text{ if } x<0 \end{matrix} $$

It is seen that $f(1)=f(2)=1$, but $1 \neq 2$.

$\therefore f$ is not one-one.

Now, as $f(x)$ takes only 3 values $(1,0$, or -1 ) for the element -2 in co-domain $\mathbf{R}$, there does not exist any $x$ in domain $\mathbf{R}$ such that $f(x)=-2$.

$\therefore f$ is not onto.

Hence, the signum function is neither one-one nor onto.

Solution

It is given that $A={1,2,3}, B={4,5,6,7}$.

$f: A \to B$ is defined as $f={(1,4),(2,5),(3,6)}$.

$\therefore f(1)=4, f(2)=5, f(3)=6$

It is seen that the images of distinct elements of $A$ under $f$ are distinct.

Hence, function $f$ is one-one.

Solution

(i) $f: \mathbf{R} \to \mathbf{R}$ is defined as $f(x)=3-4 x$.

Let $x_1, x_2 \in \mathbf{R}$ such that $f(x_1)=f(x_2)$.

$\Rightarrow 3-4 x_1=3-4 x_2$

$\Rightarrow-4 x_1=-4 x_2$

$\Rightarrow x_1=x_2$

$\therefore f$ is one-one.

For any real number $(y)$ in $\mathbf{R}$, there exists $\frac{3-y}{4}$ in $\mathbf{R}$ such that

$f(\frac{3-y}{4})=3-4(\frac{3-y}{4})=y$.

$\therefore f$ is onto.

Hence, $f$ is bijective.

(ii) $f: \mathbf{R} \to \mathbf{R}$ is defined as

$ f(x)=1+x^{2} . $

Let $x_1, x_2 \in \mathbf{R}$ such that $f(x_1)=f(x_2)$.

$\Rightarrow 1+x_1^{2}=1+x_2^{2}$

$\Rightarrow x_1^{2}=x_2^{2}$

$\Rightarrow x_1= \pm x_2$

$\therefore f(x_1)=f(x_2)$ does not imply that $x_1=x_2$.

For instance,

$ f(1)=f(-1)=2 $

$\therefore f$ is not one-one.

Consider an element -2 in co-domain $\mathbf{R}$.

It is seen that $f(x)=1+x^{2}$ is positive for all $x \in \mathbf{R}$.

Thus, there does not exist any $x$ in domain $\mathbf{R}$ such that $f(x)=-2$.

$\therefore f$ is not onto.

Hence, $f$ is neither one-one nor onto.

Solution

$f: A \times B \to B \times A$ is defined as $f(a, b)=(b, a)$.

Let $(a_1, b_1),(a_2, b_2) \in A \times B$ such that $f(a_1, b_1)=f(a_2, b_2)$

$\Rightarrow(b_1, a_1)=(b_2, a_2)$

$\Rightarrow b_1=b_2$ and $a_1=a_2$

$\Rightarrow(a_1, b_1)=(a_2, b_2)$

$\therefore f$ is one-one.

Now, let $(b, a) \in B \times A$ be any element.

Then, there exists $(a, b) \in A \times B$ such that $f(a, b)=(b, a)$. [By definition of $f$ ] $\therefore f$ is onto.

Hence, $f$ is bijective.

Solution

$ f(n)=\begin{matrix} \frac{n+1}{2}, & \text{ if } n \text{ is odd } \\ \frac{n}{2}, & \text{ if } n \text{ is even }\\ \end{matrix} \text{ for all } n \in \mathbf{N}. $

It can be observed that:

$f(1)=\frac{1+1}{2}=1$ and $f(2)=\frac{2}{2}=1 \quad[$ By definition of $f]$

$\therefore f(1)=f(2)$, where $1 \neq 2$.

$\therefore f$ is not one-one.

Consider a natural number $(n)$ in co-domain $\mathbf{N}$.

Case I: $n$ is odd

$\therefore n=2 r+1$ for some $r \in \mathbf{N}$. Then, there exists $4 r+1 \in \mathbf{N}$ such that

$ f(4 r+1)=\frac{4 r+1+1}{2}=2 r+1 $

Case II: $n$ is even

$\therefore n=2 r$ for some $r \in \mathbf{N}$. Then, there exists $4 r \in \mathbf{N}$ such that $f(4 r)=\frac{4 r}{2}=2 r$.

$\therefore f$ is onto.

Hence, $f$ is not a bijective function.

Solution

$A=\mathbf{R}-{3}, B=\mathbf{R}-{1}$

$f: A \to B$ is defined as $f(x)=(\frac{x-2}{x-3})$.

Let $x, y \in$ A such that $f(x)=f(y)$.

$\Rightarrow \frac{x-2}{x-3}=\frac{y-2}{y-3}$

$\Rightarrow(x-2)(y-3)=(y-2)(x-3)$

$\Rightarrow x y-3 x-2 y+6=x y-3 y-2 x+6$

$\Rightarrow-3 x-2 y=-3 y-2 x$

$\Rightarrow 3 x-2 x=3 y-2 y$

$\Rightarrow x=y$

$\therefore f$ is one-one.

Let $y \in B=\mathbf{R}-{1}$. Then, $y \neq 1$.

The function $f$ is onto if there exists $x \in A$ such that $f(x)=y$.

Now,

$ \begin{aligned} & f(x)=y \\ & \Rightarrow \frac{x-2}{x-3}=y \\ & \Rightarrow x-2=x y-3 y \\ & \Rightarrow x(1-y)=-3 y+2 \\ & \Rightarrow x=\frac{2-3 y}{1-y} \in A \quad[y \neq 1] \end{aligned} $

Thus, for any $y \in B$, there exists $\frac{2-3 y}{1-y} \in A$ such that $f(\frac{2-3 y}{1-y})=\frac{(\frac{2-3 y}{1-y})-2}{(\frac{2-3 y}{1-y})-3}=\frac{2-3 y-2+2 y}{2-3 y-3+3 y}=\frac{-y}{-1}=y$.

$\therefore f$ is onto.

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

Solution

$f: \mathbf{R} \to \mathbf{R}$ is defined as $f(x)=x^{4}$.

Let $x, y \in \mathbf{R}$ such that $f(x)=f(y)$.

$\Rightarrow x^{4}=y^{4}$

$\Rightarrow x= \pm y$

$\therefore f(x_1)=f(x_2)$ does not imply that $x_1=x_2$.

For instance,

$ f(1)=f(-1)=1 $

$\therefore f$ is not one-one.

Consider an element 2 in co-domain $\mathbf{R}$. It is clear that there does not exist any $x$ in domain $\mathbf{R}$ such that $f(x)=2$.

$\therefore f$ is not onto.

Hence, function $f$ is neither one-one nor onto.

The correct answer is D.

Solution

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

Let $x, y \in \mathbf{R}$ such that $f(x)=f(y)$.

$\Rightarrow 3 x=3 y$

$\Rightarrow x=y$

$\therefore f$ is one-one.

Also, for any real number $(y)$ in co-domain $\mathbf{R}$, there exists $\frac{y}{3}$ in $\mathbf{R}$ such that $f(\frac{y}{3})=3(\frac{y}{3})=y$.

$\therefore f$ is onto.

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

The correct answer is A.