Chapter 1: Relations and Functions

Ans:  Given that, f: ${{\text{R}}^*} \to {R_*}$ is defined by ${\text{f}}(x) = \dfrac{1}{x}$.

Consider $x,\,\,y \in R*$ such that ${\text{f}}(x) = {\text{f}}(y)$

$ \Rightarrow \dfrac{1}{x} = \dfrac{1}{y}$

$ \Rightarrow x = y$

Thus, ${\text{f}}$ is one-one.

It is clear that for $y \in R*$, there exists $x = \dfrac{1}{y} \in R*[$ as $y \ne 0]$ such that

${\text{f}}(x) = \dfrac{1}{{\left( {\dfrac{1}{y}} \right)}} = {\text{y}}$

Thus, ${\text{f}}$ is onto.

Therefore, the given function ${\text{f}}$ is one-one and onto.

Now, consider function ${\text{g}}:{\text{N}} \to {{\text{R}}_*}$ defined by ${\text{g}}(x) = \dfrac{1}{x}$

We have, $g\left( {{x_1}} \right) = g\left( {{x_2}} \right)\quad  \Rightarrow \, = \dfrac{1}{{{x_1}}} = \dfrac{1}{{{x_2}}}$

$ \Rightarrow {x_1} = {x_2}$

Thus, \[{\text{g}}\] is one-one.

It is clear that ${\text{g}}$ is not onto as for $1.2 \in  = {{\text{R}}_*}$, there does not exist any $x$ in ${\text{N}}$ such that 

${\text{g}}(x)$ $ = \dfrac{1}{{1.2}}$

Therefore, the function ${\text{g}}$ is one-one but not onto.

Ans: Here, ${\text{f}}\,{\text{:}}\,{\text{N}} \to {\text{N}}$ is given by ${\text{f}}(x) = {x^2}$

For $x,y \in N,\,\,$

${\text{f}}(x) = {\text{f}}(y) \Rightarrow {x^2} = {y^2} \Rightarrow x = y$

Thus, ${\text{f}}$ is injective.

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

Thus, ${\text{f}}$ is not surjective.

Therefore, the function ${\text{f}}$ is injective but not surjective.

Ans: Here, ${\text{f}}:{\text{Z}} \to {\text{Z}}$ is given by ${\text{f}}(x) = {x^2}$

It is seen that ${\text{f}}( - 1) = {\text{f}}(1) = 1$, but $ - 1 \ne 1$.

Thus, ${\text{f}}$ is not injective.

Now, $ - 2 \in {\text{Z}}$. But, there does not exist any element ${\text{x}} \in {\text{Z}}$ such that

$f(x) =  - 2$ or ${x^2} =  - 2$

Thus, ${\text{f}}$ is not surjective.

Therefore, the function ${\text{f}}$ is neither injective nor surjective.

Ans: Here, f: ${\text{R}} \to {\text{R}}$ is given by ${\text{f}}(x) = {x^2}$

It is seen that ${\text{f}}( - 1) = {\text{f}}(1) = 1$, but $ - 1 \ne 1$.

Thus, ${\text{f}}$ is not injective.

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

or ${x^2} =  - 2$.

Thus, ${\text{f}}$ is not surjective.

Therefore, the function ${\text{f}}$ is neither injective nor surjective.

Ans: Here, ${\text{f}}:{\text{N}} \to {\text{N}}$ given by ${\text{f}}(x) = {x^3}$

For $x,y \in N,$

$f(x) = f(y) \Rightarrow {x^3} = {y^3} \Rightarrow x = y$

Thus, ${\text{f}}$ is injective.

Now, $2 \in {\text{N}}$. But, there does not exist any element $x \in {\text{N}}$ such that

$f(x) = 2$ or ${x^3} = 2$

Thus, ${\text{f}}$ is not surjective

Therefore, function ${\text{f}}$ is injective but not surjective.

Ans: Here, ${\text{f}}\,{\text{:}}\,{\text{Z}} \to {\text{Z}}$ is given by $f(x) = {x^3}$

For $x,y \in {\text{Z}},$

${\text{f}}(x) = {\text{f}}(y) \Rightarrow {x^3} = {y^3} \Rightarrow x = y$

Thus, ${\text{f}}$ is injective.

Now, $2 \in {\text{Z}}$. But, there does not exist any element $x \in {\text{Z}}$ such that

${\text{f}}(x) = 2$ or ${x^3} = 2$

Thus, ${\text{f}}$ is not surjective.

Therefore, the function ${\text{f}}$ is injective hut not surjective.

Ans: Here, ${\text{f}}:{\text{R}} \to {\text{R}}$ is given by, ${\text{f}}(x) = [x]$

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

Then, ${\text{f}}(1.2) = {\text{f}}(1.9)$, but $1.2 \ne 1.9$

Thus, ${\text{f}}$ is not one-one.

Consider $0.7 \in {\text{R}}$

It is known that ${\text{f}}(x) = [x]$ is always an integer. 

Thus, there does not exist any element $x \in {\text{R}}$ such that ${\text{f}}(x) = 0.7$

Therefore, ${\text{f}}$ is not onto.

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

Ans: Here, ${\text{f}}:{\text{R}} \to {\text{R}}$ is given by \[{\text{f}}(x) = |x| = \left\{ {\begin{array}{*{20}{l}} x&{{\text{ if }}x \geqslant 0} \\ { - x}&{{\text{ if }}x < 0} \end{array}} \right.\]

It is clear that ${\text{f}}( - 1) = | - 1| = 1$ and ${\text{f}}(1) = |1| = 1$

Now, ${\text{f}}( - 1) = {\text{f}}(1)$, but $ - 1 \ne 1$

Thus, ${\text{f}}$ is not one-one.

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

It is known that ${\text{f}}(x) = |x|$ is always non-negative. 

Thus, there does not exist any element $x$ in domain ${\text{R}}$ such that ${\text{f}}(x) = |x| =  - 1$

Thus, ${\text{f}}$ is not onto.

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

Ans: Here, f: ${\text{R}} \to {\text{R}}$, given by ${\text{f}}(x) = \left\{ {\begin{array}{*{20}{l}} 1&{{\text{ if }}x > 0} \\ {0,}&{{\text{ if }}x = 0} \\ { - 1,}&{{\text{ if }}x < 0} \end{array}} \right.$

It is seen that ${\text{f}}(1) = {\text{f}}(2) = 1$, but $1 \ne 2$.

Thus, ${\text{f}}$ is not one-one.

Now, as ${\text{f}}(x)$ takes only \[3\] values $(1,\,\,0$, or $ - 1)$ for the element $ - 2$ in co-domain

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

Thus, ${\text{f}}$ is not onto

Therefore, the Signum function is neither one-one nor onto.

Ans: Given that, ${\text{A}} = \{ 1,2,3\} $

${\text{B}} = \{ 4,5,6,7\} $

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

Thus, ${\text{f}}(1) = 4,\,\,{\text{f}}(2) = 5,\,\,{\text{f}}(3) = 6$

It is seen that the images of distinct elements of ${\text{A}}$ under ${\text{f}}$ are distinct.

Therefore, the function ${\text{f}}$ is one-one.

Ans: Here, f: ${\text{R}} \to {\text{R}}$ is defined as ${\text{f}}(x) = 3 - 4x$.

Let ${x_1},\,\,{x_2} \in {\text{R}}$ such that ${\text{f}}\left( {{x_1}} \right) = {\text{f}}\left( {{x_2}} \right)$

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

$ \Rightarrow  - 4{x_1} =  - 4{x_2}$

$ \Rightarrow {x_1} = {x_2}$

Thus, ${\text{f}}$ is one-one.

For any real number \[\left( y \right)\] in ${\text{R}}$ such that ${\text{f}}\left( {\dfrac{{3 - y}}{4}} \right) = 3 - 4\left( {\dfrac{{3 - y}}{4}} \right) = y$

Thus, ${\text{f}}$ is onto.

Therefore, the function ${\text{f}}$ is bijective.

Ans: Here, ${\text{f}}:{\text{R}} \to {\text{R}}$ defined by ${\text{f}}(x) = 1 + {x^2}$

Let ${x_1},\,\,{x_2} \in {\text{R}}$ such that ${\text{f}}({x_1}) = {\text{f}}\left( {{x_2}} \right)$

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

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

$ \Rightarrow {x_1} = {x_2}$

Thus, ${\text{f}}\left( {{x_1}} \right) = {\text{f}}\left( {{x_2}} \right)$ does not imply that ${x_1} = {x_2}$.

For example, ${\text{f}}(1) = {\text{f}}( - 1) = 2$

Therefore, ${\text{f}}$ is not one-one.

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

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

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

Therefore, ${\text{f}}$ is not onto.

Hence, the function ${\text{f}}$ is neither one-one nor onto.

Ans: ${\text{f}}\,{\text{:}}\,{\text{A}} \times {\text{B}} \to {\text{B}} \times {\text{A}}$ is defined as ${\text{f}}(a,b) = (b,a)$

Let $\left( {{a_1},{b_1}} \right),\,\,\left( {{a_2},{b_2}} \right) \in {\text{A \times B}}$ such that ${\text{f}}\left( {{a_1},{b_1}} \right) = {\text{f}}\left( {{a_2},{b_2}} \right)$

$ \Rightarrow \left( {{b_1},{a_1}} \right) = \left( {{b_2},{a_2}} \right)$

$ \Rightarrow {b_1} = {b_2}$ and \[{a_1} = {a_2}\]

$ \Rightarrow \left( {{a_1},\;{b_1}} \right) = \left( {{a_2},\;{b_2}} \right)$

Thus, ${\text{f}}$ is one-one.

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

Then, there exists $(a,b) \in {\text{A B}}$ such that $f(a, b)=(b, a) .[$By definition of \[{\text{f}}]\]

Thus, ${\text{f}}$ is onto.

Therefore, the function ${\text{f}}$ is bijective.

Ans: ${\text{f}}:{\text{N}} \to {\text{N}}$ is defined as ${\text{f}}(n) = \left\{ {\begin{array}{*{20}{l}} {\dfrac{{n + 1}}{2},}&{{\text{ if }}n\,\,{\text{is}}\,{\text{odd }}} \\ {\dfrac{n}{2},}&{{\text{ if }}n{\text{ is even }}} \end{array}} \right.$ for all $n \in {\text{N}}$

It can be observed that

$f(1) = \dfrac{{1 + 1}}{2} = 1$ and $f(2) = \dfrac{2}{2} = 1$   (By definition of ${\text{f}}(n)$)

$f(1) = f(2)$, where $1 \ne 2$

Thus, ${\text{f}}$ is not one-one.

Consider a natural number \[\left( n \right)\] in co-domain \[{\text{N}}.\]

Case 1: $n$ is odd

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

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

Case 2: $n$ is even

Thus, $n = 2r$ for some $r \in {\text{N}}$. Then, there exists $4r \in {\text{N}}$ such that

${\text{f}}(4r) = \dfrac{{4r}}{2} = 2r$

Therefore, ${\text{f}}$ is onto.

Hence, the function \[{\text{f}}\]is not a bijective function.

Ans: ${\text{A}} = {\text{R}} - \{ 3\} ,{\text{B}} = {\text{R}} - \{ 1\} $ and ${\text{f}}:{\text{A}} \to {\text{B}}$ defined by ${\text{f}}(x) = \left( {\dfrac{{x - 2}}{{x - 3}}} \right)$

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

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

By cross multiplication,

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

Expand brackets,

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

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

Thus, ${\text{f}}$ is one-one.

Let $y \in {\text{B}} = {\text{R}} - \{ 1\} .$ Then, $y \ne 1$.

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

Now, $f(x) = y$

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

By cross multiplication,

$ \Rightarrow x - 2 = xy - 3y \Rightarrow x(1 - y) =  - 3y + 2$

$ \Rightarrow x = \dfrac{{2 - 3y}}{{1 - y}} \in {\text{A}}$

$[y \ne 1]$

Thus, for any $y \in {\text{B}}$, there exists $\dfrac{{2 - 3y}}{{1 - y}} \in {\text{A}}$ such that

${\text{f}}\left( {\dfrac{{2 - 3y}}{{1 - y}}} \right) = \dfrac{{\left( {\dfrac{{2 - 3y}}{{1 - y}}} \right) - 2}}{{\left( {\dfrac{{2 - 3y}}{{1 - y}}} \right) - 3}}$

Take LCM,

$ = \dfrac{{2 - 3y - 2 + 2y}}{{2 - 3y - 3 + 3y}}$

$ = \dfrac{{ - y}}{{ - 1}} = y$

Thus, ${\text{f}}$ is onto.

Therefore, the function ${\text{f}}$ is one-one and onto.