1. Show that the function ${\text{f}}:{{\mathbf{R}}_*} \to {{\mathbf{R}}_*}$ defined by ${\text{f(}}x) = \dfrac{1}{x}$ is one-one and onto, where ${{\mathbf{R}}_*}$ is the set of all non-zero real numbers. Is the result true, if the domain ${{\mathbf{R}}_*}$ is replaced by ${\text{N}}$ with co-domain being same as ${{\mathbf{R}}_*}$? 

 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.

2. Check the injectivity and surjectivity of the following functions: (i) ${\text{f}}:{\text{N}} \to {\text{N}}$ given by ${\text{f}}(x) = {x^2}$ 
 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.

(ii) ${\text{f}}:{\text{Z}} \to {\text{Z}}$ given by ${\text{f}}(x) = {x^2}$ 
 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.

(iii) f: ${\text{R}} \to {\text{R}}$ is given by ${\text{f}}(x) = {x^2}$ 
 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.

(iv) ${\text{f}}:{\text{N}} \to {\text{N}}$ given by ${\text{f}}(x) = {x^3}$ 
 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.

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

3. Prove that the Greatest Integer Function ${\text{f}}:{\text{R}} \to {\text{R}}$ given by ${\text{f}}(x) = [x]$, is neither one one nor onto, where \[[x]\]denotes the greatest integer less than or equal to $x$. 
 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.

4. Show that the Modulus Function ${\text{f}}:{\text{R}} \to {\text{R}}$ given by ${\text{f}}(x) = |x|$, is neither one-one nor  onto, where \[|x|\] is $x$, if ${\text{x}}$ is positive or 0 and $|x|$ is $ - x$, if $x$ is negative. 
 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.

5. Show that the Signum Function 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.$ 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.

6. Let ${\text{A}} = \{ 1,2,3\} ,\,\,{\text{B}} = \{ 4,5,6,7\} $ and let ${\text{f}} = \{ (1,4),\,\,(2,5),\,\,(3,6)\} $ be a function from ${\text{A}}$ to ${\text{B}}$.  Show that ${\text{f}}$ is one-one. 
 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.

7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your 
 Answer. (i) ${\text{f}}\,{\text{:}}\,{\text{R}} \to {\text{R}}$ defined by ${\text{f}}(x) = 3 - 4x$ 
 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.

(ii) ${\text{f}}\,{\text{:}}\,{\text{R}} \to {\text{R}}$ is defined as ${\text{f}}(x) = 1 + {x^2}$ 
 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.

8. Let ${\text{A}}$ and ${\text{B}}$ be sets. Show that ${\text{f}}:{\text{A}} \times {\text{B}} \to {\text{B}} \times {\text{A}}$ such that $(a,b) = (b,a)$ is bijective function. 
 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.

9. Let f: ${\text{N}} \to {\text{N}}$ be defined by ${\text{f}}(n) = \left\{ {\begin{array}{*{20}{l}}

{\dfrac{{n + 1}}{2},}&{{\text{ if }}n{\text{ is odd }}} \\

{\dfrac{n}{2},}&{{\text{ if }}n{\text{ is even }}}

\end{array}} \right.$ for all $n \in {\text{N}}$ State whether the function ${\text{f}}$ is bijective. Justify your 
 Answer. 
 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.

10. Let ${\text{A}} = {\text{R}} - \{ 3\} $ and ${\text{B}} = {\text{R}} - \{ 1\} .$ Consider the function f: ${\text{A}} \to {\text{B}}$ defined by ${\text{f}}(x)$  $ = \left( {\dfrac{{x - 2}}{{x - 3}}} \right).$ Is f one-one and onto? Justify your 
 Answer. 
 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.

11. Let ${\text{f}}:{\mathbf{R}} \to {\mathbf{R}}$ be defined as ${\text{f}}(x) = {x^4}$. Choose the correct 
 Answer. (A) ${\text{f}}$ is one-one onto (B) ${\text{f}}$ is many-one onto (C) ${\text{f}}$ is one-one but not onto (D) ${\text{f}}$ is neither one-one nor onto 
 Ans: Here, ${\text{f}}:{\mathbf{R}} \to {\mathbf{R}}$ is defined as ${\text{f}}(x) = {x^4}$. Let $x,y \in R$ such that ${\text{f}}(x) = {\text{f}}(y)$. $ \Rightarrow {x^4} = {y^4}$ $ \Rightarrow x =  \pm y$ ${\text{f}}(x) = {\text{f}}(y)$ does not imply that $x = y$ For example, ${\text{f}}(1) = {\text{f}}( - 1) = 1$ Thus, ${\text{f}}$ 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 ${\text{R}}$ such that ${\text{f}}(x) = 2$ Thus, ${\text{f}}$ is not onto. Hence, the function ${\text{f}}$ is neither one-one nor onto. The correct 
 Answer is \[{\text{D}}.\]

12. Let ${\text{f}}\,{\text{:}}\,{\text{R}} \to {\text{R}}$ be defined as ${\text{f}}(x) = 3x$. Choose the correct 
 Answer. (A) f is one – one and onto  (B) f is many – one and onto  (C) f is one – one but not onto  (D) f is neither one – one nor onto 
 Ans: Here, ${\text{f}}:{\text{R}} \to {\text{R}}$ is defined as ${\text{f}}(x) = 3x$. Let $x,\,\,y \in {\text{R}}$ such that ${\text{f}}(x) = {\text{f}}(y)$. $ \Rightarrow 3x = 3y$ $ \Rightarrow x = y$ Thus, ${\text{f}}$ is one-one. Now, for any real number \[\left( y \right)\] in co-domain ${\text{R}}$, there exists \[\dfrac{y}{3}\] in ${\text{R}}$ such that ${\text{f}}\left( {\dfrac{y}{3}} \right) = 3\left( {\dfrac{y}{3}} \right) = y$ Thus, ${\text{f}}$ is onto. Hence, the function ${\text{f}}$ is one-one and onto. Therefore, the correct 
 Answer is \[{\text{A}}.\]