NCERT Solutions Class 12 Mathematics Chapter 1- Relation and Function

Q.1 Determine whether each of the following relations are reflexive, symmetric and transitive:

(i ) Relation R in the set A = {1, 2, 3…13, 14}
defined as R = {(x, y): 3x − y = 0}

(ii) Relation R in the set N of natural numbers
defined as R = {(x, y): y = x + 5 and x < 4}

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as

R = {(x, y): y is divisible by x}

(iv) Relation R in the set Z of all integers defined as

R = {(x, y): x − y is as integer}

(v) Relation R in the set A of human beings in a
town at a particular time given by

(a) R = {(x, y): x and y work at the same place}

(b) R = {(x, y): x and y live in the same locality}

(c) R = {(x, y): x is exactly 7 cm taller than y}

(d) R = {(x, y): x is wife of y}

(e) R = {(x, y): x is father of y}

Ans.

i) A = {1, 2, 3 … 13, 14}

R = {(x, y): 3x − y = 0 or y=3x}

∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}

Given relation R is not reflexive because
(1, 1), (2, 2), (3,3)… (14, 14) ∉ R.

Also, R is not a symmetric relation as
(2, 6) ∈R, but (6,2) ∉ R.

Also, R is not transitive as (1, 3), (3, 9) ∈R,
but (1, 9) ∉ R.

Hence, R is neither reflexive, nor symmetric, nor transitive.

\begin{array}{l}\text{(ii) R}=\left\{\left(x,y\right):y=x+\text{5 and}x<\text{4}\right\}\\ \text{R}=\left\{\left(\text{1},\text{6}\right),\left(\text{2},\text{7}\right),\left(\text{3},\text{8}\right)\right\}\\ \text{Since}\left(\text{1},\text{1}\right)\notin \text{R}.\\ \therefore \text{R is not reflexive}.\\ \left(\text{1},\text{6}\right)\in \text{R But}\left(\text{6},\text{1}\right)\notin \text{R}.\\ \therefore \text{R is not symmetric}.\\ \text{Now},\text{since there is no pair in R such that}\left(a,b\right)\text{and}\\ \left(b,c\right)\in \text{R},\text{then}\left(a,c\right)\text{cannot belong to R}.\\ \therefore \text{R is not transitive}.\\ \text{Hence},\text{R is neither reflexive},\text{nor symmetric},\text{nor transitive}.\\ \text{(iii) A = {1, 2, 3, 4, 5, 6}}\\ \text{R = {(x, y): y is divisible by x}}\\ \text{Since every number is divisible by itself}.\\ \therefore \text{(x, x)}\in \text{R}\\ \text{Hence R is reflexive}\text{.}\\ \text{Now,}\\ \text{(2, 4)}\in \text{R [as 4 is divisible by 2]}\\ \text{But,}\\ \text{(4, 2)}\notin \text{R}\text{. [as 2 is not divisible by 4]}\\ \therefore \text{R is not symmetric}\text{.}\\ \text{Let (x, y), (y, z)}\in \text{R}\text{. Then, y is divisible by x and z is divisible}\\ \text{by y}\text{.}\\ \therefore \text{z is divisible by x}\text{.}\\ \text{Hence (x, z)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is reflexive and transitive but not symmetric}\text{.}\\ \text{(iv) R = {(x, y): x}-\text{y is an integer}}\\ \text{For every x}\in \text{Z, (x, x)}\in \text{R as x}-\text{x = 0 which is an integer}\text{.}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Now, for every x, y}\in \text{Z if (x, y)}\in \text{R, then x}-\text{y is an integer}\text{.}\\ \text{For every (y,x)}\in \text{R by definition}\\ \left(y-x\right)=-\left(x-y\right)\text{is also an integer}\text{.}\\ \Rightarrow \left(y-x\right)\text{is also an integer}\text{.}\\ \therefore \text{(y, x)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Now,}\\ \text{Let (x, y) and (y, z)}\in \text{R, where x, y, z}\in \text{Z}\text{.}\\ \Rightarrow \text{(x}-\text{y) = integer}\\ \Rightarrow \text{(y}-\text{z) = integer}\\ \text{On adding both, we get}\\ \left(x-z\right)=\text{integer}\text{.}\\ \therefore \left(x,z\right)\in R\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is reflexive, symmetric, and transitive}\text{.}\\ \\ \text{(v) (a) R = {(x, y): x and y work at the same place}}\\ \text{Every individual worker belongs to itself}\\ \Rightarrow \text{(x, x)}\in \text{R}\\ \therefore \text{R is reflexive}\text{.}\\ \text{If (x, y)}\in \text{R, then x and y work at the same place}\text{.}\\ \text{Similarly y and x work at the same place}\text{.}\\ \therefore \text{(y, x)}\in \text{R}\text{.}\\ \text{Hence R is symmetric}\text{.}\\ \text{Now, let (x, y), (y, z)}\in \text{R}\\ \Rightarrow \text{x and y work at the same place and y and z work at}\\ \text{the same place}\text{.}\\ \therefore \text{x and z also work at the same place}\text{.}\\ \therefore \text{(x, z)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is reflexive, symmetric and transitive}\text{.}\\ \text{(b) R = {(x, y): x and y live in the same locality}}\\ \text{Every individual belongs to itself}\\ \text{Thus (x, x)}\in \text{R}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Since (x, y)}\in \text{R it means x and y live in the same locality}\text{.}\\ \text{Which implies y and x also live in the same locality}\text{.}\\ \therefore \text{(y, x)}\in \text{R}\\ \text{Hence R is symmetric}\text{.}\\ \text{Now, let (x, y)}\in \text{R and (y, z)}\in \text{R}\text{.}\\ \Rightarrow \text{x and y live in the same locality and y and z live in the same}\\ \text{locality}\text{.}\\ \text{Which implies that x and z also live in the same locality}\text{.}\\ \therefore \text{(x, z)}\in \text{R}\\ \text{Hence R is transitive}\text{.}\\ \text{Hence, R is reflexive, symmetric and transitive}\text{.}\\ \text{(c) R = {(x, y): x is exactly 7 cm taller than y}}\\ âˆ\mu \text{An individual height can be equal to itself but cannot}\\ \text{be more by 7 cm}\\ \text{Hence (x, x)}\notin \text{R}\\ \therefore \text{R is not reflexive}\text{.}\\ \text{Now, let (x, y)}\in \text{R}\text{.}\\ \Rightarrow \text{x is exactly 7 cm taller than y}\\ \text{Then y should be shorter than x}\\ \therefore \text{(y, x)}\notin \text{R}\\ \therefore \text{R is not symmetric}\text{.}\\ \text{Again,}\\ \text{Let (x, y), (y, z)}\in \text{R}\text{.}\\ \Rightarrow \text{x is exactly 7 cm taller than y and y is exactly 7 cm}\\ \text{taller than z}\text{.}\\ \\ \text{Which implies x is exactly 14 cm taller than z}\text{.}\\ \therefore \text{(x, z)}\notin \text{R}\\ so,\text{R is not transitive}\text{.}\\ \text{Hence, R is neither reflexive nor symmetric nor transitive}\text{.}\\ \text{(d) R = {(x, y): x is the wife of y}}\\ \text{Since x cannot be the wife of herself}\text{.}\\ \therefore \text{(x, x)}\notin \text{R}\\ \text{Then, R is not reflexive}\text{.}\\ \text{Let}\left(x,y\right)\in R\\ which\text{implies that x is wife of y}\text{.}\\ i.e.,\text{y is husband of x}\text{.}\\ \text{so, y can not be wife of x}\text{.}\\ \Rightarrow \left(y,x\right)\notin R\\ Then,\text{R is not symmetric}\text{.}\\ \text{Let}\text{(x, y), (y, z)}\in \text{R}\\ \Rightarrow \text{x is the wife of y and y is the wife of z}\text{.}\\ \text{A husband can never become wife of anyone}.\\ so,\left(x,z\right)\notin R\\ Thus,\text{​}\text{R is not transitive}\text{.}\\ \text{Hence, R is neither reflexive nor symmetric nor transitive}\text{.}\\ \text{(e) R = {(x, y): x is the father of y}}\\ \text{Since, x cannot be the father of himself}\text{.}\\ \text{So,}\left(x,x\right)\notin R\\ Then,\text{R is not reflexive}\text{.}\\ \text{Now, let (x, y)}\in \text{R}\text{.}\\ \\ \text{If x is the father of y}\text{.}\\ \text{Then y cannot be the father of x}\text{.}\\ \text{so,}\left(y,x\right)\notin R\\ Then,\text{R is not symmetric}\text{.}\\ \text{Again, let (x, y)}\in \text{R and (y, z)}\in \text{R}\text{.}\\ \Rightarrow \text{x is father of y and y is father of z}\text{.}\\ \Rightarrow \text{x can not be father of z}\text{.}\\ \Rightarrow \text{x will be grandfather of z}\\ \therefore \left(x,z\right)\notin R\\ \therefore R\text{is not transitive}\text{.}\\ \text{Hence, R is neither reflexive nor symmetric not transitive}\text{.}\end{array}

Q.2

$\begin{array}{l}\mathrm{Show}\mathrm{}\mathrm{that}\mathrm{}\mathrm{the}\mathrm{}\mathrm{relation}\mathrm{}\mathrm{R}\mathrm{}\mathrm{in}\mathrm{}\mathrm{the}\mathrm{}\mathrm{set}\mathrm{}\mathrm{R}\mathrm{}\mathrm{of}\mathrm{}\mathrm{real}\mathrm{}\mathrm{numbers},\mathrm{}\\ \mathrm{defined}\mathrm{}\mathrm{as}\mathrm{}\mathrm{R}=\mathrm{}\{\mathrm{}(\mathrm{a},\mathrm{b}):\mathrm{\hspace{0.17em}}\mathrm{a}\le {\mathrm{b}}^2\}\mathrm{\hspace{0.17em}}\mathrm{is}\mathrm{neithe}\mathrm{rreflexive}\mathrm{nor}\\ \mathrm{symmetric}\mathrm{nor}\mathrm{transitive}.\end{array}$

Ans.

\begin{array}{l}\text{We have R = {(a, b): a}\le b^{\text{2}}\text{} where a, b}\in \text{R}\\ \text{We can see that}\frac{\text{1}}{3}\le {\left(\frac{\text{1}}{3}\right)}^{\text{2}}\text{is not valid}\text{. So,}\left\{\frac{\text{1}}{3}\text{,}\frac{\text{1}}{3}\right\}\notin R\\ Then,\text{R is not reflexive}\text{.}\\ \text{Since,}\left(-2,3\right)\in Rand\text{​}-{\text{2<3}}^{\text{2}}\\ But{\text{3}}^{\text{2}}\text{is not less than}-\text{2}\text{.}\\ \end{array}

So, ( 3,−2 )∉R

$\begin{array}{l}\mathrm{Thus},\mathrm{R}\mathrm{is}\mathrm{not}\mathrm{symmetric}\mathrm{.}\\ \mathrm{Again},\mathrm{let}(5,-6),(-6,2)\in \mathrm{R}\\ \mathrm{As}\mathrm{\hspace{0.17em}}5<{(-6)}^2\mathrm{and}-6{<2}^{\mathrm{2}}\\ \mathrm{But}5\mathrm{is}\mathrm{not}\mathrm{less}\mathrm{than}4\mathrm{.}\\ \mathrm{So},(5,2)\notin \mathrm{R}\\ \mathrm{Thus},\mathrm{R}\mathrm{is}\mathrm{not}\mathrm{transitive}\mathrm{.}\\ \mathrm{Therefore},\mathrm{R}\mathrm{is}\mathrm{neither}\mathrm{reflexive}\mathrm{nor}\mathrm{symmetric}\mathrm{nor}\mathrm{transitive}.\end{array}$

Q.3 Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.

Ans.

Given A = {1, 2, 3, 4, 5, 6}. A relation R is defined
on A as: R = {(a, b): b = a + 1}

\begin{array}{l}\therefore \text{R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}}\\ \text{We can find (1,1)}\notin \text{R, where 1}\in \text{A}\text{.}\\ \therefore \text{R is not reflexive}\text{.}\\ \text{It can be observed that (1, 2)}\in \text{R, but (2, 1)}\notin \text{R}\text{.}\\ \therefore \text{R is not symmetric}\text{.}\\ \text{Now, (1, 2), (2, 3)}\in \text{R but}\left(1,3\right)\notin R.\\ \therefore \text{R is not transitive}\text{.}\\ \text{Hence, R is neither reflexive nor symmetric nor transitive}\text{.}\end{array}

Q.4

$\begin{array}{l}\mathbf{Show}\mathbf{}\mathrm{}\mathbf{that}\mathbf{}\mathrm{}\mathbf{the}\mathbf{}\mathrm{}\mathbf{relation}\mathbf{}\mathrm{}\mathbf{R}\mathbf{}\mathrm{}\mathbf{in}\mathrm{}\mathbf{R}\mathrm{\hspace{0.17em}}\mathbf{defined}\mathrm{}\mathbf{as}\mathrm{}\mathbf{R}=\mathrm{}\{(\mathbf{a},\mathbf{b}):\mathrm{\hspace{0.17em}}\mathrm{a}\le \mathrm{b}\}\mathrm{\hspace{0.17em}}\\ \mathrm{is}\mathrm{reflexive},\mathrm{symmetric}\mathrm{or}\mathrm{transitive}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have}\\ \mathrm{R}=\; \{(\mathrm{a},\mathrm{b});\mathrm{a}\le \mathrm{b}\}\\ \mathrm{We}\mathrm{can}\mathrm{see}\mathrm{that}(\frac{\mathrm{1}}{2}\mathrm{,}\frac{\mathrm{1}}{2}\mathrm{)}\in \mathrm{R}\mathrm{as}\frac{\mathrm{1}}{2}=\frac{\mathrm{1}}{2}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{.}\\ \mathrm{Since}(1,2)\in \mathrm{R}\mathrm{as}2\mathrm{is}\mathrm{greater}\mathrm{than}1\mathrm{.}\\ \mathrm{but}1\mathrm{​}\mathrm{is}\mathrm{not}\mathrm{greater}\mathrm{than}2,\mathrm{so}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(2,1)\notin \mathrm{R}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{not}\mathrm{symmetric}\mathrm{.}\\ \mathrm{Again},\mathrm{let}(\mathrm{a},\mathrm{b}),\; (\mathrm{b},\mathrm{c})\in \mathrm{R}\mathrm{.}\\ \mathrm{Then}\mathrm{according}\mathrm{to}\mathrm{condition},\\ \mathrm{a}\le \mathrm{b}\mathrm{and}\mathrm{b}\le \mathrm{c}\hspace{0.17em}\hspace{0.17em}\Rightarrow \mathrm{a}\le \mathrm{c}\\ \Rightarrow (\mathrm{a},\mathrm{c})\in \mathrm{R}\end{array}$

∴R is transitive.

\text{Hence,R is reflexive and transitive but not symmetric}\text{.}

Q.5

$\begin{array}{l}\mathrm{Check}\mathrm{whether}\mathrm{the}\mathrm{relation}\mathrm{R}\mathrm{in}\mathrm{R}\mathrm{defined}\mathrm{as}\mathrm{R}=\left\{\right(\mathrm{a},\mathrm{b}):\mathrm{a}\le {\mathrm{b}}^3\}\\ \mathrm{is}\mathrm{reflexive},\mathrm{symmetric}\mathrm{or}\mathrm{transitive}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have}\\ \mathrm{R}=\; \{(\mathrm{a},\mathrm{b});\mathrm{a}\le {\mathrm{b}}^{\mathrm{3}}\mathrm{\}}\\ \mathrm{We}\mathrm{can}\mathrm{see}\mathrm{that}\frac{\mathrm{1}}{2}\le {\left(\frac{\mathrm{1}}{2}\right)}^3\mathrm{\hspace{0.17em}}\end{array}$

so,( 1 2 , 1 2 )∉ R ∴ R is not reflexive. Since ( 1,2 )∈ R as 1<2 3 But ( 2,1 )∉ R as  2 is not smaller than 1 3 . ∴R is not symmetric. Again, let (5, 5 2 ), ( 5 2 , 5 4 ) ∈ R. Then according to condition, 5< ( 5 2 ) 3 and 5 2 < ( 5 4 ) 3   ⇒ (5, 5 4 )∉R  as  5> ( 5 4 ) 3 ∴ R is not transitive. Hence,R is neither reflexive nor symmetric nor transitive.

Q.6 Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Ans.

\begin{array}{l}Since,\\ A=\left\{1,2,3\right\}\\ \text{A relation R on A is defined as R = {(1, 2), (2, 1)}}\text{.}\\ \text{Since (1, 1), (2, 2), (3, 3)}\notin \text{R}\text{.}\\ \therefore \text{R is not reflexive}\text{.}\\ \text{Now, as (1, 2)}\in \text{R and (2, 1)}\in \text{R,}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Now, (1, 2) and (2, 1)}\in \text{R}\\ But\left(1,1\right)\notin R\\ \therefore R\text{is not transitive}\text{.}\\ \text{Hence, R is symmetric but neither reflexive nor transitive}\text{.}\end{array}

Q.7 Show that the relation R in the set A of all the books in a library of a college, given by
R = {(x, y): x and y have same number of pages} is an equivalence relation.

Ans.

\begin{array}{l}\text{Set A is the set of all books in the library of a college}\text{.}\\ \text{R = {x, y): x and y have the same number of pages}}\\ \text{since (x, x)}\in \text{R as x and x has the same number of pages}\text{.}\end{array}

∴R is reflexive.

\begin{array}{l}\text{If (x, y)}\in \text{R where x and y have the same number of pages}\text{.}\\ \Rightarrow \text{y and x have the same number of pages}\text{.}\\ \Rightarrow \text{(y, x)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Again, let (x, y)}\in \text{R and (y, z)}\in \text{R}\text{.}\\ \Rightarrow \text{x and y and have the same number of pages and y and z}\\ \text{have the same number of pages}\text{.}\\ \Rightarrow \text{x and z have the same number of pages}\text{.}\\ \therefore \text{(x, z)}\in \text{R}\\ \text{Therefore, R is transitive}\text{.}\\ \text{Since, R is reflexive, symmetric and transitive, so R is an}\\ \text{equivalence relation}\text{.}\end{array}

Q.8 Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b): |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

Ans.

\begin{array}{l}Since,\text{A = {1, 2, 3, 4, 5}}\\ R=\left\{\left(a,b\right):\left|a-b\right|iseven\right\}\\ For\text{any element a}\in \text{A, we have}\left|a-a\right|=0,which\text{is even number}\text{.}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Let (a, b)}\in \text{R}\text{.}\end{array}

    | a−b | is even.

\begin{array}{l}\Rightarrow \left|-\left(b-a\right)\right|iseven.\\ \Rightarrow \left|b-a\right|iseven.\\ \Rightarrow \left(b,a\right)\in R\\ So,\text{R is symmetric}\text{.}\\ \text{Again, let}\left(a,b\right)\in Rand\left(b,c\right)\in R.\\ \Rightarrow \left|b-a\right|is\text{even number and}\left|c-b\right|is\text{also even number}\text{.}\\ \Rightarrow \left(a-b\right)iseven\text{and}\left(b-c\right)\text{is even}\text{.}\\ \Rightarrow \left(a-c\right)=\left(a-b\right)+\left(b-c\right)is\text{even}\text{.}\\ \Rightarrow \left|a-c\right|is\text{even}\text{.}\\ \Rightarrow \left(a,c\right)is\text{even}\text{.}\\ \Rightarrow \left(a,c\right)\in R\\ \therefore R\text{is transitive}\text{.}\\ \text{Hence, R is an equivalence relation}\text{.}\\ \text{Now, all elements of the set {1, 3, 5} are related to each}\\ \text{other as the modulus of difference between any two elements}\\ \text{of this set is even}\text{.}\\ \text{Similarly, all elements of the set {2, 4} are related to each}\\ \text{other as the difference between the two elements is even}\text{.}\\ \text{Also, no element of the subset {1, 3, 5} can be related to any}\\ \text{element of {2, 4} as all elements of {1, 3, 5} are odd and all}\\ \text{elements of {2, 4} are even}\text{.}\text{The modulus of difference between}\\ \text{odd and even is again odd}.\end{array}

Q.9

Ans.

\begin{array}{l}Here,A=\left\{x\in Z,0\le x\le 12\right\}=\left\{0,1,2,3,4,5,6,7,8,9,10,11,12\right\}\\ \left(i\right)R=\left\{\left(a,b\right):\left|a-b\right|\text{is a multiple of 4}\right\}\\ For\text{any element a}\in \text{A,}\\ \text{we have}\left(a,a\right)\in Ras\left|a-a\right|=0whichis\text{multiple of 4}\text{.}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Let}\left(a,b\right)\in Ras\text{​}\left|a-b\right|=multipleof4\\ and\left|b-a\right|=\left|-\left(a-b\right)\right|\\ =\left|a-b\right|=\text{multiple of 4}\\ \text{So,}\left(b,a\right)\in R\\ Therefore,R\text{is symmetric}\text{.}\\ Let\text{​}\text{​}\left(a,b\right),\left(b,c\right)\in R\\ \Rightarrow \left|a-b\right|=multipleof4and\left|c-b\right|=multipleof4\\ \Rightarrow \left(a,b\right)\text{is multiple of 4}and\left(b,a\right)\text{is multiple of 4}\end{array}

( a−c )=( a−b )+( b−c ) is multiple of 4

\begin{array}{l}\Rightarrow \left|\left(a-c\right)\right|\text{is also multiple of 4}\\ \Rightarrow \left(a,c\right)\in R\\ \therefore R\text{is transitive}\text{.}\\ \text{Hence, R is equivalence relation}\text{.}\\ \text{Let x be an element of A such that (x,1)}\in \text{R}\\ \text{Then}\\ \text{|x}-\text{1| is a multiple of 4}\\ \Rightarrow \text{|x}-\text{1|}=0,4,8,12\\ \Rightarrow x=1,5,9\left[13\text{is not a part of set}\text{A}\right]\\ \text{Hence the set of all elements of A which are related to 1}\\ \text{is {1,5,9}}.\\ \left(ii\right)\text{R = {(a, b): a = b}}\\ \text{For any element a}\in \text{A, we have (a, a)}\in \text{R, since a = a}\text{.}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Now, let (a, b)}\in \text{R}\text{.}\\ \Rightarrow \text{a}=\text{b}\\ \Rightarrow \text{b}=\text{a}\\ \left(b,a\right)\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Let (a, b)}\in \text{R and (b, c)}\in \text{R}\\ \Rightarrow \text{a = b and b = c}\\ \Rightarrow \text{a = c}\\ \Rightarrow \text{(a, c)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is an equivalence relation}\text{.}\end{array} \begin{array}{l}\text{The elements in R that are related to 1 will be elements}\\ \text{from set A which are equal to 1}\text{.}\\ \text{Hence, the set of elements related to 1 is {1}}\text{.}\end{array}

Q.10 Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

Ans.

\begin{array}{l}\text{Let O denote the origin in the given plane}\text{. Then}\\ \text{R = {(P, Q): OP=OQ}}\\ \text{We can observe that for any point P in set A we have}\\ \text{OP=OP}\\ \Rightarrow \text{(P,P)}\in \text{R}\end{array}

Thus (P,P)∈R for all P∈A

\begin{array}{l}\text{So R is reflexive}\\ \text{Now,}\\ \text{Let (P, Q)}\in \text{R}\text{.}\\ \Rightarrow \text{OP=OQ}\\ \Rightarrow \text{OQ=OP}\\ \Rightarrow \left(Q,P\right)\in \text{R}\\ \therefore R\text{is symmetric}\text{.}\\ \text{Again, let}\left(P,Q\right),\left(Q,S\right)\in \text{R}\\ \Rightarrow \text{OP=OQ and OQ=OS}\\ \Rightarrow \text{OP=OS}\\ \Rightarrow \left(P,S\right)\in R\\ \therefore R\text{is transitive}\text{.}\\ \text{Thus, R is an equivalence relation}\text{.}\end{array} \begin{array}{l}Again,\text{Let P be a fixed point in set A and Q be a point in set A}\\ \text{such that (P,Q)}\in \text{R}\text{. Then}\\ \Rightarrow \text{OP=OQ}\\ \Rightarrow \text{Q moves in the plane in such a way that its distance from}\\ \text{the origin (0,0) is always}\text{equal and is equal to OP}\\ \Rightarrow \text{Locus of Q is a circle with centre at the origin and radius OP}\text{.}\end{array}

Q.11 Show that the relation R defined in the set A of all triangles as R = {(T₁, T₂): T₁ is similar to T₂}, is equivalence relation. Consider three right angle triangles T₁ with sides 3, 4, 5, T₂ with sides 5, 12, 13 and T₃ with sides 6, 8, 10. Which triangles among T₁, T₂ and T₃ are related?

Ans.

\begin{array}{l}Since,\text{every triangle is similar to itself}\text{.}\\ {\text{R = {(T}}_{\text{1}}{\text{, T}}_{\text{2}}{\text{): T}}_{\text{1}}{\text{is similar to T}}_{\text{2}}\text{}}\\ \text{(T,T)}\in \text{R for all T}\in \text{A}\\ \therefore \text{R is Reflexive}.\\ {\text{Let (T}}_{\text{1}}{\text{, T}}_{\text{2}}\text{)}\in \text{R,}\\ \Rightarrow {\text{T}}_{\text{1}}{\text{is similar to T}}_{\text{2}}\text{.}\\ \Rightarrow {\text{T}}_{\text{2}}{\text{is similar to T}}_{\text{1}}\text{.}\\ \Rightarrow {\text{(T}}_{\text{2}}{\text{, T}}_{\text{1}}\text{)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Now,}\\ {\text{(T}}_{\text{1}}{\text{,T}}_{\text{2}}{\text{,T}}_{\text{3}}\text{)}\in {\text{A such that (T}}_{\text{1}}{\text{,T}}_{\text{2}}\text{)}\in {\text{R and (T}}_{\text{2}}{\text{,T}}_{\text{3}}\text{)}\in \text{R}\\ \Rightarrow {\text{T}}_{\text{1}}{\text{is similar to T}}_{\text{2}}\text{and}{\text{T}}_{\text{2}}{\text{is similar to T}}_{\text{3}}\text{.}\\ \Rightarrow {\text{T}}_{\text{1}}{\text{is similar to T}}_{\text{3}}\text{.}\\ \Rightarrow {\text{(T}}_{\text{1}}{\text{, T}}_{\text{3}}\text{)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Thus, R is an equivalence relation}\text{.}\\ {\text{In Triangles T}}_{\text{1}}{\text{and T}}_{\text{3}}\text{we observe that the corresponding angles}\\ \text{are equal and the}\text{corresponding sides are proportional}\\ \text{i}\text{.e}\text{.}\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}\\ Hence,{\text{T}}_{\text{1}}\text{is related to}{\text{T}}_{\text{3}}\text{.}\end{array}

Q.12 Give an example of a relation. Which is

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

Ans.

\begin{array}{l}\text{(i) Let A = {1,2,3}}\text{.}\\ \text{Define a relation R on A as R = {(1, 2), (2, 1)}}\text{.}\\ \text{Relation R is not reflexive as (1,1), (2, 2), (3, 3)}\notin \text{R}\text{.}\\ \text{Now, as (1, 2)}\in \text{R and also (2, 1)}\in \text{R, R is symmetric}\text{.}\\ \Rightarrow \text{(1, 2), (2, 1)}\in \text{R, but (1, 1)}\notin \text{R}\\ \therefore \text{R is not transitive}\text{.}\\ \text{Hence, relation R is symmetric but neither reflexive or transitive}\text{.}\\ \text{(ii) Consider a relation R in R defined as:}\\ \text{R = {(x, y): x < y}}\\ \text{Since x cannot be less than x}.\\ \therefore \text{R is not reflexive}\text{.}\\ Now,\text{(1, 2)}\in \text{R}\text{(as1 < 2)}\\ \text{But, 2 is not less than 1}\text{.}\\ \therefore \text{(2, 1)}\notin \text{R}\\ \therefore \text{R is not symmetric}\text{.}\\ \\ \text{Again, let (a, b), (b, c)}\in \text{R}\text{.}\\ \Rightarrow \text{a < b and b < c}\\ \Rightarrow \text{a < c}\\ \Rightarrow \left(a,c\right)\in R\\ \therefore R\text{is transitive}\text{.}\\ \text{Hence, relation R is transitive but neither reflexive nor symmetric}\text{.}\\ \left(iii\right)\text{Let A = {2, 4, 6}}\text{.}\\ \text{Define a relation R on A as:}\\ \text{A = {(2,2), (4,4), (6,6), (2,4), (4,2), (4,6), (6,4)}}\\ \text{Relation R is reflexive as (2,2), (4,4), (6,6)}}\in \text{R}\text{.}\\ \text{Relation R is symmetric as (2,4), (4,2), (4,6), (6,4)}\in \text{R}\text{.}\\ \text{Relation R is not transitive since (2,4), (4,6)}\in \text{R, but (2,6)}\notin \text{R}\text{.}\\ \text{Hence, relation R is reflexive and symmetric but not transitive}\text{.}\\ \left(iv\right)\text{A = {1, 2, 3, 4, 5, 6}}\\ \text{R = {(x, y): y is divisible by x}}\\ \text{Since every number is divisible by itself}\\ \therefore \text{(x, x)}\in \text{R}\\ \text{Hence R is reflexive}\text{.}\\ \text{Now,}\\ \text{(2, 4)}\in \text{R}\text{[as 4 is divisible by 2]}\\ \text{But,}\\ \text{(4, 2)}\notin \text{R}\text{.}\text{[as 2 is not divisible by 4]}\\ \therefore \text{R is not symmetric}\text{.}\\ \text{Let (x, y), (y, z)}\in \text{R}\text{.}\\ \\ \text{Then, y is divisible by x and z is divisible by y}\text{.}\\ \therefore \text{z is divisible by x}\text{.}\\ \text{Hence (x, z)}\in \text{R}\text{.}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is reflexive and transitive but not symmetric}\text{.}\\ \text{(v) Let A = {1, 2}}\text{.}\\ \text{Define a relation R on A as:}\\ \text{R = {(1,2), (2,1), (1,1)}}\\ \text{Relation R is not reflexive as (2,2)}\notin \text{R}\text{.}\\ \text{Relation R is symmetric as (1,2)}\in \text{R and (2,1)}\in \text{R}\text{.}\\ \text{It is seen that (1,2), (2,1)}\in \text{R}\text{. Also, (1,1)}\in \text{R}\text{.}\\ \therefore \text{The relation R is transitive}\text{.}\\ \text{Hence, relation R is symmetric and transitive but not reflexive}\text{.}\end{array}

Q.13 Show that the relation R defined in the set A of all polygons as R = {(P₁, P₂): P₁ and P₂ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Ans.

\begin{array}{l}Since,\\ {\text{R = {(P}}_{\text{1}}{\text{, P}}_{\text{2}}{\text{): P}}_{\text{1}}{\text{and P}}_{\text{2}}\text{have same the number of sides}}\\ {\text{Since (P}}_{\text{1}}{\text{, P}}_{\text{1}}\text{)}\in \text{R polygon with same number of sides belongs}\\ \text{to itself}\text{.}\\ \therefore \text{R is Reflexive}\text{.}\\ {\text{Let (P}}_{\text{1}}{\text{, P}}_{\text{2}}\text{)}\in \text{R}\text{.}\\ \Rightarrow {\text{P}}_{\text{1}}{\text{and P}}_{\text{2}}\text{have the same number of sides}\text{.}\\ \Rightarrow {\text{P}}_{\text{2}}{\text{and P}}_{\text{1}}\text{have the same number of sides}\text{.}\\ \Rightarrow {\text{(P}}_{\text{2}}{\text{, P}}_{\text{1}}\text{)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Again, let}\left(P_1,P_2\right),\left(P_2,P_3\right)\in \text{R}\\ \Rightarrow {\text{P}}_{\text{1}}{\text{and P}}_{\text{2}}\text{have the same number of sides}{\text{. Also, P}}_{\text{2}}{\text{and P}}_{\text{3}}\text{have}\\ \text{the same number of sides}\text{.}\\ \Rightarrow {\text{P}}_{\text{1}}{\text{and P}}_{\text{3}}\text{have the same number of sides}\text{.}\\ \Rightarrow {\text{(P}}_{\text{1}}{\text{, P}}_{\text{3}}\text{)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Therefore, R is an equivalence relation}\text{.}\\ \text{Hence, the set of all elements in A related to triangle T is the}\\ \text{set of all triangles}\text{.}\end{array}

Q.14 Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L₁, L₂): L₁ is parallel to L₂}. Show that R is
an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Ans.

\begin{array}{l}Since,\\ {\text{R = {(L}}_{\text{1}}{\text{, L}}_{\text{2}}{\text{): L}}_{\text{1}}{\text{is parallel to L}}_{\text{2}}\text{}}\\ {\text{(L}}_{\text{1}}{\text{,L}}_{\text{1}}\text{)}\in \text{R as every line is parallel to itself}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Now,}\\ {\text{Let (L}}_{\text{1}}{\text{, L}}_{\text{2}}\text{)}\in \text{R}\text{.}\\ \Rightarrow {\text{L}}_{\text{1}}{\text{is parallel to L}}_{\text{2}}\text{.}\\ \Rightarrow {\text{L}}_{\text{2}}{\text{is parallel to L}}_{\text{1}}\text{.}\\ \Rightarrow {\text{(L}}_{\text{2}}{\text{, L}}_{\text{1}}\text{)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ {\text{Let (L}}_{\text{1}}{\text{, L}}_{\text{2}}{\text{), (L}}_{\text{2}}{\text{, L}}_{\text{3}}\text{)}\in \text{R}\text{.}\\ \Rightarrow {\text{L}}_{\text{1}}{\text{is parallel to L}}_{\text{2}}{\text{. And L}}_{\text{2}}{\text{is parallel to L}}_{\text{3}}\text{.}\\ \Rightarrow {\text{L}}_{\text{1}}{\text{is parallel to L}}_{\text{3}}\text{.}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is an equivalence relation}\text{.}\\ \text{The set of all lines related to the line y = 2x + 4 must be parallel}\\ \text{line only and we know that in parallel lines’ equations only the}\\ \text{constant value changes the co-efficient of x and y remains same}\\ \therefore \text{set of parallel lines is y=2x+}\text{c, where c can be any constant}\text{.}\end{array}

Q.15

$\begin{array}{l}\mathrm{Let}\mathrm{R}\mathrm{be}\mathrm{the}\mathrm{relation}\mathrm{in}\mathrm{the}\mathrm{set}\{1,2,3,4\}\mathrm{given}\mathrm{by}\\ \mathrm{R}=\left\{\right(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2\left)\right\}.\mathrm{}\end{array}$

Choose the correct answer.

$\begin{array}{l}\left(\mathrm{A}\right)\mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{and}\mathrm{symmetric}\mathrm{but}\mathrm{not}\mathrm{transitive}.\\ \left(\mathrm{B}\right)\mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{and}\mathrm{transitive}\mathrm{but}\mathrm{not}\mathrm{symmetric}.\\ \left(\mathrm{C}\right)\mathrm{R}\mathrm{is}\mathrm{symmetric}\mathrm{and}\mathrm{transitive}\mathrm{but}\mathrm{not}\mathrm{reflexive}.\\ \left(\mathrm{D}\right)\mathrm{R}\mathrm{is}\mathrm{an}\mathrm{equivalence}\mathrm{relation}.\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{B}\right)\mathrm{is}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{Since}\{(1,1),(2,2),(3,3),(4,4)\}\in \mathrm{R}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{reflexive}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{Now}(1,2)\in \mathrm{R}\mathrm{but}(2,1)\notin \mathrm{R}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{not}\mathrm{symmetric}\\ \mathrm{And}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(1,2)\mathrm{and}(2,2)\in \mathrm{R}\Rightarrow (1,2)\in \mathrm{R}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{transitive}\\ \mathrm{Hence}\mathrm{the}\mathrm{given}\mathrm{Relation}\mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{and}\mathrm{transitive}\\ \mathrm{but}\mathrm{not}\mathrm{symmetric}\end{array}$

Q.16

$\begin{array}{l}\mathrm{Let}\mathrm{R}\mathrm{be}\mathrm{the}\mathrm{relationin}\mathrm{theset}\hspace{0.17em}\hspace{0.17em}\mathrm{N}\hspace{0.17em}\hspace{0.17em}\mathrm{given}\mathrm{by}\\ \mathrm{R}=\{(\mathrm{a},\mathrm{b}):\mathrm{a}=\mathrm{b}-2,\hspace{0.17em}\hspace{0.17em}\mathrm{b}>6\}.\\ \mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}.\\ \left(\mathrm{A}\right)(2,4)\in \mathrm{R}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{B}\right)(3,8)\in \mathrm{R}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{C}\right)(6,8)\in \mathrm{R}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\mathrm{D}\right)(8,7)\in \mathrm{R}\end{array}$

Ans.

$\begin{array}{l}\mathrm{R}=\; \{(\mathrm{a},\mathrm{b}):\mathrm{a}=\mathrm{b}-2,\mathrm{b}>\; 6\}\\ \mathrm{Now},\mathrm{since}\mathrm{b}>\; 6,\; (2,\; 4)\in \mathrm{R}\\ \therefore \mathrm{Option}\left(\mathrm{A}\right)\mathrm{cannot}\mathrm{be}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \mathrm{Also}\mathrm{by}\mathrm{putting}8\mathrm{we}\mathrm{will}\mathrm{not}\mathrm{get}3\mathrm{as}\mathrm{difference}\end{array}$ $\begin{array}{l}\therefore \mathrm{Option}\left(\mathrm{B}\right)\mathrm{is}\mathrm{also}\mathrm{not}\mathrm{the}\mathrm{correct}\mathrm{option}\\ \mathrm{And},\mathrm{as}8\ne \mathrm{7}-2\\ \therefore (8,\; 7)\notin \mathrm{R}\\ \mathrm{Now},\mathrm{consider}(6,\; 8)\mathrm{.}\\ \mathrm{We}\mathrm{have}8\; >\; 6\mathrm{and}\mathrm{also},\; 6\; =\; 8-2.\\ \therefore (6,\; 8)\in \mathrm{R}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{option}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{C}\right)\mathrm{.}\end{array}$

Q.17

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{function}\mathrm{f}:\mathrm{R}{}_{*}\to {\mathrm{R}}_{*}\mathrm{definedby}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\hspace{0.17em}\hspace{0.17em}\mathrm{is}\\ \mathrm{one}–\mathrm{one}\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{onto},\mathrm{\hspace{0.17em}}\mathrm{where}{\mathrm{R}}_{*}\mathrm{\hspace{0.17em}}\mathrm{is}\mathrm{these}\mathrm{t}\mathrm{of}\mathrm{all}\mathrm{non}–\mathrm{zero}\\ \mathrm{realnumbers}.\mathrm{Is}\mathrm{there}\mathrm{sulttrue},\mathrm{if}\mathrm{the}\mathrm{domain}\mathrm{R}{}_{*}\mathrm{is}\hspace{0.17em}\hspace{0.17em}\\ \mathrm{replaced}\mathrm{by}\mathrm{Nwithco}–\mathrm{domain}\mathrm{being}\mathrm{same}\mathrm{as}{\mathrm{R}}_{*}?\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{x},\mathrm{y}\in {\mathrm{R}}_{\mathrm{*}}\mathrm{\hspace{0.17em}}\mathrm{such}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ \mathrm{one}-\mathrm{one}:\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1}{\mathrm{x}}=\frac{1}{\mathrm{y}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\mathrm{y}\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\mathrm{f}:{\mathrm{R}}_{\mathrm{*}}\to {\mathrm{R}}_{\mathrm{*}}\mathrm{\hspace{0.17em}}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Onto}:\end{array}$

Let f( x )=y where y be any element of R * ( Co−domain )

$\begin{array}{l}\Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1}{\mathrm{x}}=\mathrm{y}\hspace{0.17em}\hspace{0.17em}\mathrm{or}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\frac{1}{\mathrm{y}}\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\mathrm{f}\left(\frac{1}{\mathrm{y}}\right)=\frac{1}{\left(\frac{1}{\mathrm{y}}\right)}=\mathrm{y}\\ \mathrm{Since}\mathrm{Range}=\mathrm{Co}–\mathrm{domain}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{function}\left(\mathrm{f}\right)\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{.}\\ \mathrm{Consider}\mathrm{function}\mathrm{f}:\mathrm{N}\to {\mathrm{R}}_{\mathrm{*}}\mathrm{\hspace{0.17em}}\mathrm{defined}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\\ \mathrm{For}\mathrm{any}\mathrm{x},\mathrm{y}\in \mathrm{N}\mathrm{we}\mathrm{see}\mathrm{that}\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{1}{\mathrm{x}}=\frac{1}{\mathrm{y}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\mathrm{y}\\ \mathrm{So}\mathrm{f}:\mathrm{N}\to {\mathrm{R}}_{\mathrm{*}}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{function}\\ \mathrm{Since}\mathrm{fractional}\mathrm{numbers}\mathrm{like}\frac{2}{3},\frac{2}{5}\mathrm{\hspace{0.17em}}\mathrm{etc}.\mathrm{\hspace{0.17em}}\mathrm{in}\mathrm{co}–\mathrm{domain}{\mathrm{R}}_{*}\mathrm{\hspace{0.17em}}\mathrm{have}\\ {\mathrm{R}}_{\mathrm{*}}\mathrm{\hspace{0.17em}}\mathrm{have}\mathrm{no}\mathrm{pre}\mathrm{image}\mathrm{in}\mathrm{\hspace{0.17em}}\mathrm{domain}\mathrm{N}.\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\mathrm{f}:\mathrm{N}\to {\mathrm{R}}_{*}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\end{array}$

Q.18(i) f: N→ N   given by  f(x)= x²

\begin{array}{l}(ii)f:Z\to Zgivenbyf(x)=x^2\\ (iii)f:R\to Rgivenbyf(x)=x^2\\ (iv)f:N\to Ngivenbyf(x)=x^3\\ (v)f:Z\to Zgivenbyf(x)=x^3\end{array}

Ans.

\begin{array}{l}\text{(i)}\text{f:N}\to \text{N is given by,}\text{f}\left(x\right)=x^2\\ \text{To check for injectivity let f}\left(x\right)=\text{f}\left(y\right)\\ \Rightarrow x^2=y^2\\ \Rightarrow x=y\left(\begin{array}{l}There\text{are no negative}\\ \text{natural numbers}\text{.}\end{array}\right)\\ \therefore f\left(x\right)\text{is injective}\text{.}\\ \text{To check for onto}:\\ \text{Let}\text{f}\left(x\right)\text{=y}\\ \Rightarrow x^2\text{=y}\\ \Rightarrow \text{x=}\sqrt{y}\\ \therefore \text{f}\left(\sqrt{y}\right)\text{=y}\\ \text{Since range is not equal to co-domain}\text{.}\\ \text{f(x) is not onto or surjective}\\ \text{(ii) f: Z}\to \text{Z is given by,}\\ \text{f(x)}={\text{x}}^{\text{2}}\\ \text{For injectivity let}\text{f(x)}=\text{f(y)}\\ \Rightarrow x^2=y^2\\ \Rightarrow x=\pm y\\ \\ \therefore \text{f is not injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{(x)}=\text{y}\\ \Rightarrow {\text{x}}^{\text{2}}=\text{y}\\ \Rightarrow \text{x}=\sqrt{\text{y}}\\ \therefore f\left(\sqrt{y}\right)=y\\ \text{Hence, function f is neither injective nor surjective}\text{.}\\ \\ \left(iii\right)f:R\to R\text{is given by, f}\left(x\right)=x^2\\ For\text{injectivity:}\\ \text{f}\left(x\right)=f\left(y\right)\\ \Rightarrow x^2=y^2\\ \Rightarrow x=\pm y\left(\begin{array}{l}x\text{can not take more than one}\\ \text{value for injectivity}\text{.}\end{array}\right)\\ \therefore f\text{is not injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{f}\left(x\right)=y\Rightarrow x^2=y\\ \Rightarrow x=\sqrt{y}\\ \therefore f\left(\sqrt{y}\right)=y\\ \text{Since range is not equal to co-domain}.\\ \therefore f\text{is not surjective}\text{.}\\ \left(iv\right)f:N\to N\text{is given by, f}\left(x\right)=x^3\\ For\text{injectivity:}\\ \text{f}\left(x\right)=f\left(y\right)\\ \\ \Rightarrow x^3=y^3\\ \Rightarrow x=y\\ \therefore f\text{is injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{f}\left(x\right)=y\Rightarrow x^3=y\\ \Rightarrow x=\sqrt[3]{y}\\ \therefore f\left(\sqrt[3]{y}\right)=y\left(y\in Nbut\sqrt[3]{y}\notin N\right)\\ \text{Since range is not equal to co-domain}.\\ \therefore f\text{is not surjective}\text{.}\\ \text{Thus, given function is injective but not surjective}\text{.}\\ \\ \left(v\right)f:Z\to Z\text{is given by, f}\left(x\right)=x^3\\ For\text{injectivity:}\\ \text{f}\left(x\right)=f\left(y\right)\\ \Rightarrow x^3=y^3\\ \Rightarrow x=y\\ \therefore f\text{is injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{f}\left(x\right)=y\Rightarrow x^3=y\\ \Rightarrow x=\sqrt[3]{y}\\ \therefore f\left(\sqrt[3]{y}\right)=y\left(y\in Zbut\sqrt[3]{y}\notin Z\right)\\ \text{Since range is not equal to co-domain}.\\ \therefore f\text{is not surjective}\text{.}\\ \text{Thus, given function is injective but not surjective}\text{.}\end{array}