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}

Q.19 Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Ans.

\begin{array}{l}Since,\\ f:R\to R\text{is given by,}\\ \text{f}\left(x\right)=\left[x\right]\\ We\text{​}\text{see that f}\left(0.1\right)=0,\text{f}\left(0.3\right)=0.\\ \therefore \text{f}\left(0.1\right)=\text{f}\left(0.3\right),but\text{0}\text{.1}\ne \text{0}\text{.3}\\ \text{So, f is not injective i}\text{.e}\text{., one-one}\text{.}\\ \text{Now, let 0}\text{.6}\in \text{R}\text{.}\\ \text{It is given that f}\left(\text{x}\right)\text{=}\left[x\right]\text{is always an integer}\text{. Thus, there does}\\ \text{not exist any element x}\in \text{R such that f}\left(\text{x}\right)\text{=0}\text{.6}\text{.}\\ \text{Then, f is not onto}\text{.}\\ \text{Therefore, the greatest integer function is neither one-one}\\ \text{nor onto}\text{.}\end{array}

Q.20

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{Modul}\mathrm{us}\mathrm{Function}f:\mathrm{R}\to \mathrm{R}\hspace{0.17em}\hspace{0.17em}\mathrm{given}\mathrm{by}\mathrm{f}\left(\mathrm{x}\right)=\left|\mathrm{x}\right|,\\ \mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{onen}\mathrm{or}\mathrm{onto},\mathrm{where}\left|\mathrm{x}\right|\mathrm{is}\hspace{0.17em}\hspace{0.17em}\mathrm{x},\mathrm{if}\mathrm{\hspace{0.17em}}\mathrm{x}\hspace{0.17em}\hspace{0.17em}\mathrm{is}\mathrm{positive}\\ \mathrm{or}0\mathrm{and}\left|\mathrm{x}\right|\mathrm{is}–\mathrm{x},\mathrm{if}\mathrm{x}\hspace{0.17em}\hspace{0.17em}\mathrm{is}\mathrm{negative}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{Modulus}\mathrm{function}\mathrm{is}\mathrm{defined}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}\mathrm{x},\mathrm{}\mathrm{if}\mathrm{x}\ge \mathrm{0}\\ -\mathrm{x},\mathrm{}\mathrm{if}\mathrm{x}<0\end{array}\\ \mathrm{Here},\mathrm{f}(-1)=|-1|=1,\mathrm{\hspace{0.17em}}\mathrm{f}\left(1\right)=\left|1\right|=1,\mathrm{}\\ \therefore \mathrm{f}(-1)=\mathrm{f}\left(1\right),\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{but}-\mathrm{1}\ne \mathrm{1}\\ \mathrm{So},\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Let}\mathrm{}-\mathrm{1}\in \mathrm{R}\mathrm{.}\\ \mathrm{Since},\mathrm{f}\left(\mathrm{x}\right)\mathrm{=}\left|\mathrm{x}\right|\mathrm{is}\mathrm{always}\mathrm{positive}.\mathrm{So},\mathrm{there}\mathrm{does}\mathrm{not}\mathrm{exist}\\ \mathrm{any}\mathrm{element}\mathrm{in}\mathrm{x}\mathrm{in}\mathrm{domain}\mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)=-1.\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{​}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Therefore},\mathrm{the}\mathrm{modulus}\mathrm{function}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}\mathrm{.}\end{array}$

Q.21

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{Signum}\mathrm{Function}\mathrm{f}:\mathrm{R}\to \mathrm{R},\mathrm{given}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1,\mathrm{\hspace{0.17em}}\mathrm{if}\mathrm{x}>0\\ \hspace{0.17em}\hspace{0.17em}0,\mathrm{\hspace{0.17em}}\mathrm{if}\mathrm{x}=0\\ –1,\mathrm{\hspace{0.17em}}\mathrm{if}\mathrm{x}<0\end{array}\\ \mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{noronto}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{Modulus}\mathrm{function}\mathrm{is}\mathrm{defined}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}1,\mathrm{}\mathrm{if}\mathrm{x}>0\\ \hspace{0.17em}\hspace{0.17em}0,\mathrm{}\mathrm{if}\mathrm{x}=0\\ -1,\mathrm{}\mathrm{if}\mathrm{x}<0\end{array}\\ \mathrm{Here},\mathrm{f}\left(1\right)=1,\mathrm{\hspace{0.17em}}\mathrm{f}\left(3\right)=1,\mathrm{but}1\ne \mathrm{3}\\ \mathrm{So},\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \\ \mathrm{Since},\mathrm{range}\mathrm{of}\mathrm{f}\mathrm{is}(1,0,-1)\mathrm{for}\mathrm{element}-2\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{R},\\ \mathrm{there}\mathrm{does}\mathrm{not}\mathrm{exist}\mathrm{any}\mathrm{value}\mathrm{in}\mathrm{domain}\mathrm{R}\mathrm{such}\mathrm{that}\\ \mathrm{f}\left(\mathrm{x}\right)=-2.\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{signum}\mathrm{function}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}\mathrm{.}\end{array}$

Q.22 Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Ans.

\begin{array}{l}\text{It is given that}\\ \text{A}=\left\{\text{1},\text{2},\text{3}\right\},\text{B}=\left\{\text{4},\text{5},\text{6},\text{7}\right\}\\ \text{f}:\text{A}\to \text{B is defined as}\text{f = {(1, 4), (2, 5), (3, 6)}}\text{.}\\ \therefore \text{f (1) = 4, f (2) = 5, f (3) = 6}\\ \text{Since every element has a unique value}\text{.}\\ \text{The given function f is one-one}\text{.}\end{array}

Q.23

\begin{array}{l}Ineachofthefollowingcases,statewhetherthefunctionis\\ one–one,ontoorbijective.Justifyyouranswer.\\ (i)f:R\to Rdefinedbyf(x)=3–4x\\ (ii)f:R\to Rdefinedbyf(x)=1+x^2\end{array}

Ans.

\begin{array}{l}\text{(i) f: R}\to \text{R is defined as f(x) = 3}-\text{4x}\text{.}\\ \text{For one-one}\\ \text{Let}\text{f(x)}=\text{f(y)}\\ \Rightarrow \text{3}-\text{4x}=\text{3}-\text{4y}\\ \\ \Rightarrow \text{x}=\text{y}\\ \therefore \text{f is one-one}\text{.}\\ \text{for onto}\\ \text{Let f(x)=y}\\ \Rightarrow \text{3}-\text{4x=y}\\ \Rightarrow \text{x=}\frac{\text{3}-y}{4}\\ \Rightarrow \text{f}\left(\frac{\text{3}-y}{4}\right)\text{=3}-\text{4}\left(\frac{\text{3}-y}{4}\right)\text{=y}\\ \text{Since Range = Co-domain}\\ \therefore \text{f is onto}\text{.}\\ \text{Hence, f is bijective}\text{.}\\ \\ \text{(ii) f: R}\to {\text{R is defined as f(x)=1+x}}^{\text{2}}\\ \text{For one-one}\\ \text{Let f(x)=f(y )}\\ \Rightarrow {\text{1+x}}^{\text{2}}={\text{1+y}}^{\text{2}}\\ \Rightarrow {\text{x}}^{\text{2}}={\text{y}}^{\text{2}}\\ \Rightarrow \text{x}=\pm \text{y}\\ \therefore \text{f is not one-one}\text{.}\\ \text{For onto:}\\ \text{Let}\text{f}\left(x\right)=y\\ \Rightarrow {\text{1+x}}^{\text{2}}=\text{y}\\ \Rightarrow \text{x}=\pm \sqrt{y-1}\\ \text{Since Range is not equal to Co-domain}\text{.}\\ \text{Then given function f(x) is not onto}\text{.}\end{array}

 

Q.24

$\begin{array}{l}\mathrm{Let}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{be}\mathrm{sets}.\mathrm{Show}\mathrm{that}\mathrm{f}:\mathrm{A}\times \mathrm{B}\to \mathrm{B}\times \mathrm{A}\hspace{0.17em}\mathrm{such}\mathrm{that}\\ \mathrm{f}(\mathrm{a},\mathrm{b})=\mathrm{f}(\mathrm{b},\mathrm{a})\hspace{0.17em}\mathrm{is}\mathrm{bijective}\mathrm{function}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{A}\times \mathrm{B}\to \mathrm{B}\times \mathrm{A}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}(\mathrm{a},\mathrm{b})\; =\; (\mathrm{b},\mathrm{a}).\hspace{0.17em}\\ \mathrm{Let}\mathrm{}({\mathrm{a}}_1,\hspace{0.17em}\hspace{0.17em}{\mathrm{b}}_1),({\mathrm{a}}_2,\hspace{0.17em}\hspace{0.17em}{\mathrm{b}}_2)\in \mathrm{A}\times \mathrm{B}\mathrm{such}\mathrm{that}\mathrm{f}({\mathrm{a}}_1,\hspace{0.17em}\hspace{0.17em}{\mathrm{b}}_1)=\mathrm{f}({\mathrm{a}}_2,\hspace{0.17em}\hspace{0.17em}{\mathrm{b}}_2)\\ \Rightarrow ({\mathrm{b}}_1,\hspace{0.17em}\hspace{0.17em}{\mathrm{a}}_1)=({\mathrm{b}}_2,\hspace{0.17em}\hspace{0.17em}{\mathrm{a}}_2)\\ \Rightarrow {\mathrm{b}}_1={\mathrm{b}}_2\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{\hspace{0.17em}}{\mathrm{a}}_1={\mathrm{a}}_2\\ \Rightarrow ({\mathrm{a}}_1,{\mathrm{b}}_1)=({\mathrm{a}}_2,{\mathrm{b}}_2)\\ \therefore \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Now},\mathrm{let}(\mathrm{b},\mathrm{a})\in \mathrm{B}\times \mathrm{A}\mathrm{be}\mathrm{any}\mathrm{element}\mathrm{.}\\ \mathrm{Then},\mathrm{there}\mathrm{exists}(\mathrm{a},\mathrm{b})\in \mathrm{A}\times \mathrm{B}\mathrm{such}\mathrm{that}\mathrm{f}(\mathrm{a},\mathrm{b})\; =\; (\mathrm{b},\mathrm{a}).\\ [\mathrm{By}\mathrm{definition}\mathrm{of}\mathrm{f}]\\ \therefore \mathrm{\hspace{0.17em}}\mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{bijective}\mathrm{.}\end{array}$

Q.25

\begin{array}{l}Letf:N\to Nbedefinedby\\ f\left(n\right)=\{\begin{array}{l}\frac{n+1}{2},ifnisodd\\ \frac{n}{2},ifniseven\end{array}foralln\in N.\\ Statewhetherthefunctionfisbijective.\\ Justifyyouranswer.\end{array}

Ans.

\begin{array}{l}Letf:N\to Nisdefinedas\\ \\ f\left(n\right)=\{\begin{array}{l}\frac{n+1}{2},ifnisodd\\ \frac{n}{2},ifniseven\end{array}foralln\in N.\\ \text{It can be observed that:}\\ f\left(1\right)=\frac{1+1}{2}=1,f\left(2\right)=\frac{2}{2}=1\left[By\text{definition of f}\right]\\ \therefore f\left(1\right)=f\left(2\right),where1\ne 2.\\ \therefore \text{f is not one-one}\text{.}\\ \text{Consider a natural number (n) in co-domain N}\text{.}\\ \text{Case I: n is odd}\\ \therefore \text{n = 2m + 1 for some m}\in \text{N}\text{. Then,there exists 4m + 1}\in \text{N}\\ \text{such that}\\ \text{f}\left(4m+1\right)=\frac{4m+1+1}{2}=2m+1\\ CaseII:\text{n is even}\\ \therefore \text{n=2}m\text{for some}m\in \text{N}\text{. Then,there exists 4}m\in \text{N such that}\\ \text{f}\left(4m\right)=\frac{4m}{2}=2m\\ \therefore \text{f is onto}\text{.}\\ \text{Hence, f is not a bijective function}\text{.}\end{array}

 

Q.26

\begin{array}{l}LetA=R-\left\{3\right\}andB=R-\left\{1\right\}.Considerthefunction\\ f:A\to Bdefinedbyf\left(x\right)=\frac{x–2}{x–3}.Isfone-oneandonto?\\ Justifyyouranswer.\end{array}

Ans.

\begin{array}{l}A=R-\left\{3\right\}andB=R-\left\{1\right\}\\ \text{f: A}\to \text{B is defined as f}\left(x\right)=\frac{x-2}{x-3}\\ Let\text{x,y}\in \text{A such that}\\ \text{f}\left(x\right)=f\left(y\right).\\ \Rightarrow \frac{x-2}{x-3}=\frac{y-2}{y-3}\\ \Rightarrow \left(x-2\right)\left(y-3\right)=\left(x-3\right)\left(y-2\right)\\ \Rightarrow \bcancel{xy}-3x-2y+\bcancel{6}=\bcancel{xy}-2x-3y+\bcancel{6}\\ \Rightarrow -3x-2y=-2x-3y\\ \Rightarrow x=y\\ \therefore f\text{if one-one}\text{.}\\ \text{y}\in \text{B}=\text{R-}\left\{1\right\}.Then,y\ne 1\\ \text{The function f is onto if there exists x}\in \text{A such that}\\ \text{f(x) = y}\\ \Rightarrow \frac{x-2}{x-3}=y\\ \Rightarrow x-2=y\left(x-3\right)\\ \Rightarrow x-2=yx-3y\\ \Rightarrow x-yx=2-3y\\ \Rightarrow x\left(1-y\right)=2-3y\\ \Rightarrow x=\frac{2-3y}{\left(1-y\right)}\in A\left[y\ne 1\right]\\ \text{Thus, for any y}\in \text{B, there exists}\frac{2-3y}{\left(1-y\right)}\in Asuchthat\\ \\ f\left(\frac{2-3y}{1-y}\right)=\frac{\frac{2-3y}{\left(1-y\right)}-2}{\frac{2-3y}{\left(1-y\right)}-3}\\ =\frac{\frac{2-3y-2\left(1-y\right)}{1-y}}{\frac{2-3y-3\left(1-y\right)}{1-y}}\\ =\frac{2-3y-2+2y}{2-3y-3+3y}\\ =\frac{-y}{-1}\\ f\left(\frac{2-3y}{1-y}\right)=y\\ \therefore f\text{is onto}\text{.}\\ \text{Hence, function f is one-one and onto}\text{.}\end{array}

Q.27

$\begin{array}{l}\mathrm{Let}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4.\mathrm{Choose}\mathrm{the}\mathrm{correct}\\ \mathrm{answer}.\\ \left(\mathrm{A}\right)\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{onto}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\mathrm{B}\right)\mathrm{fismany}–\mathrm{one}\mathrm{onto}.\\ (\mathrm{C})\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{but}\mathrm{not}\mathrm{onto}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{D}\right)\mathrm{f}\mathrm{is}\mathrm{n}\mathrm{either}\mathrm{one}–\mathrm{one}\\ \mathrm{noronto}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4.\\ \mathrm{Let}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right).\\ \mathrm{Then},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{x}}^4={\mathrm{y}}^4\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}=\pm \mathrm{y}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\mathrm{\hspace{0.17em}}\mathrm{but}\mathrm{\hspace{0.17em}}\mathrm{x}\ne \mathrm{y}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Consider}\mathrm{an}\mathrm{element}3\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{R}.\mathrm{It}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{there}\\ \mathrm{does}\mathrm{not}\mathrm{exist}\mathrm{any}\mathrm{x}\mathrm{in}\mathrm{domain}\mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}(\mathrm{x})\; =\; 3\mathrm{.}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$

Q.28

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{defined}\mathrm{asf}\left(\mathrm{x}\right)=3\mathrm{x}.\mathrm{Choose}\mathrm{the}\mathrm{correct}\\ \mathrm{answer}.\\ \left(\mathrm{A}\right)\mathrm{f}\hspace{0.17em}\hspace{0.17em}\mathrm{isone}–\mathrm{one}\mathrm{onto}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\mathrm{B}\right)\mathrm{f}\hspace{0.17em}\hspace{0.17em}\mathrm{is}\mathrm{many}–\mathrm{one}\mathrm{on}\mathrm{to}\\ \left(\mathrm{C}\right)\mathrm{f}\hspace{0.17em}\hspace{0.17em}\mathrm{isone}–\mathrm{one}\mathrm{but}\mathrm{not}\mathrm{onto}\left(\mathrm{D}\right)\mathrm{f}\mathrm{\hspace{0.17em}}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{Rbedefinedasf}\left(\mathrm{x}\right)=3\mathrm{x}.\\ \mathrm{Let}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}(\mathrm{x})\; =\mathrm{f}(\mathrm{y})\mathrm{.}\\ \Rightarrow 3\mathrm{x}=3\mathrm{y}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Also},\mathrm{for}\mathrm{any}\mathrm{real}\mathrm{number}\left(\mathrm{y}\right)\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{R},\mathrm{there}\mathrm{exists}\\ \frac{\mathrm{y}}{3}\hspace{0.17em}\hspace{0.17em}\mathrm{in}\mathrm{R}\mathrm{such}\mathrm{that}\hspace{0.17em}\hspace{0.17em}\mathrm{f}\left(\frac{\mathrm{y}}{3}\right)=3\left(\frac{\mathrm{y}}{3}\right)=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \\ \mathrm{Therefore},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$

Q.29

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\{1,3,4\}\to \{1,2,5\}\mathrm{and}\mathrm{g}:\{1,2,5\}\to \{1,3\}\mathrm{be}\mathrm{given}\\ \mathrm{by}\mathrm{f}=\left\{\right(1,2),(3,5),(4,1\left)\right\}\mathrm{and}\mathrm{g}=\left\{\right(1,3),(2,3),(5,1\left)\right\}.\\ \mathrm{Write}\mathrm{down}\mathrm{go}\mathrm{f}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\{1,3,4\}\to \{1,2,5\}\mathrm{and}\mathrm{g}:\{1,2,5\}\to \{1,3\}\mathrm{be}\mathrm{given}\\ \mathrm{by}\mathrm{f}=\left\{\right(1,2),(3,5),(4,1\left)\right\}\mathrm{and}\mathrm{g}=\left\{\right(1,3),(2,3),(5,1\left)\right\}.\\ \mathrm{Write}\mathrm{down}\mathrm{go}\mathrm{f}.\end{array}$

Q.30

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\{1,3,4\}\to \{1,2,5\}\mathrm{and}\mathrm{g}:\{1,2,5\}\to \{1,3\}\mathrm{be}\mathrm{given}\\ \mathrm{by}\mathrm{f}=\left\{\right(1,2),(3,5),(4,1\left)\right\}\mathrm{and}\mathrm{g}=\left\{\right(1,3),(2,3),(5,1\left)\right\}.\\ \mathrm{Write}\mathrm{down}\mathrm{go}\mathrm{f}.\end{array}$

Ans.

\begin{array}{l}\text{Let}\left(\left(\text{f}+\text{g}\right)\text{oh}\right)\left(\text{x}\right)=\left(\text{foh}\right)\left(\text{x}\right)+\left(\text{goh}\right)\left(\text{x}\right)\\ =\text{f}\left\{\text{h}\left(\text{x}\right)\right\}+\text{g}\left\{\text{h}\left(\text{x}\right)\right\}\\ =\left(\text{foh}\right)\left(\text{x}\right)+\left(\text{goh}\right)\left(\text{x}\right)\\ \\ =\left\{\left(\text{f}+\text{g}\right)\text{oh}\right\}\left(\text{x}\right)\\ =\left\{\left(\text{foh}\right)+\left(\text{goh}\right)\right\}\left(\text{x}\right)\forall x\in R\\ Hence,\left(f+g\right)oh=\left(\text{foh}\right)+\left(\text{goh}\right).\\ Now,\\ \left\{\left(f.g\right)oh\right\}\left(x\right)=\left(f.g\right)\left(h\left(x\right)\right)\\ =f\left(h\left(x\right)\right).g\left(h\left(x\right)\right)\\ =\left(foh\right)\left(x\right).\left(goh\right)\left(x\right)\\ =\left\{\left(foh\right).\left(goh\right)\right\}\left(x\right)\\ \therefore \left\{\left(f.g\right)oh\right\}\left(x\right)=\left\{\left(foh\right).\left(goh\right)\right\}\left(x\right)\forall x\in R\\ Hence,\left(f.g\right)oh=\left(foh\right).\left(goh\right)\end{array}

Q.31

\begin{array}{l}Findgofandfog,if\\ (i)f\left(x\right)=\left|x\right|andg\left(x\right)=\left|5x–2\right|\\ \left(ii\right)f\left(x\right)=8x^{3}andg\left(x\right)=x^{\frac{1}{3}}\end{array}

Ans.

\begin{array}{l}\left(i\right)f\left(x\right)=\left|x\right|andg\left(x\right)=\left|5x-2\right|\\ \therefore \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(\left|x\right|\right)\\ =\left|5\left|x\right|-2\right|\\ \left(fog\right)\left(x\right)=f\left(g\left(x\right)\right)\\ =f\left(\left|5x-2\right|\right)\\ \\ =\left|\left|5x-2\right|\right|\\ =\left|5x-2\right|\\ \left(ii\right)f\left(x\right)=8x^{3}andg\left(x\right)=x^{\frac{1}{3}}\\ \therefore \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(8x^3\right)\\ ={\left(8x^3\right)}^{\frac{1}{3}}\\ =2x\\ \left(fog\right)\left(x\right)=f\left(g\left(x\right)\right)\\ =f\left(x^{\frac{1}{3}}\right)\\ =8{\left(x^{\frac{1}{3}}\right)}^3\\ =8x\end{array}

Q.32

\begin{array}{l}Iff\left(x\right)=\frac{4x+3}{6x–4},x\ne \frac{2}{3},showthatfof\left(x\right)=x,forallx\ne \frac{2}{3}.\\ Whatistheinverseoff?\end{array}

Ans.

\begin{array}{l}It\text{is given that}f\left(x\right)=\frac{4x+3}{6x–4},then\\ fof\left(x\right)=f\left(f\left(x\right)\right)\\ =f\left(\frac{4x+3}{6x–4}\right)\\ =\frac{4\left(\frac{4x+3}{6x–4}\right)+3}{6\left(\frac{4x+3}{6x–4}\right)-4}\\ =\frac{\left(\frac{16x+12+18x-12}{6x-4}\right)}{\left(\frac{24x+18-24x+16}{6x–4}\right)}\\ =\frac{34x}{34}\\ fof\left(x\right)=x,\text{for all x}\ne \frac{2}{3}.\\ \Rightarrow fof=I\\ \text{Hence, the given function f is invertible and the inverse of f}\\ \text{is f itself}\text{.}\end{array}

Q.33

$\begin{array}{l}\mathrm{State}\mathrm{with}\mathrm{reason}\mathrm{whether}\mathrm{following}\mathrm{functions}\hspace{0.17em}\mathrm{have}\mathrm{inverse}\\ \left(\mathrm{i}\right)\mathrm{f}:\{1,2,3,4\}\to \left\{10\right\}\mathrm{with}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{f}=\left\{\right(1,10),(2,10),(3,10),(4,10\left)\right\}\\ \left(\mathrm{ii}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{g}:\{5,6,7,8\}\to \{1,2,3,4\}\mathrm{with}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{g}=\left\{\right(5,4),(6,3),(7,4),(8,2\left)\right\}\\ \left(\mathrm{iii}\right)\mathrm{h}:\{2,3,4,5\}\to \{7,9,11,13\}\mathrm{with}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{h}=\left\{\right(2,7),(3,9),(4,11),(5,13\left)\right\}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{f}:\; \{1,\; 2,\; 3,\; 4\}\to \left\{10\right\}\mathrm{defined}\mathrm{as}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{f}=\; \{(1,\; 10),\; (2,\; 10),\; (3,\; 10),\; (4,\; 10)\}\\ \mathrm{From}\mathrm{the}\mathrm{given}\mathrm{definition}\mathrm{of}\mathrm{f},\mathrm{we}\mathrm{can}\mathrm{see}\mathrm{that}\mathrm{f}\mathrm{is}\mathrm{a}\mathrm{many}\\ \mathrm{one}\mathrm{function}\mathrm{as}:\mathrm{f}(1)\; =\mathrm{f}(2)=\mathrm{f}(3)\; =\mathrm{f}(4)\; =\; 10\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{does}\mathrm{not}\mathrm{have}\mathrm{an}\mathrm{inverse}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{\hspace{0.17em}}\mathrm{g}:\; \{5,\; 6,\; 7,\; 8\}\to \{1,\; 2,\; 3,\; 4\}\mathrm{defined}\mathrm{as}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{g}=\; \{(5,\; 4),\; (6,\; 3),\; (7,\; 4),\; (8,\; 2)\}\\ \mathrm{From}\mathrm{the}\mathrm{given}\mathrm{definition}\mathrm{of}\mathrm{g},\mathrm{it}\mathrm{is}\mathrm{seen}\mathrm{that}\mathrm{g}\mathrm{is}\mathrm{a}\mathrm{many}\mathrm{one}\\ \mathrm{function}\mathrm{as}:\mathrm{g}(5)\; =\mathrm{g}(7)\; =\; 4\mathrm{.}\\ \therefore \mathrm{g}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Hence},\mathrm{function}\mathrm{g}\mathrm{does}\mathrm{not}\mathrm{have}\mathrm{an}\mathrm{inverse}\mathrm{.}\\ \left(\mathrm{iii}\right)\mathrm{h}:\; \{2,\; 3,\; 4,\; 5\}\to \{7,\; 9,\; 11,\; 13\}\mathrm{defined}\mathrm{as}:\\ \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{h}=\; \{(2,\; 7),\; (3,\; 9),\; (4,\; 11),\; (5,\; 13)\\ \mathrm{It}\mathrm{is}\mathrm{seen}\mathrm{that}\mathrm{all}\mathrm{distinct}\mathrm{elements}\mathrm{of}\mathrm{the}\mathrm{set}\{2,\; 3,\; 4,\; 5\}\\ \mathrm{have}\mathrm{distinct}\mathrm{images}\mathrm{under}\mathrm{h}\mathrm{.}\\ \therefore \mathrm{Function}\mathrm{h}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Also},\mathrm{h}\mathrm{is}\mathrm{onto}\mathrm{since}\mathrm{for}\mathrm{every}\mathrm{element}\mathrm{y}\mathrm{of}\mathrm{the}\mathrm{set}\\ \{7,\; 9,\; 11,\; 13\},\mathrm{there}\mathrm{exists}\mathrm{an}\hspace{0.17em}\hspace{0.17em}\mathrm{element}\mathrm{x}\mathrm{in}\mathrm{the}\mathrm{set}\\ \{2,\; 3,\; 4,\; 5\}\hspace{0.17em}\hspace{0.17em}\mathrm{such}\mathrm{that}\mathrm{h}(\mathrm{x})\; =\mathrm{y}\mathrm{.}\\ \mathrm{Thus},\mathrm{h}\mathrm{is}\mathrm{a}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{function}.\mathrm{Hence},\mathrm{h}\mathrm{has}\mathrm{an}\mathrm{inverse}\mathrm{.}\end{array}$

Q.34

\begin{array}{l}Showthatf:[-1,1]\to R,givenbyf\left(x\right)=\frac{x}{x+2}isone–one.\\ Findtheinverseofthefunctionf:[-1,1]\to Rangef.\end{array}

Ans.

\begin{array}{l}\text{f: [}-\text{1, 1]}\to \text{R is given as f}\left(x\right)=\frac{x}{x+2}\\ Let\text{f}\left(x\right)=f\left(y\right)\\ \Rightarrow \frac{x}{x+2}=\frac{y}{y+2}\\ \Rightarrow x\left(y+2\right)=y\left(x+2\right)\\ \Rightarrow xy+2x=xy+2y\\ \Rightarrow x=y\\ \therefore f\text{is one-one function}\text{.}\\ \text{It is clear that f: [}-\text{1, 1]}\to \text{Range f is onto}\text{.}\\ \therefore \text{f: [}-\text{1, 1]}\to \text{Range f is one-one and onto and therefore,}\\ \text{the inverse of the function:}\\ \text{f: [}-\text{1, 1]}\to \text{Range f exists}\text{.}\\ \text{Let g: Range f}\to \text{[}-\text{1, 1] be the inverse of f}\text{.}\\ \text{Let y be an arbitrary element of range f}\text{.}\\ \text{Since f: [}-\text{1, 1]}\to \text{Range f is onto, we have:}\\ \text{f}\left(x\right)=yfor\text{some x}\in \left[-1,1\right]\\ \Rightarrow y=\frac{x}{x+2}\\ \Rightarrow xy+2y=x\\ \Rightarrow 2y=x-xy\\ =x\left(1-y\right)\\ \Rightarrow x=\frac{2y}{1-y},y\ne 1\\ \text{Now, let us define g: Range f}\to \text{[}-\text{1, 1] as}\\ g\left(y\right)=\frac{2y}{1-y},y\ne 1.\\ Now,\\ \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(\frac{x}{x+2}\right)\\ =\frac{2\left(\frac{x}{x+2}\right)}{1-\left(\frac{x}{x+2}\right)}\\ =\frac{2x}{x+2-x}\\ =\frac{2x}{2}\\ \left(gof\right)\left(x\right)=x\\ \left(fog\right)\left(y\right)=f\left(g\left(y\right)\right)\\ =f\left(\frac{2y}{1-y}\right)\\ =\frac{\left(\frac{2y}{1-y}\right)}{\frac{2y}{1-y}+2}\\ =\frac{2y}{2y+2-2y}\\ =\frac{2y}{2}\\ \left(fog\right)\left(y\right)=y\\ \therefore go{f}^{-1}=I_{\left[-1,1\right]}andfo{g}^{-1}=I_{Rangef}\\ \therefore f^{-1}=g\\ \Rightarrow f^{-1}\left(y\right)=\frac{2y}{y-1},y\ne 1.\\ \Rightarrow f^{-1}\left(x\right)=\frac{2x}{x-1},x\ne 1.\end{array}

Q.35

$\begin{array}{l}\mathrm{Consider}\mathrm{f}:\mathrm{R}\to \mathrm{Rgiven}\mathrm{by}\mathrm{f}\left(\mathrm{x}\right)=4\mathrm{x}+3.\mathrm{Show}\mathrm{that}\mathrm{f}\mathrm{is}\\ \mathrm{invertible}.\mathrm{Find}\mathrm{the}\mathrm{inverse}\mathrm{off}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{f}(\mathrm{x})\; =\; 4\mathrm{x}+\; 3\\ \mathrm{One}–\mathrm{one}:\\ \mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\mathrm{.}\\ \Rightarrow 4\mathrm{x}+\; 3=4\mathrm{y}+\; 3\\ \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}\mathrm{\hspace{0.17em}}4\mathrm{x}=4\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}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{a}\mathrm{one}–\mathrm{one}\mathrm{function}\mathrm{.}\\ \mathrm{Onto}:\\ \mathrm{For}\mathrm{y}\in \mathrm{R},\\ \mathrm{let}\mathrm{y}=\; 4\mathrm{x}+\; 3\mathrm{.}\\ \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{\hspace{0.17em}}\mathrm{x}=\frac{\mathrm{y}-3}{4}\in \mathrm{R}\\ \\ \mathrm{Therefore},\mathrm{for}\mathrm{any}\mathrm{y}\in \mathrm{R},\mathrm{there}\mathrm{exists},{\mathrm{f}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Let}\mathrm{us}\mathrm{define}\mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{by}\mathrm{g}\left(\mathrm{y}\right)=\frac{\mathrm{y}-3}{4}\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\left(\mathrm{gof}\right)\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \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{\hspace{0.17em}}=\mathrm{g}\left(4\mathrm{x}+3\right)\\ \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{\hspace{0.17em}}=\frac{4\mathrm{x}+3-3}{4}\\ \left(\mathrm{gof}\right)\left(\mathrm{x}\right)=\mathrm{x}\\ \mathrm{\hspace{0.17em}}\left(\mathrm{fog}\right)\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{y}\right)\right)\\ \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{\hspace{0.17em}}=\mathrm{f}\left(\frac{\mathrm{y}-3}{4}\right)\\ \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{\hspace{0.17em}}=4\left(\frac{\mathrm{y}-3}{4}\right)+3\\ \mathrm{\hspace{0.17em}}\left(\mathrm{fog}\right)\left(\mathrm{y}\right)=\mathrm{y}\\ \therefore \mathrm{gof}=\mathrm{fog}={\mathrm{I}}_{\mathrm{R}}\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{invertible}\mathrm{and}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{given}\mathrm{by}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{f}}^{-1}\left(\mathrm{y}\right)=\mathrm{g}\left(\mathrm{y}\right)\\ \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{\hspace{0.17em}}=\frac{\mathrm{y}-3}{4}.\\ \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}{\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{\mathrm{x}-3}{4}\end{array}$

Q.36

\begin{array}{l}Considerf:R_{+}\to [4,\infty )givenbyf(x)=x^2+4.Showthatfis\\ invertiblewiththeinverse{f}^{–1}ofgivenfby,where{R}^{+}istheset\\ ofallnon–negativerealnumbers.\end{array}

Ans.

\begin{array}{l}{\text{f: R}}_{\text{+}}\to \text{[4,}\infty {\text{) is given as f(x) = x}}^{\text{2}}\text{+ 4}\text{.}\\ \text{One-one:}\\ \text{Let f(x) = f(y)}\\ \Rightarrow {\text{x}}^{\text{2}}\text{+ 4}={\text{y}}^{\text{2}}\text{+ 4}\\ \Rightarrow {\text{x}}^{\text{2}}=y^2\\ \Rightarrow x=y\left[asx=y\in R_{+}\right]\\ \therefore fisone-one\text{function}.\\ Onto:\\ \text{For y}\in \text{[4,}\infty \text{),}\\ \text{let y}={\text{x}}^{\text{2}}\text{+ 4}\\ \text{x}=\sqrt{y-4}\ge 0\\ \text{Therefore, for any y}\in \text{R, there exists}\text{x}=\sqrt{y-4}\ge 0,\text{such that}\\ \text{f}\left(x\right)=f\left(\sqrt{y-4}\right)\\ ={\left(\sqrt{y-4}\right)}^2+4\\ =y-4+4\\ \text{f}\left(x\right)=y\\ \therefore f\text{is onto}\text{.}\\ {\text{Thus, f is one-one and onto and therefore, f}}^{\text{-1}}\text{exists}\text{.}\\ \text{Let us define g: [4,}\infty \text{)}\to {\text{R}}_{{}^{\text{+}}}\text{by,}\\ \text{g}\left(y\right)=\sqrt{y-4}\\ Now,\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(x^2+4\right)\\ =\sqrt{\left(x^2+4\right)-4}\\ \left(gof\right)\left(x\right)=x\\ \left(fog\right)\left(y\right)=f\left(g\left(y\right)\right)\\ =f\left(\sqrt{y-4}\right)\\ ={\left(\sqrt{y-4}\right)}^2+4\\ =y-4+4\\ \left(fog\right)\left(y\right)=y\\ \therefore gof=fog=I_{R^{+}}.\\ \text{Hence, f is invertible and the inverse of f is given by}\\ f^{-1}\left(y\right)=g\left(y\right)=\sqrt{y-4}\\ \Rightarrow f^{-1}\left(x\right)=\sqrt{x-4}\end{array}

Q.37

\begin{array}{l}Considerf:R_{+}\to [–5,\infty )givenbyf(x)=9x^2+6x–5.\\ Showthatfisinvertiblewith{f}^{-1}\left(y\right)=\left(\frac{\left(\sqrt{y+6}\right)-1}{3}\right).\end{array}

Ans.

\begin{array}{l}{\text{f: R}}_{\text{+}}\to \text{[}-\text{5,}\infty {\text{) is given as f(x) = 9x}}^{\text{2}}\text{+ 6x}-\text{5}\text{.}\\ \text{Let y be an arbitrary element of [}-\text{5,}\infty \text{)}\text{.}\\ \text{Let}\text{y}={\text{9x}}^{\text{2}}\text{+ 6x}-\text{5}\\ \Rightarrow {\text{9x}}^{\text{2}}\text{+ 6x}-\left(\text{5+y}\right)=\text{0}\\ \Rightarrow x=\frac{-6\pm \sqrt{6^2-4\times 9\times -\left(5+y\right)}}{2\times 9}\\ =\frac{-6\pm \sqrt{36+36\left(5+y\right)}}{18}\\ =\frac{-6\pm 6\sqrt{1+5+y}}{18}\\ x=\frac{-1\pm \sqrt{6+y}}{3}\\ =\frac{\sqrt{6+y}-1}{3}\left[âˆ\mu y\ge -6\Rightarrow y+6\ge 0\right]\\ \therefore \text{f is onto, thereby range f = [}-\text{5,}\infty \text{)}\text{.}\\ \text{Let us define g: [}-\text{5,}\infty \text{)}\to {\text{R}}_{\text{+}}\text{as}\\ g\left(y\right)=\frac{\sqrt{\left(y+6\right)}-1}{3}\\ Now,\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left({\text{9x}}^{\text{2}}\text{+ 6x}-\text{5}\right)\\ =\frac{\sqrt{\left({\text{9x}}^{\text{2}}\text{+ 6x}-\text{5}\right)+6}-1}{3}\\ =\frac{\sqrt{\left({\text{9x}}^{\text{2}}\text{+ 6x}+1\right)}-1}{3}\\ =\frac{\sqrt{{\left(3x+1\right)}^2}-1}{3}\\ =\frac{3x+1-1}{3}=x\\ And,\\ \left(fog\right)\left(y\right)=f\left(g\left(y\right)\right)\\ =f\left(\frac{\sqrt{\left(y+6\right)}-1}{3}\right)\\ =\text{9}{\left(\frac{\sqrt{\left(y+6\right)}-1}{3}\right)}^{\text{2}}\text{+ 6}\left(\frac{\sqrt{\left(y+6\right)}-1}{3}\right)-\text{5}\\ =9\left(\frac{y+6-2\sqrt{y+6}+1}{9}\right)+2\sqrt{\left(y+6\right)}-2-5\\ =y+6-2\sqrt{y+6}+1+2\sqrt{\left(y+6\right)}-2-5\\ \left(fog\right)\left(y\right)=y\\ \therefore gof=I_{R}andfog=I_{\left[-5,\infty \right]}\\ \text{Hence, f is invertible and the inverse of f is given by}\\ f^{-1}\left(y\right)=g\left(y\right)=\frac{\sqrt{y+6}-1}{3}.\end{array}

Q.38

$\text{Let f :X→Y be an invertible function.Show that f has unique inverse.}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{X}\to \mathrm{Y}\mathrm{be}\mathrm{an}\mathrm{invertible}\mathrm{function}\mathrm{.}\\ \mathrm{Also},\mathrm{suppose}\mathrm{f}\mathrm{has}\mathrm{two}\mathrm{inverses}(\mathrm{say}{\mathrm{g}}_{\mathrm{1}}\mathrm{and}{\mathrm{g}}_{\mathrm{2}}\mathrm{)}\\ \mathrm{Then},\mathrm{for}\mathrm{all}\mathrm{y}\in \mathrm{Y},\mathrm{we}\mathrm{have}:\\ {\mathrm{fog}}_{\mathrm{1}}\left(\mathrm{y}\right)={\mathrm{I}}_{\mathrm{Y}}\left(\mathrm{y}\right)={\mathrm{fog}}_2\left(\mathrm{y}\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{f}\left({\mathrm{g}}_1\left(\mathrm{y}\right)\right)=\mathrm{f}\left({\mathrm{g}}_2\left(\mathrm{y}\right)\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{g}}_1\left(\mathrm{y}\right)={\mathrm{g}}_2\left(\mathrm{y}\right)\left[\mathrm{f}\mathrm{is}\mathrm{invertible}\Rightarrow \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\right]\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{g}}_1={\mathrm{g}}_2\left[\mathrm{g}\mathrm{is}\mathrm{one}–\mathrm{one}\right]\\ \mathrm{Hence},\mathrm{f}\mathrm{has}\mathrm{a}\mathrm{unique}\mathrm{inverse}\mathrm{.}\end{array}$

Q.39

$\begin{array}{l}\mathrm{Consider}\mathrm{f}:\{1,2,3\}\to \{\mathrm{a},\mathrm{b},\mathrm{c}\}\mathrm{given}\mathrm{by}\mathrm{f}\left(1\right)=\mathrm{a},\mathrm{f}\left(2\right)=\mathrm{b}\\ \mathrm{and}\mathrm{f}\left(3\right)=\mathrm{c}.\mathrm{Find}{\mathrm{f}}^{–1}\mathrm{and}\mathrm{show}\mathrm{that}{\left({\mathrm{f}}^{–1}\right)}^{–1}=\mathrm{f}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Function}\mathrm{f}:\; \{1,\; 2,\; 3\}\to \{\mathrm{a},\mathrm{b},\mathrm{c}\}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{f}(1)\; =\mathrm{a},\mathrm{f}(2)\; =\mathrm{b},\mathrm{and}\mathrm{f}(3)\; =\mathrm{c}\\ \mathrm{If}\mathrm{we}\mathrm{define}\mathrm{g}:\; \{\mathrm{a},\mathrm{b},\mathrm{c}\}\to \{1,\; 2,\; 3\}\mathrm{as}\mathrm{g}(\mathrm{a})\; =\; 1,\mathrm{g}(\mathrm{b})\; =\; 2,\\ \mathrm{g}(\mathrm{c})\; =\; 3,\mathrm{then}\mathrm{we}\mathrm{have}:\\ \left(\mathrm{fog}\right)\left(\mathrm{a}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{a}\right)\right)=\mathrm{f}\left(1\right)=\mathrm{a}\\ \left(\mathrm{fog}\right)\left(\mathrm{b}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{b}\right)\right)=\mathrm{f}\left(2\right)=\mathrm{b}\\ \left(\mathrm{fog}\right)\left(\mathrm{c}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{c}\right)\right)=\mathrm{f}\left(3\right)=\mathrm{c}\\ \mathrm{and}\\ \left(\mathrm{gof}\right)\left(1\right)=\mathrm{g}\left(\mathrm{f}\left(1\right)\right)=\mathrm{g}\left(\mathrm{a}\right)=1\\ \left(\mathrm{gof}\right)\left(2\right)=\mathrm{g}\left(\mathrm{f}\left(2\right)\right)=\mathrm{g}\left(\mathrm{b}\right)=2\\ \left(\mathrm{gof}\right)\left(3\right)=\mathrm{g}\left(\mathrm{f}\left(3\right)\right)=\mathrm{g}\left(\mathrm{c}\right)=3\\ \therefore \mathrm{gof}={\mathrm{I}}_{\mathrm{X}}\mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{Y}},\mathrm{where}\mathrm{X}=\{1,2,3\}\mathrm{and}\mathrm{Y}=\{\mathrm{a},\mathrm{b},\mathrm{c}\}.\\ \mathrm{Thus},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{exists}\mathrm{and}{\mathrm{f}}^{–1}=\mathrm{g}\mathrm{.}\\ \therefore {\mathrm{f}}^{-1}:\{\mathrm{a},\mathrm{b},\mathrm{c}\}\to \{1,2,3\}\mathrm{is}\mathrm{given}\mathrm{by},\\ {\mathrm{f}}^{-\mathrm{1}}\left(\mathrm{a}\right)=1,{\mathrm{f}}^{-\mathrm{1}}\left(\mathrm{b}\right)=2,{\mathrm{f}}^{-\mathrm{1}}\left(\mathrm{c}\right)=3\\ \mathrm{The}\mathrm{inverse}\mathrm{of}{\mathrm{f}}^{–1}\mathrm{}\Rightarrow \mathrm{The}\mathrm{inverse}\mathrm{of}\mathrm{g}\\ \mathrm{We}\mathrm{define},\mathrm{h}:\{1,2,3\}\to \{\mathrm{a},\mathrm{b},\mathrm{c}\}\hspace{0.17em}\hspace{0.17em}\mathrm{as}\\ \mathrm{h}(1)\; =\mathrm{a},\mathrm{h}(2)\; =\mathrm{b},\mathrm{h}(3)\; =\mathrm{c},\mathrm{then}\mathrm{we}\mathrm{have}:\\ \left(\mathrm{goh}\right)\left(1\right)=\mathrm{g}\left(\mathrm{h}\left(1\right)\right)=\mathrm{g}\left(\mathrm{a}\right)=1\\ \left(\mathrm{goh}\right)\left(2\right)=\mathrm{g}\left(\mathrm{h}\left(2\right)\right)=\mathrm{g}\left(\mathrm{b}\right)=2\\ \left(\mathrm{goh}\right)\left(3\right)=\mathrm{g}\left(\mathrm{h}\left(3\right)\right)=\mathrm{g}\left(\mathrm{c}\right)=3\\ \mathrm{and}\\ \left(\mathrm{hog}\right)\left(\mathrm{a}\right)=\mathrm{h}\left(\mathrm{g}\left(\mathrm{a}\right)\right)=\mathrm{h}\left(1\right)=\mathrm{a}\\ \left(\mathrm{hog}\right)\left(\mathrm{b}\right)=\mathrm{h}\left(\mathrm{g}\left(\mathrm{b}\right)\right)=\mathrm{h}\left(2\right)=\mathrm{b}\\ \left(\mathrm{hog}\right)\left(\mathrm{c}\right)=\mathrm{h}\left(\mathrm{g}\left(\mathrm{c}\right)\right)=\mathrm{h}\left(3\right)=\mathrm{c}\\ \therefore \mathrm{goh}={\mathrm{I}}_{\mathrm{X}}\mathrm{\hspace{0.17em}}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{hog}={\mathrm{I}}_{\mathrm{Y}},\mathrm{\hspace{0.17em}}\mathrm{where}\mathrm{\hspace{0.17em}}\mathrm{X}=\{1,2,3\}\mathrm{and}=\{\mathrm{a},\mathrm{b},\mathrm{c}\}.\\ \mathrm{Thus},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{g}\mathrm{exists}\mathrm{and}\mathrm{\hspace{0.17em}}{\mathrm{g}}^{-1}=\mathrm{h}\Rightarrow {\left({\mathrm{f}}^{-\mathrm{1}}\right)}^{-\mathrm{1}}=\mathrm{h}\mathrm{.}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{noted}\mathrm{that}\mathrm{h}=\mathrm{f}\mathrm{.}\\ \mathrm{Hence},{\left({\mathrm{f}}^{–1}\right)}^{–1}=\mathrm{f}\mathrm{.}\end{array}$

Q.40

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{X}\to \mathrm{Y}\mathrm{be}\mathrm{aninvertible}\mathrm{function}.\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}{\mathrm{f}}^{–1}\mathrm{is}\mathrm{f},\mathrm{i}.\mathrm{e}.,{\left({\mathrm{f}}^{–1}\right)}^{–1}=\mathrm{f}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{X}\to \mathrm{Y}\mathrm{be}\mathrm{an}\mathrm{invertible}\mathrm{function}\mathrm{.}\\ \mathrm{Then},\mathrm{there}\mathrm{exists}\mathrm{a}\mathrm{function}\mathrm{g}:\mathrm{Y}\to \mathrm{X}\mathrm{such}\mathrm{that}\mathrm{gof}={\mathrm{I}}_{\mathrm{X}}\\ \mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{Y}}\mathrm{.}\\ \mathrm{Here},{\mathrm{f}}^{-1}=\mathrm{g}\mathrm{.}\\ \mathrm{Now},\mathrm{gof}={\mathrm{I}}_{\mathrm{X}}\mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{Y}}\\ \Rightarrow {\mathrm{f}}^{–1}\mathrm{of}={\mathrm{I}}_{\mathrm{X}}\mathrm{\hspace{0.17em}}\mathrm{and}{\mathrm{fof}}^{–1}={\mathrm{I}}_{\mathrm{Y}}\\ \mathrm{Hence},{\mathrm{f}}^{-\mathrm{1}}:\mathrm{Y}\to \mathrm{X}\mathrm{is}\mathrm{invertible}\mathrm{and}\mathrm{f}\mathrm{is}\mathrm{the}\mathrm{inverse}\mathrm{of}{\mathrm{f}}^{-\mathrm{1}}\\ \mathrm{i}.\mathrm{e}.,\; ({\mathrm{f}}^{-\mathrm{1}}{\mathrm{)}}^{-1}=\mathrm{f}\mathrm{.}\end{array}$

Q.41

$\begin{array}{l}\mathrm{If}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{given}\mathrm{by},\mathrm{f}\left(\mathrm{x}\right)={(3-{\mathrm{x}}^3)}^{\frac{1}{3}}\mathrm{\hspace{0.17em}}\mathrm{then}\mathrm{f}\mathrm{of}\left(\mathrm{x}\right)\mathrm{is}\\ \left(\mathrm{A}\right){\mathrm{x}}^{\frac{1}{3}}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{B}\right){\mathrm{x}}^3\left(\mathrm{C}\right)\mathrm{x}\left(\mathrm{D}\right)(3-{\mathrm{x}}^3).\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{as}\hspace{0.17em}\hspace{0.17em}\mathrm{f}\left(\mathrm{x}\right)={(3-{\mathrm{x}}^3)}^{\frac{1}{3}}\\ \therefore \mathrm{fof}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{f}\left\{{(3-{\mathrm{x}}^3)}^{\frac{1}{3}}\right\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={[3-{\left\{{(3-{\mathrm{x}}^3)}^{\frac{1}{3}}\right\}}^3]}^{\frac{1}{3}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\{3-(3-{\mathrm{x}}^3)\}}^{\frac{1}{3}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={(3-3+{\mathrm{x}}^3)}^{\frac{1}{3}}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{fof}\left(\mathrm{x}\right)=\mathrm{x}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\left(\mathrm{C}\right)\mathrm{.}\\ \end{array}$

Q.42

$\begin{array}{l}\mathrm{Let}\hspace{0.17em}\hspace{0.17em}\mathrm{f}:\mathrm{R}-\{-\frac{4}{3}\}\to \mathrm{R}\mathrm{be}\mathrm{a}\mathrm{function}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)=\frac{4\mathrm{x}}{3\mathrm{x}+4}.\\ \mathrm{The}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{map}\mathrm{g}:\mathrm{R}\mathrm{ange}\mathrm{f}\to \mathrm{R}-\{-\frac{4}{3}\}\mathrm{given}\mathrm{by}\\ \left(\mathrm{A}\right)\mathrm{\hspace{0.17em}}\mathrm{g}\left(\mathrm{y}\right)=\frac{3\mathrm{y}}{3-4\mathrm{y}}\left(\mathrm{B}\right)\mathrm{\hspace{0.17em}}\mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{4-3\mathrm{y}}\\ \left(\mathrm{C}\right)\mathrm{\hspace{0.17em}}\mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{3-4\mathrm{y}}\left(\mathrm{D}\right)\mathrm{g}\left(\mathrm{y}\right)=\frac{3\mathrm{y}}{4-3\mathrm{y}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{\hspace{0.17em}}\mathrm{f}:\mathrm{R}-\{-\frac{4}{3}\}\to \mathrm{R}\mathrm{\hspace{0.17em}}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)=\frac{4\mathrm{x}}{3\mathrm{x}+4}.\\ \mathrm{Let}\mathrm{\hspace{0.17em}}\mathrm{y}\mathrm{be}\mathrm{an}\mathrm{arbitrary}\mathrm{element}\mathrm{of}\mathrm{Range}\mathrm{f}\mathrm{.}\\ \mathrm{Then},\mathrm{there}\mathrm{exists}\mathrm{x}\in \mathrm{R}-\{-\frac{4}{3}\}\mathrm{\hspace{0.17em}}\mathrm{such}\mathrm{\hspace{0.17em}}\mathrm{that}\mathrm{\hspace{0.17em}}\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right).\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=\frac{4\mathrm{x}}{3\mathrm{x}+4}\\ \Rightarrow 3\mathrm{xy}+4\mathrm{y}=4\mathrm{x}\\ \Rightarrow -3\mathrm{xy}+4\mathrm{x}=4\mathrm{y}\\ \Rightarrow \mathrm{\hspace{0.17em}}\mathrm{x}(4-3\mathrm{y})=4\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}\mathrm{\hspace{0.17em}}\mathrm{x}=\frac{4\mathrm{y}}{4-3\mathrm{y}}\\ \mathrm{Let}\mathrm{us}\mathrm{define}\mathrm{g}:\mathrm{Range}\hspace{0.17em}\hspace{0.17em}\mathrm{f}\to \mathrm{R}-\{-\frac{4}{3}\}\hspace{0.17em}\hspace{0.17em}\mathrm{as}\mathrm{\hspace{0.17em}}\mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{4-3\mathrm{y}}\mathrm{\hspace{0.17em}}\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\left(\mathrm{gof}\right)\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \\ \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{\hspace{0.17em}}=\mathrm{g}\left(\frac{4\mathrm{x}}{3\mathrm{x}+4}\right)\\ \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{\hspace{0.17em}}=\frac{4\left(\frac{4\mathrm{x}}{3\mathrm{x}+4}\right)}{4-3\left(\frac{4\mathrm{x}}{3\mathrm{x}+4}\right)}\\ \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{\hspace{0.17em}}=\frac{16\mathrm{x}}{12\mathrm{x}+16-12\mathrm{x}}\\ \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{\hspace{0.17em}}=\mathrm{x}\\ \mathrm{And},\mathrm{\hspace{0.17em}}\left(\mathrm{fog}\right)\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{y}\right)\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{f}\left(\frac{4\mathrm{y}}{4-3\mathrm{y}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{4\left(\frac{4\mathrm{y}}{4-3\mathrm{y}}\right)}{3\left(\frac{4\mathrm{y}}{4-3\mathrm{y}}\right)+4}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{16\mathrm{y}}{12\mathrm{y}+16-12\mathrm{y}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{16\mathrm{y}}{16}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{y}\\ \therefore \mathrm{gof}={\mathrm{I}}_{\mathrm{R}-\{-\frac{4}{3}\}}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{fog}={\mathrm{I}}_{\mathrm{Range}\mathrm{\hspace{0.17em}}\mathrm{f}}\\ \mathrm{Thus},\mathrm{g}\mathrm{is}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{i}.\mathrm{e}.,{\mathrm{f}}^{–1}=\mathrm{g}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{the}\mathrm{map}\mathrm{g}:\mathrm{Range},\hspace{0.17em}\hspace{0.17em}\mathrm{R}\to \mathrm{R}-\{-\frac{4}{3}\},\\ \mathrm{which}\mathrm{is}\mathrm{given}\mathrm{by}\hspace{0.17em}\hspace{0.17em}\mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{4-3\mathrm{y}}.\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{B}\mathrm{.}\end{array}$

Q.43

Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this.

(i) On Z⁺, define * by a * b = a − b

(ii) On Z⁺, define * by a * b = ab

(iii) On R, define * by a * b = ab²

(iv) On Z⁺, define * by a * b = |a − b|

(v) On Z⁺, define * by a * b = a

Ans.

\begin{array}{l}\left(i\right){\text{On Z}}^{\text{+}}\text{, * is defined by a * b = a}-\text{b}\text{.}\\ \text{Here,}\left(\text{1,2}\right)\in Z^{+}\\ so,\text{1*2}=1-2=-1\notin Z^{+}\\ Thus,*\text{is not a binary operation}\text{.}\\ \left(ii\right){\text{On Z}}^{\text{+}}\text{, * is defined by a*b = ab}\text{.}\\ \text{Here, a,b}\in {\text{Z}}^{\text{+}}\text{and ab}\in {\text{Z}}^{\text{+}}\\ \\ \text{This means that * carries each pair (a, b) to a unique element}\\ {\text{a * b = ab in Z}}^{\text{+}}\text{.}\\ \text{Therefore, *}\text{is a binary operation}\text{.}\\ \left(iii\right){\text{On R, * is defined by a * b = ab}}^{\text{2}}\text{.}\\ \text{It is seen that for each a, b}\in \text{R, there is a unique element}\\ {\text{ab}}^{\text{2}}\text{in R}\text{.}\\ \text{This means that * carries each pair (a, b) to a unique}\\ {\text{element a * b = ab}}^{\text{2}}\text{in R}\text{.}\\ \text{Therefore, * is a binary operation}\text{.}\\ \left(iv\right){\text{On Z}}^{\text{+}}\text{, * is defined by a*b = |a}-\text{b|}\text{.}\\ \text{It is seen that for each a, b}\in {\text{Z}}^{\text{+}}\text{, there is a unique element}\\ \text{|a}-{\text{b| in Z}}^{\text{+}}\text{.}\\ \text{This means that * carries each pair (a, b) to a unique element}\\ \text{a*b =}\text{|a}-{\text{b| in Z}}^{\text{+}}\text{.}\\ \text{Therefore, * is a binary operation}\text{.}\\ \left(v\right){\text{On Z}}^{\text{+}}\text{, * is defined by a*b = a}\text{.}\\ {\text{* carries each pair (a, b) to a unique element a * b = a in Z}}^{\text{+}}\text{.}\\ \text{Therefore, * is a binary operation}\text{.}\end{array}

Q.44

$\begin{array}{l}\mathrm{For}\mathrm{each}\mathrm{binary}\mathrm{operation}*\mathrm{defined}\mathrm{below},\mathrm{determine}\\ \mathrm{whether}*\mathrm{is}\mathrm{commutative}\mathrm{or}\mathrm{associative}.\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\mathrm{i}\right)\mathrm{On}\mathrm{Z},\mathrm{definea}*\mathrm{b}=\mathrm{a}-\mathrm{b}\\ \mathrm{\hspace{0.17em}}\left(\mathrm{ii}\right)\mathrm{}\mathrm{On}\mathrm{Q},\mathrm{definea}*\mathrm{b}=\mathrm{ab}+1\\ \left(\mathrm{iii}\right)\mathrm{OnQ},\mathrm{definea}*\mathrm{b}=\frac{\mathrm{ab}}{2}\\ \mathrm{}\left(\mathrm{iv}\right){\mathrm{OnZ}}^{+},\mathrm{definea}*\mathrm{b}=2^{\mathrm{ab}}\\ \left(\mathrm{v}\right){\mathrm{OnZ}}^{+},\mathrm{definea}*\mathrm{b}={\mathrm{a}}^{\mathrm{b}}\\ \left(\mathrm{vi}\right)\mathrm{\hspace{0.17em}}\mathrm{OnR}-\{-1\},\mathrm{definea}*\mathrm{b}=\frac{\mathrm{a}}{\mathrm{b}+1}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{On}\mathrm{Z},\; *\mathrm{is}\mathrm{defined}\mathrm{by}\mathrm{a}*\mathrm{b}=\mathrm{a}-\mathrm{b}\mathrm{.}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{It}\mathrm{can}\mathrm{be}\mathrm{observed}\mathrm{that}\\ \mathrm{\hspace{0.17em}}1*\; 2\; =\; 1-2=\; 1\mathrm{and}2\; *\; 1\; =\; 2-1=\; 1\mathrm{.}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1*2\ne 2*\; 1;\mathrm{where}1,\; 2\in \mathrm{Z}\\ \mathrm{Hence},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{commutative}\mathrm{.}\\ \mathrm{Also}\mathrm{we}\mathrm{have}:\\ (1*\; 2)\; *\; 3\; =\; (1-2)*3\; =-1*\; 3\; =-\mathrm{1}-3=-\mathrm{4}\\ 1*\; (2\; *\; 3)\; =\; 1\; *\; (2-3)\; =\; 1\; *-1=\; 1-(-1)\; =\; 2\\ \therefore (1*\; 2)\; *\; 3\ne 1*\; (2\; *\; 3)\; ;\mathrm{where}1,\; 2,\; 3\in \mathrm{Z}\\ \mathrm{Hence},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{\hspace{0.17em}}\mathrm{On}\mathrm{Q},\; *\mathrm{is}\mathrm{defined}\mathrm{by}\mathrm{a}*\mathrm{b}=\mathrm{ab}+\; 1\mathrm{.}\\ \mathrm{It}\mathrm{is}\mathrm{known}\mathrm{that}:\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{ab}=\mathrm{ba}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \Rightarrow \mathrm{ab}+\; 1\; =\mathrm{ba}+\; 1\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}\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{\hspace{0.17em}}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{b}=\mathrm{a}*\mathrm{b}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{commutative}\mathrm{.}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{observed}\mathrm{that}:\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(1*\; 2)\; *\; 3\; =\; (1\; \times \; 2\; +\; 1)\; *\; 3\; =\; 3\; *\; 3\; =\; 3\; \times \; 3\; +\; 1\; =\; 10\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1*\; (2\; *\; 3)\; =\; 1\; *\; (2\; \times \; 3\; +\; 1)\; =\; 1\; *\; 7\; =\; 1\; \times \; 7\; +\; 1\; =\; 8\\ \therefore (1*\; 2)\; *\; 3\ne 1*\; (2\; *\; 3)\; ;\mathrm{where}1,\; 2,\; 3\in \mathrm{Q}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{iii}\right)\mathrm{On}\mathrm{Q},\; *\mathrm{is}\mathrm{defined}\mathrm{by}\mathrm{a}*\mathrm{b}=\frac{\mathrm{ab}}{2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{It}\mathrm{is}\mathrm{known}\mathrm{that}:\\ \mathrm{ab}=\mathrm{ba}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{ab}}{2}=\frac{\mathrm{ba}}{2}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \Rightarrow \mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{b}=\mathrm{b}*\mathrm{a}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{commutative}\mathrm{.}\\ \mathrm{For}\mathrm{all}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{we}\mathrm{have}:\\ (\mathrm{a}*\mathrm{b})*\mathrm{c}=\left(\frac{\mathrm{ab}}{2}\right)*\mathrm{c}=\frac{\left(\frac{\mathrm{ab}}{2}\right)\mathrm{c}}{2}=\frac{\mathrm{abc}}{4}\\ \mathrm{a}*(\mathrm{b}*\mathrm{c})=\mathrm{a}*\left(\frac{\mathrm{bc}}{2}\right)=\frac{\mathrm{a}\left(\frac{\mathrm{bc}}{2}\right)}{2}=\frac{\mathrm{abc}}{4}\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}(\mathrm{a}*\mathrm{b})*\mathrm{c}=\mathrm{a}*(\mathrm{b}*\mathrm{c})\forall \mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{iv}\right)\mathrm{\hspace{0.17em}}\mathrm{On}{\mathrm{Z}}^{\mathrm{+}},*\mathrm{is}\mathrm{defined}\mathrm{by}\mathrm{a}*\mathrm{b}=\mathrm{ab}\mathrm{.}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{observed}\mathrm{that}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1*2={\mathrm{1}}^{\mathrm{2}}=1\mathrm{and}2*1={\mathrm{2}}^{\mathrm{2}}=4\\ \therefore \hspace{0.17em}\hspace{0.17em}1*2\ne 2*1;\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{where}\hspace{0.17em}\hspace{0.17em}1,2\in {\mathrm{Z}}^{\mathrm{+}}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{commutative}\mathrm{.}\\ \mathrm{For}\mathrm{Associative}\mathrm{Law}:\\ (3*4)*5=\left(3^2\right)*5\\ ={\left(3^2\right)}^2\\ =81\\ 3*(4*5)=3*\left(4^2\right)\\ =3^2\\ =9\\ \therefore (3*4)*5\ne 3*(4*5)\hspace{0.17em}\hspace{0.17em}\mathrm{where}2,\; 3,\; 4\in {\mathrm{Z}}^{\mathrm{+}}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{vi}\right)\mathrm{\hspace{0.17em}}\mathrm{On}\mathrm{R},\; *-\mathrm{\{}-1\}\mathrm{is}\mathrm{defined}\mathrm{by}\hspace{0.17em}\hspace{0.17em}\mathrm{a}*\mathrm{b}=\frac{\mathrm{a}}{\mathrm{b}+1}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{observed}\mathrm{that}\hspace{0.17em}\hspace{0.17em}\\ 2*3=\frac{2}{3+1}\\ =\frac{2}{4}\\ =\frac{1}{2}\\ \mathrm{and}\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{\hspace{0.17em}}3*2=\frac{3}{2+1}\\ =\frac{3}{3}\\ =1\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2*3\ne 3*2\; ;\mathrm{where}2,\; 3\in \mathrm{R}-\mathrm{\{}-1\}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{commutative}\mathrm{.}\\ \mathrm{For}\mathrm{​}\mathrm{Associative}\mathrm{Law}:\\ (2*3)*4=\left(\frac{2}{3+1}\right)*4=\left(\frac{2}{4}\right)*4=\left(\frac{1}{2}\right)*4=\frac{\left(\frac{1}{2}\right)}{4+1}=\frac{1}{10}\\ 2*(3*4)=2*\left(\frac{3}{4+1}\right)=2*\left(\frac{3}{5}\right)=\frac{2}{\left(\frac{3}{5}\right)+1}=\frac{2}{\frac{8}{5}}=\frac{5}{4}\\ \therefore (2*3)*4\ne 2*(3*4);\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{where}2,3,4\in \mathrm{R}-\{-1\}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\end{array}$

Q.45

$\begin{array}{l}\mathrm{Consider}\mathrm{the}\mathrm{binary}\mathrm{operation}\wedge \mathrm{on}\mathrm{theset}\{1,2,3,4,5\}\\ \mathrm{definedby}\hspace{0.17em}\hspace{0.17em}\mathrm{a}\wedge \mathrm{b}=\min \{\mathrm{a},\mathrm{b}\}.\mathrm{Write}\mathrm{the}\mathrm{operation}\mathrm{table}\\ \mathrm{of}\mathrm{the}\mathrm{operation}\wedge .\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{binary}\mathrm{operation}\wedge \mathrm{on}\mathrm{the}\mathrm{set}\{\mathrm{1},2,3,4,5\}\mathrm{is}\mathrm{defined}\mathrm{as}\\ \mathrm{a}\wedge \mathrm{b}=\min \{\mathrm{a},\mathrm{b}\}\forall \mathrm{a},\mathrm{b}\in \{1,\; 2,\; 3,\; 4,\; 5\}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}\mathrm{table}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{operation}\wedge \mathrm{can}\mathrm{be}\\ \mathrm{given}\mathrm{as}:\\ \begin{array}{c}\end{array}\end{array}$

^	1	2	3	4	5
1	1	1	1	1	1
2	1	2	2	2	2
3	1	2	3	3	3
4	1	2	3	4	4
5	1	2	3	4	5

Q.46

$\begin{array}{l}\mathrm{Conside}\mathrm{ra}\mathrm{binary}\mathrm{operation}*\mathrm{on}\mathrm{the}\mathrm{set}\{1,2,3,4,5\}\mathrm{given}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{by}\mathrm{the}\mathrm{following}\mathrm{multiplication}\mathrm{\hspace{0.17em}}\mathrm{table}\hspace{0.17em}\hspace{0.17em}(\mathrm{Table}\mathrm{\hspace{0.17em}}1.2)\\ \left(\mathrm{i}\right)\mathrm{Compute}(2*3)*4\mathrm{and}2*(3*4)\\ \left(\mathrm{ii}\right)\mathrm{Is}*\mathrm{commutative}?\\ \left(\mathrm{iii}\right)\mathrm{Compute}(2*3)*(4*5)\\ \begin{array}{}\end{array}\end{array}$

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	2	2	2
3	1	2	3	3	3
4	1	2	3	4	4
5	1	2	3	4	5

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{\hspace{0.17em}}(2*3)*4=1*4=1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2*(3*4)=2*1=1\\ \left(\mathrm{ii}\right)\mathrm{\hspace{0.17em}}\mathrm{For}\mathrm{every}\mathrm{a},\mathrm{b}\in \{1,\; 2,\; 3,\; 4,\; 5\},\mathrm{we}\mathrm{have}\mathrm{a}*\mathrm{b}=\mathrm{b}*\mathrm{a}.\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{commutative}\mathrm{.}\\ \left(\mathrm{iii}\right)(2*\; 3)\; =\; 1\mathrm{and}(4\; *\; 5)\; =\; 1\\ \therefore (2*\; 3)\; *\; (4\; *\; 5)\; =\; 1\; *\; 1\; =\; 1\end{array}$

Q.47

Let *′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a *′ b = H.C.F. of a and b. Is the operation *′ same as the operation * defined in Exercise 4 above? Justify your answer.

Ans.

\begin{array}{l}\text{The binary operation *’R on the set {1, 2, 3 4, 5} is defined as}\\ \text{a *’ b = H}\text{.C}\text{.F of a and b}\text{.}\\ \text{The operation table for the operation *’ can be given as:}\\ \end{array}

*’	1	2	3	4	5
1	1	1	1	1	1
2	1	2	1	2	1
3	1	1	1	1	1
4	1	2	1	4	1
5	1	1	1	1	1

We see that the operation tables for the operations * and *’ are the same. Thus, the operation *’ is same as the operation*.

Q.48

$\begin{array}{l}\mathrm{Let}*\mathrm{be}\mathrm{the}\mathrm{binary}\mathrm{operation}\mathrm{on}\mathrm{Ngiven}\mathrm{by}\mathrm{a}*\mathrm{b}=\mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}\mathrm{a}\\ \mathrm{and}\mathrm{b}.\mathrm{Find}\\ \left(\mathrm{i}\right)5*7,20*16\\ \left(\mathrm{ii}\right)\mathrm{Is}*\mathrm{commutative}?\\ \left(\mathrm{iii}\right)\mathrm{Is}*\mathrm{associative}?\\ \left(\mathrm{iv}\right)\mathrm{Find}\mathrm{the}\mathrm{identity}\mathrm{of}*\mathrm{in}\mathrm{N}\\ \left(\mathrm{v}\right)\mathrm{Which}\mathrm{element}\mathrm{so}\mathrm{f}\mathrm{Narein}\mathrm{vertible}\mathrm{f}\mathrm{or}\mathrm{the}\mathrm{operation}*?\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{binary}\mathrm{operation}*\mathrm{on}\mathrm{N}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{a}*\mathrm{b}=\mathrm{L}.\mathrm{C}.\mathrm{M}.\\ \mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{.}\\ (\mathrm{i})\; 5\; *\; 7\; =\mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}5\mathrm{and}7\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=35\\ \mathrm{\hspace{0.17em}}20*\; 16\; =\mathrm{L}.\mathrm{C}.\mathrm{M}\mathrm{of}20\mathrm{and}16\; =\; 80\\ \left(\mathrm{ii}\right)\mathrm{It}\mathrm{is}\mathrm{known}\mathrm{that}:\\ \mathrm{L}.\mathrm{C}.\mathrm{M}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}=\mathrm{L}.\mathrm{C}.\mathrm{M}\mathrm{of}\mathrm{b}\mathrm{and}\mathrm{a}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\forall \mathrm{a},\mathrm{b}\in \mathrm{N}\mathrm{.}\\ \therefore \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}\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}\hspace{0.17em}\hspace{0.17em}\mathrm{a}*\mathrm{b}=\mathrm{b}*\mathrm{a}\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{commutative}\mathrm{.}\\ \\ \left(\mathrm{iii}\right)\mathrm{For}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{N},\mathrm{we}\mathrm{have}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(\mathrm{a}*\mathrm{b})\; *\mathrm{c}=\; (\mathrm{L}.\mathrm{C}.\mathrm{M}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b})\; *\mathrm{c}\\ \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{LCM}\mathrm{of}\mathrm{a},\mathrm{b},\mathrm{and}\mathrm{c}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}*\; (\mathrm{b}*\mathrm{c})=\mathrm{a}*\; (\mathrm{LCM}\mathrm{of}\mathrm{b}\mathrm{and}\mathrm{c})\\ \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{L}.\mathrm{C}.\mathrm{M}\mathrm{of}\mathrm{a},\mathrm{b},\mathrm{and}\mathrm{c}\\ \therefore \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(\mathrm{a}*\mathrm{b})\; *\mathrm{c}=\mathrm{a}*\; (\mathrm{b}*\mathrm{c})\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{iv}\right)\mathrm{It}\mathrm{is}\mathrm{known}\mathrm{that}:\\ \mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}\mathrm{a}\mathrm{and}1\; =\mathrm{a}=\mathrm{L}.\mathrm{C}.\mathrm{M}.1\mathrm{and}\mathrm{a}\mathrm{}\forall \mathrm{a}\in \mathrm{N}\\ \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}\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{\hspace{0.17em}}\mathrm{a}*\; 1\; =\mathrm{a}=\; 1\; *\mathrm{a}\mathrm{}\forall \mathrm{a}\in \mathrm{N}\\ \mathrm{Thus},\; 1\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{of}*\mathrm{in}\mathrm{N}\mathrm{.}\\ \left(\mathrm{v}\right)\mathrm{An}\mathrm{element}\mathrm{a}\mathrm{in}\mathrm{N}\mathrm{is}\mathrm{invertible}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{the}\\ \mathrm{operation}*\mathrm{\hspace{0.17em}}\mathrm{if}\mathrm{there}\mathrm{exists}\mathrm{an}\mathrm{element}\mathrm{b}\mathrm{in}\hspace{0.17em}\hspace{0.17em}\mathrm{N},\mathrm{such}\\ \mathrm{that}\mathrm{a}*\mathrm{b}=\mathrm{e}=\mathrm{b}*\mathrm{a}\mathrm{.}\\ \mathrm{Here},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{e}=\; 1\\ \mathrm{This}\mathrm{means}\mathrm{that}:\\ \mathrm{L}.\mathrm{C}.\mathrm{M}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}=\; 1\; =\mathrm{L}.\mathrm{C}.\mathrm{M}\mathrm{of}\mathrm{b}\mathrm{and}\mathrm{a}\\ \mathrm{This}\mathrm{case}\mathrm{is}\mathrm{possible}\mathrm{only}\mathrm{when}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{are}\mathrm{equal}\mathrm{to}1\mathrm{.}\\ \mathrm{Thus},\; 1\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{invertible}\mathrm{element}\mathrm{of}\mathrm{N}\mathrm{with}\mathrm{respect}\mathrm{to}\\ \mathrm{the}\mathrm{operation}*\mathrm{.}\end{array}$

Q.49

$\begin{array}{l}\mathrm{Is}*\mathrm{defined}\mathrm{on}\mathrm{the}\mathrm{set}\{1,2,3,4,5\}\mathrm{by}\mathrm{a}*\mathrm{b}=\mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}\mathrm{a}\\ \mathrm{and}\mathrm{b}\mathrm{a}\mathrm{binary}\mathrm{operation}?\mathrm{Justify}\mathrm{your}\mathrm{\hspace{0.17em}}\mathrm{answer}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{operation}*\mathrm{on}\mathrm{the}\mathrm{set}\mathrm{A}=\; \{1,\; 2,\; 3,\; 4,\; 5\}\mathrm{is}\mathrm{defined}\mathrm{as}\\ \mathrm{a}*\mathrm{b}=\mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}.\mathrm{Then},\mathrm{the}\mathrm{operation}\mathrm{table}\mathrm{for}\mathrm{the}\mathrm{given}\\ \mathrm{operation}*\mathrm{can}\mathrm{be}\mathrm{givenas}:\\ \end{array}$

*	1	2	3	4	5
1	1	2	3	4	5
2	2	2	6	4	10
3	3	6	3	12	15
4	4	4	12	4	20
5	5	10	15	20	5

It can be observed from the obtained table that:

3 * 2 = 2 * 3 = 6∈A, 5 * 2 = 2 * 5 = 10∈A,

3 * 4 = 4 * 3 = 12∈A

3 * 5 = 5 * 3 = 15∈A, 4 * 5 = 5 * 4 = 20∉A

Hence, the given operation * is not a binary operation.

Q.50

Let * be the binary operation on N defined by
a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?

Ans.

\begin{array}{l}\text{The binary operation * on N is defined as:}\\ \text{a * b = H}\text{.C}\text{.F}\text{. of a and b}\\ Since,\\ \text{H}\text{.C}\text{.F}\text{. of a and b}=\text{H}\text{.C}\text{.F}\text{. of b and a}\forall \text{a, b}\in \text{N}\text{.}\\ \therefore \text{a*b}=\text{b*a}\\ Thus,\text{the operation * is commutative}\text{.}\\ \text{For a, b, c}\in \text{N, we have:}\\ \text{(a * b)* c}=\text{(H}\text{.C}\text{.F}\text{. of a and b) * c}\\ =\text{H}\text{.C}\text{.F}\text{. of a, b, and c}\\ \\ \text{a *(b * c)}=\text{a *(H}\text{.C}\text{.F}\text{. of b and c)}\\ =\text{H}\text{.C}\text{.F}\text{. of a, b, and c}\\ \therefore \text{(a*b)*c}=\text{a*(b*c)}\\ \text{Thus, the operation * is associative}\text{.}\\ Let\text{an element e}\in \text{N will be the identity for the operation}\\ \text{* if a*e}=\text{a}=\text{e* a}\forall \text{a}\in \text{N}\text{.}\\ \text{But this relation is not true for all a}\in \text{N}\text{.}\\ \text{Thus, the operation * does not have any identity in N}\text{.}\end{array}

Q.51

$\begin{array}{l}\mathrm{Let}*\mathrm{be}\mathrm{a}\mathrm{binary}\mathrm{operation}\mathrm{on}\mathrm{the}\mathrm{set}\hspace{0.17em}\hspace{0.17em}\mathrm{Q}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{of}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{rational}\\ \mathrm{numbers}\mathrm{as}\mathrm{follows}:\\ \left(\mathrm{i}\right)\mathrm{a}*\mathrm{b}=\mathrm{a}–\mathrm{b}\left(\mathrm{ii}\right)\mathrm{a}*\mathrm{b}={\mathrm{a}}^2+{\mathrm{b}}^2\\ (\mathrm{iii})\mathrm{a}*\mathrm{b}=\mathrm{a}+\mathrm{ab}\left(\mathrm{iv}\right)\mathrm{a}*\mathrm{b}={\left(\mathrm{a}–\mathrm{b}\right)}^2\\ \left(\mathrm{v}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{b}=\frac{\mathrm{ab}}{4}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\mathrm{vi}\right)\mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{b}={\mathrm{ab}}^2\\ \mathrm{Find}\mathrm{which}\mathrm{of}\mathrm{the}\mathrm{binary}\mathrm{operation}\mathrm{s}\mathrm{are}\mathrm{commutative}\\ \mathrm{and}\mathrm{which}\mathrm{are}\mathrm{as}\mathrm{sociative}.\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{\hspace{0.17em}}\mathrm{Since},\hspace{0.17em}\hspace{0.17em}\mathrm{a}*\mathrm{b}=\mathrm{a}-\mathrm{b}\mathrm{}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ =-(\mathrm{b}-\mathrm{a})\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ =-\mathrm{b}*\mathrm{a}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{So},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{cummutative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{then}\\ \hspace{0.17em}\hspace{0.17em}(\mathrm{a}*\mathrm{b})*\mathrm{c}=(\mathrm{a}-\mathrm{b})*\mathrm{c}\\ =\mathrm{a}-\mathrm{b}-\mathrm{c}\\ \\ \mathrm{and}\mathrm{\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{\hspace{0.17em}}\mathrm{a}*(\mathrm{b}*\mathrm{c})=\mathrm{a}*(\mathrm{b}-\mathrm{c})\\ =\mathrm{a}-(\mathrm{b}-\mathrm{c})\\ =\mathrm{a}-\mathrm{b}+\mathrm{c}\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(\mathrm{a}*\mathrm{b})*\mathrm{c}\ne \mathrm{a}*(\mathrm{b}*\mathrm{c})\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{Since},\hspace{0.17em}\hspace{0.17em}\mathrm{a}*\mathrm{b}={\mathrm{a}}^2+{\mathrm{b}}^2\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ ={\mathrm{b}}^2+{\mathrm{a}}^2\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ =\mathrm{b}*\mathrm{a}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{So},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{cummutative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{then}\\ \hspace{0.17em}\hspace{0.17em}(\mathrm{a}*\mathrm{b})*\mathrm{c}=({\mathrm{a}}^2+{\mathrm{b}}^2)*\mathrm{c}\\ ={({\mathrm{a}}^2+{\mathrm{b}}^2)}^2+{\mathrm{c}}^2\\ \mathrm{and}\mathrm{\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{\hspace{0.17em}}\mathrm{a}*(\mathrm{b}*\mathrm{c})=\mathrm{a}*({\mathrm{b}}^2+{\mathrm{c}}^2)\\ ={\mathrm{a}}^2+{({\mathrm{b}}^2+{\mathrm{c}}^2)}^2\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(\mathrm{a}*\mathrm{b})*\mathrm{c}\ne \mathrm{a}*(\mathrm{b}*\mathrm{c})\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{iii}\right)\mathrm{Since},\hspace{0.17em}\hspace{0.17em}\mathrm{a}*\mathrm{b}=\mathrm{a}+\mathrm{ab}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{b}*\mathrm{a}=\mathrm{b}+\mathrm{ba}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \therefore \mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{b}\ne \mathrm{b}*\mathrm{a}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{So},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{cummutative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{then}\\ \\ \hspace{0.17em}\hspace{0.17em}(\mathrm{a}*\mathrm{b})*\mathrm{c}=(\mathrm{a}+\mathrm{ab})*\mathrm{c}\\ =(\mathrm{a}+\mathrm{ab})+(\mathrm{a}+\mathrm{ab})\mathrm{c}\\ =\mathrm{a}+\mathrm{ab}+\mathrm{ac}+\mathrm{abc}\\ \mathrm{and}\mathrm{\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{\hspace{0.17em}}\mathrm{a}*(\mathrm{b}*\mathrm{c})=\mathrm{a}*(\mathrm{b}+\mathrm{bc})\\ =\mathrm{a}+\mathrm{a}(\mathrm{b}+\mathrm{bc})\\ =\mathrm{a}+\mathrm{ab}+\mathrm{abc}\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(\mathrm{a}*\mathrm{b})*\mathrm{c}\ne \mathrm{a}*(\mathrm{b}*\mathrm{c})\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{iv}\right)\mathrm{Since},\hspace{0.17em}\hspace{0.17em}\mathrm{a}*\mathrm{b}={(\mathrm{a}-\mathrm{b})}^2\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\{-(\mathrm{b}-\mathrm{a})\}}^2\\ \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}\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}\hspace{0.17em}\hspace{0.17em}={(\mathrm{b}-\mathrm{a})}^2\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{b}*\mathrm{a}\\ \therefore \mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{b}\ne \mathrm{b}*\mathrm{a}\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{\hspace{0.17em}}\hspace{0.17em}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{So},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{cummutative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{then}\\ \hspace{0.17em}\hspace{0.17em}(\mathrm{a}*\mathrm{b})*\mathrm{c}=\left\{{(\mathrm{a}-\mathrm{b})}^2\right\}*\mathrm{c}\\ ={\{{(\mathrm{a}-\mathrm{b})}^2-\mathrm{c}\}}^2\\ ={({\mathrm{a}}^2-2\mathrm{ab}+{\mathrm{b}}^2-\mathrm{c})}^2\\ \mathrm{and}\\ \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{\hspace{0.17em}}\mathrm{a}*(\mathrm{b}*\mathrm{c})=\mathrm{a}*{(\mathrm{b}-\mathrm{c})}^2\\ ={\{\mathrm{a}-{(\mathrm{b}-\mathrm{c})}^2\}}^2\\ \\ ={(\mathrm{a}-{\mathrm{b}}^2+2\mathrm{bc}-{\mathrm{c}}^2)}^2\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(\mathrm{a}*\mathrm{b})*\mathrm{c}\ne \mathrm{a}*(\mathrm{b}*\mathrm{c})\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{v}\right)\mathrm{Since},\hspace{0.17em}\hspace{0.17em}\mathrm{a}*\mathrm{b}=\frac{\mathrm{ab}}{4}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{\mathrm{ba}}{4}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{b}*\mathrm{a}\\ \therefore \mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{b}=\mathrm{b}*\mathrm{a}\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}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{So},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{cummutative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{then}\\ \hspace{0.17em}\hspace{0.17em}(\mathrm{a}*\mathrm{b})*\mathrm{c}=\left(\frac{\mathrm{ab}}{4}\right)*\mathrm{c}\\ =\frac{\left(\frac{\mathrm{ab}}{4}\right)\mathrm{c}}{4}\\ =\frac{\mathrm{abc}}{16}\\ \mathrm{and}\mathrm{\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{\hspace{0.17em}}\mathrm{a}*(\mathrm{b}*\mathrm{c})=\mathrm{a}*\left(\frac{\mathrm{bc}}{4}\right)\\ =\frac{\mathrm{a}\left(\frac{\mathrm{bc}}{4}\right)}{4}\\ =\frac{\mathrm{abc}}{16}\\ \\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(\mathrm{a}*\mathrm{b})*\mathrm{c}=\mathrm{a}*(\mathrm{b}*\mathrm{c})\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{associative}\mathrm{.}\\ \\ \left(\mathrm{vi}\right)\mathrm{Since},\hspace{0.17em}\hspace{0.17em}\mathrm{a}*\mathrm{b}={\mathrm{ab}}^2\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{b}*\mathrm{a}\mathrm{\hspace{0.17em}}={\mathrm{ba}}^2\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \therefore \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{b}\ne \mathrm{b}*\mathrm{a}\mathrm{\hspace{0.17em}}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{So},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{cummutative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{then}\\ \hspace{0.17em}\hspace{0.17em}(\mathrm{a}*\mathrm{b})*\mathrm{c}=\left({\mathrm{ab}}^2\right)*\mathrm{c}\\ =\left({\mathrm{ab}}^2\right){\mathrm{c}}^2\\ ={\mathrm{ab}}^2{\mathrm{c}}^2\\ \mathrm{and}\mathrm{\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{\hspace{0.17em}}\mathrm{a}*(\mathrm{b}*\mathrm{c})=\mathrm{a}*\left({\mathrm{bc}}^2\right)\\ =\mathrm{a}{\left({\mathrm{bc}}^2\right)}^2\\ ={\mathrm{ab}}^2{\mathrm{c}}^4\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(\mathrm{a}*\mathrm{b})*\mathrm{c}\ne \mathrm{a}*(\mathrm{b}*\mathrm{c})\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{operations}\mathrm{defined}\mathrm{in}(\mathrm{ii}),\; (\mathrm{iv}),\; (\mathrm{v})\mathrm{are}\\ \mathrm{commutative}\mathrm{and}\mathrm{the}\mathrm{operation}\mathrm{defined}\mathrm{in}\left(\mathrm{v}\right)\mathrm{is}\mathrm{associative}\mathrm{.}\end{array}$

Q.52

$\begin{array}{l}\mathrm{Let}*\mathrm{be}\mathrm{a}\mathrm{binary}\mathrm{operation}\mathrm{on}\mathrm{the}\mathrm{set}\hspace{0.17em}\hspace{0.17em}\mathrm{Q}\hspace{0.17em}\hspace{0.17em}\mathrm{of}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{rational}\\ \mathrm{numbersas}\mathrm{follows}:\\ \left(\mathrm{i}\right)\mathrm{a}*\mathrm{b}=\mathrm{a}–\mathrm{b}\left(\mathrm{ii}\right)\mathrm{a}*\mathrm{b}={\mathrm{a}}^2+{\mathrm{b}}^2\\ \left(\mathrm{iii}\right)\mathrm{a}*\mathrm{b}=\mathrm{a}+\mathrm{ab}\left(\mathrm{iv}\right)\mathrm{a}*\mathrm{b}={\left(\mathrm{a}–\mathrm{b}\right)}^2\\ \left(\mathrm{v}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{b}=\frac{\mathrm{ab}}{4}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\mathrm{vi}\right)\mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{b}={\mathrm{ab}}^2\end{array}$

Find which of the operations given above has identity.

Ans.

$\begin{array}{l}\mathrm{An}\mathrm{element}\mathrm{e}\in \mathrm{Q}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{for}\mathrm{the}\\ \mathrm{operation}*\mathrm{if}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{e}=\mathrm{a}=\mathrm{e}*\mathrm{a},\forall \mathrm{a}\in \mathrm{Q}\mathrm{.}\\ \mathrm{However},\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{such}\mathrm{element}\mathrm{e}\in \mathrm{Q}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{each}\\ \mathrm{of}\mathrm{the}\mathrm{six}\mathrm{operations}\mathrm{satisfying}\mathrm{the}\mathrm{\hspace{0.17em}}\mathrm{above}\mathrm{condition}\mathrm{.}\\ \mathrm{Thus},\mathrm{none}\mathrm{of}\mathrm{the}\mathrm{six}\mathrm{operations}\mathrm{in}\mathrm{above}\mathrm{question}\mathrm{has}\mathrm{identity}\mathrm{.}\end{array}$

Q.53

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\mathrm{N}\times \mathrm{Nand}*\mathrm{be}\mathrm{the}\mathrm{binary}\mathrm{operation}\mathrm{on}\mathrm{A}\mathrm{defined}\mathrm{by}\\ (\mathrm{a},\mathrm{b})*(\mathrm{c},\mathrm{d})=(\mathrm{a}+\mathrm{c},\mathrm{b}+\mathrm{d})\\ \mathrm{Show}\mathrm{that}*\mathrm{is}\mathrm{commutative}\mathrm{and}\mathrm{associative}.\mathrm{Find}\mathrm{the}\mathrm{}\\ \mathrm{identity}\mathrm{element}\mathrm{for}*\mathrm{onA},\mathrm{if}\mathrm{any}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{A}=\mathrm{N}\times \mathrm{N}\\ *\mathrm{is}\mathrm{a}\mathrm{binary}\mathrm{operation}\mathrm{on}\mathrm{A}\mathrm{and}\mathrm{is}\mathrm{defined}\mathrm{by}:\\ (\mathrm{a},\mathrm{b})\; *\; (\mathrm{c},\mathrm{d})\; =\; (\mathrm{a}+\mathrm{c},\mathrm{b}+\mathrm{d})\\ \mathrm{Let}(\mathrm{a},\mathrm{b}),\; (\mathrm{c},\mathrm{d})\in \mathrm{A}\\ \mathrm{Then},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\in \mathrm{N}\\ \mathrm{Since},\\ (\mathrm{a},\mathrm{b})\; *\; (\mathrm{c},\mathrm{d})\; =\; (\mathrm{a}+\mathrm{c},\mathrm{b}+\mathrm{d})\\ \mathrm{and}\\ \mathrm{\hspace{0.17em}}(\mathrm{c},\mathrm{d})\; *\; (\mathrm{a},\mathrm{b})\; =\; (\mathrm{c}+\mathrm{a},\mathrm{d}+\mathrm{b})\\ \mathrm{}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}+\mathrm{c},\mathrm{b}+\mathrm{d})\\ [\mathrm{Addition}\mathrm{is}\mathrm{commutative}\mathrm{in}\mathrm{the}\mathrm{set}\mathrm{of}\mathrm{natural}\mathrm{numbers}]\\ \therefore (\mathrm{a},\mathrm{b})*(\mathrm{c},\mathrm{d})\; =\; (\mathrm{c},\mathrm{d})*(\mathrm{a},\mathrm{b})\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{commutative}\mathrm{.}\\ \mathrm{For}\mathrm{associativity}:\\ \mathrm{Now},\mathrm{let}(\mathrm{a},\mathrm{b}),\; (\mathrm{c},\mathrm{d}),\; (\mathrm{e},\mathrm{f})\in \mathrm{A}\\ \mathrm{Then},\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e},\mathrm{f}\in \mathrm{N}\\ \\ \mathrm{So},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left\{\right(\mathrm{a},\mathrm{b})\; *\; (\mathrm{c},\mathrm{d}\left)\right\}\mathrm{*}(\mathrm{e},\mathrm{f})=\left\{\right(\mathrm{a}+\mathrm{c},\mathrm{b}+\mathrm{d}\left)\right\}*(\mathrm{e},\mathrm{f})\\ \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{\hspace{0.17em}}\hspace{0.17em}=(\mathrm{a}+\mathrm{c}+\mathrm{\hspace{0.17em}}\mathrm{e},\mathrm{b}+\mathrm{d}+\mathrm{f}\hspace{0.17em})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(\mathrm{a},\mathrm{b})\; *\left\{\right(\mathrm{c},\mathrm{d})*(\mathrm{e},\mathrm{f})\}=(\mathrm{a},\mathrm{b})\; *\left\{\right(\mathrm{c}+\mathrm{e},\mathrm{d}+\mathrm{f}\left)\right\}\\ \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{\hspace{0.17em}}\hspace{0.17em}=(\mathrm{a}+\mathrm{c}+\mathrm{e},\mathrm{b}+\mathrm{d}+\mathrm{f})\\ \therefore \left\{\right(\mathrm{a},\mathrm{b})\; *\; (\mathrm{c},\mathrm{d}\left)\right\}\mathrm{*}(\mathrm{e},\mathrm{f})=(\mathrm{a},\mathrm{b})\; *\left\{\right(\mathrm{c},\mathrm{d})*(\mathrm{e},\mathrm{f})\}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{associative}\mathrm{.}\\ \mathrm{An}\mathrm{element}\mathrm{e}=({\mathrm{e}}_1+{\mathrm{e}}_2)\in \mathrm{A}\mathrm{\hspace{0.17em}}\mathrm{will}\mathrm{be}\mathrm{an}\mathrm{identity}\mathrm{element}\mathrm{for}\mathrm{the}\\ \mathrm{operation}*\mathrm{if}\\ \mathrm{a}*\mathrm{e}=\mathrm{a}=\mathrm{e}*\mathrm{a}\forall \mathrm{a}=({\mathrm{a}}_1,{\mathrm{a}}_2)\in \mathrm{A}\\ \mathrm{i}.\mathrm{e}.,\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}({\mathrm{a}}_1+{\mathrm{e}}_1,{\mathrm{a}}_2+{\mathrm{e}}_2)=({\mathrm{a}}_1,{\mathrm{a}}_2)=({\mathrm{e}}_1+{\mathrm{a}}_1,{\mathrm{e}}_2+{\mathrm{a}}_2)\\ \mathrm{which}\mathrm{is}\mathrm{not}\mathrm{true}\mathrm{for}\mathrm{any}\mathrm{element}\mathrm{in}\mathrm{A}\mathrm{.}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{does}\mathrm{not}\mathrm{have}\mathrm{any}\mathrm{identity}\mathrm{element}\mathrm{.}\end{array}$

Q.54

$\begin{array}{l}\mathrm{State}\mathrm{whether}\mathrm{the}\mathrm{following}\mathrm{statements}\mathrm{are}\mathrm{true}\mathrm{or}\mathrm{false}.\\ \mathrm{Justify}.\\ \left(\mathrm{i}\right)\mathrm{For}\mathrm{an}\mathrm{arbitrary}\mathrm{binary}\mathrm{operation}*\mathrm{on}\mathrm{a}\mathrm{set}\mathrm{N},\\ \mathrm{a}*\mathrm{a}=\mathrm{a}“\mathrm{a}*\mathrm{N}.\\ \left(\mathrm{ii}\right)\mathrm{If}*\mathrm{is}\mathrm{acommutative}\mathrm{binary}\mathrm{operationon}\mathrm{N},\mathrm{then}\\ \mathrm{a}*(\mathrm{b}*\mathrm{c})=(\mathrm{c}*\mathrm{b})*\mathrm{a}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Define}\mathrm{an}\mathrm{operation}*\mathrm{on}\mathrm{N}\mathrm{as}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{a}*\mathrm{b}=\mathrm{a}+\mathrm{b}\forall \mathrm{a},\mathrm{b}\in \mathrm{N}\\ \mathrm{Then},\mathrm{in}\mathrm{particular},\mathrm{for}\mathrm{b}=\mathrm{a}=\; 5,\mathrm{we}\mathrm{have}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}5*\; 5\; =\; 5\; +\; 5=10\ne \mathrm{5}\\ \mathrm{Therefore},\mathrm{statement}\left(\mathrm{i}\right)\mathrm{is}\mathrm{false}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{R}.\mathrm{H}.\mathrm{S}.=\; (\mathrm{c}*\mathrm{b})\; *\mathrm{a}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{b}*\mathrm{c})\; *\mathrm{a}[*\mathrm{is}\mathrm{commutative}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{a}*\; (\mathrm{b}*\mathrm{c}\left)\right[\mathrm{Again},\mathrm{as}*\mathrm{is}\mathrm{commutative}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}\\ \therefore \mathrm{a}*\; (\mathrm{b}*\mathrm{c})\; =\; (\mathrm{c}*\mathrm{b})\; *\mathrm{a}\\ \mathrm{Therefore},\mathrm{statement}\left(\mathrm{ii}\right)\mathrm{is}\mathrm{true}\mathrm{.}\end{array}$

Q.55

$\begin{array}{l}\mathrm{Consider}\mathrm{a}\mathrm{binary}\mathrm{operation}*\mathrm{on}\mathrm{N}\hspace{0.17em}\hspace{0.17em}\mathrm{defined}\mathrm{as}\mathrm{a}*\mathrm{b}={\mathrm{a}}^3+{\mathrm{b}}^3.\\ \mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}.\\ (\mathrm{A})\mathrm{Is}*\mathrm{both}\mathrm{a}\mathrm{ssociative}\mathrm{and}\mathrm{commutative}?\\ \left(\mathrm{B}\right)\mathrm{Is}*\mathrm{commutative}\mathrm{but}\mathrm{not}\mathrm{associative}?\\ \left(\mathrm{C}\right)\mathrm{Is}*\mathrm{associative}\mathrm{but}\mathrm{not}\mathrm{commutative}?\\ \left(\mathrm{D}\right)\mathrm{Is}*\mathrm{neither}\mathrm{commutative}\mathrm{nor}\mathrm{associative}?\end{array}$

Ans.

\begin{array}{l}{\text{On N, the operation * is defined as a*b = a}}^{\text{3}}{\text{+ b}}^{\text{3}}\text{.}\\ \text{For, a, b}\in \text{N, we have:}\\ \text{a*b}={\text{a}}^{\text{3}}{\text{+ b}}^{\text{3}}={\text{b}}^{\text{3}}{\text{+ a}}^{\text{3}}=b*a\text{[Addition is commutative in N]}\\ \text{Therefore, the operation * is commutative}\text{.}\\ \text{For associative:}\\ Since,2,3,4\in N\\ So,\left(2*3\right)*4=\left(2^3+3^3\right)*4\\ =35*4\\ ={35}^3+4^3\\ =42939\\ and2*\left(3*4\right)=2*\left(3^3+4^3\right)\\ =2*\left(91\right)\\ =2^3+{91}^3\\ =8+753571\\ =753579\\ \therefore \left(2*3\right)*4\ne 2*\left(3*4\right);\text{where 2,3,4}\in \text{N}\\ \text{Therefore, the operation * is not associative}\text{.}\\ \text{Hence, the operation * is commutative, but not associative}\text{.}\\ \text{Thus, the correct answer is B}\text{.}\end{array}

Q.56

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)=10\mathrm{x}+7.\mathrm{Find}\mathrm{the}\mathrm{function}\\ \mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{such}\mathrm{that}\mathrm{go}\mathrm{f}=\mathrm{fog}={\mathrm{I}}_{\mathrm{R}}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}(\mathrm{x})\; =\; 10\mathrm{x}+\; 7\mathrm{.}\\ \mathrm{One}–\mathrm{one}:\\ \mathrm{Let}\mathrm{f}(\mathrm{x})\; =\mathrm{f}(\mathrm{y}),\mathrm{where}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{.}\\ \Rightarrow 10\mathrm{x}+\; 7\; =\; 10\mathrm{y}+\; 7\\ \Rightarrow \mathrm{x}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{a}\mathrm{one}–\mathrm{one}\mathrm{function}\mathrm{.}\\ \mathrm{Onto}:\\ \mathrm{For}\mathrm{y}\in \mathrm{R},\mathrm{let}\mathrm{y}=10\mathrm{x}+\; 7\mathrm{.}\\ \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{\hspace{0.17em}}\mathrm{x}=\frac{\mathrm{y}-7}{10}\in \mathrm{R}\\ \mathrm{Then},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{y}-7}{10}\right)[\mathrm{y}\in \mathrm{R}]\\ \\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=10\left(\frac{\mathrm{y}-7}{10}\right)+7\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \mathrm{Therefore},\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{.}\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{an}\mathrm{invertible}\mathrm{function}\mathrm{.}\\ \mathrm{For}\mathrm{inverse}:\\ \mathrm{Let}\mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{as}\mathrm{g}\left(\mathrm{y}\right)=\frac{\mathrm{y}-7}{10}\\ \mathrm{So},\\ \mathrm{gof}\left(\mathrm{x}\right)=\mathrm{g}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}=\mathrm{g}(10\mathrm{x}+\; 7)=\frac{10\mathrm{x}+\; 7-7}{10}=\mathrm{x}\\ \mathrm{and}\\ \mathrm{fog}\left(\mathrm{y}\right)=\mathrm{f}\left\{\mathrm{g}\left(\mathrm{y}\right)\right\}=\mathrm{f}\left(\frac{\mathrm{y}-7}{10}\right)=10\left(\frac{\mathrm{y}-7}{10}\right)+7=\mathrm{y}\\ \therefore \mathrm{gof}={\mathrm{I}}_{\mathrm{R}}\mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{R}}\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{function}\mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{defined}\mathrm{as}\\ \mathrm{g}\left(\mathrm{y}\right)=\frac{\mathrm{y}-7}{10}\end{array}$

Q.57

$\begin{array}{l}\mathrm{Letf}:\mathrm{W}\to \mathrm{W}\mathrm{bede}\mathrm{fine}\mathrm{d}\mathrm{as}\mathrm{f}\left(\mathrm{n}\right)=\mathrm{n}-1,\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{odd}\mathrm{and}\\ \mathrm{f}\left(\mathrm{n}\right)=\mathrm{n}+1,\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{even}.\mathrm{Show}\mathrm{tha}\mathrm{tf}\mathrm{is}\mathrm{invertible}.\mathrm{Find}\\ \mathrm{the}\mathrm{inverse}\mathrm{off}.\mathrm{Here},\mathrm{W}\mathrm{is}\mathrm{the}\mathrm{set}\mathrm{of}\mathrm{all}\mathrm{whole}\mathrm{numbers}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{f}:\mathrm{W}\to \mathrm{W}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{n}\right)\mathrm{=}\{\begin{array}{l}\mathrm{n}-1,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{odd}\\ \mathrm{n}+1,\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{even}\end{array}\\ \mathrm{One}-\mathrm{one}:\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{f}\left(\mathrm{n}\right)=\mathrm{f}\left(\mathrm{m}\right)\\ \Rightarrow \mathrm{n}-1=\mathrm{m}+1[\mathrm{If}\mathrm{n}\mathrm{is}\mathrm{odd}\mathrm{and}\mathrm{m}\mathrm{is}\mathrm{even}\mathrm{.}]\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{n}-\mathrm{m}=2,\\ \mathrm{which}\mathrm{is}\mathrm{impossible}\mathrm{.}\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{f}\left(\mathrm{n}\right)=\mathrm{f}\left(\mathrm{m}\right)\\ \Rightarrow \mathrm{n}+1=\mathrm{m}-1[\mathrm{If}\mathrm{n}\mathrm{is}\mathrm{even}\mathrm{and}\mathrm{m}\mathrm{is}\mathrm{odd}\mathrm{.}]\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{n}-\mathrm{m}=-2,\\ \mathrm{which}\mathrm{is}\mathrm{also}\mathrm{impossible}\mathrm{.}\\ \mathrm{Now},\mathrm{if}\mathrm{both}\mathrm{n}\mathrm{and}\mathrm{m}\mathrm{are}\mathrm{odd},\mathrm{then}\mathrm{we}\mathrm{have}:\\ \mathrm{\hspace{0.17em}}\mathrm{f}\left(\mathrm{n}\right)=\mathrm{f}\left(\mathrm{m}\right)\\ \Rightarrow \mathrm{n}-1=\mathrm{m}-1\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{n}=\mathrm{m}\\ \mathrm{Again},\mathrm{if}\mathrm{both}\mathrm{n}\mathrm{and}\mathrm{m}\mathrm{are}\mathrm{even},\mathrm{then}\mathrm{we}\mathrm{have}:\\ \mathrm{f}(\mathrm{n})\; =\mathrm{f}(\mathrm{m})\\ \Rightarrow \mathrm{n}+\; 1=\mathrm{m}+\; 1\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{n}=\mathrm{m}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Onto}:\\ \mathrm{For}\mathrm{any}\mathrm{odd}\mathrm{number},\; 2\mathrm{r}+1\in \mathrm{N}\\ \mathrm{f}(2\mathrm{r}+1)=2\mathrm{r}+1-1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2\mathrm{r}\in \mathrm{N}\\ \mathrm{For}\mathrm{any}\mathrm{even}\mathrm{number},\; 2\mathrm{r}\in \mathrm{N}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{f}\left(2\mathrm{r}\right)=2\mathrm{r}+1\in \mathrm{N}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{function}\mathrm{.}\\ \mathrm{Therefore},\mathrm{f}\mathrm{is}\mathrm{invertible}\mathrm{function}\mathrm{.}\\ \mathrm{Let}\mathrm{us}\mathrm{define}\mathrm{g}:\mathrm{W}\to \mathrm{W}\mathrm{as}:\\ \mathrm{g}\left(\mathrm{m}\right)=\{\begin{array}{l}\mathrm{m}+1,\hspace{0.17em}\hspace{0.17em}\mathrm{if}\mathrm{m}\mathrm{is}\mathrm{even}\\ \mathrm{m}-1,\hspace{0.17em}\hspace{0.17em}\mathrm{if}\mathrm{m}\mathrm{is}\mathrm{odd}\end{array}\mathrm{}\\ \mathrm{Now},\mathrm{when}\mathrm{n}\mathrm{is}\mathrm{odd}\\ \mathrm{gof}\left(\mathrm{n}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{n}\right)\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{g}(\mathrm{n}-1)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{n}-1+1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{n}\\ \mathrm{And},\mathrm{when}\mathrm{n}\mathrm{is}\mathrm{even}:\\ \mathrm{gof}\left(\mathrm{n}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{n}\right)\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{g}(\mathrm{n}+1)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{n}+1-1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{n}\\ \mathrm{When}\mathrm{m}\mathrm{is}\mathrm{odd}:\\ \mathrm{gof}\left(\mathrm{m}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{m}\right)\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{g}(\mathrm{m}-1)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{m}-1+1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{m}\\ \mathrm{And},\mathrm{when}\mathrm{n}\mathrm{is}\mathrm{even}:\\ \mathrm{gof}\left(\mathrm{m}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{m}\right)\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{g}(\mathrm{m}+1)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{m}+1-1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{m}\\ \therefore \mathrm{gof}={\mathrm{I}}_{\mathrm{W}}\mathrm{\hspace{0.17em}}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{gof}={\mathrm{I}}_{\mathrm{W}}\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{invertible}\mathrm{and}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{given}\mathrm{by}{\mathrm{f}}^{–1}=\mathrm{g},\\ \mathrm{which}\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{f}.\mathrm{Hence},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{f}\mathrm{itself}\mathrm{.}\end{array}$

Q.58

${\text{If f : R → R is defined by f(x)=x}}^2–3\mathrm{x}+2,\mathrm{find}\mathrm{f}\left(\mathrm{f}\right(\mathrm{x}\left)\right).$

Ans.

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-3\mathrm{x}+2.\\ \mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{f}({\mathrm{x}}^2-3\mathrm{x}+2)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={({\mathrm{x}}^2-3\mathrm{x}+2)}^2-3({\mathrm{x}}^2-3\mathrm{x}+2)+2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{x}}^4+9{\mathrm{x}}^2+4-6{\mathrm{x}}^3-12\mathrm{x}+4{\mathrm{x}}^2-3{\mathrm{x}}^2+9\mathrm{x}-6+2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{x}}^4-6{\mathrm{x}}^3+10{\mathrm{x}}^2-3\mathrm{x}\end{array}$

Q.59

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{function}\mathrm{f}:\mathrm{R}\to \mathrm{\{}\mathrm{x}\in \mathrm{R}:–1<\mathrm{x}<1\}\mathrm{defined}\mathrm{by}\mathrm{}\\ \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{1+\left|\mathrm{x}\right|},\mathrm{x}\in \mathrm{R}\mathrm{}\mathrm{isone}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{function}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{f}:\mathrm{R}\to \{\mathrm{x}\in \mathrm{R}:–1<\mathrm{x}<1\}\mathrm{isdefinedasf}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{1+\left|\mathrm{x}\right|},\mathrm{x}\in \mathrm{R}.\\ \mathrm{Ler}\hspace{0.17em}\hspace{0.17em}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(\mathrm{y}),\mathrm{where}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{.}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{x}}{1+\left|\mathrm{x}\right|}=\frac{\mathrm{y}}{1+\left|\mathrm{y}\right|}\\ \mathrm{If}\mathrm{x}\mathrm{is}\mathrm{positive}\mathrm{and}\mathrm{y}\mathrm{is}\mathrm{negative},\mathrm{then}\mathrm{we}\mathrm{have}:\\ \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{\mathrm{x}}{1+\mathrm{x}}=\frac{\mathrm{y}}{1-\mathrm{y}}\\ \Rightarrow \mathrm{x}(1-\mathrm{y})=\mathrm{y}(1+\mathrm{x})\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}-\mathrm{xy}=\mathrm{y}+\mathrm{xy}\\ \Rightarrow \mathrm{x}-\mathrm{y}=2\mathrm{xy}\\ \mathrm{Since}\mathrm{x}\mathrm{is}\mathrm{positive}\mathrm{and}\mathrm{y}\mathrm{is}\mathrm{negative}:\\ \mathrm{x}>\mathrm{y}\Rightarrow \mathrm{x}-\mathrm{y}>0\\ \mathrm{But},\; 2\mathrm{xy}\mathrm{is}\mathrm{negative}\mathrm{.}\\ \mathrm{Then},2\mathrm{xy}\ne \mathrm{x}-\mathrm{y}\\ \mathrm{Thus},\mathrm{the}\mathrm{case}\mathrm{of}\mathrm{x}\mathrm{being}\mathrm{positive}\mathrm{and}\mathrm{y}\mathrm{being}\mathrm{negative}\mathrm{can}\mathrm{not}\\ \mathrm{be}\mathrm{considered}\mathrm{.}\\ \mathrm{Similarly},\mathrm{the}\mathrm{case}\mathrm{of}\mathrm{x}\mathrm{being}\mathrm{negative}\mathrm{and}\mathrm{y}\mathrm{being}\mathrm{positive}\\ \mathrm{also}\mathrm{can}\mathrm{not}\mathrm{be}\mathrm{considered}\mathrm{.}\\ \therefore \mathrm{x}\mathrm{and}\mathrm{y}\mathrm{have}\mathrm{to}\mathrm{be}\mathrm{either}\mathrm{positive}\mathrm{or}\mathrm{negative}\mathrm{.}\\ \mathrm{When}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{are}\mathrm{both}\mathrm{positive},\mathrm{we}\mathrm{have}:\\ \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}\frac{\mathrm{x}}{1+\mathrm{x}}=\frac{\mathrm{y}}{1+\mathrm{y}}\\ \Rightarrow \mathrm{x}+\mathrm{xy}=\mathrm{y}+\mathrm{xy}\\ \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{\hspace{0.17em}}\mathrm{x}=\mathrm{y}\\ \mathrm{When}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{are}\mathrm{both}\mathrm{negative},\mathrm{we}\mathrm{have}:\\ \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}\frac{\mathrm{x}}{1-\mathrm{x}}=\frac{\mathrm{y}}{1-\mathrm{y}}\\ \Rightarrow \mathrm{x}-\mathrm{xy}=\mathrm{y}-\mathrm{xy}\\ \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{\hspace{0.17em}}\mathrm{x}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{y}\in \mathrm{R}\mathrm{such}\mathrm{that}-1<\mathrm{y}<\; 1\mathrm{.}\\ \mathrm{If}\mathrm{y}\mathrm{is}\mathrm{negative},\mathrm{then}\mathrm{there}\mathrm{exists}\mathrm{x}=\frac{\mathrm{y}}{1+\mathrm{y}}\in \mathrm{R}\mathrm{such}\mathrm{that}\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{y}}{1+\mathrm{y}}\right)=\frac{\frac{\mathrm{y}}{1+\mathrm{y}}}{1+\left|\frac{\mathrm{y}}{1+\mathrm{y}}\right|}=\frac{\frac{\mathrm{y}}{1+\mathrm{y}}}{1+\frac{-\mathrm{y}}{1+\mathrm{y}}}=\mathrm{y}\\ \mathrm{If}\mathrm{y}\mathrm{is}\mathrm{positive},\mathrm{then}\mathrm{there}\mathrm{exists}\mathrm{x}=\frac{\mathrm{y}}{1+\mathrm{y}}\in \mathrm{R}\mathrm{\hspace{0.17em}}\mathrm{such}\mathrm{\hspace{0.17em}}\mathrm{that}\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{y}}{1-\mathrm{y}}\right)=\frac{\frac{\mathrm{y}}{1-\mathrm{y}}}{1+\left|\frac{\mathrm{y}}{1-\mathrm{y}}\right|}=\frac{\frac{\mathrm{y}}{1-\mathrm{y}}}{1+\frac{\mathrm{y}}{1-\mathrm{y}}}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{.}\end{array}$

Q.60

${\text{Show that the function f : R→ R given by f(x)=x}}^3\mathrm{is}\mathrm{injective}.$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{as}\mathrm{f}(\mathrm{x})\; ={\mathrm{x}}^{\mathrm{3}}\mathrm{.}\\ \mathrm{Suppose}\mathrm{f}(\mathrm{x})\; =\mathrm{f}(\mathrm{y}),\mathrm{where}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{.}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{x}}^{\mathrm{3}}={\mathrm{y}}^{\mathrm{3}}\mathrm{}...(1)\\ \mathrm{Now},\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{show}\mathrm{that}\mathrm{x}=\mathrm{y}\mathrm{.}\\ \mathrm{Suppose}\mathrm{x}\ne \mathrm{y},\mathrm{their}\mathrm{cubes}\mathrm{will}\mathrm{also}\mathrm{not}\mathrm{be}\mathrm{equal}\mathrm{.}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{x}}^{\mathrm{3}}\ne {\mathrm{y}}^{\mathrm{3}}\\ \mathrm{However},\mathrm{it}\mathrm{will}\mathrm{be}\mathrm{a}\mathrm{contradiction}\mathrm{to}\mathrm{equation}(1)\mathrm{.}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\mathrm{y}\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{injective}\mathrm{.}\end{array}$

Q.61

$\begin{array}{l}\mathrm{Give}\mathrm{examples}\mathrm{of}\mathrm{two}\mathrm{functions}\mathrm{f}:\mathrm{N}\to \mathrm{Z}\mathrm{and}\mathrm{g}:\mathrm{Z}\to \mathrm{Z}\mathrm{such}\\ \mathrm{that}\mathrm{g}\mathrm{\hspace{0.17em}}\mathrm{o}\mathrm{\hspace{0.17em}}\mathrm{f}\mathrm{is}\mathrm{injective}\mathrm{but}\mathrm{g}\mathrm{is}\mathrm{not}\mathrm{injective}.\\ (\mathrm{Hint}:\mathrm{Consider}\mathrm{f}(\mathrm{x})=\mathrm{x}\mathrm{and}\mathrm{g}(\mathrm{x})=\left|\mathrm{x}\right|)\end{array}$

Ans.

$\begin{array}{l}\mathrm{Define}\mathrm{f}:\mathrm{N}\to \mathrm{Z}\mathrm{as}\mathrm{f}(\mathrm{x})\; =\mathrm{x}\mathrm{and}\mathrm{g}:\mathrm{Z}\to \mathrm{Z}\mathrm{as}\mathrm{g}(\mathrm{x})\; =\left|\mathrm{x}\right|\\ \mathrm{Let}\mathrm{us}\mathrm{first}\mathrm{show}\mathrm{that}\mathrm{g}\mathrm{is}\mathrm{not}\mathrm{injective}\mathrm{.}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{observed}\mathrm{that}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{g}(-1)\; =|-1|=1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{g}(1)\; =\left|1\right|=1\\ \therefore \mathrm{g}(-1)\; =\mathrm{g}(1),\mathrm{but}-1\ne \mathrm{1}\\ \therefore \mathrm{g}\mathrm{is}\mathrm{not}\mathrm{injective}\mathrm{.}\\ \mathrm{Now},\mathrm{gof}:\mathrm{N}\to \mathrm{Z}\mathrm{is}\mathrm{defined}\mathrm{as}\\ \mathrm{gof}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{g}\left(\mathrm{x}\right)=\left|\mathrm{x}\right|\\ \mathrm{Let}\mathrm{x},\mathrm{y}\in \mathrm{N}\mathrm{such}\mathrm{that}\mathrm{gof}(\mathrm{x})\; =\mathrm{gof}(\mathrm{y})\mathrm{.}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}=\mathrm{y}\\ \mathrm{Since}\mathrm{x}\mathrm{and}\mathrm{y}\in \mathrm{N},\mathrm{both}\mathrm{are}\mathrm{positive}\\ \therefore \left|\mathrm{x}\right|=\left|\mathrm{y}\right|\Rightarrow \mathrm{x}=\mathrm{y}\\ \mathrm{Hence},\mathrm{}\mathrm{gof}\mathrm{is}\mathrm{injective}\mathrm{.}\end{array}$

Q.62

$\begin{array}{l}\mathrm{Given}\mathrm{example}\mathrm{s}\mathrm{of}\mathrm{two}\mathrm{functions}\mathrm{f}:\mathrm{N}\to \mathrm{N}\mathrm{and}\mathrm{g}:\mathrm{N}\to \mathrm{N}\mathrm{such}\\ \mathrm{that}\mathrm{go}\mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{but}\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}.\\ (\mathrm{Hint}:\mathrm{Consider}\mathrm{f}(\mathrm{x})=\mathrm{x}+1\mathrm{and}\mathrm{\hspace{0.17em}}\mathrm{g}\left(\mathrm{x}\right)=\{\begin{array}{l}\mathrm{x}–1,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{ifx}>1\\ \mathrm{x}+1,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{ifx}<1\end{array}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Define}\mathrm{f}:\mathrm{N}\to \mathrm{N}\mathrm{by},\hspace{0.17em}\hspace{0.17em}\mathrm{f}(\mathrm{x})\; =\mathrm{x}+\; 1\\ \mathrm{And},\mathrm{g}:\mathrm{N}\to \mathrm{N}\mathrm{by},\hspace{0.17em}\hspace{0.17em}\mathrm{g}\left(\mathrm{x}\right)=\{\begin{array}{l}\mathrm{x}-1,\hspace{0.17em}\hspace{0.17em}\mathrm{if}\mathrm{x}>1\\ 1,\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{if}\mathrm{x}=1\end{array}\\ \mathrm{We}\mathrm{first}\mathrm{show}\mathrm{that}\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}:\\ \mathrm{For}\mathrm{this},\mathrm{consider}\mathrm{element}1\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{N}.\mathrm{It}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{this}\\ \mathrm{element}\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{image}\mathrm{of}\mathrm{any}\mathrm{of}\mathrm{the}\mathrm{elements}\mathrm{in}\mathrm{domain}\mathrm{N}\mathrm{.}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Now},\mathrm{gof}:\mathrm{N}\to \mathrm{N}\mathrm{is}\mathrm{defined}\mathrm{by},\\ \mathrm{gof}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{g}(\mathrm{x}+1)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{x}+1-1\left[\mathrm{x}\in \mathrm{N}\Rightarrow (\mathrm{x}+1)>1\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{x}\\ \mathrm{Then},\mathrm{it}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{for}\mathrm{y}\in \mathrm{N},\mathrm{there}\mathrm{exists}\mathrm{x}=\mathrm{y}\in \mathrm{N}\mathrm{such}\mathrm{that}\\ \mathrm{gof}(\mathrm{x})\; =\mathrm{y}\mathrm{.}\\ \therefore \mathrm{\hspace{0.17em}}\mathrm{gof}\mathrm{is}\mathrm{onto}\mathrm{.}\end{array}$

Q.63

$\begin{array}{l}\mathrm{Given}\mathrm{an}\mathrm{on}\mathrm{empty}\mathrm{set}\mathrm{X},\mathrm{consider}\mathrm{P}\left(\mathrm{X}\right)\mathrm{which}\mathrm{is}\mathrm{the}\mathrm{set}\mathrm{ofall}\mathrm{sub}\mathrm{sets}\mathrm{of}\mathrm{X}.\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{Define}\mathrm{the}\\ \mathrm{relation}\mathrm{Rin}\mathrm{P}\left(\mathrm{X}\right)\mathrm{as}\mathrm{follows}:\mathrm{For}\mathrm{sub}\mathrm{sets}\mathrm{A},\mathrm{Bin}\mathrm{P}\left(\mathrm{X}\right),\mathrm{ARB}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}\mathrm{A}\subset \mathrm{B}.\mathrm{Is}\mathrm{Ran}\\ \mathrm{equivalence}\mathrm{relation}\mathrm{on}\mathrm{P}\left(\mathrm{X}\right)?\mathrm{Justify}\mathrm{you}\hspace{0.17em}\mathrm{answer}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Since}\mathrm{every}\mathrm{set}\mathrm{is}\mathrm{a}\mathrm{subset}\mathrm{of}\mathrm{itself},\mathrm{ARA}\mathrm{for}\mathrm{all}\mathrm{A}\in \mathrm{P}\left(\mathrm{X}\right)\mathrm{.}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{reflexive}.\\ \mathrm{Let}\mathrm{ARB}\Rightarrow \mathrm{A}\subset \mathrm{B}\mathrm{.}\\ \mathrm{For}\mathrm{instance},\mathrm{if}\mathrm{A}=\; \{3,\; 4\}\mathrm{and}\mathrm{B}=\; \{3,\; 4,\; 5\},\mathrm{then}\mathrm{it}\mathrm{cannot}\mathrm{be}\\ \mathrm{implied}\mathrm{that}\mathrm{B}\mathrm{is}\mathrm{related}\mathrm{to}\mathrm{A}\mathrm{.}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{not}\mathrm{symmetric}\mathrm{.}\\ \mathrm{Further},\mathrm{if}\mathrm{ARB}\mathrm{and}\mathrm{BRC},\mathrm{then}\mathrm{A}\subset \mathrm{B}\mathrm{and}\mathrm{B}\subset \mathrm{C}\mathrm{.}\\ \Rightarrow \mathrm{A}\subset \mathrm{C}\\ \Rightarrow \mathrm{ARC}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{transitive}\mathrm{.}\\ \mathrm{Hence},\mathrm{R}\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{equivalence}\mathrm{relation}\mathrm{since}\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{symmetric}\mathrm{.}\end{array}$

Q.64

$\begin{array}{l}\mathrm{Given}\mathrm{a}\mathrm{non}–\mathrm{empty}\mathrm{set}\mathrm{X},\mathrm{consider}\mathrm{the}\mathrm{binary}\mathrm{operation}*:\\ \mathrm{P}\left(\mathrm{X}\right)\times \mathrm{P}\left(\mathrm{X}\right)\to \mathrm{P}\left(\mathrm{X}\right)\mathrm{given}\mathrm{by}\mathrm{A}*\mathrm{B}=\mathrm{A}\cap \mathrm{B}\hspace{0.17em}\hspace{0.17em}\forall \mathrm{A},\mathrm{Bin}\mathrm{P}\left(\mathrm{X}\right),\\ \mathrm{where}\mathrm{P}\left(\mathrm{X}\right)\mathrm{is}\mathrm{the}\mathrm{power}\mathrm{set}\mathrm{of}\mathrm{X}.\mathrm{Show}\mathrm{that}\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{identity}\\ \mathrm{element}\mathrm{for}\mathrm{this}\mathrm{operation}\mathrm{and}\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{invertible}\mathrm{element}\\ \mathrm{in}\mathrm{P}\left(\mathrm{X}\right)\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{the}\mathrm{operation}*.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}\mathrm{that}:\\ \mathrm{P}(\mathrm{X})\times \mathrm{P}(\mathrm{X})\to \mathrm{P}(\mathrm{X})\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{A}*\mathrm{B}=\mathrm{A}\cap \mathrm{B}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\forall \mathrm{A},\mathrm{B}\in \mathrm{P}\left(\mathrm{X}\right)\\ \mathrm{Since},\mathrm{A}\cap \mathrm{X}=\mathrm{A}=\mathrm{X}\cap \mathrm{A}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\forall \mathrm{A},\mathrm{B}\in \mathrm{P}\left(\mathrm{X}\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{A}*\mathrm{X}=\mathrm{A}=\mathrm{X}*\mathrm{A}\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}\forall \mathrm{A},\mathrm{B}\in \mathrm{P}\left(\mathrm{X}\right)\\ \mathrm{Thus},\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{binary}\mathrm{operation}*\mathrm{.}\\ \mathrm{Now},\mathrm{an}\mathrm{element}\hspace{0.17em}\hspace{0.17em}\mathrm{A}\in \mathrm{P}\left(\mathrm{X}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{is}\mathrm{invertible}\mathrm{if}\mathrm{there}\mathrm{exists}\mathrm{B}\in \mathrm{P}\left(\mathrm{X}\right)\\ \mathrm{such}\mathrm{that}\\ \mathrm{A}*\mathrm{B}=\mathrm{X}=\mathrm{B}*\mathrm{A}[\mathrm{As}\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{.}]\\ \mathrm{i}.\mathrm{e}.,\hspace{0.17em}\hspace{0.17em}\mathrm{A}\cap \mathrm{B}=\mathrm{X}=\mathrm{B}\cap \mathrm{A}\\ \mathrm{This}\mathrm{case}\mathrm{is}\mathrm{possible}\mathrm{only}\mathrm{when}\mathrm{A}=\mathrm{X}=\mathrm{B}\mathrm{.}\\ \mathrm{Thus},\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{invertible}\mathrm{element}\mathrm{in}\mathrm{P}\left(\mathrm{X}\right)\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{the}\\ \mathrm{given}\mathrm{operation}*\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{result}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.65

Find the number of all onto functions from the set {1, 2, 3, …, n) to itself.

Ans.

\begin{array}{l}\text{Onto functions from the set {1, 2, 3,}\mathrm{\dots }\text{,n} to itself is simply a}\\ \text{permutation on n symbols 1, 2,}\mathrm{\dots }\text{, n}\text{.}\\ \text{Thus, the total number of onto maps from {1, 2,}\mathrm{\dots }\text{, n} to itself}\\ \text{is the same as}\text{the total number of permutations on n symbols}\\ \text{1, 2,}\mathrm{\dots }\text{, n, which is n}\text{.}\end{array}

Q.66

$\begin{array}{l}\mathrm{Let}\mathrm{S}=\{\mathrm{a},\mathrm{b},\mathrm{c}\}\mathrm{and}\mathrm{T}=\{1,2,3\}.\mathrm{Find}{\mathrm{F}}^{–1}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{functions}\mathrm{F}\mathrm{from}\mathrm{S}\mathrm{to}\mathrm{T},\mathrm{if}\mathrm{it}\mathrm{exists}.\\ \left(\mathrm{i}\right)\mathrm{F}=\left\{\right(\mathrm{a},3),(\mathrm{b},2),(\mathrm{c},1\left)\right\}\mathrm{\hspace{0.17em}}\left(\mathrm{ii}\right)\mathrm{F}=\left\{\right(\mathrm{a},2),(\mathrm{b},1),(\mathrm{c},1\left)\right\}\end{array}$

Ans.

$\begin{array}{l}Given:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S=\; \{a,b,c\},T=\; \{1,\; 2,\; 3\}\\ \left(i\right)F:S\to Tisdefinedas:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}F=\; \{(a,\; 3),\; (b,\; 2),\; (c,\; 1)\}\\ \Rightarrow F(a)\; =\; 3,F(b)\; =\; 2,F(c)\; =\; 1\\ Therefore,F^{-1}:T\to Sisgivenby\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{F}^{-1}\hspace{0.17em}=\{(3,a),\; (2,b),\; (1,c)\}.\\ \left(ii\right)F:S\to Tisdefinedas:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}F=\; \{(a,\; 2),\; (b,\; 1),\; (c,\; 1)\}\\ SinceF(b)\; =F(c)\; =\; 1,Fisnotone–one.\\ Therefore,Fisnotinvertiblei.e.,F^{–1}doesnotexist.\end{array}$

Q.67

$\begin{array}{l}\mathrm{Consider}\mathrm{the}\mathrm{binary}\mathrm{operations}*:\mathrm{R}\times \mathrm{R}\to \mathrm{ando}:\mathrm{R}\times \mathrm{R}\to \mathrm{R}\\ \mathrm{defined}\mathrm{as}\mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{b}=|\mathrm{a}-\mathrm{b}|\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{aob}=\mathrm{a},\forall \mathrm{a},\mathrm{b}\in \mathrm{R}.\mathrm{Show}\mathrm{that}*\mathrm{is}\\ \mathrm{commutative}\mathrm{but}\mathrm{notassociative},\mathrm{o}\mathrm{is}\mathrm{associative}\mathrm{but}\mathrm{not}\\ \mathrm{commutative}.\mathrm{Further},\mathrm{show}\mathrm{that}\hspace{0.17em}\hspace{0.17em}\forall \mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{R},\\ \mathrm{a}*\left(\mathrm{boc}\right)=(\mathrm{a}*\mathrm{b})\mathrm{o}(\mathrm{a}*\mathrm{c}).[\mathrm{If}\mathrm{it}\mathrm{is}\mathrm{so},\mathrm{we}\mathrm{say}\mathrm{that}\mathrm{the}\\ \mathrm{operation}*\mathrm{distributes}\mathrm{over}\mathrm{the}\mathrm{operationo}].\\ \mathrm{Does}\mathrm{odistributeover}*?\mathrm{Justif}\mathrm{y}\mathrm{your}\mathrm{answer}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}*:\mathrm{R}\times \mathrm{R}\to \mathrm{and}\mathrm{o}:\mathrm{R}\times \mathrm{R}\to \mathrm{R}\mathrm{isdefined}\mathrm{as}\\ \mathrm{a}*\mathrm{b}=|\mathrm{a}-\mathrm{b}|\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{\hspace{0.17em}}\mathrm{aob}=\mathrm{a}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\forall \mathrm{a},\mathrm{b}\in \mathrm{R}\mathrm{.}\\ \mathrm{For}\mathrm{a},\mathrm{b}\in \mathrm{R},\mathrm{we}\mathrm{have}:\\ \mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{b}=|\mathrm{a}-\mathrm{b}|\\ \mathrm{b}*\mathrm{a}=|\mathrm{b}-\mathrm{a}|=|-(\mathrm{a}-\mathrm{b})|=|\mathrm{a}-\mathrm{b}|\\ \therefore \mathrm{a}*\mathrm{b}=\mathrm{b}*\mathrm{a}\\ \therefore \mathrm{The}\mathrm{operation}*\mathrm{is}\mathrm{commutative}\mathrm{.}\\ \mathrm{For}\mathrm{associative},\mathrm{we}\mathrm{see}\mathrm{that}\\ (2*3)*4=\left(|2-3|\right)*4\\ =1*4\\ =|1-4|=3\\ \mathrm{\hspace{0.17em}}2*(3*4)=2*|3-4|\\ \hspace{0.17em}\hspace{0.17em}=2*1\\ \hspace{0.17em}\hspace{0.17em}=|2-1|=1\\ (2*3)*4\ne \mathrm{\hspace{0.17em}}2*(3*4)[\mathrm{where}\hspace{0.17em}\hspace{0.17em}2,3,4\in \mathrm{R}]\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \mathrm{Now},\mathrm{consider}\mathrm{the}\mathrm{operation}\mathrm{o}\mathrm{:}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{noticed}\mathrm{that}1\mathrm{o}2=\; 1\mathrm{and}2\mathrm{o}1=\; 2\mathrm{.}\\ \therefore \mathrm{1}\mathrm{o}\mathrm{2}\ne \mathrm{2}\mathrm{o}1(\mathrm{where}1,\; 2\in \mathrm{R})\\ \therefore \mathrm{The}\mathrm{operation}\mathrm{o}\mathrm{is}\mathrm{not}\mathrm{commutative}\mathrm{.}\\ \mathrm{Let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{R}.\mathrm{Then},\mathrm{we}\mathrm{have}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{aob}\right)\mathrm{oc}=\mathrm{aoc}=\mathrm{a}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ao}\left(\mathrm{boc}\right)=\mathrm{aob}=\mathrm{a}\\ \Rightarrow \left(\mathrm{aob}\right)\mathrm{oc}=\mathrm{ao}\left(\mathrm{boc}\right)\\ \therefore \mathrm{The}\mathrm{operation}\mathrm{o}\mathrm{is}\mathrm{associative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{R},\mathrm{then}\mathrm{we}\mathrm{have}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}*\left(\mathrm{boc}\right)=\mathrm{a}*\mathrm{b}=|\mathrm{a}-\mathrm{b}|\\ (\mathrm{a}*\mathrm{b})\mathrm{o}(\mathrm{a}*\mathrm{c})=\left(|\mathrm{a}-\mathrm{b}|\right)\mathrm{o}\left(|\mathrm{a}-\mathrm{c}|\right)=|\mathrm{a}-\mathrm{b}|\\ \therefore \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}*\left(\mathrm{boc}\right)=(\mathrm{a}*\mathrm{b})\mathrm{o}(\mathrm{a}*\mathrm{c})\\ \mathrm{Now},\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}2\mathrm{o}(3*4)=2\mathrm{o}\left(|3-4|\right)=2\mathrm{o}1=2\\ \mathrm{and}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(2*3)\mathrm{o}(2*4)=\left(|2-3|\right)\mathrm{o}\left(|2-4|\right)=1\\ \therefore \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}\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{2}\mathrm{o}(3*4)\ne (2*3)\mathrm{o}(2*4)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(\mathrm{where}1,\; 2,\; 3\in \mathrm{R})\\ \therefore \mathrm{The}\mathrm{operation}\mathrm{o}\mathrm{does}\mathrm{not}\mathrm{distribute}\mathrm{over}*\mathrm{.}\end{array}$

Q.68

$\begin{array}{l}\mathrm{Given}\mathrm{a}\mathrm{non}–\mathrm{empty}\mathrm{set}\mathrm{X},\mathrm{let}*:\mathrm{P}\left(\mathrm{X}\right)\times \mathrm{P}\left(\mathrm{X}\right)\to \mathrm{P}\left(\mathrm{X}\right)\mathrm{be}\mathrm{defined}\mathrm{as}\\ \mathrm{A}*\mathrm{B}=(\mathrm{A}-\mathrm{B})\cup (\mathrm{B}-\mathrm{A}),\forall \mathrm{A},\mathrm{B}\in \mathrm{P}\left(\mathrm{X}\right).\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{empty}\\ \mathrm{se}\mathrm{t}\varnothing \mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{for}\mathrm{the}\mathrm{operation}\mathrm{\hspace{0.17em}}*\mathrm{and}\mathrm{all}\mathrm{thee}\mathrm{lements}\mathrm{A}\mathrm{of}\\ \mathrm{P}\left(\mathrm{X}\right)\mathrm{are}\mathrm{invertible}\mathrm{with}{\mathrm{A}}^{–1}=\mathrm{A}.(\mathrm{Hint}:(\mathrm{A}-\varnothing )\cup (\varnothing –\mathrm{A})=\mathrm{A}\\ \mathrm{and}(\mathrm{A}–\mathrm{A})\cup (\mathrm{A}–\mathrm{A})=\mathrm{A}*\mathrm{A}=\varnothing ).\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}*:\mathrm{P}\left(\mathrm{X}\right)\times \mathrm{P}\left(\mathrm{X}\right)\to \mathrm{P}\left(\mathrm{X}\right)\mathrm{is}\mathrm{defined}\mathrm{as}\\ \mathrm{A}*\mathrm{B}=\; (\mathrm{A}-\mathrm{B})\cup (\mathrm{B}-\mathrm{A})\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\forall \mathrm{A},\mathrm{B}\in \mathrm{P}(\mathrm{X})\mathrm{.}\\ \mathrm{Let}\mathrm{A}\in \mathrm{P}\left(\mathrm{X}\right).\mathrm{Then},\mathrm{we}\mathrm{have}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{A}*\varnothing =(\mathrm{A}-\varnothing \mathrm{)}\cup \mathrm{(}\varnothing -\mathrm{A})=\mathrm{A}\cup \varnothing =\mathrm{A}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\varnothing *\mathrm{A}=\mathrm{(}\varnothing -\mathrm{A})\cup (\mathrm{A}-\varnothing \mathrm{)}=\varnothing \cup \mathrm{A}=\mathrm{A}\\ \therefore \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{A}*\varnothing =\mathrm{A}=\varnothing *\mathrm{A}.\hspace{0.17em}\hspace{0.17em}\forall \mathrm{A}\in \mathrm{P}\left(\mathrm{X}\right)\\ \mathrm{Thus},\varnothing \mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{operation}*\mathrm{.}\\ \mathrm{Now},\mathrm{an}\mathrm{element}\mathrm{A}\in \mathrm{P}\left(\mathrm{X}\right)\mathrm{will}\mathrm{be}\mathrm{invertible}\mathrm{if}\mathrm{there}\mathrm{exists}\\ \mathrm{B}\in \mathrm{P}\left(\mathrm{X}\right)\mathrm{such}\mathrm{that}\\ \mathrm{A}*\mathrm{B}=\varnothing =\mathrm{B}*\mathrm{A}.(\mathrm{As}\varnothing \mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{element})\\ \mathrm{Now},\mathrm{we}\mathrm{observed}\mathrm{that}\\ \mathrm{\hspace{0.17em}}\mathrm{A}*\mathrm{A}=(\mathrm{A}-\mathrm{A})\cup (\mathrm{A}-\mathrm{A})=\varnothing \cup \varnothing =\varnothing \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\forall \mathrm{A}\in \mathrm{P}\left(\mathrm{X}\right).\\ \mathrm{Hence},\mathrm{all}\mathrm{the}\mathrm{elements}\mathrm{A}\mathrm{of}\mathrm{P}\left(\mathrm{X}\right)\mathrm{are}\mathrm{invertible}\mathrm{with}{\mathrm{A}}^{–1}=\mathrm{A}\mathrm{.}\\ \end{array}$

Q.69

$\begin{array}{l}\mathrm{Define}\mathrm{a}\mathrm{binary}\mathrm{operation}*\mathrm{on}\mathrm{the}\mathrm{set}\{0,1,2,3,4,5\}\hspace{0.17em}\hspace{0.17em}\mathrm{as}\\ \mathrm{a}*\mathrm{b}=\{\begin{array}{l}\mathrm{a}+\mathrm{b},\hspace{0.17em}\hspace{0.17em}\mathrm{ifa}+\mathrm{b}<\mathrm{\hspace{0.17em}}6\\ \mathrm{a}+\mathrm{b}–6,\hspace{0.17em}\hspace{0.17em}\mathrm{ifa}+\mathrm{b}\ge 6\end{array}\\ \mathrm{Show}\mathrm{that}\mathrm{zero}\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{for}\mathrm{this}\mathrm{operation}\mathrm{and}\mathrm{each}\\ \mathrm{element}\mathrm{a}\ne 0\mathrm{of}\mathrm{these}\mathrm{t}\mathrm{is}\mathrm{invertible}\mathrm{with}6–\mathrm{a}\mathrm{being}\mathrm{the}\\ \mathrm{inverse}\mathrm{of}\mathrm{a}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{X}=\; \{0,\; 1,\; 2,\; 3,\; 4,\; 5\}\mathrm{.}\\ \mathrm{The}\mathrm{operation}*\mathrm{on}\mathrm{X}\mathrm{is}\mathrm{defined}\mathrm{as}:\\ \mathrm{a}*\mathrm{b}=\{\begin{array}{l}\mathrm{a}+\mathrm{b},\hspace{0.17em}\hspace{0.17em}\mathrm{if}\mathrm{a}+\mathrm{b}<6\\ \mathrm{a}+\mathrm{b}-6,\hspace{0.17em}\hspace{0.17em}\mathrm{if}\mathrm{a}+\mathrm{b}\ge \mathrm{6}\end{array}\\ \mathrm{An}\mathrm{element}\mathrm{e}\in \mathrm{X}\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{for}\mathrm{the}\mathrm{operation}*,\\ \mathrm{if}\mathrm{\hspace{0.17em}}\mathrm{a}*\mathrm{e}=\mathrm{a}=\mathrm{e}*\mathrm{a}\forall \mathrm{a}\in \mathrm{X}.\\ \mathrm{For}\mathrm{a}\in \mathrm{X},\mathrm{we}\mathrm{see}\mathrm{that}:\\ \mathrm{a}*0=\mathrm{a}+0=\mathrm{a}\left[\mathrm{a}\in \mathrm{X}\Rightarrow \mathrm{a}+0<6\right]\\ 0*\mathrm{a}=0+\mathrm{a}=\mathrm{a}\left[\mathrm{a}\in \mathrm{X}\Rightarrow 0+\mathrm{a}\ge 6\right]\\ \therefore \mathrm{a}*0=0*\mathrm{a}\forall \mathrm{a}\in \mathrm{X}\\ \mathrm{Thus},\; 0\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{operation}*\mathrm{.}\\ \mathrm{An}\mathrm{element}\mathrm{a}\in \mathrm{X}\mathrm{is}\mathrm{invertible}\mathrm{if}\mathrm{there}\mathrm{exists}\mathrm{b}\in \mathrm{X}\mathrm{such}\mathrm{that}\\ \mathrm{a}*\mathrm{b}=\; 0\; =\mathrm{b}*\mathrm{a}\mathrm{.}\\ \mathrm{i}.\mathrm{e}.,\{\begin{array}{l}\mathrm{a}+\mathrm{b}=0=\mathrm{b}+\mathrm{a},\mathrm{if}\mathrm{\hspace{0.17em}}\mathrm{a}+\mathrm{b}<6\\ \mathrm{a}+\mathrm{b}-6=0=\mathrm{b}+\mathrm{a}-6,\mathrm{if}\mathrm{\hspace{0.17em}}\mathrm{a}+\mathrm{b}\le 6\end{array}\\ \Rightarrow \mathrm{a}=-\mathrm{b}\mathrm{or}\mathrm{b}=6-\mathrm{a}\\ \mathrm{But},\mathrm{X}=\; \{0,\; 1,\; 2,\; 3,\; 4,\; 5\}\mathrm{and}\mathrm{a},\mathrm{b}\in \mathrm{X}.\mathrm{Then},\mathrm{a}\ne -\mathrm{\hspace{0.17em}}\mathrm{b}\mathrm{.}\\ \therefore \mathrm{\hspace{0.17em}}\mathrm{b}=\; 6-\mathrm{a}\mathrm{is}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{a}\forall \mathrm{a}\in \mathrm{X}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{an}\mathrm{element}\mathrm{a}\in \mathrm{X},\mathrm{a}\ne 0\mathrm{is}6-\mathrm{a}\\ \mathrm{i}.\mathrm{e}.,{\mathrm{a}}^{-\mathrm{1}}=\; 6-\mathrm{a}\mathrm{.}\end{array}$

Q.70

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\{–1,0,1,2\},\mathrm{B}=\{-4,-2,0,2\}\mathrm{and}\mathrm{f},\mathrm{g}:\mathrm{A}\to \mathrm{B}\mathrm{be}\\ \mathrm{functions}\mathrm{defined}\mathrm{by}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-\mathrm{x},\mathrm{x}\in \mathrm{A}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\\ \mathrm{g}\left(\mathrm{x}\right)=2|\mathrm{x}-\frac{1}{2}|-1,\mathrm{\hspace{0.17em}}\mathrm{x}\in \mathrm{A}.\mathrm{Are}\mathrm{f}\mathrm{and}\mathrm{g}\mathrm{equal}?\\ \mathrm{Justify}\mathrm{your}\mathrm{answer}.\\ (\mathrm{Hint}:\mathrm{One}\mathrm{may}\mathrm{note}\mathrm{that}\mathrm{two}\mathrm{function}\mathrm{f}:\mathrm{A}\to \mathrm{B}\mathrm{and}\mathrm{g}:\mathrm{A}\to \mathrm{B}\\ \mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{a}\right)=\mathrm{g}\left(\mathrm{a}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\forall \mathrm{a}\in \mathrm{A},\mathrm{are}\mathrm{called}\mathrm{equal}\mathrm{functions}).\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{A}=\{-1,\hspace{0.17em}\hspace{0.17em}0,\hspace{0.17em}\hspace{0.17em}1,\hspace{0.17em}\hspace{0.17em}2\},\mathrm{}\mathrm{B}=\{-\mathrm{\hspace{0.17em}}4,-2,\mathrm{}0,\mathrm{}2\}\mathrm{.}\\ \mathrm{Also},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{f},\mathrm{g}:\mathrm{A}\to \mathrm{B}\hspace{0.17em}\hspace{0.17em}\mathrm{are}\mathrm{defined}\mathrm{by}\\ \mathrm{f}(\mathrm{x})\; ={\mathrm{x}}^{\mathrm{2}}-\mathrm{x},\mathrm{x}\in \mathrm{A}\mathrm{and}\mathrm{g}\left(\mathrm{x}\right)=2|\mathrm{x}-\frac{1}{2}|-1,\mathrm{\hspace{0.17em}}\mathrm{x}\in \mathrm{A}.\\ \mathrm{It}\mathrm{is}\mathrm{observed}\mathrm{that}:\\ \mathrm{f}(-1)={(-1)}^{\mathrm{2}}-(-1)=1+1=2\\ \mathrm{g}(-1)=2|(-1)-\frac{1}{2}|-1=2\left(\frac{3}{2}\right)-1\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=3-1=2\Rightarrow \mathrm{f}(-1)=\mathrm{g}(-1)\\ \mathrm{f}\left(0\right)={\left(0\right)}^2-\left(0\right)=0\\ \mathrm{g}\left(0\right)=2|0-\frac{1}{2}|-1\\ =1-1=0\\ \Rightarrow \mathrm{f}\left(0\right)=\mathrm{g}\left(0\right)\\ \mathrm{f}\left(1\right)={\left(1\right)}^2-\left(1\right)=0\\ \mathrm{g}\left(1\right)=2|1-\frac{1}{2}|-1\\ =1-1=0\Rightarrow \mathrm{f}\left(1\right)=\mathrm{g}\left(1\right)\\ \mathrm{f}\left(2\right)={\left(2\right)}^2-\left(2\right)=2\\ \mathrm{g}\left(2\right)=2|2-\frac{1}{2}|-1\\ =3-1=2\\ \Rightarrow \mathrm{f}\left(2\right)=\mathrm{g}\left(2\right)\\ \therefore \mathrm{f}\left(\mathrm{a}\right)=\mathrm{g}\left(\mathrm{a}\right)\forall \mathrm{a}\in \mathrm{A}\\ \mathrm{Hence},\mathrm{the}\mathrm{functions}\mathrm{f}\mathrm{and}\mathrm{g}\mathrm{are}\mathrm{equal}\mathrm{.}\end{array}$

Q.71

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\{1,2,3\}.\mathrm{Then}\mathrm{number}\mathrm{of}\mathrm{relations}\mathrm{containing}(1,2)\\ \mathrm{and}(1,3)\mathrm{which}\mathrm{are}\mathrm{reflexive}\mathrm{and}\mathrm{\hspace{0.17em}}\mathrm{symmetric}\mathrm{but}\mathrm{not}\\ \mathrm{transitive}\mathrm{is}\\ \left(\mathrm{A}\right)1\left(\mathrm{B}\right)2\left(\mathrm{C}\right)3\left(\mathrm{D}\right)4\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{set}\mathrm{is}\mathrm{A}=\; \{1,\; 2,\; 3\}\mathrm{.}\\ \mathrm{The}\mathrm{smallest}\mathrm{relation}\mathrm{containing}(1,\; 2)\mathrm{and}(1,\; 3)\mathrm{which}\mathrm{\hspace{0.17em}}\mathrm{is}\\ \mathrm{reflexive}\mathrm{and}\mathrm{symmetric},\mathrm{but}\mathrm{not}\mathrm{transitive}\mathrm{is}\mathrm{given}\mathrm{by}:\\ \mathrm{R}=\; \{(1,\; 1),\; (2,\; 2),\; (3,\; 3),\; (1,\; 2),\; (1,\; 3),\; (2,\; 1),\; (3,\; 1)\}\\ \mathrm{This}\mathrm{is}\mathrm{because}\mathrm{relation}\mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{as}(1,\; 1),\; (2,\; 2),\; (3,\; 3)\in \mathrm{R}\mathrm{.}\\ \mathrm{Relation}\mathrm{R}\mathrm{is}\mathrm{symmetric}\mathrm{since}(1,\; 2),\; (2,\; 1)\in \mathrm{R}\mathrm{and}\\ (1,\; 3),\; (3,\; 1)\in \mathrm{R}.\mathrm{But}\mathrm{relation}\mathrm{R}\mathrm{is}\mathrm{not}\mathrm{transitive}\mathrm{as}\\ (3,\; 1),\; (1,\; 2)\in \mathrm{R},\mathrm{but}(3,\; 2)\notin \mathrm{R}\mathrm{.}\\ \mathrm{Now},\mathrm{if}\mathrm{we}\mathrm{add}\mathrm{any}\mathrm{two}\mathrm{pairs}(3,\; 2)\mathrm{and}(2,\; 3)\; (\mathrm{or}\mathrm{both})\mathrm{to}\\ \mathrm{relation}\mathrm{R},\mathrm{then}\mathrm{relation}\mathrm{R}\mathrm{will}\mathrm{become}\hspace{0.17em}\hspace{0.17em}\mathrm{transitive}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{desired}\mathrm{relations}\mathrm{is}\mathrm{one}\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$

Q.72 Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is

(A) 1 (B) 2 (C) 3 (D) 4

Ans.

\begin{array}{l}\text{It is given that A = {1, 2, 3}}\text{.}\\ \text{The smallest equivalence relation containing (1, 2) is given by,}\\ {\text{R}}_{\text{1}}\text{= {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}}\\ \text{Now, we are left with only four pairs i}\text{.e}\text{., (2, 3), (3, 2), (1, 3),}\\ \text{and (3, 1)}\text{.}\\ {\text{If we odd any one pair [say (2, 3)] to R}}_{\text{1}}\text{, then for symmetry}\\ \text{we must add (3, 2)}\text{. Also, for transitivity we are required to}\\ \text{add (1, 3) and (3, 1)}\text{.}\\ {\text{Hence, the only equivalence relation (bigger than R}}_{\text{1}}\text{) is the}\\ \text{universal relation}\text{.This shows that the total number of}\\ \text{equivalence relations containing (1, 2) is two}\text{.}\\ \text{The correct answer is}B\text{.}\end{array}

Q.73

$\begin{array}{l}\mathrm{Letf}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{the}\mathrm{Signum}\mathrm{Function}\mathrm{defined}\mathrm{as}\\ \mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}1,\hspace{0.17em}\hspace{0.17em}\mathrm{ifx}>0\\ 0,\mathrm{if}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=0\\ –1,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ifx}<0\end{array}\\ \mathrm{Also},\mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{the}\mathrm{Greatest}\mathrm{Integer}\mathrm{Function}\mathrm{given}\mathrm{by}\\ \mathrm{g}\left(\mathrm{x}\right)=\left[\mathrm{x}\right],\mathrm{where}\left[\mathrm{x}\right]\mathrm{is}\mathrm{great}\mathrm{est}\mathrm{integer}\mathrm{\hspace{0.17em}}\mathrm{less}\mathrm{than}\mathrm{o}\mathrm{requal}\\ \mathrm{tox}.\mathrm{Then}\mathrm{does}\mathrm{fog}\mathrm{and}\mathrm{go}\mathrm{f}\mathrm{co}\mathrm{incide}\mathrm{in}(0,1]?\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}:\mathrm{\hspace{0.17em}}\mathrm{f}:\mathrm{R}\to \mathrm{R}\\ \mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}1,\hspace{0.17em}\hspace{0.17em}\mathrm{if}\mathrm{x}>0\\ 0,\hspace{0.17em}\hspace{0.17em}\mathrm{if}​\hspace{0.17em}\mathrm{x}=0\\ -1,\hspace{0.17em}\hspace{0.17em}\mathrm{if}\mathrm{x}<1\end{array}\\ \mathrm{Also},\mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{g}(\mathrm{x})\; =\; [\mathrm{x}],\mathrm{where}[\mathrm{x}]\mathrm{is}\mathrm{the}\mathrm{greatest}\\ \mathrm{integer}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}\mathrm{x}.\mathrm{Now},\mathrm{let}\mathrm{x}\in (0,\; 1]\mathrm{.}\\ \mathrm{Then},\mathrm{we}\mathrm{have}:\\ [\mathrm{x}]\; =\; 1\mathrm{if}\mathrm{x}=\; 1\mathrm{and}[\mathrm{x}]\; =\; 0\mathrm{if}0\; <\mathrm{x}<\; 1\mathrm{.}\\ \therefore \mathrm{fog}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)=\mathrm{f}\left(\left[\mathrm{x}\right]\right)=\{\begin{array}{l}\mathrm{f}\left(1\right),\hspace{0.17em}\hspace{0.17em}\mathrm{if}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=1\\ \mathrm{f}\left(0\right),\hspace{0.17em}\hspace{0.17em}\mathrm{if}\mathrm{\hspace{0.17em}}\mathrm{x}\in (0,1)\end{array}=\{\begin{array}{l}1,\hspace{0.17em}\hspace{0.17em}\mathrm{if}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=1\\ 0,\hspace{0.17em}\hspace{0.17em}\mathrm{if}\mathrm{\hspace{0.17em}}\mathrm{x}\in (0,1)\end{array}\\ \mathrm{gof}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{g}\left(1\right)[\mathrm{x}>0]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left[1\right]=1\\ \mathrm{gof}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{g}\left(1\right)[\mathrm{x}>0]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left[1\right]=1\\ \mathrm{Thus},\mathrm{when}\mathrm{x}\in (0,1),\mathrm{we}\mathrm{have}\mathrm{fog}\left(\mathrm{x}\right)=0\mathrm{and}\mathrm{gof}\left(\mathrm{x}\right)=1.\\ \mathrm{Hence},\mathrm{fog}\mathrm{and}\mathrm{gof}\mathrm{do}\mathrm{not}\mathrm{coincide}\mathrm{in}(0,\; 1]\mathrm{.}\end{array}$

Q.74 Number of binary operations on the set {a, b} are

(A) 10 (B) 16 (C) 20 (D) 8

Ans.

$\begin{array}{l}\mathrm{A}\mathrm{binary}\mathrm{operation}*\mathrm{on}\{\mathrm{a},\mathrm{b}\}\mathrm{is}\mathrm{a}\mathrm{function}\mathrm{from}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\{\mathrm{a},\mathrm{b}\}\times \{\mathrm{a},\mathrm{b}\}\to \{\mathrm{a},\mathrm{b}\}\\ \mathrm{i}.\mathrm{e}.,\; *\mathrm{is}\mathrm{a}\mathrm{function}\mathrm{from}\{(\mathrm{a},\mathrm{a}),\; (\mathrm{a},\mathrm{b}),\; (\mathrm{b},\mathrm{a}),\; (\mathrm{b},\mathrm{b})\}\to \{\mathrm{a},\mathrm{b}\}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{binary}\mathrm{operations}\mathrm{on}\mathrm{the}\mathrm{set}\{\mathrm{a},\mathrm{b}\}\\ \mathrm{is}{2}^{\mathrm{4}}\mathrm{i}.\mathrm{e}.,\; 16\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{B}\mathrm{.}\end{array}$