NCERT Exemplar Class 12 Maths Solutions Chapter 1 Relations and Functions

NCERT Exemplar Class 12 Maths Solutions Chapter 1

Question:1

Let A = {a, b, c} and the relation R be defined on A as follows:

R = {(a, a), (b, c), (a, b)}.

Then, write the minimum number of ordered pairs to be added in R to make R reflexive and transitive.

Answer:

Here, R = {(a, a), (b, c), (a, b)}

The minimum number of ordered pairs to be added to make R as reflexive is (b, b) and (c, c) to R. Whereas, to make R transitive, the minimum number of ordered pairs to be added is (a, c) to R.

Therefore, we need 3 ordered pairs to add to R to make it reflexive and transitive.

Question:2

Let D be the domain of the real-valued function f defined by $f(x)=\sqrt{25-x^2}$ . Then, write D.

Answer:

$f(x)=\sqrt{25-x^2}$
Therefore, the function can be defined as:
$25-x^2\ge 0$

$$\begin{gathered} or,x^2\le 25 \\ -5\le x\le 5 \end{gathered}$$

Therefore, the given function has a domain of [-5, 5].

Question:3

Let f, g:$R\to R$ be defined by f(x) = 2x + 1 and $g(x)=x^2-2,\mathrm{\forall }x\in R,$ respectively. Then, find gof.

Answer:

Here,

f(x) = 2x + 1
$g(x)=x^2-2,\mathrm{\forall }x\in R$
Therefore,

$gof=g(f(x))=g(2x+1)$

$=(2x+1{)}^2-2$

$=4x^2+4x+1-2$

$=4x^2+4x-1$

Question:4

. Let f: $R\to R$ be the function defined by $f(x)=2x-3,\mathrm{\forall }x\in R$. write $f^{-1}$.

Answer:

Here, $f(x)=2x-3,\mathrm{\forall }x\in R$
Let’s say,
$y=2x-3$

$or,x=(y+3)/2$

Therefore,
$f^{-1}(x)=(x+3)/2$

Question:5

If $A=\{a,b,c,d\}$ and the function $f=\{(a,b),(b,d),(c,a),(d,c)\}$ , write $f^{-1}$.

Answer:
Here, $A=\{a,b,c,d\}$ and the function $f=\{(a,b),(b,d),(c,a),(d,c)\}$
Therefore, $f^{-1}=\{(b,a),(d,b),(a,c),(c,d)\}$

Question:6

If $f:R\to R$ is defined by $f(x)=x^2-3x+2$, write $f(f(x))$.

Answer:

Here, $f(x)=x^2-3x+2$

Therefore,
$$\begin{gathered} f(f(x))=f(x^2-3x+2) \\ =(x^2-3x+2{)}^2-3(x^2-3x+2)+2, \end{gathered}$$

$=x^4+9x^2+4-6x^3+4x^2-12x-3x^2+9x-6+2$

$=x^4-6x^3+10x^2-3x$
Similarly,
$f(f(x))=x^4-6x^3+10x^2-3x$

Question:7

Is $g=\{(1,1),(2,3),(3,5),(4,7)\}$ a function? If g is described by $g(x)=\alpha x+\beta$ , then what value should be assigned to $\alpha and\beta$ .

Answer:

Here,$g=\{(1,1),(2,3),(3,5),(4,7)\}$
Here, each and every element of a domain has a unique image. Therefore, g is a function.
Also, $g(x)=\alpha x+\beta$
Therefore,
$$\begin{gathered} g(1)=\alpha (1)+\beta =1 \\ \alpha +\beta =1\dots \dots ..(1) \end{gathered}$$
Similarly,
$g(2)=\alpha (2)+\beta =2\alpha +\beta =3\dots \dots ..(2)$
By solving (1) and (2), we get
$\alpha =2and\beta =-1$
Therefore, $g(x)=2x-1$

Question:8

Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.
(i) $\{(x,y):\text{x is a person, y is the mother of x}\}.$
(ii) $\{(a,b):\text{a is a person, b is an ancestor of a}\}.$

Answer:
(i)Here, $\{(x,y):\text{x is a person, y is the mother of x}\}.$
Therefore, each person (x) has only one biological mother.
Hence, the given set of ordered pairs makes a function.
Therefore, there are more than one person who may have the same mother. Hence, the function is many-one and surjective.
(ii) Here, $\{(a,b):\text{a is a person, b is an ancestor of a}\}.$
It’s seen that any person ‘a’ has more than one ancestor.
Therefore, it is not a function.

Question:9

If the mappings f and g are given by $f=\{(1,2),(3,5),(4,1)\}$ and $g=\{(2,3),(5,1),(1,3)\}$ , write $fog.$

Answer:
Here,

$f=\{(1,2),(3,5),(4,1)\}$ and $g=\{(2,3),(5,1),(1,3)\}$
Therefore,
$fog(2)=f(g(2))=f(3)=5$

$fog(5)=f(g(5))=f(1)=2$

$fog(1)=f(g(1))=f(3)=5$
So, we can write that:
$fog=\{(2,5),(5,2),(1,5)\}$

Question:10

Let C be the set of complex numbers. Prove that the mapping $f:C\to R$ is given by $f(z)=|z|,\mathrm{\forall }z\in C$, is neither one-one nor onto.

Answer:

Here, $f:C\to R$ is given by $f(z)=|z|,\mathrm{\forall }z\in C$
If we assume $z=4+3i$
Then,
$f(4+3i)=|4+3i|=\sqrt{(}{4}^2+3^2)=\sqrt{2}5=5$
Similarly, for $z=4-3i$
$f(4-3i)=|4-3i|=\sqrt{(}{4}^2+3^2)=\sqrt{2}5=5$
Therefore, it is clear that f(z) is many-one.
So, $|z|\ge 0,\mathrm{\forall }z\in C,$
However, in the question R is the co-domain given.
Hence, f(z) is not onto. So, f(z) is neither one-one nor onto.

Question:11

Let the function $f:R\to R$ be defined by $f(x)=cosx,\mathrm{\forall }x\in R.$ Show that f is neither one-one nor onto.

Answer:
It is given that, $f:R\to R,f(x)=cosx,\mathrm{\forall }x\in R$
So, we can write:
$$\begin{gathered} f(x_1)=f(x_2) \\ or,cos{x}_1=cos{x}_2 \end{gathered}$$
Hence, $x_1=2n\pi \pm x_2,wheren\in Z$
It is understandable that for any value of $x_{1}and{x}_2$, the above equation has an infinite number of solutions.
Therefore, f(x) is a many one function.
We know the range of cos x is [-1, 1] and it is a subset of the given co-domain R.
Hence, the given function is not onto.

Question:12

Let $X=\{1,2,3\}andY=\{4,5\}$ . Find whether the following subsets of $X\times Y$ are functions from X to Y or not.
(i) $f=\{(1,4),(1,5),(2,4),(3,5)\}$
(ii) $g=\{(1,4),(2,4),(3,4)\}$
(iii) $h=\{(1,4),(2,5),(3,5)\}$
(iv) $k=\{(1,4),(2,5)\}.$

Answer:

Here, $X=\{1,2,3\}andY=\{4,5\}$
Therefore, $X\times Y=\{(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)\}$
(i) $f=\{(1,4),(1,5),(2,4),(3,5)\}$
Here, f(1) = 4 and again f(1) = 5.
Therefore, f is not a function here.
As a result, there is no unique of pre- image ‘1’.
(ii) $g=\{(1,4),(2,4),(3,4)\}$
We can clearly see that g is a function. Here in g, each element of the given domain has a unique image at the given range
(iii) $h=\{(1,4),(2,5),(3,5)\}$
It’s clear that h is a function of each pre-image that has a unique image.
Again, h(2) = h(3) = 5
Therefore, the function h is also many-one.
(iv) $k=\{(1,4),(2,5)\}$
Here, ‘3’ does not have any image under the mapping. Therefore, k is not a function.

Question:13

If functions $f:A\to Bandg:B\to A\text{ satisfies }gof=IA$, then show that f is one-one and g is onto.

Answer:

Here, it is given:
$f:A\to Bandg:B\to A\text{ satisfies }gof=IA$
It’s clear here that the function ‘g’ is inverse of ‘f’.
So, ‘f’ has to be both one-one as well as onto.
As a result, ‘g’ is both one-one and onto.

Question:14

Let $f:R\to R$ be the function defined by $f(x)=1/(2-cosx)\mathrm{\forall }x\in R$. Then, find the range off.

Answer:

Here, $f(x)=1/(2-cosx)\mathrm{\forall }x\in R$
Let’s say,
$$\begin{gathered} y=1/(2-cosx) \\ \therefore 2y-ycosx=1 \\ Or,cosx=(2y-1)/y \\ Or,cosx=2-1/y \end{gathered}$$
As the range of cos x is, $-1\le cosx\le 1$
Therefore,
$$\begin{gathered} -1\le 2-1/y\le 1 \\ -3\le -1/y\le -1 \\ 1\le -1/y\le 3 \\ 1/3\le y\le 1 \end{gathered}$$
Hence, the range of the given function is [1/3, 1].

Question:15

Let n be a fixed positive integer. Define a relation R in Z as follows: $\mathrm{\forall }a,b\in Z$, aRb if and only if a - b is divisible by n. Show that R is an equivalence relation.

Answer:

Here, we have to a relation R in Z as follows: $\mathrm{\forall }a,b\in Z$, aRb if and only if a - b is divisible by n.
Here, aRa $\Rightarrow (a-a)$ is divisible by n and this is true for all integers.
Therefore, R is reflective.
For aRb, $aRb\Rightarrow (a-b)$ is also divisible by n.
or, - (b - a) is divisible by n.
or, (b - a) is divisible by n
Hence, we can write it as bRa.
Therefore, R is symmetric.
For aRb , (a - b) is divisible by n.
For bRc, (b - c) is divisible by n.
Hence, (a - b) + (b-c) is divisible by n.
Or, (a-c) is divisible by n. This can be expressed as aRc.
Therefore, R is transitive.
So, R is an equivalence relation.

Question:16

If A = {1, 2, 3 }, define relations on A which have properties of being:

(a) reflexive, transitive but not symmetric

(b) symmetric but neither reflexive nor transitive

(c) reflexive, symmetric, and transitive.

Answer:

Here, $A=\{1,2,3\}.$
(i) Assume $R_1=\{(1,1),(1,2),(1,3),(2,3),(2,2),(1,3),(3,3)\}$
Here, (1, 1), (2, 2) and (3, 3) $\in R_1.R_1$ is reflexive.
$(1,2)\in R_1,(2,3)\in R_1\Rightarrow (1,3)\in R_1$
. Hence,$R_1$ is transitive.
Now, $(1,2)\in R_1\Rightarrow (2,1)\notin R_1.$
Therefore, $R_1$ is not symmetric.
(ii) Let say, $R_2=\{(1,2),(2,1)\}$
So, $(1,2)\in R_2,(2,1)\in R_2$
Therefore, $R_2$ is symmetric,
$(1,1)\notin R_2$. Therefore, $R_2$ is not reflexive.
$(1,2)\in R_2$, $(2,1)\in R_2$ but $(1,1)\notin R_2$. Hence, $R_2$ is not transitive.
(iii) Let $R_3$ = $\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}$
$R_3$ is reflexive as $(1,1)(2,2)and(3,3)\in R_3$
$R_3$ is symmetric as $(1,2),(1,3),(2,3)\in R_3$ $\Rightarrow (2,1),(3,1),(3,2)\in R_3$
Therefore, $R_3$ is reflexive, symmetric and transitive.

Question:17

Let R be relation defined on the set of natural number N as follows:
$R=\{(x,y):x\in N,y\in N,2x+y=41\}$ . Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric, and transitive.

Answer:

Here, $R=\{(x,y):x\in N,y\in N,2x+y=41\}$
So, the domain $D=\{1,2,3,\dots ..,20\}$
And the Range $=\{1,3,5,\dots ..,39\}$
Here, $(2,2)\notin Ras2\times 2+2\ne 41$. Therefore, R is not reflexive.
Again, $(1,39)\in Rbut(39,1)\notin R$. So, R is not symmetric.
Again, $(11,19)\notin R,(19,3)\notin R;but(11,3)\notin R$. So, Further R is not transitive.
Therefore, R is neither reflexive nor symmetric and nor transitive.

Question:18

Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:

(a) an injective mapping from A to B

(b) a mapping from A to B which is not injective

(c) a mapping from B to A.

Answer:

Here, $A=\{2,3,4\},B=\{2,5,6,7\}$
(i) Assume, $f:A\to B\text{ denote a mapping}$
Then, $f=\{(x,y):y=x+3\}or$
$f=\{(2,5),(3,6),(4,7)\}$ , this is an injective mapping.
(ii) Again assume, $g:A\to B$ denotes a mapping: $g=\{(2,2),(3,2),(4,5)\}$ . Hence, it is not an injective mapping.
(iii) Again assume, h: $B\to A$ denotes a mapping: $h=\{(2,2),(5,3),(6,4),(7,4)\}$ . Hence, it is a mapping from B to A.

Question:19

Give an example of a map

(i) which is one-one but not onto

(ii) which is not one-one but onto

(iii) which is neither one-one nor onto.
Answer:
(i) Say, $f:N\to N,$ be a mapping defined by $f(x)=x^2$
If, $f(x_1)=f(x_2)$
Then, $x_1^2=x_2^2$
So,$x_1=x_2(as,x_1+x_1\text{cannot be }0)$
$f(x_1)=f(x_2)$, hence, f(x) is one-one.
However, ‘f’ is not onto, as for $1\in N$, therefore, there is no existence of x in N : f(x) = 2x + 1.
(ii) Let $f:R\to [0,\mathrm{\infty })$, is a mapping that is defined by $f(x)=|x|$
Then, we can conclude that f(x) is not a one-one as f(2) and f(-2) are the same here.
But $|x|\ge 0$, so the range is $[0,\mathrm{\infty }].$
Therefore, f(x) is onto.
(iii) Assume, $f:R\to R$ , be a mapping which is defined by $f(x)=x^2$
Then we can say that f(x) is not one-one as f(1) and f (-1) are the same.
The range of f(x) is $[0,\mathrm{\infty })$.
Therefore, f (x) is neither one-one nor onto.

Question:20

Let $A=R-\{3\},B=R-\{1\}$ . Let $f:A\to B$ be defined by $f(x)=x-2/x-3\mathrm{\forall }x\in A$ . Then show that f is bijective.

Answer:

Here,

$A=R-\{3\},B=R-\{1\}$
$f:A\to B$ be defined by $f(x)=x-2/x-3\mathrm{\forall }x\in A$
Hence, $f(x)=(x-3+1)/(x-3)=1+1/(x-3)$
Say $f(x1)=f(x2)$
$1+\frac{1}{x_1-3}=1+\frac{1}{x_2-3}$
$$\begin{gathered} \frac{1}{x_1-3}=\frac{1}{x_2-3} \\ {x}_1=x_2 \end{gathered}$$
So, f (x) is an injective function.
$If,y=(x-2)/(x-3)$

$Or,x-2=xy-3y$

$Or,x(1-y)=2-3y$

$Or,x=(3y-2)/(y-1)$

$Or,y\in R-\{1\}=B$
So, f (x) is onto or subjective.
Therefore, f(x) is a bijective function.

Question:21

Let A = [-1, 1]. Then, discuss whether the following functions defined on A are one-one, onto, or bijective:
(i) $f(x)=x/2$ (ii) $g(x)=|x|$
(iii) $h(x)=x|x|$ (iv) $k(x)=x^2$

Answer:

Here, A = [-1, 1]
(i) $f:[-1,1]\to [-1,1],$

$f(x)=x/2$
If, $f(x_1)=f(x_2)$
$x_1/2=x_2/2$
So, f(x) is one-one.
Also, $x\in [-1,1]$
$x/2=f(x)=[-1/2,1/2]$
Hence, the range is a subset of co-domain A
So, f(x) is not onto.
Therefore, f (x) is not bijective.
(ii)
$g(x)=|x|$

$\text{Let}g(x_1)=g(x_2)$

$|x_1|=|x_2|$

$x_1=\pm x_2$
So, g(x) is not one-one.
Also, $g(x)=|x|\ge 0,\text{for all real x}$
Hence, the range is [0, 1], which is subset of co-domain ‘A’
So, f(x) is not onto.
Therefore, f(x) is not bijective.
(iii)
$h(x)=x|x|$

$$\begin{gathered} \text{Let }h(x_1)=h(x_2) \\ {x}_1|x_1|=x_2|x_2| \end{gathered}$$

$$\begin{gathered} \text{If }{x}_1,x_2>0 \\ {x}_1^2=x_2^2 \\ {x}_1^2-x_2^2=0 \\ (x_1-x_2)(x_1+x_2)=0 \\ {x}_1=x_2(as{x}_1+x_2\ne 0) \end{gathered}$$

$\text{Again },x_1,x_2<0,and{x}_1=x_2$

$\text{Therefore },x_{1}and{x}_2\text{of opposite sign},x_1\ne x_2.$
Hence, f(x) is one-one.
$\text{For x}\in [0,1],f(x)=x_2\in [0,1]$

$\text{For x}<0,f(x)=-x_2\in [-1,0)$
Hence, the range is [-1, 1].
So, h(x) is onto.
Therefore, h(x) is bijective.
(iv)
$k(x)=x^2$

$\text{Let }k(x_1)=k(x_2)$

$x_1^2=x_2^2$

$x_1=\pm x_2$
Therefore, k(x) is not one-one.

Question:22

Each of the following defines a relation on N:
(i) x is greater than y, $x,y\in N$
(ii) x + y = 10, $x,y\in N$
(iii) x y is square of an integer $x,y\in N$
(iv) x + 4y = 10 $x,y\in N$
Determine which of the above relations are reflexive, symmetric, and transitive.

Answer:

(i)Here, x is greater than y; $x,y\in N$
If $(x,x)\in R$, then x > x, that does not satisfy for any $x\in N$.\
Therefore, R is not reflexive.
Say,$(x,y)\in R$
$Or,xRy$

$Or,x>y$
$\Rightarrow y>x$ For any $x,y\in N$, the above condition is not true.
Hence, R is not symmetric.
Again, xRy and yRz
$Or,x>yandy>z$

$Or,x>z$

$Or,xRz$
Hence, R is transitive.
(ii) x + y = 10;$x,y\in N$
Thus,
$R=\{(x,y);x+y=10,x,y\in N\}$

$R=\{(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1)\}$
Therefore, $(1,1)\notin R$
So, R is not reflexive.
Again, $(x,y)\in R$

$\Rightarrow (y,x)\in R$
Therefore, R is symmetric.
And, $(1,9)\in R,(9,1)\in R,but(1,1)\notin R$
Therefore, R is not transitive.
(iii)Here, xy is square of an integer $x,y\in N$
$R=\{(x,y):\text{xy is a square of an integer}x,y\in N\}$
So, $(x,x)\in R,\mathrm{\forall }x\in N$
For any $x\in N,x^2$ is an integer.
Thus, R is reflexive.
If $(x,y)\in R$

$\Rightarrow (y,x)\in R$
So, R is symmetric.
Again, if xy and yz both are square of an integer.
Then, $xy=m^{2}andyz=n^2\text{for some m, n}\in Z$
$$\begin{gathered} x=m^2/yandz=n^2/y \\ xz=m^2{n}^2/y^2 \end{gathered}$$, this must be the square of an integer.
Therefore, R is transitive.
(iv) x + 4y = 10; $x,y\in N$
$R=\{(x,y):x+4y=10;x,y\in N\}$

$$\begin{gathered} R=\{(2,2),(6,1)\} \\ Here,(1,1)\notin R \end{gathered}$$
Hence, R is not symmetric.
$(x,y)\in R$

$$\begin{gathered} \Rightarrow x+4y=10 \\ And,(y,z)\in R \end{gathered}$$

$\Rightarrow y+4z=10$

$Or,x-16z=-30$

$Or,(x,z)\notin R$
Therefore, R is not transitive.

Question:23

Let A = {1, 2, 3, … 9} and R be the relation in $A\times A$ defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in $A\times A$. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].

Answer:

$Here,A=\{1,2,3,\dots 9\}and(a,b)R(c,d)$

$$\begin{gathered} ifa+d=b+cfor(a,b),(c,d)\in A\times A. \\ Say(a,b)R(a,b) \end{gathered}$$

$Therefore,a+b=b+a,\mathrm{\forall }a,b\in A.$
This must be true for any $a,b\in A.$
Hence, R is reflexive.
Say, (a, b) R (c, d)
Then,
$$\begin{gathered} a+d=b+c \\ c+b=d+a \\ (c,d)R(a,b) \end{gathered}$$
Therefore, R is symmetric.
Let $(a,b)R(c,d),and,(c,d)R(e,f)$
$$\begin{gathered} a+d=b+c,and,c+f=d+e \\ a+d=b+c,and,d+e=c+f \end{gathered}$$

$Or,(a+d)-(d+e=(b+c)-(c+f)$

$Or,a-e=b-f$

$Or,a+f=b+e$

$Or,(a,b)R(e,f)$
So, R is transitive.
Therefore, R is an equivalence relation.

Question:24

Using the definition, prove that the function $f:A\to B$ is invertible if and only if f is both one-one and onto.

Answer:

Say, $f:A\to B$ be many-one functions.
If, $f(a)=p\text{ and }f(b)=p$
Then $f^{-1}(p)=aand{f}^{-1}(p)=b$
In this case, we have two images ‘a and b’ for one pre-image ‘p’. This is because the inverse function is not defined here.
However, to be one-one, f must be invertible.
Say, $f:A\to B$ is not onto function.
$B=\{p,q,r\}and\{p,q\}$ is the range of f.
There is no pre-image for the image r, which will have no image in set A.
And, f must be onto to be invertible.
Thus, to be both one-one and onto f must be invertible
If f is a bijective function, then $f=X\to Y$ is invertible.

Question:25

Functions $f,g:R\to R$ are defined, respectively, by $f(x)=x^2+3x+1,g(x)=2x-3$, find
(i) fog (ii) gof (iii) fof (iv) gog

Answer:

Here,
$f(x)=x^2+3x+1,g(x)=2x-3$

$$\begin{gathered} (i)fog=f(g(x)) \\ =f(2x-3) \\ =(2x-3{)}^2+3(2x-3)+1 \end{gathered}$$$$\begin{gathered} =4x^2+9-12x+6x-9+1 \\ =4x^2-6x+1 \end{gathered}$$

$$\begin{gathered} (ii)gof=g(f(x)) \\ =g(x^2+3x+1) \\ =2(x^2+3x+1)-3 \end{gathered}$$

$=2x^2+6x-1$

$$\begin{gathered} (iii)fof=f(f(x)) \\ =f(x^2+3x+1) \\ =(x^2+3x+1)2+3(x^2+3x+1)+1 \end{gathered}$$

$=x4+9x^2+1+6x^3+6x+2x^2+3x^2+9x+3+1$

$$\begin{gathered} =x4+6x^3+14x^2+15x+5 \\ (iv)gog=g(g(x)) \end{gathered}$$

$$\begin{gathered} =g(2x-3) \\ =2(2x-3)-3 \\ =4x-6-3 \\ =4x-9 \end{gathered}$$

Question:26

Let $\ast$ be the binary operation defined on Q. Find which of the following binary operations are commutative
$$\begin{gathered} (i)a\ast b=a-b\mathrm{\forall }a,b\in Q \\ (ii)a\ast b=a^2+b^2\mathrm{\forall }a,b\in Q \\ (iii)a\ast b=a+ab\mathrm{\forall }a,b\in Q \\ (iv)a\ast b=(a-b{)}^2\mathrm{\forall }a,b\in Q \end{gathered}$$

Answer:

Here, $\ast$ is a binary operation defined on Q.
(i)
$$\begin{gathered} a\ast b=a-b,\mathrm{\forall }a,b\in Qandb\ast a=b-a \\ So,a\ast b\ne b\ast a \end{gathered}$$
Hence, $\ast$ is not commutative.
(ii)
$$\begin{gathered} a\ast b=a^2+b^2 \\ b\ast a=b^2+a^2 \end{gathered}$$
Therefore, $\ast$ is commutative.
(iii)
$$\begin{gathered} a\ast b=a+ab \\ b\ast a=b+ab \\ Hence,a+ab\ne b+ab \end{gathered}$$
Therefore, $\ast$ is not commutative.
(iv)
$$\begin{gathered} a\ast b=(a-b{)}^2,\mathrm{\forall }a,b\in Q \\ b\ast a=(b-a{)}^2 \\ As,(a-b{)}^2=(b-a{)}^2 \end{gathered}$$
Therefore, $\ast$ is commutative.

Question:27

Let $\ast$ be binary operation defined on R by $a\ast b=1+ab,\mathrm{\forall }a,b\in R$. Then the operation $\ast$ is
(i) commutative but not associative
(ii) associative but not commutative
(iii) neither commutative nor associative
(iv) both commutative and associative

Answer:

(i) Here,

$\ast$ is a binary operation defined on R by $a\ast b=1+ab,\mathrm{\forall }a,b\in R$
So, $a\ast b=ab+1=b\ast a$
Therefore, $\ast$ is a commutative binary operation.
$Now,a\ast (b\ast c)=a\ast (1+bc)=1+a(1+bc)=1+a+abc$

$Again,(a\ast b)\ast c=(1+ab)\ast c=1+(1+ab)c=1+c+abc$

$Therefore,a\ast (b\ast c)\ne (a\ast b)\ast c$
Hence, $\ast$ is not associative.
Therefore, $\ast$ is commutative but not associative.

Question:28

Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to $b\mathrm{\forall }a,b\in T$. Then R is
(A) reflexive but not transitive (B) transitive but not symmetric
(C) equivalence (D) none of these

Answer:

(C) equivalence
Here aRb, if a is congruent to b, $\mathrm{\forall }a,b\in T.$
So, in aRa, a is congruent to a. This must always be true.
Therefore, R is reflexive.
Say, $aRb\Rightarrow a\sim b$
$b\sim a\Rightarrow bRa$
Therefore, R is symmetric.
Say, aRb and bRc
$$\begin{gathered} a\sim bandb\sim c \\ a\sim c\Rightarrow aRc \end{gathered}$$
Therefore, R is transitive.
Therefore, R is an equivalence relation.

Question:29

Consider the non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is

(A) symmetric but not transitive (B) transitive but not symmetric

(C) neither symmetric nor transitive (D) both symmetric and transitive

Answer:

(B) transitive but not symmetric

If, aRb means a is brother of b.

Then, it does not mean b is also a brother of a. Because, b can be a sister of a too.

Therefore, R is not symmetric.

If, aRb implies that a is the brother of b.

and bRc implies that b is the brother of c.

Therefore, a must be the brother of c.

Hence, R is transitive.