NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

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NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions: Exercise Questions

These NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions have been prepared by Careers360 experts to make learning simpler and to help you score better in exams.

Question1(i): 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 $R = \{(x,y): 3x - y = 0\}$

Answer:

$A = \{1, 2, 3, \ldots, 13, 14\}$

$R = \{(x, y) : 3x - y = 0\} = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$

Since $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), \ldots, (14, 14) \notin R$, so $R$ is not reflexive.

Since $(1, 3) \in R$ but $(3, 1) \notin R$, so $R$ is not symmetric.

Since $(1, 3), (3, 9) \in R$ but $(1, 9) \notin R$, so $R$ is not transitive.

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

Question 1(ii): Determine whether each of the following relations are reflexive, symmetric and
transitive:

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

Answer:

$R = \{(x,y): y = x + 5 \;\textup{and}\;x<4\}$ $= \left \{ \left ( 1,6 \right ),\left ( 2,7 \right ),\left ( 3,8 \right ) \right \}$

Since, $\left ( 1,1 \right ) \notin R$

So $R$ is not reflexive.

Since, $\left ( 1,6 \right )\in R$ but $\left ( 6,1 \right )\notin R$

So $R$ is not symmetric.

Since there is no pair in $R$ such that $\left ( x,y \right ),\left ( y,x \right )\in R$, this is not transitive.

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

Question1(iii): Determine whether each of the following relations are reflexive, symmetric and
transitive:

(iii) Relation R in the set $A = \{1,2,3,4,5,6\}$ as $R = \{(x,y) : y \; \textup{is} \; divisible \; by\; x\}$

Answer:

$A = \{1,2,3,4,5,6\}$

$R = \left \{ \left ( 2,4 \right ),\left ( 3,6 \right ),\left ( 2,6 \right ),\left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right ),\left ( 6,6 \right )\right \}$

Any number is divisible by itself and $\left ( x,x \right ) \in R$. So it is reflexive.

$\left ( 2,4 \right ) \in R$ but $\left ( 4,2 \right ) \notin R$ .Hence,it is not symmetric.

$\left ( 2,4 \right ),\left ( 4,4 \right ) \in R$, and 4 is divisible by 2, and 4 is divisible by 4.

Hence, it is transitive.

Hence, it is reflexive and transitive but not symmetric.

Question 1(iv): Determine whether each of the following relations are reflexive, symmetric and
transitive:

(iv). Relation R in the set Z of all integers defined as $R = \{(x,y): x - y \;is\;an\;integer\}$

Answer:

$R = \{(x,y): x - y \;is\;an\;integer\}$

For $x \in Z$ , $\left ( x,x \right ) \in R$ as $x-x = 0$ which is an integer.

So, it is reflexive.

For $x,y \in Z$ , $\left ( x,y \right ) \in R$ and $\left ( y,x \right ) \in R$ because $x-y \, \, and \, \, y-x$ are both integers.

So, it is symmetric.

For $x,y,z \in Z$ , $\left ( x,y \right ),\left ( y,z \right ) \in R$ as $x-y \, \, and \, \, y-z$ are both integers.

Now, $x-z = \left ( x-y \right )+\left ( y-z \right )$ is also an integer.

So, $\left ( x,z \right ) \in R$ and hence it is transitive.

Hence, it is reflexive, symmetric and transitive.

Question 1 (v): Determine whether each of the following relations are reflexive, symmetric and
transitive:

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

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

Answer:

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

$\left ( x,x \right )\in R$ ,so it is reflexive

$\left ( x,y \right )\in R$ means $x \;and\; y\;work\;at\;the\;same\;place$ .

$y \;and\; x\;work\;at\;the\;same\;place$ i.e. $\left ( y,x \right )\in R$ so it is symmetric.

$\left ( x,y \right ),\left ( y,z \right )\in R$ means $x \;and\; y\;work\;at\;the\;same\;place$ also $y \;and\; z\;work\;at\;the\;same\;place$ .It states that $x \;and\; z\;work\;at\;the\;same\;place$ i.e. $\left ( x,z \right )\in R$ .So, it is transitive.

Hence, it is reflexive, symmetric and transitive.

Question 1 (v): Determine whether each of the following relations are reflexive, symmetric and
transitive:

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

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

Answer:

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

$\left ( x,x \right )\in R$ as $x$ and $x$ is same human being.So, it is reflexive.

$\left ( x,y \right )\in R$ means $x\;and\;y\;live\;in\;the\;same\;locality$ .

It is same as $y\;and\;x\;live\;in\;the\;same\;locality$ i.e. $\left ( y,x \right )\in R$ .

So, it is symmetric.

$\left ( x,y \right ),\left ( y,z \right )\in R$ means $x\;and\;y\;live\;in\;the\;same\;locality$ and $y\;and\;z\;live\;in\;the\;same\;locality$ .

It implies that $x\;and\;z\;live\;in\;the\;same\;locality$ i.e. $\left ( x,z \right )\in R$ .

Hence, it is reflexive, symmetric and transitive.

Question 1 (v): Determine whether each of the following relations are reflexive, symmetric and
transitive:

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

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

Answer:

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

$\left ( x,y\right )\in R$ means $x\;is\;exactly\;7\;cm\;taller\;than\;y$ but $x\;is\;not\;\;taller\;than\;x$ i.e. $\left ( x,x \right )\notin R$ .So, it is not reflexive.

$\left ( x,y\right )\in R$ means $x\;is\;exactly\;7\;cm\;taller\;than\;y$ but $y\;is\;not\;\;taller\;than\;x$ i.e $\left ( y,x \right )\notin R$ .So, it is not symmetric.

$\left ( x,y\right ),\left ( y,z \right )\in R$ means $x\;is\;exactly\;7\;cm\;taller\;than\;y$ and $y\;is\;exactly\;7\;cm\;taller\;than\;z$ .

$x\;is\;exactly\;14\;cm\;taller\;than\;z$ i.e. $\left ( x,z \right )\notin R$ .

Hence, it is not reflexive, symmetric, or transitive.

Question 1 (v): Determine whether each of the following relations are reflexive, symmetric and
transitive:

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

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

Answer:

$R = \{(x, y) : x\;is\;wife\;of\;y\}$

$\left ( x,y \right ) \in R$ means $x\;is\;wife\;of\;y$ but $x\;is\;not\, wife\;of\;x$ i.e. $\left ( x,x \right ) \notin R$ .

So, it is not reflexive.

$\left ( x,y \right ) \in R$ means $x\;is\;wife\;of\;y$ but $y\;is\;not\, wife\;of\;x$ i.e. $\left ( y,x \right ) \notin R$ .

So, it is not symmetric.

Let, $\left ( x,y \right ),\left ( y,z \right ) \in R$ means $x\;is\;wife\;of\;y$ and $y\;is\;wife\;of\;z$ .

This case is not possible, so it is not transitive.

Hence, it is not reflexive, symmetric or transitive.

Question 1(iv): Determine whether each of the following relations are reflexive, symmetric, and Transitive:

(iv). Relation R in the set Z of all integers defined as R={(x,y):x−y is an integer}

Answer:

R={(x,y):x−y is an integer}

For x∈Z, (x,x)∈R as x−x=0, which is an integer.

So, it is reflexive.

For x,y∈Z, (x,y)∈R and (y,x)∈R because x−y and y−x are both integers.

So, it is symmetric.

For x,y,z∈Z, (x,y),(y,z)∈R as x−y and y−z are both integers.

Now, x−z=(x−y)+(y−z) is also an integer.

So, (x,z)∈R and hence it is transitive.

Hence, it is reflexive, symmetric, and transitive.

Question 2: Show that the relation R in the set R of real numbers is defined as
$R={(a,b): a≤b^2}$ is neither reflexive nor symmetric nor transitive.

Answer:

$R=\left\{(a, b): a \leq b^2\right\}$
Taking

$\left(\frac{1}{2}, \frac{1}{2}\right) \notin R$

and

$\left(\frac{1}{2}\right)>\left(\frac{1}{2}\right)^2$
So, R is not reflexive.
Now,
$(1,2) \in R$ because $1<4$.
But, $4 \nless 1$, i.e. 4 is not less than 1
So, $(2,1) \notin R$
Hence, it is not symmetric.
$(3,2) \in R$ and $(2,1.5) \in R$ as $3<4$ and $2<2.25$
Since $(3,1.5) \notin R$ because $3 \nless 2.25$
Hence, it is not transitive.
Thus, we can conclude that it is neither reflexive, nor symmetric, nor transitive.

Question 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.

Answer:

$R$ defined on the set $\{1, 2, 3, 4, 5, 6\}$

$R = \{(a, b) : b = a + 1\}$

$R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$

Since $\{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\} \notin R$, so it is not reflexive.

$\{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\} \in R$ but $\{(2, 1), (3, 2), (4, 3), (5, 4), (6, 5)\} \notin R$, so it is not symmetric.

$\{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\} \in R$ but $\{(1, 3), (2, 4), (3, 5), (4, 6)\} \notin R$, so it is not transitive.

So, it is not transitive.

Hence, it is neither reflexive, nor symmetric, nor transitive

Question 4: Show that the relation R in R defined as $R = \{(a, b) : a \leq b\}$ , is reflexive and

transitive but not symmetric.

Answer:

$R = \{(a, b) : a \leq b\}$

As $\left ( a,a \right )\in R$ so it is reflexive.

Now we take an example

$\left ( 2,3 \right )\in R$ as $2< 3$

But $\left ( 3,2 \right )\notin R$ because $2 \nless 3$ .

So, it is not symmetric.

Now if we take, $\left ( 2,3 \right )\in R\, \, and\, \, \left ( 3,4 \right )\in R$

Than, $\left ( 2,4 \right )\in R$ because $2< 4$

So, it is transitive.

Hence, we can say that it is reflexive and transitive but not symmetric.

Question 5: Check whether the relation R in R defined by $R = \{(a, b) : a \leq b^3 \}$ is reflexive,
symmetric or transitive.

Answer:

$R = \{(a, b) : a \leq b^3 \}$

$\left ( \frac{1}{2},\frac{1}{2} \right )\notin R$ because $\frac{1}{2}\nleqslant (\frac{1}{2}) ^{3}$

So, it is not symmetric

Now, $\left ( 1,2 \right ) \in R$ because $1< 2^{3}$

but $\left ( 2,1 \right )\notin R$ because $2\nleqslant 1^{3}$

It is not symmetric

$\left ( 3,1.5 \right ) \in R\, \, and \, \, \left ( 1.5,1.2 \right ) \in R$ as $3< 1.5^{3} \, \, and \, \, 1.5< 1.2^{3}$ .

But, $\left ( 3,1.2 \right )\notin R$ because $3 \nleqslant 1.2^{3}$

So it is not transitive

Thus, it is neither reflexive, nor symmetric, nor transitive.

Question 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.

Answer:

Let A= $\{1, 2, 3\}$

$R = \{(1, 2), (2, 1)\}$

We can see $\left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right )\notin R$ so it is not reflexive.

As $\left ( 1,2 \right )\in R \, and \, \left ( 2,1 \right )\in R$ so it is symmetric.

$(1, 2) \in R \, and\, (2, 1)\in R$

But $(1, 1)\notin R$ so it is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

Question 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 the same number of pages}, is an equivalence relation?

Answer:

A = all the books in the library of a college

$R=\{(x, y): x$ and $y$ have same number of pages $\}$

$(x, x) \in R$ because x and x have same number of pages so it is reflexive.

Let $(x, y) \in R$ means x and y have the same number of pages.

Since y and x have the same number of pages, so $(y, x) \in R$.

Hence, it is symmetric.

Let $(x, y) \in R$ means x and y have the same number of pages.

And $(y, z) \in R$ means y and z have the same number of pages.

This states, x and z also have same number of pages i.e. $(x, z) \in R$

Hence, it is transitive.

Thus, it is reflexive, symmetric, and transitive, i.e. it is an equivalence relation.

Question 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 $\{2, 4\}$ .

Answer:

$A = \{1, 2, 3, 4, 5\}$

$R = \{(a, b) : |a - b| \;is\;even\}$

$R=\left \{ \left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right ),\left ( 1,3 \right ),\left ( 2,4 \right ),\left ( 3,5 \right ),\left ( 3,1 \right ),\left ( 5,1 \right ),\left ( 4,2 \right ),\left ( 5,3 \right )\right \}$

Let there be $a\in A$ then $(a,a)\in R$ as $\left | a-a \right |=0$ which is even number. Hence, it is reflexive

Let $(a,b)\in R$ where $a,b\in A$ then $(b,a)\in R$ as $\left | a-b \right |=\left | b-a \right |$

Hence, it is symmetric

Now, let $(a,b)\in R \, and\, (b,c)\in R$

$\left | a-b \right | \, and \, \left | b-c \right |$ are even number i.e. $(a-b)\, and\,(b-c)$ are even

then, $(a-c)=(a-b)+(b-c)$ is even (sum of even integer is even)

So, $(a,c)\in R$ . Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive, i.e. it is an equivalence relation.

The elements of $\{1, 3, 5\}$ are related to each other because the difference of odd numbers gives an even number, and in this set all numbers are odd.

The elements of $\{2, 4\}$ are related to each other because the difference of even numbers is even, and in this set, all numbers are even.

The element of $\{1, 3, 5\}$ is not related to $\{2, 4\}$ because a difference of odd and even numbers is not even.

Question 9(i): Show that each of the relation R in the set $A = \{x \in Z : 0 \leq x \leq 12\}$, given by

(i) $R = \{(a, b) : |a - b|\; is\;a\;multiple \;of\; 4\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer:

$A = \{x \in Z : 0 \leq x \leq 12\}$

$A=\left \{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \right \}$

$R = \{(a, b) : |a - b|\; is\;a\;multiple \;of\; 4\}$

For $a\in A$ , $(a,a)\in R$ as $\left | a-a \right |=0$ which is multiple of 4.

Hence, it is reflexive.

Let, $(a,b)\in R$ i.e. $\left | a-b \right |$ is multiple of 4.

then $\left | b-a \right |$ is also multiple of 4 because $\left | a-b \right |$ = $\left | b-a \right |$ i.e. $(b,a)\in R$

Hence, it is symmetric.

Let, $(a,b)\in R$ i.e. $\left | a-b \right |$ is multiple of 4 and $(b,c)\in R$ i.e. $\left | b-c \right |$ is multiple of 4 .

$( a-b )$ is multiple of 4 and $(b-c)$ is multiple of 4

$(a-c)=(a-b)+(b-c)$ is multiple of 4

$\left | a-c \right |$ is multiple of 4 i.e. $(a,c)\in R$

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive, i.e. it is an equivalence relation.

The set of all elements related to 1 is $\left \{1,5,9 \right \}$

$\left | 1-1 \right |=0$ is multiple of 4.

$\left | 5-1 \right |=4$ is multiple of 4.

$\left | 9-1 \right |=8$ is multiple of 4.

Question 9(ii): Show that each of the relations R in the set $A = \{x \in Z : 0 \leq x \leq 12\}$ , given by

(ii) $R = \{(a, b) : a = b\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer:

$A = \{x \in Z : 0 \leq x \leq 12\}$

$A=\left \{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \right \}$

$R = \{(a, b) : a = b\}$

For $a\in A$ , $(a,a)\in R$ as $a=a$

Hence, it is reflexive.

Let, $(a,b)\in R$ i.e. $a=b$

$a=b$ $\Rightarrow$ $b=a$ i.e. $(b,a)\in R$

Hence, it is symmetric.

Let, $(a,b)\in R$ i.e. $a=b$ and $(b,c)\in R$ i.e. $b=c$

$\therefore$ $a=b=c$

$a=c$ i.e. $(a,c)\in R$

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive, i.e. it is an equivalence relation.

The set of all elements related to 1 is {1}

Question 10(i): Give an example of a relation.

(i) Which is Symmetric but neither reflexive nor transitive.

Answer:

Let

$A = \left \{ 1,2,3 \right \}$

$R = \left \{ \left ( 1,2 \right ),\left ( 2,1 \right )\right \}$

$\left ( 1,1 \right ),\left ( 2,2 \right ),(3,3) \notin R$ so it is not reflexive.

$(1,2)\in R$ and $(2,1)\in R$ so it is symmetric.

$(1,2)\in R \, and\, (2,1)\in R$ but $(1,1)\notin R$ so it is not transitive.

Hence, symmetric but neither reflexive nor transitive.

Question 10(ii): Give an example of a relation.

(ii) Which is transitive but neither reflexive nor symmetric.

Answer:

Let

$R = \left \{ \left ( x,y \right ): x> y \right \}$

Now for $x\in R$ , $(x,x)\notin R$ so it is not reflexive.

Let $(x,y) \in R$ i.e. $x> y$

Then $y> x$ is not possible i.e. $(y,x) \notin R$ . So it is not symmetric.

Let $(x,y) \in R$ i.e. $x> y$ and $(y,z) \in R$ i.e. $y> z$

we can write this as $x> y> z$

Hence, $x> z$ i.e. $(x,z)\in R$ . So it is transitive.

Hence, it is transitive but neither reflexive nor symmetric.

Question 10 (iii): Give an example of a relation.

(iii) Which is Reflexive and symmetric but not transitive.

Answer:

Let

$A = \left \{ 1,2,3 \right \}$

Define a relation R on A as

$R= \left \{ (1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2) \right \}$

If $x\in A$ , $(x,x)\in R$ i.e. $\left \{ (1,1),(2,2),(3,3)\right \} \in R$ . So it is reflexive.

If $x,y\in A$ , $(x,y)\in R$ and $(y,x)\in R$ i.e. $\left \{(1,2),(2,1),(2,3),(3,2) \right \}\in R$ . So it is symmetric.

$(x,y)\in R$ and $(y,z)\in R$ i.e. $(1,2)\in R$ . and $(2,3)\in R$

But $(1,3)\notin R$, So it is not transitive.

Hence, it is Reflexive and symmetric but not transitive.

Question 10 (iv): Give an example of a relation.

(iv) Which is Reflexive and transitive but not symmetric.

Answer:

Let there be a relation R in R

$R=\left \{ (a,b):a\leq b \right \}$

$(a,a)\in R$ because $a=a$

Let $(a,b)\in R$ i.e. $a\leq b$

But $(b,a)\notin R$ i.e. $b\nleqslant a$

So it is not symmetric.

Let $(a,b)\in R$ i.e. $a\leq b$ and $(b,c)\in R$ i.e. $b\leq c$

This can be written as $a\leq b\leq c$ i.e. $a\leq c$ implies $(a,c)\in R$

Hence, it is transitive.

Thus, it is Reflexive and transitive but not symmetric.

Question 10 (v): Give an example of a relation.

(v) Which is Symmetric and transitive but not reflexive.

Answer:

Let there be a relation A in R

$A= \left \{ 1,2 \right \}$

$R=\left \{ (1,2),(2,1),(2,2)\right \}$

$(1,1)\notin R$ So R is not reflexive.

We can see $(1,2)\in R$ and $(2,1)\in R$

So it is symmetric.

Let $(1,2)\in R$ and $(2,1)\in R$

Also $(2,2)\in R$

Hence, it is transitive.

Thus, it is symmetric and transitive but not reflexive.

Question 11: 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 the same as the distance of the point Q from the origin}, is an equivalence relation. Further, shows that the set of all points related to a point P≠(0,0) is the circle passing through P with the origin as the centre.

Answer:

$R=\{(P, Q)$: distance of the point $P$ from the origin is the same as the distance of the point $Q$ from the origin $\}$

The distance of point P from the origin is always the same as the distance of the same point P from another origin, i.e. $(P, P) \in R$

$\therefore \mathrm{R}$ is reflexive.

Let $(P, Q) \in R$, i.e. the distance of the point P from the origin is the same as the distance of the point Q from the origin.

This is the same as the distance of point Q from the origin, same as the distance of point P from the origin, i.e. $(Q, P) \in R$

$\therefore \mathrm{R}$ is symmetric.

Let $(P, Q) \in R \quad$ and $(Q, S) \in R$

i.e. distance of the point P from the origin is the same as the distance of the point $Q$ from the origin, and also the distance of the point Q from the origin is the same as the distance of the point S from the origin.

We can say that the distance of points $P, Q, S$ from the origin is the same. This means a distance of point P from the origin is the same as the distance of point S from the origin, i.e. $(P, S) \in R$

$\therefore \mathrm{R}$ is transitive.

Hence, $R$ is an equivalence relation.

The set of all points related to a point $P \neq(0,0)$ are points whose distance from the origin is the same as the distance of point $P$ from the origin.

In other words, we can say there be a point $0(0,0)$ as the origin and the distance between point 0 and point $P$ be $k=O P$; then the set of all points related to $P$ is at a distance $k$ from the origin.

Hence, this set of points forms a circle with the centre as the origin, and this circle passes through point $P$.

Question 12: Show that the relation R defined in the set A of all triangles as R={(T1, T2): T1 is similar to T2} is an 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 T ₁ , T _2, and T ₃ are related?

Answer:

$R=\left\{\left(T_1, T_2\right): T_1 \text { is similar to } T_2\right\}$

All triangles are similar to themselves, so it is reflexive.

Let,

$\left(T_1, T_2\right) \in R$ i.e. $\mathrm{T}_1$ is similar to $\mathrm{T}_2$

$\mathrm{T}_1$ is similar to T 2 ; T 2 is similar to T 1 , i.e. $\left(T_2, T_1\right) \in R$

Hence, it is symmetric.

Let,

$\left(T_1, T_2\right) \in R$ and $\left(T_2, T_3\right) \in R$ i.e. $\mathrm{T}_1$ is similar to $\mathrm{T}_2$ and $\mathrm{T}_2$ is similar to $\mathrm{T}_3$.

$\Rightarrow \mathrm{T}_1$ is similar to $\mathrm{T}_3$ i.e. $\left(T_1, T_3\right) \in R$

Hence, it is transitive,

Thus, $R=\left\{\left(T_1, T_2\right): T_1\right.$ is similar to $\left.T_2\right\}$, is equivalence relation.

Now, we see the ratio of the sides of triangles T1 and T3 as shown

$\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}$

i.e. ratios of sides of T1 and T3 are equal. Hence, $\mathrm{T}_1$ and $\mathrm{T}_3$ are related.

Question 13: Show that the relation R defined in the set A of all polygons as R={(P1, P2): P1 and P2 have the 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?

Answer:

$R=\left\{\left(P_1, P_2\right): P_1\right.$ and $P_2$ have same number of sides $\}$

The same polygon has the same number of sides with itself,i.e. $\left(P_1, P_2\right) \in R$, so it is reflexive.

Let,

$\left(P_1, P_2\right) \in R$ i.e. $P_1$ have same number of sides as $\mathrm{P}_2$

$\mathrm{P}_1$ have same number of sides as $\mathrm{P}_2$ is same as $\mathrm{P}_2$ have same number of sides as $\mathrm{P}_1$ i.e. $\left(P_2, P_1\right) \in R$

Hence, it is symmetric.

Let,

$\left(P_1, P_2\right) \in R$ and $\left(P_2, P_3\right) \in R$ i.e. $\mathrm{P}_1$ have same number of sides as $\mathrm{P}_2$ and $\mathrm{P}_2$ have same number of sides as $\mathrm{P}_3$

$\Rightarrow \mathrm{P}_1$ have same number of sides as $\mathrm{P}_3$ i.e. $\left(P_1, P_3\right) \in R$

Hence, it is transitive,

Thus, $R=\left\{\left(P_1, P_2\right): P_1\right.$ and $P_2$ have same number of sides $\}$, is an equivalence relation.

The elements in A related to the right angle triangle $T$ with sides 3,4 and 5 are those polygons that have 3 sides.

Hence, the set of all elements in A related to the right-angle triangle T is a set of all triangles.

Question 14: Let $L$ be the set of all lines in $X Y$ plane and $R$ be the relation in $L$ defined as $R=\{(L 1, L 2): L 1$ is parallel to $L 2\}$. Show that $R$ is an equivalence relation. Find the set of all lines related to the line $y=2 x+4$.

Answer:

$R=\left\{\left(L_1, L_2\right): L_1\right.$ is parallel to $\left.L_2\right\}$

All lines are parallel to themselves, so it is reflexive.

Let,

$\left(L_1, L_2\right) \in R$ i.e. $\mathrm{L}_1$ is parallel to T 2.

$\mathrm{L}_1$ is parallel to L 2 is the same as L 2 is parallel to $\mathrm{L}_1$, i.e. $\left(L_2, L_1\right) \in R$

Hence, it is symmetric.

Let,

$\left(L_1, L_2\right) \in R$ and $\left(L_2, L_3\right) \in R$ i.e. $L_1$ is parallel to $L 2$ and $L 2$ is parallel to $L_3$.

$\Rightarrow \mathrm{L}_1$ is parallel to $\mathrm{L}_3$ i.e. $\left(L_1, L_3\right) \in R$

Hence, it is transitive,

Thus, $R=\left\{\left(L_1, L_2\right): L_1\right.$ is parallel to $\left.L_2\right\}$, is equivalence relation.

The set of all lines related to the line $y=2x+4$. Are lines parallel to $y=2x+4$?

Here, Slope $=m=2$ and constant $=c=4$

It is known that the slope of parallel lines is equal.

Lines parallel to this $(y=2 x+4$.) line are $y=2 x+c, c \in R$

Hence, the set of all parallel lines to $y=2x+4$. are $y=2 x+c$.

Question 15: Let R be the relation in the set A= {1,2,3,4}

given by $R = \{(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)\}$ . Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(C) R is symmetric and transitive but not reflexive.

(D) R is an equivalence relation.

Answer:

A = {1,2,3,4}

$R = \{(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)\}$

For every $a \in A$ there is $(a,a) \in R$ .

$\therefore$ R is reflexive.

Given, $(1,2) \in R$ but $(2,1) \notin R$

$\therefore$ R is not symmetric.

For $a,b,c \in A$ there are $(a,b) \in R \, and \, (b,c) \in R$ $\Rightarrow$ $(a,c) \in R$

$\therefore$ R is transitive.

Hence, R is reflexive and transitive but not symmetric.

The correct answer is option B.

Question 16: Let R be the relation in the set N given by $R = \{(a, b) : a = b - 2, b > 6\}$ . Choose the correct answer.

(A) $(2, 4) \in R$
(B) $(3,8) \in R$
(C) $(6,8) \in R$
(D) $(8,7) \in R$

Answer:

$R = \{(a, b) : a = b - 2, b > 6\}$

(A) Since, $b< 6$ so $(2, 4) \notin R$

(B) Since, $3\neq 8-2$ so $(3,8) \notin R$

(C) Since, $8> 6$ and $6=8-2$ so $(6,8) \in R$

(d) Since, $8\neq 7-2$ so $(8,7) \notin R$

The correct answer is option C.

Question 1: Show that the function f: R∗⟶R∗ defined by f(x)=1/x is one-one and onto, where R _∗ is the set of all non-zero real numbers. Is the result true if the domain R _∗ is replaced by N, with the co-domain being the same as R _∗ ?

Answer:

Given, $f: R_* \longrightarrow R_*$ is defined by $f(x)=\frac{1}{x}$.
One - One :

$f(x)=f(y)$

$\frac{1}{x}=\frac{1}{y}$

$x=y$

$\therefore \mathrm{f}$ is one-one.
Onto:
We have $y \in R_*$, then there exists $x=\frac{1}{y} \in R_* \quad($ Here $y \neq 0)$ such that

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

$\therefore$ f is onto.
Hence, the function is one-one and onto.

If the domain $\mathrm{R}_*$ is replaced by N with co-domain being same as $\mathrm{R}_*$ i.e. $g: N \longrightarrow R_*$ defined by

$g(x)=\frac{1}{x}$

$g\left(x_1\right)=g\left(x_2\right)$

$\frac{1}{x_1}=\frac{1}{x_2}$

$x_1=x_2$

$\therefore \mathrm{g}$ is one-one.
For $1.5 \in R_*$,
$g(x)=\frac{1}{1.5}$ but there does not exists any x in N.
Hence, function g is one-one but not onto.

Question 2 (i): Check the injectivity and surjectivity of the following functions:

(i) f:N→N given by $f(x)=x^2$

Answer:

$f: N \rightarrow N$

$f(x)=x^2$
One-one:
$x, y \in N$ then $f(x)=f(y)$

$x^2=y^2$

$x=y$

$\therefore \mathrm{f}$ is one- one i.e. injective.
For $3 \in N$ there is no x in N such that $f(x)=x^2=3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is injective but not surjective.

Question 2 (ii): Check the injectivity and surjectivity of the following functions:

(ii) f:Z→Z given by $f(x)=x^2$

Answer:

$f: Z \rightarrow Z$

$f(x)=x^2$

One-one:

For $-1,1 \in Z$ then $f(x)=x^2$

$f(-1)=(-1)^2$

$f(-1)=1$ but $-1 \neq 1$

$\, therefore, \mathrm{f}$ is not one- one i.e. not injective.
For $-3 \in Z$ there is no x in Z such that $f(x)=x^2=-3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is neither injective nor surjective

Question 2 (iii): Check the injectivity and surjectivity of the following functions:

(iii) f:R→R given by $f(x)=x^2$

Answer:

$f: R \rightarrow R$

$f(x)=x^2$
One-one:
For $-1,1 \in R$ then $f(x)=x^2$

$f(-1)=(-1)^2$

$f(-1)=1 \text { but }-1 \neq 1$

$\, therefore, \mathrm{f}$ is not one- one i.e. not injective.
For $-3 \in R$ there is no x in R such that $f(x)=x^2=-3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is not injective and not surjective.

Question 2 (iv): Check the injectivity and surjectivity of the following functions:

(iv) f:N→N given by $f(x)=x^3$

Answer:

$f: N \rightarrow N$

$f(x)=x^3$

One-one:
$x, y \in N$ then $f(x)=f(y)$

$x^3=y^3$

$x=y$

$\therefore \mathrm{f}$ is one- one i.e. injective.
For $3 \in N$ there is no x in N such that $f(x)=x^3=3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is injective but not surjective.

Question 2 (v): Check the injectivity and surjectivity of the following functions:

(v) f:Z→Z given by $f(x)=x^3$

Answer:

$f: Z \rightarrow Z$

$f(x)=x^3$

One-one:
For $(x, y) \in Z$ then $f(x)=f(y)$

$x^3=y^3$

$x=y$

$\therefore \mathrm{f}$ is one- one i.e. injective.
For $3 \in Z$ there is no x in Z such that $f(x)=x^3=3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is injective but not surjective.

Question 3: 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.

Answer:

$
\begin{aligned}
& f: R \rightarrow R \\
& f(x)=[x]
\end{aligned}
$
One-one:
For $1.5,1.7 \in R$
then $f(1.5)=[1.5]=1$ and $f(1.7)=[1.7]=1$
but $1.5 \neq 1.7$
$\therefore \mathrm{f}$ is not one-one, i.e. not injective.
For $0.6 \in R$ there is no $x$ in $R$ such that $f(x)=[0.6]$
$\therefore \mathrm{f}$ is not onto, i.e. not surjective.
Hence, f is not injective but not surjective.

Question 4: Show that the Modulus Function: R → R, given by f(x)=|x], is neither one-one nor onto, where |x| is if x is positive or 0 and |x| is −x if x is negative.

Answer:

$f: ~ R \rightarrow R$

$
f(x)=|x|
$
$
f(x)=|x|=x \text { if } x \geq 0 \text { and }-x \text { if } x<0
$
One-one:

For $-1,1 \in R$ then $f(-1)=|-1|=1$

$
\begin{aligned}
& f(1)=|1|=1 \\
& -1 \neq 1
\end{aligned}
$

$\therefore \mathrm{f}$ is not one-one, i.e. not injective.
For $-2 \in \mathrm{R}$,

We know $f(x)=|x|$ is always positive there is no $x$ in $R$ such that $f(x)=|x|=-2$
$\therefore \mathrm{f}$ is not onto, i.e. not surjective.
Hence, $f(x)=|x|$ is neither one-one nor onto.

Question 5: Show that the Signum Function f: R→R, given by f(x)={1 if x>0 0 if x=0 −1 if x<0, is neither one-one nor onto.

Answer:

$f: R \rightarrow R$ is given by

$f(x)=\left\{\begin{array}{cl}1 & \text { if } x>0 \\ 0 & \text { if } x=0 \\ -1 & \text { if } x<0\end{array}\right.$

As we can see $f(1)=f(2)=1$, but $1 \neq 2$
So it is not one-one.
Now, $\mathrm{f}(\mathrm{x})$ takes only 3 values $(1,0,-1)$ for the element -3 in codomain $R$, there does not exists x in domain $R$ such that $f(x)=-3$.

So it is not onto.
Hence, the signum function is neither one-one nor onto.

Question 6: 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.

Answer:

$
\begin{aligned}
& A=\{1,2,3\} \\
& B=\{4,5,6,7\} \\
& f=\{(1,4),(2,5),(3,6)\} \\
& f: A \rightarrow B \\
& \therefore f(1)=4, f(2)=5, f(3)=6
\end{aligned}
$
Every element of $A$ has a distant value in $f$.
Hence, it is one-one.

Question 7 (i): In each of the following cases, state whether the function is one-one, onto, or bijective. Justify your answer.

(i) f:R→R defined by f(x)=3−4x

Answer:

$\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$

$
f(x)=3-4 x
$
Let there $\mathrm{be}(\mathrm{a}, \mathrm{b}) \in \mathrm{R}$ such that $\mathrm{f}(\mathrm{a})=\mathrm{f}(\mathrm{b})$

$
\begin{aligned}
& 3-4 a=3-4 b \\
& -4 a=-4 b \\
& a=b
\end{aligned}
$

$\therefore \mathrm{f}$ is one-one.
Let there be $y \in, y=3-4 x$

$
\begin{aligned}
& x=(3-y) 4 \\
& f(x)=3-4 x
\end{aligned}
$
Putting value of $x, f(3-4y)=3-4(3-4y)$

$
f(3-4y)=y
$

$\therefore \mathrm{f}$ is onto.
$f$ is both one-one and onto; hence, $f$ is bijective.

Question 7 (ii): In each of the following cases, state whether the function is one-one, onto, or bijective. Justify your answer.

(ii) f:R→R defined by $f(x)=1+x^2$

Answer:

$f: R \rightarrow R$

$f(x)=1+x^2$

Let there be $(a, b) \in R$ such that $f(a)=f(b)$

$\begin{gathered}1+a^2=1+b^2 \\ a^2=b^2 \\ a= \pm b\end{gathered}$

For $f(1)=f(-1)=2$ and $1 \neq-1$
$\therefore f$ is not one-one.
Let there be $-2 \in R \quad(-2$ in codomain of R$)$

$f(x)=1+x^2=-2$

There does not exists any x in domain R such that $f(x)=-2$
$\therefore \mathrm{f}$ is not onto.
Hence, f is neither one-one nor onto.

Question 8: Let A and B be sets. Show that f: A×B→B×A such that f(a,b)=(b, a) is a bijective function.

Answer:

$\mathrm{f}: \mathrm{A} \times \mathrm{B} \rightarrow \mathrm{B} \times \mathrm{A}$

$
f(a, b)=(b, a)
$
Let $(a_1, b_1),(a_2, b_2) \in A \times B$
such that $\mathrm{f}(\mathrm{a} 1, \mathrm{~b} 1)=\mathrm{f}(\mathrm{a}_2, \mathrm{~b}_2)$
$(b_1,a_1)=(b_2,a_2)$
$\Rightarrow \mathrm{b}_1=\mathrm{b}_2$ and $\mathrm{a}_1=\mathrm{a}_2$
$⇒ (a_1,b_1)=(a_2,b_2)$
$\therefore \mathrm{f}$ is one- one
Let $(\mathrm{b}, \mathrm{a}) \in \mathrm{B} \times \mathrm{A}$
then there exists $(a, b) \in A \times B$ such that $f(a, b)=(b, a)$
$\therefore \mathrm{f}$ is onto.

Hence, it is bijective.

Question 9: Let $f: N \rightarrow N$ be defined by $ f(n)=\left\{\begin{array}{cl} \frac{n+1}{2} & \text { if } n \text { is odd } \\ \frac{n}{2} & \text { if } n \text { is even } \end{array} \text { for all } n \in N .\right.$ State whether the function f is bijective. Justify your answer.

Answer:

$f: N \rightarrow N, n \in N$

$f(n)=\left\{\begin{array}{cc}\frac{n+1}{2} & \text { if } n \text { is odd } \\ \frac{n}{2} & \text { if } n \text { is even }\end{array}\right.$

Here we can observe,

$f(2)=\frac{2}{2}=1 \quad$ and $\quad f(1)=\frac{1+1}{2}=1$
As we can see $f(1)=f(2)=1$ but $1 \neq 2$
$\therefore \mathrm{f}$ is not one-one.
Let, $n \in N \quad$ (N=co-domain)
case1 $n$ be even
For $r \in N, \quad n=2 r$
then there is $4 r \in N$ such that $f(4 r)=\frac{4 r}{2}=2 r$
case2 n be odd
For $r \in N, n=2 r+1$
then there is $4 r+1 \in N$ such that $f(4 r+1)=\frac{4 r+1+1}{2}=2 r+1$
$\therefore \mathrm{f}$ is onto.
$f$ is not one-one but onto
Hence, the function $f$ is not bijective

Question 10: Let $A=R-\{3\}$ and $B=R-\{1\}$. Consider the function $f: A \rightarrow B$ defined by $f(x)=x-2 / x-3$. Is $f$ one-one and onto? Justify your answer.

Answer:

$\begin{aligned} & A=R-\{3\} \\ & B=R-\{1\} \\ & f: A \rightarrow B \\ & f(x)=\left(\frac{x-2}{x-3}\right)\end{aligned}$

Let $a, b \in A$ such that $f(a)=f(b)$

$\begin{gathered}\left(\frac{a-2}{a-3}\right)=\left(\frac{b-2}{b-3}\right) \\ (a-2)(b-3)=(b-2)(a-3) \\ a b-3 a-2 b+6=a b-2 a-3 b+6 \\ -3 a-2 b=-2 a-3 b \\ 3 a+2 b=2 a+3 b \\ 3 a-2 a=3 b-2 b \\ a=b\end{gathered}$

$\therefore \mathrm{f}$ is one-one.

Let, $b \in B=R-\{1\} \quad$ then $b \neq 1$

$\begin{aligned} & a \in A \text { such that } f(a)=b \\ & \begin{array}{c}\left(\frac{a-2}{a-3}\right)=b \\ (a-2)=(a-3) b \\ a-2=a b-3 b \\ a-a b=2-3 b \\ a(1-b)=2-3 b \\ a=\frac{2-3 b}{1-b} \quad \in A\end{array}\end{aligned}$

For any $b \in B$ there exists $a=\frac{2-3 b}{1-b} \quad \in A$ such that

$\begin{aligned} f\left(\frac{2-3 b}{1-b}\right) & =\frac{\frac{2-3 b}{1-b}-2}{\frac{2-3 b}{1-b}-3} \\ & =\frac{2-3 b-2+2 b}{2-3 b-3+3 b} \\ & =\frac{-3 b+2 b}{2-3} \\ & =b\end{aligned}$

$\therefore \mathrm{f}$ is onto
Hence, the function is one-one and onto.

Question 11: Let f: R→R be defined as $f(x)=x^4$. Choose the correct answer.

(A) f is one-one onto

(B) f is many-one onto

(C) f is one-one but not onto

(D) f is neither one-one nor onto.

Answer:

$\begin{aligned} & f: R \rightarrow R \\ & f(x)=x^4\end{aligned}$

One-one:
For $a, b \in R$ then $f(a)=f(b)$

$\begin{aligned} & a^4=b^4 \\ & a= \pm b\end{aligned}$

$\therefore f(a)=f(b)$ does not imply that $a=b$
example : $f(2)=f(-2)=16$ and $2 \neq-2$
$\therefore \mathrm{f}$ is not one- one
For $2 \in R$ there is no x in R such that $f(x)=x^4=2$
$\therefore \mathrm{f}$ is not onto.
Hence, $f$ is neither one-one nor onto.
Hence, option D is correct.

Question 12: Let f: R→R be defined as f(x)=3x. Choose the correct answer.

(A) f is one-one onto

(B) f is many-one onto

(C) f is one-one but not onto

(D) f is neither one-one nor onto.

Answer:

f: $\mathrm{R} \rightarrow \mathrm{R}$

$
f(x)=3 x
$
One - One :
Let $(x, y) \in R$

$
\begin{aligned}
& f(x)=f(y) \\
& 3 x=3 y \\
& x=y
\end{aligned}
$

$\therefore \mathrm{f}$ is one-one.
Onto:
If we have $y \in R$, then there exists $x=\frac{y}{3} \in R$ such that

$
f(\frac{y}{3})=3 \times \frac{y}{3}=y
$

$\therefore \mathrm{f}$ is onto.
Hence, the function is one-one and onto.

Question 1: Show that the function f:R→{x∈R:−1<x<1} defined by f(x)=x/(1+|x|) x∈R is one one and onto function.

Answer:

The function $f: R \rightarrow\{x \in R:-1<x<1\}$ defined by

$f(x)=\frac{x}{1+|x|}, x \in R$

One-one:

Let $f(x)=f(y) \quad, x, y \in R$

$\frac{x}{1+|x|}=\frac{y}{1+|y|}$

It is observed that if x is positive and y is negative.

$\frac{x}{1+x}=\frac{y}{1+y}$

Since x is positive and y is negative.

$x>y \Rightarrow x-y>0 \quad \text { but } 2 \mathrm{xy} \text { is negative. }$

$x-y \neq 2 x y$

Thus, the case of x being positive and y being negative is removed.

The same happens in the case where y is positive, and x is negative, so this case is also removed.

When x and y both are positive:

$\begin{gathered}f(x)=f(y) \\ \frac{x}{1+x}=\frac{y}{1+y} \\ x(1+y)=y(1+x) \\ x+x y=y+x y \\ x=y\end{gathered}$

When $x$ and $y$ both are negative:

$\begin{aligned} & f(x)=f(y) \\ & \frac{x}{1-x}=\frac{y}{1-y} \\ & x(1-y)=y(1-x) \\ & x-x y=y-x y \\ & \quad x=y\end{aligned}$

$\therefore \mathrm{f}$ is one-one.

Onto:

Let $y \in R$ such that $-1<y<1$

If y is negative, then $x=\frac{y}{y+1} \in R$

$f(x)=f\left(\frac{y}{y+1}\right)=\frac{\frac{y}{1+y}}{1+\left|\frac{y}{1+y}\right|}=\frac{\frac{y}{1+y}}{1+\frac{-y}{1+y}}=\frac{y}{1+y-y}=y$

If y is positive, then $x=\frac{y}{1-y} \in R$

$f(x)=f\left(\frac{y}{1-y}\right)=\frac{\frac{y}{1-y}}{1+\left|\frac{y}{1-y}\right|}=\frac{\frac{y}{1-y}}{1+\frac{-y}{1-y}}=\frac{y}{1-y+y}=y$

Thus, $f$ is onto.

Hence, $f$ is one-one and onto.

Question 2: Show that the function f: R→R given by $f(x)=x^3$ is injective.

Answer:

f: R→R

$f(x)=x^3$

One-one:

Let $f(x)=f(y)x,y∈R$

$x^3=y^3$

We need to prove $x=y$. So,

• Let $x≠y,$ then their cubes will not be equal, i.e. $x^3≠y^3$

• It will contradict the given condition of cubes being equal.

• Hence, $x=y$, and it is one-one, which means it is injective

Question 3: Given a nonempty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A⊂B Is R an equivalence relation on P(X)? Justify your answer.

Answer:

Given a nonempty set X, consider $\mathrm{P}(\mathrm{X})$, which is the set of all subsets of X.

Since every set is a subset of itself, ARA for all $A \in P(x)$

$\therefore \mathrm{R}$ is reflexive.

Let $A R B \Rightarrow A \subset B$

This is not the same as $B \subset A$

If $A=\{0,1\}$ and $B=\{0,1,2\}$

then we cannot say that $B$ is related to $A$.

$\therefore \mathrm{R}$ is not symmetric.

If $A R B$ and $B R C$, then $A \subset B$ and $B \subset C$

this implies $A \subset C=A R C$

$\therefore \mathrm{R}$ is transitive.

Thus, R is not an equivalence relation because it is not symmetric.

Question 4: Find the number of all onto functions from the set {1,2,3,...,n} to itself.

Answer:

The number of all onto functions from the set {1,2,3,...,n} to itself is the permutations of n symbols 1,2,3,4,5...............n.

Hence, permutations on n symbols 1,2,3,4,5...............n = n

Thus, the total number of all onto maps from the set {1,2,3,...,n} to itself is the same as the permutations on n symbols 1,2,3,4,5...............n, which is n.

Question 5: Let $A=\{-1,0,1,2\}, B=\{-4,-2,0,2\}$ and $f, g: A \rightarrow B$ be functions defined by $f(x)=x^2-x, x \in A$ and $g(x)=2\left|x-\frac{1}{2}\right|-1, x \in A$. Are $f$ and $g$ equal?

Justify your answer. (Hint: One may note that two functions $f: A \rightarrow B$ and $g: A \rightarrow B$ such that $f(a)=g(a) \forall a \in A$, are called equal functions).

Answer:

Given :

$A=\{-1,0,1,2\}, B=\{-4,-2,0,2\}$

$f, g: A \rightarrow B$ are defined by $f(x)=x^2-x, x \in A$ and $g(x)=2\left|x-\frac{1}{2}\right|-1, x \in A$.

It can be observed that.

$\begin{aligned} & f(-1)=(-1)^2-(-1)=1+1=2 \\ & g(-1)=2\left|-1-\frac{1}{2}\right|-1=2\left|\frac{-3}{2}\right|-1=3-1=2 \\ & \quad f(-1)=g(-1)\end{aligned}$

$\begin{aligned} & f(0)=(0)^2-(0)=0+0=0 \\ & g(0)=2\left|0-\frac{1}{2}\right|-1=2\left|\frac{-1}{2}\right|-1=1-1=0 \\ & \quad f(0)=g(0)\end{aligned}$

$\begin{aligned} & f(1)=(1)^2-(1)=1-1=0 \\ & g(1)=2\left|1-\frac{1}{2}\right|-1=2\left|\frac{1}{2}\right|-1=1-1=0 \\ & \quad f(1)=g(1)\end{aligned}$

$\begin{aligned} & f(2)=(2)^2-(2)=4-2=2 \\ & g(2)=2\left|2-\frac{1}{2}\right|-1=2\left|\frac{3}{2}\right|-1=3-1=2 \\ & \quad f(2)=g(2)\end{aligned}$

$\therefore f(a)=g(a) \forall a \in A$

Hence, $f$ and $g$ are equal functions.

Question 6: Let A={1,2,3}. The number of relations containing (1,2) and (1,3) which are reflexive and symmetric but not transitive is

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

Answer:

A={1,2,3}

The smallest relations containing (1,2) and (1,3), which are reflexive and symmetric but not transitive, are given by

R={(1,1),(2,2),(3,3),(1,2),(1,3),(2,1),(3,1)}

(1,1),(2,2),(3,3)∈R , so relation R is reflexive.

(1,2),(2,1)∈R and (1,3),(3,1)∈R , so relation R is symmetric.

(2,1),(1,3)∈R but (2,3)∉R , so realation R is not transitive.

Now, if we add any two pairs (2,3) and (3,2) to relation R, then relation R will become transitive.

Hence, the total number of the desired relation is one.

Thus, option A is correct.

Question 7: Let A={1,2,3}. The number of equivalence relations containing (1,2) is
(A) 1
(B) 2
(C) 3
(D) 4

Answer:

A={1,2,3}

The number of equivalence relations containing (1,2) is given by

R={(1,1),(2,2),(3,3),(1,2),(2,1)}

We are left with four pairs (2,3), (3,2), (1,3),(3,1).

(1,1),(2,2),(3,3)∈R , so relation R is reflexive.

(1,2),(2,1)∈R and (2,3),(3,2)∉R , so relation R is not symmetric.

(1,3),(3,1)∉R , so realation R is not transitive.

Hence, the equivalence relation is bigger than R, which is the universal relation.

Thus, the total number of equivalence relations containing (1,2) is two.

Thus, option B is correct.

Relations and Functions Class 12 NCERT Solutions: Exercise-wise

Exercise-wise NCERT Solutions of Statistics Class 12 Maths Chapter 1 are provided in the links below.

• Relations And Functions Class 12 NCERT Solutions Exercise 1.1

• Relations And Functions Class 12 NCERT Solutions Exercise 1.2

• Relations And Functions Class 12 NCERT Solutions Miscellaneous Exercise

Class 12 Maths NCERT Chapter 1: Extra Question

Question: Let an operation $\ast$ on the set of natural numbers N be defined by $a\ast b= a^{b}\cdot$
Find (i) whether $\ast$ is a binary or not, and (ii) if it is a binary, then is it commutative or not.

Solution:

(i) As $a^b \in N$ for all $a, b \in N$
ie $a * b \in N \forall a, b \in N$
Hence, $*$ is binary.

(ii) As $1^2 \neq 2^1$
so $1 * 2 \neq 2 * 1$

Hence, $\ast$ is not commutative.

Also read,

• NCERT Exemplar Class 12 Maths Solutions Chapter 1 Relations and Functions
• NCERT Notes Class 12 Maths Chapter 1 Relations and Functions

Relations and Functions Class 12 NCERT Solutions: Topics

Relations and functions are an integral part of mathematics, and the NCERT Class 12 textbook discusses the following Maths topics.

• Introduction
• Types of Relations
• Types of Functions
• Composition of Functions and Invertible Functions

Class 12 Maths NCERT Chapter 1, Relations and Functions: Important Formulae

Relations:

A relation $R$ is a subset of the Cartesian product of $A \times B$, where $A$ and $B$ are non-empty sets.
$R^{-1}$, the inverse of relation $R$, is defined as:

$
R^{-1}=\{(b, a):(a, b) \in R\}
$

• Domain of $R=$ Range of $R^{-1}$
• Range of $R=$ Domain of $R^{-1}$

Functions:

A relation $f$ from set A to set B is a function if every element in A has one and only one image in B.

$
A \times B=\{(a, b): a \in A, b \in B\}
$
If $(a, b)=(x, y)$, then $a=x$ and $b=y$
$n(A \times B)=n(A)$ * $n(B)$, where $n(A)$ is the cardinality (number of elements) of set $A$.
$\mathrm{A} \times \phi=\phi$ (where $\phi$ is the empty set)
A function $f: A \rightarrow B$ is denoted as:
$ f(x)=y$
This means $(x,y)∈f$.

Algebra of functions:

• $(f+g)(x)=f(x)+g(x)$
• $(f-g)(x)=f(x)-g(x)$
• $(f⋅g)(x)=f(x)⋅g(x)$
• $(k⋅f)(x)=k⋅f(x)$, where $k∈R$
• $(f / g)(x)=f(x) / g(x)$, where $g(x) \neq 0$

Approach to Solve Questions of Relations and Functions Class 12

Relations and Functions can feel like a puzzle of definitions, diagrams, and properties—but with the right approach, every question starts to make perfect sense. Using these approaches, students can tackle the Relations and Functions Class 12 Chapter 1 Question Answers with greater confidence.

• Recognising the type: Build a strong foundation for different kinds of relations, such as reflexive, symmetric, transitive, and equivalence.
  Familiarise yourself with the properties of One-One, onto, and bijective functions.
• Efficient usage of diagrams and graphs: Visual representation of the properties of relations and functions helps to clarify the types.
  Use tables for small sets for binary operations to verify identity, inverse, etc.
• Domain, codomain, range: Grasp these concepts thoroughly to identify whether a relation is a function and determine its behaviour.
  Also, practice extracting the domain and range from relations using ordered pairs or graphs.
• Elimination method: For MCQ-type questions, cancel out the extreme options by checking the basic definitions.
• Shortcut tricks: Smart tricks and shortcuts are always handy during the exam to save time. Write those tricks in a notebook and revise them from time to time. Here are some tricks.
  - If a relation is reflexive and only uses equality or divisibility, it may be an equivalence.
  - For transitive relations, check in pairs that share a common middle.
  - Use the formula 2n2 to find the total relations over a set of n elements.
  - During the domain checking, avoid the values that cause division by zero or negative square roots.

What Extra Should Students Study Beyond the NCERT for JEE?

Here is a comparison list of the concepts in Statistics that are covered in JEE and NCERT, to help students understand what extra they need to study beyond the NCERT for JEE:

Why are Class 12 Maths Chapter 12 Relations and Functions Question Answers Important?

This chapter helps you understand how different mathematical quantities are connected with each other. It explains how relations and functions describe relationships between sets and variables. These Class 12 Maths Chapter 1 Relations and Functions question answers make these concepts easier to understand through step-by-step explanations and examples. Here are some more points on why these question answers are important.

• These solutions help us learn how to identify relations and functions and understand their properties using logical reasoning.
• Students learn to analyse domain, range, and types of functions, which builds a strong base for higher mathematics.
• Practising Class 12 Maths Chapter 1 Relations and Functions question answers improves analytical and reasoning skills needed for competitive exams.
• It also helps students understand advanced topics like inverse trigonometric functions and their applications in different mathematical fields.

NCERT Solutions for Class 12 Maths: Chapter Wise

For the convenience of students, Careers360 provides complete NCERT Class 12 Maths Solutions together in one location. Simply click the links below to access.

NCERT Solutions for Class 12 Subject-wise

Here, you can find the NCERT Solutions for other subjects as well.

• NCERT solutions class 12 chemistry
• NCERT solutions for class 12 physics
• NCERT solutions for class 12 biology

Class-wise NCERT Solutions

Here, you can find the NCERT Solutions for classes 9 to 11.

• NCERT solutions for class 11
• NCERT solutions for class 10
• NCERT solutions for class 9

NCERT Books and NCERT Syllabus

As students step into a new class, they must first explore the latest syllabus to identify the chapters included. Below are the links to the most recent syllabus and essential reference books.

• NCERT Books Class 12 Maths
• NCERT Syllabus Class 12 Maths
• NCERT Books Class 12
• NCERT Syllabus Class 12