NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions

• Exercise 2.1: This exercise introduces the concept of relations, their types, and their properties. It includes finding the domain and range of a relation and representing it in different ways.

• Exercise 2.2: This exercise covers the concept of functions and their types, along with their domain and range. It also includes a graphical representation of functions and the method of finding the inverse of a function.

• Exercise 2.3: This exercise discusses the composition of functions, including the definition, types, and properties. It also covers the concept of one-to-one and onto functions and their composition.

• Miscellaneous Exercise: This exercise covers various topics like finding the domain and range of a function, solving equations, and proving identities using functions. It includes a mix of conceptual and numerical problems.

Access NCERT Solutions for Class 11 Maths Chapter 2 - Relation and Functions

Exercise 2.1

1. If $\left( \dfrac{x}{3}+1,y-\dfrac{2}{3} \right)=\left( \dfrac{5}{3},\dfrac{1}{3} \right)$ , find the values of  $\text{x}$ and $y$ .

Ans: We are provided with the fact that $\left( \dfrac{x}{3}+1,y-\dfrac{2}{3} \right)=\left( \dfrac{5}{3},\dfrac{1}{3} \right)$ 

These are ordered pairs which are equal with each other, then the corresponding elements should also be equal to each other.

Thus, we will have, $\dfrac{x}{3}+1=\dfrac{5}{3}$ 

And also $y-\dfrac{2}{3}=\dfrac{1}{3}$ 

Now, we will try to simplify the given equations and find our needed values.

$\dfrac{x}{3}+1=\dfrac{5}{3} $ 

 $\Rightarrow \dfrac{x}{3}=\dfrac{5}{3}-1$ 

Simplifying further,

$\Rightarrow \dfrac{x}{3}=\dfrac{5-3}{3}=\dfrac{2}{3}$ 

$\Rightarrow x=2$

So, we have the value of $x$ as $2$ .

Again, for the second equation,

 $y-\dfrac{2}{3}=\dfrac{1}{3}$ 

$\Rightarrow y=\dfrac{2}{3}+\dfrac{1}{3}$ 

And, after more simplification,

$\Rightarrow y=\dfrac{1+2}{3}=\dfrac{3}{3}$

$\Rightarrow y=1$

So, we have the value of $x$ and $y$ as $2$ and $1$ respectively.

2. If the set A has 3 elements and the set B={3,4,5} , then find the number of elements in $(A\times B)$ ?

Ans: We are provided with the fact that the set $A$ has $3$ elements and the set $B$ is given as $\{3,4,5\}$ .

So, the number of elements in set $B$ is $3$ .

Thus, the number of elements in $(A\times B)$ will be,

= Number of elements in $A\,\times $ Number of elements in $B$ 

$=3\times 3=9$ 

So, the number of elements in $(A\times B)$ is $9$ .

3. If $G=\{7,8\}$ and $H=\{5,4,2\}$ , find $G\times H$ and $H\times G$ .

Ans: We have the sets $G=\{7,8\}$ and $H=\{5,4,2\}$ .

The Cartesian product of two non-empty sets $A$ and $B$ is defined as $A\times B=\{(a,b):a\in A\,and\,\,b\in B\}$ 

So, the value of $G\times H$ will be, 

$G\times H=\{(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)\}$ 

And similarly the value of $H\times G$ will be,

$H\times G=\{(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)\}$ 

4. State whether each of the following statement are true or false. If the statement is false, rewrite the given statement correctly.

(i) If $P=\{m,n\}$ and $Q=\{n,m\}$ , then $P\times Q=\{(m,n),(n,m)\}$ .

Ans: The statement is False.

We have the value as, $P=\{m,n\}$ and $Q=\{n,m\}$.

Thus, $P\times Q=\{(m,m),(m,n),(n,m),(n,n)\}$

(ii) If $A$ and $B$ are non-empty sets, then $A\times B$ is a non-empty set of ordered pairs $(x,y)$ such that $x\in A$ and $y\in B$.

Ans: The statement is True.

(iii) If $A=\{1,2\},B=\{3,4\}$ , then $A\times \{B\cap \varnothing \}=\varnothing $ .

Ans: The statement is True.

We know, $B\cap \varnothing =\varnothing $ 

Thus, we have, $A\times \{B\cap \varnothing \}=A\times \varnothing    =\varnothing $.

5. If $A=\{-1,1\}$ , find $A\times A\times A$ .

Ans: For any non-empty set $A$ , the set $A\times A\times A$ is defined by, 

$A\times A\times A=\{(p,q,r):p,q,r\in A\}$  

Now, we are provided with the fact that, $A=\{-1,1\}$ 

Thus, 

$A\times A\times A=\left\{ (-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1) \right\}$

6. If $A\times B=\{(a,x),(a,y),(b,,x),(b,y)\}$ . Find $A$ and $B$ .

Ans: We are provided with the fact that $A\times B=\{(a,x),(a,y),(b,x),(b,y)\}$ 

On the other hand, the Cartesian product of two non-empty sets $A$ and $B$ is defined as $A\times B=\{(a,b):a\in A\,and\,\,b\in B\}$ 

As we can see, $A$ is the set of all the first elements and $B$ is the set of all the second elements.

So, we will have, $A=\{a,b\}$ and $B=\{x,y\}$ .

7. Let $A=\{1,2\},B=\{1,2,3,4\},C=\{5,6\}$ and $D=\{5,6,7,8\}$ . Verify that

(i) $A\times (B\cap C)=(A\times B)\cap (A\times C)$ 

Ans: We are provided with 3 sets and we have to prove $A\times (B\cap C)=(A\times B)\cap (A\times C)$ 

To start with, we will have, $B\cap C=\varnothing $ , as there are no elements in common between these sets.

Thus, we have, $A\times (B\cap C)=A\times \varnothing =\varnothing $ 

For the right hand side, we have,

$A\times B=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}$ 

And similarly,

$A\times C=\{(1,5),(1,6),(2,5),(2,6)\}$ 

Again, we can see there are no elements in common between these sets. So, we have, $(A\times B)\cap (A\times C)=\varnothing $ 

So, we get, L.H.S = R.H.S.

(ii) $A\times C$ is a subset of $B\times D$ 

Ans: Again, we are to verify, $A\times C$ is a subset of $B\times D$ 

So, we have, $A\times C=\{(1,5),(1,6),(2,5),(2,6)\}$ 

And similarly,

$B\times D=\{(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)\}$ 

We can easily see, every element of $A\times C$ is an element of $B\times D$ . So, $A\times C$ is a subset of $B\times D$ .

8. Let $A=\{1,2\}$ and $B=\{3,4\}$ . Write $A\times B$ . How many subsets will $A\times B$ have? List them.

Ans: We are provided with the fact that $A=\{1,2\}$ and $B=\{3,4\}$ 

Thus, we have, $A\times B=\{(1,3),(1,4),(2,3),(2,4)\}$ 

So, the set $A\times B$ has 4 elements.

Now, this is also known to us that, if a set $A$ has $n$ elements, then the number of subsets of $A$is ${{2}^{n}}$ .

We can thus conclude that, $A\times B$will have ${{2}^{4}}=16$ subsets.

Now, noting down the subsets of $A\times B$ we get,

$\varnothing ,\{(1,3)\},\{(1,4)\},\{(2,3)\},\{(2,4)\},\{(1,3),(1,4)\}, $

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

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

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

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

 $\{(1,3),(1,4),(2,3),(2,4)\} $ 

9. Let $A$ and $B$ be two sets such that $n(A)=3$ and $n(B)=2$ . If $(x,1),(y,2),(z,1)$ are in $A\times B$ , find $A$ and $B$ , where $x,y$ and $z$ are distinct elements.

Ans: We are provided with the fact that $n(A)=3$ and $n(B)=2$ ; and\[(x,1),(y,2),(z,1)\] are in $A\times B$ .

We also know that, $A$ is the set of all the first elements and $B$ is the set of all the second elements.

So, we can conclude, $A$ having elements $x,y,z$ and $B$ having elements $1,2$ .

Thus, we get, $n(A)=3,n(B)=2$.

So, $A=\{x,y,z\},B=\{1,2\}$ .

10. The Cartesian product $A\times A$ has 9 elements among which are found $(-1,0)$ and $(0,1)$ . Find the set $A$ and the remaining elements of $A\times A$ .

Ans: We are provided with, $n(A\times A)=9$ .

We also know that, if $n(A)=a,n(B)=b$ , then $n(A\times B)=ab$ 

As it is given that, $n(A\times A)=9$

It can be written as,

 $n(A)\times n(A)=9$ 

 $\Rightarrow n(A)=3$ 

And it is also given that $(-1,0),(0,1)$ are the two elements of $A\times A$ .

Again, the fact is also known that, $A\times A=\{(a,a):a\in A\}$ . And also $-1,0,1$ are the elements of $A$ .

Also, $n(A)=3$ , implies $A=\{-1,0,1\}$ .

So, $(-1,-1),(-1,1),(0,-1),(0,0),(1,-1),(1,0),(1,1)$ are the remaining elements of $A\times A$ .

Exercise  2.2

1. Let $A=\{1,2,3,......,14\}$ . Define a relation $R$ from $A$ to $A$ by $R=\{(x,y):3x-y=0\}$ , where $x,y\in A$ . Write down its domain, codomain and range.

Ans: We are given with the relation $R$ from $A$ to $A$ as, $R=\{(x,y):3x-y=0\}$ where $x,y\in A$ .

So, we can write $R$ as, $R=\{(1,3),(2,6),(3,9),(4,12)\}$ .

Thus, the domain of $R$ is, $\{1,2,3,4\}$ .

And similarly, the range of $R$ is, $\{3,6,9,12\}$ .

And also, the codomain of $R$ is, $A=\{1,2,3,......,14\}$ .

2. Define a relation R on the set N of natural numbers by R=(x,y) : y=x+5x is a natural number less than 4; x,y in N . Depict this relationship using roster form. Write down the domain and the range.

Ans: We are given with the fact that, R=(x,y): y=x+5,x is a natural number less than 4; x,y in N.

We have the value of $x$ as $1,2,3$ are it must be less than 4.

So, the relation $R$ will look like, $R=\{(1,6),(2,7),(3,8)\}$ 

The domain of $R$ will be, $=\{1,2,3\}$ 

And similarly, the range of $R$ will be, $=\{6,7,8\}$ .

3. A={1,2,3,5} and B={4,6,9} . Define a relation R from A to B by R=(x,y):the difference between x and y is odd; x in A,y in B  . Write R in roster form.

Ans: We are provided with the fact that $A=\{1,2,3,5\}$ and $B=\{4,6,9\}$ .

We are also given that, R=(x,y):the difference between x and y is odd; x in A,y in B

Simply, writing down according to the given condition,

$R=\{(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)\}$ 

4. The given figure shows a relationship between the sets $P$ and $Q$ .

Write this  relation 

(i) In set-builder form

Ans: From the given figure in the problem, we have, $P=\{5,6,7\},Q=\{3,4,5\}$ 

Now, writing the relation in the set-builder form, 

$(i)R=\{(x,y):y=x-2;x\in P\}$ 

And in another form,

$R=\{(x,y):y=x-2;x\in 5,6,7\}$ 

 (ii) In roster form

What is its domain and range?

Ans: From the given figure in the problem, we have, $P=\{5,6,7\},Q=\{3,4,5\}$ 

And again, in roster form,

$(ii)R=\{(5,3),(6,4),(7,5)\}$ 

Where the domain of $R$ is$\{5,6,7\}$ and range of $R$ is$\{3,4,5\}$ .

5. Let A={ 1,2,3,4,6 } . Let R be the relation on A defined by (a,b):a,b in A,b is exactly divisible by a .

(i) Write $R$ in roster form.

Ans: We are provided with the fact that,$A=\{1,2,3,4,6\},R=(a,b):a,b in A,b is exactly divisible by a

Using the conditions given in the problem, we get,

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

 $ (2,4),(2,6),(3,3),(3,6),(4,4),(6,6)\}$ 

And this is the roster form of the relation.

(ii) Find the domain of R

Ans: We can clearly see, the domain of $R$ is, $\{1,2,3,4,6\}$ 

(iii) Find the range of R.

Ans: And similarly, the range of $R$ is, $\{1,2,3,4,6\}$ 

6. Determine the domain and range of the relation $R$ defined by $R=\{(x,x+5):x\in \{0,1,2,3,4,5\}\}$ 

Ans: We are provided with the fact, $R=\{(x,x+5):x\in \{0,1,2,3,4,5\}\}$ 

Using the condition given,

We can clearly write that, $R=\{(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)\}$ 

And this is our needed relation.

Now, it can be clearly observed, that the domain of $R$ is, $\{(x:x\in (0,1,2,3,4,5)\}$. And similarly, the range of $R$ is, $\{(y:y\in (5,6,7,8,9,10)\}$.

7. Write the relation$ R=(x,{x}^{3})$: x is a prime number less than 10 in roster form.

Ans: We are provided with the fact that,

$R=\{(x,{{x}^{3}}):x\,is\,a\,prime\,number\,less\,than\,10\}$ . 

We know, the prime numbers less than $10$ are $2,3,5,7$ .

Thus, the relation can be written as, 

$R=\{(2,8),(3,27),(5,125),(7,343)\}$ 

8. Let $A=\{x,y,z\}$ and $B=\{1,2\}$ . Find the number of relations from $A$ to $B$ .

Ans: The facts provided to us are, $A=\{x,y,z\}$ and $B=\{1,2\}$ .

Now, we will try to find out the Cartesian product of these to sets, $A\times B=\{(x,1),(x,2),(y,1),(y,2),(z,1),(z,2)\}$ 

Thus, we see, the number of elements in $A\times B$ is $6$ .

So, the number of subsets, ${{2}^{6}}$ .

Then, the number of relations from $A$ to $B$ is ${{2}^{6}}$ .

9. Let $R$ be the relation on $Z$ defined by $R=\{(a,b):a,b\in Z,a-b\,is\,an\,integer\}$ . Find the domain and range of $R$ .

Ans: The relation is given as, $R=\{(a,b):a,b\in Z,a-b\,is\,an\,integer\}$ .

And, we know the fact that, the difference of two given integers in always an integer.

Thus, it can be concluded that, Domain of $R$ is $Z$ and similarly, the range of $R$ is also $Z$ .

Exercise 2.3

1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) $\{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)\}$ 

Ans: We have the given relation as, $\{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)\}$ .

Thus, we can see, the domain of the relation consists of $\{2,5,8,11,14,17\}$ and range is $\{1\}$ .

And we also have, every element of the domain is having their unique images, then it is a function.

 (ii) $\{(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)\}$ 

Ans: We have our given relation, $\{(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)\}$.

Thus, we have our domain as, $\{2,4,6,8,10,12,14\}$ and range as, $\{1,2,3,4,5,6,7\}$ .

Every element of the domain is having their unique images, so this is a function.

(iii) $\{(1,3),(1,5),(2,5)\}$ 

Ans: Our given relation is, $\{(1,3),(1,5),(2,5)\}$.

From the domain of the relation the element $1$ is having two different images $3,5$ .

So,every element of the domain is not having their unique images. So, this is not a function.

2. Find the domain and range of the following real function:

$(i)\,f(x)=-\left| x \right|$ 

Ans: We have the given function as, $f(x)=-\left| x \right|$ .

It is also know that,$\left| x \right|=\left\{ \begin{matrix} x,if\,x\ge 0 \\ -x,if\,x<0 \\ \end{matrix} \right.$

Thus, $f(x)=-\left| x \right|=\left\{ \begin{matrix} -x,if\,x\ge 0 \\ x,if\,x<0 \\ \end{matrix} \right.$

As the function is a real function, the domain of the function is $R$ .

And again, we can see that the function is giving values of all real numbers except positive ones.

So, the range of the function is, $(-\infty ,0]$ .

$(ii)\,f(x)=\sqrt{9-{{x}^{2}}}$ 

Ans: The function is given as, 

$f(x)=\sqrt{9-{{x}^{2}}}$ 

We can clearly see that the function is well defined for all the real numbers which are greater than or equal to $-3$ and less than or equal to $3$ , thus, the domain of the function is, $\{x:-3\le x\le 3\}$ or $[-3,3]$ .

And for such value of $x$ , the value of the function will always be between $0$ and $3$.

Thus, the range is, $\{x:0\le x\le 3\}$ or $[0,3]$ .

3. A function is defined by $f(x)=2x-5$ 

(i) $f(0)$ 

Ans: We have the given function as, $f(x)=2x-5$ 

So, the value of,

$f(0)=2\times 0-5=-5$ 

(ii) $f(7)$ 

Ans: We have the given function as, $f(x)=2x-5$ 

So, the value of,

$f(7)=2\times 7-5=14-5=9$ 

(iii) $f(-3)$ 

Ans: We have the given function as, $f(x)=2x-5$ 

So, the value of,

$f(-3)=2\times (-3)-5=-6-5=-11$

4. The function $'t'$ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t(C)=\dfrac{9C}{5}+32$ . Find

(i) $t(0)$ 

Ans: We have our given function as, $t(C)=\dfrac{9C}{5}+32$ .

Thus, to find our needed values of the function we just have to put the values in the given function and simplify it.

So, we get now,

\[t(0)=\dfrac{9\times 0}{5}+32=32\] 

(ii) $t(28)$ 

Ans: We have our given function as, $t(C)=\dfrac{9C}{5}+32$ .

Thus, to find our needed values of the function we just have to put the values in the given function and simplify it.

So, we get now,

 $t(28)=\dfrac{9\times 28}{5}+32$

 $\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{252+160}{5}$

 $\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{412}{5}$

 $\,\,\,\,\,\,\,\,\,\,\,\,=82.4 $ 

(iii) $t(-10)$ 

Ans: We have our given function as, $t(C)=\dfrac{9C}{5}+32$ .

Thus, to find our needed values of the function we just have to put the values in the given function and simplify it.

So, we get now,

$t(-10)=\dfrac{9\times (-10)}{5}+32$

$\,\,\,\,\,\,\,\,\,\,\,\,=-18+32$

$\,\,\,\,\,\,\,\,\,\,\,\,=14$

(iv) The value of $C$, when $t(C)=212$ .

Ans: We have our given function as, $t(C)=\dfrac{9C}{5}+32$ .

For this problem, we are given that, $t(C)=212$ .

So, it can be written as, $\dfrac{9C}{5}+32=212$ 

Simplifying further,

$\dfrac{9C}{5}=212-32$

 $\Rightarrow \dfrac{9C}{5}=180$

 $\Rightarrow C=\dfrac{900}{9}=100 $

Thus, it can be said that, for $t(C)=212$, the value of $t$ is $100$ .

5. Find the range of each of the following functions:

(i)  $f(x)=2-3x,x\in R,x>0$ .

Ans: We have the given function as, $f(x)=2-3x,x\in R,x>0$ 

Let us try to write the value of the given function in a tabular form as,

We can now see, it can be seen that the elements of the range is less than 2.

So, the range will be, $f=(-\infty ,2)$ 

Alternative way to solve

Let us take, $x>0$ 

We can again go forward by writing, 

$3x>0 $

$\Rightarrow 2-3x<2$

 $\Rightarrow f(x)<2$ 

So, the range of $f$ is $(-\infty ,2)$ 

(ii) $f(x)={{x}^{2}}+2,x\,is\,a\,real\,number.$

Ans: We have our given function that,$f(x)={{x}^{2}}+2$ 

Let us try to write the value of the given function in a tabular form as,

So, we see that the range of the function $f$ is the set of all numbers which are greater than or equal to 2.

Thus, we can conclude that the range of the function is, $[2,\infty )$ .

Alternative Method:

Let $x$ be any real number. So,

${{x}^{2}}>0$ 

After further simplification,

${{x}^{2}}+2\ge 2$

 $\Rightarrow f(x)\ge 2$

Thus, the range of the function is $=[2,\infty )$ 

(iii) $f(x)=x,x$ is a real number .

Ans: We have our given function as, $f(x)=x,x\,is\,a\,real\,\,number$ .

Now, we can clearly have, that the range of the function is the set of all the numbers.

So, the range of the function will be, $R$ .

Miscellaneous Exercise

1. The relation $f$  is defined by $f(x)=\left\{ \begin{matrix} {{x}^{2}},0\le x\le 3 \\ 3x,3\le x\le 10 \\ \end{matrix} \right.$

And the relation $g$ is defined by \[g(x)=\left\{ \begin{matrix} {{x}^{2}},0\le x\le 2 \\ 3x,2\le x\le 10 \\ \end{matrix} \right.\]

Show that $f$ is a function and $g$ is not a function.

Ans: According to the problem, we have the function $f$ as,

$f(x)=\left\{ \begin{matrix} {{x}^{2}},0\le x\le 3 \\ 3x,3\le x\le 10 \\ \end{matrix} \right.$

We can see that, for $x=3$ ,

$f(x)={{3}^{2}}\,=9$ from the first given condition.

And again, $f(x)=3\times 3=9$ from the second condition.

But now, 

\[g(x)=\left\{ \begin{matrix} {{x}^{2}},0\le x\le 2 \\ 3x,2\le x\le 10 \\ \end{matrix} \right.\] 

We can see that, for $x=2$ ,

$f(x)={{2}^{2}}\,=4$ from the first given condition.

And again, $f(x)=3\times 2=6$ from the second condition.

Thus, the domain of the relation $g$ is having two different images from a single element.

So, it can be concluded that the relation is not a function.

2. If $f(x)={{x}^{2}}$ , find $\dfrac{f(1.1)-f(1)}{(1.1-1)}$ .

Ans: We have the function, $f(x)={{x}^{2}}$ .

So, we will have, 

\[\dfrac{f(1.1)-f(1)}{(1.1-1)}\] equaling to,  

\[\dfrac{{{(1.1)}^{2}}-{{1}^{2}}}{1.1-1}\] , putting the values.

After further simplification,

$\dfrac{1.21-1}{0.1} $

$ =\dfrac{0.21}{0.1}$

$=2.1 $ 

3. Find the domain of the function $f(x)=\dfrac{{{x}^{2}}+2x+1}{{{x}^{2}}-8x+12}$

Ans: According to the problem, we have the given function as,

$f(x)=dfrac{{{x}^{2}}+2x+1}{{{x}^{2}}-8x+12}$.

Let us try to simplify the given function and bring it to a form where we can analyze the problem.

The denominator can be factorized as,

${{x}^{2}}-8x+12$ 

 $={{x}^{2}}-6x-2x+12$ 

$=x(x-6)-2(x-6) $ 

$=(x-2)(x-6)$ 

So, we see that the function is defined for every real numbers except $6,2$ .

Thus, the domain of the function will be, $R-\{2,6\}$ 

4. Find the domain and the range of the real function $f$ defined by $f(x)=\sqrt{(x-1)}$.

Ans: We have the given function as, $f(x)=\sqrt{(x-1)}$ .

Clearly, the term inside the root sign must be non-negative.

So, the function is valid for all values of $x\ge 1$ .

Thus, the domain of the function will be, $[1,\infty )$ .

Now, again, for $x\ge 1$, the value of the function will always be greater than or equal to zero.

So, the range of the function is, $[0,\infty )$ .

5. Find the domain and the range of the real function $f$ defined by $f(x)=\left| x-1 \right|$.

Ans: The function which is given is, $f(x)=\left| x-1 \right|$ .

We can clearly see that, the function is well defined for all the real numbers.

Thus, it can be concluded that, the domain of the function is $R$ .

And for every $x\in R$ , the function gives all non-negative real numbers.

So, the range of the function is the set of all non-negative real numbers. i.e, $[0,\infty )$ .

6. Let $f=\left\{ \left( x,\dfrac{{{x}^{2}}}{1+{{x}^{2}}} \right):x\in R \right\}$ be a function from $R$ to $R$. Determine the range of $f$ .

Ans: We have our given function as,$f=\left\{ \left( x,\dfrac{{{x}^{2}}}{1+{{x}^{2}}} \right):x\in R \right\}$

Expressing it by term to term, we are getting, 

$f=\left\{ \left( 0,0 \right),\left( \pm 0.5,\dfrac{1}{5} \right),\left( \pm 1,\dfrac{1}{2} \right),\left( \pm 1.5,\dfrac{9}{13} \right),\left( \pm 2,\dfrac{4}{5} \right),\left( 3,\dfrac{9}{10} \right),\left( 4,\dfrac{16}{17} \right),.... \right\}$ 

And we also know, the range of $f$ is the set of all the second elements. We can also see that the terms are greater than or equal to $0$ but less than $1$ .

So, the range of the function is, $[0,1)$ .

7. Let $f,g:R\to R$ be defined, respectively by $f(x)=x+1,g(x)=2x-3$ . Find $f+g,f-g$ and $\dfrac{f}{g}$ .

Ans: We have the functions defined as, $f,g:R\to R$is defined as, $f(x)=x+1,g(x)=2x-3$ .

Thus, the function

 $(f+g)(x)=f(x)+g(x)$ 

 $=(x+1)+(2x-3)$ 

$=3x-2$ 

So, the function $(f+g)(x)=3x-2$ .

 Again, the function, 

$(f-g)(x)=f(x)-g(x)$ 

$=(x+1)-(2x-3)$ 

$=-x+4$

So, the function $(f-g)(x)=-x+4$.

Similarly, 

$\left( \dfrac{f}{g} \right)(x)=\dfrac{f(x)}{g(x)}$ where $g\left( x \right)\ne 0$ and also $x\in R$ .

Now, putting the values, 

$\left( \dfrac{f}{g} \right)(x)=\dfrac{x+1}{2x-3}$ 

where, 

$2x-3\ne 0$ 

$\Rightarrow x\ne \dfrac{3}{2}$ 

8. Let $f=\left\{ \left( 1,1 \right),\left( 2,3 \right),\left( 0,-1 \right),\left( -1,-3 \right) \right\}$ be a function from $Z$ to $Z$ defined by $f(x)=ax+b$ , for some integers $a,b$ . Determine $a,b$ 

Ans: We have the given function as,  $f=\left\{ \left( 1,1 \right),\left( 2,3 \right),\left( 0,-1 \right),\left( -1,-3 \right) \right\}$ and also $f(x)=ax+b$ .

As, $(1,1)\in f$ , we get, 

$a\times 1+b=1$

$\Rightarrow a+b=1$

And again, $(0,-1)\in f$ , from this we can get,

$a\times 0+b=-1$

$\Rightarrow b=-1$ 

Putting this value in the first equation, we have,

$a-1=1$

$\Rightarrow a=2$

So, the value of $a$ and $b$ are respectively, $2,-1$ .

9. Let $R$ be a relation from $N$ to $N$ defined by $R=\left\{ (a,b):a,b\in      N\,and\,a={{b}^{2}} \right\}$ . Are the following true? Justify your answer in each case.

(i) $\left( a,a \right)\in R$ , for all $a\in N$ .

Ans: We are given our relation as, $R=\left\{ (a,b):a,b\in N\,and\,a={{b}^{2}} \right\}$ 

Let us take, $2\in N$ .

But we have, $2\ne {{2}^{2}}=4$ 

So, the statement that $\left( a,a \right)\in R$ , for all $a\in N$is not true.

(ii) $\left( a,b \right)\in R$ , implies $\left( b,a \right)\in R$ 

Ans:

We are given our relation as, $R=\left\{ (a,b):a,b\in N\,and\,a={{b}^{2}} \right\}$ 

Let us take, $(9,3)\in N$ . 

We have to check if, $\left( 3,9 \right)\in N$or not.

But, the condition of the relation says,  $R=\left\{ (a,b):a,b\in N\,and\,a={{b}^{2}} \right\}$and ${{9}^{2}}\ne 3$ .

So, the statement $\left( a,b \right)\in R$ , implies $\left( b,a \right)\in R$ is not true.

(iii) $\left( a,b \right)\in R,\left( b,c \right)\in R$ implies $\left( a,c \right)\in R$ .

Ans: We are given our relation as, $R=\left\{ (a,b):a,b\in N\,and\,a={{b}^{2}} \right\}$ 

Now, let us take, $\left( 9,3 \right)\in R,\left( 16,4 \right)\in R$ .

We have to check if, $\left( 9,4 \right)\in N$or not.

Thus can also easily see, $9\ne {{4}^{2}}=16$ .

So, the given statement $\left( a,b \right)\in R,\left( b,c \right)\in R$ implies $\left( a,c \right)\in R$ .

10. Let $A=\{1,2,3,4\},B=\{1,5,9,11,15,16\}$ and $f=\left\{ \left( 1,5 \right),\left( 2,9 \right),\left( 3,1 \right),\left( 4,5 \right),(2,11) \right\}$ . Are the following true? Justify your answer in each case.

(i) $f$ is a relation from $A$ to $B$ .

Ans: We are provided with two sets, $A=\{1,2,3,4\},B=\{1,5,9,11,15,16\}$ 

Thus, the Cartesian product of these two sets will be,

$A\times B=${ $\left( 1,1 \right),\left( 1,5 \right),\left( 1,9 \right),\left( 1,11 \right),\left( 1,15 \right),\left( 1,16 \right),$ 

$\left( 2,1 \right),\left( 2,5 \right),\left( 2,9 \right),\left( 2,11 \right),\left( 2,15 \right),\left( 2,16 \right), $

 $\left( 3,1 \right),\left( 3,5 \right),\left( 3,9 \right),\left( 3,11 \right),\left( 3,15 \right),\left( 3,16 \right), $

$\left( 4,1 \right),\left( 4,5 \right),\left( 4,9 \right),\left( 4,11 \right),\left( 4,15 \right),\left( 4,16 \right) $

 And it is also given that, 

$f=\left\{ \left( 1,5 \right),\left( 2,9 \right),\left( 3,1 \right),\left( 4,5 \right),(2,11) \right\}$

A relation from a non-empty set $A$ to a non-empty set $B$is a subset of the Cartesian product $A\times B$ .

Thus, it can be easily checked that $f$ is a relation from $A$ to B.

(ii) f is a function from $A$ to $B$ .

Ans: We are provided with two sets, $A=\{1,2,3,4\},B=\{1,5,9,11,15,16\}$ 

Thus, the Cartesian product of these two sets will be,

$A\times B=\{\left( 1,1 \right),\left( 1,5 \right),\left( 1,9 \right),\left( 1,11 \right),\left( 1,15 \right),\left( 1,16 \right),$ 

$\left( 2,1 \right),\left( 2,5 \right),\left( 2,9 \right),\left( 2,11 \right),\left( 2,15 \right),\left( 2,16 \right), $

 $\left( 3,1 \right),\left( 3,5 \right),\left( 3,9 \right),\left( 3,11 \right),\left( 3,15 \right),\left( 3,16 \right), $

$\left( 4,1 \right),\left( 4,5 \right),\left( 4,9 \right),\left( 4,11 \right),\left( 4,15 \right),\left( 4,16 \right)$

And it is also given that, 

$f=\left\{ \left( 1,5 \right),\left( 2,9 \right),\left( 3,1 \right),\left( 4,5 \right),(2,11) \right\}$

If we check carefully, we see that the first element $2$ is providing us two different value of the image $9,11$ .

So, it can be concluded that $f$ is not a function from $A$ to B.

11. Let $f$ be the subset of $Z\times Z$ defined by $f=\left\{ \left( ab,a+b \right):a,b\in Z \right\}$ . If $f$ a function from $Z$ to $Z$. Justify your answer.

Ans: Our given relation $f$ is defined as $f=\left\{ \left( ab,a+b \right):a,b\in Z \right\}$.

We also know that a relation will be called a function from $A$ to $B$ if every element of the set $A$ has unique images in set $B$ .

Let us take 4 elements, $2,6,-2,-6\in Z$ .

So, for the first two elements, 

$\left( 2\times 6,2+6 \right)\in f$ 

 $\Rightarrow \left( 12,8 \right)\in f$

And for the last two elements,

$\left( -2\times -6,-2+-6 \right)\in f$

 $\Rightarrow \left( 12,-8 \right)\in f$

So, it is clearly visible that one single element $12$ having two different images $8,-8$. Thus, the relation is not a function.

12. Let $A=\left\{ 9,10,11,12,13 \right\}$ and let $f:A\to N$ be defined by $f(n)=$ the highest prime factor of $n$ . Find the range of $f$ .

Ans: We have our given set as, $A=\left\{ 9,10,11,12,13 \right\}$ and the relation is given as $f(n)=$ the highest prime factor of $n$.

The prime factor of $9$ is $3$ .

The prime factors of 10 is $2,5$ .

The prime factor of $11$ is $11$ .

The prime factor of 12 is $2,3$ .

The prime factor of $13$ is $13$ .

Thus, it can be said, 

$f(9)=$ the highest prime factor of $9=3$ .

$f(10)=$ the highest prime factor of $10=5$ .

$f(11)=$ the highest prime factor of $11=11$ .

$f(12)=$ the highest prime factor of $12=3$ .

$f(13)=$ the highest prime factor of $13=13$ .

Now, the range of the function will be, $\left\{ 3,5,11,13 \right\}$ .

NCERT Solutions for Class 11 Maths Chapter 2 PDF

Overview of Deleted Syllabus for CBSE Class 11 Maths - Relations and Functions

Class 11  Maths Chapter 2: Exercises Breakdown

Conclusion

In relations and functions class 11, understanding the basics of relations and functions is crucial. Focus on the types of relations (reflexive, symmetric, transitive) and types of functions (one-one, onto, bijective). Grasping the concepts of function composition and inverse functions is also important.

Practicing the exercises will help reinforce these concepts. From previous years' question papers, around 3-4 questions are asked on this topic, highlighting its significance in exams. For detailed solutions and practice, refer to Vedantu's resources to solidify your understanding and perform well in exams.

Other Study Material for CBSE Class 11 Maths Chapter 2

Chapter-Specific NCERT Solutions for Class 11 Maths 

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Important Related Links for CBSE Class 11 Maths